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Tag 05G8

Chapter 10: Commutative Algebra > Section 10.15: Cayley-Hamilton

Lemma 10.15.4. 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 10.15.3 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. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 2719–2725 (see updates for more information).

    \begin{lemma}
    \label{lemma-fun}
    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.
    \end{lemma}
    
    \begin{proof}[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 \ref{lemma-charpoly-module-ideal}
    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.
    \end{proof}
    
    \begin{proof}[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.
    
    \medskip\noindent
    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.
    \end{proof}

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