Timeline for For a finite-type $\mathbb{Z}$-algebra $A$, is the intersection of all ideals $I$ such that $A/I$ is finite and local necessarily zero?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2022 at 19:32 | vote | accept | Owen Biesel | ||
| Sep 15, 2022 at 7:03 | comment | added | Kapil | If $X\to Y$ is a morphism of finite type schemes (over $k=\mathbb{Z}$ or a field $k$) such that $X(A)\to Y(A)$ is a bijection for all "finite" $k$-algebras $A$ (either finite rings for $\mathbb{Z}$ or finite dimensional as $k$ vector spaces), then $X\to Y$ is an isomorphism. | |
| Sep 14, 2022 at 17:52 | answer | added | Luc Guyot | timeline score: 4 | |
| Sep 13, 2022 at 22:36 | comment | added | Benjamin Steinberg | Maybe helpful mathoverflow.net/questions/57515/… | |
| Sep 13, 2022 at 20:29 | comment | added | YCor | More generally, every noetherian (commutative) ring is residually local artinian (the intersection of $I$ such that $A/I$ is local artinian is reduced to zero). Combined with the fact that f.g. commutative rings that are fields are finite. The proof consists of taking $x\neq 0$ and a maximal ideal $I$ not containing $x$, and show that $A/I$ is artinian. Namely a noetherian ring whose intersection of nonzero ideals is nonzero is artinian. This is easy using associated ideals. | |
| Sep 13, 2022 at 20:23 | answer | added | Peter Kropholler | timeline score: 5 | |
| Sep 13, 2022 at 19:56 | history | asked | Owen Biesel | CC BY-SA 4.0 |