# Why the equality $\mathcal O_L/\mathfrak p\mathcal O_L=(\mathcal O_K/\mathfrak p)[\overline\alpha]$ holds?

$$\newcommand{\O}{\mathcal O}$$ $$\newcommand{\p}{\mathfrak p}$$ $$\newcommand{\a}{\alpha}$$ $$\newcommand{\N}{\operatorname{N}_{L/K}}$$ $$\newcommand{\ol}{\overline}$$ 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?

• 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 $\mathfrak{p} O_K[\alpha]$. – Watson Oct 23 '18 at 14:37
• @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?) – josephz Oct 24 '18 at 5:31
• 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). – Watson Oct 24 '18 at 6:41
• 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). – Watson Nov 2 '18 at 20:14