Splitting Field over a Field

This is indeed an interesting question. And, yes, there is an intrinsic (or conceptual) characterization of splitting fields of polynomials: they are called normal extensions.

More precisely, given a finite dimensional field extension $F\subset K$ there exists a polynomial $P(X)\in F[X]$ such that $K$ is the splitting field $K=\text{Split}_F(P)$ of $P$ iff the extension $F\subset K$ is normal.
Of course I have to tell you what a normal extension is!

Definition An algebraic extension $F\subset K$ is normal iff every irreducible polynomial $f(X)\in F[X]$ which has a root in $K$ actually is a product of linear factors in $K[X]$.

You then have the generalization of the equivalence above to algebraic but not necessarily finite dimensional extensions $F\subset K$:
$K$ is normal over $F$ iff $K$ is the splitting field $K=\text{Split}_F((P_i)_{i\in I})$ of a (maybe infinite) family $(P_i)_{i\in I}$ of polynomials $P_i(X)\in F(X)$.

For example an algebraic closure $F^{alg}$ of a completely arbitrary field $F$ is clearly normal over $F$ and it is just as clearly a splitting field for the family of all polynomials in $F[X]$ !

Well $\mathbb{C}$, as you may know, is algebraicaly closed. So, every polynomial with coefficients in $\mathbb{R}$ splits in $\mathbb{C}$. Is $\mathbb{C}$ the splitting field of a particular polynomial over $\mathbb{R}$? Sure, just take the polynomial $x^2 +1$ in $\mathbb{R}[x]$, and $\mathbb{C}$ will be the splitting field of this polynomial.

The same cannot be said of $\mathbb{R}$ over $\mathbb{Q}$, because if $\mathbb{R}$ were the splitting field of a particular polynomial with coefficients in $\mathbb{Q}$, then $\mathbb{R}$ would have to be algebraic over $\mathbb{Q}$ (in fact, it would have to be a finite dimensional vector space over $\mathbb{Q}$), but it is not because elements like $\pi$ are not algebraic over $\mathbb{Q}$.