ANSWERS TO QUESTIONS ASKED BY STUDENTS in Math 114, Spring 2001, taught from Ian Stewart's "Galois Theory". _________________________________________________________________ You ask about the inclusion Q \subset Q(gamma,delta) \subset Q(alpha,beta,gamma,delta) on p.xxviii, saying that you would expect Q, as the field of all rationals, to contain both these fields of rational expressions. I think the problem involves the phrases "rationals" and "rational expressions". The field of "rationals" consists of the rational numbers, i.e., elements m/n where m and n are integers. The field of "rational expressions" in, say, gamma and delta consists of all elements that can be gotten using integers, gamma and delta, addition, subtraction, multiplication and division. So, for example, if gamma = sqrt 5 and delta = - sqrt 5, then (3 + 7 sqrt 5) / (9 - sqrt 5) is a member of Q(gamma, delta), but not a member of Q. _________________________________________________________________ You ask about the relation between the Euclidean algorithm and the Chinese Remainder Theorem. Well, if R is a ring, and a, b elements of R with hcf = 1, then we have the implications: (R satisfies Euclidean algorithm) || \/ (Every ideal of R is principal) || \/ (aR + bR = R) /\ || \/ (Every pair of congruences x == u (mod a), x == v (mod b) has a solution in R) (Here "==" stands for "congruent to".) But neither of the first two implications is reversible: There are rings R which do not satisfy the Euclidean algorithm for any "Euclidean norm function" (was that concept introduced in your previous classes?) but which have the property that every ideal is principal, and in a ring in which not every ideal is principal, there may still be interesting examples of elements a and b such that aR + bR = 1, so that by the last implication, one has the "Chinese Remainder Theorem" for those elements. So the Chinese Remainder Theorem applies to a much wider class of cases than the Euclidean algorithm. If you want further references for one or another aspect of this, let me know. (Examples of rings in which every ideal is principal, but which don't satisfy any Euclidean algorithm, actually take a bit of work to come up with. If you look on my web page, http://www.math.berkeley.edu/~gbergman and click on "Handouts for graduate algebra classes", one of the items you will find there is "A principal ideal domain that is not Euclidean, developed as a series of exercises". It is intended for students who have had the first few weeks of Math 250A; you can look at it and see how much you can follow.) _________________________________________________________________ You ask why the definition of a field as a ring such that F \ {0} is an abelian group under multiplication implies that ab = 0 => a = 0 or b = 0. Well, consider two elements a, b of F neither of which is 0. This says both lie in F \ {0}, hence since this is a group under multiplication, it must contain their product. And to say that ab is in F \ {0} is to say that ab is nonzero. (This proves the statement in contrapositive form: If it is not true that a=0 or b=0, then it is not true that ab=0.) _________________________________________________________________ You ask what examples authors like Stewart wish to include when they allow rings not to have 1. I think the main justification generally given is that this allows results proved about rings to be applied to ideals, which generally don't contain 1. But I find that fairly specious; the things one wants to know about ideals can generally be seen by looking at them as subsets of the rings they are ideals in. But there are some nonunital rings that arise naturally. For instance, the ring of "strictly upper triangular nxn matrices", i.e., matrices of the form / 0 * * * \ | 0 0 * * | where each "*"s represents an arbitrary element of the | 0 0 0 * | base ring (e.g., the real numbers), and the "0"s \ 0 0 0 0 / represent 0. For another example, if V is an infinite-dimensional vector-space, then the set of all linear transformation V -> V whose ranges are finite-dimensional forms an interesting ring without 1. And similarly, if we take all sequences (a_0, a_1, a_2, ... , a_n, ...), say of real numbers, and make these a ring (with 1!) by componentwise operations, i.e., defining (a_0, a_1, a_2, ... , a_n, ...) + (b_0, b_1, b_2, ... , b_n, ...) = (a_0+b_0, a_1+b_1, a_2+b_2, ... , a_n+b_n, ...) and likewise (a_0, a_1, a_2, ... , a_n, ...) x (b_0, b_1, b_2, ... , b_n, ...) = (a_0 b_0, a_1 b_1, a_2 b_2, ... , a_n b_n, ...) then the set of sequences with only finitely many nonzero terms forms a nonunital subring. In each of these cases, the subring in question is an ideal of a larger ring with unit. (In the last two cases, that ring is fairly obvious: just drop the "finiteness" condition from the description given. In the matrix case, it is the ring of upper triangular matrices, not required to be "strictly" upper triangular; i.e., may have arbitrary entries on the diagonal.) But these nonunital examples are of some ring-theoretic interest for themselves, not just as ideals of larger rings. So I don't say that nonunital rings shouldn't be looked at. Just that they are not an important enough concept to muddy the waters with in a beginning course. You also ask how Stewart gets the first display on p.3. The second of the two arguments you indicated seems the most straightforward: use the fact that x |-> I+x is a homomorphism, and apply it to the equation ar+bn=1, rewritten ar = 1-bn. I assume this is what Stewart meant. Your other argument involves writing "I + 1 - bn", and requires one to decide what one means by that expression. If you mean to set up a definition in which one can add any set of elements of a ring to any other (and regard 1 and -bn as abbreviations for the corresponding one-element sets), then you need to ask whether such addition of sets is associative, so that you can safely write expressions without parentheses. It is, but one needs to prove it. Alternatively, you may mean "I + (1-bn)", i.e., the coset of the ideal I containing the element 1-bn. In that case, the argument is essentially the same as the first one. All this shows one reason I don't like the "I+x" notation for elements of a quotient ring R/I ! _________________________________________________________________ You ask about the equation (I + r)(I + s) = I + rs on p.2. That equation is the _definition_ of multiplication in a factor ring, so it is not something one can or should prove. _________________________________________________________________ Yes, what Stewart calls an "hcf" is what is often called a "gcd". He defined the term on p.11, lines 3-4. This may be a difference between British and American usages; at least a difference of which is used more often in which country. _________________________________________________________________ You both asked about the statement near the middle of p.21, that "f(t) obviously irreducible if and only if f(t+1) is irreducible". I intended to speak about it in class, but ran out of time! Basically, the point is that one can not only substitute for the indeterminate t in polynomials f(t)\in K[t] any element alpha \in K, getting homomorphisms K[t] ---> K; one can substitute any element alpha of any commutative ring R containing K, getting homomorphisms K[t] --> R. In particular, taking R = K[t] and alpha = t+1, one gets a homomorphism K[t] --> K[t] taking f(t) to f(t+1). This has an inverse, the homomorphism taking f(t) to f(t-1). Hence it is an isomorphism, from which one can easily deduce that an element f(t) of K[t] is irreducible if and only if the image of that element under this homomorphism, i.e., f(t+1), is irreducible. _________________________________________________________________ You ask about the meaning of "highest coefficient" in the discussion before the last example on p.21. Stewart means "coefficient of the highest power of t". So, for instance, in the display earlier on the page, beginning "9f(t)", the highest coefficient of the polynomial on the right is 2 . As to the "graph" of a polynomial over Q (p.24, top), one simply regards Q as a subset of R, and marks points on the graph in the same was as one does in R. Since Q is dense in R, there is no real way of seeing the difference between such a graph and the graph of a function R --> R, unless one artificially draws it with a "dotted line". But the interpretation of such a graph is tricky: The graphs of t^2 - 1 and t^2 - 2 both seem to have zeroes, but over Q, the former does, while the latter does not. So Stewart mentions the idea of graphing functions on Q only in parenthesis. _________________________________________________________________ You ask what Stewart means in Theorem 2.3 by "unique up to order and constant factors." He means that given two factorizations of f, one can be gotten from the other by rearranging the terms ("order") and multiplying them by constant factors. ("Constant" meaning "belonging to K".) E.g., two factorizations of (x - 1/4) are (x + 1/2)(x - 1/2) and (2x - 1)(x/2 + 1/4). If we take the first factorization, reverse the order of the factors, and then multiply the first factor by 2 and the second by its inverse, 1/2, we get the second factorization; so these count as "the same up to order and constant factors". _________________________________________________________________ You ask whether Proposition 2.4 holds for more general rings than Z. Yes! The proof works with Z replaced by any unique factorization domain, and Q by its field of fractions. (Was "unique factorization domain" defined in your 113?) But it is not true over an arbitrary integral domain. For instance, letting R be the domain Z[2i] = {m + 2ni | m, n \in Z} that I mentioned today, one finds that t^2 + 1 is irreducible in R[t], essentially because -1 is not a square in that ring; but the field of fractions of R[t] is Q(i), so t^2 + 1 = (t+i)(t-i) there. _________________________________________________________________ You ask how, in the proof of Lemma 2.2, one can have a divisor of f equal to kf. I hope what I said in class clarified this: For k\in K - {0}, k is invertible, hence f = (kf) (k^-1). You also ask why f|haf and f|hbg imply f|h. This is because Stewart has just written h as the sum of haf and hbg, and if two elements are divisible by f, so is their sum. (Here I think you should have followed the general guideline: If you don't see the reason for the assertion about something, check what has just been proved about that something, and see whether that is what the author is using.) As for where Eisenstein's Criterion is applicable, I will talk more about it Wednesday. There is definitely a simpler way to show that it is applicable to t^16 + ... + 1 than trial and error! If p is any prime, and we let f(t) = t^p-1 + t^p-2 + ... + t + 1, then Stewart shows on p.168 that it is irreducible by Eisenstein's criterion using that prime p: Go to the proof of Lemma 17.9 on that page and ignore statement of the Lemma and the second sentence of the proof; the first sentence and the remainder of the proof should make sense. As to whether it is applicable to "a large percentage of irreducibles" -- I doubt it. Nevertheless, when one needs an example for some purpose, it often turns up that one can concoct one that Eisenstein's criterion is applicable to; so it is very useful in that way. _________________________________________________________________ You ask about the argument at the top of p.20, that since p divides the sum h_0 g_i+j + ... + h_j g_i + ... h_i+j g_0, it must divide h_j g_i. It is certainly true, as you say, that p can divide a sum without dividing any of the summands. But if it divides a sum, and divides _all_but_one_ of the summands, then it must divide that one summand as well. (For that one summand can be written as the whole sum, minus all the other summands; and since each of these terms is divisible by p, the resulting element will be too.) Since Stewart has shown why all the other summands are divisible by p, he can conclude that h_j g_i is too. _________________________________________________________________ You ask whether R needs to be an integral domain in the definition of a zero of a polynomial. No, one can make the _definition_ without assuming R a domain. But for most of the properties of the concept that one wants to prove, it must be assumed a domain. And Stewart does assume it a domain, in fact, a field, when he proves things about it. _________________________________________________________________ You ask how to find the multiplicative inverse of p + q (cube-root 2) + r (cube-root 2)^2. I started to send you an "easy" answer, then realized that it was not so easy. I hope to find time to talk about it on Friday. Anyway, Here is another argument which, though perhaps not what Stewart had in mind, does work: Consider the set V of elements A + B (cube-root 2) + C (cube-root 2)^2 (A, B, C \in Q) as a vector space over Q. It is finite-dimensional, since it is spanned by {1, cube-root 2, (cube-root 2)^2}. (If we had an easy way of showing at this point that those elements were linearly independent, we could say precisely that it was 3-dimensional; but we can in any case say it has dimension _< 3.) Now if p + q (cube-root 2) + r (cube-root 2)^2 is nonzero, then multiplication by that element is a one-to-one linear map V --> V. A one-to-one linear map of a finite-dimensional vector-space to itself is invertible (Math 110!), in particular, onto, so there is some element of V which this map takes to 1; i.e., an element which when multiplied by p + q (cube-root 2) + r (cube-root 2)^2 gives 1; and that is what we were looking for. _________________________________________________________________ You ask what is meant by an "inclusion map". If T is a set and S a subset, then the "inclusion map" of S into T means the map i: S --> T defined by i(s) = s for all s in S. So it is like an identity map, but instead of going from a set to itself, it goes from a set to a set containing it. Unlike an identity map, it is not in general onto. One thinks of it as "including" S in T. You also ask whether the inclusion maps given as examples of field extensions on p.30 are "any different from his previous examples ...", but I don't know what previous examples you mean. _________________________________________________________________ You ask why the field described at the top of p.31 should have the property stated at the bottom of p.30, of being the _smallest_ subfield of K containing X. First, I hope you are clear on what being the "smallest" such subfield means: it means that it is a subfield of K containing X which is contained in every subfield of K that contains X. To see that the field described at the top of p.31 has the latter property, note that any subfield of K is closed under the field operations; hence if it contains the elements of X, it will also contain every element that can be gotten from the members of X using those operations. Clear now? _________________________________________________________________ You ask how to see that statement #2 at the top of p.31 is equivalent to the definition of the subfield of K generated by X. When you have a question like this, you should let me know how far you have been able to get on it, and, if possible, what difficulty you have going further, so that I know what problems to address. You need to show that the set in question is equal to the intersection of all subfields of K containing X. To show two sets equal, one must show that each is contained in the other; equivalently, that every element of the first is contained in the second and that every element of the second is contained in the first. (Did you at least reach the stage of seeing that there were two inclusions to be proved? I will assume you did.) Did you examine each of these inclusions in turn? If you could see that one was true (e.g., that the intersection was contained in the set described at the top of p.31), but not the other, you should say this in your question, and if possible, say what difficulties you encountered when examining the other needed inclusion. If you looked at each inclusion and could not see why _either_ was true, you should say that! As I wrote in the course information sheet, if you asked such a question at my office hours, I would question you to see what you understood and what needed to be explained. But that is very time-consuming to do by e-mail, so you should do as much of it as you can in posing your question. So let me know about these points, and I will then do my best to help. _________________________________________________________________ You ask for an example of a simple transcendental extension, noting that on p.32, Stewart says that K(t), the field of rational expressions, is such an example, but that you don't know what that means. A basic survival technique in mathematics courses is: If you don't know what a phrase or symbol that the author uses means, check the index and/or table of notation, and see whether he has defined it! (In general, I recommend first looking in the preceding paragraphs of the section you are reading. But in this case, you wouldn't have found the phrase there.) Looking in the index under "rational expression", you would have found a reference to p.9. Looking in the Symbol index (pp.197-198), you wouldn't have found K(t), but you would have found R(t), with the gloss "field of rational expressions", also referring you to p.9. So look on p.9! Does this answer your question? Of