at the bottom of p.203. When you're unsure about the meaning of a symbol, you should always first look in the index of symbols on pp. 471-473! It is arranged according to the page on which the symbol is introduced, so you should start from the page where you found the symbol you are wondering about, and work back. If you look for "< ... >" starting from p.203 and working back, you will find for p.203 itself the symbol F[x]/

,
but going up just two lines, it shows that means what you get if
your start with F[x], and form in it the set of cosets of the
additive subgroup .
In the future, I hope you will remember to check the index of symbols
first. If after doing so you still have a question, then you can pose
it as your question of the day.
----------------------------------------------------------------------
You ask whether, in Proposition 4.3.4, p.204, any of the polynomials
a(x), b(x), c(x), d(x) can be zero.
Certainly!
----------------------------------------------------------------------
You ask about the statement in Example 4.3.1, p.205, that every
congruence class in R[x]/ in Def.4.3.2, p .203; R/ker(phi) in
Theorem 5.2.5, p.244, and R/I in Theorem 5.3.5, p.253).
I hope that what I said in class made it clear that these are
all the same usage. On the one hand, in each case we take the
additive group of the ring, and form the factor-group (Def. 3.8.6,
p.169) of that group by the subgroup in question, getting a new
additive group (so all these instances of "/" come from the use of "/"
in group theory), and we then make that additive group into a ring by
defining an appropriate multiplication. On the other hand, the
constructions F[x]/ and R/ker(phi) are special cases of
the construction R/I, since and ker(phi) are ideals of
the rings F[x] and R respectively.
So basically, we have one construction "G/N", for groups, and a
construction "R/I" for rings that starts with that group-theoretic
construction, and adds more structure to it.
----------------------------------------------------------------------
You ask about the statement after the display in the middle of p.246
that "If deg(pi-hat(f(x)))=deg(f(x)), then this gives a nontrivial
factorization of pi-hat(f(x)) in Z_n[x]".
The display gives a factorization of pi-hat(f(x)); the question
is whether it is a nontrivial factorization, i.e., a factorization
into polynomials of smaller degree. The polynomials pi-hat(g(x))
and pi-hat(h(x)) on the right-hand side have degrees less than or
equal to the degrees of g(x) and h(x) respectively, which are less
than the degree of f(x) (since f(x)=g(x)h(x) was assumed a
nontrivial factorization of f(x)). So if the left-hand side has the
same degree as f(x), we have the desired inequality of degrees. That
is what the authors are pointing out.
----------------------------------------------------------------------
You ask about the relationship between the last two displays
before Example 5.2.13 on p.247.
You need to read the words of the book carefully. Although the last
of those displays is introduced by "It then follows ...", they are not
saying that it follows from the preceding display. That preceding
display is given as the justification for the assertion two lines
above it, that "an element (a_1,...,a_n) ... is a unit if and only
if each component a_i is a unit ...". (The fact that that display
is a justification of this can be seen from the fact that it is
introduced by the words "This can be shown by observing ...".) So
when, after that display, the authors say "It then follows easily ...",
then mean that the final display follows easily from the statement they
have just justified, that (a_1,...,a_n) is a unit if and only if each
a_i is a unit.
Does the argument make sense now?
----------------------------------------------------------------------
You ask what, precisely, an n-tuple (used in the definition of
direct sum, p.247) is.
The short answer is: a list of length n. I assume you're familiar
with the phrase "ordered pair" for a "length-2 list", (a,b). Similarly,
a length-3 list (a,b,c) used to be, and is still sometimes called an
"ordered triple", and a list of length 4 and "ordered quadruple". For
higher numbers, the forms ended in "-tuple" (e.g., quintuple), from
which mathematicians generalized to "n-tuple", and then extended this
usage backwards, so that "triple" is now usually replaced by "3-tuple",
"quintuple" by "5-tuple", etc..
Set-theorists, who don't want to introduce more "primitive notions"
than are needed, introduce a way of regarding these "lists" in terms
of other mathematical notions: Namely, they define an n-tuple
(a_1,...,a_n) to be a function on the set {1,...,n}; so its value
at 1 is called a_1, etc.. (Actually, for their own reasons, they
prefer to use {0,...,n-1} and write (a_0,...,a_{n-1}); but if a
tuple is written (a_1,...,a_n), as most mathematicians prefer to
do, then set-theorists would agree that it is a function on {1,...,n}.)
