Let $A_{1},A_{2}$ be discrete valuation rings whose fraction fields are isomorphic. Let $A_{i}^{\mathrm{sh}}$ be the strict henselization of $A_{i}$, and let $K_{i}$ be the fraction field of $A_{i}^{\mathrm{sh}}$. Are $K_{1},K_{2}$ isomorphic?
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asked Oct 17, 2018 at 2:42
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7$\begingroup$ For the localization $A_i$ of the ring $\mathbb{Z}$ at the prime ideal $p_i\mathbb{Z}$, the fraction field of $A_i$ equals $\mathbb{Q}$. But if $p_1$ and $p_2$ are distinct prime integers, then the fraction fields $K_1$ and $K_2$ of the strict Henselizations are not isomorphic (just consider the groups of roots of unity in each field). $\endgroup$– Jason StarrCommented Oct 17, 2018 at 2:48
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1$\begingroup$ @JasonStarr Thanks very much. Do you also know of any equicharacteristic examples? $\endgroup$– Minseon ShinCommented Oct 17, 2018 at 2:53
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3$\begingroup$ A non-separably closed field can have at most one henselian valuation (see Endler, Valuation Theory, (26.6)). So any isomorphism between $K_1$ and $K_2$ must respect the valuations and, in particular, the residue fields must be isomorphic. $\endgroup$– Laurent Moret-BaillyCommented Oct 17, 2018 at 9:57
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5$\begingroup$ So you get an example by taking the following valuations on $K:=\mathbb{Q}(x,y)$: the $y$-adic valuation $v_1$, with residue field $\mathbb{Q}(x)$, and the valuation obtained by fixing $h\in\mathbb{Q}[[t]]$ transcendental over $\mathbb{Q}(t)$ and embedding $K$ into $\mathbb{Q}(\!(t)\!)$ by $f(x,y)\mapsto f(t,h(t))$. The valuation $v_2$ induced by the $t$-adic valuation has residue field $\mathbb{Q}$. The strict henselizations of $v_1$, $v_2$ thus have non-isomorphic residue fields, namely $\overline{\mathbb{Q}(x)}$ and $\overline{\mathbb{Q}}$, hence are not isomorphic by the above comment. $\endgroup$– Laurent Moret-BaillyCommented Oct 17, 2018 at 10:00
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$\begingroup$ @LaurentMoret-Bailly Thanks very much. $\endgroup$– Minseon ShinCommented Oct 17, 2018 at 23:20
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