**Proof.**
If $R \to S$ is finite, then $R \to S/\mathfrak m$ is finite by Lemma 10.7.3. Conversely, assume $R \to S/\mathfrak m$ is finite. As $S$ has finite length over itself (Lemma 10.52.6) we can choose a filtration

\[ 0 \subset I_1 \subset \ldots \subset I_ n = S \]

by ideals such that $I_ i/I_{i - 1} \cong S/\mathfrak m$ as $S$-modules. Thus $S$ has a filtration by $R$-submodules $I_ i$ such that each successive quotient is a finite $R$-module. Thus $S$ is a finite $R$-module by Lemma 10.5.3.

If $R \to S$ is of finite type, then $R \to S/\mathfrak m$ is of finite type by Lemma 10.6.2. Conversely, assume that $R \to S/\mathfrak m$ is of finite type. Choose $f_1, \ldots , f_ n \in S$ which map to generators of $S/\mathfrak m$. Then $A = R[x_1, \ldots , x_ n] \to S$, $x_ i \mapsto f_ i$ is a ring map such that $A \to S/\mathfrak m$ is surjective (in particular finite). Hence $A \to S$ is finite by part (1) and we see that $R \to S$ is of finite type by Lemma 10.6.2.

If $R \to S$ is essentially of finite type, then $R \to S/\mathfrak m$ is essentially of finite type by Lemma 10.53.2. Conversely, assume that $R \to S/\mathfrak m$ is essentially of finite type. Suppose $S/\mathfrak m$ is the localization of $R[x_1, \ldots , x_ n]/I$. Choose $f_1, \ldots , f_ n \in S$ whose congruence classes modulo $\mathfrak m$ correspond to the congruence classes of $x_1, \ldots , x_ n$ modulo $I$. Consider the map $R[x_1, \ldots , x_ n] \to S$, $x_ i \mapsto f_ i$ with kernel $J$. Set $A = R[x_1, \ldots , x_ n]/J \subset S$ and $\mathfrak p = A \cap \mathfrak m$. Note that $A/\mathfrak p \subset S/\mathfrak m$ is equal to the image of $R[x_1, \ldots , x_ n]/I$ in $S/\mathfrak m$. Hence $\kappa (\mathfrak p) = S/\mathfrak m$. Thus $A_\mathfrak p \to S$ is finite by part (1). We conclude that $S$ is essentially of finite type by Lemma 10.53.2.
$\square$

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