# Extending homomorphism from from splitting field of a polynomial to another field where that polynomial splits into linear factor

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

As $$F$$ is minimal splitting field I guess there is an embedding of $$F$$ inside $$K$$.

I have no idea that I could write.

By definition, if $$\alpha_1,\dots,\alpha_n$$ are the roots of $$f$$ in $$F$$, then $$F=k(\alpha_1,\dots,\alpha_n)$$ is generated by them. Therefore any homomorphism to $$K$$ extending the identity on $$k$$ is uniquely determined by the images of these roots. Moreover, $$f$$ has roots in $$K$$. Hence, mapping each $$\alpha_i$$ into a root of $$f$$ in $$K$$ gives you an embedding. More precisely,
1. you should actually check that mapping $$\alpha_i$$ into a root of $$f$$ in $$K$$ is a necessary condition for $$j\colon F\to K$$ to be a homomorphism extending the identity on $$k$$.
2. Note that if $$f$$ is reducible over $$k$$ you cannot map $$\alpha_i$$ into any root of $$f$$ in $$K$$, but you have to map it into a root with same minimal polynomial over $$k$$, i.e. a root of the same irreducible factor of $$f$$ as $$\alpha_i$$. You should then check that $$j\colon F\to K$$ defined in this fashion is indeed a homomorphism extending the identity on $$k$$.
As $$K$$ contians all roots of $$f$$, the roots generate a splitting field $$E\subset K$$ of $$f$$ by definition. As splitting fields are unique up to isomorphism, the composition $$F\xrightarrow{\sim}E\subset K$$ is a desired embedding.