To complicate things further, they define "function" to be a certain
sort of set of ordered pairs; so this means that they have to start
with a notion of pair before they can construct that of n-tuple. So
set-theorists have two notions of "ordered pair": one that they define
in a different way, and one that is the n=2 case of their definition
of n-tuple.
But you don't have to know any of these details! The idea that
an ordered n-tuple is an "ordered list", written in parentheses,
is all you need. If you want to see some of the additional details,
you can take Math 135.
----------------------------------------------------------------------
You ask about the relation between direct sums as in Def.5.2.9, p.247,
and direct sums of vector spaces.
Whether one is dealing with rings, abelian groups, or vector spaces,
there is a distinction between "internal direct sums" and "external
direct sums" which our text does not go into (but which I sketched
in class for groups). If one takes objects (groups etc.) X_1, ...,
X_n, and forms from them an object consisting of ordered n-tuples of
elements, with operations defined componentwise, this is an "external"
direct sum of the X_i. But one may also start with a known object X,
and find certain subsets X_i such that every element of X can be
written uniquely as a sum of elements, one from each X_i; and such
that each X_i can be given a structure of the same sort (ring,
abelian group, vector space), so that this association of each element
of X with an n-tuple of elements of the X_i constitutes an
isomorphism between X and the (external) direct sum of those
structures. Then X is called the "internal direct sum" of the X_i.
The conditions for this to hold were found for groups in Theorem 7.1.3,
p.319, although for groups written non-additively the authors speak of
"direct products" rather than "direct sums"; and they don't actually
introduce the terminology of "internal" and "external", and they only
consider there the case n=2. But I hope that result bridges for you
the gap between what you saw in 110 and what the authors are defining
here.
----------------------------------------------------------------------
You ask how the statement on p.247 that (a_1,...,a_n) is a unit if
and only if each a_i is a unit implies that R^x is isomorphic to
the direct product of the groups R^x_i.
Well, strictly speaking, it proves more than that the groups are
isomorphic: It shows that they are equal. I.e., it shows that an
element (a_1,...,a_n) belongs to R^x if and only if it belongs
to the product-set written on the right in the last display; and
if you look at the definition of the group operations on the two sides,
you will see that they are the same.
I hope this helps.
----------------------------------------------------------------------
You ask whether the characteristic (p.248, Def. 5.2.10) of an
integral domain should have to be zero, since if n . 1 = 0,
with 1 not equal to 0, the factor n must equal 0.
When one writes "n . 1" in the definition of characteristic, the
"n" is a member of Z, while the "1" is a member of R; so we
are not dealing with a product of two elements of R, which is the
situation that the definition of integral domain applies to.
(Actually, for any ring R and any positive integer n, if we
write 1_R for the multiplicative identity of R, and n_R for
the result of adding 1_R to itself n times, then for any
element x\in R, the element "n . x" can easily be shown to be the
product n_R x within the ring R. Now if n is a multiple of the
characteristic of R, then n_R = 0; so the fact that n . x
will also equal zero becomes, in the ring R, 0 x = 0, which is
not a violation of the possibility of R being an integral domain.)
----------------------------------------------------------------------
Regarding the reference on p.248, paragraph before Prop.5.2.11,
you say you aren't sure what results from Chapter 3 the authors
are referring to.
It is the display on p.97 (the additive-notation analogs of the
laws of exponents). In the proof of Proposition 5.2.11, the
authors verify that these conditions imply that phi is a ring
homomorphism.
----------------------------------------------------------------------
You ask about the meaning of the phrase "integral domain" in
Proposition 5.2.11, p.248.
When you run into a term whose meaning you aren't sure of, the
first thing to do is look in the index. If you can't find the term
there, or if the page that it leads you to does not clear up your
uncertainty, then you should ask about it. If you check for "integral
domain" in the index, you will see that it has nothing to do with
integers. (There is a historical connection with integers, which I
sketch briefly in the answer to a student's question that I have put
on the "answers" web page; but the present-day meaning is simply the
one defined in the book.)
