# Homomorphism of splitting field to its closure

Let $$k$$ be a field, $$f(x)\in k[x]$$, and let $$F$$ be the splitting field of $$f(x)$$ over $$k$$. Let $$k\subseteq K$$ be an extension such that $$f(x)$$ splits as a product of linear factors over $$K$$. Prove that there is a homomorphism $$F\to K$$ extending the identity on $$k$$.

This is a question from Aluffi's Algebra Chapter 0.

It looks to me as if here, $$K$$ is just the algebraic closure of $$F$$. Thus, a homomorphism would just be the inclusion. Is this all there is to it or am I missing something?

• $K$ is just any extension of $k$ in which the polynomial splits. It need not be the algebraic closure of $F$. For example, you could have $k=\mathbb{Q}$, $f(x) = x^2-2$, and $K=\mathbb{Q}[\sqrt[4]{2}]$. $K$ is not the algebraic closure of $F$. Or it could be a field that does not obviosuly include $\mathbb{Q}[\sqrt{2}]$. For example, it could be $\mathbb{Q}[x]/(x^4-10x^2+1,x^2-3)$. – Arturo Magidin Apr 11 at 22:48
• @ArturoMagidin I see your point. So it is not necessary that $F \subseteq K$, correct? How else can I construct a homomorphism, or in other words, where can I map the roots of $f$ to in $K$? – blanchey Apr 11 at 23:02
• Well, the polynomial $f(x)$ is supposed to split in $K$, so there are some natural candidates... – Arturo Magidin Apr 12 at 1:43