Lemma 15.111.5. Let $A$ be a valuation ring. Let $A^ h$, resp. $A^{sh}$ be its henselization, resp. strict henselization. Then

are extensions of valuation rings which induce bijections on value groups, i.e., which are weakly unramified.

Lemma 15.111.5. Let $A$ be a valuation ring. Let $A^ h$, resp. $A^{sh}$ be its henselization, resp. strict henselization. Then

\[ A \subset A^ h \subset A^{sh} \]

are extensions of valuation rings which induce bijections on value groups, i.e., which are weakly unramified.

**Proof.**
Write $A^ h = \mathop{\mathrm{colim}}\nolimits (B_ i)_{\mathfrak q_ i}$ where $A \to B_ i$ is étale and $\mathfrak q_ i \subset B_ i$ is a prime ideal lying over $\mathfrak m_ A$, see Algebra, Lemma 10.151.7. Then Lemma 15.111.4 tells us that $(B_ i)_{\mathfrak q_ i}$ is a valuation ring and that the induced map

\[ (A \setminus \{ 0\} )/A^* \longrightarrow ((B_ i)_{\mathfrak q_ i} \setminus \{ 0\} ) / (B_ i)_{\mathfrak q_ i}^* \]

is bijective. By Algebra, Lemma 10.49.5 we conclude that $A^ h$ is a valuation ring. It also follows that $(A \setminus \{ 0\} )/A^* \to (A^ h \setminus \{ 0\} )/(A^ h)^*$ is bijective. This proves the lemma for the inclusion $A \subset A^ h$. To prove it for $A \subset A^{sh}$ we can use exactly the same argument except we replace Algebra, Lemma 10.151.7 by Algebra, Lemma 10.151.13. Since $A^{sh} = (A^ h)^{sh}$ we see that this also proves the assertions of the lemma for the inclusion $A^ h \subset A^{sh}$. $\square$

Your email address will not be published.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)