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
18 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2022 at 19:32 | vote | accept | Owen Biesel | ||
| Sep 19, 2022 at 15:17 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Mentions that quotient rings built in the proof of Claim 5 are also Artinian (in keeping with YCor's comment)
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| Sep 17, 2022 at 14:28 | comment | added | Luc Guyot | @OwenBiesel I just added an alternative proof of Claim 5 (it might be closer to what YCor envisioned). | |
| Sep 17, 2022 at 14:26 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Add an alternative proof of Claim 5 (Matlis' injective module theory)
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| Sep 16, 2022 at 16:39 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Minor change
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| Sep 16, 2022 at 15:08 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Expands the last sentence in the proof of Claim 1 and add several references
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| Sep 16, 2022 at 14:47 | comment | added | Luc Guyot | @OwenBiesel Thanks for your feedback. This is what we get indeed. As we have assumed that $M^{c + 1} \neq \{0\}$, we know that $M^{c + 1}$ contains $\overline{x}$. The same holds for $M^c$. Let's see if I can improve this last line. | |
| Sep 16, 2022 at 13:48 | comment | added | Owen Biesel | Thank you, this is very helpful! I follow you right up until the last line of the proof of Claim 5: letting $n = c+1$, don't we get $M^{c+1} \cap \overline{R}\overline{x} = M(M^c \cap \overline{R}\overline{x})$? Why does that mean $\overline{R}\overline{x} = M\overline{R}\overline{x}$? | |
| Sep 15, 2022 at 22:05 | comment | added | Luc Guyot | @OwenBiesel I finally got it. (I am willing to improve the presentation if anything remains obscure). | |
| Sep 15, 2022 at 22:03 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Completes the proof of Claim 1
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| Sep 15, 2022 at 20:41 | comment | added | Luc Guyot | @OwenBiesel Dear Owen, the fact that $\overline{x}^2 = 0$ is correct (but it doesn't seem to be useful eventually). Hopefully, the changes that I just made in my tentative proof of Claim 1 will make this clearer. Note indeed that we have $\overline{x} \in M$, by construction, i.e. simply by means of our assumptions on $I$ and $x$. The (removed) error was in the reasoning right after this point. I'll come back to this incomplete proof in a day or two. | |
| Sep 15, 2022 at 20:29 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Removes flawed reasoning in the proof of Claim 1. The proof is still incomplete at the moment.
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| Sep 15, 2022 at 13:39 | comment | added | Owen Biesel | I really appreciate you expanding on how the ideas connect together! At the time of writing you're mentioning there's an error in your proof of Claim 1: is it in deducing that $\overline x^2 = 0$? I don't see why that would follow, since the Jacobson radical of a ring can contain non-nilpotents in general. | |
| Sep 15, 2022 at 4:53 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Warns on erroneous proof of Claim 1
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| Sep 14, 2022 at 19:24 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Add further references (classic CA results)
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| Sep 14, 2022 at 19:05 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Removes superfluous assumption
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| Sep 14, 2022 at 18:17 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Fixes incorrect introductory sentence
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| Sep 14, 2022 at 17:52 | history | answered | Luc Guyot | CC BY-SA 4.0 |