Winter 2023 Algebra

Proof.

(a) Suppose . Let . Consider the natural projection homomorphism . We have for some and for some . Therefore, and where . We see that

(b) The center of a finite -group is non-trivial. Therefore . In the second case, is abelian. Suppose . We know that is normal in . We have . Since every group of order is cyclic, applying the first part shows that is abelian. (So in fact, is not possible.)

(c) No. Recall that , which has order . By Cauchy's theorem, has an automorphism of order . Consider where is define by sending a generator of to . Since is a nontrivial homomorphism, this semidirect product is nonabelian.

Proof.

(a) A function with is called a norm on . The integral domain is a Euclidean domain if there is a norm on such that for any two elements and of with there exist elements and in with

(b) Suppose , with . We prove that there are , such that

If then take .

Assume . We argue by induction on . Let . If , take and . Suppose . Write

and

Then the polynomial

is of degree less than . By induction, there exist polynomials , with

Let

We have

completing the proof.

Proof.

By Sylow III, and . This implies . Let . We divide the argument into two cases.

Suppose . By Sylow III,

This implies $|N_G(P)| = |G|/8 = 28.

Suppose . Then . Consider . Since , it contains a subgroup of order 4. By the lattice isomorphism theorem for groups, there is a subgroup with of order $|P|\cdot|\bar{H}|=28.

The only claim we need to justify here is that a group of order contains a subgroup of order for each . We prove this by induction on . For this statement is trivial. Suppose this is true for groups of order at most . We know that the center of a nontrivial -group is nontrivial. Applying Cauchy's theorem to shows that contains a normal subgroup of order . By induction, contains a subgroup of order . By the lattice isomorphism theorem for groups, there exists a subgroup of containing of order . This completes the proof.

Proof.

Suppose that is a ring homomorphism for which . So fixes every element of .

We have for some . We have

but we also have

Therefore, we must have and . We see that either or , and that either way, there are no solutions to .

Proof.

If is an ideal of , then the submodules of are of the form for any ideal . This makes it clear that is simple if and only if is maximal. In particulat, is simple for any maximal ideal .

Conversely, suppose that is a simple module, and fix . The submodule of generated by is nonzero, so we must have . By the freeness of the module , the assignment extends to a homomorphism , which is surjective because its image is a submodule containing . It following that where the ideal (submodule) is the kernel of . By the discussion above, must be maximal.

Proof.

Let be a matrix over whose eigenvalues are and . The roots of its minimal polynomial must be exactly and , and its degree mst be at most 3. We have the following options:

• : This is satisfied by .
• : This is satisfied by .
• : This is satisfied by .

Proof.

The element is a root of the polynomial

which is irreducible over by Eisenstein's criterion. Thus is the irreducible polynomial of over . Let be a primitive 3rd root of unity. Any Galois extension that contains must contian the remaining roots for of . Thus we may take to be the splitting field of the separable polynomial :

Proof.

Not true. If and is the primitive third root of unity then but so .

Proof.

: Say is a splitting field of . Assume is irreducible in and is a root of . Work in some field such that and all roots of are in . Let be another root of . Since is irreducible in , there is an -isomorphism . The field is a splitting field of over . Let be a field extending which contains all roots of and be the splitting field of over . Then there is an isomorphism extending . Since is the identity on and is a splitting field of over , the field is a splitting field of over . It follows that . Since , we conclude . This means that all roots of are in , so splits in into linear factors.

: Assume , otherwise there is nothing to prove. Let . Since is finite, the minimal polynomial splits into linear factors in , so all roots of are in . Let be the result of adjoining all roots of to . If then is the splitting field of over . Otherwise there is some , and similarly as above we obtain and . If then is the splitting field of over . In this manner we proceed inductively to construct and as above. Since

there is some number such that . Then the field is the splitting field for .

Proof.

We use the theorem on hte correspondence between bilinear maps and homomorphisms. The map defined by is easily seen to be -bilinear. By the correspondence theorem, there is a unique module homomorphism such that for every simple tensor . We show that is a ring isomorphism. That is a ring homomorphism follows from the general theory, so we focus on showing that is a bijection.

To see that is surjective, notice that if then .

To see that is injective if suffices to show the kernel of is trivial. For this, observe that any is of the form

Since is a homomorphism, we have