Lemma 1

Let A be a commutative ring, M a finitely generated A-module, and Ann(M) its annihilator. Let $\mathfrak{a}$ be an ideal of A. The following properties are equivalent:

(i) $\mathfrak{a}M = M$.

(ii) Ann(M) $+\mathfrak{a}= A$.

(iii) There exists an element $a$ of $\mathfrak{a}$ such that $am=m$ for every $m\in M$.

(i) $\Rightarrow$ (ii): Let $(x_1, . . . , x_n)$ be a generating family of the A-module M. If $\mathfrak{a}M = M$, then each $x_i$ can be written as $\sum^n_{j=1}c_{ij}x_j$, where the $c_{ij}$ belong to $\mathfrak{a}$. Denote the matrix $(c_{ij})$ by C and the column matrix with entries $x_1, . . . , x_n$ by X. We have ($I_n-C$)X = 0. Let $d$ be the determinant and V the cofactor matrix of the matrix $I_n-C$. By formula (26) of III, §8, No. 6, p. 532, we have $dX=^tV$($I_n-C$)X = 0, so that $d\in$ Ann(M). On the other hand, since the $c_{ij}$ belong to $\mathfrak{a}$, we have $d\equiv$ 1(mod $\mathfrak{a}$) (III, §8, No. 5, p. 529, (18)), and therefore $1\in$ Ann(M) $+\mathfrak{a}$.

(ii) $\Rightarrow$ (iii): Assuming that (ii) holds, there exist $a\in \mathfrak{a}$ and $b\in$ Ann(M) such that $a+b= 1$. We then have $am=m$ for every $m\in M$.

(iii) $\Rightarrow$ (i): This is clear.