# Given a valuation on a field , does it always extend to a valuation on every extension field

Let $$(G,+,\ge)$$ be a totally ordered abelian group. Let $$K\subseteq L$$ be an extension of fields. Let $$v : K\setminus \{0\} \to G$$ be a valuation (https://en.wikipedia.org/wiki/Valuation_(algebra)) .

Then does there exist a valuation $$\bar v : L \setminus \{0\} \to G$$ such that $$\bar v|_{K \setminus \{0\}}=v$$ ?

• Not in general; you have to allow extensions of $G$. – Lord Shark the Unknown Jan 7 at 4:43
• If $a \in L$ let $f(X) = \prod_{j=1}^n (X-a_j) \in K[X]$ its minimal polynomial then $v(a) = \frac{v(f(0))}{n}$ is a valuation on $L$. – reuns Jan 7 at 7:06

No, not even for finite extensions. For example, consider a field $$K$$ with a valuation $$v : K \setminus \lbrace 0 \rbrace \twoheadrightarrow \mathbb{Z}$$. Let $$a \in K$$ be such that $$v(a) = 1$$ and consider the quadratic extension $$L = K[\sqrt{a}]$$. Then $$v$$ cannot extend to a valuation $$\bar{v} : L \setminus \lbrace 0 \rbrace \to \mathbb{Z}$$, because then what would $$v(\sqrt{a})$$ be?
Concrete example: $$K = \mathbb{Q}$$, $$p$$ a prime number, $$v_p$$ the $$p$$-adic valuation, $$a = p$$.
However, if you also allow extensions of $$G$$, then the answer is always positive (assuming Axiom of Choice), this is Chevalley's Extension Theorem.