The Stacks project

Proof. We reduce the question to the well-known Cayley-Hamilton theorem from linear algebra in several steps:

1. If $\phi :S \rightarrow R$ is a ring morphism and $b_{ij}$ are inverse images of the $a_{ij}$ under this map, then it suffices to show the statement for $S$ and $(b_{ij})$ since $\phi $ is a ring morphism.

2. If $\psi :R \hookrightarrow S$ is an injective ring morphism, it clearly suffices to show the result for $S$ and the $a_{ij}$ considered as elements of $S$.

3. Thus we may first reduce to the case $R = \mathbf{Z}[X_{ij}]$, $a_{ij} = X_{ij}$ of a polynomial ring and then further to the case $R = \mathbf{Q}(X_{ij})$ where we may finally apply Cayley-Hamilton.

Proof. Choose a surjective $R$-module map $R^{\oplus n} \to M$, given by $(a_1, \ldots , a_ n) \mapsto \sum a_ ix_ i$ for some generators $x_ i \in M$. Choose $(a_{i1}, \ldots , a_{in}) \in R^{\oplus n}$ such that $\varphi (x_ i) = \sum a_{ij} x_ j$. In other words the diagram

is commutative where $A = (a_{ij})$. By Lemma 10.16.1 there exists a monic polynomial $P$ such that $P(A) = 0$. Then it follows that $P(\varphi ) = 0$. $\square$

Proof. Choose a surjective $R$-module map $R^{\oplus n} \to M$, given by $(a_1, \ldots , a_ n) \mapsto \sum a_ ix_ i$ for some generators $x_ i \in M$. Choose $(a_{i1}, \ldots , a_{in}) \in I^{\oplus n}$ such that $\varphi (x_ i) = \sum a_{ij} x_ j$. In other words the diagram

is commutative where $A = (a_{ij})$. By Lemma 10.16.1 the polynomial $P(t) = \det (t\text{id}_{n \times n} - A)$ has all the desired properties. $\square$

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. $\square$