----------------------------------------------------------------------
You ask whether subrings of principal ideal domains (p.252) are
also principal ideal domains.
Nope. They are integral domains, of course; and the book hasn't
yet pointed out examples of integral domains that aren't principal
ideal domains, so they can't address the question. However, an example
of an integral domain that isn't a principal ideal domain is Z[x]. For
instance, the ideal of all polynomials whose constant terms are even
can be shown not to be principal. But it is a subring of Q[x], which
is a principal ideal domain by Example 5.3.2.
----------------------------------------------------------------------
You ask why the factor group R/I referred to on p.253 line 3
is abelian.
If phi: G --> H is a group homomorphism, and G is abelian,
then its image, phi(G), is also abelian. You should be able
to prove this quickly from the definition of homomorphism. Now
a factor-group G/N is a homomorphic image of G (under the
"natural projection" pi); so a factor-group of an abelian group
G is abelian. This applies, in particular, when G is the
additive group of a ring R.
----------------------------------------------------------------------
Sorry I didn't have time to talk in class about the part of the
proof of Prop.5.3.7 (p.254) that you asked about, where the authors
say "We must show that this correspondence preserves ideals".
What they mean is that since we know that the map described gives a
bijection between additive subgroups J of R containing I and
additive subgroups K of R/I, to get the result claimed it suffices
to show that for additive subgroups J and K that correspond in that
way, J is an ideal of R if and only if K is an ideal of R/I.
Since an ideal is an additive group that is closed under multiplication
by elements of the ring in question, all one has to show is that if
J is closed under such multiplication, then so is K, and vice
versa. That is the computation they give; let me know if you find
any difficulty with it.
----------------------------------------------------------------------
You ask about the use of "x^2 + I = -1 + 2x +I" in getting the first
display in Example 5.3.3, p.255.
Where the authors say "we can use the formula "x^2 + I = -1 + 2x +I
to simplify products", what they mean is that they can use that
formula if needed. For the first display, as you observe, it is
not needed; grouping terms is enough. For the second display it
is needed.
----------------------------------------------------------------------
You ask about the equation x^2 + I = -1 + 2x + I in Example 5.3.3,
p.255.
Remember that a + I = b + I is equivalent to a-b \in I. So what
the authors are doing here is finding a "nice" polynomial which differs
from x^2 by a member of I = is a field.
We know this in two ways: On the one hand, from Prop.5.3.9(a) on the
preceding page, since Theorem 5.3.10 (also on that page) shows that
will then be maximal. And on the other hand, by Theorem 4.3.6.
----------------------------------------------------------------------
You ask how we know, in the 4th-from-last line of Example 5.3.9,
p.258, that phi_u(F[x]) is a field.
It is, as you guessed, because it is isomorphic to F[x]/ker(phi_u),
which is a field by Proposition 5.3.9(a). (Any ring that is isomorphic
to a field is a field.)
----------------------------------------------------------------------
Regarding the quotient-field construction (Theorem 5.4.4, p.264)
you ask, "Are there any examples that are not like fractions and
their operations?"
In the context of this course, no: Theorem 5.4.6 shows that any
field F containing an isomorphic image of D contains a field
isomorphic to Q(D), and by construction, the structure of Q(D)
is "like fractions and their operations". Of course, F may
contain elements not in the image of Q(D) as well, just as the field
of real numbers has elements other than rationals; but I assume that
things that "don't come from D" aren't what you're asking about.
One situation where your question has more interesting answers is
in noncommutative ring theory. The noncommutative analog of a field
is called a "skew field" (or "division algebra"). The case that
was studied earliest was the one where a construction "like fractions
and their operations" did work. (Though one has to distinguish between
fractions of the form a b^-1 and those of the form b^-1 a. Some
rings allow one sort, some allow the other, and some allow both.) But
there are rings that can be embedded in fields where no such expression
is possible -- e.g., where an element x y^-1 z can't be converted
into a form a b^-1 or b^-1 a; where an element u v^-1 w + x y^-1 z
can't be "brought to a common denominator"; where an element such
as (u v^-1 w + x y^-1 z)^-1 can't be expressed without using
inverses-of-expresions-containing-inverses. I have a paper on the
subject; it requires much more background than Math 113 to read, but
if you want to take a look at it, it is at
/~gbergman/papers/sfd_frm_mtrd.ps .
