ANSWERS TO QUESTIONS ASKED BY STUDENTS in Math H113, Spring 2009, taught from Dummit and Foote's "Abstract Algebra", 3rd Edition. ---------------------------------------------------------------------- Regarding the discussion of well-definedness at the bottom of p.1 and the top of p.2, you ask when we have to worry about proving that a function is well-defined. In general, if the definition of the function involves some choice, one needs to prove that the values assigned by the function do not depend on that choice. In the book's example of a function on A = A_1 \cup A_2, it is not obvious how a choice is involved; but in applying that definition to an element, one implicitly says "since x\in A = A_1 \cup A_2, let it be contained in A_i, and then define f(x) as ..."; but this means choosing one of the A_i that contains it (if it is in both). I say more about this on p.10 of the handout on Sets and Logic (though that page wasn't in today's reading). Anyway, when cases of this sort come up in the course, it will be pointed out very explicitly. If you give a construction that doesn't involve a choice, there's no need to say anything about verifying well-definedness. ---------------------------------------------------------------------- Regarding Proposition 1, p.2, you ask "Why can the map f be injective iff f has a left inverse and surjective iff f has a right inverse? Why not injective iff it has a right inverse and surjective iff it has a left one?" Well, if these ideas are new to you, I suggest you experiment. Take a very simple example of a function that is 1-1 but not onto (for example, the function from \{0, 1\} to \{0, 1, 2\} that takes every element to itself), and see whether you can construct a right inverse to it, and whether you can construct a left inverse to it. This means taking the definition of right inverse (respectively, left inverse; but it's best to think about one at a time), and seeing what it really means in terms of how the function is to act on elements; then trying to get an example with that behavior. Hopefully, you will find that you can construct a left inverse, but cannot construct a right inverse. Then think about why; and whether the reasons you could do one but not the other are not particular to this case, but apply whenever a function is one-to-one but not onto. If you run into difficulties along the way, ask about them. If you don't get this e-mail till after class, the question will be moot, because I will have discussed the proof in lecture; but the above indicates the approach I feel you should have taken. I also notice that in saying "Why can the map f be injective iff f has a left inverse ...", you don't refer to the additional condition that the domain be nonempty. You might have just omitted it for simplicity; but if it was because you hadn't corrected the errata on the homework sheet -- please remember to do so! ---------------------------------------------------------------------- You ask about definition (4), p.3, noting that the indexing map i |-> A_i might not be 1-1, and asking whether one should not therefore replace "i\neq j" in part (b) by "A_i\neq A_j". Well, if P is a partition of A, we can always choose a one-to-one indexing of the members of P, and then (b) will hold. So the book's definition should be taken as meaning that a partition of A is a set of subsets of A which can be indexed so as to satisfy (a) and (b). Personally, I prefer to avoid indexing sets except in situations where it improves clarity. So I would prefer to define a partition of A as a set P of nonempty subsets of A such that \bigcup_{E\in P} E = A, and such that if E, F\in P and E \neq F, then E\cap F = \emptyset. This is like your version of the definition, but without the subscripts. ---------------------------------------------------------------------- You note that the word "partition" had a different meaning in Math 104 from the one given in this course (p.3, statement (4)), and ask whether there are likely to be many such cases. I'm not sure what the meaning you saw in 104 was. Most likely it was the statement that a "partition" of an interval [a,b] means a sequence of points a = a_0 _< a_1 _< ... _< a_n = b. Note that the purpose of those points was to divide the interval up into subintervals [a_0, a_1], [a_1, a_2], ..., [a_{n-1}, a_n]. This is very nearly a partition in the sense our book gives. It fails to be so in that successive intervals have a single point in common, rather than being completely disjoint; but since the concept of integration is based on the lengths of intervals, and these intersections have length zero, from the point of view of what is being done in 104, that distinction is inconsequential. It is also true that a partition in that sense doesn't give an arbitrary decomposition of the interval into (almost) disjoint subsets, but only those decompositions whose component sets are intervals. So in summary, the two definitions are variants of the same idea, though differently formulated. Moreover, in a different treatment of the 104 material, a partition might be defined as a family of half-open intervals [a_0, a_1), ... , [a_{n-1}, a_n) whose union gives the half-open interval [a_0, a_n); so in this case, it would precisely fall under the definition we had. Anyway, throughout mathematics, you will find that different authors develop the same concept using slightly different formalisms. How often you are likely to see the same word used in (H)113 and in 104 with apparently unrelated meanings I'm not sure; probably not very often. ---------------------------------------------------------------------- You ask about the formula dl = ab on p.4, relating positive integers a and b, their g.c.d. "d", and their l.c.m. "l". That's a question I would ordinarily like to answer in lecture; but in this reading, the text is rushing through so much important stuff that I will almost surely not have time; so I'll answer here. There are different ways to look at this. The quickest is to use the formula for the g.c.d. that the authors give two pages later, at the very bottom of p.6, and the corresponding formula for the l.c.m. mentioned right at the top of p.7. Combining these with the general observation min(x,y) + max(x,y) = x + y, the desired result follows immediately. For another approach, consider any positive integer x, and let us look at divisibility relations with a and b. Note that if x divides a, then ab/x = (a/x)b is a multiple of b; in fact, if and only if; and similarly, x divides b if and only if ab/x is a multiple of a. Hence x is a common divisor of a and b if and only if ab/x is a common multiple of a and b. So if x is the greatest common divisor of a and b, then ab/x will be the least among all common multiples of a and b that can be written ab/x. Can every common multiple of a and b be so written? No. (There are common multiples of a and b that are larger than ab, and they certainly can't.) However, I say that the least common multiple is among those that can. This uses the fact, noted in the book, that the least common multiple is not merely least in size; it divides all common multiples of a and b. Now ab is a common multiple of a and b, so the least common multiple of a and b is a divisor of ab; hence it can be written ab/x for some positive integer x. Thus, the least common multiple of a and b is a common multiple of a and b having the form ab/x, and being least among all common multiples of a and b, it is certainly among all such common multiples of a and b having this form; so the x involved must be the largest among the common divisors of a and b. So in the symbols the book uses, l = ab/d, which is equivalent to the formula dl = ab that they give. ---------------------------------------------------------------------- You ask how one proves assertion (1) on p.7, that the Euler \phi-function is "multiplicative". That's quite nontrivial. It is based on two facts: (1) if a and b are relatively prime, then for every combination of an integer i with 0 _< i < a and an integer j with 0 _< j < b, there is one and only one integer k with 0 _< k < ab whose residue on division by a is i and whose residue on division by b is j (this fact is called the Chinese Remainder Theorem; in a more general form it is given on p.265 of the text), and (2) an integer i is relatively prime to ab if and only if it is relatively prime to both a and b. Putting these facts together, one finds that the pairs of integers (i,j) with 0 _< i < a and 0 _< j < b such that i is relatively prime to a and j to b, of which there are exactly \phi(a)\phi(b), are in bijective correspondence with the integers k with 0 _< k < ab that are relatively prime to ab, of which there are exactly \phi(ab). so \phi(a)\phi(b) = \phi(ab). As an example, you can take a = 2, b = 5, so ab = 10, and classify the integers 0,...,9 according to their residues on division by 2 and 5; and see that you get all 10 possible combinations, each once; and that those that are relative prime to 10 are the ones with 1st term 1 (not 0) and 2nd term 1,2,3,4 (not 0); so the total number is 1 . 