Theorem 3

Let K be a commutative field, A a K-algebra, and M and N modules of finite length. Let $K'$ be a nonzero commutative K-algebra such that the $A_{(K')}$-modules $M_{(K')}$ and $N_{(K')}$ are isomorphic. Then the A-modules M and N are isomorphic.

a) First, suppose that the algebra $K'$ has finite degree $d$ over K. Then the A-module $M_{(K')}$ is isomorphic to $M^d$ and the A-module $N_{(K')}$ is isomorphic to $N^d$, so that the A-modules $M^d$ and $N^d$ are isomorphic. By Theorem 2, d), the A-modules M and N are isomorphic.

b) Now, suppose that the K-algebra $K'$ is generated by finitely many elements. Choose a maximal ideal $\mathfrak{m}$ of $K'$, and set $K''= K'/\mathfrak{m}$. By Hilbert’s Nullstellensatz (VIII, p. 462, Corollary 1 of Theorem 1)$, K''$ is a finite-degree extension of K. By extension of scalars from $K'$ to $K''$, we deduce from the $A_{(K')}$-linear isomorphism $M_{(K')}\rightarrow N_{(K')}$ an $A_{(K'')}$-linear isomorphism $M_{(K'')}\rightarrow N_{(K'')}$. By part a) of the proof, the A-modules M and N are isomorphic.

c) Finally, let us treat the general case. Let $u: M_{(K')}\rightarrow N_{(K')}$ be an isomorphism of $A_{(K')}$-modules and $v: N_{(K')}\rightarrow M_{(K')}$ the inverse isomorphism. Denote by $\mathscr{E}$ the set of K-subalgebras of $K'$ generated by finitely many elements. If E is such a subalgebra, then $A_{(E)}$ is identified with a subring of $A_{(K')}$, and $M_{(E)}$ and $N_{(E)}$ are identified with $A_{(E)}$-submodules of $M_{(K')}$ and $N_{(K')}$, respectively (II, §7, No. 7, p. 306); moreover, $M_{(K')}$ and $N_{(K')}$ are unions of right directed families $(M_{(E)})_{E\in\mathscr{E}}$ and $(N_{(E)})_{E\in\mathscr{E}}$, respectively. The A-modules M and N are of finite length, hence finitely generated; let S be a finite generating subset of the A-module M and T a finite generating subset of the A-module N. There exists a K-algebra $E\in \mathscr{E}$ such that we have $u(1\otimes s)\in N_{(E)}$ for every $s\in S$ and $v(1\otimes t)\in M_{(E)}$ for every $t\in T$. It follows by linearity that we have $u(M_{(E)})\subset N_{(E)}$ and $v(N_{(E)})\subset M_{(E)}$. The maps $u$ and $v$ then induce inverse bijections from $M_{(E)}$ to $N_{(E)}$ and from $N_{(E)}$ to $M_{(E)}$. These bijections are clearly $A_{(E)}$-linear. Thus, the $A_{(E)}$-modules $M_{(E)}$ and $N_{(E)}$ are isomorphic. By part b) of the proof, the A-modules M and N are isomorphic.