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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] $?

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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.

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

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