course, if you haven't seen the construction of K(t) before, you need to do a lot of thinking to picture the field described on that page (lines 3-6) and maybe you will have questions about that, which I will be glad to help with. But at least you will be further along than if you were still stuck on Stewart's example. I hope you'll remember that basic technique, "Look in the index and/or table of notations", in the future! _________________________________________________________________ You ask whether, if x and y are transcendental, Q[x] must equal Q[y]. No. What Stewart says is that a simple transcendental extension of a field K is unique _up_to_isomorphism_. That means that Q(x) will be isomorphic to Q(y). (And the same will be true of Q[x] and Q[y].) But not that they will be equal. This is related to what I said in class, that we will often "identify" two isomorphic objects _when_this_is_not_likely_to_lead_to_confusion. When we are studying the internal algebraic properties of a simple transcendental extension of Q, we can speak as though there were just one such object, because they all have the same internal algebraic properties. But if we are looking at the structure of R, we can't safely identify different transcendental subfields, because they consist of different elements of R. _________________________________________________________________ You ask about Stewart's statement that the element t\in K(t) is transcendental because "If ... p(t) = 0, then p = 0 ..." (p.35). Well, if p(t) = Sigma a_i t^i is an element of K[t], then for any alpha in an extension field L of K, Stewart has defined p(alpha) to mean Sigma a_i alpha^i; and we say that alpha "satisfies" the polynomial p if p(alpha) = 0. Now in the special case where L = K(t) and we take alpha = t, we see that on substituting alpha for t as defined above, we get back our original polynomial p(t) (which shows, incidentally, that the definition of p(alpha) and the notation p(t) are consistent). Hence in this case the map "substituting alpha for t" is the inclusion of K[t] in K(t), and this clearly has kernel {0}. _________________________________________________________________ When mathematicians say "identify X with Y", the meaning is something inbetween "pretend X is Y" and "assume X is Y". With that in mind, I hope what Stewart says at the top of p.34 makes more sense. An "inclusion map" means the map from a subset S to a set T that contains it, defined by f(s) = s for all s in S. In other words, it is like an identity map, but if its codomain is larger than its range we can't call it the identity, so we think of it as "including" S in T". _________________________________________________________________ You ask, first, why the situation two lines before the lower Definition on p.35 is "contrary to the definition". It is, as you guessed, because p is defined as having lowest degree among monic polynomials satisfied by alpha, but what is described there would be a monic polynomial of lower degree satisfied by alpha. You also ask whether, when one writes "m(alpha) = 0", this means the zero of the original field, or the zero of the extension field. It means the zero of the extension field, insofar as one distinguishes these. (After all, it is only in the extension field that one can evaluate m(alpha).) In general, one identifies K with its image in the extension field, and then one no longer has to make this distinction. But when one constructs K[t]/I as consisting of cosets of I, then one has to distinguish the two zeroes, and one is looking at the zero of the extension field. You ask how, in constructing a factor-ring R/I, one proves that the map pi(r) = r + I is well-defined. There is no trouble with that! Well-definedness needs to be proved only when a definition requires making a choice, and that definition does not. Where one does have to prove well-definedness is for the _operations_ of R/I. E.g., one defines the product of r+I and s+I to be rs + I. But given an element x \in R/I, there are in general many elements r_1, r_2, ... such that x = r_i + I = r_2 + I = ... , so the definition of multiplication just mentioned requires choosing one such expression for r, and one for s. Then one has to prove that changing either of these choices doesn't change the coset rs + I. I suggest you try that verification; it's not very hard. If you have trouble with it, ask about it. If phi: R --> S is a homomorphism whose kernel contains I, you want to know how one gets a homomorphism psi: R/I --> S such that psi pi = phi. Well, to see how we _have_ to define it, we consider what that equation means when applied to an element r of R. It means psi(r+I) = phi(r). So one uses that equation to define psi. Namely, every element of R/I has the form r+I, and we define the result of applying psi to that element r+I to be phi(r). Again, we must prove well-definedness. Again, I suggest you try, and ask me if you have trouble with it. You say that "the same Psi is actually an injective homomorphism from R/Ker(Phi) -> R'". Not in general. But if you take I = Ker(phi) (rather than letting I be any ideal contained in ker(phi)) then this is true. Incidentally, when we write Greek letters in e-mail, "Phi", "Pi" etc. refer to the capital Greek letters. But the symbols generally used for maps are the lower-case letters, so it is best to call them "phi", "pi" etc.. (See the handout on the Greek alphabet for the difference in appearance between the capital and lower-case forms.) _________________________________________________________________ You ask whether in the Definition on p.40, "monomorphism" could be replaced by "homomorphism". First, have you noted the definition of monomorphism? Do you realize that it _is_ a homomorphism, but just with one additional condition, namely, of being one-to-one? Second, have you noted Lemma 3.3 on p.36, which shows that any homomorphism whose domain is a field is a monomorphism unless it is the homomorphism that sends every element to 0 ? With these two facts in mind, you can see that the map i that Stewart starts with is simply any homomorphism of fields, other than the one that sends every element to 0. We are not interested in the map that sends everthing to zero, because it messes other things up (e.g., it won't satisfy f(x^-1) = f(x)^-1), so by assuming i a monomomorphism, Stewart is really saying "a homomorphism other than the bad one that we don't want to deal with." As for the i-hat of his conclusion, in calling it a monomorphism, he is saying that it is a homomorphism, just as you would like, but he is also saying more: that it is one-to-one (which, since K[t] and L[t] are not fields, is more than just saying it is not the zero map). As I mentioned in class, I don't like the fact that Stewart uses the term "monomorphism" as a way of saying "field homomorphism, other than the zero map". I believe that the best definition to use for a homomorphism of rings with 1 is to require it to take 1 to 1 as well as respecting addition and multiplication. If we put this into our definition, then the zero map is not a homomorphism of fields, and in most of the places where Stewart says "monomorphism", one could simply say "homomorphism", and the one-one-ness will follow by Lemma 3.3. But since he has written the book, and chosen what definitions to use, we must accept that "monomorphism" is the appropriate term for him to use for what he is talking about. _________________________________________________________________ You ask whether, where Stewart writes "... either [M:L] = infinity or [L:K] = infinity", they can both be infinite. Certainly! "Or" is not exclusive. (If the prerequisite for a course is "math 110 or math 113", someone who has taken both has certainly satisfied the prerequisites for the course. In many situations, context makes "or" exclusive, e.g., "It must be an animal or a plant", but that is not a consequence of the meaning of "or".) You also ask why, in the proof of Lemma 4.4, "any algebraic extension K(alpha_1,..., alpha_s) : K is finite", when Stewart has just said that not every algebraic extension is finite. Although not every algebraic extension is finite, he is saying here that every algebraic extension _of_the_form_ K(alpha_1,..., alpha_s) : K is so. An algebraic extension of that form is one that can be gotten by adjoining _finitely_many_ algebraic elements alpha_1,..., alpha_s to K. (The extenson A:Q discussed at the bottom of the page is an example which cannot be so obtained.) You suggested that the proper interpretation of what Stewart meant might be "a polynomial extension of finite degree". But by the definition at the bottom of p.47, that is exactly the same as a finite extension. So "finite degree" is what we are trying to prove here, not what we are assuming. You based that guess on the fact that Stewart said "let n = [L:K] ...". But notice that he says this after the word "Conversely". Saying "Conversely" is a signal that he is about to prove the _converse_ of what he has just proved; so the old conclusion becomes the new hypothesis, and vice versa. So in the half of the proof after that word, the condition of being a finite extension, equivalently, having "finite degree, is assumed, and the condition of "having the form K(alpha_1,..., alpha_s)" is then to be proved. But the reverse is true before that word. I hope I don't sound as though I am being over-critical of your question; my aim is to point out some things about how to read mathematical writing. _________________________________________________________________ You ask what Stewart means by the "representative polynomials" in the last paragraph of p.37. He means the polynomials chosen in the preceding sentence -- one representative from each coset, namely, the unique polynomial in that coset that has degree < the degree of m. Whenever one has a one-to-one correspondence between two sets, any operations on one of them can be used to induce "corresponding" operations on the other. (E.g., since we have a one-to-one correspondence between integers 1, 2, 0, -1 etc. and number-words "one", "two", "zero", "minus one" etc., the operations on the integers, "1+1=2" etc. induce operations on the number-words, "one + one = two" etc..) Now here we have a one-to-one correspondence between cosets and representative polynomials (one representative polynomial in each coset); hence the operations on cosets induce operations on the representative polynomials. You wrote "I think I got the general idea, but ...". If you were at my office hours, I could question you on the general idea you had, and know what I needed to clarify or correct. Since we aren't face to face when you send in a question of the day, it would help if you indicated what you had been able to figure out or guess for yourself, rather than having me fish blindly. _________________________________________________________________ You ask how Stewart gets the condition that phi|_K is the identity, at the end of the proof of Theorem 3.6. Well, he has given a precise definition of how phi acts on any element. Try applying that definition to the case where phi is applied to an element of K, and see what it gives. Write me again if you have difficult with this. _________________________________________________________________ You ask about Stewart's statement in the proof of Theorem 3.8, that every element of k(alpha) can be written x_0 + ... + x_n alpha^n where n = deg(m) - a, and ask how we can assert the exact equality "n = deg(m) - 1" rather than just "n < deg(m)". If we were requiring alpha_n to be nonzero, as in the definition of the degree of a polynomial, then we could only say "n < deg(m)". But we are not requiring that; so if we have an expression with fewer terms, we just fill in zero terms to get the indicated form. _________________________________________________________________ You ask what Stewart means by saying that the operation which turns L into a vector space over K "simply forgets some of the structure." Well, if we know the structure of L as an extension of K, we know a lot about it -- it is a set, with operations of addition and multiplication, and a map of K into it, and combining the multiplication with the map of K into it, we can define multiplication of elements of L by elements of K, and see that L is a K-vector-space. But what we know about it is more than a structure of K-vector-space, since we have a way of multiplying elements of L by other elements of L even when neither element comes from K. If we forget about how to multiply elements of L by general elements of L, and just remember how to multiply them by elements of K, and how to add them, are simply left with the K-vector-space structure. So this K-vector-space structure is gotten by forgetting some of the structure. As an example, Q(sqrt 2) and Q(sqrt 3) are non-isomorphic field extensions of Q -- we saw at the end of class today that Q(sqrt 2) does not contain any element which when multiplied by itself gives 3, while Q(sqrt 3) does. However, if we forget how to multiply members of these extension fields by each other, and simply remember their K-vector-space structures, then we see that they are isomorphic as K-vector-spaces: the correspondence p + q sqrt 2 <--> p + q sqrt 3 is a vector-space isomorphism, though not a field isomorphism. So in regarding them a Q-vector-spaces, we have forgotten so much of the structure that we can no longer tell them apart. _________________________________________________________________ You ask how the uniqueness clause of lemma 3.7 implies that, in the situation of Proposition 4.3 the set {1, alpha, (alpha)^2, ..., (alpha)^(n-1)} is linearly independent. A set of elements {x_1, ..., x_n} of a vector space V is linearly dependent if and only if the expression for 0 as a linear combination of these vectors is _non_unique. For "0x_1 + ... + 0x_n" is always an expression for 0 as a linear combination of these elements, while linear dependence of these elements means the existence of an expression a_1 x_1 + ... + a_n x_n for 0 in which not all a_i are zero; i.e., which is different from "0x_1 + ... + 0x_n". So if we know that every element of the field can be expressed uniquely as a linear combination of 1, alpha, (alpha)^2, ..., (alpha)^(n-1), then in particular, 0 can be expressed uniquely, so that set is linearly independent. (In fact, it is not hard to show, using the above and a few similar arguments, that a subset B of a vector space V is a basis if and only if every element of V can be expressed uniquely as a linear combination of members of B. Using this criterion, we see that Lemma 3.7 is precisely equivalent to the statement that {1, alpha, (alpha)^2, ..., (alpha)^(n-1)} is a basis of K(alpha) over K.) _________________________________________________________________ You ask whether in a vector space over a general field there is a danger "that a vector could be linearly dependent with itself (ie if there are multiple scalars that result in the same answer after multiplication?)" I guess you mean, "If x is a nonzero vector space over a field K, could there exist scalars a not-equal-to b in K such that ax = bx?" No. The equation ax = bx is equivalent to (a-b) x = 0; and since K is a field and a and b are distinct, K contains an inverse to a-b. Multiplying the above equation by that inverse, we get the relation x = 0. This contradicts the assumption that x is nonzero. (However, if one replaces the field K by a general ring R, then the analog of a vector space over K is what is called a "module" over R, and what you ask about can definitely happen in that context, since nonzero elements of R need not be invertible. For instance, a module over the ring Z of integers is the same as an abelian group, and in abelian groups, relations n x = 0 are well-known. For another example, if R is the ring of n x n matrices over a field, then one can apply these matrices to column vectors of height n over that field, so the set of such vectors forms an R-module. In this module we have A x = 0 whenever the vector x belongs to the null space of the matrix A.) (There's also a situation in which something like what you referred to _appears_ to be happening, but really isn't. If K is a field of nonzero characteristic p, and n is a multiple of p, then for any element x of any K-vector-space, nx = 0. But here n and 0 are not distinct elements of K, so there is no contradiction.) _________________________________________________________________ You ask whether it makes sense to talk about C(t) as extension of R(t), and ask for [C(t):R(t)]. Yes. The inclusion of R in C induces an inclusion of R[t] in C[t], as noted in the "Definition" (which isn't really a definition) on p.40. This in turn induces a map of fields, R(t) --> C(t), which can be thought of as an inclusion (sending each rational function f(t)/g(t) in R(t) to the same rational function f(t)/g(t) in C(t); although strictly speaking, looking at rational functions f(t)/g(t) as certain equivalence classes, they are not the same things.) So one can ask what [C(t):R(t)] is. In fact, it is 2, just like [C:R]. Can you see how to prove this? _________________________________________________________________ I hope that the discussion in class answered your question about the relation between ruler and compass constructions, and subfields. The ruler and compass operations apply to points of the plane; the subfields are subfields of R. So the points of the plane are not elements of the fields in question; it is the two coordinates of each point