I would like to verify the validity of the following line of thought:
Let $K \subset E$ be a field extension. Let $K[x]$ be the polynomial ring over $x$ and denote $K(x)$ its field of quotients (the rational functions over $K$).
Now assume $f(x)=g(x)h(x)$ with $f(x),g(x) \in K[x]$ and $h(x) \in E[x]$ and all $f$, $g$, $h$ are different than zero. Then $\frac{f(x)}{g(x)} = h(x)$ in $E(x)$. So $h(x) \in K(x)$. But $h(x)$ is a polynomial, so $h(x) \in K[x]$.
Is there any other simpler argument than that?
Added:
Let me be more specific. Let $x^p - a \in K[x]$, with $p$ being prime and $char(K)=p$. Let $b$ be the unique root of $x^p - a$ in its splitting field $K(b)$. Then $x^p - a=(x-b)^p \in K(b)[x].$ Let $m_b(x) \in K[x]$ be the minimal polynomial of $b$ over $K$. Then $m_b(x)$ divides $(x-b)^p$ and so $m_b(x)=(x-b)^l$ and $l$ is the smallest positive integer for which $(x-b)^l \in K[x]$. Let $p = lq+r$ where $0 \ge
Thanks.