Let $E/F$ be algebraic extension, $f(x)\in E[x]$, then $\exists g(x)\in E[x]$ s.t. $f(x)g(x)\in F[x]$

Question

Let $E/F$ be algebraic extension, $f(x)\in E[x]$, then $\exists g(x)\in E[x]$ s.t. $f(x)g(x)\in F[x]$

This is a problem in my test of Galois theory course (teaching field extensions now), but I totally have no idea. Could you please give me some hints?

Answer 1

Let $\alpha_1,\ldots,\alpha_n$ be the roots of $f$ listed possibly with repeats. Let $m_i(x)$ be the minimal polynomials of the $\alpha_i$ over $F$. Then $f(x)|M(x) = m_1(x)\cdots m_n(x)\in F[x]$. Now let $p_{ij}(x)\in E[x]$ be the irreducible factors of $m_i(x)$ in $E[x]$. Then by unique factorization there is a subset of the $p_{ij}(x)$, $S=\{p_{ij}(x)\}_{i\in I, j\in J}$ so that

$$f(x) = \prod_{p_{ij}\in S}p_{ij}(x)$$

So let $\displaystyle g(x) =\prod_{p_{ij}\not\in S}p_{ij}(x)$ and $f(x)g(x)=M(x)$ as desired.

Answer 2

Here is another proof. Let $E\subseteq L$ be such that $L/F$ is Galois. Consider all $\sigma\in Gal(L/F)$.

Then $$\prod_{\sigma} \sigma(f(x)) \in F[x]$$

so let $$g(x)=\prod_{\sigma\neq 1} \sigma(f(x)).$$