Suppose that we have irreducible polynomials over a field, $f,g\in F[x]$, and that $g$ divides $f$ in $L[x]$ where $L$ is a field containing splitting fields of $f$ and $g$ over $F$. Can we say that $g$ divides $f$ in $F[x]$?
I've got this question from the following proof: "If there is finite group $G\subseteq \text{Aut }E$ satisfying $F=\text{Inv}(G)$ then $E/F$ is a Galois' extension."
Considerations:
- $f\in F[x]$ irreducible;
- $r$ such that $f(r)=0$;
- $\mathcal{O}_r=\left\{\sigma(r):\: \sigma\in G\right\}=\left\{r_1,...,r_m\right\}$;
- $g=\prod (x-r_i)$.
The author concludes that $g$ divides $f$ in $F[x]$ immediately after showing that $g\in F[x]$. But I don't consider that so trivial. I proved that it's true because if $f=gh$ for some $h\in L[x]$ then $\widehat{\sigma}(h)=h$ for any $\sigma\in G$ where $\widehat{\sigma}\:\colon E[x]\to E[x]$ is the unique ring homomorphism mapping $a\mapsto \sigma(a)$ and $x\mapsto x$, hence $h\in F[x]$.
Is there any forward way?
One way is the following. Remember polynomial division. It works over any field $K$.
So if $a(x),b(x)\in K[x]$, $b(x)\neq0$, then there exists unique polynomials $q(x),r(x)\in K[x]$ such that
At the point you reached you know that when $K=L$ you have $r(x)=0$. Possibly I should temporarily write $r_L(x)=0$. Because $F\subseteq L$, whichever solution $(q_F(x),r_F(x))$ worked over $F$, will also work over $L$. Therefore $r_F(x)$ must be equal to $r_L(x)$, and the claim follows.
A consequence of this is that Euclid's algorithm runs with exactly the same intermediate steps in $F[x]$ and $L[x]$ for any pair of inputs $f(x),g(x)\in F[x]$. In other words, $\gcd(f(x),g(x))$ means the same thing in both $F[x]$ and $L[x]$.