4 = \phi(2)\phi(5). ---------------------------------------------------------------------- You ask how the coefficients -7 and 2 in the expression for 1 as a linear combination of 60 and 17 in the second example on p.11 were obtained. See point (7) on p.5. If you have trouble with it, let me know where you run into difficulty. ---------------------------------------------------------------------- You ask whether, in the definition of groups on pp.16-17, there is a reason for requiring the operations to be associative, but not necessarily commutative. If X is any mathematical object, then the bijective maps X --> X that exactly preserve the given mathematicsl structure can be composed, and under this composition they will satisfy the associative identity and the other conditions for being a group; so those conditions apply to a class of entities appearing in an important way throughout mathematics. Moreover, from these conditions, one can deduce a great deal, as we shall see in the next few weeks. So the concept of "A set with a composition operation satisfying the associative law, and having an identity element and inverses to every element" constitutes a useful and rich sort of mathematical structure, and is given a name. It is easy to come up with operations that satisfy the commutative law but not the associative law; e.g., the real numbers with the operation x * y = x^3 + y^3. But these turn out to be much less important. Moreover, from the commutative law there is little more that one can deduce, while associative law, as I said, has very strong consequences. Sets with a binary operation that is not necessarily associative are nevertheless studied. They are sometimes called "groupoids" (though that usage conflicts with another sense of that word); and those that satisfy various additional conditions are given special names by the people who work with them: "loops", "quasigroups", etc.. If we do assume associativity, but drop or weaken the other two conditions in the definition of a group, we get concepts of intermediate importance, called (depending on how much one assumes) "semigroups" or "monoids". I hope to say a few words about these at some point. ---------------------------------------------------------------------- Regarding property (4) on p.25, you ask "Given statement (3), shouldn't this follow from the fact that |r| = n?" Yes. Even without statement (3), it follows from the fact you quote, by Prop.2 (1), p.20. I guess the authors wanted to record this step on the way to getting the statement that the elements listed in the next line are all distinct (i.e., the word "uniquely" in the line following that). But a more important step toward that statement which they left out is that r^i \neq s r^j for any j. (This can be obtained by noting that if we had equality, then multiplying on the right by r^{-j}, we'd get a contradiction to (3).) ---------------------------------------------------------------------- You ask about the statement on p.26, line 6 that \{r,s\} is a generating set for D_{2n} "by property (4)". I looked, and it took me a moment to figure out what the authors meant. They don't mean the statement "sr^i not-= sr^j ..." with which (4) begins, but the display on the next line: "D_2n = \{...\}". Do you see that this indeed tells us that \{r,s\} is a generating set for D_{2n}? ---------------------------------------------------------------------- You ask whether you have correctly summarized the result given on pp.38 (bottom) to 39. The first statement in your summary is correct: > a) If all the relations in the presentation of G can be satisfied by > replacing the elements of G by some arbitrary elements of H, then > there is a homomorphism from G to H -- assuming that where you wrote "the elements of G" you meant "the given generators of G". But in your second statement, > b) If the order of G and H are equal plus the above condition, then G > and H are isomorphic? ... you've left out two key conditions: that H be generated by the elements the book calls r_1,...,r_m, so that the homomorphism \phi will be surjective, and that H be finite. It is easy to give examples where isomorphism fails if surjectivity is not assumed. For example, suppose we write Z_6 as , and map it to S_3 by sending a to (1 2) (which has 6th power equal to the identity because it actually has square equal to the identity). These groups have the same order, but the map is neither one-to-one nor onto, and the groups are non-isomorphic. For an example where the orders are the same, but infinite, and the groups are non-isomorphic, let G = Z x Z (see the definition of direct product of groups, p.18, item (6)), which can be presented as , where a = (1,0), b = (0,1) (and we are using multiplicative notation in the presentation, though the group is written additively), and let H = Z. I don't know whether you have seen the fact that Z x Z and Z have the same cardinality ("order"); if not, I can show this. Anyway, we can map \{a,b\} to H by sending a to 1 and b to 0. This gives a surjective map \phi: Z x Z -> Z, but it is not injective, and Z x Z and Z are not isomorphic, since the first group cannot be generated by a single element, while the second can. ---------------------------------------------------------------------- Regarding the analogy between generating sets for groups and bases of vector spaces mentioned in the second paragraph on p.39, with the comment "(here there are no relations to satisfy)" you ask "What would happen if we were to put relations on a basis of a vector space?" If we present a vector space by generators and relations, then each relation can be solved to give an expression for one of our generators in terms of the rest; so we can drop one member from our list of generators, and the remaining elements will still generate the vector space. When we have dropped enough generators so that none of those left can be expressed in terms of the rest, what we have left is a basis. (In this discussion, I'm assuming finitely many generators and finitely many relations, for simplicity.) Vector spaces have much less variety in their possible structures than groups do: Up to isomorphism, there is just one vector space of each dimension over a given field. So one can always describe a vector space by a set of generators subject to no relations; i.e., a basis. Nevertheless, the concept of a vector space presented by generators and relations can be of use. E.g., there might be some vector space that is most natural to describe as . Even though it can also be described as having basis w, x, y, there may be a concept one is studying that is symmetric in w, x, y and z, so that choosing three of them to form a basis and expressing the fourth in terms of those three would be less natural than using all four as generators, and remembering that there is a linear dependence relation among them. Also, one generalization of vector space is the concept of a module; and for modules, presentations by generators and relations are as natural as for groups. Modules are in the book, and one can read about them as soon as one has the concept of ring; but unfortunately, there isn't time to work them into H113. ---------------------------------------------------------------------- Regarding item (i) on p.42, you ask whether a permutation of A means a bijection from A to A. Did you look in the index of the book? That's always the first thing to do when you come on a word whose meaning you're not sure of. If you look for "permutation" in the index, the first reference is to p.3, and if you turn to p.3, the first line answers your question. > I had problems at first at understanding why is the action > on a set a permutation of the set. ... It isn't. An action of G on A is a _homomorphism_ from G to the group S_A of permutations of A. However, if what you mean is that for each element g\in G, the map A -> A induced by g under the action is a permutation, then that is correct. ---------------------------------------------------------------------- Regarding the first two paragraphs on