MATH 250B: COMMUTATIVE ALGEBRA HOMEWORK #2

JOHN VOIGHT

Some solutions are omitted or sketched.

Problem XIII.25. Let a_{11}, …, a_{1n} be elements from a principal ideal ring, and assume that they generate the unit ideal. Show that there exists a matrix (a_{ij}) with this given first row, and whose determinant is equal to 1.

Solution. This problem is false for n = 1!

Let R be our principal ideal ring. Recall that we denote by gcd(x_{1}, …, x_{n}) any element x such that (x) = (x_{1}, …, x_{n}); such an element is well-defined up to a unit R^{∗}.

We prove the result by induction, namely, we show: given a_{11}, …, a_{1n}, there exists a matrix A = (a_{ij}) such that:

(i) A has the given first row; (ii) det A = gcd(a_{1j}); and (iii) If b_{j} is the minor obtained from the matrix by removing the first row and

jth column, then gcd(b_{1}, …, b_{n}) = 1. First, the base case, n = 2. Given b_{1}, b_{2} such that a_{11}b_{1} + a_{12}b_{2} = gcd(a_{11}, a_{12}),

we take the matrix

A = (\begin{matrix}a_{11} & a_{12} \\ −b_{2} & b_{1} \end{matrix}) .

Note that gcd(b_{1}, b_{2}) = 1 by the same equation, for gcd(b_{1}, b_{2}) gcd(a_{11}, a_{12}) divides both sides.

Now for the general case: Let gcd(a_{11}, …, a_{1(n−1)}) = g, and write gr + a_{1n}s = gcd(g, a_{1n}), (as before gcd(r, s) = 1). By inductive assumption, there exists an (n − 1) × (n − 1) matrix A = (a_{ij}) with first row a_{11}, …, a_{1(n−1)} with det A = g and gcd(b_{j}) = 1. Let b_{1}c_{1} + ⋯ + b_{n−1}c_{n−1} = 1; consider the matrix

(\begin{matrix}a_{11} & … & a_{1(n−1)} & a_{1n} \\ sc_{1} & … & sc_{n−1} & −r \\ a_{21} & … & a_{2(n−1)} & 0 \\ ⋮ & ... & ⋮ & ⋮ \\ a_{(n−1)1} & … & a_{(n−1)(n−1)} & 0 \end{matrix}) .

Note A is obtained from this matrix by removing the second row and last column. By expanding about the last column, we see that the determinant of this matrix is

a_{1n}(sc_{1}b_{1} + ⋯ + sc_{n−1}b_{n−1}) + r det A = sa_{1n} + rg = gcd(a_{11}, …, a_{1n}).

Date: February 6, 2003.

XIII: 25–28, XIV: 3, 6, 8, 9, 13, 14, 15, 18, 20, 23.

This matrix has the given first row and the correct determinant. To show (iii), we note that the minor obtained by removing the first row and last column is equal to s, whereas the minor obtained by removing the first row and jth column is equal to −rb_{j}, and gcd(−rb_{j}, s) = gcd(r, s) = 1, since gcd(b_{j}) = 1 by inductive hypothesis. This completes the proof.

Problem XIII.26. Let A be a commutative ring, and J = (x_{1}, …, x_{r}) an ideal. Let c_{ij} ∈ A and let

y_{i} = ∑_{j=1}^{r} c_{ij}x_{j}.

Let J′ = (y_{1}, …, y_{r}), and let D = det(c_{ij}). Show that DJ ⊂ J′.

Solution. Let C̃ be the adjoint matrix so that CC̃ = DI. Note that

C̃ (y_{1} ; ⋮ ; y_{n}) = C̃C (x_{1} ; ⋮ ; x_{n}) = D (x_{1} ; ⋮ ; x_{n}) .

Thus each term Dx_{i} = ∑ c̃_{ij}y_{j}, hence DJ ⊂ J′.