that we are regarding as field elements. And it is not the ruler and compass constructions that give the field operations; rather, Stewart _defines_ K_i as the _field_ obtained by adjoining certain elements to another field; so it is the definition of "field obtained by adjoining elements to another field" that gives closure under the field operations. _________________________________________________________________ You ask about "constructing pi", as in Stewart's discussion in the first two lines of p.57. Stewart is not saying that one _can_ construct pi. He is saying that _if_ one could square the circle, then one would be able to construct pi! Reread the Theorem at the bottom of p.56 and the two sentences at the top of p.57 carefully. If you still have difficulty (which at least will not be the difficulty of thinking that Stewart is saying we really _can_ construct pi), let me know (or come to office hours) and I'll try to help. _________________________________________________________________ You ask how, near the end of the proof of Theorem 5.2, Stewart concludes that [K_0(x):K_0] is a power of 2. He has pointed out that [K_n:K_0(x)] [K_0(x):K_0] = [K_n:K_0], and shown that [K_n:K_0] is a power of two. Hence [K_n:K_0(x)] is a divisor of a power of two. But every divisor of a power of two is a power of two! _________________________________________________________________ You ask how to construct parallel lines with ruler and compass. Given a line L and a point P not on L, drop a perpendicular L' from P to L, then erect a perpendicular L'' to L' at P. Then L'' will be parallel to L, and pass through P. | | | ------------*----------------- L'' |P | |L' | | L | ------------*----------------- | _________________________________________________________________ You ask in what sense Plato considered the line and circle the only "perfect" geometric figures. I wondered about that myself. My guess is that the condition they satisfy that he considered "perfection" is perfect symmetry -- every point is "just like" every other, since one can rotate a circle to carry any point to any other, and translate a line with the same result. This is supported by the interest of the Greeks in regular polygons and polyhedra; these are polygons and polyhedra which have the property that every vertex, every edge, and (in the case of polyhedra) every face corresponds to every other vertex, edge and face under motions that preserve the figures. But that's just my guess. _________________________________________________________________ You ask for a hint on how to do Exercise 5.11. OK, my hint is: First figure out how, using the weakened operations of that exercise, to perform the construction "Given points A, B, C, construct a line through A parallel to BC." Then, given points P_0, Q_0, Q_1, if you want to draw a circle centered at P_0 and with radius Q_0 Q_1, construct a parallelogram three of whose vertices are P_0, Q_0, Q_1, and note that the desired circle can be gotten using the weakened operation described, and two verticles of that parallelogram. _________________________________________________________________ You ask what the point is, on p.57, of discussing a lot of possible kinds of geometric construction. The idea of limiting constructions to ruler and compass is a historical accident; one could ask what the point is of spending so much effort on what can be done using that specific set of tools. We do so because of the historical interest, but it is certainly also of interest to ask "What if we changed the set of tools, or the precise description of what we are allowed to do with them?" Stewart records some results on that subject. They aren't relevant to the rest of this course; just some notes that the reader interested in the topic could follow up. _________________________________________________________________ You ask what Stewart means on p.57 by "linearly independent circles". Good question! I don't know. One guess is that if the three circles have equations (x-a_1)^2 + (y-b_1)^2 = r_1^2, (x-a_2)^2 + (y-b_2)^2 = r_2^2 and (x-a_3)^2 + (y-b_3)^2 = r_3^2, then the functions (x-a_1)^2 + (y-b_1)^2 - r_1^2, (x-a_2)^2 + (y-b_2)^2 - r_2^2 and (x-a_3)^2 + (y-b_3)^2 - r_3^2 should be linearly independent in the space of polyomial functions on R^2. Another possibility is that he means something like that the three centers should not lie on a line. I'll include this in the comments I send Stewart at the end of the Semester. If you're very interested in knowing the answer, I could ask whether a colleague knows, or e-mail Stewart about this particular question now. _________________________________________________________________ You ask what the connection implied on p.71 between the method of studying field extensions in terms of their automorphism groups, and the Erlanger Programm is. Well, the Erlanger Programm was to study geometries in terms of their groups of "automorphisms", so in a loose way the ideas are the same. I can't claim that they are exactly the same, but I will note that one of the differences is only apparent: Stewart in this paragraph goes from field structure to group of automorphisms, but his description of the Erlanger Programm goes the other way, from the group to the geometric structure (the "structure" being interpreted as whatever is invariant under the given group). However, as we see later in this chapter, in Galois Theory one also goes from a group H to what is invariant under the group, namely, the subfield H^dagger. _________________________________________________________________ You ask how the relation H \subset G implies that the fixed field of G is a subset of the fixed field of H (p.74). If x is an element of the fixed field of G, that means it is fixed by the actions of _all_ elements of G. Since H is a subset of G, x will in particular be fixed by all elements of H, i.e., it will belong to the fixed subfield of H. _________________________________________________________________ You ask what the "K" in the next-to-last line before the exercises on p.75 is. It should say "Q". _________________________________________________________________ In connection with Stewart's statement on p.75, last line before exercises, "So in this case the Galois correspondence is a bijection", you ask what he means by "the Galois correspondence." That sentence is definitely poorly worded. What he wants to say is, essentially, "So in this case, the maps *: {intermediate fields} --> {subgroups} and dagger: {subgroups} --> {intermediate fields} are inverse to one another, and so give a bijection between the set of subgroups and the set of intermediate fields." As to whether this must be the case whenever the extension is obtained by adjoining finitely many elements -- no; example "2." in the middle of p.73 is a counterexample. _________________________________________________________________ Concerning Stewart's answer to Ex.5.10(h), you ask why pi is not transcendental over R Because it is a zero of the polynomial t - pi, which has coefficients in R. (Note that this polynomial does not have coefficients in Q, so this argument does not contradict the result "pi is transcendental over Q" which Stewart quotes.) _________________________________________________________________ You ask why when Stewart adjoins a zero "zeta" of t^2 + t + 1 to Z_2 on p.81 he gets a field with four elements, rather than just the three elements {0, 1, zeta}. For this, you have to remember the meaning that Stewart has defined for the word "adjoin". If you look it up in the index, you will see that it is defined on p.31, where the field "obtained from K by adjoining Y" is defined to be, not the set K \union Y, but the _field_generated_by_ that set. Now the 3-element set that you refer to, {0, 1, zeta}, would not be a field; it would not be closed under addition. So to get the field it generates, we have to bring in the sum 1 + zeta. There's a lot more to be said about this example; in particular, that the exact set of elements in such a field is described by Lemma 3.7. But the first step is to clarify what you misunderstood. If after looking at this construction in the light of the above answer you have further questions, let me know. _________________________________________________________________ You ask why Stewart says in part 3 of the Example on p.80 that t^2 - 2t - 2 splits over Q(sqrt 3). Well, in part 2 of that example, he has already observed that the zeros of t^2 - 2t - 2 in C are 1 +- sqrt 3. (These are obtained using the quadratic formula. We haven't developed that formula as of this point in this course, but we can always use it and check by hand that the results we get, when substituted into the polynomial, are indeed zeros). Since 1 +- sqrt 3 lie in Q(sqrt 3), we can write t^2 - 2t - 2 = (t - (1+sqrt 3)) (t - (1-sqrt 3)) over that field. Even if you hadn't noticed the connection with the preceding example, you should probably have asked yourself "What are the zeros of t^2 - 2t - 2 in C ?" When you figured out the answer, you would have seen why Stewart had made his claim. _________________________________________________________________ You ask why on p.81, in the next-to-last sentence of section 8.1, Stewart says that t^2 + t + 1 doesn't split in any smaller field than Z_2 (zeta). Well, any subfield of Z_2 (zeta) must contain its prime subfield Z_2. (To be a field it must contain 0 and 1, and these always generate the prime subfield of a field -- in this case, they comprise it.) And if the given polynomial is to split over that subfield, the subfield must contain the zeroes of the polynomial, which are zeta and 1+zeta. So, as it contains Z_2 and contains zeta, such a subfield must contain the field they generate, Z_2 (zeta). _________________________________________________________________ You asked a question about Chapter 6. But Chapter 6 is not in our reading! If you look at the list of readings you will see that it is the one chapter that we skip entirely. Well, I've looked at the computation you asked about, equation (6.1) on p.62. When Stewart writes "Integrating by parts", he is summarizing a more complicated computation. In the preceding integral, write the integrand, (1 - x^2)^n cos(alpha x)dx, as u dv where u = (1 - x^2)^n and dv = cos(alpha x)dx. When you apply the formula for integration by parts, note that assuming n >_ 1, the difference between the values of the term uv at -1 and +1 is 0, because the factor u is 0 at both ends. This leaves us with the integral of v du. This differs in form from the original expression because it involves a sine rather than a cosine, so we repeat the process. Since Stewart has assumed n >_ 2, the "uv" term in the result still contributes nothing, while the integral is now of approximately the desired form. Precisely, we have (1-x^)^(n-2) times an expression involving a constant term and a multiple of x^2. Writing this in terms of a constant and a multiple of (1-x^2), we see that it becomes a linear combination of (1-x^2)^(n-2) and (1-x^2)^(n-1), so the integral becomes a linear combination of I_(n-2) and I_(n-1). I haven't checked the details of the computation. If you want to work it through, and have further difficulties, let me know. But please keep track of the sequence of readings for this course, and remember that your question has to be on the current reading. If you want to include with it some question(s) about past material, or even material we have skipped, that's OK; but it doesn't remove the requirement of asking a question about the reading for the day. The reading for Wednesday is #9. _________________________________________________________________ You ask about the classification of finite fields. This is done in Chapter 16. The result of the classification is much simpler than the situation for finite groups. As you can see from the chart on p.xii, Chapter 16 can be read immediately after Chapter 8; so though we will be reading the book in order, you can look at that chapter at the end of this week if you wish. If you do and have any questions, let me know. _________________________________________________________________ You ask what the "zeta" on p.81, line 4 is. It is a zero of the irreducible polynomial t^2 + t + 1, adjoined to Z_2 via the construction of Theorem 3.5. How could you have seen that Stewart meant this? Note that he wrote two lines earlier "we must go back to the basic construction of the splitting field". This construction was given in the proof of Theorem 8.1. Following that proof, you find the relevant step to be "Using Theorem 3.5 we adjoin sigma_1 to K ...". And looking back at the proof of Theorem 3.5, you would see how that is done. _________________________________________________________________ You ask about the last paragraph (middle of p.83) of the proof of Theorem 8.4, and why Stewart assumes there that theta_1\in L. He wants to prove that L is normal, i.e., that if it contains one zero theta_1 of f, then it contains every other zero, theta_2 of f. So in this paragraph, he introduces the assumption theta_1\in L, and shows that (as a result of the preceding computations) this implies that theta_2\in L. _________________________________________________________________ You ask whether, if K is a field, and alpha, beta are elements of an extension field which are linearly independent over K, then we must have have K(alpha, beta) = K(alpha + beta). No. For a counterexample, let K = Q alpha = sqrt 2 beta = sqrt 3 - sqrt 2. Then Q(alpha, beta) = Q(sqrt 2, sqrt 3), but Q(alpha + beta) = Q(sqrt 3). But in this case, something _like_ what you stated is true, as one sees in proving part (b) of Exercise 4.19. _________________________________________________________________ I'm not sure whether you cleared up to your own satisfaction the question of why we can rearrange generators, e.g., why if K is a field, and theta_1, alpha_1, ..., alpha_n are members of some field M containing K, then (1) K(theta_1) (alpha_1, ..., alpha_n) = (2) K(theta_1, alpha_1, ..., alpha_n) = (3) K(alpha_1, ... , alpha_n, theta_1) = (4) K(alpha_1, ... , alpha_n) (theta_1). Intuitively, (1)-(4) are four descriptions of the subfield of M generated by the set (5) K \union {theta_1, alpha_1, ..., alpha_n}. (Recall that this subfield can be described as the least subfield of M that contains the above set (5), as the intersection of all subfields of M that contain (5), or as the set of elements of M that can be obtained from the elements of (5) by field operations.) In the case of the fields numbered (2) and (3) above, the statement that they are the subfield of M generated by (5) is equivalent to their definition (p.31). To establish that the remaining cases can also be so described, you should verify for yourself that (1) and (4) are fields which contain the set (5), and that any subfield of M which contains the set (5) must contain (1) and (4). Therefore they, too, are equal to subfield of M generated by the set (5). _________________________________________________________________ You ask what Stewart means on p.82, a few lines before the Theorem, about an extension having "well-behaved" or "badly behaved" Galois group according to whether "the Galois correspondence is a bijection". The Galois correspondence actually consists of a pair of maps, "*" and "dagger", so it would be better if rather than saying it "is a bijection", Stewart said that those maps were inverse to one another (and so give a bijective correspondence). Anyway, you say that he can't mean what he seems to, because for Q(sqrt 2) : Q, the correspondence is a bijection, but "the extension is not normal" because (x^2 - 2)(x^2 -3) has one root in that extension, but doesn't completely split. Well, in fact that extension is normal -- if you look at the definition at the top of the page, you will see that it says that every polynomial _irreducible_over__K_ with a root in L splits over L. But (x^2 - 2)(x^2 -3) is not irreducible over Q, as shown by the fact that you have factored it! _________________________________________________________________ You ask for examples of non-normal simple extensions that are significantly different from the book's. If f is any irreducible polynomial over Q of degree > 1 but having only one real root alpha, Q(alpha) will be simple, but it won't be normal because it will contain the root alpha of f, but not the other roots, since they're not real. One can get lots of examples of this sort which one can prove irreducible using Eisenstein's criterion; e.g., anything of the form t^3 + a t + b where a and b are integers, a is positive (so that the polynomial is everywhere increasing, and can only have one real root), and a is divisible by some prime that also divides, b, but whose square doesn't. (E.g., a = 100, b = any of 2, 5, 6, 10, 14, 15, 18, 20, ... ). Of course, not all non-normal simple extensions have this form; but because we can use elementary properties of the real numbers to test for number of real roots, and Eisenstein's criterion to prove irreducibility, it gives lots of easy examples. _________________________________________________________________ You asked why Proposition 4.3 was applicable on p.83; specifically, why theta_1 and theta_2 both have f as their minimal polynomial. Remember that the minimal polynomial of any element alpha is the monic common divisor of all polynomials having alpha as a zero. Since f has theta_1 as a zero, it must be divisible by the minimal polynomial of theta_1; but since it is irreducible, it has (up to units) no