Lemma 10.155.16 (Tate). Let $R$ be a ring. Let $x \in R$. Assume

$R$ is a normal Noetherian domain,

$R/xR$ is a domain and N-2,

$R \cong \mathop{\mathrm{lim}}\nolimits _ n R/x^ nR$ is complete with respect to $x$.

Then $R$ is N-2.

[Theorem 23.1.3, EGA]

Lemma 10.155.16 (Tate). Let $R$ be a ring. Let $x \in R$. Assume

$R$ is a normal Noetherian domain,

$R/xR$ is a domain and N-2,

$R \cong \mathop{\mathrm{lim}}\nolimits _ n R/x^ nR$ is complete with respect to $x$.

Then $R$ is N-2.

**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

\[ \mathfrak q = \{ f \in S \mid f^ q \in \mathfrak p\} = yS \]

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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #2749 by Ariyan Javanpeykar on

Comment #2863 by Johan on

There are also: