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
\xymatrix{ 0 \ar[r] & M' \ar[r]\ar[d]^{\varphi \mid _{M'}} & M\ar[d]^\varphi \ar[r] & M/M' \ar[d]^{\varphi \bmod M'} \ar[r] & 0 \\ 0 \ar[r] & M' \ar[r] & M \ar[r] & M/M' \ar[r] & 0, }
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
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