divisors other than itself and 1. So it must (up to units) equal the minimal polynomial of theta_1 -- and similarly, that of theta_2. So strictly speaking, it is not f itself, but the monic polynomial that one gets by dividing f by its leading coefficient that is the minimal polynomial of theta_1. Anyway, by the same argument, the same polynomial is the minimal polynomial of theta_2, so Prop.4.3 (and likewise, Theorem 3.8) is applicable. _________________________________________________________________ You ask what the concept of separability is good for. As I noted on Wednesday, inseparability represents an obstacle to our goal of describing intermediate fields of an extension L:K as fixed fields determined by subgroups of the Galois group: If L = K(alpha) where alpha is a pth root of an element of K, then all the roots of the minimal polynomial of alpha over K are the same, so no automorphism of K(alpha) over K can move alpha, so we can't distinguish K(alpha) from K in terms of what automorphisms fix its elements. Although I didn't go into the case of inseparable elements with more complicated minimal polynomials than t^p - a, these have similar problems. As we will see, the two conditions "separable" and "normal" are what we need to make Galois theory work. Galois theory will turn out to give strong information about the structures of field extensions, so it will not be surprising to learn that inseparable extensions differ from separable ones in further ways; but the above should be good enough as a starting point. _________________________________________________________________ You ask why e^{2 pi i/5} etc. are zeroes of t^4 + t^3 + t^2 + t + 1. To answer that, I need to know how much you know about complex exponentiation. Do you know where e^{a+bi} lies on the complex plane? Do you know how to multiply complex numbers expressed in terms of their angles with respect to the real axis and their magnitudes? ... As I say on the course information sheet, if you asked such a question at my office hours, I would question you to see what you understood and what needed to be explained. But that is very time-consuming to do by e-mail. If you had either said that you didn't understand how to raise e to an imaginary power, or said that you understood how to define such an exponential, but didn't see what special properties that power of e would have, or said that you saw some special properties of that element (and said what they were) but that you didn't see how they led to its being a zero of the polynomial -- then I would have known where to start! So let me know how much you know about the topic, and I'll go from there! _________________________________________________________________ You ask about Stewart's writing "tau = v(u) / w(u)" just before the last display on p.84. He is using the fact that tau has been assumed to lie in the field of fractions K of the polynomial ring K_0 [u]. Any element of the field of fractions can be written as a fraction with numerator and denominator in the ring. _________________________________________________________________ You ask how Stewart deduces near the beginning of the proof of Proposition 8.6 that Df = 0. f and Df share a common factor of degree >_ 1 (Lemma 8.5). This means a common factor in K[t]. (Stewart doesn't make that explicit in the statement of the Lemma, but is clear from the proof.) But since f is irreducible, the only factors of f in K[t] are (up to multiplication by units) f and 1. Since 1 does not have degree >_ 1, the common factor must be f; so Df must be divisible by f. But by the properties of formal differentiation, Df has degree < deg f. What multiples of f have smaller degree than deg f? Only 0! _________________________________________________________________ You ask whether Galois theory can be extended to infinite field extensions and groups. I hope I will find time to say a little about this in class in a few days, but in case I don't, here are some brief facts. For non-algebraic extensions, I don't think there is any nice theory. E.g., consider the simple transcendental extension Q(u):Q. This has an automorphism sending u to u+1 (and hence r(u) to r(u+1) for every rational function r). Let H be the cyclic group generated by that automorphism. It is not hard to show that the fixed field of that automorphism, and hence of the group H, is Q. But the full group of automorphisms fixing Q is much larger than H -- it is noncommutative, and consists of automorphisms taking u to (au+b)/(cu+d) for all rational numbers a, b, c, d with ad-bc nonzero. If one starts with a larger transcendental extension, things are even worse -- one can get examples similar to the above, but where the full group of automorphisms is so big no one even knows how to generated it. For infinite separable normal _algebraic_ extensions, however, there is a nice theory, though things are not as straightforward as in the finite case. As an example, let L be the field generated over Q by the square roots of all positive integers, equivalently, by the square roots of all prime numbers. Note that an automorphism of L over Q will send every square root of a positive integer to itself or its negative, and will be determined by what it does to the square roots of the primes (since every positive integer is a product of primes). So if alpha is an automorphism of L over Q, we can specify alpha by specifying the set S_alpha of prime numbers p such that alpha(sqrt p) = - sqrt p. One can prove that for every set A of primes there exists an automorphism alpha such that S_alpha = A; so automorphisms of L correspond to subsets of the set of all primes. It is not hard to see that the automorphisms corresponding to _finite_ sets of primes form a subgroup H of the Galois group G, and that the fixed field of H is just Q. Hence H-dagger-* is the whole group G, which is again much bigger than H. However, in this case there is a natural way to get H-dagger-* from H. Note that every element alpha of G can be "approximated arbitrarily closely" by members of H, meaning that for every finite subset of L, there is a member of H which acts the same on all members of that finite subset as alpha does. This has the consequence that G is the _closure_ of H in a certain topology; and this is representative of the general situation in infinite Galois theory: The operators * and dagger give one a bijective correspondence between intermediate fields and _closed_ subgroups of the Galois group with respect to a certain natural topology. (I have a write-up on the subject in the "graduate course handouts" page on my web page -- the first item under the "250B" list -- but it assumes a lot of advanced material.) _________________________________________________________________ You ask whether the introduction of the concept of linear independence of monomorphisms means that the set of such monomorphisms forms a vector space. No. The set of _all_ maps from K to L forms a vector space, and the monomorphisms are a _subset_ (not a subspace!) of that vector space. In fact (as Stewart proves), they are a linearly independent subset, and a nonempty linearly independent subset of a vector space can never be a subspace. (Although a sum of homomorphisms will be a map f such that f(x+y) = f(x) + f(y), it will not satisfy f(xy) = f(x) f(y). You can check this for the sum of the identity map of C and the complex-conjugation map; or the sum of any homomorphism with itself.) _________________________________________________________________ You ask for other examples of the "trick" used to prove Lemma 9.1. I looked in my file on the subject, which reminded me that one such example comes up at a more elementary level than this: The linear independence of eigenvectors corresponding to distinct eigenvalues of a linear transformation. Another case of the method occurs in Stewart's proof of Theorem 9.4, or, in my handout, in Proof 1 of Lemma 9.6. But as Proof 2 of that lemma shows, this can really be done without an independent application of the trick. For the application of similar methods to skew fields see Jan Treur, Separate zeros and Galois extensions of skew fields, J. Algebra 120 (1989), 392-405. For a different sort of application, see section 30 of "Cogroups and co-rings in categories of associative rings", by myself and A. Hausknecht, where we get a result equivalent to something called "Sweedler's pre-dual of the Jacobson-Bourbaki Theorem". The "trick" is used on the lower part of p.159. I don't remember whether it occurs similarly in the paper of Sweedler's we refer to, nor in the results of Jacobson and Bourbaki that Sweedler was "predualizing"; but you can follow the references and see. Lemmas 5.9.1 and 5.9.2 and Theorem 5.9.3 of P.M.Cohn's book "Free Rings and Their Relations" (2nd edition) use the same method. _________________________________________________________________ You ask why Stewart says, in the first sentence on p.89, that the Fundamental Theorem of Galois Theory will require separability and normality. Separability and normality are not actually needed for the equality H-dagger-* = H that Stewart mentions in this sentence. They are, however, needed for the other half of the Fundamental Theorem, the equation M*-dagger = M. _________________________________________________________________ You ask how the proof of Lemma 9.1 on p.90 uses the assumption that the lambda_i are nonzero. Because the zero homomorphism is excluded, the least n for which an equation (9.1) holds must be greater than 1 (since a single nonzero homomorphism can't satisfy such an equation). Hence one can choose distinct homomorphisms lambda_1 and lambda_n to finish the proof. _________________________________________________________________ You ask why Stewart says, in the proof of Lemma 9.1 on p.90, that without loss of generality we can assume all a_i nonzero. Because if any of them are zero, we can drop those terms from the equation, and get a true equation of the same form (9.1), but with no nonzero coefficients. Note that this is not quite the same equation we had to begin with. (It is a linear dependence relation among a smaller set of lambda's.) This is in fact what mathematicians generally mean when they say something is true without loss of generality: that if we have a case in which it is not true, we can modify it (usually in some straightforward way) to get a case in which it is true, and such that proving what we want about that modified case will establish what we wanted to begin with. (In this case, it will give a contradiction.) _________________________________________________________________ You ask how Stewart deduces from the final display on p.84 that f is irreducible. He began that paragraph by assuming that f was not irreducible, namely that it factored f = gh. From this, he obtained the final display, which led to a contradiction, since the degree of the first term is a multiple of p, while that of the second term is one more than a multiple of p, hence they cannot cancel, hence the equation cannot hold. This contradiction shows that the assumption that f was reducible was false. _________________________________________________________________ As you say, the map g_i |-> g g_i of Stewart's Lemma 9.3 will not be a homomorphism. You ask whether it would be an automorphism if G was abelian and g was idempotent. The trouble with this is that in a group, the only idempotent element is the identity. For if x = x^2, then multiplying that equation by x^-1, we get e = x. However, there are several related ways to get homomorphisms or automorphisms. (By the way, the word for "homomorphism into itself" is "endomorphism", so I will use that below.) (a) If we replace "group" by "monoid" (defined like a group, but without the assumption that every element has an inverse), then we can have nonidentity idempotent elements, and it is true that in an abelian monoid, multiplying by such an element is a endomorphism. (More generally, in any monoid multiplying by an idempotent element in the center is a endomorphism.) But it still won't be an automorphism, because whenever x is an idempotent element other than e, multiplication by x is not one-to-one: x.x = x.e. (b) If we take two elements g and h, and multiply by one on the right and the other on the left, we still get a bijection, g_i |-> g g_i h. In a group, this will be an automorphism if g = h^-1. This automorphism is called conjugation by h; you've probably seen it in 113. (c) If we take a group, and define on it a 3-ary operation = x y^-1 z, and then forget the original group operation, then the resulting system of a set together with this 3-ary operation is called a "heap". Now the original operation g_i |-> g g_i, though not a homomorphism of groups, is a homomorphism of heaps: = g . _________________________________________________________________ You ask on what basis Stewart says on p.93, line 7, that k is nonzero. I think that if you look and see how Stewart gets "k", you will have the answer to that question. Stewart is not very explicit about that point; so perhaps your question should really have been, "What is k, and what are the elements z_i \in K_0 ?" In fact, if you look at lines 3 and 4 of this page, you will see that the z_i are the elements y_i y_1^-1, so k is y_1. And this is nonzero by the way Stewart chose to arrange the y's on the middle of the preceding page, with all the nonzero ones first. I hope that the version of this proof that I gave in my handout is clearer! _________________________________________________________________ You ask how, in my handout on Theorem 9.4, I determine the K-dimension of the space of K_0-linear maps K --> K to be m. I do this in two ways. One is to first determine its dimension as a K_0-vector-space to be m^2, and then note that its dimension as a K-vector-space is 1/m of that dimension. The other, which I sketch in parenthesis, is based on the fact that every such map is determined by an m-tuple of elements of K. Assuming that it is the first argument that you are asking about, is what you want to know why the space in question has K_0-dimension m^2, or why its K-dimension is 1/m times its K_0-dimension, or both? As I note on the class information sheet, if you asked this in my office hours, I would question you to see what you understood, and what I needed to clarify. But this is much harder to do by e-mail, so you should try to be very explicit about what you understand and precisely what you are asking. So please let me know, and I will then answer your question. _________________________________________________________________ You ask about the meaning of "larger" and "smaller" in the second sentence of the third paragraph of the handout on Theorem 9.4. In general, mathematicians use "larger" to mean "containing the other" and "smaller" to mean "contained in the other". That is what I mean here. However, it happens that what I say here also makes sense with "larger" and "smaller" interpreted as "having larger cardinality" and "having smaller cardinality", because, as I note in the next sentence, the method of proof will just be based on the orders of the groups. (Perhaps I should be more precise about how mathematicians "generally" use "larger" and "smaller". "Containing" and "contained in" are a common case; another is "containing something isomorphic to" and its reverse, "embeddable in", as when one calls a vector space of higher dimension "larger" than one of lower dimension, even if neither contains the other. One can in fact use the words with reference to any partial or total order relation on a set, as when referring to a "larger" or "smaller" real number. So it would be more accurate to say that the words refer to whatever partial ordering is natural to the situation; not in general to cardinality alone.) _________________________________________________________________ You ask why, in proving Theorem 8.11 of my addendum to Chapter 8, it is sufficient to show that the set of separable elements is closed under the field operations. Well, that means that the set of elements of L separable over K forms a subfield of L. Now we are assuming L is generated over K by a set X of elements that are separable over K. The statement that X generates L over K means that the only subfield of L containing K and X is L itself; so in particular, the subfield of elements of L separable over K, since it contains K and X, is L itself; i.e., L is separable over K. _________________________________________________________________ You ask whether, in Lemma 9.1, the elements of G, being proved linearly independent, form a vector space, or a basis of a vector space. As I said in class, they don't form a vector space, because the sum or product of two homomorphisms is not in general a homomorphism. Rather, they form a _subset_ of the K-vector space of all K_0-linear maps K --> K. Dedekind's result proves that this subset is a linearly independent one. As to whether this make it a basis "of a vector space" -- well, in a vector space V, if X is a linearly independent set, then X will be a basis _of_the_subspace_of__V__spanned_by__X. But it won't necessarily be a subspace of V itself, unless it spans V. _________________________________________________________________ You ask where Stewart gets his "four candidates for Q-automorphisms" at the bottom of p.93; specifically, how he excludes other possible arrangements of the coefficients p, q, r and s. Stewart's four candidates are not gotten by looking at possible permutations of the