Let $F$ be a field, and let $f(x)\in F[x]$ be a polynomial of prime degree. Suppose for every field extension $K$ of $F$

Question

Let $F$ be a field, and let $f(x)\in F[x]$ be a polynomial of prime degree. Suppose for every field extension $K$ of $F$ that if $f$ has a root in $K$, then $f$ splits over $K$. Prove that either $f$ is irreducible over $F$ or $f$ has a root (and hence splits) in $F$.

Suppose $f$ is reducible over $F$ then there exists a $a$ in $F$ such that $f(x)=(x-a)g(x)$ which means that $f$ has a root in $F$. I can not understand what I am doing wrong, someone could help me please (where do I have to use the polynomial has prime degree?)? Thank you.

Answer 1

Hint: Think about the degree of each irreducible factor of $f$, by considering the degree of the field extension obtained by adjoining a root.

Stronger hint:

Let $K$ be a splitting field of $f$ over $F$. Can you prove that the degree of every irreducible factor of $f$ is $[K:F]$?

Full proof:

Let $K$ be a splitting field of $f$ over $F$. If $g$ is any irreducible factor of $f$, then $K'=F[x]/(g(x))$ is a field extension of $F$ with a root of $g$ and hence a root of $f$, and therefore $f$ splits over $K'$. Thus $K'$ is a splitting field of $f$ over $F$ and is isomorphic to $K$ over $F$. Thus $\deg g=[K':F]=[K:F]$. Since this is true of every irreducible factor of $f$, $\deg f=n[K:F]$ where $n$ is the number of irreducible factors of $f$ (counting multiplicity). Since $\deg f$ is prime, this means we either have $[K:F]=1$ (i.e., $f$ has a root in $F$) or $n=1$ (i.e., $f$ is irreducible).