p.43, you write > The book states that "the actions of a group G on a set A and the > homomorphisms from G into the symmetric group S_A are in bijective > correspondence". The next statement talks about left and right > actions. Does this statement hold in either case? Yes and no. Recall that the key condition in the definition of a left action is g(ha) = (gh)a. In a right action, the corresponding condition is (ah)g = a(hg). Thus, when we apply h to an element, and then apply g to the result, in a left action this is equivalent to applying gh to the element, while in a right action it is equivalent to applying hg to the element. That is the only essential difference between left and right actions. Now the definition of composition of permutations that we use is based on writing the permutations to the left of their arguments: (\sigma\tau)(x) = \sigma(\tau (x)). If we are given a _right_ action of G on A, each g\in G will still induce a permutation \phi(g) of A; but since we compose permutations under the conventions of left actions, and that disagrees with the behavior of right actions, we get the rule \phi(gh) = \phi(h)\phi(g), which does not match the definition of a homomorphism. This can be handled in various ways. Some authors consider right actions instead of left actions to begin with, and treat permutations by the right-action conventions, writing x \sigma for the image of x under \sigma; and they compose permutations so that first applying \sigma and then \tau is equivalent to applying \sigma\tau. For them, right, rather than left actions correspond to homomorphisms G --> S_A. Another approach is to consider a right action to correspond to an "antihomomorphism" G --> S_A, meaning a map \phi that satisfies \phi(gh) = \phi(h)\phi(g). An antihomomorphism G --> S_A can be looked at as a homomorphism from the "opposite group", G^{op}, to S_A, where G^{op} is defined to have the same set of elements as G, but the modified operation g * h = h . g. Finally, one can use the fact that every group is "antiisomorphic" to itself by the map g |-> g^{-1}, and so associate to every antihomomorphism \psi: G --> H the homomorphism g |-> \psi(g)^{-1}. Then a right action of G on A corresponds to a homomorphism \phi: G -> S_A via the formula ag = \phi(g)^{-1}(a). (This last trick works for groups, but it doesn't for semigroups, which differ from groups in that elements are not required to have inverses. E.g., the maps from a set into itself, not restricted to permutations, form a semigroup. In the theory of semigroups, antihomomorphisms are taken seriously, whereas in group theory, they are seldom discussed because they can be reduced to homomorphisms by the "inverse" trick noted above.) This is probably more than you wanted to know! Any group theorist knows how to switch between right and left actions when desired. But in a first introduction to the subject, it's best to stick to one sort of action to avoid confusion. (However, after writing the above, I looked in the index and found that the authors do say more about right actions on pp.128-129.) ---------------------------------------------------------------------- Regarding the proofs that centralizers, normalizers, etc. are subgroups of a group G (pp.49-51), you write, "for the axiom about H being nonempty, it seems that every proof uses the identity to prove this axiom. Are there situations where it would be easier to use a non-identity element?" The reason for the condition of nonemptiness in the criterion for a subset to be a subgroup is to guarantee that H contains the identity, one of the conditions in the definition of a group. This is stated as "nonemptiness" so as to look more general than "contains 1"; but the presence of 1 is such a natural condition that it is hard to think of a situation where it would be easier to prove nonemptiness by a different route! Good observation. ---------------------------------------------------------------------- Regarding the observation on p.50 that Z(G) = C_G(G), you ask why we have two notations for the same object. I assume that Z(G) was the earlier notation; the center is a very useful tool in analyzing the structure of a group. The relative concept of the centralizer C_G(Z) of a subset A of a group came later (and got the letter "C" from English "centralizer" in contrast to the "Z" from German "Zentrum"). In theory, one could then have abandoned "Z(G)" and just written the center of G as C_G(G); but I think the center is so important, and people were so accustomed to having the brief symbol Z(G) for it, that C_G(G) would have been felt an unnecessarily roundabout way of expressing it, and wasn't adopted. ---------------------------------------------------------------------- Concerning the comment at the top of p.70 that the subgroup lattice of D_{16} is not a planar graph, you ask what the condition of having a planar subgroup lattice says about the structure of a group. A Google search for "subgroup lattice" planar gives 584 results; so people have looked at the question of characterizing groups with this property. I don't think the question is a really natural one, but it can be approached as follows: It is known that a finite graph is planar if and only if it doesn't contain copies of any of a certain small list of subgraphs. So one figures out all ways that such graphs can appear in the subgroup lattice of a group, how that is reflected in the group structure, and which groups this excludes. ---------------------------------------------------------------------- You ask why Theorem 3 on p.77 is not just a trivial application of Proposition 2 on the preceding page, and in particular, why the authors say that it is a generalization of that proposition. Where they say "generalization", a better word would have been "strengthening". The theorem is stronger than the proposition in that it says how cosets are multiplied, in a way that does not depend on knowing the map \phi. (The proof uses \phi, but the assertion of the result does not.) I find this section of Dummit and Foote to be very plodding, moving one little step at a time (e.g., here, between "we can define the operation on cosets by using the operation of \phi(G)" to "we can define the operation on cosets by letting uK o vK = (uv) K"). But they may have written it this way based on experience, that too many students get lost when this topic is presented in a concise way. ---------------------------------------------------------------------- Regarding problems like Exercises 2 and 3 on p.95, you ask how one can find all subgroups of a given group. The method is always one of trying out all possibilities. As what we know of group structure grows, the class of possibilities we have to try out shrinks. Suppose G is a finite group of order n. Based on the definition of "subgroup", we could consider all 2^n subsets of G, and test each of them for the properties defining a subgroup. After we learned about the subgroup generated by a subset, we could streamline that process in various ways. E.g., we could take some element g\in G, and divide subgroups H into those that contain g and those that do not. In the former, we know that we have the whole cyclic subgroup generated by g, and also that as soon as a subgroup contains another element h, it will also contain all elements that can be expressed as words in g and h. In the latter, we know that if a subgroup contains an element h, then it cannot contain, say, anything of the form h^i g h^j, since from that and h we could get g. So there are much fewer cases to consider. Now that we have Lagrange's theorem, we can limit attention to subsets whose order is a divisor m of n, and know that a subgroup H of such an order m can only contain elements whose orders are divisors of m. So problems like exercises 2 and 3 on p.95, which you ask about, can be done relatively easily. ---------------------------------------------------------------------- You ask whether, in contrast with Theorem 22(1) (p.103) infinite groups always or sometimes have composition series. Well, before asking such a question, you should see how far you can get on it by straightforward approaches. What infinite groups do you know? The most obvious example is Z. Have you asked yourself whether Z has a composition series? If you do, you will (hopefully) find that the answer is "No". So that eliminates the possibility that all infinite groups have composition series, and the remaining possibilities are just that some do, or that none do. Note that if an infinite group has a composition series, then at least one of the