# Why is any non-archimedean field Huber?

Here a non-archimedean field means a field $k$ whose topology is induced from a non-archimedean norm $| \cdot |: k \to \mathbb{R}_{\geq 0}$. As a reminder, a ring $A$ is adic if there is an ideal $I \subset A$, called the ideal of definition, such that $(I^n)_n$ forms a system of neighborhood of zero, and a ring $A$ is called Huber if there is an open adic subring $A_0 \subset A$ such that $A_0$ has a finitely generated ideal of definition.

In this case of a non-archidedean field $k$, what would $k_0$ be? I want to guess $k$ or $\mathcal{O}_k$, but can't show either is adic with finitely generated ideal of definition.

• $k_0$ can't be $k$ because a field has no nontrivial ideals. The only possibility is ${\cal O}_k$ which is clearly adic, with $I$ the maximal ideal. But $I$ is only finitely generated if the valuation is discrete, so my guess is that $k$ is Huber if and only if the valuation is discrete. Are you sure this hypothesis is not mentioned in whatever place you saw this? – Felipe Voloch Apr 15 '18 at 5:26
• I don't think this hypothesis is mentioned. I was looking at mathi.uni-heidelberg.de/~gqpaspi1geom/manuscripts/AS.pdf and in their example 1.5.(2), it says that $(k, \mathcal{o}_k)$ is a Huber pair whenever $k$ is a non-archimedean field. – Aaron Johnson Apr 15 '18 at 6:07
• Discrete or not, $k_0:=\mathscr{O}_k$ is a height 1 valuation ring, so it is adic with $I=t k_0$, where $t$ is any element with $0<\vert t\vert<1$. – Laurent Moret-Bailly Apr 15 '18 at 8:22