2025 Spring Qualifying Exam in Algebra (AI-generated)

Proof.

The elements contained in every conjugate of form the core

In particular, , so . Put . Since is proper, .

If , then . If , then

Thus in all cases the required bound holds.

Proof.

Let

Because , the map defines an automorphism of order of . Hence

Both cyclic factors have odd order. The product is not abelian because , and therefore

Proof.

Conjugation by any permutes the Sylow -subgroups. Therefore

Thus .

Now let be a -subgroup, and let be any Sylow -subgroup. Since is normal, is a subgroup. It is a -group because

is a power of . The maximality of the Sylow subgroup therefore implies , so . This holds for every Sylow -subgroup, and hence .

Proof.

Fix a quotient map

The correspondence theorem identifies ideals of containing with ideals of .

(a) Let . Then is a maximal ideal of . It is nonzero because is not maximal in . Choose . The additive group is finite of order . Since

is a quotient of , it is finite. Finally,

so is finite.

(b) The Gaussian primes and are not associates, and both have norm . Thus

are distinct maximal ideals and

Set . Then and have all the required properties.

Proof.

Define a homomorphism

by for and . It is surjective because every element equals . Also , so induces a surjective map

Conversely, define

This is well defined. Indeed, if , then for some . In the quotient, , so multiplying that equality by gives . Direct calculation shows that is a ring homomorphism and that and are inverse maps. Hence the rings are isomorphic.

Proof.

Choose generators of . Let

generate . We claim that

generate . Given , its coset can be written as

Therefore , so it is an -linear combination of the . Thus is an -linear combination of the displayed finite set, proving that is finitely generated.

Proof.

The minimal polynomial of divides

Put . The polynomial is irreducible over , so its two roots lie in . Every element of is a square in : the latter group is cyclic of order , and its subgroup of order is generated by the tenth power of a generator, which is a square. Therefore both roots of have square roots in , and splits there.

It remains to check that has no repeated root. In characteristic ,

The value is not a root of . If , then

Thus , so is separable. The minimal polynomial of therefore splits into distinct linear factors over , which proves that is diagonalizable over that field.

Proof.

The subgroups of are the trivial subgroup, , the unique subgroup of order , and three subgroups of order generated by transpositions.

By the Galois correspondence, corresponds to one field with

The three order- subgroups correspond to three distinct fields satisfying

Together with the endpoints and , these are all intermediate fields.

An intermediate extension is Galois exactly when the corresponding subgroup is normal in . The normal subgroups are

Thus, including the endpoints as the question does, exactly three intermediate fields are Galois over : , , and . If only strict intermediate fields are counted, the answer is one.

Proof.

Let be a root in an algebraic closure. In characteristic ,

so is the only root of . Let be the minimal polynomial of over . Since divides , over the algebraic closure it has the form

for some . If , then the coefficient of in is . Because is nonzero in , this coefficient being in would imply . That would make a root of in , contrary to the hypothesis. Therefore , so the minimal polynomial of has the same degree as . Hence is irreducible.

Proof.

We have

Since , it is a root of the cyclotomic polynomial . Thus is a primitive tenth root of unity and

Every cyclotomic extension of is Galois: it is the splitting field of in characteristic zero. Therefore is Galois. Its Galois group is