Corollary 2

Let A be a commutative ring and B an A-algebra. Let M be a simple B-module that is a finitely generated A-module, and let $\mathfrak{m}$ be the annihilator of the A-module M. Then $\mathfrak{m}$ is a maximal ideal of A, and M is a finite-dimensional vector space over the field $A/\mathfrak{m}$.

Since the ring A is commutative, the annihilator of a simple A-module is a maximal ideal of A (VIII, p. 46, Proposition 1). By Corollary 1 applied to the A-module M, there exists a maximal ideal $\mathfrak{a}$ of A such that $\mathfrak{a}M\not= M$. But $\mathfrak{a}M$ is a submodule of the simple B-module M; we therefore have $\mathfrak{a}M = 0$ and consequently $\mathfrak{a}\subset \mathfrak{m}$. Since the ideal $\mathfrak{m}$ is distinct from A, it is equal to $\mathfrak{a}$, so that $\mathfrak{m}$ is maximal. The module M is a finitely generated module over the field $A/\mathfrak{m}$; the last assertion follows.