coefficients. (There is no reason why an automorphism should permute the coefficients. For instance, if tau is a zero of t^2 - t - 1 in an extension of Q, then the nontrivial automorphism of Q(tau) sends p + q tau to (p+q) - q tau .) Rather, they are gotten by taking the four possibilities for alpha(omega) that Stewart has found in the preceding sentence, namely omega, omega^2, omega^3 and omega^4, and figuring out, using the definition of homomorphism, how a homomorphism having each of these properties must act. For instance, consider alpha_2, defined by alpha_2(omega) = omega^2. Using the definition of homomorphism, we see that alpha_2(omega^n) = (omega^2)^n = omega^(2n). This immediately gives alpha_2(omega^2) = omega^4. It also gives alpha_2(omega^3) = omega^6, but omega^6 is not among the terms occurring in (9.7). But since omega^5 = 1, we see that omega^6 = omega, so alpha_2(omega^3) = omega. Similarly, alpha_2(omega^4) = omega^2; hence alpha_2 has the form shown in the final display. The key to seeing what Stewart must mean is the phrase that begins the sentence, "This gives". It shows that the possible values for alpha must somehow be determined by the list of four possiblities, with which the preceding sentence ends. I.e., one must be able to deduce from the fact that alpha(omega) has one of those forms the form that alpha(p + ... + t omega^4) has. One sees that the definition of "homomorphism" allows one to do so. _________________________________________________________________ You ask where the first equality in the proof of Corollary 9.5 (p.93) comes from. It comes from the tower law. At this point, you need the make the tower law an automatic part of your way of looking at degrees of extensions, so that whenever you see two successive field extensions, you are automatically conscious that their degrees are related in the way that law states! Do you have any question about how the tower law gives that equality? _________________________________________________________________ You ask what I mean, in the second sentence of the first proof of Lemma 9.6, by "Choose (y_1, ..., y_N) \in V - {0} to minimize the number of y_i which are nonzero;" specifically, whether I am dropping elements which are zero out of the set. I'm not sure what you mean. The first question is whether by "elements which are zero" you mean "elements of V which are zero" or "elements y_i which are zero". In V, there is, of course, just one element which is zero, namely the vector 0 = (0,...,0), and in writing "V - {0}" I mean "all elements of V which are not zero", i.e., I am excluding (you can say "dropping") 0 from the set of elements we consider. If you mean "elements y_i which are zero", then no, I am not dropping them. Rather, I am saying that from all vectors (y_1, ..., y_N) \in V - {0}, we are to choose one for which the number of y_i that are nonzero is the smallest. For instance, if N=6, and we look at all elements of V - {0}, and ask how many nonzero components each such element has, there might be some with 2 nonzero components, some with 4, some with 5 and some with all 6 components nonzero. In that case, choosing an element (y_1, ..., y_N) \in V - {0} to minimize the number of y_i which are nonzero means choosing an element (y_1, ..., y_N) for which the number of nonzero components is 2 (rather than 4, 5, or 6). Perhaps you took the word "minimize" to imply making some change in the N-tuple. But it doesn't; it just means "make the minimal choice". Does this help? Is this what your problem was? _________________________________________________________________ You ask about Stewart's application of Theorem 10.1 in the last sentence of section 10.1 (p.98). To discover what he is doing, look carefully at the statement of that Theorem. It concerns a K-monomorphism from a subextension of a normal extension field into that field. Now look at the situation in the proof of the Proposition. The only normal extension field under discussion is L, so to apply the Theorem, Stewart must be considering a K-monomorphism from some subextension of L into L. But tau is described as an isomorphism K(alpha) --> K(beta). So where does he have a K-monomorphism into L ? Answer: He must be using the fact that K(beta) is contained in L to regard tau as a map into L. Do you see that, regarding tau in that way, the application of the Theorem gives the desired result? _________________________________________________________________ You ask whether there were other examples of inseparable irreducible polynomials than those given on p.84. Well, Proposition 8.6 shows you that any inseparable irreducible polynomial has to have certain features in common with that example, namely that it has to be over a field of nonzero characteristic p, and that the exponents of all powers of t have to be multiples of p. With those restrictions, you should be able to see that one can do a lot of variations on the particular example on p.84. First of all, though Stewart assumes K_0 = Z_p in that example, he doesn't do anything that requires that particular field; he merely needs characteristic p; so one can start with K_0 any field of that characteristic. Secondly, the polynomial need not have the very simple form shown. It is easy to show that t^(p^n) - u works just as well. Variants such as t^2p - t^p - u also work; however, with the tools we have so far, it is harder to show them irreducible, so Stewart just gave the simplest case. _________________________________________________________________ As you suggest, at the very end of the proof of Theorem 10.1 on p.97, when Stewart says "Therefore sigma is an automorphism of L", this is synonymous with the statement in the preceding sentence that it is an isomorphism of L with itself. I think he wanted to get the word "automorphism" in there so that he could combine it with the observation in this sentence that its restriction to K is the identity, to get the final conclusion that it is a K-automorphism of L. _________________________________________________________________ You ask whether the material we've been looking at is related to the concept of "lifting" in topology. I can see only a very loose connection. When one has an interesting map of sets f: X --> Y (especially where X and Y consist of complicated structures of some sort), then given an element y\in Y, if we can find an element x\in X such that f(x) = y, we may call it a "lifting" of y to an element of X. So in topology, if we have a map of spaces m: S --> T, and let X, Y be the set of all paths in S and all paths in T respectively, then a map f: X --> Y is induced, and given a path y in T one can try to "lift" it to a path x in S. In the present chapter, given a subfield M of a field L, we can let X be the set of automorphisms of L, Y the set of monomorphisms M --> L, and f: X --> Y the operation of restricting an automorphism of L to M to get a monomorphism M --> L. In that sense, Theorem 10.1 concerns the "lifting" of monomorphisms M --> L to automorphisms of L. _________________________________________________________________ You say in your answer to your pro-forma question that near the end of the first paragraph of the proof of Theorem 10.3, p.98, "since N is a splitting field for f and f splits in P, therefore P = N". But this does not follow directly from the definition of splitting field. A splitting field for f is a _smallest_ field in which f splits. So the facts that N is a splitting field for f and f splits in P imply that P _contains_ N. Combining this with the assumption we have made that P is a subfield of N, we conclude that P = N. _________________________________________________________________ You ask why Stewart's proof of Theorem 10.3 starts by constructing a splitting field for f over L, rather than a splitting field for f over K. We want the field we get to contain L. If we constructed a splitting field for f over K, we could, with some work, show that it contained a subfield isomorphic to L, and so by making identifications, we could assume without loss of generality that it contained L; but Stewart's way of doing it has it containing L from the start. _________________________________________________________________ You ask whether the normal closure of an extension L:K is "truly unique", and not just unique up to isomorphism. Actually, there are _three_ kinds of uniqueness one could ask about, and although the normal closure is not "truly unique", it is unique in the third sense. First, why is it not truly unique? Because if N is a normal closure of L:K, then one can create an extension of L isomorphic to N but not equal to N, and it will also be a normal closure. Loosely speaking, just take a different set of elements in one-to-one correspondence with the elements of N, and give them the same field structure as that of N. What is the third kind of uniqueness? If a given extension field M of L contains a normal closure N of L over K, then N is the _only_ normal closure of L over K in M. So, for instance, given any algebraic extension L:K lying within the field C of complex numbers, there is a unique subfield N of C which is a normal closure of L:K. _________________________________________________________________ You ask about the equalities m(alpha) = tau(m(alpha)) = m(tau(alpha)) in the display on p.99 The first holds, as you say, because m(alpha) = 0, and tau, as a homomorphism, must send 0 to 0. The second holds because tau is a K-homomorphism, and m has coefficients in K. Specifically, if m(t) = Sigma a_i t^i, then m(alpha) = Sigma a_i alpha^i; now apply tau to this, and use the fact that tau(a_i) = a_i. _________________________________________________________________ You ask whether one always form normal closures as Stewart does at the top of p.99 by "adjoining the missing zeroes". Yes, although this language of Stewart's was vague, because he was just giving an intuitive summary of what he had done. Note that in that example, he was working with subfields of C, and since every polynomial splits over C, the "missing zeroes" could all be found there. In general, we may not be given an extension containing our L over which every polynomial splits; so we have to "create" the missing zeroes as in the proof of Theorem 3.5. You ask in particular about the case of an extension of infinite degree. Well, first recall that my errata to Stewart involved the insertion of the word "algebraic" before "extension" in the Definition at the top of p.82. So even when we are talking about infinite extensions, in discussing normality we must restrict attention to infinite _algebraic_ extensions. For such extensions, one can show using the Axiom of Choice that the process of adjoining zeroes as in the proof of Theorem 3.5 can be "performed infinitely many times", and will give a field with the properties of the normal closure. But that is outside the scope of Math 114. _________________________________________________________________ You asked about the relation between the results in Exercise "7.8" and De Morgan's laws. There's a formal similarity: In a Boolean ring (or if you only had this special context, in the set of all subsets of a set X), the operation of _complement_ interchanges unions and intersections; and here, the operators * and dagger interchange least upper bound (= subobject generated) and intersection. But it's hard to get more in the way of relationship, since in one case one is looking at one map (complementation) from a set to itself, and here we are looking at two maps between two different sets. _________________________________________________________________ You ask how one could have discovered an ingenious proof like that of Theorem 10.6. Well, one thing I suspect is that the shortest route to the proof that one could come up with in retrospect is not the way it actually developed historically. And I don't know the historical development myself. In particular, there is the question of whether the original statement of the result looked anything like the present formulation; and how one could have come up with the present statement. However, we'll have to ignore these historical questions. Taking the result as it stands, I look at it as follows. We want to count the possible ways of mapping L into N; intuitively, of "fitting a copy of L into N". How to we study other questions of this sort? Suppose, for instance, that we have a cube of side 1, and we also have a right triangle with two sides of length 1, and want to see how many ways we can paste that right triangle onto the cube, so that the right angle goes onto one of the vertices, and the adjacent sides go to two sides of the cube. We can first ask where the right angle is to be placed; there are 8 possibilities. Once we have placed it, we can choose one of the adjacent edges of the triangle, and ask where it can go; there are three possibilities, the three edges of the cube that come out of that vertex. And once we have placed this, the remaining edge can go to either of the other two other edges coming from that vertex. So altogether there are 8 x 3 x 2 = 48 ways of placing the right triangle on the cube. Now we return to our field L, and the problem of counting the ways of embedding it in a normal closure N. If we write L = K(alpha_1,..., alpha_m), then we can count the choices for where alpha_1 will go, for each such choice count the possibilities for where alpha_2 will go, and so forth. The number of choices at each stage is degree of the element alpha_i we are looking at over the extension generated by the preceding terms; i.e., the degree of the extension K(alpha_1,..., alpha_i) : K(alpha_1,..., alpha_(i-1)). The product of these numbers is the degree of the whole extension K:L, so that degree is the number of embeddings. In saying "count the choices at each step", I am referring to an inductive process without using the words "By induction". Stewart does formalize the result as an induction; the process of doing so renders a proof simpler, but sometimes makes it look more mysterious. As I mentioned in class, he had a choice of using the inductive hypothesis "at the bottom", "at the top", or "in the middle", and chose to use it "at the top" (applying it to L:K(alpha), rather than, say L' : K where L = L'(alpha)). And instead of asking "In how many ways can we extend a given partial homomorphism?", he said, in effect, "Let's take a particular extension of our partial homomorphism, and see in how many ways we can modify it." One has many choices like this in writing up a proof, even when the underlying idea is the same. I don't know whether these comments answer your question ... . _________________________________________________________________ You ask about the line "Hence we may assume" on the middle of p.101. I meant to talk about that, but forgot to put it on my list! What he means is, "If the result is true whenever M is normal, then it must be true in all cases -- because if we have any old M, then just take its normal closure, and apply the result for normal extensions to that closure, instead of to M, and the result for M will follow. Hence, it is enough to prove the result for normal M, since as we have just seen, the general result will follow from this case." This is the kind of reasoning that is frequently meant when mathematicians say "Without loss of generality, we may assume ...". _________________________________________________________________ You ask about the sentence preceding the statement of Theorem 10.10. This is a continuation of the ideas of the preceding paragraph, i.e., the proof of Theorem 10.9. In proving Theorem 10.6, where the extension was assumed separable, Stewart used the fact that the number of zeroes of the minimal polynomial f of alpha in a splitting field was exactly the degree of f. In proving Theorem 10.9, he uses the weaker statement that, without the assumption of separability, the number of zeroes could still be asserted to be _< the degree. In this sentence, he observes that if alpha is in fact _inseparable_, then the number of zeroes will be _less_than_ the degree. In each case, the same calculation that was used in the proof of Theorem 10.6 is followed, but with a different statement about the relation between the degree and the number of zeroes. _________________________________________________________________ You ask what Stewart means by "order-reversing" on p.104, Theorem 11.1, statement 2. Stewart understands the set of intermediate fields and the set of subgroups of G to be partially ordered by inclusion; so "order-reversing" means "inclusion-reversing", i.e., if the intermediate field M is contained in the intermediate field M', then the subgroup M* will _contain_ the subgroup M'*, and similarly for the operation dagger. (These are properties we have seen before; he is just describing them by a brief phrase here. Incidentally, have you seen the concept of "partially ordered set", or "partial ordering" on a set? If not, my statement above that Stewart understands the set of intermediate fields and the set of subgroups to be partially ordered by inclusion won't mean anything to you; but hopefully what follows it will still be clear.) _________________________________________________________________ You ask why, on p.111, Stewart can say the fixed field of U is "clearly" Q(i sqrt 2). Actually, all the fixed fields except those of C and E can be gotten fairly easily using the following observation: If H is any subgroup of G other than C or E, then _either_ H has no element in which sigma