I have some problems solving this exercise:
Let $L$ be the splitting field of $f(x) = x^6 +x^4 +x^2 + 1 ∈ K[x]$ when
(a) $K = \mathbb{Q}$,
(b) $K = \mathbb{F}_5$.
Determine the degree $[L : K]$, the isomorphism type of the group $\mathrm{Aut}(L/K)$ and the effect of the group elements on a basis of the K-vector space $L$.
Determine the degree $[L : K]$: for a) we find the roots and we have that the roots that are not in $\mathbb{Q}$ are $\pm i,\frac{\sqrt{2}}{2}\pm\sqrt[4]{\frac{1}{4}}$
so we have that the splitting field $L$ is $\mathbb{Q}(i, \sqrt{2}, \sqrt[4]{\frac{1}{4}})$
We have $[\mathbb{Q}(i, \sqrt{2})]=[\mathbb{Q}(i, \sqrt{2}):\mathbb{Q}(\sqrt{2})]\cdot[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=4$
and therefore $[\mathbb{Q}(i, \sqrt{2}, \sqrt[4]{\frac{1}{4}}):\mathbb{Q}]= [\mathbb{Q}(i, \sqrt{2}, \sqrt[4]{\frac{1}{4}}): \mathbb{Q}(i, \sqrt{2})][\mathbb{Q}(i, \sqrt{2}):\mathbb{Q}]=4 \cdot 4=16$
for b) we have $f(3)=f(2)=0$, and therefore $f(x)=(x-3)(x-2)(x^4+1)$ and $x^4+1$ is irreducible this means that the splitting field of $f(x)$ is the same of $g(x)=x^4+1$. Let $a$ be root of $g(x)$.
$a^4=-1 \Rightarrow a^8=1$
this means that $\mathbb{F}_{5^2}$ is the smallest field that contains all root of $g$ (since $8|5^2=1$).$\Rightarrow L=\mathbb{F}_{5^2}, [L:K]=2$
So far everything right?
I don't know what to do with the rest: I don't really understand what "the effect of the group elements on a basis of the K-vector space $L$" means and I've never been good at finding types of isomorphism. Can you help me?
a) $\sqrt[4]{1/4} = 1/\sqrt{2} \in \Bbb Q(\sqrt{2})$, so in the splitting field is indeed $L =\Bbb Q(i, \sqrt{2})$. One of its possible $\Bbb Q$-bases is $\{1,i,\sqrt{2},i\sqrt{2}\}$. The $K$-minimal polynomials of $i$ and $\sqrt{2}$ are respectively $x^4+1$ and $x^2-2$. A $K$-endomorphism of $L$ is specified by action on the roots of these minimal polynomials* (as it has to conserve the minimal polynomials in $K$). You can't send $i \mapsto \pm 1$, since the map would not then be injective. Thus any automorphism you have sends $i \mapsto \pm i$ and $\sqrt2 \mapsto \pm\sqrt2$, i.e. you have at most $4$ automorphisms.
Take specific maps $\sigma$: $i \mapsto -i$, $\sqrt2\mapsto\sqrt2$ and $\tau$: $i \mapsto i$, $\sqrt2\mapsto-\sqrt2$, and then write their action on an arbitrary element of $L$:
$$\sigma(a+bi+c\sqrt2+di\sqrt2) = a-bi+c\sqrt2-di\sqrt2; \quad \tau(\ldots) = \ldots$$
($a,b,c,d\in K$). These isomorphisms are involutive, and they commute (check these two facts). So you can conclude for the isomorphism type of $Aut(L/K)$.
The effect of $Aut(L/K)$ on a basis would, uh, depend on the basis. In my basis above, it acts by diagonal matrices with elements $\pm 1$, we can write these out, — not sure what exactly they want you to say.
*Note that for a general $K(\alpha,\beta,\gamma,\ldots)$, you would also want to check that your map defined on $\alpha, \beta, \gamma, \ldots$ is well-defined (let alone bijective). You can use the constructive version of the primitive element theorem to avoid this, but that would mean a lot more calculations.
b) $x^4+1 = (x^2 + 2) (x^2 - 2)$ is not irreducible. (The quadratic factors are.) If $\alpha^2 = 2$, then $(2\alpha)^2 = -2$, and so $K(\alpha) = L$, so you obtain the basis and $Aut(L/K)$.