A splitting field over $\Bbb{F_{5}}$ I have a few troubles with this problem, I would appreciate  for some help, thanks
Let $f(x)=3x^4+3x^3+x+1$, find a splitting field $L$ of $f(x)$ over $\Bbb{F_{5}}$ 
That's what I got:
It is clearly  that $f(x)=3x^4+3x^3+x+1$ =$(x+1)(3x^3+1)$, then $f(-1)=0$ $\in \Bbb{Q}$ and his others 3 roots are "$ \sqrt[3]{\frac{-1}{3}}$,$\frac{-1}{\sqrt[3]{3}}$,$-\frac{-1^{(2/3)}}{\sqrt[3]{3}}$"
Note that $i,\sqrt[3]{3},\sqrt[6]{3} \notin \Bbb{Q}$ then $L=\Bbb{Q}(i,\sqrt[3]{3},\sqrt[6]{3})$  is an splitting field of $f(x)$ Over $\Bbb{Q}$, with $deg(Irr(i+\sqrt[3]{3}+\sqrt[6]{3}))=12$
Then $|Gal(L/ \Bbb{Q})|=[L:\Bbb{Q}]$ this because $L/\Bbb{Q}$ is  a Normal and separable extension
$\Rightarrow$ $Gal(L/\Bbb{Q}) \simeq G_{12}$ where $G_{12}$ is a group of order 12, the groups of order 12 are the 5 next;    $ \lbrace \Bbb{Z_{12}}, A_{4},\Bbb{D_{12}},\Bbb{Z_{6}}\times \Bbb{Z_{2}}$, $\Bbb{Dic_{12}} \rbrace$, 
Here is my firts question ¿which one of these is the indicated?,obviously if I'm right till this point
Now it's clear that $f(x)$ its irreducible over $\Bbb{F_{5}}$=$\lbrace {0,1,2,3,4 \rbrace}$ with $f(0)=6,f(1)=8,f(2)=3,f(3)=328,f(4)=193$
Then let $L_{2}$ an splitting field of $f(x)$ over $\Bbb{F_5}$, as $\Bbb{F_5}$ is finite,There must be a correspondence between $Gal(L_{2}/\Bbb{F_5})$ and $Gal(L/\Bbb{Q})$ here is where I am stuck.
I would appreciate any help, and please tell me if I am well or if my procedure is wrong, in any case it would help me to improve, greetings and thanks.
 A: I spotted a few problems in your attempt (props for posting the details as that makes it easier to gauge where you are in your studies).
The factor $x+1$ can pretty much be ignored as it won't affect neither the splitting fields nor the Galois groups. You are really only looking at the splitting fields of the other factor $p(x)=3x^3+1$. This cubic has as its complex zeros $x_j=\omega^j\root3\of{-1/3}$, where $j=0,1,2$ and $\omega=(-1+i\sqrt3)/2$ is a primitive third root of unity. So a splitting field $L$ of $p(x)$ over $\Bbb{Q}$ is $L=\Bbb{Q}(\root3\of3,\omega)$. In other words, this is the same splitting field that the polynomial $x^3-3$. The Galois group is isomorphic to $S_3$, and $[L:\Bbb{Q}]=6$. The arguments proving those facts are undoubtedly familiar to you from the standard example of the splitting field of $x^3-2$ over $\Bbb{Q}$ so I won't rehash them. You seem to have made the mistake of including $i$ and $\root6\of3$ into the splitting field. That is not unnatural given that $\sqrt{-3}$ and $\root3\of3$ are in there, but you cannot actually get the real sixth root of $3$ by combining those (but you do get a sixth root of $-3$).

Over the field $\Bbb{F}_5$ a splitting field $L_2$ is much smaller. You apparently missed (due to numerical errors) that $p(2)=25=0\in\Bbb{F}_5$, so we get a factorization
$$
p(x)=(x-2)(3x^2+x+2)\in\Bbb{F}_5[x].
$$
That quadratic is irreducible over $\Bbb{F}_5$. To see that you can either verify that it has no zeros in $\Bbb{F}_5$, or you can check that the discriminant
$$\Delta=b^2-4ac=1^2-4\cdot3\cdot 2=-23=2$$
has no square root in $\Bbb{F}_5$. I outlined a different way of reaching the same conclusion in the comments: because $\Bbb{F}_5^*$ is cyclic of order $4$ the element $-1/3$ has a single cube root in there.
Anyway, over $\Bbb{F}_5$ the splitting field is thus $L_2=\Bbb{F}_5(\sqrt\Delta)=\Bbb{F}_{25}$. Therefore $Gal(L_2/\Bbb{F}_5$ is cyclic of order two.

You were also interested in the relation between these two Galois groups,
$G_1=Gal(L/\Bbb{Q})$ and $G_2=Gal(L_2/\Bbb{F}_5)$. A relation can be described using concepts of algebraic number theory. Let $\mathcal{O}_L$ be the ring of algebraic integers in $L$. It turns out that in that ring the ideal $(5)$ can be written as a product of three prime ideals
$$
(5)=\mathfrak{p}_1\mathfrak{p}_2\mathfrak{p}_3.
$$
All those prime ideals $\mathfrak{p}_j, j=1,2,3,$ have $5$ as an element, and all the quotient rings $\mathcal{O}_l/\mathfrak{p}_j$ are actually isomorphic to $L_2$.  Furthermore, if $\sigma\in G_1$ then $\sigma$ must also permute the three prime ideals. It then turns out that the point stabilizers of that action are all isomorphic to $G_2$. If $\sigma(\mathfrak{p}_1)=\mathfrak{p}_1$, then $\sigma$ induces an automorphism $\overline{\sigma}$ of $\mathcal{O}/\mathfrak{p}_1\simeq L_2$. A non-trivial fact is that the mapping
$\sigma\mapsto \overline{\sigma}$ is surjective onto $G_2$. The kernel of this mapping measures the ramification of a prime. The fact that five is not a factor of the discriminant of $p(x), \Delta=-972,$ is the key here. It implies that $p=5$ is unramified. This allows us to identify $G_2$ as a subgroup of $G_1$.
A: $3x^4+3x^3+x+1 = 3 (x + 1) (x + 3) (x^2 + 2 x + 4)$ in $\Bbb{F}_{5}[x]$.
Therefore, the splitting field of $f$ is the same as the splitting field of $g(x)=x^2 + 2 x + 4$.
Since $g$ is irreducible in $\Bbb{F}_{5}[x]$ and has degree $2$, its splitting field is $\Bbb{F}_{5}[x]/(g) \cong \Bbb{F}_{25}$.