appears with odd exponent, _or_ H contains sigma^2. Consider these two cases separately: (1) Suppose H has no element in which sigma appears with odd exponent. Then every element of H, when applied to any of the basis elements i^m xi^n, (m = 0, 1, n = 0, 1, 2, 3) changes it either to itself or its negative. (Elements in which sigma has odd exponent send xi to +-i xi, and hence permute some of these basis elements, along with sign-changes.) It follows that the only linear combinations of the given basis elements that are fixed under H are those in which the basis elements that get their signs changed by some members of H have coefficient 0. In each case, if we list these basis elements we get an immediate description of H^dagger. (2) Suppose H contains sigma^2. This element fixes those basis elements in which xi has even exponent, and changes the sign of those in which xi has odd exponent. Hence every element of H^dagger involves only basis elements in which xi has even exponent. Now for basis elements in which xi has even exponent, _every_ element of G either leaves the element fixed or changes it to its negative; in particular, this is true of every element of H, so as in (1) we see that H^dagger will consist of all linear combinations of a certain subset of our basis. _________________________________________________________________ At first I was puzzled by your asking why someone might think that the situation Stewart refers to at the top of p.115 implied that each G_i was normal in G. Then I saw that, as you indicate, Stewart just refers there to "(13.1)". Well, that's not what he means -- he means "(13.1) together with condition 1 following it". Yet another item for me to write him that he should clarify! Thanks for pointing it out. Do you see that (13.1) together with condition 1 might lead people to think that all G_i are normal in G ? And can you see that, as he points out, it really does not follow? _________________________________________________________________ You ask whether on p.116, in the proof of statement 2 of Theorem 13.2, the normality statements follow by writing "G_i N / N =~ G_i". No -- that isn't true in general! After all, suppose we have G_i = N. Then G_i N = NN = N, so in that case G_i N / N is trivial. Rather, the relation G_i+1 N / N <| G_i N / N is a case of the relation "A/H <| G/H" in part 2 of Lemma 13.1; in this case with N in the role of H, G_i+1 N in the role of A, and G_i N in the role of G. _________________________________________________________________ You ask about the first step in the last display of the proof of part 2 of Theorem 13.2 on p.116, namely G_(i+1)N/G_iN = (G_(i+1))(G_iN)/G_iN. Because G_i is a subgroup of G_i+1, we can write G_i+1 = G_i+1 G_i. Make that substitution in the numerator of the left-hand side, and (after an application of associativity) you get the right-hand side. However, the end of Stewart's proof of part 2 is rather ugly. Here is the way I would finish it: After getting the group into the form G_i+1 N / G_i N, consider the maps G_i+1 --> G_i+1 N --> G_i+1 N / G_i N, where the first is the inclusion and the second is the quotient map. I claim the composite of these maps is surjective. Indeed, every element x \in G_i+1 N / G_i N is a coset of an element g n with g \in G_i+1, n \in N. But since N is contained in the group we are dividing out by, the coset of g n is the same as the coset of g, so x is the image of g, proving surjectivity. Moreover, the kernel of the composite map contains G_i. Hence the image of the composite map is isomorphic to a factor-group of G/G_i; and a factor-group of an abelian group is abelian. _________________________________________________________________ You ask about Stewart's hint to Exercise 11.4 p.107, and how maps that move elements of C can be relevant to the study of automorphism of C(t) over C, which must keep all elements of C fixed. The idea is to consider elements of C(t) intuitively as "functions" on the Riemann sphere (each of which will be undefined at a finite set of points; e.g., the function t is undefined at infinity, while 1/t, though defined at infinity, is undefined at 0). Thus "C" appears in several guises in this problem: as a subfield of C(t), as the set of values these functions take on, and as comprising all but one point of the domain-set of these functions. In terms of the first and second of these viewpoints, we are not interested in "moving" elements of C, but from the third point of view, we can observe, for instance, that the automorphism of C(t) that takes t to t^-1 corresponds to composing each element r(t) \in C(t), regarded as a function, with the map of the Riemann sphere into itself that takes each point z to z^-1 (counting 0 and infinity as inverses). Likewise, each of the elements of this group is given by composition with a particular transformation of the Riemann sphere into itself. It takes some work to make this viewpoint rigorous; for instance, if one multiplies the elements t^2, t^-1 \in C(t), regarded as functions, one gets a function which is undefined at 0 because t^-1 is undefined there; but one wants the result to be the function t; so one needs some operation of "filling in missing values". But ignoring the question of how to formulate things precisely, one can use heuristically the idea sketched above to help one picture this group of automorphisms. This, I think, is all Stewart intends by his hint. _________________________________________________________________ You comment in connection with the list of subgroups on p.110 that A and B are isomorphic subgroups of G, with A normal but not B, so that "normality is not preserved under isomorphism". This true in the sense that you state it, namely that a group G can have two isomorphic subgroups, one of which is normal in G and the other isn't. But the natural way to look at normality is as a property of a pair (G, H) consisting of a group G and a subgroup H; and one should consider two such pairs (G_1, H_1) and (G_2, H_2) isomorphic if there is an isomorphism phi: G_1 -> G_2 such that phi(H_1) = H_2. For that formulation, normality is preserved under isomorphism; i.e., if (G_1, H_1) and (G_2, H_2) are isomorphic group-and-subgroup pairs, then H_1 <| G_1 if and only if H_2 <| G_2. _________________________________________________________________ You ask about Stewart's statement on p.117, second paragraph, that the class of soluble subgroups is closed under extensions, by Theorem 13.2. Well, if G is an extension of a solvable group A by a solvable group B, then, as discussed in the first sentence of that paragraph, it has a subgroup N isomorphic to A such that G/N is isomorphic to B. Being isomorphic to A, N will be solvable, and being isomorphic to B, G/N will be solvable, so by part 3 of the Theorem, G will be solvable. _________________________________________________________________ You ask about the isomorphisms A_4 / V =~ C_3 and S_4 / A_4 =~ C_2 on p.115. Since |A_4| = 12 and |V| = 3, |A_4 / V| = 3, and any group of order 3 is isomorphic to C_3. The analogous argument gives the description of the other factor-group. One can, of course, in each case work out the list of cosets and their multiplication tables, and verify that these are isomorphic to the multiplication tables of C_3 and C_2; but the above shortcut will do in this case. _________________________________________________________________ You ask, in connection with Example 4, p.115, "How does a group of degree 4 have a subgroup of order 12"? You might better ask, "I know what is meant by the order of a group, but what is meant by its `degree'?" What Stewart means by "degree" is "the number n of elements such that G is represented as a group of permutations of that many elements". It would be better if he defined the term, but I think he is taking for granted that whether one knows what "degree" means for a general group or not, one can deduce from the phrase he uses, "The symmetric group S_4 of degree 4", that in this context "degree" means the subscript n appearing on the groups S_n. _________________________________________________________________ You ask why Stewart writes G_0 = N in the proof of point 3 of Theorem 13.2. He has assumed that G/N is solvable, so that it has a chain of subgroups, starting with 1 and ending with G/N, with certain properties. Now any subgroup of G/N has the form H/N where H is a subgroup of G containing N. Hence, instead of giving the i-th subgroup in the abovementioned chain a name like "F_i", he takes advantage of the above fact to write it "G_i/N", where G_i is a subgroup of G containing N. In particular, since the first of these subgroups of G/N is 1, when we write it G_0 / N, the numerator "G_0" must be N. (Clearly, 1 = G_0/N <==> G_0 = N.) _________________________________________________________________ You ask why in step 2 of Stewart's proof of Theorem 13.4 (p.119) we can assume N contains (t^-1 x t)x^-1 . We have assumed it contains x. Hence as it is a normal subgroup, it contains all conjugates of x, including t^-1 x t. Multiplying together the elements t^-1 x t and x^-1, we see that it contains the element you asked about. _________________________________________________________________ I hope my lecture answered most of your questions. Regarding the reference to "(1)" at the top of p.118, he means "(13.1)"; but unfortunately, not just the line of text at the bottom of p.114 labeled with that number, but also the conditions "1." and "2." at the top of the next page. This seemed too complicated to give as an "erratum to the text", but I will point it out to Stewart, and am explaining it to those who ask. _________________________________________________________________ You ask why Stewart's 4 cases in the proof of Theorem 13.4 really cover all possibilities. Although he states them in the form "Suppose N contains ...", and asserts that these four cases exhaust all possible nontrivial normal subgroups N, more is true: Every nonidenity element x \in A_n is of one of the four forms indicated; i.e., when it is written as a product of disjoint cycles, either one of these cycles has length >_ 4, or at least two have length 3, or one has length 3 and the rest have length 2, or all have length 2. When stated in this form, is it clear? _________________________________________________________________ You ask about the statement on p.117 that "every element of G generates a cyclic subgroup". If x is an element of G, then the subgroup that it generates is a group generated by one element (namely, x), and a group generated by one element is called a "cyclic group". So that group is a cyclic subgroup of G. You're not the only one who asked this. Is there something about the way you saw "cyclic (sub)group" defined previously that made it hard to see this? _________________________________________________________________ You ask how one proves that a normal subgroup is a union of conjugacy classes. It follows from the definitions. The statement that a subgroup N of G is normal means that for every x \in N and every g \in G, the conjugate g x g^-1 lies in N. In other words, whenever x is in N, all elements of G conjugate to x are in N. In other words, for every element x of N, all the other members of its conjugacy class are also in N. In other words, N is a union of conjugacy classes. These observations don't even require that we are looking at a subgroup of G; just a subset that is closed under conjugating by all elements of G. _________________________________________________________________ You ask how step 2 on p.119 uses step 1 on the preceding page, when they involve different permutations "t". The argument given in step 2 shows that if N contains a permutation whose cycle decomposition includes two 3-cycles, then N contains a 5-cycle, and step 1 showed that if N contains a permutation whose decomposition includes an n-cycle for any n >_ 4 (so in particular, if it contains a 5-cycle) then it contains a 3-cycle. Putting those two results together, we conclude that if N contains a permutation whose cycle decomposition includes two 3-cycles, then N contains a 3-cycle. We don't have to talk about whether we "use a different t", because we are not saying "by the proof of step 1", just "by (the result proven in) step 1"; and the result proven doesn't specify any t. _________________________________________________________________ You ask about formal manipulations involving multiplication of subsets of groups by elements of the groups. All the statements you wrote were correct: (ab)B = a(bB), aA = aB <=> A = B, and aAb = B <=> aA = Bb^-1. They can be verified quickly from the definitions. One can also note that multiplication of subsets by elements defines an "action" of G (in the sense I spoke about in class) on the set of all subsets of G. Your first equation is one of the two identities one has to verify to show this, while the second equation follows from the properties of such an action. (In any set on which a group G acts, whenever ax = ay one can multiply this equation by a^-1 and get x = y.) Your third implication is an instance of the fact that we also have a right action of G on the set of subsets of G, and the right and left actions respect one another: a(Xb) = (aX)b. (After one learns that when a group G acts on the left a set, one has ax = ay ==> x = y, one has to be careful to note that ax = bx does not imply a = b; rather, this holds whenever a^-1 b is in the isotropy subgroup of x.) _________________________________________________________________ You write that you're having trouble understanding Stewart's manipulations with permutations. Aside from reviewing your 113 text, here are two points to bear in mind: First, remember that Stewart, for some reason, is composing his permutations as though written on the right of their arguments. (This was something mentioned in the list of errata. Did you note it in your text?) So, for instance, (12)(23) = (132) in his notation, because he means "first interchange 1 and 2 and then interchange 2 and 3", rather than the more common interpretation "first interchange 2 and 3 and then interchange 1 and 2", which makes the product (123). Second, whenever he mentions a factor that "doesn't move the elements named" (in step 1, "bc...", in step 2, "y", in step 3, "p" and in step 4 again "p"), one should take it into account in one's calculations, but it doesn't make any difference in the end. For instance, in step 2, when he forms t^-1 x t x^-1, then on the one hand, one has to verify that on n \in {1,2,3,4}, this element indeed behaves as (14)(23); but what if n is not in {1,2,3,4}? Then applying t^-1 leaves it unchanged (still n), applying x takes it to p(n), applying t leaves p(n) unchanged, and applying x^-1 takes p(n) back to n; so it is unchanged. Thus, t^-1 x t x^-1 is precisely (14)(23). _________________________________________________________________ You ask why Stewart chooses the cycles that he does in the proof of Theorem 13.4. I guess you mean the cycles t (and not the choice of cases 1-4 for the cycle-decomposition of x). I hope you recall the discussion I gave of conjugation last Wednesday. In particular, if x is a permutation, then t^-1 x t will be a permutation that acts "like" x, but where the details about _which_ elements are at which points of which cycles gets modified by the action of t. (For a more detailed statement see the handout on the same result accessible throught my web page.) Hence, if t moves _few_ elements, then t^-1 x t will agree with x in what it does on most elements, so (t^-1 x t) x^-1 will agree with the identity in what it does on most elements, i.e., it will move few elements, bringing us close to what we want, a lone 3-cycle. So the strategy is to choose t that simultaneously (a) moves few elements, (b) belongs to A_n (so we can't use a 2-cycle), (c) can be expressed in terms of the form we have assumed x has (e.g., in case 1, in terms of a_1 , ... , a_m ) so that we can compute (t^-1 x t) x^-1 explicitly, and (d) doesn't actually commute with x, so that (t^-1 x t) x^-1 gives us a nonidentity element of N. In each of Stewart's cases (1)-(4), if you look for an element satisfying all of the above conditions, you are likely to come up with something close to what Stewart gives. (It takes a little experience to see when (d) will hold. The condition is equivalent to saying that when you modify x by conjugation by t, you don't just get x again. If you understand how to compute easily the conjugate of a permutation by another permutation -- again see the beginning of my online note if that isn't clear -- this is probably the easiest form of that condition to test). If you have time, experiment and see! _________________________________________________________________ You ask whether in Lemma 13.6 (p.120) one can replace "(12)" with "(1m)" for arbitrary m \in {2,...,m}. Nope! Notice that if we label the 4 vertices of a square consecutively as "1, 2, 3, 4" and regard the symmetry group D_8 as a group of permutations of those four vertices, then D_8 contains the rotation (1234) and the reflection (13); but it is not all of S_4. It is an interesting exercise to figure out for which values of m S_n _is_ generated by (12...n) and (1m) ! _________________________________________________________________ You ask how Stewart uses the First Isomorphism Theorem to get "|MT| = |M| |T| / |M \intersect T|" near the bottom of p.123. Since A is abelian, all subgroups are normal. Hence we can apply the First Isomorphism Theorem with