Is the Splitting Field necessarily a subset of a field where the polynomial splits? Just a very basic theoretical question that has been puzzling me.
Let $f(x)$ be a polynomial with coefficients in a field $F$. Let $K$ be the splitting field of $f$ over $F$.
Say $f$ splits in a field extension $L$ of $F$.
Can we conclude that $K\subseteq L$?
Intuitively, I know that $K$ is supposed to be the 'smallest' field over which $f$ splits. However, I am worried that there are some elements in $K$ that are not in $L$. Can that be the case?
Thanks for enlightenment!
 A: If $L_1$, $L_2$ are two extensions of $k$ such that $f$  splits in both $L_1$ and $L_2$, and if $K_i$ are the subfields of $L_i$ generated by the roots of $f$ in $L_i$, then there exist an isomorphisms taking $K_1$ to $K_2$ and fixing $k$. So any such $K$ can be called "the" splitting field of $f$. 
$\bf{Added:}$ Inspired by  reading the exhaustive answer of @Thomas Andrews: I want to add here the following: if $k \subset L$ and $f$ splits completely in $L$, $K$ is the field generated by the roots, then $K$ is the unique subfield of $L$ that contains $k$ and is isomorphic to $L$ ( this follows from the uniqueness of the decomposition of a polynomial in linear factors). Therefore, if in an extension the polynomial splits completely, the splitting field "is in there", and is unique.  
A: Let $p(x)\in F[x]$ be your polynomial.
if $K$ and $L$ are subfields of the same field, so $F\subseteq K,L\subseteq F_2$, then yes, $K$ is a subfield of $L$. This is true because the polynomial cannot have more than $\deg p$ roots, so all roots in $F_2$ have to be in $K$, and therefore $K$ must be in $L$. (You have to deal with repeated roots here, but it is not hard to fix that argument. The splitting field for $p$ is the same as the splitting field for a $q$ which does not have repeated roots.)
You might take $F_2=\overline{F},$ the algebraic closure of $F$, for example. Then given a field $L$ with $F\subseteq L\subseteq \overline{F}$ in which your polynomial splits, we know $L$ contains the splitting field.
However, depending on how splitting fields are defined, you might not have such an $F_2$. Then all you know is that $K$ is isomorphic to a subfield of $L$. The image of $K$ in $L$ is always the same, but the isomorphism is usually not unique. 
