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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 $\bar v|_{K \setminus \{0\}}=v$ ?

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    $\begingroup$ Not in general; you have to allow extensions of $G$. $\endgroup$ – Lord Shark the Unknown Jan 7 at 4:43
  • $\begingroup$ 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 $L$. $\endgroup$ – reuns Jan 7 at 7:06
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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.

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