composition factors must be infinite; so there must be an infinite simple group. Conversely, if there is an infinite simple group, then it will have a (one-step) composition series. So the question comes down to whether there exist any infinite simple groups. That is not easy to answer from what you have learned, and if you had gotten that far, you could have asked that as your question. (If not, you could at least have indicated why not all infinite groups had compositions series, and asked whether some did.) The answer is yes, there are infinite simple groups; but these are not easy to describe in a few words at this point. (But if you are interested in the question -- or don't see some of the other points I have asserted above -- you can come by office hours, and I can show you.) ---------------------------------------------------------------------- You ask whether the definition of transitivity of group actions, as defined on p.115, has anything to do with transitivity of binary relations? Only in the very vague sense, that "transitivity" refers to the possibility of getting from one place to another. So the condition on a binary relation says "If you can get from X to Y and from Y to Z, then you can get from X to Z", while the definition of a transitive action is "You can get from anywhere to anywhere else". They aren't analogous statements; just vaguely of the same sort. ---------------------------------------------------------------------- You ask about the comment preceding Proposition 10 on p.125 about conjugation in matrix groups representing change of basis. Here is my way of looking at it. A choice of ordered basis \{v_1,...,v_n\} of an n-dimensional vector space V can be thought of as a choice of an isomorphism i: V =~ F^n: Given a basis, we get a way of representing vectors by the n-tuples of their coordinates, i.e., of associating to each vector a member of F^n, while given an isomorphism between V and F^n, we get a basis of V from the standard basis of F^n. Now if i is such an isomorphism, then a linear operator T on V corresponds to a linear operator i T i^{-1} on F^n. If we think of n x n matrices as symbols for linear operators on F^n, then this is the representation of T with respect to the indicated basis. If we have, not only the basis corresponding to the isomorphism i, but also another basis, corresponding to an isomorphism j: V =~ F^n, then j i^{-1} will be a linear operator on F^n which clearly converts the representation i v of a vector v with respect to the first basis into the representation j v of v with respect to the second. Writing j i^{-1} as a matrix P, this is the change-of-basis matrix. Now how do we convert the representation of a linear transformation T: V -> V in terms of the old basis to its representation in terms of the new basis? The former, as noted above, is i T i^{-1}; the latter is therefore j T j^{-1}, and to get from one to the other, we conjugate by j i^{-1} = P. Hence the "P^{-1} A P" formula of linear algebra. Conjugation _also_ has interpretations that don't have to do with change of basis. If a given tranformation T takes certain vectors v_1, v_2, ... to certain other vectors w_1, w_2, ..., then for any operator U, the transformation U T U^{-1} will take U v_1, U v_2, ... to U w_1, U w_2, ...; in other words, it will show a behavior "just like T, but with different vectors in the corresponding roles, related to the old vectors by U". In particular, if the given vector space is F^n, then this becomes a statement about how a conjugate of a given matrix behaves. The contrast between this and the change-of-basis interpretation of conjugation is an instance of the contrast between "alias and alibi" which I spoke about briefly in class a few weeks ago. ---------------------------------------------------------------------- Regarding the paragraph on p.134 after the definition of inner automorphisms, you ask how Inn(G) =~ G/Z(G) follows "by Corollary 15". The second statement of the Corollary says that G/Z(G) is isomorphic to "a subgroup of Aut(G)". From the context, we know that it is the subgroup of automorphisms given by conjugation by elements of G. The definition in the middle of p.134 gives that subgroup the name Inn(G), so the statement of the Corollary becomes "G/Z(G) =~ Aut(G)". Our authors frequently understand the results they state as theorems etc. to include more than they actually put into words. I think one should generally put such facts into words; but one has to get used to "reading between the lines" when authors don't do this. ---------------------------------------------------------------------- > ... p.135, Proposition 16 ... > > Why is \Psi clearly injective? In a situation like this, you should show me how far you have been able to get by straightforward reasoning, and where you got stuck. Straightforward reasoning would begin "Suppose \Psi_a and \Psi_b are different automorphisms. We want to show that the congruence classes of a (mod n) and b (mod n) are different ...", then look at how a and b are determined by the automorphism, and try to determine whether their being different automorphisms implies that they come from different congruence classes. Show me where this reasoning gets you, and where you run into difficulty. ---------------------------------------------------------------------- Regarding the proof on p.150 of the simplicity of A_n, you ask why \lambda_k \in G_i, and why G_i is isomorphic to A_{n-1}. Each G_i is the stabilizer of i in G. Now if \lambda_k = (p q)(r s), then since n > 4, p, q, r and s can't be all the members of \{1,...,n\}. If we let i be some element of that set other than p, q, r, s, it will be fixed by \lambda_k = (p q)(r s), i.e., \lambda_k\in G_i. Finally, because G_i is the group of even permutations of a set of n-1 elements, namely, the members of \{1,...,n\} other than i, it will be isomorphic to the group of permutations of any set of n-1 elements; in particular, to A_{n-1}. ---------------------------------------------------------------------- Regarding Theorem 5 on p.161, you ask > Is a subgroup of a direct product a direct product of subgroups? > That is, if P _< A \times B, is P=C \times D, where C _< A, D _< B? No. For instance, on p.155, Example (3), it is shown that the group E_{p^2} = Z_p x Z_p has p+1 subgroups of order p; but only two of these have the form you describe: Z_p x 1 and 1 x Z_p. If one asks whether every subgroup of a direct product is isomorphic to a direct product of subgroups, then for finitely generated abelian groups the answer is yes; but there are counterexamples for nonabelian groups. For instance, the subgroup of S_3 x S_3 consisting of all elements (\sigma,\tau) such that \epsilon(\sigma) = \epsilon(\tau) (where \epsilon is the function defined on p.108, Def.(1)) is a group of order 18 which is not a direct product of nontrivial subgroups; hence certainly not a direct product of subgroups of S_3. ---------------------------------------------------------------------- You ask about the number of partitions of an integer N, as defined at the top of p.162. The properties of this number as a function of N have been studied by number theorists. There are numerous results about it (e.g., how fast it grows as N increases; what it is congruent to modulo various integers); and I'm sure shortcuts to computing these numbers have been found; but I believe there is no expression for this function in closed form. Cf. http://en.wikipedia.org/wiki/Partition_function_(number_theory) ---------------------------------------------------------------------- Regarding Example (2) on p.178, you ask, > How does "so that x^2 acts as the identity on H" follow from "Define > \phi again by mapping x to inversion."