By an absolute value on a field, I mean this.
I noticed that many fields of characteristic $0$ have infinitely many non-equivalent absolute values. For fields algebraic over $\mathbb Q$, every absolute value of $\mathbb Q$ extends to at least one and, of course, extensions of non-equivalent ones are non-equivalent. Purely transcendental extensions of arbitrary fields have infinitely many absolute values anyway.
But what happens in between?
Does every field of characteristic $0$ have infinitely many non-equivalent absolute values?
A possible counterexample that comes to my mind is $(\prod_{p \text{ prime}} \mathbb F_p) /\mathcal U$, an ultraproduct over all finite prime fields.
Thank you for your help!
Best Answer
Let $F$ be a field of characteristic $0$, so $F$ contains $\mathbf Q$. If we can show for each prime $p$ that the $p$-adic absolute value on $\mathbf Q$ extends to an absolute value on $F$ in at least one way, then $F$ has infinitely many inequivalent absolute values. The following general theorem takes care of this.
Theorem. If $K$ is a field and $L$ is an arbitrary extension field of $K$, then each non-Archimedean absolute value on $K$ extends to an absolute value on $L$.
I am not claiming the extension of the absolute value from $K$ to $L$ is unique at all. That's already apparent in concrete cases like extending $|\cdot|_5$ from $\mathbf Q$ to $\mathbf Q(i)$, which can be done in two ways using $|\cdot|_{1+2i}$ and $|\cdot|_{1-2i}$.
Proof. By Zorn's lemma and transcendence bases, it suffices to show each non-Archimedean value on a field $M$ can be extended to an absolute value on (i) all finite extensions of $M$ and (ii) the purely transcendental extension $M(x)$. The first case is treated in books on valuation theory. For the second case, when $|\cdot|$ is a non-Archimedean absolute value on $M$, extend it to $M[x]$ by $|\sum_{i=0}^n a_ix^i| = \max |a_i|$ for $a_i \in M$. This turns out to be a non-Archimedean absolute value on $M[x]$. The only subtle issue is multiplicativity of $|\cdot|$ on $M[x]$, and that property is proved as Theorem $8$ here in the case $M = \mathbf Q$, with the reasoning there going through in the general case. This absolute value on $M[x]$ restricts to the original absolute value on $M$ (the constants of $M[x]$). This non-Archimedean absolute value on $M[x]$ extend to an absolute value on the fraction field $M(x)$ by multiplicativity: $|f(x)/g(x)| := |f(x)|/|g(x)|$ for $f(x)$ and $g(x)$ in $M[x]$.