SOME SOLUTIONS TO HMK PROBLEMS.
Hungerford: I.9: 2.
1) Show that G is the internal direct product of its p-Sylows S(p_i).
This is elementary.
2) Assume the statement to be proved is ok for abelian p-groups.
3) Let a in G be of maximal order. Using 1) a= a_1+...+a_l, in a unique way,
with a_i in S(p_i).
By 2), < a_i > is a factor of S(p_i). Check that the internal direct sum, H,
of the < a_i > in G is well-defined and a factor of G. Check that < a >=H
(use the fact that we are dealing with distinct primes).
This step 3) reduces the problem to proving that the assumption 2) is true.
This is done in 4)
4) If G= < a > we are done. Assume G is not < a >. Then there is y in G of order p,
such that Y:= < y > not equal to A:= < a > (check!). The image of a in G/Y has still maximal order.
By induction on |G|: there is M' in G/Y with G/Y= M'x ((AY)/Y). M' is M/Y for some M in G
containing Y: so there is M in G with G/Y= M/Y x (AY)/Y.
In particular: AY \cap M=Y.
Since Y is in M: AY\cap M=(A \cap M)Y=Y.
If A\cap M= e, we are done, we have that < A,M >=G etc.
IF not A\cap M=Y and Y < A, a contradiction.
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Prove: The direct product A over a set I of copies of Z is a free abelian group
iff |I|=finite. (Take for granted that every subgroup of a free abelian
group is free abelian; we proved this for finite rank; the general proof is
in the textbook; see also Fuchs, infinite abelian groups, 14.5.)
1) If |I| is finite ok. If |I| is infinite it is enough to assume that
I is countable (why?), e.g. I=N the natural numbers.
2) Assume A is free.
We define a subgroup H of A which cannot be a free abelian group.
Fix a prime number p. Denote the elements of A by sequences in Z.
Let H:={ {a_i}| for every m>0, p^m divides a_j for all but finitely many
indices j in N}.
H is uncountable.
H is free over an uncountable set, J.
Consider the subgroup pH. It is also free over the same set.
The quotient H/pH is uncountable.
Show that every coset in H/pH contains at least one element of
the weak direct product of Z over N.
This shows that H/pH is countable.
This is a contradiction.
For more see, Fuchs' book, especially 19.2.
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III: 3: 8. Check: Abstract algebra, Dummit-Foote, QA162.D85, pages 279-..
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III.6.5: Let R be commutative with identity.
Prove that g=a_0+ a_1x^1+ ... + a_nx^n \in R[x] is invertible
iff a_0 is a unit and the a_1, ..., a_n are nilpotent.
First check that without loss of generality you may assume that a_0=1.
<-- Let f=1-g. Check that f is nilpotent (just expand).
Then the infinite sum 1+f+f^2+f^3+... ( = 1/(1-f)!!!) is actually finite
and inverts 1-f which is g.
--> Consider the natural embedding of R[x] into R[[x]] (power series).
It is a ring homomorphisms sending 1 to 1 and units to units.
Note that g is invertible in R[[x]]
(either by III.5.9 or, better) by just taking the geometric series
1+f+f^2+f^3+... in R[[x]]. Of course you must check that it
indeed defines an element in R[[x]].
Since we are assuming that g is invertible in R[x] and inverses are unique,
the geometric series
stops after finitely many powers, i.e. f is nilpotent.
Now check, by induction (descending on n) that a_i,
i=1,...,n is nilpotent (you will need the easy fact that
if a and b are nilpotent, so is a+b).
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Ash's book: 3.4 n.8:
This is reaaly the culmination of a series of exercises on F(E^p).
You need only consider char(F)=p. Let E/F.
First, show (exercise number 3!) that if a is in E and it is separable over
F(a^p), then it is in F(a^p).
Next, show that if E/F is finite, then it is separable iff E=F(E^p)
(this is exercise 7!).
The --> direction can be proved as follows:
let a be in E. It is separable over F so it is over F(a^p). Then it is in
F(a^p) by ex 3.
For <-- argue as follows:
let a be inseparable with m(a,F) in F[x^p] (we showed this in class)
= f_0+ ... f_rx^{rp}. The a^{sp}, s=0,..., r are dependent.
By the minimality of m(a,F) the a^{s}, s=0,...,rp-1$ are independent.
Then a^s, s=0, ..., r are independent: r is less than (rp-1).
Then a^{sp}, s=0, ... r are independent (this is exercise 6! do not get upset:
6 follows from 5 and they are both easy).
Finally: Since E/K is separable, then E=K(E^p). Since K/F is separable,
then K=F(K^p). We have E=K(E^p)=F(K^p)(E^p)=F(E^p), since K is in E.
So E/F is separable.
Note that the statement is true in general: i.e. in any characteristic and
also fon not necessarely finite extensions.