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

Question

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.

Answer 1

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 $k$.

Answer 2

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.