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] $?
Best Answer
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: