Timeline for A finitely generated $\mathbb{Z}$-algebra that is a field has to be finite
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Apr 9, 2023 at 20:45 | comment | added | LSpice | @QingLiu's answer referenced by @KeenanKidwell. | |
| Apr 9, 2023 at 20:43 | history | edited | LSpice | CC BY-SA 4.0 |
Typo, while this is on the front page
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| Apr 9, 2021 at 2:13 | comment | added | MAS | @DanielLitt, if we replace $\mathbb{Z}$ by $\mathbb{Z}_p$ where $\mathbb{Z}_p$ is the ring of $p$-adic integers, that is, if $R$ be a finitely generated $\mathbb{Z}_p$-algebra, does also the same result holds ? | |
| Oct 18, 2012 at 16:03 | comment | added | Daniel Litt | It doesn't bother me! | |
| Oct 18, 2012 at 7:24 | comment | added | Qing Liu | @aglearner: I am happy that you found my answer helpfull, but please don't unaccept Daniel's answer. | |
| Oct 14, 2012 at 13:36 | comment | added | aglearner | Daniel, sorry I decided to accept now Qing's answer, since I understand it better after one year an half. | |
| Mar 7, 2011 at 11:44 | comment | added | Keenan Kidwell | @Daniel I guess it depends on what you call the Nullstellensatz (or classical Nullstellensatz, or whatever), but the result (which I've also heard called Zariski's lemma) isn't weaker than the (classical) "Weak" Nullstellensatz (which in turn implies the "strong" version about radicals of ideals). | |
| Mar 7, 2011 at 7:58 | comment | added | Daniel Litt | @Keenan, Anthony: Actually this weaker result goes by the name of Zariski's lemma, I believe (see e.g. math.lsa.umich.edu/~hochster/615W10/supNoeth.pdf) | |
| Mar 7, 2011 at 2:36 | comment | added | Keenan Kidwell | @Anthony Yes, the Nullstellensatz (for fields) is being used here. Once you know that $\mathfrak{m}\cap\mathbb{Z}$ is not zero, you infer that $R/\mathfrak{m}$ is a finitely generated algebra over $\mathbb{F}_p$. Since $R/\mathfrak{m}$ is a field, the Nullstellensatz implies that it is a finite extension of $\mathbb{F}_p$. | |
| Mar 6, 2011 at 21:54 | comment | added | Daniel Litt | $\mathbb{F}_p(x)$ is not finitely generated as an $\mathbb{F}_p$-algebra; in particular, all the irreducibles need to be inverted. See e.g. mathreference.com/ag,fgaf.html | |
| Mar 6, 2011 at 21:44 | comment | added | Anthony Quas | An ignorant question: "$R/\mathfrak{m}$ is finitely generated over $\mathbb F_p$. But all finite field extensions of $\mathbb F_p$ are still finite, completing the proof". I'm missing something here. How did finitely generated extension get to be the same as finite field extension? (in particular isn't $\mathbb F_p(x)$ a finitely generated extension but not a finite extension)? | |
| Mar 6, 2011 at 1:41 | comment | added | aglearner | Keenan, thanks for pointing to this exercise! | |
| Mar 6, 2011 at 1:13 | comment | added | Daniel Litt | @Keenan Kidwell: Of course; thanks for making that explicit. | |
| Mar 6, 2011 at 1:07 | comment | added | Keenan Kidwell | Perhaps it's worth emphasizing that what makes this argument work is the non-trivial fact (proved in Qing Liu's answer) that if $f:\mathbb{Z}\rightarrow\mathbb{R}$ is of finite type, then the pre-image of a maximal ideal of $R$ under $f$ is non-zero. This is a particular case of the general Nullstellensatz on Jacobson rings (found in, e.g., Eisenbud's book). A proof of the statement about fields finitely generated as rings using the usual Nullstellensatz is outlined in Exercise 6 of the Noetherian rings chapter of Atiyah and Macdonald. | |
| Mar 6, 2011 at 0:16 | comment | added | aglearner | Daniel, thanks for the proof of the lemma! | |
| Mar 6, 2011 at 0:15 | vote | accept | aglearner | ||
| Oct 14, 2012 at 13:32 | |||||
| Mar 6, 2011 at 0:08 | history | answered | Daniel Litt | CC BY-SA 2.5 |