My following task is:
Let $\mathbb{K}$ be a field und be $\mathbb{K}^{\prime}$ a subfield of $\mathbb{K} . \mathbb{K}[t]$ and $\mathbb{K}^{\prime}[t]$ are polynomial rings over the respective fields in an indefinite. Prove: $f, g \in \mathbb{K}^{\prime}[t]$ and $q \in \mathbb{K}[t]$ with $f=q g,$ then $q \in \mathbb{K}^{\prime}[t]$ follows.
My idea to prove this is:
A subfield is a subset of a field and it always contains the elements {0,1}. It must also be completed in respect of (+,-,*,:). Furthermore, there must be a reciprocal value for each element.
If f,g are elements of K', f,g are also elements of K. If now f = q*g and g={0,1} (because the 0/1 must be in the subfield), then f=q. And since f is in K', q must also be in K'.
Could someone tell me if this is correct and if not, what is missing or wrong, please. Thank you.
How about this:
We assume that $g\neq 0$, since otherwise the question becomes trivial.
The polynomial ring $\mathbb{K}^{\prime}[t]$ is a Euclidean domain, which means it has a division algorithm: there exist $q_0, r\in \mathbb{K}^{\prime}[t]$ such that $f = q_0 g + r$, with $\deg r < \deg g$. We are given that $f = q g$ with $q\in \mathbb{K}[t]$, so combining these equalities yields $$ q_0 g + r = q g. $$ In particular, $g$ divides $r$ in $\mathbb{K}[t]$. Since $\deg r < \deg g$, we must have $r = 0$. Therefore $q_0 g = q g$. Since $g\neq 0$, we have $q = q_0\in \mathbb{K}^{\prime}[t]$.