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This is an exercise from Nathan Jacobson's Lectures in Abstract Algebra Vol 3 from the chapter on Artin-Schreier Theory.

Let $L/F$ be algebraically closed and $F$ be formally real. Show that $L/F$ contains a real closed field $K/F$ such that $L=K(\sqrt{-1})$. In particular, show that every algebraically closed field $L$ of characteristic $0$ contains a real closed field $K$ with $L=K(\sqrt{-1}) $.

Here is my approach. Let us consider the collection $\mathcal {C} $ of all subfields in extension $L/F$ which are formally real. By using Zorn's lemma there is a maximal element in $\mathcal {C} $ which implies the existence of a formally real field $K$ with $F\subseteq K\subseteq L$ such that there is no formally real field $M$ with $K\subset M\subseteq L$.

Next we show that $K$ is real closed. If that's not the case then there is an algebraic extension $M/K$ which is formally real. But since $L$ is algebraically closed we can assume $M\subseteq L$ and this contradicts the maximal nature of $K$. So $K$ is real closed.

This technique of getting a maximal field is used in same manner by Jacobson in one of his theorems in this chapter.

Now using another theorem (by Artin, also proved in text) we conclude that $K(\sqrt{-1})$ is algebraically closed and since $\sqrt{-1}\in L$ we have $K(\sqrt{-1})\subseteq L$.

The challenge now is to show that $K(\sqrt{-1})=L$. I think the existence of an element $a\in L\setminus K(\sqrt{-1})$ should lead to the existence of some formally real field $M$ with $K\subset M\subseteq L$ and thus contradict the maximal nature of $K$. Somehow I am not able to figure out the existence of such an $M$.

I need some hints here. Or maybe the approach needs to be discarded and I need to start afresh.

Some further thoughts here. I see this exercise as a way to construct $\mathbb {R} $ given $\mathbb{C} $ and somehow this means extracting real and imaginary parts of a complex number algebraically without any idea of real numbers. This looks rather mysterious to me as we have been taught the traditional approach of creating $\mathbb {C} $ from $\mathbb {R} $ and not the reverse.

Paramanand Singh
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  • Re: the last paragraph, note that $\mathbb{C}$ has lots of non-Archimedean real closed subfields, and a maximal real closed subfield might be such. See the discussion here, especially Eric Wofsey's comments. – Noah Schweber Sep 13 '21 at 04:18
  • Not sure why this got a -1, incidentally. – Noah Schweber Sep 13 '21 at 04:38
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    @NoahSchweber: no need to worry about downvotes! Most of my questions in field theory get a downvote so I am used to it. I will read the answer you linked a bit later. Thanks for the same. – Paramanand Singh Sep 13 '21 at 05:12

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Since $a$ is transcendental over $K$, you really only have one choice $M=K(a)$ you can make. Order $M$ by "leading term": $$ \frac{\sum_{i=0}^n p_i a^i}{\sum_{j=0}^m q_j a^j}>_M0\iff \frac{p_n}{q_m}>_K0 $$ where of course $p_i,q_j\in K$, $p_nq_m\neq 0$ (so $a$ is "infinitely positive" and $1/a$ is "infinitesimally positive"). Easy to check $(M,>_M)$ is indeed an ordered field, so $M\leq L$ is formally real, contradicting maximality of $K$.

user10354138
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    Wow! This is so cool. It appears I somehow missed the fact that $a$ being outside the algebraically closed field $K(\sqrt{-1})$ must be transcendental over $K$. +1 – Paramanand Singh Sep 13 '21 at 08:20