Timeline for Hensel's lemma, Bezout's identity, and the integers
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 23, 2021 at 2:55 | vote | accept | Pace Nielsen | ||
| Oct 23, 2021 at 2:54 | comment | added | Pace Nielsen | John Voight says: Yes, this is a consequence of the Chebotarev density theorem, which implies that the trivial class in (the finite abelian group) Pic(R) is represented by infinitely many primes of degree 1: i.e., there are infinitely many principal primes pp = pi.R in Pic(R). Taking norms, we get Nm(pp) = p = +/- Nm(pi). We have Nm(pi) = pi.mu with mu in R from Galois theory... Adjusting signs we get p = pi.mu with pi, mu in R. We can't have pi a unit (its norm is p), and taking norms, Nm(p) = p^d = p.Nm(mu) so as long as d = deg q(x) > 1, we have that mu is also not a unit. | |
| Oct 23, 2021 at 0:46 | comment | added | Pace Nielsen | Now $a\mathcal{O}_K$ and $b\mathcal{O}_K$ are relatively prime, since $p$ is not ramified. Thus, $ax+by=1$ for some $x,y\in \mathcal{O}_K$. Hence $a(mx)+b(my)=m$ with $mx,my\in \mathcal{O}$. Thus, $\langle a,b,q\rangle_{\mathbb{Z}[x]}\supseteq \langle p,m\rangle_{\mathbb{Z}[x]}=\langle 1\rangle_{\mathbb{Z}[x]}=\mathbb{Z}[x]$. So, the only question left is whether there are primes that have nontrivial factorizations in $\mathcal{O}$ (rather than just in $\mathcal{O}_K$). I expect that is true, and well known to experts. | |
| Oct 23, 2021 at 0:42 | comment | added | Pace Nielsen | Let $K=\mathbb{Q}[x]/(q)$. Let $\mathcal{O}_{K}$ be the full ring of integers, and let $\mathcal{O}=\mathbb{Z}[x]/(q)$, which is an order in the ring of integers. Let $m=[\mathcal{O}_K:\mathcal{O}]$. Let $p\in \mathbb{Z}$ be prime, and relatively prime to $m$, and not ramified in $\mathcal{O}_K$. These conditions only disqualify finitely many primes. Assume that $p=ab$ for some (nonunits) $a,b\in \mathcal{O}$. (They remain nonunits in $\mathcal{O}_K$, since any algebraic integer unit $u$ has norm $\pm 1$, and hence has an inverse that is a $\mathbb{Z}$-linear combination of powers of $u$.) | |
| Oct 22, 2021 at 20:32 | comment | added | Pace Nielsen | Unfortunately, generally we don't have a Dedekind domain. (For instance, try $q(x)=x^2-8$.) | |
| Oct 22, 2021 at 20:17 | comment | added | François Brunault | @PaceNielsen Here, this is because $\mathcal{P}+I=\mathcal{O}$. More generally, in a Dedekind domain $R$, you just need two coprime ideals $I,J$, this implies $I+J=R$. | |
| Oct 22, 2021 at 20:15 | history | edited | François Brunault | CC BY-SA 4.0 |
added 69 characters in body
|
| Oct 22, 2021 at 20:15 | comment | added | Pace Nielsen | In general, $\mathbb{Z}[x]/(q)$ won't be the full ring of integers, but that can probably be overcome. However, I still have a question. After Johan's comment, I also was thinking about showing $\langle a,b,q\rangle = \mathbb{Z}[x]$. How do you show that equality in general? | |
| Oct 22, 2021 at 19:44 | history | answered | François Brunault | CC BY-SA 4.0 |