2015 Spring Qualifying Exam in Algebra (AI-generated)

Problem 1.

Prove that every finite group of order greater than has a nontrivial automorphism.

Proof.

If is nonabelian, then some inner automorphism is nontrivial.

Suppose is abelian. The inversion map is an automorphism. If it is nontrivial, we are done. If it is the identity, then every element satisfies , so is a vector space over . Since , its dimension is at least , and a linear automorphism interchanging two basis vectors is nontrivial. Thus in all cases has a nonidentity automorphism.

Problem 2.

(a) Define a UFD.

(b) Define a PID.

Proof.

(a) A unique factorization domain is an integral domain in which every nonzero nonunit is a product of irreducibles and this factorization is unique up to reordering and multiplication of factors by units.

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

is both a PID and a UFD.
, for any field , is a UFD but not a PID; the ideal is not principal. There is no example that is a PID but not a UFD because every PID is a UFD.
is neither. The equality

gives inequivalent factorizations into irreducibles, so it is not a UFD; since every PID is a UFD, it is not a PID either.

Problem 3.

Let be an indeterminate.

(a) Prove that is not Galois.

(b) Find its Galois closure and determine the Galois group abstractly and by explicit automorphisms.

Proof.

Put and . The polynomial is irreducible in , for example by Eisenstein's criterion at the prime in . Thus .

(a) The roots are

The field embeds in a real function field and does not contain , so it does not contain all roots. Hence it is not normal and therefore not Galois.

(b) The Galois closure is

Since , . Define

and

Then has order , has order , and

The eight automorphisms are and for . Therefore

Problem 4.

Let be a commutative ring with identity. An element is nilpotent if some positive power is zero.

(a) Prove that every nilpotent element lies in every prime ideal.

(b) If every element of is either nilpotent or a unit, prove that has a unique prime ideal.

Proof.

(a) Let and let be prime. Since , repeated use of primality gives .

(b) Let be the set of nilpotent elements. In a commutative ring, is an ideal: if , then , and every multiple of a nilpotent is nilpotent. By part (a), lies in every prime ideal.

If , the hypothesis says that is a unit. Hence every nonzero class in is a unit, so is a field. Thus is maximal, and in particular prime. Since is contained in every prime ideal and is itself maximal, it is the unique prime ideal.

Problem 5.

Write

so has order .

(a) Prove that every subgroup of is normal in .

(b) If with odd, prove that .

Proof.

(a) Every subgroup of the cyclic group has the form . Conjugation by fixes it, and conjugation by sends to , which generates the same subgroup. Since generate , the subgroup is normal.

(b) The element has order and is central. The subgroup

is isomorphic to because has order . Since is odd, , and hence

Their orders multiply to , and the first factor is central. Therefore

has order . Isomorphic groups have isomorphic centers, so the groups are not isomorphic.

Problem 6.

Suppose are primes. Prove that no group of order is simple.

Proof.

Let have order . Sylow's theorems give

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

If , then divides . Since , this forces , hence . Two consecutive integers greater than cannot both be prime, so the only possibility is .

It remains to consider groups of order . The number of Sylow -subgroups is or . If it is , that subgroup is normal. If it is , the four Sylow -subgroups contribute nonidentity elements, leaving exactly four elements including the identity. Every Sylow -subgroup has order and must be this remaining set, so it is unique and normal. Thus no group of order is simple.

Problem 7.

Determine the maximal ideals of:

(a) ;

(b) .

Proof.

(a) Since

with relatively prime factors, the Chinese remainder theorem gives

Its maximal ideals are the images of and , namely

(b) The discriminant is

which is not a square in . Thus is irreducible, the quotient is a field, and its only maximal ideal is the zero ideal.

Problem 8.

Find two matrices with the same characteristic and minimal polynomials but different Jordan canonical forms.

Proof.

Take nilpotent matrices

and

Both have characteristic polynomial . In each case the largest Jordan block has size , so both have minimal polynomial . Their Jordan block partitions are and , which are different; hence the matrices are not similar.

Problem 9.

(a) Define a perfect field.

(b) Give an example of a perfect field.

Proof.

(a) A field is perfect if every algebraic extension of is separable. Equivalently, every irreducible polynomial over is separable. In characteristic , this is equivalent to surjectivity of the Frobenius map .

(b) Every field of characteristic is perfect; for example, is perfect. Every finite field is also perfect.

Problem 10.

(a) Classify the conjugacy classes of .

(b) Construct its character table.

Proof.

(a) In a symmetric group, conjugacy classes are determined by cycle type. Thus the three classes are

of sizes .

(b) The irreducible representations are the trivial representation, the sign representation, and the two-dimensional standard representation. Their character table is

The squared degrees sum to , so the table is complete.