Problem XIII.27. Let L be a free module over ℤ with basis e_{1}, …, e_{n}. Let M be a free submodule of the same rank, with basis u_{1}, …, u_{n}. Let u_{i} = ∑ c_{ij}e_{j}. Show that the index (L : M) is given by the determinant:

(L : M) = | det(c_{ij})|.

Solution. By Theorem III.7.8, there exists a basis f_{1}, …, f_{n} of L and integers a_{1}, …, a_{n} such that a_{1}f_{1}, …, a_{n}f_{n} is a basis of M. Write f_{i} = ∑ p_{ij}e_{j} and a_{i}f_{i} = ∑ q_{ij}u_{j}; then the matrix P = (p_{ij}) is a change of basis for L, hence invertible hence det P ∈ ℤ^{∗} = {±1}; similarly, Q = (q_{ij}) has | det Q| = 1. Then

(L : M) = |a_{1} … a_{n}| = | det(QCP^{−1})| = | det C|.

Problem XIV.3. Let k be a commutative ring, and let M, M′ be square n × n matrices in k. Show that the characteristic polynomials of MM′ and M′M are equal.

Solution. We prove this fact for “generic” matrices A, A′, from which the result follows. Let F be the prime subfield of k, and consider the polynomial ring R = F [x_{ij}, x′_{ij}] for 1 ≤ 1, j ≤ n. Let A = (x_{ij}), A′ = (x′_{ij}) be matrices of indeterminates, and let K be the field of fractions of R, namely, K = F (x_{ij}, x′_{ij}). The matrices M, M′ are invertible over K, since det A, det A′ ≠ 0 (they are homogeneous polynomials of degree n). Therefore over K, we have

det(tI − AA′) = det (A^{−1}(tI − AA′)A) = det(tI − A′A)

so the characteristic polynomials p_{AA′}(t) = p_{A′A}(t). But already p_{AA′}(t), p_{A′A}(t) ∈ R[t] (clear from the definition as a determinant); we have a map

R[t] → k[t]
x_{ij} ↦ m_{ij}
x′_{ij} ↦ m′_{ij}

which sends p_{AA′}(t) ↦ p_{MM′}(t) and p_{A′A}(t) ↦ p_{M′M}(t), hence p_{MM′}(t) = p_{M′M}(t).

Problem XIV.6. Let A be a nilpotent endomorphism of a finite-dimensional vector space E over the field k. Show that tr(A) = 0.

Solution. The minimal polynomial of A is of the form t^{r} since X^{m} = 0 for some m. Therefore the characteristic polynomial p_{A}(t) = t^{n} where n = dim E. Then tr A is the coefficient of t^{n−1} in p_{A}(t), namely, tr A = 0.

Problem XIV.8. Let E be a finite-dimensional vector space over a field k. Let A ∈ Aut_{k}(E). Show that the following conditions are equivalent:

(a) A = I + N, with N nilpotent;

(b) There exists a basis of E such that the matrix of A with respect to this basis has all its diagonal elements equal to 1 and all elements above the diagonal equal to zero.

(c) All roots of the characteristic polynomial of A (in the algebraic closure of k) are equal to 1.

Solution. If A = I + N, then by Exercise 9, there exists a basis of E such that N is strictly upper triangular; let M give such a change of basis, i.e. let M^{−1}NM be strictly upper triangular. If we let P = P^{−1} be the matrix with all anti-diagonal elements equal to 1 and all other elements zero, then P^{−1}(M^{−1}NM)P is strictly lower triangular (P interchanges ith and (n − i)th rows and columns). Then

P^{−1}M^{−1}A(MP) = I + (MP)^{−1}N(MP)

so A in this basis has diagonal elements (given by the identity matrix) equal to one, and all elements above the diagonal equal to zero. This shows (a) ⟹ (b).

The statement (b) ⟹ (c) is clear, since if we write A in this basis, A − tI is lower triangular so det(A − tI) = (t − 1)^{n}, the product of its diagonal elements.

