Here is my books definition of a splitting field:
Note that it uses the word: smallest:
In the last converse part of this theorem. I see that the field E created is a field that contains F(this is by hypothesis of the theorem). And it also contains all the zeros of the given polynomials. But I do not see why it is the smallest such field? I mean, how is it ensured that we do not get any "extra" elements in our field?
The proof is from "A first course in Abstract Algebra", by John B. Fraleigh, seventh edition, chapter 50 "Splitting fields".
As in the proof, consider the set $S$ of all irreducoble polynoials having at least one zero in $E$. Assume the splitting field $K$ of this set of polynomials is strictly smaller than $E$. Let $\alpha\in E\setminus K$. As $\alpha$ is algebraic over $F$, there exists a polynomial $f\in F[x]$ with $f(\alpha)=0$. WE may assume wlog that $f$ is irreducible. Then $f\in S$ and hence all its roots - including $\alpha$ - are in $K$, contradiction.