Proposition 2

Let L be a commutative ring and V a free L-module of finite dimension $m$. Suppose that every L-module M such that $M^m$ is isomorphic to $L^m$ is isomorphic to L. Then every automorphism of the L-algebra End$_L(V)$ is an inner automorphism.

Set B = End$_L(V)$. Let $u$ be an automorphism of the L-algebra B. We view V as a left B-module; let $u_*(V)$ be the left B-module associated with $u$, with external law $(b, v)\mapsto u(b)(v)$ (II, §1, No. 13, p. 221). Let $(e_1, . . . , e_m)$ be a basis of the L-module V; given elements $v_1, . . . , v_m$ of V, there exists a unique element $b$ of B such that we have $b(e_i) =v_i$ for $1\leqslant i\leqslant m$. In other words, the element $e= (e_1, . . . , e_m)$ of $V^m$ provides a basis of the B-module $V^m$. Since $u$ is an automorphism, $e$ also gives a basis of the B-module $u_*(V^m) =u_*(V)^m$, which is therefore isomorphic to $V^m$. The $(B$, L)-bimodule V is invertible (VIII, p. 102, Example 1). By Theorem 2, b) of VIII, p. 103, there consequently exists an L-module M such that the B-module $u_*(V)$ is isomorphic to $V\otimes_LM$. The B-modules $V\otimes_LL^m$ and $V\otimes_LM^m$, respectively isomorphic to $V^m$ and $u_*(V)^m$, are therefore isomorphic. By loc. cit., the L-modules $L^m$ and $M^m$ are isomorphic. Given the assumption, M is isomorphic to L; consequently, the B-module $u_*$(V), which is isomorphic to $V\otimes_LM$, is isomorphic to V. Let $h$ be a B-module isomorphism from V to $u_*(V)$; it is, in particular, an automorphism of the L-module V, that is, an invertible element of B. For $b$ in B and $v$ in V, we have $h(b(v)) =u(b)(h(v))$ and therefore $u(b) =hbh^{-1}$.

The conditions of Proposition 2 are, in particular, satisfied when the commutative ring L is a principal ideal domain (VII, §3, p. 15, Corollary 3) or is Artinian (VIII, p. 37, Theorem 2, d)) or local (VIII, p. 36, Corollary 6).