MATH 250B: COMMUTATIVE ALGEBRA HOMEWORK #4

JOHN VOIGHT

Problem 1. Let E be a finite-dimensional vector space over a field k. Let x_{1}, …, x_{p} be elements of E such that x_{1} ∧ ⋯ ∧ x_{p} ≠ 0, and similarly y_{1} ∧ ⋯ ∧ y_{p} ≠ 0. If c ∈ k and

x_{1} ∧ ⋯ ∧ x_{p} = cy_{1} ∧ ⋯ ∧ y_{p}

show that x_{1}, …, x_{p} and y_{1}, …, y_{p} generate the same subspace.

Solution. This reduces to the fact: z_{1} ∧ ⋯ ∧ z_{r} = 0 if and only if the elements z_{i} are linearly dependent. After all, a dependence relation, say z_{r} = ∑_{i} a_{i}z_{i}, allows us to write

z_{1} ∧ ⋯ ∧ z_{r} = _{∑i}z_{1} ∧ ⋯ ∧ (a_{i}z_{i}) = 0

by the alternating property.

This implies that the elements x_{i} ∈ E are linearly independent, as are the y_{j} ∈ E; suppose that the element y_{j} is not in the span of the x_{i}, which is to say, the elements x_{1}, …, x_{p}, y_{j} are linearly independent. Then x_{1} ∧ ⋯ ∧ x_{p} ∧ y_{j} ≠ 0, but

cy_{1} ∧ ⋯ ∧ y_{p} ∧ y_{j} = 0

a contradiction. Therefore the subspace generated by the y_{j} is contained in the span of the x_{i}; since both are spaces of dimension p, they are equal.

Problem 2. Let E be a free module of dimension n over the commutative ring R. Let f : E → E be a linear map. Let α_{r}(f) = tr ⋀^{r}(f), where ⋀^{r}(f) is the endomorphism of ⋀^{r}(E) into itself induced by f. We have

α_{0}(f) = 1,  α_{1}(f) = tr(f),  α_{n}(f) = det f,

and α_{r}(f) = 0 if r > n. Show that

det(1 + f) = ∑_{r≥0} α_{r}(f).

[Hint: As usual, prove the statement when f is represented by a matrix with variable coefficients over the integers.] Interpret the α_{r}(f) in terms of the coefficients of the characteristic polynomial of f.

Solution. First assume that R = k is an algebraically closed field. Then we can choose a basis e_{1}, …, e_{n} for E such that f is given in Jordan canonical form (see Theorem 2.4 in §XIV, for example), with diagonal elements α_{1}, …, α_{n} (not necessarily distinct). Note the choice of basis does not affect the value of det(1 + f). Then

Date: February 20, 2003.

XIX: 1–4.

det(1 + f) = ∏_{i=1}^{n}(1 + α_{i}) = ∑_{I⊂{1,...,n}} α_{I}

where for I ⊂ S = {1, …, n} we denote α_{I} = ∏_{i∈I} α_{i}.

On the other hand, a basis for ⋀^{r} E is given by e_{i_{1}} ∧ ⋯ ∧ e_{i_{r}} for i_{1} < ⋯ < i_{r}; by definition

⋀^{r}(f)(e_{i_{1}} ∧ ⋯ ∧ e_{i_{r}}) = f(e_{i_{1}}) ∧ ⋯ ∧ f(e_{i_{r}})

so since f in this basis is lower triangular,

tr ⋀^{r}(f) = ∑_{i_{1}<···<i_{r}}α_{i_{1}} … α_{i_{r}}.

Matching expressions, we obtain the desired result. Since by definition the characteristic polynomial of f is P_{f}(t) = det(tI−f), we see that det(1+f) = (−1)^{n}P_{f}(−1). From the Jordan decomposition, we have

P_{f}(t) = ∏_{i}(t − α_{i}) = ∑_{r=0}^{n} a_{r}t^{r} = ∑_{r=0}^{n}(−1)^{r}t^{r} ∑_{\begin{matrix}I⊂S \\ #I=n−r \end{matrix}} α_{I};

matching these two, we find that α_{r}(f) = (−1)^{r}a_{n−r}.

