Is the value group of an algebraically closed valued field divisible? Since Every existentially closed abelian group is divisible, I'm trying to show the value group is existentially closed but I don't know how... Even I don't know if the claim holds or not...
2026-03-25 11:20:08.1774437608
Is the value group of an algebraically closed valued field divisible?
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Yes. If $b$ is a root of $x^n-a$, then $n\cdot v(b) = v(a)$.