Exercise 49, § 1

Let $A$ be a ring, $E$ a finitely generated projective $A$-module, and $u \in \mathrm{End}_A(E)$. If $E$ is a free $A$-module, one denotes by $\det_E(u)$ the determinant of $u$. In general, one can choose an $A$-module $F$ such that $E \oplus F$ is a free $A$-module of finite type, and one puts

a) Show that $\det_E(u)$ is independent of the choice of $F$, that $\det(1_E) = 1$, and that $\det_E(u \circ v) = \det_E(u) \cdot \det_E(v)$ for any $u, v$ in $\mathrm{End}_A(E)$. b) Suppose that $L$ is a direct factor submodule of $E$ stable under $u$, and denote by $u_L \in \mathrm{End}_A(L)$ and $u_{E/L} \in \mathrm{End}_A(E/L)$ the endomorphisms defined by $u$. Then one has $\det_E(u) = \det_L(u_L) \cdot \det_{E/L}(u_{E/L})$. Let $T$ be an indeterminate, $E[T] = A[T] \otimes_A E$, and identify $u$ with $1_{A[T]} \otimes_A u \in \mathrm{End}_{A[T]}(E[T])$. Put

(characteristic polynomial of $u$) and

where 1 denotes $1_{E[T]}$ (cf. A, III, p. 107).

c) One has $\chi_E(0) = 1$. Suppose that E is locally free of constant rank r. One has $\overline{\chi}_E(1_E) = (1 + T)^r$. If $\sigma_r : A[T, T^{-1}] \to A[T, T^{-1}]$ denotes the mapping $(\sigma_r f)(T) = T^r \cdot f((-T)^{-1})$ one has $\sigma_r(\sigma_r(f)) = (-1)^r f$ and $\sigma_r(\overline{\chi}_E(u)) = \chi_E(u)$.

d) Let $\alpha : A \to A'$ be a ring homomorphism. One has $\overline{\chi}_{A' \otimes_A E}(\mathrm{Id}_{A'} \otimes_A u) = \alpha(\overline{\chi}_E(u))$, where $\alpha$ also denotes the homomorphism $A[T] \to A'[T]$ defined by $\alpha$.

e) One has $\overline{\chi}_E(u) \in \Lambda(A)$. For every $n \in J$, one has

(It is enough to verify the formulas locally on the spectrum of A, hence one may suppose E equal to $A'$. Let $(u_{ij})_{1 \leq i,j \leq r}$ be the matrix of $u$, let $(X_{ij})_{1 \leq i,j \leq r}$ be a family of indeterminates, put $B = \mathbf{Z}[(X_{ij})]$, and let $v$ be the endomorphism of $B'$ having matrix $X = (X_{ij})_{1 \leq i,j \leq r}$. It is enough to verify the formulas for $B'$ and $v \in \mathrm{End}_B(B')$. One can embed B in an algebraically closed field C, and it is enough to verify the formulas for $C'$ and $w = \mathrm{Id}_C \otimes_B v$. Let $a_1, ..., a_r$ be the eigenvalues of $w$, so that $\overline{\chi}_{C'}(w^n) = \prod_i (1 + a_i T)$. We have $\psi_n^C(\overline{\chi}_{C'}(w)) = \prod_{i=1}^r (1 + a_i^n T) = \overline{\chi}_{C'}(w^n)$. Moreover $\chi_{\Lambda^n(C')}(\Lambda^n(w)) = \prod_H (1 + a_H T)$, where H runs through the set of subsets with n elements of $\{1 ..., r\}$, and where $a_H = \prod_{h \in H} a_h$. We can now apply Exercise 46, d), p. 59, to show that $\lambda_n^C(\overline{\chi}_{C'}(w)) = \overline{\chi}_{\Lambda^n(C')}(\Lambda^n(w))$.)

f) We have

in other words $L_n(\overline{\chi}_E(u)) = \operatorname{Tr}(u^n)$ for $n \in J$. (Reason as in e).)

g) Let $E'$ be a finitely generated projective A-module, and $u' \in \mathrm{End}_A(E')$. We have, in the ring $\Lambda(A)$,

(Reason as in e).)

h) Let $k$ be an algebraically closed field of non-zero characteristic $p$, E a finite-dimensional vector space over $k$ of dimension finie $n$, $u$ an endomorphism of E. We shall denote by $\tilde{\chi}_E(u)$ the image of $\overline{\chi}_E(u)$ in $W(k)$ by the canonical homomorphisms (Exercise 43, p. 57 and Exercise 39, p. 53):

Prove that if $\alpha_1, ..., \alpha_n$ are the eigenvalues of the endomorphism $u$, one has