Lemma 15.123.5. 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 15.123.5. 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 10.155.7. Then Lemma 15.123.4 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 10.50.5 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 10.155.7 by Algebra, Lemma 10.155.11. Since $A^{sh} = (A^ h)^{sh}$ we see that this also proves the assertions of the lemma for the inclusion $A^ h \subset A^{sh}$. $\square$

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