Timeline for Can you constructively prove a univariate polynomial algebra over a Jacobson ring is itself Jacobson?
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| Aug 22 at 21:27 | vote | accept | Arrow | ||
| Jun 12 at 2:27 | answer | added | Ryota Kuroki | timeline score: 3 | |
| Dec 12, 2021 at 0:28 | comment | added | Pace Nielsen | By the way, a nonconstructive proof appears explicitly on pages 67--69 of T.Y. Lam's book "A First Course in Noncommutative Ring Theory" (among many other places). This answers question #2, but the given proofs definitely use prime ideals. | |
| Dec 12, 2021 at 0:24 | comment | added | Pace Nielsen | @Arrow How are you defining $R$ to be a Jacobson ring constructively? | |
| Dec 11, 2021 at 23:32 | comment | added | Pace Nielsen | @darijgrinberg If $R$ is a commutative ring, then the Jacobson radical of $R[x]$ is equal to the nilradical of $R[x]$, whether or not the same is true for $R$. This is a theorem of E. Snapper. For a nice proof see Theorem 5.1 in Lam's "First Course in Noncommutative Rings". | |
| Dec 11, 2021 at 21:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 12, 2021 at 11:45 | comment | added | Andrej Bauer | @DavidESpeyer: how is your "shenanigans" limited to constructive mathematics? Your argument and example works equally well if excluded middle is assumed. Also, there is no question about what "for all ideals" means in the constructive world. | |
| Nov 11, 2021 at 21:10 | comment | added | darij grinberg | You might not need the "for every ideal $I$" part. It seems that if the Jacobson radical of $R$ is the nilradical of $R$, then the same holds for $R\left[x\right]$. At least, I seem to conclude this from Lemmas 2J and 3J in S. A. Amitsur, Radicals of polynomial rings, Canadian Journal of Mathematics, 8, 355--361, doi:10.4153/cjm-1956-040-9. Am I misreading Amitsur? (This still doesn't answer the question for a constructive proof, but hopefully simplifies the question.) | |
| Nov 7, 2021 at 20:51 | comment | added | Arrow | Dear @DavidESpeyer, I didn't consider such "pathologies". I am happy to restrict to ideals which we know to be finitely generated. What interests me is the "equational" approach. | |
| Nov 7, 2021 at 15:00 | comment | added | David E Speyer | It seems to me that there might be issues with what "for every ideal $I$" means in the constructive world. For example, let $R$ be a field $k$ and let $I \subset k[x]$ be generated by $x^n$ for all even integers $n \geq 4$ which are not sums of two primes. Then, if the Goldbach conjecture is true, $I = (0)$ so $x$ is in neither the nil-radical nor the Jacobson radical of $k[x]/I$ but, if the Goldbach conjecture is false, then $x$ is in both the nil-radical and the Jacobson radical of $k[x]/I$. Do you allow this sort of shenanigans? | |
| Nov 6, 2021 at 8:45 | comment | added | Arrow | Dear @PaceNielsen I am interested in both. I have edited the question. | |
| Nov 6, 2021 at 8:45 | history | edited | Arrow | CC BY-SA 4.0 |
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| Nov 5, 2021 at 14:00 | history | edited | Arrow | CC BY-SA 4.0 |
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| Nov 5, 2021 at 13:43 | history | edited | Arrow | CC BY-SA 4.0 |
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| Nov 5, 2021 at 13:34 | history | edited | Arrow | CC BY-SA 4.0 |
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| Oct 8, 2021 at 8:26 | history | edited | Arrow | CC BY-SA 4.0 |
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| Oct 8, 2021 at 8:25 | comment | added | Arrow | @AurélienDjament yes. I have added this to the question body. | |
| Oct 8, 2021 at 5:54 | comment | added | Aurélien Djament | You probably assume your ring commutative, else nilpotent elements have no reason to form an ideal. Is it what you mean? | |
| Oct 7, 2021 at 17:02 | history | asked | Arrow | CC BY-SA 4.0 |