Timeline for Prove that $\Bbb C[x,y]/(x^3+y^3-1)$ is not a UFD
Current License: CC BY-SA 4.0
38 events
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| Nov 15, 2023 at 21:01 | vote | accept | user108580 | ||
| Nov 10, 2023 at 23:16 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 10, 2023 at 23:07 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 10, 2023 at 20:43 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 10, 2023 at 20:35 | comment | added | GH from MO | @Balazs Unfortunately, the short proof in the "Added" section was not complete, but we have now a very nice argument by user516477. | |
| Nov 10, 2023 at 20:32 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 10, 2023 at 17:32 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 10, 2023 at 17:26 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 10, 2023 at 17:16 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 10, 2023 at 13:00 | comment | added | Chris Wuthrich | With all explicit proofs, I have some doubts. They should not work for $x^2+y^2-1$. They should not work when the curve has a singularity. $(x,y-1)=(y-1)^3$ is because this is an inflection point on the curve, but singular curves could have an inflection point.... | |
| Nov 10, 2023 at 10:56 | comment | added | Balazs | This is great. What I find remarkable about this last proof is that it is basically the same proof as the insolubility of the Fermat cubic over ${\mathbb Z}$, except that it has a different conclusion; note that proof would run in ${\mathbb Z}[\omega]$ which... is a UFD. | |
| Nov 10, 2023 at 6:52 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 10, 2023 at 6:44 | comment | added | Zach Teitler | It is very nice! Thank you. | |
| Nov 10, 2023 at 6:26 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 10, 2023 at 3:38 | comment | added | GH from MO | @ZachTeitler In every Dedekind domain, it is true for any nonzero ideals $I$ and $J$ that $\gcd(I,J)^n=\gcd(I^n,J^n)$. But here is a more direct proof. The ideal $(x,y-1)^3$ contains $x^3$ and $(y-1)^3$, hence it also contains $(x^3,(y-1)^3)=(y-1)$. On the other hand, $(x,y-1)^3$ is clearly generated by the four elements $x^3$, $x^2(y-1)$, $x(y-1)^2$, $(y-1)^3$, each of which lies in $(y-1)$. So $(x,y-1)^3$ is contained in $(y-1)$, and we are done. | |
| Nov 10, 2023 at 3:25 | comment | added | Zach Teitler | When you say $(x,y-1)^3 = (x^3,(y-1)^3)$ can you please give a hint why that is true? Perhaps it is obvious, but I am not seeing why the $x^2(y-1)$ and $x(y-1)^2$ are apparently redundant. | |
| Nov 10, 2023 at 2:50 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 9, 2023 at 14:36 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 9, 2023 at 8:02 | comment | added | François Brunault | Maybe one can show directly here, using divisors on $E$, that the maximal ideal $\mathfrak{m}$ of $R$ associated to $P=(1,0)$ is not principal. | |
| Nov 9, 2023 at 7:09 | comment | added | Jesse Elliott | ..or, in the spirit of the OP's question, an example of an irreducible that is not prime. | |
| Nov 9, 2023 at 7:08 | history | edited | Daniel Asimov | CC BY-SA 4.0 |
Changed commas to colons for the points of ℙ(ℂ^3).
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| Nov 9, 2023 at 6:34 | comment | added | Jesse Elliott | I feel like, while it's useful to determine the class group, this answer uses a sledgehammer to circumvent a more basic question: what's an example of non-unique factorization in the given Dedekind domain. | |
| Nov 9, 2023 at 6:32 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 9, 2023 at 6:23 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 8, 2023 at 20:29 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 8, 2023 at 20:19 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 8, 2023 at 19:54 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 8, 2023 at 19:46 | comment | added | GH from MO | @ChrisWuthrich Thanks for pointing out my mistake. I fixed it, please have a look. | |
| Nov 8, 2023 at 19:46 | history | undeleted | GH from MO | ||
| Nov 8, 2023 at 19:46 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 8, 2023 at 15:11 | history | deleted | GH from MO | via Vote | |
| Nov 8, 2023 at 9:59 | comment | added | Chris Wuthrich | @GHfromMO But there are three points at infinity for this curve. I know you are trying to take the proof of what the Jacobian of the projective closure is and get the minimal part of it to show that the ring is non PID. But I believe you need to be more careful here. | |
| Nov 8, 2023 at 9:55 | comment | added | Jesse Elliott | It might be nice to say what its class group is, as that would give information as to exactly how non-UFD it is. | |
| Nov 7, 2023 at 23:48 | comment | added | GH from MO | @ChrisWuthrich $O$ is the point at infinity, i.e. the neutral element, of the projectivization of $E$ (which is also an abelian group). | |
| Nov 7, 2023 at 23:41 | comment | added | Chris Wuthrich | What is $O$ here? | |
| Nov 7, 2023 at 17:01 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 7, 2023 at 16:45 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 7, 2023 at 16:40 | history | answered | GH from MO | CC BY-SA 4.0 |