I want to find a splitting field of $x^{6}-3$ over $\mathbb{F}_{7}$. I learned that Finite field containing $\mathbb{F}_{7}$ is the form of $\mathbb{F}_{7^m}$ and it is normal extension. So I've tried to find smallest m containing a single root of $x^{6}-3$. If $x^{6}-3$ is irreducible in $\mathbb{F}_{7}$, then $\mathbb{F}_{7^m}$ is splitting field of $x^{6}-3$. But I don't know where I should start. Even, I cannot prove that $x^{6}-3$ is irreducible. For polynomial of degree less than 3, there is a method to determine whether it is irreducible or not. But this is not the case. And the try using Gauss' Lemma and Eisenstein is failed, because I cannot find ring $R$ of which field fraction is $\mathbb{F}_{7}$. Is it wrong approach for this kind of question?
Field Theory – Finding a Splitting Field of x^6-3 Over F7
field-theoryfinite-fieldssplitting-field
Related Solutions
First reduce the $\zeta$'s: you have $$\mathbb{Q}(\zeta_{77}, \zeta_{15}) = \mathbb{Q}(\zeta_{lcm(77,15)}) = \mathbb{Q}(\zeta_{1155}).$$ This is an extension of $\mathbb{Q}$ of degree $\varphi(1155) = \varphi(77)\cdot \varphi(15) = 480.$
What is the intersection $\mathbb{Q}(\zeta_{1155}) \cap \mathbb{Q}(\sqrt[15]{5})$?
Subfields of cyclotomic fields are abelian (the converse is also true), that is, they are Galois with abelian Galois groups. However, the nontrivial subfields of $\mathbb{Q}(\sqrt[15]{5})$ - $\mathbb{Q}(\sqrt[3]{5})$, $\mathbb{Q}(\sqrt[5]{5})$ and $\mathbb{Q}(\sqrt[15]{5})$ - are not Galois. So you have $\mathbb{Q}(\zeta_{1155}) \cap \mathbb{Q}(\sqrt[15]{5}) = \mathbb{Q}$ and $$[\mathbb{Q}(\zeta_{1155}, \sqrt[15]{5}) : \mathbb{Q}] = 480 \cdot 15 = 7200.$$
Note that the lattice of fields of the shape $$\Bbb F_{\displaystyle 2^r}$$ corresponds to the lattice of the $r$-values w.r.t. division. The field $$\Bbb F_{32}=\Bbb F_{2^5}$$ intersects (in a common embedding) the fields $\Bbb F_{2^k}$ for $k=1,2,3,4$ only in $\Bbb F_2$, in the prime field.
The polynomial $$ f=X^4+X^3+1\in \Bbb F_2[X]$$ is irreducible.
To see these, note that there is no root of it in $\Bbb F_2$. The only possibility to factor it would be as a product of two irreducible polynomials of degree two. But there is only one such irreducible polynomial, it is reciprocal, $X^2+X+1$, its square is reciprocal, $X^4+X^2+1$, but it is not our polynomial.
Form here, the splitting field of $f$ over $\Bbb F_2$ is $\Bbb F_{2^4}\cong \Bbb F_2[X]/(f)$.
The minimal field containing $\Bbb F_{2^4}$ and $\Bbb F_{2^5}$ is
$$\Bbb F_{\displaystyle 2^{4\cdot 5}}
=
\Bbb F_{\displaystyle 2^{20}}
\ ,
$$
which is the splitting field of $f$ considered as a polynomial over $\Bbb F_5$.
Later EDIT:
Let us split the polynomial $T^4 + T^3 +1 \in F[T]$ over the field $F=\Bbb F_2[X]/(f)=\Bbb F_2(a)$, where $a=X$ modulo $(f)$ is the generator of $F$, and the minimal relation over the prime field is $a^4+a^3+1=0$.
First of all, $a$ is a root in $F$ of $T^4 + T^3 +1$.
The multiplicative order of $a$ is $2^4-1=15$, it generates the cyclic multiplicative group $F_{16}^\times$.
The Frobenius morphism ($u\to u^2$) applied on the relation $a^4+a^3+1=0$ then gives: $$ \begin{aligned} 0 &=a^4+a^3+1\\ 0 &=(a^2)^4+(a^2)^3+1\\ 0 &=(a^4)^4+(a^4)^3+1\\ 0 &=(a^8)^4+(a^8)^3+1\ . \end{aligned} $$ So we have $T^4+T^3+1=(T-a)(T-a^2)(T-a^4)(T-a^8)$.
Computer checks:
sage: var('x');
sage: F.<a> = GF(2^4, modulus=x^4+x^3+1)
sage: F
Finite Field in a of size 2^4
sage: a.minpoly()
x^4 + x^3 + 1
sage: R.<T> = PolynomialRing(F)
sage: (T-a) * (T-a^2) * (T-a^4) * (T-a^8)
T^4 + T^3 + 1
sage: factor(T^4+T^3+1)
(T + a) * (T + a^2) * (T + a^3 + 1) * (T + a^3 + a^2 + a)
sage: a, a^2, a^4, a^8
(a, a^2, a^3 + 1, a^3 + a^2 + a)
Best Answer
$x^6-3$ has no roots in $\mathbb{F}_7$, since $3$ is not a quadratic residue $\!\!\pmod{7}$. Moreover, in $\mathbb{F}_7[x]$: $$ \gcd\left(x^6-3,x^{49}-x\right) = \gcd\left(x^6-3,x^{48}-1\right) = \gcd\left(x^6-3,3^8-1\right) = 1 $$ hence there is no quadratic irreducible polynomial over $\mathbb{F}_7$ that is a divisor of $x^6-3$.
In a similar way: $$ \gcd\left(x^6-3,x^{343}-x\right) = \gcd\left(x^6-3,x^{342}-1\right) = \gcd\left(x^6-3,-1\right) = 1 $$ hence there is no irreducible polynomial over $\mathbb{F}_7$ that divides $x^6-3$, hence $x^6-3$ is an irreducible polynomial over $\mathbb{F}_7$ and its splitting field is isomorphic to $\mathbb{F}_{7^6}$.