How much do you need to prove when finding splitting fields? When asked to find a splitting field for a polynomial over some field, how much do you need to prove ? For example, do you need to prove that the element in the extension is not in the original field? Specifically Am I going into too much detail here;  splitting field of $(x^2-3)(x^2-5)$ over $Q(\sqrt{ 2})$. Am I thinking of this correctly? ? Or can I just say for example for $x^2 − 2$ the roots are $\{±\sqrt{2}\}$
; hence, a splitting field is $\Bbb Q(\sqrt{2})$.
 A: The splitting field of a polynomial $p(x) \in F[x]$ for some field $F$ is the minimal field extension $K/F$ over which $p$ splits entirely into linear factors. So, knowing this, to prove $K$ is in fact the splitting field of $p$, you need to prove two things: all roots of $p$ lie in $K$ and any field extension $K'/F$ containing all roots of $p$ necessarily has $K \subseteq K'$. 
Note that this definition is dependent upon the field $F$. For instance, for the polynomial you gave: $p(x)= (x^2-3)(x^2-5)$, the splitting field of $p$ over $\mathbb{R}$ is precisely $\mathbb{R}$, while the splitting field over $\mathbb{Q}$ is $\mathbb{Q}( \sqrt{3}, \sqrt{5})$. 
So for your first question: "do I need to prove that some element in the extension is not in the original field?" Well, no, not necessarily, consider the above case when the field is $\mathbb{R}$. 
So for example let $p(x)= x^2-2$. If we consider $p(x) \in \mathbb{R}[x]$, a proof that the splitting field is $\mathbb{R}$ could look like this:
Note that $p(x)= (x-\sqrt{2})(x+\sqrt{2})$ and $\pm \sqrt{2} \in \mathbb{R}$, so $p$ splits over $\mathbb{R}$. Moreover, any field extension of $\mathbb{R}$ contains $\mathbb{R}$ by definition, so $\mathbb{R}$ is in fact the splitting field. 
On the other hand, let's look at $p(x) \in \mathbb{Q}[x]$:
Note that $p(x)= (x-\sqrt{2})(x+\sqrt{2})$ and $\pm \sqrt{2} \in \mathbb{Q}(\sqrt{2})$, so $p$ splits over $\mathbb{Q}(\sqrt{2})$. Moreover, suppose $K/\mathbb{Q}$ is a field extension over which $p$ splits entirely. Then $\pm \sqrt{2} \in K$ and $\mathbb{Q} \subseteq K$, so $\mathbb{Q}(\sqrt{2}) \subseteq K$. Since $K$ was arbitrary, $\mathbb{Q} (\sqrt{2})$ is the splitting field of $p$ over $\mathbb{Q}$. 
Now in this last argument, proving that $\mathbb{Q}(\sqrt{2}) \neq \mathbb{Q}$ is a different question, and this amounts to proving that $\sqrt{2}$ is not rational. In fact, I could also say the splitting field is $\mathbb{Q}(\sqrt{2}, - \sqrt{2})$. The point is that adjoining both roots of $p$ to the base field is not necessarily an incorrect answer, but it is certainly not the most efficient way of giving the splitting field and, in the above case, we may remove $-\sqrt{2}$ because this is already an element of $\mathbb{Q}(\sqrt{2})$. 
