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 straightforward to see that this implies a_ i = a_{n - i} as desired.
\square
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