2014 Spring Qualifying Exam in Algebra (AI-generated)

Proof.

(a) No. The exponent of the first group is

whereas the exponent of the second is

The exponent is an isomorphism invariant, so the groups are not isomorphic.

(b) Factor

Because is relatively prime to , Fermat's little theorem applies modulo each of these primes. Moreover,

Thus

The three primes are pairwise relatively prime, so the Chinese remainder theorem gives

Proof.

Let

Then , and therefore

Thus is a root of

This polynomial is Eisenstein at , so it is irreducible and

Put . Since

we have . Hence all four roots

of lie in , so the extension is Galois.

There is an automorphism with . It sends

to , and consequently

Thus

so has order . The Galois group has order and contains an element of order , hence it is cyclic.

Proof.

A unit cannot belong to any proper ideal, so .

Conversely, suppose is not a unit. Then the principal ideal is proper and is contained in some maximal ideal. Since has only one maximal ideal, , so . Thus every element outside is a unit, proving

Proof.

The units of are precisely the nonzero constants. Both and are irreducible in this ring. Indeed, every nonconstant element has lowest nonzero exponent at least . A factorization of into two nonunits would therefore have lowest exponent at least , and a factorization of into two nonunits would have lowest exponent at least , both impossible.

Now

These are factorizations into irreducibles. The elements and are not associates because their quotient is not a unit and does not even belong to the ring. Hence factorization is not unique, so is not a UFD.

Proof.

Since , the number of Sylow -subgroups satisfies

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

If , the distinct Sylow -subgroups contribute

nonidentity elements. Exactly elements remain, including the identity. Every Sylow -subgroup has order and contains no nonidentity element of order , so it must equal this remaining set. It is therefore unique and normal. In either case the group is not simple.

Proof.

The statement is false. Let

The subgroup is normal in . Let

Since is abelian, . But conjugation by a -cycle permutes the three order- subgroups of , so is not normal in . Normality is therefore not transitive.

Proof.

Ideals of correspond to ideals of containing , via

If is maximal, the only ideals containing it are and , so the quotient has only the ideals and . A nonzero commutative ring with identity having no nontrivial ideals is a field: for , the ideal is nonzero and hence is the whole ring, so is invertible.

Conversely, if is a field, it has no proper nonzero ideals. Therefore there is no ideal strictly between and , so is maximal.

Proof.

Similar matrices plainly have the same characteristic and minimal polynomials.

Conversely, similarity classes over a field are classified by invariant factors

Their product is the characteristic polynomial, and the largest one is the minimal polynomial. In dimension , these two polynomials determine the whole list:

• If the minimal polynomial has degree , it is the sole invariant factor and equals the characteristic polynomial.
• If it has degree , there are exactly two invariant factors. The larger is the minimal polynomial , and the other is the uniquely determined linear polynomial , where is the characteristic polynomial.
• If it has degree , the matrix is scalar, and its similarity class is uniquely determined.

Thus equal characteristic and minimal polynomials give equal invariant factors and hence similar matrices.

Proof.

(a) A group is a set with an associative binary operation, an identity element, and an inverse for every element.

(b) A ring is an abelian group under addition equipped with an associative multiplication that distributes over addition. Under the convention used here, it also has a multiplicative identity.

(c) An integral domain is a nonzero commutative ring with identity and no zero divisors.

(d) An -module is an abelian group with a scalar multiplication satisfying the usual distributive, associative, and identity axioms.

(e) An -module homomorphism is an additive map satisfying

for all and .

Proof.

Let be a finite field of characteristic . The Frobenius map

is injective because implies , hence . An injective map from a finite set to itself is surjective. Therefore every element of is a th power.

For a field of characteristic , surjectivity of Frobenius is equivalent to perfection: an irreducible polynomial with zero derivative would be a polynomial in , and taking th roots of its coefficients would make it a th power, contradicting irreducibility. Thus every irreducible polynomial over is separable, so every finite field is perfect.