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Let $L/K$ be an extension of number fields, $\mathfrak p\subseteq\mathcal O_K$ be a prime ideal, $\mathfrak{q}$ a prime lying above $\mathfrak p$. Here $L_\mathfrak q/K_\mathfrak p$ is a finite extension of non-Archimedean local fields with ring of integers $\mathcal O_{L_\mathfrak q}$ and $\mathcal O_{K_\mathfrak p}$ respectively.

I have already learnt that $L_\mathfrak q/K_\mathfrak p$ is unramified if and only $L_\mathfrak q=K_\mathfrak p[\zeta_{q^n-1}]$ with $n=[L_\mathfrak q:K_\mathfrak p]$ and $q=\#\mathcal O_{K_\mathfrak p}/\mathfrak pO_{K_\mathfrak p}$.

Here the $m$-th primitive root of unity is denoted by $\zeta_m$.

I wonder whether in this case we have $\mathcal O_{L_\mathfrak q}=\mathcal O_{K_\mathfrak p}[\zeta_{q^n-1}]$ or not.

Always we have $\mathcal O_{L_\mathfrak q}=\mathcal O_{K_\mathfrak p}[\alpha]$ for some $\alpha$.

The proof of this proposition is stated as follows.

Pick $\bar\alpha\in\mathcal O_{L_\mathfrak q}/\mathfrak q\mathcal O_{L_\mathfrak q}$ as a primitive element with monic minimal polynomial $\bar f\in\mathcal O_{K_\mathfrak p}/\mathfrak p\mathcal O_{K_\mathfrak p}[X]$. We obtain $\alpha$ and $f$ such that $\alpha\equiv\bar\alpha\mod\mathfrak q\mathcal O_{L_\mathfrak q}$ and $f\equiv\bar f\mod \mathfrak p\mathcal O_{K_\mathfrak p}$. Pick $\pi\in\mathfrak q\setminus\mathfrak q^2$. We know one of $f(\alpha)$ and $f(\alpha+\pi)$ is of $\mathfrak q$-adic valuation 1 ($\mathfrak q$-adic valuation means $\nu_\mathfrak q$ here). Suppose $f(\alpha)$, and we can check $f(\alpha)^i\alpha^j$ forms an integral basis for $0\leq i<e$ and $1\leq j\leq f$ (as in the proof of local fundamental identity in Neukirch's book). Here $e$ refers to the ramification degree and $f$ refers to the inertia degree.

I want to apply the same procedure, but I have some trouble.

Kango Mochizuki
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  • Sorry, I understand how to do. This is immediate form the proof of the local fundamental identity (the construction of an integral basis). Since e=1, things are obvious. – Kango Mochizuki Apr 12 '25 at 11:16

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$\def\p{\mathfrak{p}}\def\q{\mathfrak{q}}$ Let $x$ be a $(q^n-1)$-th root of unity in $L_\q$ such that $L_\q=K_\p(x)$. Then the minimal polynomial of $x$, call it $f(t)$, divides $t^{q^n-1}-1$, which is separable mod $\p$ since $q^n-1$ is prime to the residue characteristic. Hence $f$ is also separable mod $\p$. Applying the well-known computation that the different of $\mathcal{O}_{K_\p}[x]$ is the principal ideal generated by $f'(x)$, we see that this different is the unit ideal $(1)$, since $x$ is not a root of $f'$ mod $\p$. Thus $\mathcal{O}_{L_\q}$ cannot be any bigger than $\mathcal{O}_{K_\p}[x]$, or else the different of $\mathcal{O}_{L_\q}$ would be non-integral (it would be $[\mathcal{O}_{L_\q}:\mathcal{O}_{K_\p}[x]]^{-2}$ where $[A:B]$ is the module index).

pzq_alex
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