We know that $F$ is a real-closed field if $F$ is not algebraically closed but $F(\sqrt{-1})$ is algebraically closed.
So I have this question
What can we say about $\operatorname{char}F$? Is it zero? Why?
2026-04-12 15:10:54.1776006654
The characteristic of real-closed fields is zero?
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By definition a real-closed field is an ordered field. The required interaction between the arithmetic and the linear order implies character zero. I do not know whether an infinite field F with char(F)> 2 and no square root of (-1) can be algebraically completed by adjoining it to F.