any two subgroups in the roles of the "A" and "H" of that theorem; let us use M and T. Then the Theorem (Lemma 13.1, part 1) says M / M \intersect T =~ MT / T. Taking the orders of both sides, we get |M| / |M \intersect T| = |MT| / |T|. Solving for |MT| now gives the equation claimed. (Intuitively, the idea is that when we multiply all elements of M by all elements of T, we get |M| |T| products; but some of those products are equal; namely, given a product x y, we can get another product (x g)(g^-1 y) equal to it for any g in M \intersect T. So to evaluate |MT| we have to divide |M| |T| by the number of such elements g, i.e., |M \intersect T|. This counting argument can be made precise, and works even when neither M nor T is normal, so that MT is just a set of elements, not necessarily a subgroup.) _________________________________________________________________ You ask how we know G_i <| G_i+1 in the proof of Corollary 13.11. By the preceding Lemma, G_i is normal in G; this automatically makes it normal in any intermediate group. _________________________________________________________________ You ask how Stewart knows, near the bottom of p.123, that the order of t^(r/p) is p. In general, if an element has order mn, then its n-th power has order m. If you recall that the order is the least integer such that this power of the element gives the identity, this should not be hard to see. _________________________________________________________________ You ask what inductive hypothesis Stewart is assuming when he says that Lemma 13.13 (p.123) will be proved "by induction on |A|". He is assuming that for all abelian groups B of orders smaller than |A|, the result of the Lemma holds. _________________________________________________________________ You ask what Stewart means by the statement in Theorem 13.12 that "All such subgroups are conjugate in G". He means that for any two such subgroups A and B, there exists an element g\in G such that g A g^-1 = B. _________________________________________________________________ You ask whether in the definition of "radical extension" on p.128, the "m" must be finite. Yes; whenever a "list" of the form "x_1, ..., x_n" is shown, this is understood to imply that n is a nonnegative integer. This is a convention you should learn to take for granted. One could easily define a not-necessarily-finite radical extension to be an extension L:K such that for every x\in L, there is a radical sequence alpha_1 , ... , alpha_m of elements of L (defined as on that page, but without the assumption that they generate L) such that x\in K(alpha_1,...,alpha_n). But for the purposes of studying what equations can be solved in radicals, it is enough to have the concept of a finite radical extension, so that is what Stewart defines "radical extension" to mean. _________________________________________________________________ Yes, the extensions you name are examples of extensions to which Lemma 14.4 applies. It is reasonable to expect an author to give examples when a new concept is introduced, and to give applications of a result proved when these applications are not obvious. But when a result concerns concepts the reader is familiar with, it is reasonable to expect the reader to apply the result to these concepts by his or herself. The obvious way to get a field in which t^n - 1 splits is to take the splitting field of that polynomial over Q, namely Q(*zeta_n), and an example of a splitting field of an equation t^n - a over such a field is Q(*zeta_n, a^1/n). So -- glad you found those examples; I hope that looking for basic examples will be an automatic part of reading mathematics for you, and that you won't be surprised at authors' taking it for granted that you will do this. _________________________________________________________________ You ask why, as stated at the bottom of p.130, we can "insert extra elements" to get all the n(i) prime. Well, as an example, suppose that at the first step, the element alpha_1 we adjoined satisfied (alpha_1)^12 = a_1, for some a_1 \in K. Then let us insert the terms (alpha_1)^6 and (alpha_1)^2 before it. Now (alpha_1)^6 has the property that its square is in K (since its square is (alpha_1)^12 = a_1), and the next term, (alpha_1)^2 has the property that its cube is in the extension we have just formed, K((alpha_1)^6), and finally K(alpha_1) has the property that its square is in the second extension, K((alpha_1)^2). So we've replaced the single step of adjoining a 12th root by successive steps of adjoining a square root, a cube root, and a square root, where the exponents 2, 3, 2 are prime. I hope the general argument is clear from this example. _________________________________________________________________ You ask why, on p.131 line 10, when Stewart sets epsilon = alpha_1/beta, he can assert that epsilon^p = 1. Remember that alpha is a zero of t^p - a for some a\in K (first line of the page), and f is the minimal polynomial of alpha_1; hence it is a divisor of t^p - a. Hence beta, being another zero of f, is also a zero of t^p - a. Hence (alpha_1/beta)^p = a/a = 1. _________________________________________________________________ You ask how one can show that Q(sqrt 2) is not isomorphic to Q(sqrt 3). That is actually a case of what I cover in the handout on "Extensions of the form K(sqrt alpha_1 , ... , sqrt alpha_n) : K". Part (iii) of the main theorem of that note describes which elements of K have square roots in L. Applying it to the case where K = Q and L = Q(sqrt 2), it says that the only rational numbers having square roots in Q(sqrt 2) are those that are squares of rational numbers, and those that are 2 times squares of rational numbers. Since 3 is neither (i.e., neither 3 nor 3/2 is a square in Q) it follows that 3 does not have a square root in Q(sqrt 2). Since it does have a square root in Q(sqrt 3), these are not isomorphic (as extensions of Q, and hence as fields). However, if you don't want to use the whole inductive proof of that theorem, you can just isolate the key argument as it applies to this case: Suppose that (a + b sqrt 2)^2 = 3. Expanding, and recalling that {1, sqrt 2} is a basis of Q(sqrt 2) : Q, we conclude that the coefficient of 1 and the coefficient of sqrt 2 must be the same on both sides of that equation. Looking at the coefficient of sqrt 2, you immediately get ab = 0, so a = 0 or b = 0, and it is easy to get a contradiction in either case. _________________________________________________________________ You ask why on p.132, on the 4th line of the proof of Theorem 14.1, N:K_0 is radical. I assume that you followed Stewart's reference to Lemma 14.2 and saw that this makes N:K radical. Now K_0 contains K, and we see that a "radical sequence" for N:K will also be a radical sequence for N:K_0 _________________________________________________________________ You ask whether the possibility that a polynomial has multiple zeroes would affect the concept of the Galois group of a polynomial as a group of permutations of the zeroes, as discussed in the last paragraph of p.132. Yes and no! If we write the factorization of the polynomial over the splitting field as (t - alpha_1) ... (t - alpha_n), then if some of these alphas are the same, we can't associate to each automorphism a specific permutation of the symbols alpha_1 , ..., alpha_n (because different _symbols_ can represent the same zero). However, if we specify that the _distinct_ roots should be alpha_1, ... , alpha_m (where m may be less than the degree of the polynomial), then we can associate to each automorphism a specific permutation of the symbols alpha_1 , ..., alpha_m, and hence a permutation of {1,...,m}. _________________________________________________________________ You ask whether the method Stewart uses to get a polynomial over Q that is not solvable by radicals in Theorem 14.8, p.134, can also be used to get such a polynomial over the real numbers. Nope! By the reasoning Stewart gives on the preceding page, every polynomial over the real numbers has a splitting field in the complex numbers. But the extension C:R has no intermediate fields other than R and C, so the splitting field of every polynomial over R is either R or C. These extensions have Galois groups Gamma(R:R) = 1 and Gamma(R:R) = Z_2, both of which are solvable. Now Stewart's proof of Lemma 14.7 is valid, step by step, with "R" in place of "Q". Why is there not a contradiction? Because that Lemma applies to an irreducible polynomial of degree p; but there is no irreducible polynomial of degree > 2 over R. In Theorem 14.8, Stewart shows that t^5 - 6t + 3 is irreducible over Q, but his proof uses Eisenstein's criterion, which we proved only for Q. (Eisenstein's criterion can be adapted to a much larger class of fields, but definitely not to all fields.) Can you see how to prove the fact I stated above, that "there is no irreducible polynomial of degree > 2 over R" ? (It follows from other facts mentioned in this e-mail.) _________________________________________________________________ You ask about the trigonometric solution of the cubic that Stewart refers to on p.135. When he calls it "well-known", he doesn't mean that most mathematicians know it, and that you are expected to know it! Just that most mathematicians know that it exists, and that it can be found in books. Anyway, the idea is as follows. Note the third-from-last display on p.56: cos(3 theta) = 4 cos^3 (theta) - 3 cos (theta). This can be used to solve certain cubics, in the following way. Suppose one has an equation to solve of the form A = 4 x^3 - 3x where A is a known real number between -1 and 1, and x is the unknown. Find in a trig table an angle phi such that cos phi = A. Then letting theta = phi/3 in the above formula about cosines, we get A = 4 cos^3 (theta) - 3 cos (theta). so cos(theta) is a solution to the equation we wanted to solve. Moreover, if we have one value of phi that satisfies cos phi = A, then obviously phi + 2pi and phi + 4pi will also have that property, but cos(phi + 2pi)/3 and cos(phi + 4pi)/3 will in general be different from cos(phi), so this method gives not one, but three solutions. In fact, it is not hard to show that the condition I had to assume to make this method work, namely -1 _< A _< 1, will hold if and only if the equation A = 4 x^3 - 3x has three real solutions. Now if we have a general cubic equation t^3 + a t^2 + b t + c with three real zeroes, we can make a linear change of variables that brings it to the form A = 4 x^3 - 3x. Hence the above method becomes applicable. _________________________________________________________________ You ask about Stewart's statement in the proof of Lemma 14.7 on p.133 that the zeros of f are distinct because the characteristic is 0. On p.83 he defined an irreducible polynomial to be separable if its zeros (in a splitting field) were distinct, and in Proposition 8.6 (p.86) he shows that any irreducible polynomial over a field of characteristic 0 is separable. That is the justification for the facts stated here. You also ask whether, when he refers to "the characteristic", he means the characteristic of Q. Whenever one field is contained in another, they have the same prime subfield (see p.3), and hence the same characteristic (p.4). So whenever we are dealing with a family of fields containing a common subfield, we can speak of "the characteristic" which will be the same for all the fields in question. _________________________________________________________________ You ask how the concept of transitivity of a permutation group (p.135) is related to that of transitivity of an equivalence relation. It isn't. But in both cases, the idea of the word "transitive" is that of being able to "get from here to there". In the equivalence relation case, it refers to the property "if you can get from X to Y and from Y to Z then you can get from X to Z". In the group case, it simply means "You can get from anywhere to anywhere else". _________________________________________________________________ You ask whether, in the term "general polynomial" introduced on p.139, the statement that the coefficients "do not satisfy any algebraic relation" simply means that they are transcendental elements. It means that and more! To see example of a polynomial whose coefficients are transcendentals, but which is _not_ a "general polynomial", let x be transcendental over Q, and consider the polynomial t^2 + at + b, where a = x and b = x^2. Each of these coefficients is transcendental over Q, but nonetheless, the coefficients satisfy an algebraic relation, namely a^2 - b = 0; so this is not a "general polynomial". On the other hand, if we form a polynomial ring Q[x,y] in two indeterminates, then t^2 + xt + y is a "general polynomial". Stewart's introduction is merely intended to give a general idea. He makes the precise definition in the middle of p.143, in the next reading. You also ask "How are general polynomials easier to work with than polynomials over Q?" I hope that the preview I gave of the next reading partly answered that. But aside from their being, in certain ways, easier to work with, their importance comes from the idea I sketched last time, that if we can find a formula for the zeros of a "general polynomial", we may be able to "substitute" values in the base field for the coefficients, and so get a formula for finding the zeros of arbitrary polynomials. _________________________________________________________________ You ask what Stewart means by a "nontrivial polynomial" in the third line of the Definition at the top of p.140. He means a _nonzero_ polynomial. (He is confusing two phrases, "nonzero polynomial" and "nontrivial polynomial equation". An equation p(t_1,...,t_n) = 0 is called "trivial" if p is itself the zero polynomial, since in that case this equation holds automatically, and doesn't tell us anything about t_1,...,t_n. If the equation is not trivial, it is called "non-trivial". So the equation is non-trivial if and only if the polynomial is nonzero, hence his mistake is easy to make.) _________________________________________________________________ You ask whether the concept of "independent indeterminates" is related to that of linear independence. Where Stewart speaks of "independent indeterminates", my handout uses the more standard phrases "algebraically independent elements". In the last couple of pages of the handout, I discuss at length the analogy between algebraic independence and linear independence. There is one concrete relation between them: alpha_1,..., alpha_n are algebraically independent if and only if the set of all (infinitely many) monomials alpha_1 ^m_1 ,..., alpha_n ^m_n are linearly independent. _________________________________________________________________ You ask what I mean by "composing with this isomorphism" near the top of p.2 of the handout. I should have made this more precise. If i: K[beta_1,...,beta_s] --> K[t_1,...,t_s] is an isomorphism, then it induces an isomorphism i': K[beta_1,...,beta_s][t_s+1] --> K[t_1,...,t_s][t_s+1]. Composing with the substitution map K[t_1,...,t_s,t_s+1] --> L we get a map K[beta_1,...,beta_s][t_s+1] --> K[t_1,...,t_s][t_s+1] --> L which will have nontrivial kernel if and only if substitution map does (since the map we are composing with is an isomorphism). _________________________________________________________________ You ask whether, in the proof of Lemma 15.1 in the handout, the fact that beta_i is algebraic over K[beta_i(1),...,beta_i(s)] doesn't follow directly from non-one-one-ness of the substitution map. Well, it follows from that non-one-one-ness _together_ with the fact that on the subring K[t_1,...,t_s], the map is one-to-one. Not being careful of that distinction is what leads to the erroneous proof of Lemma 15.2 in Stewart, so I take pains to make clear how one-one-ness on the subring is used in proving the result of this lemma. _________________________________________________________________ You ask how the inequalities that Stewart obtains in the proof of Lemma 15.3 (p.142) lead to the conclusion that F = K(s_1,...,s_n), and exclude the possibility that it is larger. Well, by the tower law, [K(t_1,...,t_n) : K(s_1,...,s_n)] = [K(t_1,...,t_n) : F] [F : K(s_1,...,s_n)]. The first term on the right-hand side is n!, and if F were larger than K(s_1,...,s_n)], the second term would be > 1, so the product would be > n!. But Stewart has shown the left-hand side _< n!, giving a contradiction. _________________________________________________________________ You ask what Stewart means by saying on p.143, 6 lines from the bottom, "The s_i are now the elementary symmetric polynomials in t_1, ... t_n." Whenever a monic polynomial t^n - a_1 t^n-1 + ... +- a_n can be factored as (t - alpha_1) ... (t - alpha_n), the coefficients a_i can be expressed as symmetric polynomials in the zeros alpha_i; e.g., a_1 = alpha_1 + ... + alpha_n. So he means here that the indeterminates s_i can be expressed in this way in terms of the zeros t_i of the polynomial having them as coefficients. Hmm, I see that this becomes confusing if one tries to write it out. In the notation of chapter 2, the general relation between coefficients and zeroes is expressed by the formulas a_i = s_i (alpha_1 ,..., alpha_n). But if we want to apply these formulas in the present situation, "s_i" stands for both the field element and the polynomial operation; so there is no good way to write what is meant. Another good reason for using "u_i" and "v_i" instead of "s_i" and "t_i". _________________________________________________________________ You ask why the "general polynomial over K" is so called, when it is not a polynomial over K. Perhaps a better name for it would be "the general polynomial of degree n _for_ K", or something of the sort. Anyway, it is a construction where, starting with any field K, and any positive integer n, one gets a certain polynomial, which is essentially uniquely determined by K and n ("essentially" meaning that if we take two versions of it, there is an isomorphism between the fields over which they are defined, making one polynomial correspond to the other). The property of being "general" (having algebraically independent elements for coefficients) which it has over K it does not have over K(s_1,...,s_n) (s_1,...,s_n are not algebraically independent over that field, since they are in it), so it wouldn't be appropriate to call it the "general polynomial over K(s_1,...,s_n)". Maybe "general polynomial over K" really is the best term; it's just a different sense of "over". A polynomial over K is not in K, but it is in something constructed from K, namely K[t]. Similarly, a general polynomial is also characterized as being in something constructed from K -- but something different from the object used in defining "polynomial over K". It's a question of whether one wants to allow the word "over" to have many related meanings, or try to coin different words for each of these meanings. _________________________________________________________________ You ask about generalizing Theorem 15.10 (p.146) to positive characteristic, if the extension is separable. Inseparability isn't the problem -- in fact, an inseparable extension arises by starting with a separable extension of characteristic p, and adjoining p-th roots, which we can consider "radicals". The problem is that in characteristic p, there cannot be a primitive p-th root of unity, i.e., an element of multiplicative order p, because the polynomial t^p - 1 factors as (t-1)^p, so in any extension field, its only roots are 1. However, one can investigate what interesting sort of element will generate a separable normal extension with Galois group of order p in that characteristic, and get a nice answer; and if we modify the definition of "radical" to include this kind of element, then the same result holds. Stewart mentions this very briefly in the Remark on p.147. If we don't make this modification, the theorem still holds for finite normal extensions whose Galois groups are solvable and of order not divisible by p. _________________________________________________________________ You ask about the implication on p.147, "phi is a monomorphism ==> Gamma(N:M) is isomorphic to a subgroup of Gamma(L:K)". A one-to-one homomorphism of algebraic objects can in general be regarded as an isomorphism of the domain with the image of the map, which is a subobject of the codomain. I.e., if we look at phi as a map from Gamma(N:M) to the subgroup phi(Gamma(N:M)) of Gamma(L:K), it will be one-to-one because phi was to begin with, and onto because phi(Gamma(N:M)) is its image; hence it will be an isomorphism. _________________________________________________________________ You ask about the strategy of pp.148-149, and in particular, why we look for elements invariant under the permutation group and its subgroups. The first thing to be aware of is that there are two different points of view here. One of them is "We form the rational function field F = K(s_1,...,s_n) generated by n independent transcendentals, consider the splitting field L over F of the polynomial t^n - s_1 t^n-1 + ... +- s_n, let t_1, ..., t_n be the zeroes of this polynomial in the splitting field, and try to figure out formulas for the t's in terms of the s's, using field operations and radicals." The other point of view is "We form the rational function field L = K(t_1,...,t_n) generated by n independent transcendentals, let the permutation group S_n act on this field by permutations of the t's, let s_1,...,s_n denote the elementary symmetric polynomials in the t's, which we have proved generate the fixed field F of the action of S_n, and (again) try to figure out formulas for the t's in terms of the s's, using field operations and radicals." Now from Lemma 15.3 and Theorem 15.6 (which you should review if you have not fixed the facts proved there in your mind) the two situations just described are exactly the same! So as you read this section, you should keep both viewpoints in mind, and if a statement doesn't seem to make sense from one point of view, see whether it does from the other. The strategy used in the section in the cases of degrees n = 2 and 3 is to find expressions in the t's which are invariant under the permutation group S_n, and from which the t's can be recovered using radicals. The fact that the expressions are invariant under S_n means that they lie in F = K(s_1,...,s_n), so that we can figure out expressions for them in terms of the coefficients s_i; the fact that the t's can be recovered from them using radicals means that when we do so, we get formulas expressing the zeros of our polynomial in radicals. (As for the choice of particular expressions to use, this is motivated by the proof of Theorem 15.8, as I indicated in class.) When we come to degree 4, the idea is the same, but we do it in two steps: From the observation {1} <| V <| S_4 we get a tower L \contains V-dagger \contains F, and we first find a nice set of generators for V-dagger over F by looking at polynomials invariant under V, then work out how to generate L over V-dagger. (We really did n = 3 in two steps too, using {1} <| A_3 <| S_3; but the steps came so close together that one could see the light at the end of the tunnel as one was coming into the tunnel.) _________________________________________________________________ You ask how, on p.150, Stewart gets the formulas y^3 + z^3 = -27 q, y^3 z^3 = -27 p^3. Well, have you tried calculating y^3 + z^3, verifying that it is a symmetric polynomial in t_1, t_2, and t_3, expressing it in terms of the elementary symmetric polynomials, and then noting what happens when, as a result of the Tschirnhausen transformation, s_1 is set to zero? As for y^3 s^3, it is easier to start with yz, which as I pointed out in class, is also symmetric, figuring out how to express it, and then cubing the result. _________________________________________________________________ You ask about Stewart's use of the phrase "Frobenius monomorphism" on p.156, and whether "monomorphism" is a term of his own invention. He definitely did not invent the word. It was invented, I believe by Bourbaki, to mean "one-to-one homomorphism", essentially because French does not have an easy way to make "one-to-one" into a modifier. Likewise, they coined "epimorphism" (in French these words end in "-isme", and "epi-" has an accent on the "e"; but I'll use them here in their English forms) to mean "onto homomorphism". They were then taken into English, even though we don't really need them. At this point the story becomes complicated. When Category Theory (which I can't explain here) was created, its creators wanted to come up with abstract category-theoretic versions of various concepts from traditional mathematics. They found a pair of mutually dual properties one of which in _most_ classical cases, characterized one-to-one maps, so that they named it "monomorphism", while the dual concept in many cases likewise characterized onto maps. In many other cases, it did not, but the classes of maps that it did characterize turned out to be of great interest nonetheless. Being dual to "monomorphism", it had to be called "epimorphism", despite the fact that it's meaning frequently didn't match Bourbaki's original meaning. Well, the category-theoretic usage has become well-established and important, so each of these words now have two common meanings, which for some classes of mathematical objects agree, and for others don't. My preference is to avoid using these two words except where one means the category-theoretic concept, as distinct from the classical concept, which one can express explicitly. (Even in French, one can express the classical concepts by the phrases "homomorphisme injectif" and "homomorphisme surjectif".) Now back to Stewart: He uses "monomorphism" in the classical sense of "one-to-one homomorphism"; but why does he continually say "monomorphism of fields" where one would simply expect "homomorphism of fields"? Every field homomorphism between fields is one-to-one _except_ the map that takes all field-elements to 0. So he uses the word "monomorphism" just to make clear that he is excluding that map. Most modern algebraists specify that unless the contrary is stated, all rings have 1 and homomorphisms are defined to take the element 1 of the domain ring to the element 1 of the codomain ring. Therefore the map between two fields taking everything to zero is not considered a homomorphism, and one can simply say "homomorphism of fields". However, for some reason I don't understand, authors of textbooks feel they are being virtuous when they define rings in a way that allows rings without 1. Hence they can't specify that homomorphisms must take 1 to 1; hence the zero map would count as a homomorphism of fields if one didn't exclude it in some other way; hence Stewart's use of "monomorphism" to achieve this. As for what the Frobenius map is called by other mathematicians: it is the "Frobenius endomorphism", where "endomorphism" means "homomorphism from an object to itself". _________________________________________________________________ You ask about GF(25) as a splitting field of t^25 - t, as in Theorem 16.4, versus the assertion in Example 2, p.159 that it is a splitting field of t^2 - 2 (in each case, over Z_5). It is both! The theorem has been proved. On the other hand, since t^2 - 2 is irreducible, its splitting field has degree 2 over Z_5, hence has t^2 = 25 elements, hence must also equal GF(25). To get some additional insight, let us note that the theorem shows that the splitting field of t^25 - t has 25 elements, hence has degree 2 over Z_5, so its zeros must all have degrees 1 and 2. Hence their minimal polynomials, the irreducible factors of t^25 - t, must have those degrees. There will be 5 factors of degree 1, namely t, t-1, t-2, t-3, t-4, hence the result of dividing by the product of those factors, a polynomial of degree 25 - 5 = 20, must be a product of quadratic factors, hence there must be exactly 10 of the latter. Each of those ten quadratic factors will have splitting field which has degree 2, hence is equal to GF(25). One of those factors must in fact be t^2 - 2 (since a zero of that polynomial belongs to GF(25), and so is a zero of t^25 - t). _________________________________________________________________ You both asked why the results of section 17.1 are stated only for subfields of the real numbers. Simply because Stewart is giving the solution to the classical problem of constructing figures by ruler and compass, and this problem was posed in the plane R^2. We could, of course, consider K^2 to be a "plane over K" for any field K; but then we'd have a whole different subject to examine. E.g., over the real numbers, a line that passes through the center of a circle intersects the circle; but if we consider the "plane" Q^2, then the circle of radius 1 about the origin in that plane and the line y = x don't intersect. On the other hand, in the "plane" C^2 every line intersects every circle, which is also very different from classical plane geometry. The question of construction by ruler and compass is not studied so much for its intrinsic importance as for its historical interest. Since its historical source involved the plane R^2, that is what Stewart considers. If we wanted to study "constructions by ruler and compass in K^2" for general K, we would first have to make precise how geometric concepts and the action of a "ruler and compass" are defined in that context. I'm sure that some people have studied geometry over general fields, but it would take us far afield to go into such a theory here. _________________________________________________________________ You ask how, on p.173, Stewart gets t^2 + 4 t cot phi - 4 from (17.4). Well, he has defined phi by the condition tan phi = 4. What does that make cot phi ? _________________________________________________________________ You ask how many vertices of a 17-gon one needs to get them all. Well, if, as on p.174, we are given the circle in which the 17-gon is to be inscribed, and we have started off by choosing a point A that is to be one of the vertices, then all we need is one other vertex -- any one -- to construct the whole thing. Once we have one such vertex, we copy the angle between it and the given vertex and propagate it around the center until, after 16 such copyings, we find ourselves back at the starting point, with all 17 vertices drawn. So Stewart gives us more vertices than are needed on p.174. Of course, if we count the starting vertex A, the answer to your question is "two". Finally, if we are not given the circle, but just some points that are somehow known to be vertices of a regular p-gon (p prime), then we need three to construct the polygon: We find the center of the circle on which they lie by intersecting perpendicular bisectors of the segments connecting one of them to the other two, then once we have the circle, we can use two of them to get them all. _________________________________________________________________ You ask about Stewart's statement on p.170, next-to-last paragraph, that we can perform the ruler-and-compass construction by "faithfully following the above theory". Well, it would be a lot of work to follow through all the arguments in preceding chapters and see how they apply to this case; but after chasing down all those arguments, the mathematical steps that they tell us to go through would be essentially what Stewart does on p.171-173 without giving details on the reasons for doing it. For instance, a key step in the preceding pages is Proposition 17.4, which shows how to obtain by successive quadratic extensions any extension contained in a normal extension of degree a power of 2. This involves finding an appropriate chain of subgroups of its Galois group; the proof used general properties of p-groups. In the approach to the 17-gon that I outlined Wednesday and commented on again today (essentially the method in Stewart, but with more motivation), we wrote down the Galois group explicitly, and saw the chain of subgroups that was needed, without having to refer to the general theory. And so on. _________________________________________________________________ You ask about Stewart's use of the term "perfect square in K" on p.180, proof of Thm 18.2 (3). He simply means "square of an element of K." _________________________________________________________________ Feedback results from 18/3 to 25/4 studyg rev thnkg wrtg *S 1{1} 1{1} 3{3} 1{1} 6{:6} 3{3} 2{2} 6{6} 1{1} 12{:12} 2{2} 1{1} 5[9] 1[2] 9[:14] 2{2} 36{36} 10{10} 2{2} 50{:50} 2 1.5 5.5[7.5] 1 10.5[13] You ask about the second sentence in Stewart's proof of Lemma 19.3, "By passing to a normal closure ...". The point is that if we can prove that a normal closure N of an extension L:K has degree a power of p, then [L:K], being a divisor of [N:K] will also be a power of p. Thus, if we can prove the stated result for all _normal_ extensions N:K, it will be true for all extensions L:K; so in proving it for an extension, we can without loss of generality assume that extension is normal. This is a common type of step in mathematical proofs: to note that if a result is true in cases where a particular condition holds, then it will be true in the general case we want, and then to say "Hence we can assume without loss of generality [in proving our result] that the particular condition holds." This can indeed be confusing if one is not prepared for it. In particular, it often involves an implicit change of notation: If Stewart had given different names, L and N, to the original extension and its normal closure, as I did above, then he would be "re-naming" N as L. It also often leaves it to the reader to verify that the appropriate relation holds; e.g., Stewart doesn't state that [L:K] is a divisor of [N:K], so that if the latter is a power of p, so is the former; he leaves that to the reader to see! If you were giving such an argument in your homework, you should make a step like that explicit; if you are writing an article or a textbook, you need to weigh whether the reader can be expected to see how to verify the required implications. Finally, as a reader of mathematics, you should note that in a "we may assume" argument, the situation generally determines the implication that must be verified for the argument to be valid; in this case "If we proved [N:K] a power of p, then [L:K] will also be a power of p", and once you think through what that implication is, you only have to find the reason that is true, in this case [L:K] is a divisor of [N:K] by the tower law, and a divisor of a power of p is a power of p". _________________________________________________________________ You ask why, on p.187 near the end of the proof of Lemma 19.3, the index if P in G is prime to p. Because P is a p-Sylow subgroup of G ! Look at the definition of "p-Sylow subgroup", and note that for p a prime "p does not divide r" is equivalent to "p is prime to r". _________________________________________________________________