ANSWERS TO QUESTIONS ASKED BY STUDENTS in Math H113, Spring 2009, taught
from Dummit and Foote's "Abstract Algebra", 3rd Edition.
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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.
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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!
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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.
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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.
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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.
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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).
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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.
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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.
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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).)
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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}?
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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.
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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.
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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.
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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.)
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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.
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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.
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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.
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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.
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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.
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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.)
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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.
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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.
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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.
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> ... 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.
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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.
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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".
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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.
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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.
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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.
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