$\newcommand{\O}{\mathcal O}$ $\newcommand{\p}{\mathfrak p}$ $\newcommand{\a}{\alpha}$ $\newcommand{\N}{\operatorname{N}_{L/K}}$ $\newcommand{\ol}{\overline}$ I saw such an argument in a note of algebraic number theory (say $L/K$ is an extension of number fields and $\O_L$, $\O_K$ are their rings of integers)

If $\p$ is a prime in $\O_K$ and $\a\in\O_L$ satisfies $\p\not\mid\N(f'(\a))$, then $\O_L/\p\O_L=(\O_K/\p)[\ol\a]$.

Here $f$ is the minimal polynomial of $\a$. This argument is left unproved as a remark, but for me it is not really so obvious. I know the inclusion $\O_K\hookrightarrow\O_L$ induces an inclusion $\O_K/\p\hookrightarrow\O_L/\p\O_L$ and $\O_K/\p$ is a field since $\O_K$ is Dedekind. From $\p\not\mid\N(f'(\a))$ we have $\ol{\operatorname{Disc}(1,\a,\cdots,\a^{n-1})}\neq 0$ ($\mathrm{Disc}$ means the discriminant). I feel that the case is a bit similar to the case of field extensions, that is, if $\O_L/\p\O_L$ is also a field, then $\operatorname{Disc}(1,\ol\a,\cdots,\ol\a^{n-1})\neq 0$ and the argument holds. I wonder how I can see the general case? Is there any way to reduce the general case to the case of them all being fields?

Thanks in advance.

  • 1
    $\begingroup$ This should be explained on page 48 of Neukirch's Algebraic number theory (possibly using this question to ensure that $\mathfrak p$ doesn't divide the discriminant of $O_K[\alpha]$). One has to show that $O_K[\alpha] \to O_L / \mathfrak{p} O_L$ is surjective of kernel $\mathfrak{p} O_K[\alpha]$. $\endgroup$ – Watson Oct 23 '18 at 14:37
  • $\begingroup$ @Watson I am sorry but, in Neukirch's book it seems that an essential argument is $\mathfrak p\mathcal O_L+\mathcal O_K[\alpha]=\mathcal O_L$ and the book used a concept of 'conductor'. How does this relate to the condition $\mathfrak p\not\mid\mathrm N_{L/K}(f'(\alpha))$, or that $\mathfrak p$ does not divide the discriminant of $\mathcal O_K[\alpha]$ (and what is the discriminant of it? Is $O_K[\alpha]$ a fractional ideal?) $\endgroup$ – josephz Oct 24 '18 at 5:31
  • $\begingroup$ As I wrote in my comment, I think that you can deal with this issue possibly using this question to ensure that $\mathfrak p$ doesn't divide the discriminant of $O_K[\alpha]$ (I haven't checked the details though). $\endgroup$ – Watson Oct 24 '18 at 6:41
  • 1
    $\begingroup$ Following proposition 2.12 in Neukirch and 2.3.6 here, we have $$N_{L/K}(f'(a)) = disc(1, a, ..., a^{n-1}) = [O_L : O_K[a]]^2 disc(O_K),$$ so if $\mathfrak p$ doesn't divide $N_{L/K}(f'(a))$, it doesn't divide the index $r := [O_L : O_K[a]]$, thus $\mathfrak p$ is coprime to $r O_K[a]$. Since $r O_K[a]$ is contained in the conductor $\mathfrak F$, we get $\mathfrak p O_L + \mathfrak F = O_L$, as needed in Neukirch (page 48). $\endgroup$ – Watson Nov 2 '18 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.