When I was considering extensions of non-Archimedean valuation $|\cdot| : K \rightarrow \mathbb{R}$, I came across the problem :

If $L/K$ is an algebraic field extension and $R$ is a valuation ring of $K$, then for $\alpha \in L$, does $N_{K(\alpha)/K}(\alpha) \in R$ imply that $\alpha$ is integral over $R$ ?


1 Answer 1


The question is a bit old, but I'll try to answer it anyway. I assume in addition that $K$ is complete, and I think by "$R$ is a valuation ring of $K$" you mean $R = \{x \in K \mid |x| \leq 1\}$. Then the answer is yes:

Let $p \in K[X]$ be the minimal polynomial of $\alpha$, then we know that

$$N_{K(x)|K}(\alpha) = (-1)^{[K(x):K]}\cdot p(0),$$

hence from $N_{K(x)|K}(\alpha) \in R$ it follows that $|p(0)| \leq 1$. This implies that $p \in R[X]$ by Hensel's Lemma and $x$ is integral over $R$.

Short proof for why $|p(0)| \leq 1 \implies p \in R[X]$:

The minimal polynomial $p$ is irreducible and monic. If $|.|$ is trivial, there is nothing to show. So suppose the valuation is non-trivial and we suppose that $p \notin R[X]$. As $|.|$ non-trivial we find $\lambda \in K^*$ s.t. $g := \lambda \cdot p \in R[X]$ where the constant coefficient of $g$ has valuation $<1$. Hence the reduction $\tilde g$ of $g$ to the residue field $\mathcal k$ (as $R$ is local ring) yields decomposition $$\tilde g = \tilde h \cdot X^r \in \mathcal k[X],$$ for $r > 0$ and $\tilde h$ coprime to $X$. Hensel's Lemma lifts the decomposition to $g = h\cdot q$ with corresponding reductions and $\deg(q) = r$. As $f$ was monic, get $\deg(\tilde g) < \deg(g)$ and with $r > 0$ we obtain

$$\deg(q) = r \leq \deg(\tilde g) < \deg(g) = \deg(p),$$ in particular $p = \lambda^{-1}hq$ with $\deg(p) > \deg(q) > 0$ in contradiction to irreducibility of $p$.


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