Finding the splitting field of $x^4+x^2-6$ over $\mathbb{Z_7}$ and $\mathbb{Z_{13}}$ Finding the splitting field of $x^4+x^2-6$ over $\mathbb{Z_7}$ and $\mathbb{Z_{13}}$.
So should I start off by just checking for roots? Or should I let $\alpha$ be a root and see if I need to add anything else after I add that? Either way, I'd appreciate it if anyone could lend some insight into the best ways to approach problems like this. Thanks!
 A: Hints:
$$x^4+x^2-6=(x^2+3)(x^2-2)$$
Now, is it possible to find square root to $\;-3\;$ in $\;\Bbb F_p\;,\;\;p=7,\,13\;$ ? What about square root of $\;2\;$ ....?
Once you have the above, check for possible quadratic extensions
Added under request: since (Legendre's Symbol):
$$\begin{pmatrix}-3\\7\end{pmatrix}=\begin{pmatrix}4\\7\end{pmatrix}=1$$ 
$\;-3\;$ is a square $\;\pmod 7\;$ , and in fact $\;x^2+3=(x-2)(x-5)\pmod7\;$ , and since also $\;(\pm3)^2=2\pmod 7\;$ , we get $\;x^2-2=(x-3)(x+3)\pmod7\;$ , so in fact
$$x^4+x^2-7=(x-2)(x-5)(x-3)(x+3)\pmod 7$$
so the quartic completey splits in $\;\Bbb F_7\;$ and this last is the splitting field of it (over $\;\Bbb F_7\;$) .
Over $\;\Bbb F_{11}\;$ . Since neither $\;-3\;$ nor $\;2\;$ are squares $\;\pmod{11}\;$ (check this!), we need a quadratic extension to add a root $\;w\;$ of $\;x^2+3\;$ to $\;\Bbb F_{11}\;$ :
$$K:=\Bbb F_{11}/\langle x^2+3\rangle\cong\Bbb F_{11}(w)$$ .
Knowing that there is one single extension (up to isomorphism) of any given degree for any finite field, show that $\;x^2-2\;$ also splits over $\;K\;$ . Fill in details.