Finally, to see that (c) ⟹ (a), we note that p_{A}(t) = (t − 1)^{n}, so p_{A}(A) = (A − I)^{n} = 0. If we let N = A − I, so that A = I − N, then N^{n} = 0, i.e. N is nilpotent.

Problem XIV.9. Let k be a field of characteristic 0 and let M be an n × n matrix in k. Show that M is nilpotent if and only if tr(M^{ν}) = 0 for 1 ≤ ν ≤ n.

Solution. If M is nilpotent, then M^{ν} is also nilpotent, hence by Exercise 6 tr(M^{ν}) = 0.

Conversely, suppose that tr(M^{ν}) = 0. If λ_{1}, …, λ_{n} ∈ \overline{k} are the (not necessarily distinct) eigenvalues of M, then λ_{1}^{ν}, …, λ_{n}^{ν} are the eigenvalues of M^{ν} (with the same multiplicities), as if v_{i} has Mv_{i} = λv_{i}, then

M^{ν}v_{i} = M^{ν−1}(Mv_{i}) = M^{ν−1}(λ_{i}v_{i}) = ⋯ = λ_{i}^{ν}v_{i}.

It suffices to show that λ_{1} = ⋯ = λ_{n} = 0, for then the characteristic polynomial of M is p_{M}(t) = ∏(t − λ_{i})^{n} = t^{n} = 0, i.e. M is nilpotent.

Therefore tr(M^{ν}) = ∑ λ_{i}^{ν}. It suffices to show that ∑ λ_{i}^{ν} = 0 for 1 ≤ ν ≤ n implies that λ_{1} = ⋯ = λ_{n} = 0. Without loss of generality, we may assume that all λ_{i} ≠ 0. Consider the Vandermonde matrix

V (λ_{1}, …, λ_{n}) = (\begin{matrix}λ_{1} & … & λ_{n} \\ ⋮ & ... & ⋮ \\ λ_{1}^{n} & … & λ_{n}^{n} \end{matrix})

Since

Λ (1 ⋮ 1) = (∑ λ_{i} ; ⋮ ; ∑ λ_{i}^{n}) = (0 ⋮ 0)

we see that det Λ = 0. But this is a Vandermonde matrix (see after Proposition XIII.4.10), hence

det Λ = λ_{1} … λ_{n} ∏_{i<j}(x_{j} − x_{i}).

Therefore, say, λ_{1} = λ_{2}, so the column vector (1, −1, 0, …, 0)^{t} is in the nullspace of Λ and is linearly independent of (1, …, 1)^{t}, so the nullspace of Λ is at least of dimension 2. This implies that all (n − 1) × (n − 1) minors of Λ must vanish, in particular, the minor obtained by deleting the first column and row,

λ_{2} … λ_{n} det V (λ_{2}, …, λ_{n}) = 0.

By induction, this implies that λ_{2} = ⋯ = λ_{n} = λ = λ_{1}. Since k is characteristic zero, the equation nλ = 0 implies λ = 0 as desired.

Problem XIV.13. Let E be a finite-dimensional vector space over a field k, and let S ∈ End_{k}(E). We say that S is diagonalizable if there exists a basis of E consisting of eigenvectors of S. The matrix of S with respect to this basis is then a diagonal matrix.

(a) If S is diagonalizable, then its minimal polynomial over k is of type

q(t) = ∏_{i=1}^{m}(t − λ_{i}),

where λ_{1}, …, λ_{m} are distinct elements of k.

(b) Conversely, if the minimal polynomial of S is of the preceding type, then S is diagonalizable. [Hint: The space can be decomposed as a direct sum of the subspaces E_{λ_{i}} annihilated by S − λ_{i}.]

(c) If S is diagonalizable, and if F is a subspace of E such that SF ⊂ F, show that S is diagonalizble as an endomorphism of F, i.e. that F has a basis consisting of eigenvectors of S.

