The definition of henselian is: A valued field $(\mathbb{K},\nu)$ is said to be Henselian if for any algebraic extension $\mathbb{L}$ of $\mathbb{K}$ there is a unique valuation $\tilde{\nu}$ on $\mathbb{L}$ such that $\mathcal{O}_{\mathbb{L}} \cap \mathbb{K} = \mathcal{O}_{\mathbb{K}}$.

But isn't it always true for any valuation v on an algebraic extension L of K that $\mathcal{O}_{\mathbb{L}} \cap \mathbb{K} = \mathcal{O}_{\mathbb{K}}$ ?

If that is the case, would it mean that there is only one possible valuation on an algebraic extension L of an henselian field K?

  • $\begingroup$ $O_K=\{ x \in K, v(x) \ge 0\}$ is a ring means the valuation is non-archimedian then $v$ extends uniquely to finite extensions as $v(\alpha)=\frac{1}{\deg(f)}v(f(0))$ with $f$ the minimal monic polynomial of $\alpha$. If $K$ is complete, looking at $(1+\pi O_K)^{1/n}$ isn't there is only one non-archimedian valuation on it ? $\endgroup$
    – reuns
    Feb 24 '19 at 7:14
  • 1
    $\begingroup$ The valuation on $\mathbb{L}$ may have nothing to do with the valuation on $\mathbb{K}$, for instance take $\mathbb{L}=\mathbb{K}$ with a non equivalent valuation. $\endgroup$
    – nombre
    Feb 24 '19 at 14:24

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.