There are also some cases where commutative fields can be constructed
in ways that don't look like fractions, even if they are isomorphic
to the construction by fractions -- e.g., the construction of the
field of rational numbers as repeating decimal expressions.
----------------------------------------------------------------------
You ask about the statement on p.264 line 3, "We are given that
ab' = ba' and cd' = dc'".
This is the translation of the first sentence of the proof (on the
preceding page), "Let [a,b] = [a',b'] and [c,d] = [c',d'] ...".
----------------------------------------------------------------------
Concerning the expression of the given homomorphism theta in
Theorem 5.4.6, p.265, as the composite of two homomorphisms, phi
and theta-hat, you ask why one would want to go from D to F
by two homomorphisms instead of one.
The central point of that theorem is the _existence_ of theta-hat;
it says that if you have a way of mapping D to F by a homomorphism
(namely, theta), this leads to a way of mapping the larger object
Q(D) to F as well (namely theta-hat). The "factorization" of
theta expressed by writing it as theta-hat phi then shows the
relationship between these two maps: theta-hat behaves in a "known"
way on the most easily described elements of Q(D), namely, those
that come from D via phi.
----------------------------------------------------------------------
You ask about the difference between Theorem 5.4.6 (p.265) and
Corollary 5.4.7 (p.266).
There are two differences. One is that in the Corollary, the
homomorphism D --> F is assumed to be an inclusion, so that
one can write elements of F in terms of elements of D without
having to use the symbol phi. More important is the assumption
that every element of F has the form ab^-1, which does not
correspond to any assumption in the original theorem. This means
that the resulting map Q(D) --> F is actually onto. Since it
is always one-to-one (as in the Theorem), this makes it an isomorphism
in the case of the Corollary.
----------------------------------------------------------------------
You ask how one concludes that theta-hat is onto, in Cor.5.4.7, p.266.
Check out the definition of theta-hat in this case. You will
see that it comes to theta-hat([a,b]) = ab^-1. (If you don't, let
me know.) So if every element of F has the form ab^-1, then every
such element is in the image of theta-hat; i.e., theta-hat is onto.
----------------------------------------------------------------------
Regarding the terms "base field" and "subfield" mentioned near the
end of the second paragraph of section 6.1, p.270, you ask whether
there is any difference between them.
Just a difference in point of view. When one is considering one
or more extensions of a given field K, and looking at what
polynomials elements of those extension fields satisfy in K[x],
what degrees such elements have over K, what dimensions those
fields have over K, etc., one calls K "the base field". One
may note in such discussion that one of those extension fields E
is a subfield of another of them, F, but one doesn't call E "the
base field" unless one changes one's point of view and starts
looking at minimal polynomials of members of F in E[x]. (So,
for instance, if one considers Q(sqrt 2) and Q(sqrt 2, sqrt 3)
as extensions of Q, one may note that the former is a subfield
of the latter, but as long as one is looking at polynomials that
the elements of these fields satisfy over Q, one is considering
Q "the base field".)
----------------------------------------------------------------------
You ask about the reasons for the last two sentences before
Def.6.1.1, p.271, relating primeness, irreducibility, and
maximality in rings K[x].
These points were discussed on p.256, second paragraph before
Def.5.3.8. Of course, the words "prime ideal" and "maximal ideal"
weren't used in that paragraph, because that paragraph was motivating
the definition of those terms. But the conditions defining those
terms were used. (See also Prop 5.3.9 on the next page.)
----------------------------------------------------------------------
You ask about the difference between u "satisfying" f(x) (p.271,
Def. 6.1.1) and being a "root" or "zero" of p(x).
No difference. I would consider the latter terms better, since
what is "satisfied" is really the equation "f( ) = 0". The book
uses "root"; I don't think it uses "zero", though that is another
term often used.
