The Stacks project

Proof. We may assume $x \not= 0$ since otherwise the lemma is trivial. Let $K$ be the fraction field of $R$. If the characteristic of $K$ is zero the lemma follows from (1), see Lemma 10.155.11. Hence we may assume that the characteristic of $K$ is $p > 0$, and we may apply Lemma 10.155.12. Thus given $K \subset L$ be a finite purely inseparable field extension we have to show that the integral closure $S$ of $R$ in $L$ is finite over $R$.

Let $q$ be a power of $p$ such that $L^ q \subset K$. By enlarging $L$ if necessary we may assume there exists an element $y \in L$ such that $y^ q = x$. Since $R \to S$ induces a homeomorphism of spectra (see Lemma 10.45.7) there is a unique prime ideal $\mathfrak q \subset S$ lying over the prime ideal $\mathfrak p = xR$. It is clear that

since $y^ q = x$. Hence $R_{\mathfrak p}$ and $S_{\mathfrak q}$ are discrete valuation rings, see Lemma 10.118.7. By Lemma 10.118.10 we see that $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is a finite field extension. Hence the integral closure $S' \subset \kappa (\mathfrak q)$ of $R/xR$ is finite over $R/xR$ by assumption (2). Since $S/yS \subset S'$ this implies that $S/yS$ is finite over $R$. Note that $S/y^ nS$ has a finite filtration whose subquotients are the modules $y^ iS/y^{i + 1}S \cong S/yS$. Hence we see that each $S/y^ nS$ is finite over $R$. In particular $S/xS$ is finite over $R$. Also, it is clear that $\bigcap x^ nS = (0)$ since an element in the intersection has $q$th power contained in $\bigcap x^ nR = (0)$ (Lemma 10.50.4). Thus we may apply Lemma 10.95.12 to conclude that $S$ is finite over $R$, and we win. $\square$