Timeline for Is a base-change of an integral domain by an extension of its base field without algebraic elements still a domain?
Current License: CC BY-SA 3.0
15 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jan 13, 2017 at 2:19 | vote | accept | Neil Epstein | ||
| Jan 13, 2017 at 0:36 | comment | added | nfdc23 | OK, I have done this now (with notation fixed up to be consistent with the question but not with my initial comment). | |
| Jan 13, 2017 at 0:36 | answer | added | nfdc23 | timeline score: 8 | |
| Jan 12, 2017 at 21:06 | comment | added | Neil Epstein | @nfdc23 Please post your comment as an answer, appropriately phrased in light of above conversation, and I'll accept it. | |
| Jan 12, 2017 at 18:55 | comment | added | Fred Rohrer | Concerning terminology, it is funny to note that Bourbaki has a section (namely A.V.4) called Extensions algébriquement closes, but does not use that precise term... | |
| Jan 12, 2017 at 18:25 | comment | added | nfdc23 | I usually say "$K$ is algebraically closed in $L$" (rather than "$L$ is an algebraically closed extension of $K$" as you put it); I think this is the usual terminology. Maybe one could also say "$K$ is relatively algebraically closed in $L$", or that may be too much of a mouthful. | |
| Jan 12, 2017 at 18:12 | comment | added | Neil Epstein | @nfdc23 I didn't mean to throw a 'curveball'; I just don't know another compact phrase that means what I meant. Do you (or someone else) know such a phrase? It would be useful to have one in searching for more information on the concept. In general if you have $A \subset B$ objects in a concrete category and a closure operation c on subobjects of $B$, to say the extension is c-closed typically means $A$ is c-closed in $B$. But maybe that meaning isn't well-known for fields. | |
| Jan 12, 2017 at 17:21 | comment | added | nfdc23 | @KevinBuzzard: Ack, I misread the question when setting my notation. (I have never heard of the phrase "algebraically closed extension" for anything other than an extension that is algebraically closed...so I didn't notice the curveball in the intended meaning.) Sorry about that! | |
| Jan 12, 2017 at 17:18 | comment | added | nfdc23 | That $K$ is algebraically closed in ${\rm{Frac}}(R)$ is a fun exercise; I don't know a reference (I heard it from somewhere years ago, and worked it out for myself). I mentioned that $R$ is Dedekind as a "hint" on this since that ensures one just has to study elements of $R$ integral over $K$, which one can control by looking in residue fields of $R$ at various maximal ideals. But as Kevin says, the Bourbaki references in the EGA reference dispose of the theoretical aspects on this and other related questions (though I don't know offhand if Bourbaki includes MacLane's example). | |
| Jan 12, 2017 at 16:53 | comment | added | Kevin Buzzard | Did you check EGA IV_2 yet? There are explicit references to Bourbaki in there -- lots of material. | |
| Jan 12, 2017 at 16:49 | comment | added | Neil Epstein | Why is $K$ algebraically closed in $L := \Frac(R)$? Do you have a reference for MacLane's example? | |
| Jan 12, 2017 at 16:48 | comment | added | Kevin Buzzard | I think Neil Epstein's $L$ is @nfdc23's $Frac(R)$ and NE's $R$ is nf's $\overline{K}$ (or $K(s^{1/p},t^{1/p})$ if you want something finitely generated over $K$). | |
| Jan 12, 2017 at 16:15 | comment | added | nfdc23 | No (but "yes" for $K=\overline{K}$ is classical). An example of MacLane is geometric irreducible over $K$ but not geometrically reduced even with $K$ algebraically closed in the fraction field of $R$ (to avoid lame examples): for $K=\mathbf{F}_p(s,t)$, $R= K[x,y]/(sx^p+ty^p-1)$ is a Dedekind domain and $K$ is algebraically closed in ${\rm{Frac}}(R)$, but $R \otimes_K \overline{K}=\overline{K}[x,y]/(h^p)$ for the linear form $h=s^{1/p}x+t^{1/p}y - 1$. In char. 0, if $K$ is algebraically closed in ${\rm{Frac}}(R)$ then $R\otimes_K K'$ is a domain for any $K'/K$; see 4.3 in EGA IV$_2$ for more. | |
| Jan 12, 2017 at 15:51 | history | asked | Neil Epstein | CC BY-SA 3.0 |