The Stacks project

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. $\square$