# Doubt in proof of "if $F$ is a real closed field then $F(\sqrt{-1})$ is algebraically closed".

I am studying properties of real closed fields from Lectures in Abstract Algebra, Vol 3 by Nathan Jacobson. He proves the following theorem :

Theorem: Let $$F$$ be an ordered field such that positive members of $$F$$ have square root in $$F$$ and every polynomial of odd degree in $$F[x]$$ has a root in $$F$$. Then $$-1$$ has no square roots in $$F$$ and $$F(\sqrt{-1})$$ is algebraically closed.

The key idea of the proof is by Gauss where it is shown that quadratic polynomials in $$K[x]$$ where $$K=F(\sqrt{-1})$$ have roots in $$K$$ so that there is no extension $$L$$ of $$K$$ of degree $$2$$.

Jacobson next shows that if $$f(x) \in F[x]$$ is of positive degree then $$f$$ has a root in $$K$$ (this is sufficient to prove that $$K$$ is algebraically closed). To do so he considers the polynomial $$g(x) =(x^2+1)f(x)$$ and its splitting field $$E$$ over $$F$$. Also it can be assumed that $$E\supseteq K$$. Further argument is based on studying the Galois group of $$E$$ over $$F$$ and it is deduced that $$E$$ of degree $$2$$ over $$F$$.

My doubt (which may be trivial) is over choice of polynomial $$g(x)$$. Why can't we instead study the splitting field of the polynomial $$f(x)\in F[x]$$ itself? Is it only to justify the assumption $$E\supseteq K$$ or something else? Can we instead work without $$g(x)$$ and study the splitting field of polynomial $$f(x)$$ as a polynomial in $$K[x]$$?

I am almost positive that your hunch is correct. The extra factor $$x^2+1$$ is there simply to make sure that we can think of the splitting field as an extension of $$K$$. A convenient way of including $$\sqrt{-1}$$.

My copy of Jacobson's Basic Algebra I is in my office (IIRC published after Lectures in Abstract Algebra), so I cannot check whether he later edited the proof.

An alternative way of organizing the proof, based on exact same ideas, would be to take an irreducible polynomial $$g(x)\in K[x]$$. Then consider the polynomial $$f(x)=g(x)\overline{g}(x)\in F[x]$$, where $$z\mapsto\overline{z}$$ is the obvious $$F$$-automorphism of $$K$$. Then proceed along the same route:

• Let $$L$$ be the splitting field of $$f$$ over $$F$$.
• Because $$f$$ is separable $$L/F$$ is Galois. Let $$G$$ be the Galois group, and let $$P\le G$$ be a Sylow $$2$$-subgroup.
• Let $$M$$ be the fixed field of $$P$$. Because $$M/F$$ is simple and $$[M:F]$$ is odd, we can conclude that we must have $$M=F$$ and, consequently $$G=P$$.
• Let $$P_m=\{1\}\unlhd P_{m-1}\unlhd\cdots\unlhd P_2\unlhd P_1\unlhd P_0=P$$ be the decomposition series. By basic properties of $$p$$-groups $$[P_{i-1}:P_i]=2$$ for all $$i$$.
• The fixed field of $$P_1$$ is a quadratic extension of $$F$$, and the quadratic formula shows that it is isomorphic to $$K$$. So we can identify it with $$K$$.
• The earlier lemma showed that $$K$$ has no quadratic extensions so $$P_2$$ cannot exist, implying that $$L\simeq_F K$$.

The way the above outline reintroduces $$K$$ as the fixed field of $$P_1$$ is not very elegant. We should justify that this reintroduction doesn't meddle with the polynomial we started with! Proving that $$[L:F]=2$$ is one way, and there are probably alternative ways of making the desired conclusions, and I may have missed the simplest way. But having that extra factor $$(x^2+1)$$ takes care of such issues.

• Thanks Jyrki for prompting me to study this interesting topic! I feel relieved to know that my hunch is correct. +1 and accept. Dec 23, 2018 at 12:23