----------------------------------------------------------------------
You ask whether I can outline the proofs that pi and e are
transcendental, mentioned on p.271.
Nope! The proofs that I have seen are those in Ian Stewart's
"Galois Theory" (chapter 6 in the second edition, chapter 24
in the third edition. If you try looking them up and have the
choice, I recommend the second edition as a better book, generally,
than the third.) They involve writing down some messy integrals
and performing complicated computations, but give no hint what
the idea behind those computations is; and without such an idea,
I can't wrap my mind around a long proof.
----------------------------------------------------------------------
Regarding the proof of Prop.6.1.2 on p.271 you ask "Why does the
fact that I is prime ideal imply that there exists only one monic
polynomial p(x) with minimal degree and p(u) = 0?"
The authors don't deduce that from the fact that I is prime --
any nonzero ideal I in K[x] has a unique monic generator;
i.e., among all the elements of least degree, only one is monic.
I thought the book made this clear, but looking back now, I can
only find it stated before this in the last sentence of Exercise 5.3.2,
p.252, which is a somewhat inconspicuous spot. In any case, I've
pointed it out several times in class. You should see whether the
reason is clear to you when you think about it. If not, ask again.
----------------------------------------------------------------------
You ask about the relationship between the polynomials p(x) in
Prop.6.1.2 and Def.6.1.3 on p.271.
The definition gives a name to the polynomial described in the
proposition. The reason the definition doesn't refer to irreducibility
or the "p(x)|f(x)" condition is that these are not required to determine
the polynomial. As the third sentence of the proposition shows, the
polynomial is uniquely determined by the "minimal degree having u as
a root" condition. So this is all that belongs in the definition, and
the proposition gives the additional facts.
----------------------------------------------------------------------
You ask why there are two different terms describing K(u_1,...,u_n)
(p.273, Def.6.1.4).
It's not too unusual to have different ways of saying the same thing.
"To generate the extension field of K" is what the elements
u_1,...,u_n are said to do; "to adjoin u_1,...,u_n to K" is
what the mathematician studying that extension is said to do; so
using each of those verbs, one gets a way of talking about the
extension.
However, the authors' wording, making these into two separate sentences,
is awkward, and I will include in my letter to them at the end of the
semester a suggestion that they shorten it.
----------------------------------------------------------------------
You ask about the difference between the expression
a_0+a_1u+...+a_nu^n on p.273, 3rd-from-last paragraph, and the
expression a_0+a_1u+....+a_n-1u^n-1 on on p.274, middle.
Since the difference concerns the relation between the degree
of the polynomial and the integer n, you have to check what
"n" means in each case. On p.273, there is no "n" in the context,
so "all elements of the form a_0+a_1u+...+a_nu^n" means all such
expressions as n ranges over _all_ natural numbers. (Although
there was an "n" in the paragraph preceding the proposition, that
has no connection with this n. That was the number of elements
being adjoined, while as the sentence before the proposition
indicates, they are beginning the study of that situation by just
looking at the case where one element is adjoined; so that "n" has
disappeared.) On p.274, on the other hand, at the beginning of
the middle paragraph, the authors let the minimal polynomial of u
be p(x) = c_0+c_1x+...+c_nx^n; so n is the degree of that minimal
polynomial; and the division algorithm allows us to reduce every
polynomial in u to one of smaller degree (or 0), i.e., to the
form a_0+a_1u+....+a_n-1u^n-1. See Prop.4.3.3, p.204, and the
condition "deg(r(x)) < deg(p(x))" there.
----------------------------------------------------------------------
You ask about the expressions a_0+a_1 u+...+a_m u^m and
a_0+a_1 u+....+a_n-1 u^n-1 in the middle paragraph on p.274.
The authors first note that any expression a_0+a_1 u+...+a_m u^m can
be reduced to a linear combination of 1, u, ..., u^n-1 using the
relation p(u) = 0. This can be described, as you do, in terms of
solving p(u) = 0 for u^n and substituting this repeatedly into
the highest-degree term of the given polynomial. However, it's more
convenient to look at it in terms of a process we have already studied,
the division algorithm for polynomials, Theorem 4.2.1, p.193. Using
that, we can write the given expression as a multiple of p(u) plus
an expression of degree < n; and since p(u) = 0, the former summand
vanishes, and we are left with the latter.
