What is the Galois group of $t^4+2$ over $\mathbb{Z}_5$ I see that $t^4+2$ is irreducible. I now want to find an extension that contains the roots of this - but I'm not sure how to do that. 
 A: You are asking for the roots of $x^4=-2=3$ over $\Bbb F_5$, and since $3$ is a primitive fourth root of unity in $\Bbb F_5$, you are asking for the smallest field that contains the sixteenth roots of unity.
In other words, you’re looking for the smallest field $\Bbb F_{5^n}$ with $16|(5^n-1)$, thus $16k=5^n-1$, thus $5^n\equiv1\pmod{16}$. The powers of $5$ are $5$, $5^2\equiv9$, $5^3\equiv-3$, $5^4\equiv-15\equiv1$.
So $\Bbb F_{5^4}=\Bbb F_{625}$ is your splitting field, of degree four over the prime field, and so your Galois group is cyclic of order four.
A: $t^4+1 = \Phi_8(t)$, hence $t^4+1$ completely splits in a finite field $\mathbb{F}_{p^k}$ as soon as $8$ divides the order of $\mathbb{F}_{p^k}^*$, i.e. $p^k-1$. In our case $p=5$, hence the splitting field of $t^4+1$ over $\mathbb{F}_5$ is $\mathbb{F}_{25}$. That also follows from the fact that over $\mathbb{F}_5$ we have
$t^4+1 = (t^2+2)(t^2+3)$ and neither $-2$ or $-3$ are quadratic residues $\!\!\pmod{5}$. However, you may notice that $t^4+1$ is reducible over $\mathbb{F}_5$, and

If $p$ is some prime, $q(t)=t^4+1$ is never an irreducible polynomial over $\mathbb{F}_p$.

About $t^4+2$: this polynomial is irreducible over $\mathbb{F}_5$, since it has no roots in $\mathbb{F}_5$ and
$$ \gcd\left(t^4+2,t^{25}-t\right) = \gcd(t^4+2,t^{24}-1)=\gcd(t^4+2,(-2)^6-1)=1.$$
It follows that the splitting field of $t^4+2$ over $\mathbb{F}_5$ is $\mathbb{F}_{5^4}$.
