2025 Fall Qials in Algebra

Proof.

Let act on the set of left cosets of by left multiplication:

Since , this action gives a homomorphism

Let

Then is a normal subgroup of , and , because any element in the kernel fixes the coset .

The image is a subgroup of , so divides . Also

which is a power of because is a -group. The largest power of dividing is . Hence

If , then the action is trivial. In particular, for every ,

so . Thus , contradicting . Therefore

Hence

Since and , we must have . Therefore is normal in .

Thus every subgroup of index in a finite -group is normal.

Proof.

Let

The group acts on by conjugation. The orbits of this action are exactly the -conjugacy classes contained in . Therefore is a union of -conjugacy classes.

(a) We show that all these -conjugacy classes have the same size.

Let . The -conjugacy class of has size

Since ,

Because is normal in ,

Thus

Therefore every -conjugacy class contained in has size

Hence is the union of some number of -conjugacy classes, all of equal size.

(b) Since the -conjugacy class of has size

and each -conjugacy class has size

we get

Now

Therefore

Hence

This proves both claims.

Proof.

Consider the determinant homomorphism

It is a group homomorphism because

Its kernel is

Since the kernel of a group homomorphism is normal, is normal in .

By the First Isomorphism Theorem,

Since every nonzero real number occurs as the determinant of an invertible real matrix, for example

we have

Therefore

The group is abelian, so is abelian.

Proof.

First consider

Define a homomorphism

by

Then is surjective and

Hence

Therefore is an integral domain.

Now consider

We show that is not an integral domain.

If , then

In ,

Neither nor is zero in , because neither nor belongs to the principal ideal generated by . Thus has zero divisors.

If , then

In ,

but . Thus has a nonzero nilpotent element, so it is not an integral domain.

Therefore is an integral domain while is not. Since being an integral domain is preserved by ring isomorphism, the two rings are not isomorphic.

Proof.

First we prove that is an ideal.

Clearly . Let . Then there exist positive integers such that

Since is commutative, the binomial theorem gives

In each term, either or . Hence each term is zero. Therefore

So .

Also, if and , with , then

Thus . Hence is an ideal of .

Now we prove that has no nonzero nilpotent elements. Suppose that is nilpotent. Then for some ,

Thus

Since is nilpotent, there exists such that

Therefore

So is nilpotent, meaning . Hence

Thus the only nilpotent element of is zero.

Proof.

Since , every element satisfies

Since , every element satisfies

For a pure tensor , we have

and

Because and are distinct primes, there exist integers such that

Therefore

Every element of is a finite sum of pure tensors, and each pure tensor is zero. Hence

Proof.

The equation

is equivalent to

Thus

Let

Then

By the rational root test, the only possible rational roots of are

A direct check shows

So has no rational root. Since is cubic, it is irreducible over .

(a) The minimal polynomial divides . Since is irreducible and is not annihilated by the constant polynomial , we must have

We use the rational canonical form theorem: if a matrix over has minimal polynomial dividing , then its invariant factors divide . Since is irreducible, every nonconstant invariant factor must equal . Therefore the characteristic polynomial of is

for some integer . Hence

Thus divides .

(b) Let be another rational matrix satisfying

Then also

The same argument shows that the minimal polynomial of is , and all nonconstant invariant factors of are equal to .

Since , both and have rational canonical form equal to the direct sum of companion blocks for . Therefore and have the same rational canonical form. By uniqueness of rational canonical form, and are similar over .

(c) For

a companion matrix is

Its characteristic polynomial is

By the Cayley-Hamilton theorem,

Therefore satisfies the required equation.

A second example is

Since

the matrix also satisfies the required equation.

Thus, for , two examples are

They were obtained from the companion matrix of the irreducible polynomial and its transpose.

Proof.

By the Fundamental Theorem of Galois Theory,

because

Now suppose first that

Then for every ,

Thus

for every . Therefore

Hence

So and lie in the same left coset of .

Conversely, suppose and lie in the same left coset of . Then

so there exists such that

Every element of fixes every element of . Therefore, for every ,

Hence

Thus the restrictions of and to are equal if and only if and lie in the same left coset of .

Proof.

We claim that

works.

First we check that has no root in . We compute:

and