Lemma 10.96.7. Let $R \to S$ be a local homomorphism of local rings $(R, \mathfrak m)$ and $(S, \mathfrak n)$. Let $R^\wedge $, resp. $S^\wedge $ be the completion of $R$, resp. $S$ with respect to $\mathfrak m$, resp. $\mathfrak n$. If $\mathfrak m$ and $\mathfrak n$ are finitely generated and $\dim _{\kappa (\mathfrak m)} S/\mathfrak mS < \infty $, then

$S^\wedge $ is equal to the $\mathfrak m$-adic completion of $S$, and

$S^\wedge $ is a finite $R^\wedge $-module.

**Proof.**
We have $\mathfrak mS \subset \mathfrak n$ because $R \to S$ is a local ring map. The assumption $\dim _{\kappa (\mathfrak m)} S/\mathfrak mS < \infty $ implies that $S/\mathfrak mS$ is an Artinian ring, see Lemma 10.52.2. Hence has dimension $0$, see Lemma 10.59.4, hence $\mathfrak n = \sqrt{\mathfrak mS}$. This and the fact that $\mathfrak n$ is finitely generated implies that $\mathfrak n^ t \subset \mathfrak mS$ for some $t \geq 1$. By Lemma 10.95.9 we see that $S^\wedge $ can be identified with the $\mathfrak m$-adic completion of $S$. As $\mathfrak m$ is finitely generated we see from Lemma 10.95.3 that $S^\wedge $ and $R^\wedge $ are $\mathfrak m$-adically complete. At this point we may apply Lemma 10.95.12 to $S^\wedge $ as an $R^\wedge $-module to conclude.
$\square$

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