2014 Fall Qualifying Exam in Algebra (AI-generated)

Proof.

(a) The symmetric group is the group of all bijections from to itself, with composition as its operation.

(b) A ring is an abelian group under addition together with an associative multiplication that distributes over addition. Under the convention used here, rings have a multiplicative identity.

(c) A ring homomorphism is a map preserving addition, multiplication, and identity:

(d) A field is a nonzero commutative ring in which every nonzero element has a multiplicative inverse.

(e) A principal ideal domain is an integral domain in which every ideal is generated by one element.

Proof.

If , the additive group is torsion-free, whereas has order . Hence the groups are not isomorphic.

Suppose . Every nonzero element of has order . But has no element of order : if , then in characteristic ,

so . Thus the two groups have different -torsion. This also covers ; when , the additive group has two elements while has one. Therefore the groups are never isomorphic.

Proof.

Write with . Suppose

Then for some . A generator of a nonzero prime ideal is a prime element and hence irreducible. Therefore either is a unit, in which case , or is a unit, in which case . There is no ideal strictly between and , so is maximal.

Proof.

Since , the number of Sylow -subgroups is or . If it is , that subgroup is normal.

Assume . The twelve Sylow -subgroups have pairwise trivial intersections and contribute

nonidentity elements. The number of Sylow -subgroups satisfies

so is , , or . The value is impossible because it would contribute additional nonidentity elements, while only remain.

If , the Sylow -subgroup is normal. If , conjugation on the four Sylow -subgroups gives a nontrivial homomorphism . Were simple, this map would be injective, but does not divide . Hence its kernel is a nontrivial proper normal subgroup. Thus is not simple in every case.

Proof.

We have

The roots are the primitive third and sixth roots of unity. They all lie in

Conversely, a root of either quadratic generates . Thus the splitting field is

and its degree over is .

Proof.

Let

and

Both matrices have characteristic polynomial . Their largest Jordan blocks have size , so both have minimal polynomial . But their Jordan partitions are and , respectively. Hence their Jordan canonical forms differ and they are not similar.

Proof.

(a) If in , then because is a domain,

The product has degree , so both polynomials are constant. Their constant values multiply to , so each lies in . Conversely, every unit of is plainly a unit in .

(b) Take and

Both are nonconstant, and

in .

Proof.

Let and let be a primitive th root of unity. The splitting field is

The polynomial is Eisenstein at , so . Also . The intersection of these fields has degree dividing both and , hence is . Therefore

Every automorphism is determined by

where and . All choices occur. Composition satisfies

which is exactly the multiplication law

Thus the stated map is a group isomorphism.

Proof.

The factor is irreducible over and occurs with multiplicity one, so it contributes the unique real companion block

For the zero eigenvalue, the Jordan-block sizes form a partition of . There are exactly three partitions: , , and . Hence the three similarity classes are represented by

and

These are pairwise nonsimilar because their minimal polynomials are respectively

Proof.

The multiplicative group is cyclic of order

If is a generator, then lies in the kernel exactly when

There are precisely residue classes modulo satisfying this condition. Therefore

By the first isomorphism theorem,