Timeline for Hensel's lemma, Bezout's identity, and the integers
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 23, 2021 at 2:55 | vote | accept | Pace Nielsen | ||
| Oct 22, 2021 at 21:08 | history | edited | Pace Nielsen | CC BY-SA 4.0 |
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| Oct 22, 2021 at 19:44 | answer | added | François Brunault | timeline score: 6 | |
| Oct 22, 2021 at 15:24 | comment | added | Pace Nielsen | @KConrad Thanks! Fixed. | |
| Oct 22, 2021 at 15:23 | history | edited | Pace Nielsen | CC BY-SA 4.0 |
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| Oct 22, 2021 at 15:16 | comment | added | Pace Nielsen | @Johan You are correct! Apparently, I made an error. I'll correct my post soon. | |
| Oct 21, 2021 at 23:34 | comment | added | KConrad | You are missing a quotient symbol / in the first sentence of the second paragraph. | |
| Oct 21, 2021 at 23:18 | comment | added | Johan | $13 = (2 + 3*x + (1 + 3*x)*(1 + x^2)))*(2 - 3*x - (1 + x^2)) \bmod (1 + x^2)^2$ according to my pari/gp | |
| Oct 21, 2021 at 22:12 | comment | added | Pace Nielsen | @GerryMyerson The difference is that the smallest positive integer that is a $\mathbb{Z}[x]$-linear combination of $3+2x$ and $3-2x$ is $6$, which doesn't divide $4$. | |
| Oct 21, 2021 at 21:39 | comment | added | Gerry Myerson | What's the difference between $13=(3+2x)(3-2x)+4(1+x^2)$ and $5=(1+2x)(1-2x)+4(1+x^2)$? | |
| Oct 21, 2021 at 21:03 | comment | added | Pace Nielsen | @LucGuyot Sorry, I thought it was obvious that I intended $\deg(q)\geq 2$. I've now made that clear. The point of my question is whether at least one prime factorization lifts. There are no (nontrivial) factorizations when $q$ is linear. So of course there can't be any such factorizations if we go from working mod $q$ to mod $q^2$. | |
| Oct 21, 2021 at 21:02 | history | edited | Pace Nielsen | CC BY-SA 4.0 |
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| Oct 21, 2021 at 19:27 | comment | added | Luc Guyot | What about $\mathbb{Z}[x]/(x^2)$? | |
| Oct 21, 2021 at 17:05 | history | asked | Pace Nielsen | CC BY-SA 4.0 |