The authors are certainly not saying that if one starts with
a_0+a_1 u+...+a_m u^m and performs this process, one will be
left with a_0+a_1 u+....+a_n-1 u^n-1. Rather, the idea is "Think
of all expressions of the form a_0+a_1 u+...+a_m u^m that represent
our given element. By the above argument, one of these will have
degree < n; so choosing that to be our expression, we can assume
m = n-1; and then our element is a_0+a_1 u+....+a_n-1 u^n-1".
----------------------------------------------------------------------
You ask about the converse to the statement of Proposition 6.2.1,
p.277, "If F is an extension field of K, then F is a vector space
over K."
Well, the wording of that proposition simplifies things a bit.
What it really means is, "If F is an extension of a field K,
and we ignore the operation of multiplying members of F by other
members of F except when the latter elements are members of K,
then the structure on F that we are left with is a structure of
vector space over K." Or briefly, given any extension field F
of K, we can strip away part of its structure, and be left with
a structure of vector space over K on the set F.
So put that way, there isn't an obvious converse. One might ask,
"Given a vector space F over a field K, must there always be
a way of extending its vector space structure to a structure of
extension field over K ?" To that, the answer is no. For instance,
because the only irreducible monic polynomials over the field of
complex numbers are the polynomials x-c, one can't construct any
extensions containing elements algebraic of degree > 1, from which
one can deduce by the methods of section 6.2 that C has no algebraic
extensions F with [F:C] > 1. So a 2- or 3-dimensional vector space
over C, for instance, can't be made into an extensions field.
----------------------------------------------------------------------
You ask whether one can prove directly the implication (2)=>(3)
of Prop.6.2.4, p.277, namely that if u\in F belongs to a finite
extension of K, then K(u) is a finite extension of K.
Certainly. Suppose u\in F belongs to a finite extension E of K.
By definition, K(u) is the _smallest_ extension of K in F
containing u, so it is contained in E; so being contained in a
finite dimensional subspace E of F, it itself must be finite
dimensional.
(I see that in the authors' very brief development of the theory
of dimensions of vector-spaces and appendix A.7, they don't explicitly
say that a subspace of a finite dimensional vector space is finite
dimensional. But that is easily deduced from the first conclusion of
Corollary A.7.9. If you want more details as to how, let me know.)
----------------------------------------------------------------------
You ask why Theorem 6.2.5 (p.278) doesn't apply the concept of the
degree of an extension to infinite extensions as well.
Because this course doesn't assume familiarity with the set theory
needed to talk about cardinalities of infinite sets. From your
Class Questionnaire, I see that you have taken Math 104; but most
students in the class haven't; and even 104 doesn't talk about
multiplying infinite cardinalities, as would be required to make sense
of the formula [F:K] = [F:E] [E:K] when the extensions are infinite.
That comes in Math 135. The formula is true for arbitrary extensions,
but we don't have the concepts to state it.
----------------------------------------------------------------------
You ask why, in examples like 6.2.2, p.279, the roots of a
polynomial f(x)\in K[x] don't have to lie in K.
It's true that in Def.4.1.10, p.189, we defined a root of a
polynomial f(x)\in K[x] to be an element of K. But that was
before we saw the idea of one field being contained in the other.
Now that we have that concept, whenever K is a subfield of F,
a polynomial f(x) over K can also be looked at as a polynomial
over F, and we can apply the Def.4.1.10 to it from that point of
view, and so ask what roots it has in F. So for instance, we
know that x^2 + 1 has no roots in R, but has two roots in C.
Much of what we have been doing from section 4.3 on is studying
the situation of a polynomial over a subfield having roots in an
extension field. The authors actually revised Def.4.1.10 to include
this case, in the first paragraph of Example 5.2.11, p.245.
----------------------------------------------------------------------
You ask about the statement in Example 6.4.3, p.290, "Then the field
is identified with the cosets of the form a +