2008 Fall Qualifying Exam in Algebra (AI-generated)

Problem 1.

Give examples, with justification, of:

(a) two nonisomorphic rings with isomorphic additive groups;

(b) a prime ideal that is not maximal in an integral domain;

Proof.

(a) Take and . Their additive groups are both , but the first ring is a field and the second has zero divisors.

(b) In , the ideal is prime because is a domain, but it is not maximal because is not a field.

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

Problem 2.

Count all subgroups, including the trivial group and the whole group, of:

(a) a cyclic group of order ;

(b) the additive group ;

Proof.

(a) A cyclic group has one subgroup for each divisor of its order. Since

the number of divisors is

(b) Subgroups are exactly vector subspaces. There is one subspace each of dimensions and . The number of lines is

and by duality the number of planes is also . Thus the total is

and three of order ,

Hence there are subgroups.

Problem 3.

Prove that there are no simple groups of order .

Proof.

The number of Sylow -subgroups is or , and the number of Sylow -subgroups is or . If either is , that Sylow subgroup is normal.

Otherwise, the Sylow -subgroups contribute nonidentity elements and the Sylow -subgroups contribute different nonidentity elements. These disjoint sets cannot fit among the nonidentity elements of . Hence some Sylow subgroup is normal, and is not simple.

Problem 4.

Determine the splitting field and Galois group of over . Give the subgroup and subfield lattices, with fixed fields identified.

Proof.

Let . The roots are , so the splitting field is

Eisenstein's criterion at gives , and is not in this real field. Hence .

Define

Then and , so

The full fixed-field correspondence is

The three order- subgroups are , , and . The first contains ; the second contains ; and the third contains . The subfield lattice reverses these inclusions.

Problem 5.

Let be finite groups of relatively prime orders.

(a) Prove that every subgroup is for subgroups .

(b) Give a counterexample without the coprime-order assumption.

Proof.

(a) Let be the image of under projection to . If , the orders of and are relatively prime. By the Chinese remainder theorem, there is an integer such that

Then

Similarly . Thus contains , while the reverse inclusion follows from the definition of the projections. Hence .

(b) In , the diagonal subgroup

is not a product of a subgroup of the first factor and a subgroup of the second.

Problem 6.

Suppose is finite and a nontrivial subgroup is contained in every nontrivial subgroup of .

(a) Prove that is a power of a prime and that has exactly elements of order .

(b) Give such an example with nonabelian of order .

Proof.

(a) If two different primes divided , Cauchy's theorem would give subgroups of orders and . Both would contain , forcing to divide both and , impossible because is nontrivial. Hence is a power of one prime .

Every subgroup of order contains . Since such a subgroup has no nontrivial proper subgroup, it follows that equals every subgroup of order . Thus there is exactly one such subgroup, and its nonidentity elements are exactly the elements of order .

(b) Take and

Every nontrivial subgroup of contains , so it contains .

Problem 7.

Find all prime ideals of

Proof.

The class is nilpotent, and every prime ideal contains every nilpotent element. Thus every prime ideal of contains . Prime ideals containing it correspond to prime ideals of

Therefore the complete list is

where ranges over the rational primes.

Problem 8.

Let be finite Galois over with cyclic Galois group. Let and suppose . Prove that

(a) is even;

(b) is odd.

Proof.

Let and . Complex conjugation restricts to an element of order . Since is not real, does not fix pointwise, so .

(a) The nontrivial coset has order in the cyclic quotient . Hence

is even.

(b) A cyclic group has a unique element of order . If were even, then would contain that unique involution , contradicting . Therefore is odd.

Problem 9.

Let be an odd prime. Show that there are exactly five groups of order up to isomorphism.

Proof.

The Sylow -subgroup is unique because its number divides and is congruent to modulo . By Schur--Zassenhaus,

There are two possibilities for .

If , an involutory automorphism is either the identity or inversion. This gives

If , the action is an involution in . Since is odd, every such operator is diagonalizable with eigenvalues in . Up to conjugacy there are three possibilities:

They give respectively

and

These five groups are pairwise nonisomorphic: the two abelian groups have different exponents, and the three nonabelian groups have Sylow -subgroups or involution actions of different types. Hence the list is complete.

Problem 10.

Let have characteristic polynomial

(a) Find all possible minimal polynomials.

(b) Find the trace and determinant.

Proof.

The polynomials

are irreducible and relatively prime over . The -primary part occurs once. The multiplicity-two -primary part has either one block for or two blocks for .

(a) Thus the possible minimal polynomials are

(b) The coefficient of in is zero, so

Since the dimension is even, the determinant equals the constant term of the characteristic polynomial:

and

Their minimal polynomials are and , so they are not conjugate.