Is $\mathbb{Q}(\sqrt{5}, \alpha)$ over $\mathbb{Q}(\alpha)$, where $\alpha$ is the real seventh root of $5$ a the splitting field? Is $\mathbb{Q}(\sqrt{5}, \alpha)$ over $\mathbb{Q}(\alpha)$, where $\alpha$ is the real seventh root of $5$ a the splitting field ?
I am claim that it is not. My reasoning is this...
What I am thinking is that the minimal polynomial of $\mathbb{Q}(\sqrt{5},\alpha)$ is $f(x)=(x^2-5)(x^7-5)$ over $\mathbb{Q}(\alpha)$. However $\mathbb{Q}(\sqrt{5},\alpha)$ doesn't contain the imaginary zeros of $x^7-5$. 
Is there a formal way of showing this?
 A: Since this field $K$ is contained in $\mathbb{R}$, it does not contain one root, $\beta$ of the polynomial $x^7 - 5$. There is a homomorphism from $K$ to $\mathbb{C}$ which sends $\alpha$ to $\beta$ and fixes $\sqrt{5}$.
This homomorphism does not map $K$ to itself, so $K$ is not a normal extension of $\mathbb{Q}$. Hence it is not a splitting field.
A: If $k$ is a field, then $K\supset k$ is a splitting field over $k$ if there is a $k$-polynomial $f\in k[X]$ such that $K$ is the field gotten from $k$ by adjoining all roots of $f$. In my world, it does not make sense to say “$K$ is a splitting field” unless you mean for $K$ to be a splitting field over $\Bbb Q$. To give an explicit example, $\Bbb C$ is a splitting field over $\Bbb R$, but $\Bbb C$ is not a splitting field over any smaller field than $\Bbb R$, because $\Bbb R$ has no subfields $k$ with $[\Bbb R:k]<\infty$.
For this reason, I say that as you seem to have stated the problem, $\Bbb Q(\sqrt5,\alpha)$ certainly is a splitting field over $\Bbb Q(\alpha)$, since the extension is quadratic.
