What is the Galois group of $x^4+1$ over $F_3$? $x^4+1$ is separable by the derivative test. Is it irreducible over $F_3$? The only way I can think of to check this is to write down all possible irreducible polynomials of degree $2$. 
If it is irreducible, then its Galois extension is its splitting field. The Galois group of finite fields is cyclic, thus we only need to calculate the degree of the extension. How?
 A: First factorize: $x^4+1 = (x^2+x-1)(x^2-x-1) = f(x)g(x)$. One can check (using root test) that both $f$ and $g$ are irreducible. Then let $\alpha$ be a root of $f(x)$ in some splitting field. Then we can see that $g(\alpha + 1) = (\alpha+1)^2 - (\alpha+1)-1 = \alpha^2 + \alpha -1 = 0$, so that $\alpha+1$ is a root of $g(x)$.
So we have $\mathbb{F}_3 (\alpha, \alpha+1) = \mathbb{F}_3 (\alpha)$ is the splitting field of $f(x)g(x) = x^4 + 1$ (Since both $f(x)$ and $g(x)$ have one root in this field extension, and they are quadratic, so they must split). Since $\alpha$ is a root of the irreducible  degree $2$ polynomial, $\mathbb{F}_3(\alpha)$ is a degree $2$ extension, and by the uniqueness of splitting fields, it is isomorphic to $\mathbb{F}_{3^2}$. The Galois group is then $\mathbb{Z} / 2\mathbb{Z}$.
A: Take a square root of $2=-1$, call it $i$, which makes for a degree two extension of $F_3$. Then
$$
(a+bi)^2=a^2-b^2+2abi=i
$$
is solved by $a=1$ and $b=-1$. Hence the polynomial factors over $F_9=F_3[i]$. Since it has no roots in $F_3$, its splitting field is indeed $F_9$. The automorphism group of $F_9$ is cyclic of order two, containing the identity and the Frobenius automorphism.
A: The roots of $f(X)=X^4 + 1$ in a separable closure of $\mathbf F_3$ verify $x^4=-1$, hence they form a cyclic group of order $8$. If the splitting field $N$ of $f(X)$ has degree $n$ over $\mathbf F_3$, then necessarily $8$ divides $3^n -1$. The only possibility is $n=2$ and $N=\mathbf F_9$ .