(d) Let S, T be endomorphisms of E, and assume that S, T commute. Assume that both S, T are diagonalizable. Show that they are simultaneously diagonalizable, i.e. there exists a basis of E consisting of eigenvectors for both S and T. [Hint: If λ is an eigenvalue of S, and E_{λ} is the subspace of E consisting of all vectors v such that Sv = λv, then T E_{λ} ⊂ E_{λ}.] Solution (sketch). For (a), let λ_{1}, …, λ_{m} be the distinct eigenvalues of S, and let v_{ij} be a basis of eigenvectors of S with eigenvalues λ_{i}. Since already S −λ_{i}I annihilates v_{ij}, we see that ∏_{i}(S−λ_{i}I) annihilates every v_{ij}, hence all of E, so ∏_{i}(S−λ_{i}I) = 0, and therefore the minimal polynomial of S divides ∏_{i}(t − λ_{i}), hence is equal to it.

For (b), as in the hint, choose a basis v_{ij} for the subspace E_{λ_{i}}; the matrix of S in this basis will be diagonal, since it is diagonal (constant!) on each subspace.

For (c), choose a basis for F and extend it to one of E. The matrix for S, since SF ⊂ F is block upper triangular; computing the minimal polynomial for S restricted to F from this we see that it must be squarefree as it divides the minimal polynomial of S.

For (d), let λ be an eigenvalue of S (such and eigenvalue exists by assumption that S is diagonalizable). In the subspace E_{λ} = {v ∈ S : Sv = λv}, we see that

S(T v) = T (Sv) = T (λv) = λ(T v)

so T E_{λ} ⊂ E_{λ}. The result now follows from part (c).

Problem XIV.15. Let E, F be finite-dimensional vector spaces over an algebraically closed field k, and let A : E → E and B : F → F be k-endomorphisms of E, F, respectively. Let

P_{A}(t) = ∏(t − α_{i})^{n_{i}}, P_{B}(t) = ∏(t − β_{j})^{m_{j}}

be the factorizations of their respective characteristic polynomials into distinct linear factors. Then

P_{A⊗B}(t) = ∏_{i,j}(t − α_{i}β_{j})^{n_{i}m_{j}}.

[Hint: Decompose E into the direct sum of subspaces E_{i}, where E_{i} is the subspace of E annihilated by some power of A−α_{i}. Do the same for F, getting a decomposition into a direct sum of subspaces F_{j}. Then show that some power of A ⊗ B − α_{i}β_{j} annihilates E_{i} ⊗ F_{j}. Use the fact that E ⊗ F is the direct sum of the subspaces E_{i} ⊗ F_{j}, and that dim_{k}(E_{i} ⊗ F_{j}) = n_{i}m_{j}.]

Solution (sketch). By the hint, it is enough to show that the characteristic polynomial of A ⊗ B on E_{i} ⊗ F_{j} is (t − α_{i}β_{j})^{n_{i}m_{j}}. By direct computation, the only eigenvalue of A ⊗ B on E_{i} ⊗ F_{j} is α_{i}β_{j}. Since we are over an algebraically closed field, this implies the claim.

Problem XIV.20.

(a) How many non-conjugate elements of GL_{2}(ℂ) are there with characteristic polynomial t^{3}(t + 1)^{2}(t − 1)?

(b) How many with characteristic polynomial t^{3} − 1001t?

Solution. Zero and zero, because the degree of the characteristic polynomial is equal to 2. What is up with this problem?

Problem XIV.23. Let E be a finite-dimensional vector space over an algebraically closed field k. Let A, B be k-endomorphisms of E which commute, i.e. AB = BA. Show that A and B have a common eigenvector. [Hint: Consider a subspace consisting of all vectors having a fixed element of k as eigenvalue.] Solution. Let λ be an eigenvector of A (λ exists because k is algebraically closed), and let E_{λ} = {v ∈ E : Av = λv}. Then for v ∈ E_{λ},

A(Bv) = B(Av) = λBv

so BE_{λ} ⊂ E_{λ}. This implies that B is well-defined as an endomorphism restricted to E_{λ}, and therefore (over the algebraically closed field k) B has an eigenvector w ∈ V, which by construction is also an eigenvector of A.