# If $\alpha$ is $R$-linear epimorphism, $\alpha$ is isomorphism.

I am having a difficult time understanding the proof of a corollary to Cayley-Hamilton theorem in Eisenbud's Commutative Algebra.

The statement is:

Let $R$ be a ring, and let $M$ be a finitely generated $R$-module. If $\alpha\colon M\to M$ is an epimorphism of $R$-modules, then $\alpha$ is an isomorphism.

For the proof, we give $M$ an $R[t]$-module structure, by letting $t\cdot m=\alpha(m)$ for each $m\in M$. Then set $I=(t)$, giving us $IM=M$. Thus we apply Cayley-Hamilton, yielding a monic polynomial $q$ such that $q(id_M)=0$ as an endomorphism. The next line I am having trouble seeing,

It follows that $(id_M-q(t)t)M=0$, or equivalently $id_M-q(\alpha)\alpha=0$.

It's clear $q(t)tM=q(t)M$, but why is $q(t)M=M$?

$\def\id{\text{id}}$ $q$ is not the polynomial that you get from the Cayley-Hamilton Theorem.
Since $tM = IM$, we can put $\phi= \id_{M}$ to get a polynomial $p \in C[t][x]$ such that
1. $p(x) = x^{n} + p_{1}x^{n-1} + \cdots + p_{n}$ with $p_{i} \in (t^{i}).$
2. $p(\id_{M}) = 0$ and hence $$\id_{M} + p_{1} + \cdots + p_{n} = 0.$$
Now, since $t^{i}$ divides $p_{i}$, we can write $$p_{1} + \cdots p_{n} = t(r(t)).$$ We set $q(t) = -r(t)$ and hence $$\id_{M} = tq(t).$$