? If \phi(x) acts by h |-> h^{-1}, then \phi(x^2) will act by h |-> (h^{-1})^{-1} = h. I.e., when one "squares" the operation of inversion, one gets the identity operation. ---------------------------------------------------------------------- You ask whether "HK", in the statement of Theorem 5.12, p.180, denotes the internal direct product of H and K, as defined on p.172. No, it doesn't! It's an instance of the general notation introduced in the Definition on p.93. As noted in the bottom paragraph of p.93, in general HK need not even be a subgroup. If at least one of H and K is normal, HK will be a subgroup (Cor.4.15, p.94). If they are both normal and intersect trivially, that subgroup will have properties that make it isomorphic to H x K, and we call it the "internal direct product" of H and K. If we merely assume H normal, and again assume H and K intersect trivially, then as shown here, it gives an "internal semidirect product". ---------------------------------------------------------------------- You ask why, in part (1) of the second Definition on p.218, they specify that R should have the property that the least normal subgroup containing R is the kernel, rather than just the subgroup generated by R. Well, that would require a lot more generators. To take an example that is a bit unnatural, but illustrates very starkly the difference, suppose we want to regard the infinite cyclic group as presented by two generators x and y and a relation making y = 1. This can be written as the quotient of the free group F(\{x,y\}) by the normal closure of \{y\}. On the other hand, the subgroup generated by \{y\} is much smaller that that normal subgroup -- it just contains the powers of y, but not elements such as x^n y x^{-n}, which also must go to the identity in a homomorphic image of F(\{x,y\}) where y goes to 1. So if our "presentations" had to use sets of relators that actually generated the kernel as a subgroup, we would have to use an infinite set of relators to get from F(\{x,y\}). To put it briefly: since we know that the kernel K of a homomorphism is a normal subgroup, it's natural to describe such a kernel by specifying elements which, together with its being a normal subgroup, yield all of K. If we just specify elements which, together with the more restricted information that it is a subgroup, yield K, we are making things hard for ourself by refusing to use information that we have. ---------------------------------------------------------------------- In connection with the concept of finitely presented groups, introduced on p.218, you ask whether a finitely presented group can be uncountable. No. One can show that a group generated by finitely many elements must be finite or countably infinity. (Namely, one can "list" the possible words in a finite number of generators, so that list is countable. In a particular group, some of these words may represent the same element. Crossing out such repetitions from the countable list leaves a finite or countable list.) So an uncountable group can't be finitely generated, and so a fortiori can't satisfy the stronger condition of being finitely presented. ---------------------------------------------------------------------- You ask about the use of free monoids in theoretical computer science, mentioned on p.220 (last paragraph before exercises). Well, just as one can define an action of a group on a set, so one can define an action of a monoid on a set. In the case of a group action, each group element acts as an invertible map of the set into itself. We got invertibility from the fact that each group element has an inverse in the group; since this is not true in a monoid, an action of a monoid on a set X corresponds to a homomorphism from the given monoid into the monoid E_X of _all_ (not necessarily invertible) maps X --> X. Now if F(S) is a free monoid on the set S, then a homomorphism F(S) -> E_X is determined by specifying in an arbitrary way the member of E_X to which each member of S is mapped. If we take for X the set of "states" of a device, and think of elements of S as "inputs" to the device, then such an action is a rule associating to each state and each input the state into which the device will go when the given input is applied to it. A device whose behavior is so determined is called a "finite state automaton", and is a basic model of a computer. Also, just as the elements of a free group can be written as words involving the generators and their inverses, so the elements of a free monoid can be written as words in the generators, without any inverses (and so without the need for rules specifying that a generator should not occur adjacent to its inverses). A "language" in the sense of theoretical computer science is a subset of the set of all strings in a set of symbols; so languages are studied using the theory of monoids. ---------------------------------------------------------------------- You ask about the application of polynomial rings to linear algebra alluded to in the last paragraph on p.233. This is too distant from the material we cover for me to talk about it in class. But I can try to sketch it roughly for you here. I've pointed out that just as we can consider "actions" of groups on sets, so one considers "actions" of rings on abelian groups. Now suppose V is a vector space over a field K. That is equivalent to saying it is an abelian group with an "action" of K. Suppose further that T: V -> V is a linear transformation. Then in the ring of endomorphisms of V, we have not only the copy of K that gives the vector space structure, but also the element T, which "respects" that structure (because it is a K-linear transformation), in other words, which commutes with the actions of all members of K. Now -- you need to think about this -- the combination of an action of K on V and this additional element T is equivalent to an action of K[x] on V, where x acts on V as T. An abelian group with an action of a ring R on it is called an "R-module". (So when R is a field, R-module means K-vector space.) For K a field, K[x] is a ring with a structure very much like that of Z, and one can prove neat results describing all finitely generated modules over rings with those properties. (We will look at rings with the properties that Z and K[x] share -- "principal ideal domains" -- in reading #26; but as I have said, we won't be able to study modules in this course.) These results allow one to decompose every finitely generated module over such a ring as a direct sum of submodules of an elementary form. When the given module has a certain property even stronger than being finitely generated -- which in the case R = Z is that of being finite, and in the case R = K[x] is that of having finite dimension over K -- then we get a result which in the R = Z case is the primary decomposition theorem for finite groups, and in the R = K[x] case is the decomposition of V into subspaces on which T acts (if K is algebraically closed) by Jordan matrices or (in general) by matrices in "rational canonical form". So that's it, in summary. Let me know if you want anything clarified. ---------------------------------------------------------------------- You ask why the definition of polynomial rings on p.234 doesn't allow expressions in which x occurs with negative as well as positive exponent: a_n x^n + ... + a_0 + ... + a_{-m} x^{-m}. The idea of a polynomial ring is that one should be able to substitute for x any element of R, or more generally, any element of a commutative ring containing R. (So, for instance, one asks whether the polynomial x^2 + 1 over the reals has a root -- not necessarily in the reals, but in some commutative ring containing the reals.) This substitution operation should be a homomorphism from R into the ring in question. If we made R[x] contain an inverse to x, then we could not define homomorphisms R[x] --> S that took x to non-invertible elements; e.g., we couldn't substitute 0 for x in polynomials over the reals, or substitute 2\in Z for x in polynomials over the integers. Nonetheless, expressions a_n x^n + ... + a_0 + ... + a_{-m} x^{-m} are of interest! For the reason indicated above, one doesn't include them in R[x]; but one makes another ring out of such expressions, written R[x, x^{-1}], and called the ring of "Laurent polynomials in the indeterminate x". This has the property that x is invertible, and that one can substitute for x any invertible element of a ring containing R. The text does not treat the ring of Laurent polynomials, but it does treat an extension thereof, in three exercises distributed over the book plus one example near the end that refers to these exercises. To say what this is, I should say first that the ring R[x] of polynomials can be regarded as a subring of the ring R[[x]] of "formal power series", defined as formal expressions a_0 + a_1 x + ... + a_n x^n + ... with no restriction that there be only finitely many nonzero terms. If in addition to the possibly infinitely many terms with nonnegative powers of x, one allows finitely many terms with negative powers of x, one has the ring R((x)) of "formal Laurent series". Although the ring of ordinary Laurent polynomials is a not very important modification of the polynomial ring, formal Laurent series are more important, because if R is a field, then R((x)) is also a field (unlike R[x], R[x,x^{-1}], and R[[x]]). You might ask "If we can allow infinitely many