Lemma 10.149.8. Let $(R, \mathfrak m, \kappa )$ be a strictly henselian local ring. Let $R \to S$ be an unramified ring map. Then

with each $R \to A_ i$ surjective and no prime of $B$ lying over $\mathfrak m$.

Lemma 10.149.8. Let $(R, \mathfrak m, \kappa )$ be a strictly henselian local ring. Let $R \to S$ be an unramified ring map. Then

\[ S = A_1 \times \ldots \times A_ n \times B \]

with each $R \to A_ i$ surjective and no prime of $B$ lying over $\mathfrak m$.

**Proof.**
First write $S = A_1 \times \ldots \times A_ n \times B$ as in Lemma 10.149.5. Now we see that $R \to A_ i$ is finite unramified and $A_ i$ local. Hence the maximal ideal of $A_ i$ is $\mathfrak mA_ i$ and its residue field $A_ i / \mathfrak m A_ i$ is a finite separable extension of $\kappa $, see Lemma 10.148.5. However, the condition that $R$ is strictly henselian means that $\kappa $ is separably algebraically closed, so $\kappa = A_ i / \mathfrak m A_ i$. By Nakayama's Lemma 10.19.1 we conclude that $R \to A_ i$ is surjective as desired.
$\square$

Your email address will not be published.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 (0)

There are also: