Let $f(x) = x^2 + 1 \in \Bbb Q[x]$. Since $x^2+1 = (x + \sqrt{-1})(x-\sqrt{-1})$, we see that $f(x)$ splits in $\Bbb C$, but a splitting field over $\Bbb Q$ is $\Bbb Q(i) = \{r + si : r,s, \in \Bbb Q\}$.
Why is this the splitting field? What makes this the smallest field extension that $f(x)$ splits in? I can't figure why this is.
Let $K$ be an extension of $\Bbb Q$ (with $j : \Bbb Q \to K$) such that $X^2+1$ splits over $K$, say $X^2+1=(X-a)(X-b)$ with $a,b \in K$. We know that $b=-a$ since $(-a)^2=a^2=1$.
Let $K' := j(\Bbb Q) \subset K$, which is isomorphic to $\Bbb Q$ (since $j$ is injective). Then there is a (unique) field morphism
$$\phi : \Bbb Q(i) \subset \Bbb C \longrightarrow K'(a,b)=K'(a) \subset K$$ which extends $j$ and such that $\phi(i) = a$.
By injectivity of $\phi$ (it is a field morphism), we get that $K$ is an extension of $\Bbb Q(i)$. Moreover, $X^2+1$ clearly splits over $\Bbb Q(i)$.
This shows that $L:=\Bbb Q(i)$ is the splitting field of $X^2+1$ over $\Bbb Q$, because $X^2+1$ splits over $L$, and if $X^2+1$ splits over some $K$, then $K$ is an extension of $L$.