The Stacks project

Proof. Omitted. $\square$

Proof. Omitted. $\square$

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 10.53.6) we can choose a filtration

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 10.54.2. 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 10.54.2. $\square$

Proof. We can write $S$ as a localization of a quotient of $R[x_1, \ldots , x_ n]$. Hence it suffices to prove the lemma in case $S = R[x_1, \ldots , x_ n]_\mathfrak q$ for some prime $\mathfrak q \subset R[x_1, \ldots , x_ n]$. If $\mathfrak q + \mathfrak m_ R R[x_1, \ldots , x_ n] \not= R[x_1, \ldots , x_ n]$ then we can find a maximal ideal $\mathfrak m$ as in the statement of the lemma with $\mathfrak q \subset \mathfrak m$ and the result is clear.

Choose a valuation ring $A \subset \kappa (\mathfrak q)$ which dominates the image of $R \to \kappa (\mathfrak q)$ (Lemma 10.50.2). If the image $\lambda _ i \in \kappa (\mathfrak q)$ of $x_ i$ is contained in $A$, then $\mathfrak q$ is contained in the inverse image of $\mathfrak m_ A$ via $R[x_1, \ldots , x_ n] \to A$ which means we are back in the preceding case. Hence there exists an $i$ such that $\lambda _ i^{-1} \in A$ and such that $\lambda _ j/\lambda _ i \in A$ for all $j = 1, \ldots , n$ (because the value group of $A$ is totally ordered, see Lemma 10.50.12). Then we consider the map

Let $\mathfrak q' \subset R[y_0, \ldots , \hat{y_ i}, \ldots , y_ n]$ be the inverse image of $\mathfrak q$. Since $y_0 \not\in \mathfrak q'$ it is easy to see that the displayed arrow defines an isomorphism on localizations. On the other hand, the result of the first paragraph applies to $R[y_0, \ldots , \hat{y_ i}, \ldots , y_ n]$ because $y_ j$ maps to an element of $A$. This finishes the proof. $\square$