Lemma 10.95.7. Let $A$ be a ring. Let $I = (f_1, \ldots , f_ r)$ be a finitely generated ideal. If $M \to \mathop{\mathrm{lim}}\nolimits M/f_ i^ nM$ is surjective for each $i$, then $M \to \mathop{\mathrm{lim}}\nolimits M/I^ nM$ is surjective.

Proof. Note that $\mathop{\mathrm{lim}}\nolimits M/I^ nM = \mathop{\mathrm{lim}}\nolimits M/(f_1^ n, \ldots , f_ r^ n)M$ as $I^ n \supset (f_1^ n, \ldots , f_ r^ n) \supset I^{rn}$. An element $\xi$ of $\mathop{\mathrm{lim}}\nolimits M/(f_1^ n, \ldots , f_ r^ n)M$ can be symbolically written as

$\xi = \sum \nolimits _{n \geq 0} \sum \nolimits _ i f_ i^ n x_{n, i}$

with $x_{n, i} \in M$. If $M \to \mathop{\mathrm{lim}}\nolimits M/f_ i^ nM$ is surjective, then there is an $x_ i \in M$ mapping to $\sum x_{n, i} f_ i^ n$ in $\mathop{\mathrm{lim}}\nolimits M/f_ i^ nM$. Then $x = \sum x_ i$ maps to $\xi$ in $\mathop{\mathrm{lim}}\nolimits M/I^ nM$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).