Lemma 42.3.3. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\varphi : M \to M$ be an endomorphism and $n > 0$ such that $\varphi ^ n = 0$ and such that $\mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Im}}(\varphi ^{n - 1})$ has finite length as an $R$-module. Then

$e_ R(M, \varphi ^ i, \varphi ^{n - i}) = 0$

for $i = 0, \ldots , n$.

Proof. The cases $i = 0, n$ are trivial as $\varphi ^0 = \text{id}_ M$ by convention. Let us think of $M$ as an $R[t]$-module where multiplication by $t$ is given by $\varphi$. Let us write $K_ i = \mathop{\mathrm{Ker}}(t^ i : M \to M)$ and

$a_ i = \text{length}_ R(K_ i/t^{n - i}M),\quad b_ i = \text{length}_ R(K_ i/tK_{i + 1}),\quad c_ i = \text{length}_ R(K_1/t^ iK_{i + 1})$

Boundary values are $a_0 = a_ n = b_0 = c_0 = 0$. The $c_ i$ are integers for $i < n$ as $K_1/t^ iK_{i + 1}$ is a quotient of $K_1/t^{n - 1}M$ which is assumed to have finite length. We will use frequently that $K_ i \cap t^ jM = t^ jK_{i + j}$. For $0 < i < n - 1$ we have an exact sequence

$0 \to K_1/t^{n - i - 1}K_{n - i} \to K_{i + 1}/t^{n - i - 1}M \xrightarrow {t} K_ i/t^{n - i}M \to K_ i/tK_{i + 1} \to 0$

By induction on $i$ we conclude that $a_ i$ and $b_ i$ are integers for $i < n$ and that

$c_{n - i - 1} - a_{i + 1} + a_ i - b_ i = 0$

For $0 < i < n - 1$ there is a short exact sequence

$0 \to K_ i/tK_{i + 1} \to K_{i + 1}/tK_{i + 2} \xrightarrow {t^ i} K_1/t^{i + 1}K_{i + 2} \to K_1/t^ iK_{i + 1} \to 0$

which gives

$b_ i - b_{i + 1} + c_{i + 1} - c_ i = 0$

Since $b_0 = c_0$ we conclude that $b_ i = c_ i$ for $i < n$. Then we see that

$a_2 = a_1 + b_{n - 2} - b_1,\quad a_3 = a_2 + b_{n - 3} - b_2,\quad \ldots$

It is straighforward to see that this implies $a_ i = a_{n - i}$ as desired. $\square$

Comment #4227 by jojo on

In the first line of the proof I think $\pi$ should be $\varphi$.

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