Timeline for Extending section of étale morphism of adic spaces
Current License: CC BY-SA 4.0
14 events
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| Mar 30, 2019 at 17:03 | comment | added | Nib | Dear gdb, I am confused: a valued field whose valuation extends uniquely to any finite extension is henselian, i.e. the valuation ring is henselian with respect to its maximal ideal. It is not true in general for higher ranks that a complete valued field is henselian; so you must be using that the local ring $\widehat{k(y)^+}$ is henselian with respect to its maximal ideal, which is not so clear (at least not to me) if the rank is $>1$. | |
| Mar 28, 2019 at 6:30 | comment | added | gdb | @Nib, I updated my answer. The key is that there is only one extension of a complete valuation on a field to a valuation on a finite extension of this field. This is rather standard for rk-$1$ valuation, and the only reference for an arbitrary valuation rings that I know is a book "Commutative Algebra" by Bourbaki. | |
| Mar 28, 2019 at 6:27 | history | edited | gdb | CC BY-SA 4.0 |
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| Mar 27, 2019 at 21:40 | comment | added | Nib | Dear gdb, assuming that $y$ has rk $1$, the fact that for an algebraic field extension $L/K$ and a fixed valuation ring $R$ of $L$, the map $R'\mapsto R'\cap K$ induces a bijection between those valuation rings of $L$ containing $R$ and those valuation rings of $K$ containing $R\cap K$, should imply that each point $x_i$ is of rk $1$ (hence maximal wrt generalization). But I do not see a way of arguing, if $y$ is not of rk $1$. | |
| Mar 25, 2019 at 15:24 | comment | added | Nib | Could you please explain how to show that the points $x_i$ do not have a common generalization? | |
| Mar 23, 2019 at 5:30 | history | edited | gdb | CC BY-SA 4.0 |
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| Mar 22, 2019 at 23:34 | comment | added | gdb | @Nib > I only know that a finite module with the natural topology is complete This also holds for noetherian complete Tate rings, this is Lemma 2.4(ii) in Huber's paper "A generalization of formal schemes and rigid analytic varieties". | |
| Mar 22, 2019 at 19:03 | comment | added | Nib | pair of definition of $A$, $B_0$ is a ring of definition of $B$ such that $f(A_0)\subset B_0$. So it is enough to show that $\mathcal{O}_Y(V)\otimes_A B$ is complete for this topology. It would be enough to show that $(\ast)$ the topology on $\mathcal{O}_Y(V)\otimes_A B$ is the natural $\mathcal{O}_Y(V)$-module topology (or it would also follow from $B_0$ being of finite presentation over $A_0$). Why is the topology the natural topology? $(\ast)$ I only know that a finite module with the natural topology is complete if the underlying ring has a noetherian ring of definition. | |
| Mar 22, 2019 at 19:00 | comment | added | Nib | If I see things correctly, then the second last bullet point reduces the last one to showing that for a rational subset $V=R(\frac{T}{s})$ of $Y$, one has $\mathcal{O}_Y(V)\otimes_A B=\mathcal{O}_X(f^{-1}(V))$. Now, $\mathcal{O}_X(f^{-1}(V))$ is a pushout of $A\to \mathcal{O}_Y(V)$, $A\to B$ (omitting the $+$-component). A pushout would also be the completion of $\mathcal{O}_Y(V)\otimes_A B$ wrt the topology defined by the open subring $\widehat{A_0[\frac{T}{s}]}\otimes_{A_0} B_0$ with ideal of definition $I^n\widehat{A_0[\frac{T}{s}]}\otimes_{A_0} B_0$, where $(A_0,I)$ is a... | |
| Mar 22, 2019 at 10:55 | vote | accept | Nib | ||
| Mar 22, 2019 at 8:54 | comment | added | gdb | Ok, you can probably just require this property in the very definition of an etale map. I guess you will then face some other obstacles. But I am not sure. | |
| Mar 22, 2019 at 8:52 | comment | added | gdb | @Nib That's correct, it is sufficient to show that $k(y) \subset k(x)$ is finite separable. A priori, etaleness guarantees only that $\widehat{k(y)} \subset \widehat{k(x)}$ is finite and separable. Does it formally imply that $k(y) \subset k(x)$ is finite and separable? But I agree that this is a question how you set up general theory. I prefer to prove that around rk-1 points all quasi-finite maps are actually finite and then use the argument from my answer. Since all points have rk-1 vertical generalization it is sufficient for the claim. I guess you can argue somehow in a different way. | |
| Mar 22, 2019 at 5:58 | comment | added | Nib | Dear gdb, thank you for the nice answer. But can't we avoid the use of Lemma 1 as follows: The morphism $f\colon X\to Y$ being étale already implies that the extension $k(y)\subset k(x)$ is finite separable; then as explained in my question (with your notation) one has $k(y)\subset k(x)\subset \widehat{k(y)}$ and hence $\widehat{k(x)}=\widehat{k(y)}$ and so one would obtain $k(y)=k(x)$ by the equivalence of categories you stated. Or am I missing something? | |
| Mar 21, 2019 at 22:46 | history | answered | gdb | CC BY-SA 4.0 |