Compute the Galois group over $F_{101}$ The problem is as follows:
Determine the Galois group of the polynomial $f(x)=x^4-2$ over the finite field with $101$ elements, $\mathbb{F}_{101}$. 
I am not really sure how to go about this, but here are some things I've thought about: 
So, I know that Finite fields have cyclic Galois group. So, if $f(x)$ were irreducible over $\mathbb{F}_{101}$ then it would follow that it's splitting field is $\mathbb{F}_{101^4}$, and hence its Galois group would by isomorphic to $\mathbb{Z}_4.$ Similarly, if it reduced into a linear and a cubic, then the Galois group would be isomorphic to $\mathbb{Z}_3,$ and so on..
But in this case, it doesn't seem like it is the intention of the problem to check how $f(x)$ factors over $\mathbb{F}_{101}$. There seem to be way too many things that could happen. If it were over a field $\mathbb{F}_p$ for a reasonably small prime $p$ then one could check the irreducibility of $f(x)$ over this field. 
The roots of $f(x)$ are certainly $\sqrt[4]{2}\zeta_4^k$, for $k=0,1,2,3,$ where $\zeta_4$ is the $4th$ primitive root of unity, so it is a root of the cyclotomic polynoial $\Phi_4(x)$. This in turn is a factor of $x^{101}-x$. So, in particular, it splits completely over $\mathbb{F}_{101}$, since this last one is the splitting field of $x^{101}-x.$ 
Hence, the splitting field of $f(x)$ over $\mathbb{F}_{101}$ is $\mathbb{F}_{101}(\sqrt[4]{2}).$ 
So, I guess, then the question is really if the following is true $$x^4\equiv 2\mod 101$$ for any $x$ in $\mathbb{F}_{101}.$
Another approach would be, i think, to try and compute the discriminant of $f(x)$ and it's cubic resolvent, but i am encountering similar issues of whether a polynomial is irreducible over this field. So, this approach seems to be even less worth pursuing. 
I'd appreciate any help, or if there is any general strategy how to approach similar problems?!
 A: Why don't you see if $2$ is quadratic residue mod $101?$ WHat about $-1?$
A: If you know the result that $2$ is a square modulo a prime $p$ if and only if $p \equiv \pm 1 \mod 8$, then you can do this with virtually no calculation. Since $101 \equiv 5 \mod 8$ (you do need to do that calculation!), $2$ is a non-square in ${\mathbb F}_{101}$. Since $4$ does not divide $102$ (more calculation!), it follows that the splitting field of $x^4-2$ must be ${\mathbb F}_{101^4}$.
To see this, let  $\beta$ be a square root of $2$ in ${\mathbb F}_{101^2}$. Let $w$ be a primitive element of ${\mathbb F}_{101^2}$ and $\beta = w^k$. So $w^{2k}=\beta^2=2 \in {\mathbb F}_{101}$, hence $102 \vert 2k$. So, if $k$ was even, we would have $102 \vert k$, but then $\beta \in {\mathbb F}_{101}$, contradiction. So $k$ is odd, and hence $\beta$ has no square root in ${\mathbb F}_{101^2}$. So the splitting field is not ${\mathbb F}_{101^2}$, and it must be ${\mathbb F}_{101^4}$.
A: Let $\alpha$ be a root of $X^4-2$ in $\overline{\mathbf{F}_{101}}$, what can you say about $\alpha^{{101}^k}$ for $k = 1,2,3$?
