# Are Valuations on algebraic extensions of an henselian field unique?

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?

• $O_K=\{ x \in K, v(x) \ge 0\}$ 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 ? Feb 24 '19 at 7:14
• 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. Feb 24 '19 at 14:24