Timeline for Example of an integral scheme which is geometrically connected but not geometrically irreducible
Current License: CC BY-SA 4.0
8 events
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| Oct 2, 2020 at 8:38 | history | edited | Jason Starr | CC BY-SA 4.0 |
corrected typo, "irrelevant"
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| Aug 26, 2017 at 0:10 | comment | added | R. van Dobben de Bruyn | Another way to argue the normal case is by using that étale extensions of normal rings are normal rings (see Tag 033C). Thus, if $X$ is normal, then so is $X_{k^{\operatorname{sep}}}$. Since the latter is also connected by assumption, it is integral. Irreducibility is not changed when passing from $k^{\operatorname{sep}}$ to $\bar k$ (but reducedness can be). | |
| Aug 24, 2017 at 22:14 | comment | added | dorebell | We have \begin{align*} X \times_{\mathrm{Spec}\ k} \mathrm{Spec}\ \overline{k} &= (X \times_{\mathrm{Spec}\ k'} \mathrm{Spec}\ k') \times_{\mathrm{Spec}\ k} \mathrm{Spec}\ \overline{k}\\ &= X \times_{\mathrm{Spec}\ k'} \mathrm{Spec}\ (k' \otimes_k \overline{k})\\ &= X \times_{\mathrm{Spec}\ k'} \left( \sqcup_{k' \hookrightarrow \overline{k}} \mathrm{Spec}\ \overline{k}\right) \\ &= \sqcup_{k' \hookrightarrow \overline{k}} \left(X \times_{\mathrm{Spec}\ k'} \overline{k}\right) \end{align*} This is not connected, since $k'/k$ is a non-trivial separable extension. | |
| Aug 24, 2017 at 22:12 | comment | added | dorebell | But an element of $K(X)$ which is algebraic over $k$ is certainly integral over $A$, so it is an element of $K(X)$. Let $k'$ be the separable closure of $k$ in $K(X)$; then we've shown that $X$ is naturally a $k'$-scheme ($k'$ doesn't depend on the choice of affine open), and that it is geometrically irreducible over $k'$ (since $k'$ is separably closed in $K(X)$). | |
| Aug 24, 2017 at 22:03 | comment | added | dorebell | I was surprised by the statement when $X$ is normal, and I couldn't find a proof online, so here's a sketch of the argument I worked out for future reference: We can work affine-locally on $X$, so assume $X = \mathrm{Spec}(A)$ for $A$ an integrally closed domain. It suffices to show that $A \otimes_k \overline{k}$ has a unique minimal prime ideal. Since field extensions are flat and $A$ injects into $K(X)$, it suffices to show that $K(X) \otimes_k \overline{k}$ has a unique minimal prime ideal. This is equivalent to $k$ being separably closed in $K(X)$. | |
| Aug 24, 2017 at 21:50 | vote | accept | dorebell | ||
| S Aug 24, 2017 at 13:29 | history | answered | Jason Starr | CC BY-SA 3.0 | |
| S Aug 24, 2017 at 13:29 | history | made wiki | Post Made Community Wiki by Jason Starr |