Lemma 10.16.4. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $\varphi : M \to M$ be a surjective $R$-module map. Then $\varphi $ is an isomorphism.

**First proof.**
Write $R' = R[x]$ and think of $M$ as a finite $R'$-module with $x$ acting via $\varphi $. Set $I = (x) \subset R'$. By our assumption that $\varphi $ is surjective we have $IM = M$. Hence we may apply Lemma 10.16.3 to $M$ as an $R'$-module, the ideal $I$ and the endomorphism $\text{id}_ M$. We conclude that $(1 + a_1 + \ldots + a_ n)\text{id}_ M = 0$ with $a_ j \in I$. Write $a_ j = b_ j(x)x$ for some $b_ j(x) \in R[x]$. Translating back into $\varphi $ we see that $\text{id}_ M = -(\sum _{j = 1, \ldots , n} b_ j(\varphi )) \varphi $, and hence $\varphi $ is invertible.
$\square$

**Second proof.**
We perform induction on the number of generators of $M$ over $R$. If $M$ is generated by one element, then $M \cong R/I$ for some ideal $I \subset R$. In this case we may replace $R$ by $R/I$ so that $M = R$. In this case $\varphi : R \to R$ is given by multiplication on $M$ by an element $r \in R$. The surjectivity of $\varphi $ forces $r$ invertible, since $\varphi $ must hit $1$, which implies that $\varphi $ is invertible.

Now assume that we have proven the lemma in the case of modules generated by $n - 1$ elements, and are examining a module $M$ generated by $n$ elements. Let $A$ mean the ring $R[t]$, and regard the module $M$ as an $A$-module by letting $t$ act via $\varphi $; since $M$ is finite over $R$, it is finite over $R[t]$ as well, and since we're trying to prove $\varphi $ injective, a set-theoretic property, we might as well prove the endomorphism $t : M \to M$ over $A$ injective. We have reduced our problem to the case our endomorphism is multiplication by an element of the ground ring. Let $M' \subset M$ denote the sub-$A$-module generated by the first $n - 1$ of the generators of $M$, and consider the diagram

where the restriction of $\varphi $ to $M'$ and the map induced by $\varphi $ on the quotient $M/M'$ are well-defined since $\varphi $ is multiplication by an element in the base, and $M'$ and $M/M'$ are $A$-modules in their own right. By the case $n = 1$ the map $M/M' \to M/M'$ is an isomorphism. A diagram chase implies that $\varphi |_{M'}$ is surjective hence by induction $\varphi |_{M'}$ is an isomorphism. This forces the middle column to be an isomorphism by the snake lemma. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)