Exercise 2.11

Proof. Let $𝔪$ be a maximal ideal of $A$ and let $\phi :A^m\to A^n$ be an isomorphism. Then $1\otimes \phi :(A∕𝔪){\text{⊗}}_A{A}^m\to (A∕𝔪){\text{⊗}}_A{A}^n$ is an isomorphism between vector spaces of dimensions $m$ and $n$ over the field $k=A∕𝔪$, hence $m=n$.

If $\phi$ is merely surjective, then surjectivity of $1\otimes \phi$ follows by the right exactness of $(A∕𝔪){\text{⊗}}_A-$ (Prop. 2.18), hence $m\ge n$.

Now suppose $\psi$ is injective, and suppose by way of contradiction that $m>n$. First identify $A^n$ with the submodule of $A^m$ generated by the first $n$ basis elements; $\psi$ then defines an injective endomorphism $\psi :A^m\stackrel{\phi }{\text{↪}}{A}^n\hookrightarrow A^m$. By the Cayley-Hamilton theorem for modules (Prop. 2.4), $\psi$ satisfies an equation of the form

for $a_i\in A$, and moreover, there is such an equation with minimal degree $k$. Since $\psi$ is injective, this implies $a_k\ne 0$, for otherwise then $\psi \circ c(\psi )=0$, hence by injectivity of $\psi$, ${\psi }^{k-1}+a_1{\psi }^{k-2}+\cdots +a_{k-1}=0$, contradicting minimality. Now note that $\psi (0,0,\dots ,0,1)$ has last coordinate $0$ by construction of $\psi$, so $c(\psi )(0,0,\dots ,0,1)$ has last coordinate $a_{k-1}$, contradicting that $c(\psi )=0$. □