Now if A is a linear map given by a matrix A = (x_{ij})_{i,j} with indeterminate coefficients over ℤ, then performing the above computation in an algebraic closure of ℚ(x_{ij})_{i,j}, we see then that the result holds for A. Since the conclusion is then an equality over ℤ[x_{ij}] (concerning the characteristic polynomial), by the homomorphism ℤ[x_{ij}] → R which takes x_{ij} ↦ f_{ij}, where f = (f_{ij}), we see that the conclusion holds for any commutative ring R.

Problem 3. Let E be a finite dimensional free module over the commutative ring R. Let E^{∨} be its dual module. For each integer r ≥ 1, show that ⋀^{r} E and ⋀^{r} E^{∨} are dual modules to each other, under the bilinear map such that

(v_{1} ∧ ⋯ ∧ v_{r}, v′_{1} ∧ ⋯ ∧ v′_{r}) ↦ det(⟨v_{i}, v′_{j}⟩)

where ⟨v_{i}, v′_{j}⟩ is the value of v′_{j} on v_{i}, as usual, for v_{i} ∈ E and v′_{j} ∈ E^{∨}.

Solution. Let f : ⋀^{r} E × ⋀^{r} E^{∨} → R be the above map; we need to show that f is nonsingular, i.e. that the R-linear map

ϕ : ^{⋀r}E → Hom_{R}(⋀^{r}E^{∨}, R)
v_{1} ∧ ⋯ ∧ v_{r} ↦ ((v′_{1} ∧ ⋯ ∧ v′_{r}) ↦ det(⟨v_{i}, v′_{j}⟩))

is an isomorphism. Let e_{1}, …, e_{n} be a basis for E, and let e′_{1}, …, e′_{n} be the duals, so that

⟨e_{i}, e′_{j}⟩ = \begin{cases}1, & i = j; \\ 0, & else. \end{cases}

A basis for ⋀^{r} E is given by wedge products e_{s} = e_{s_{1},...,s_{r}} = e_{s_{1}} ∧ ⋯ ∧ e_{s_{r}}. Then the wedges e′_{s_{1},...,s_{r}} = e′_{s_{1}} ∧ ⋯ ∧ e′_{s_{r}} form a basis for ⋀^{r} E^{∨}, and a basis for Hom_{R}(⋀^{r}E^{∨}, R) is given by the characteristic functions defined on basis vectors as

χ_{s}(e_{t}) = χ_{s_{1},...,s_{r}}(e_{t}) = \begin{cases}1, & e_{t} = e_{s}; \\ 0, & else. \end{cases}

Inside this notational morass, we now easily compute that in fact ϕ(e_{s}) = χ_{s}, since ϕ(e_{s})(e_{s}) = 1 = det(⟨e_{s_{i}}, e_{s_{j}}⟩)_{i,j}), the determinant of the identity matrix, and otherwise ϕ(e_{s})(e_{t}) = 0. This shows that ϕ is an isomorphism.

Problem 4. Notation being as in the preceding exercise, let F be another R-module which is free, finite dimensional. Let A : E → F be a linear map. Relative to the bilinear map of the preceding exercise, show that the transpose of ⋀^{r} A is ⋀^{r}(^{t}A), i.e. is equal to the rth alternating product of the transpose of A.

Solution. It suffices to verify (see Chapter XIII, §5) that for all e_{s} ∈ ⋀^{r} E and f_{s} ∈ ⋀^{r} F^{∨} that

⟨Ae_{s}, f_{t}⟩ = ⟨e_{s},^{t}Af_{t}⟩.

But after all,

⟨^{⋀r}(A)(e_{s}), f_{t}⟩ = det(⟨Ae_{s_{i}}, f_{t_{j}}⟩) = det(⟨e_{s_{i}},^{t}Af_{t_{j}}⟩) = ⟨e_{s},^{⋀r}(^{t}A)(f_{t})⟩;

the middle equality follows from the fact that ^{t}A : F → E is the transpose of A, and by the preceding exercise, ⟨, ⟩ is nonsingular.