positive-exponent terms together with finitely many negative-exponent terms, why can't we allow infinitely many terms of both sorts?" The answer is that we then can't define multiplication. E.g., you can see what happens if you try to square ... + x^{-2} + x^{-1} + 1 + x + x^2 + ... . ---------------------------------------------------------------------- In connection with the observation at the top of p.236 that the units of M_n(R) are the invertible n x n matrices over R, you ask whether it is common for a ring to have so many units. Assuming that by "R" you mean (as on the page in question) any ring (and not specifically the real numbers), then how common units are among members of M_n(R) depends on R. If R is commutative, then a matrix over R is a unit if and only if its determinant is a unit of R; so they are a lot more common if R is a field such as the rationals than if it is, say, the integers, even though there are infinitely many in both cases. Among rings, Z is at the extreme low end, in having only finitely many units. If you take Z[\sqrt D], where D is a nonsquare integer, it will have infinitely many units if D > 0, finitely many if D < 0. (You might look for examples of units of infinite order in Z[2] and Z[5].) In algebraic extensions of Z more complicated than quadratic extensions, one invariably gets an infinite group of units. If r is a rational number that is not an integer, then Z[r], i.e., the subring of Q generated by Z and r, will also have an infinite group of units. (E.g., Z[3/2] contains 3/2 - 1 = 1/2, so 2 is a unit, and clearly has infinite order, so <2> is an infinite subgroup of the group of units of that ring.) ---------------------------------------------------------------------- You ask why the Second Isomorphism Theorem for Rings (p.246) does not have a condition analogous to the statement that B _< N_G(A) in the Second Isomorphism Theorem for Groups (p. 97). Good point. In stating the theorem for groups, I think the authors decided to make the statement as strong as they could, despite the fact that this made for a less clear statement than the usual formulation, and that the "added strength" can be recovered from the usual statement by using it appropriately. The usual version of that theorem is that if A and B are subgroups of G, with B normal in G, then AB is a subgroup, and AB/B =~ A/(A \intersect B). This corresponds exactly to the Second Isomorphism Theorem for Rings. To see that the authors' "stronger" version is implied by the above version, note that if B is not necessarily normal in G, but A lies in N_G(B), as in their statement, then one can apply the above version with N_G(B) in place of G, and get the desired conclusions. Inversely, if one wanted to state the Second Isomorphism Theorem for Rings in a manner parallel to the authors' version of the theorem for groups, one would define the "idealizer" of a subring A of R as \{ r\in R | rA \subset A and Ar \subset A \}, (which one can see is a subring containing A), and then assume, not that B is an ideal of R, but that it is contained in the idealizer of A. Idealizers are in fact used in noncommutative ring theory; though usually not for arbitrary subrings, but for one-sided ideals: if I is, say, a left ideal of R, then its idealizer of A is the largest subring of R in which A is a 2-sided ideal. However, we won't be going deeply enough into the theory of noncommutative rings in this course to use that concept. ---------------------------------------------------------------------- You ask about the use of the phrase "finite sums" in defining the product ideal IJ in the definition on p.247. The point of "finite" is not to exclude infinite sums -- infinite sums simply aren't part of the definition of a ring, so there's no way they could occur. Rather, the authors write "finite sums" so that one won't think that "sums" simply means expressions a_1 b_1 + a_2 b_2, but will understand that any finite number of summands is allowed. ---------------------------------------------------------------------- You ask whether there is an analog of Corollary 10, p.253, for group homomorphisms. Yes. Any nontrivial group homomorphism whose domain is a simple group is injective (where "nontrivial" means that it is not the homomorphism that sends everything to 1). Same proof. ---------------------------------------------------------------------- In connection with the fact that commutative simple rings are fields, (Prop.9(2), p.253, you ask whether general simple rings are close to fields in properties. Not very close. I mentioned in class on Monday that to construct the right or left ideal generated by a single element a of a ring R (and hence in the commutative case, the ideal generated by a), one just had to take all elements ra, respectively ar; but that to get the 2-sided ideal generated by a, one might need arbitrarily long expressions r_1 a s_1 + ... + r_k a s_k. So in the commutative case, to say that every nonzero element generates R as an ideal (which is equivalent to R being simple) means that for every nonzero a there is an r such that ar = 1; i.e., an inverse. But in the noncommutative case, it says that for each a there is a k, which may depend on a, and elements r_1, ..., s_k such that 1 = r_1 a s_1 + ... + r_k a s_k; a much weaker condition. The example in the paragraph you refer to of M_n(F) is already pretty different from a field, but there a fixed k will do, namely n. There are examples where the required k is unbounded as a function of a. ---------------------------------------------------------------------- You ask about the comment before the Examples on p.906 that the representation of an I-tuple depends on a choice of ordering of I. I think the authors' point is that the formal definition of an element of a cartesian product set, as a function on the index set I, does not depend on a choice of ordering of I; that choice only comes in when we decide to write elements of the cartesian product using a string of symbols. We can do so if I is reasonably small; e.g., finite, or, using "...", if it is countably infinite. In doing so, we have to decide what order to put the symbols in; but the abstract definition of an I-tuple doesn't involve any order. (Occasionally, we put symbols in a non-linear array; for instance, the entries of a matrix.) ---------------------------------------------------------------------- You ask how one can prove, as asserted in the middle of p.909, that one can show that if ZF is consistent, then so are ZF with the Axiom of Choice (equivalently Zorn's Lemma) and also ZF with the negation of that axiom. Set theory is far from my field; but my understanding is that assuming ZFC is true, one can construct a "model" of set theory satisfying ZF but not C (= the axiom of Choice, equivalent to Zorn's Lemma), and conversely, that assuming ZF is true but not Choice, one can construct a "model" of ZFC. The "model" in either case would consist of a certain subclass of the sets which are to be called the "sets" of the model; a relation among these which is called "membership" (even if it is not the same as membership in the given theory), etc.; and using the axioms that one assumes, one proves that the new structure satisfies the modified axioms. Hence if one set of axioms is consistent, so is the other. (This is like the way one approaches various sorts of non-Euclidean geometry, by constructing a region of the Euclidean plane, and a certain set of curves therein, which one treats as the lines of the new geometry, and a certain metric which one calls the "distance" function of the new geometry, and then verifies that it satisfies appropriate non-Euclidean axioms; hence these axioms are consistent.) ---------------------------------------------------------------------- You ask whether the concept of prime ideal (p.255) and the construction of rings of fractions (pp.260-263) work for noncommutative rings. The answer to both questions is that in the noncommutative context, we have more than one generalization of the concept. For the concept of prime ideal, the situation is relatively simple. In the commutative case, the condition that an ideal P be prime can be expressed equivalently either as saying that (1) for any two elements x, y, if neither is in P, then their product is not in P, or (2) for any two ideals I, J, if neither is contained in P, then their product is not contained in P. (Proving one direction of that equivalence is a problem in the homework I will give out today.) In the noncommutative case, these are two different conditions: Ideals satisfying (2) are called prime; those satisfying (1) are "completely prime". For P to be completely prime is equivalent to R/P being a ring without zero-divisors. The condition that P be prime means that in R/P, a product of nonzero ideals is nonzero. It's nice to have a completely prime ideal, but one may not be able to find them; e.g., M_n(Q) has none. Prime ideals may not be as nice, but they have existence properties analogous to those in the commutative case; e.g., every maximal ideal is prime. The situation for rings of fractions is more complicated. The construction given in the book doesn't work without commutativity: the definitions that they give of addition and multiplication aren't even well-defined. There is another approach to the construction in the commutative case, described in an unassigned problem on the sheet I will give out today, involving maps defined on certain sorts of principal ideals in R. If for "ideal" one substitutes "left ideal" or "right ideal", then there is a nice class of noncommutative rings for which the construction works -- the "left Ore rings" and the "right Ore rings" respectively. For rings which are not necessarily left or right Ore, there are variants which use non-principal ideals; but when the ring is indeed not Ore, the resulting "rings of fractions" are really ghastly -- generally much messier than the original rings, rather than nicer. Finally, are two entirely different approaches, one based on adjoining inverses by "generators and relations"; the other based on adjoining inverses not to certain elements of the ring, but to certain square matrices over it. ---------------------------------------------------------------------- Regarding the construction of rings of fractions (p.261, Theorem 15) you ask > What can we say about the cardinality of Q? It is not |R| x |D| > because 1 = d/d = d^2 / d^2. Then what is it? It will equal both |R| and (assuming R infinite) |R| x |D|. Infinite cardinalities are not as simple as finite cardinalities -- a set and a proper subset, or a set and its image under a non-one-to-one map, _can_ have the same cardinality. (E.g., Z and 2Z have the same cardinality, since the map n |-> 2n is a bijection; but 2Z is a proper subset of Z, and it is also the image of Z under the non-one-to-one map sending each even integer 2n to itself, and each odd integer 2n-1 to the next even integer, 2n.) For any infinite cardinal \kappa, one has \kappa x \kappa = \kappa. Hence for any infinite ring R and any D, we have |R| _< |Q| _< |R| x |D| _< |R| x |R| = |R|. Since the first and last term above are equal, all the terms are equal. This leaves only the case of finite R. One can show that any finite integral domain is a field, hence R = Q (though for any D with more than one element, |R| _< |R| x |D|). ---------------------------------------------------------------------- In connection with the definition of "norm" on p.270, and my comment in the handout that this is just one instance of a wide class of uses of the term, you write, > ... You mentioned ... that norms are "not necessarily > integer-valued". What are the restrictions, if any, for the > set that is mapped into? ... What I meant was that "norm" is not a single mathematical concept, but a useful general term for a function used in a certain way; the functions so used and so named in different contexts can be very different. For instance, in H110 we saw that in an inner product space, the "norm" of a vector x was defined to be ||x|| = sqrt{}. There is a more general concept of a "normed vector space", a real or complex vector space with a nonnegative-real-valued function "|| ||" which is not necessarily determined by an inner product, but has some of the properties of functions defined in that way. The "field norm", also mentioned in the homework errata, has a very different sort of codomain. So the answer is simply "whatever sort of set one finds useful". > ... Why are norms only defined over domains? In the context of this chapter, because the norms considered are generalizations of Euclidean norms, used to prove that rings have certain properties very much like those of Z, which non-domains don't have. > Is it possible to "measure" the elements of a ring with > zero divisors? The term is used in connection with certain rings with zero divisors, for very different purposes. For instance, if V is a normed vector space and R is the ring of "bounded operators" on V, that is, linear maps T: V --> V for which there exists a nonnegative real number c such that for all x\in V one has ||Tx|| _< c ||x||, then the least such c is called "the norm of T". This is a very important tool. Rings of bounded operators, despite being rings, are more in the domain of functional analysis than of algebra. I don't recall any uses of the term "norm" applied in algebra to rings that are not domains; but it has probably been so used. ---------------------------------------------------------------------- You ask for examples of rings in which greatest common divisors do not exist, as mentioned on p.274, second sentence before Prop.2 Let k be any field, and let k[x^2, x^3] be the subring of k[x] generated by k, x^2, and x^3; i.e., the ring of all polynomials a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 such that a_1 = 0. Note that in this ring: x^m divides x^n if and only if n = m or n >_ m+2. But x^m does not divide x^{m+1} (because x is not in the ring). Now note that a=x^5 and b=x^6 have x^2 and x^3 as common divisors; but no larger power of x is a common divisor of these elements in the ring, from which you can see that they have no greatest common divisor. (The statement "gcd's don't exist" means that, as in the above example, there are _some_ pairs of elements that have no gcd, not that _no_ pair has a gcd.) ---------------------------------------------------------------------- You ask whether there are noncommutative analogs of the concept of principal ideal domains (p.279). Yes! One can look at the condition that right ideals, left ideals, or 2-sided ideals be principal. The condition that 2-sided ideals be principal is not very useful, because the 2-sided ideal generated by a single element a does not consist of expressions of a nice form like xa or ax; rather, it involves arbitarily large sums x_1 a y_1 + ... + x_n a y_n; so to know that for an ideal I there exists an a such that every element of I has this form doesn't give us a fact that we can easily use. On the other hand, the conditions that right and/or left ideals are all principal are similar in their strength to the PID condition in the commutative case (and incidentally imply the above condition that every 2-sided ideal is principal, since if I is a 2-sided ideal, it is in particular a left ideal, and if as such we write it Ra, then a will certainly generate it as a 2-sided ideal.) There is a standard easy class of examples, of which our text gives a particular case as Exercise 13 on p.302. But they make a choice that makes that example seem more complicated than it needs to. The general example is based on a field F and a one-to-one endomorphism \alpha of f (i.e., a one-to-one homomorphism F --> F); the ring R one constructs from this will always have left ideals principal, and will also have right ideals principal if and only if \alpha is onto (hence is an automorphism of F). The point of their exercise is to show that a noncommutative ring can have the principal ideal property on one side but not on the other, for which they need a field with an endomorphism \alpha that is not surjective. To get this, they use the fact that in a field of characteristic p, the p-th power map x |-> x^p is always an endomorphism, and say "Let F be a field of characteristic p in which not every element is a p-th power". The much easier way to get an example is to take any field K, form the field of rational function K(t) (i.e., the field of fractions of K[t]), and let \alpha be the map that sends r(t) to r(t^2). (I am writing "t" rather than "x" for the indeterminate because a different "x" is used in the construction.) So you can look at that exercise; if you have any questions, let me know. Note also that if you merely want to get a noncommutative PID, then you can use a nontrivial endomorphism of F without requiring that it be non-surjective; for instance, you can let F be the field of complex numbers, and \alpha be complex conjugation. There are other more subtle noncommutative generalizations of the property of being a principal ideal domain. One of them is the condition that every (say) left ideal I of R have a "basis", i.e., a subset X such that every element of I can be written uniquely as a left linear combination of elements of X with coefficients in R. If R is commutative, then given two nonzero elements a and b of I, the element 0a + ab = ba + 0b has nonunique representation as a left linear combination of them, so the only way an ideal can have such a "basis" is if the basis consists of one element, i.e., the ideal is principal. But in the noncommutative case, this isn't so, and one gets interesting examples of rings with this property, called "free ideal rings" (abbreviated "firs"). My PhD thesis concerned these. ---------------------------------------------------------------------- In connection with the result on p.280 that the gcd of two elements of two elements of a PID is a linear combination of those elements, you ask about gcd's of larger sets, in particular infinite sets. I answered this in class, but you weren't there! Basically: The ideal generated by a set A will always consist of linear combinations of elements of A; so if R is a PID, the generator of the ideal (A) will constitute a gcd of A, and will indeed be a linear combination of elements of A. Note that even if A is infinite, every linear combination of its elements is a finite linear combination; so the gcd of A will equal the gcd of some finite subset. As an example: suppose R is the integers, and A is the set of all perfect numbers (integers which are equal to the sum of their proper factors). A lot is known about even perfect numbers; the first two are 6 and 28; but it is unknown whether there are any odd ones. So the gcd of the set of perfect numbers will be a divisor of gcd(6,28) = 2, i.e., will equal 1 or 2, depending on whether there are any odd perfect numbers. Whatever the value is, it can be expressed as a linear combination of finitely many perfect numbers; but there is no finite test which one can be sure will tell us whether it is 1 or 2. ---------------------------------------------------------------------- Regarding Proposition 13 on p.287, you ask whether we can avoid using unit factors u and v by just using variant forms of some primes -- > For instance, if R is the integers, if a=3 and b=-3, let > p_1=2, > p_2=-2, > p_3=3, > p_4=-3 If you did this, then Proposition 13 would not be true. E.g., with your values of a and b, we would have a = p_3^1 p_4^0, b = p_3^0 p_4^1 and since no prime appears with positive exponent in both expressions, we would get d = p_3^0 p_4^0 = 1; but that is not the gcd of 3 and -3. As I discussed in class, where the authors say the primes are "distinct", they should say they are "distinct up to unit factors", i.e., that none of them is an associate of any other. ---------------------------------------------------------------------- You ask about the use of the words "variable" vs. "indeterminate" both of which occur on pp.295-296. The term "variable" for the x in R[x] is not logical in the context of ring theory, because we are not considering polynomials f(x) as functions. We can't, because we don't have a fixed domain and codomain. E.g., if we thought of an element of (Z/3Z)[x] as a function Z/3Z -> Z/3Z, then x^3 - x = x(x-1)(x-2) would be zero at all elements of Z/3Z, and hence would equal the zero function; yet when we adjoin a square root of -1 to that ring, getting Z[i]/(3), then that field has 9 elements, so not all of them are roots of x(x-1)(x-2), and the polynomial would have changed from the zero function to a nonzero one. The construction of the complex numbers is based on adjoining a root to x^2+1 \in R[x]; if we regarded this as a function R -> R only, how would we interpret the statement that f(i) = 0 in C? Finally, one is interested in applying polynomials to matrices, as in Math 110, so again it would be too limiting to regard them as functions on the given field only. So we regard polynomials as more abstract entities (from which we can obtain functions); and instead of speaking of the x as a "variable", we introduce the new word "indeterminate", meaning, historically, "something that doesn't have a particular value". However, some people carry the word "variable" from other fields into algebra, and use it to mean indeterminate. The practice of extending the meaning of an old term to a new situation (e.g., extending "cyclic" from the case of finite cyclic groups, which "look like" cycles, to infinite cyclic groups, which don't) is long established, so this use of "variable" is not unreasonable if an author makes the usage clear, and the reader keeps in mind that in this context, it doesn't have the same meaning as in the case of functions. Generally speaking, modern usage prefers "indeterminate", and the authors introduce that term to begin with; but, as I say in the notes on today's reading (on the current homework sheet), in this section they sometimes slip back to calling them "variables". ---------------------------------------------------------------------- You point out that Corollary 8 on p.305 says that for an arbitrary set of indeterminates, "R is a UFD => R[...] is", but doesn't state the reverse implication, and ask whether that reverse implication can fail. No! The implication "R[...] is a UFD => R is" is the easy direction in these "<=>" results, and the authors presumably took it for granted and didn't think to mention it here; but you should be able to verify easily that it is true. If not, ask for details. ---------------------------------------------------------------------- You ask what the authors mean by the statement on p.518 (end of paragraph after Examples) that the property of being real, which distinguishes one cube root of 2 from the two others, involves "continuous operations". As I said in class on Wednesday, the reference to "continuous operations" here seems to be mistake. My best guess as to what the authors mean is that the real cube root of 2 is distinguished from the complex cube roots of 2 by _limit_ properties; namely the former is a limit of rational numbers, while the latter are not. Their point, in either case, is that limit properties (or continuity properties) are not part of algebra, so in algebraic terms, i.e., in terms expressible using addition, multiplication, and equalities (but not the absolute value function, or the "<" relation), these elements are indistinguishable. ---------------------------------------------------------------------- You ask about the relationship between minimal polynomials in linear algebra (Math (H)110) and in the study of field extensions (p.520). If R is any ring with 1 containing a field F whose elements commute with all elements of R, and \alpha is any element of R which, together with F, generates a subring F[\alpha] finite-dimensional over F, then one can see that the infinitely many powers of \alpha must be linearly dependent over F (otherwise F[\alpha] would be infinite-dimensional), hence \alpha satisfies a polynomial. The set of polynomials that it satisfies will form an ideal I of F[x], and F[\alpha] will be isomorphic to F[x]/I by the First Isomorphism Theorem. Moreover, since F[x] is a PID, I will be a principal ideal generated by some polynomial m(x). This will be unique if we take it to be monic, and is then called the minimal polynomial of \alpha. These observations can, in particular, be applied either when R is the ring of linear operators on a finite-dimensional F-vector-space V (we consider R to contain F in the guise of the ring of scalar multiples of the identity operator) and \alpha is any member of this ring, or when R is a field extension of F and \alpha any algebraic element thereof. The main difference is that when R is a field extension of F, it has no zero-divisors, so m(x) is necessarily irreducible. On the other hand, in linear algebra, the best case is when F is algebraically closed, so that m(x) factors into linear factors. There are also minor differences: In (H)110, we didn't have the language of ideals, Euclidean domains, and PID's, so the fact that the set of polynomials satisfied by \alpha consisted of all multiples of m(x) had to be proved ad hoc (Friedberg, Insel and Spence, Theorem 7.12(a), p. 516. But note that the method, used there, the Euclidean algorithm, is the same one used in this course to show F[x] is a Euclidean domain). On the other hand, in this course, which doesn't have (H)110 as prerequisite, we can't assume familiarity with the determinant function, so we can't get study analog of the relation between minimal and characteristic polynomials. ----------------------------------------------------------------------