2017 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

Suppose is a group of order . Prove that is not simple.

Proof.

Since , Sylow's theorems give

Thus or . If , the unique Sylow -subgroup is normal.

Suppose . Distinct Sylow -subgroups intersect trivially, so their nonidentity elements account for

elements of . Exactly elements remain, including the identity. Every Sylow -subgroup has order and contains no nonidentity element of order , so it must be precisely this remaining set. Hence the Sylow -subgroup is unique and normal. In either case has a nontrivial proper normal subgroup, so it is not simple.

Problem 2.

Prove that the additive group is isomorphic to the multiplicative group

Proof.

Define

Then , so is a homomorphism. It is surjective because every element of the unit circle is for some , and

The first isomorphism theorem therefore gives

Problem 3.

Let be an integral domain. A nonzero nonunit is prime if implies or , and it is irreducible if implies that or is a unit.

(a) Show that every prime element is irreducible.

(b) Show that in a UFD every irreducible element is prime.

Proof.

(a) Suppose is prime and . Since , primality gives or . If , write . Then

Cancellation in the domain gives , so is a unit. Similarly, if , then is a unit. Hence is irreducible.

(b) Let be irreducible in a UFD and suppose . Then . Factor into irreducibles. Uniqueness of factorization implies that the irreducible is associate to a factor occurring in the factorization of or of . Hence or , so is prime.

Problem 4.

Let and be finite abelian groups whose orders are relatively prime. Show that

Proof.

Let and . For every pure tensor ,

and similarly . Choose integers with . Then

Pure tensors generate the tensor product, so .

Problem 5.

Let

be the dihedral group of order .

(a) Compute its center.

(b) Compute its commutator subgroup.

Proof.

(a) A rotation commutes with exactly when , or . No reflection commutes with . Thus

(b) The defining relation gives

Also is abelian. Hence

They have total size .

Problem 6.

Let be a commutative ring with identity and let be an -module. Show that if is finitely generated as an -module, then is finitely generated.

Proof.

Let

generate . For any , express

Comparing first coordinates gives

Thus generate , so is finitely generated.

Problem 7.

(a) Find whose splitting field has Galois group .

(b) Find whose splitting field has Galois group .

Proof.

(a) Take

Its splitting field is , a quadratic extension, so its Galois group is .

(b) Take

It is irreducible by Eisenstein's criterion. Its discriminant is , which is not a square in . Hence the Galois group of its splitting field is .

The splitting field of has Galois group and has a unique quadratic subfield, namely . Therefore

The two Galois extensions are linearly disjoint, so the splitting field of has Galois group

Problem 8.

Suppose is a perfect field and is nonconstant. Show that is a direct product of fields if and only if is separable.

Proof.

The perfectness assumption is necessary: without it, an irreducible inseparable polynomial gives a quotient that is a field although the polynomial is not separable.

Factor

where and the are distinct monic irreducibles. By the Chinese remainder theorem,

This is a product of fields exactly when every : if , the class of is a nonzero nilpotent in the corresponding factor; if , every factor is a field.

Over a perfect field every irreducible polynomial is separable, so is separable exactly when it has no repeated irreducible factor, that is, exactly when every . This proves the equivalence.

Problem 9.

Let be prime.

(a) Show that all matrices of order exactly have the same characteristic polynomial, and find it.

(b) Show that they all have the same minimal polynomial, and find it.

Proof.

Since and the characteristic is ,

Thus the minimal polynomial of divides

The only eigenvalue is , so the characteristic polynomial of the matrix is

The minimal polynomial has degree at most and is a positive power of . It cannot equal , because that would imply , whose order is , not . Therefore

Problem 10.

Let and let .

(a) Show that is the splitting field of .

(b) Find a generator of .

Proof.

Let , so in . The polynomial has no root in , so .

Set

Then

and hence

Therefore has order , so it generates the cyclic group .

The roots of satisfy . They are exactly the odd powers

They all lie in , and has no root in , so is its splitting field.