Proposition 9

Let K be a field, L an algebraic extension of K and $\mathfrak{f}$ an absolute value on K. Then $\mathfrak{f}$ can be extended to an absolute value on L.

Suppose first that there exists a valuation $v$ on $K$ with real values such that $f(x) = e^{-v(x)}$. There exists a valuation $v'$ on $L$ whose restriction to $K$ is equivalent to $v$ ($§ 3$, no. 3, Proposition 5). Then $v'$ is of height 0 or 1 (no. 1, Corollary 2 to Proposition 1) and therefore may be assumed to have real values. The restriction of the mapping $x \mapsto e^{-v'(x)}$ to $K$ is an absolute value equivalent to $f$ and hence of the form $f^s$ with $s > 0$ (General Topology, Chapter IX, $§ 3$, no. 2, Proposition 5). We conclude that

is an absolute value on $L$ extending $f$.

Suppose now that $f$ is not ultrametric. Then $K$ is identified with a subfield of $\mathbf{C}$ such that $f(x) = |x|^s$ where $0 \leq s \leq 1$ ($§ 6$, no. 4, Theorem 2). As $\mathbf{C}$ is algebraically closed, $L$ is identified with a subfield of $\mathbf{C}$ and the absolute value $x \mapsto |x|^s$ extends $f$.