Proving Divisibility of Polynomials in Field Extensions

[mathmari]

Hey!

Let $K/F$ be a field extension, $f,g\in F[X]$. I want to show that if $g\mid f$ in $K[X]$, then $g\mid f$ also in $F[X]$.

Suppose that $g\mid f$ in $K[X]$. Then $f=g\cdot h$, where $h\in K[X]$. We have to show that $h\in F[X]$.
Could you give me some hints how we could show that? (Wondering)

 

[caffeinemachine]

Hey!

Let $K/F$ be a field extension, $f,g\in F[X]$. I want to show that if $g\mid f$ in $K[X]$, then $g\mid f$ also in $F[X]$.

Suppose that $g\mid f$ in $K[X]$. Then $f=g\cdot h$, where $h\in K[X]$. We have to show that $h\in F[X]$.
Could you give me some hints how we could show that? (Wondering)

Hint: Use division algorithm for the polynomials $f$ and $g$ in $F[x]$.

 

[mathmari]

Hint: Use division algorithm for the polynomials $f$ and $g$ in $F[x]$.

Applying the division algorithm for the polynomials $f$ and $g$ in $F[x]$, we have $$f=gq+r$$ where $q,r\in F[x]$ with $\deg g>\deg r$.

We have that $g\mid f$ in $K[x]$, so $f=gh$, where $h\in K[x]$.

Therefore, we have the following:
$$r=f-gq=gh-gq=g(h-q)$$
We have that $g\neq 0$.
So, if $r\neq 0$ then $h-q\neq 0$. We have that $$\deg r=\deg (g(h-q))=\deg g+\deg (h-q)>\deg r+\deg (h-q) \Rightarrow \deg (h-q)<0$$ a contradiction.

Therefore, it must be $r=0$, i.e., $f=gq$ in $F[x]$, i.e., $g\mid f$ in $F[x]$.

Is everything correct? Could I improve something? (Wondering)