I have to find a splitting field of $x^{6}-3$ over $\mathbb{F}_{7}$ 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? 
 A: $x^6-3$ 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 $\mathbb{F}_{7^6}$.
A: Yet another approach, which works because of the special nature of your polynomial $X^6-3$. Notice first that $3$ is a primitive element of $\Bbb F_7$, that is, it generates the cyclic group of order six consisting of the nonzero elements of this field. So any $\lambda$ with $\lambda^6=3$ must be a primitive $36$-th root of unity in its field.
Now you want the smallest extension of $\Bbb F_7$ having thirty-sixth roots of unity, in other words, you want the smallest $m$ such that $36|(7^m-1)$, in other words, smallest $m$ with $7^m\equiv1\pmod{36}$, in other words, you want the period of $7$ in the group $(\Bbb Z/36\Bbb Z)^*$. Well, the powers of $7$ in $\Bbb Z/36\Bbb Z$ are $1,7,13,19,25,31,1,\cdots$, so $m=6$ here. 
A: I will answer this with more elementary techniques, because I'm not sure of what facts you are familiar with.  If $x^6-3$ factors, then it has a factor of degree $1$, $2$, or $3$ by a degree argument.  
We can check that $x^6-3$ does not have factor of degree $1$ because it doesn't have a root in $\mathbb{F}_7$.  This can be done by plugging in all $7$ values in $\mathbb{F}_7$ or by Fermat's little theorem because $a^6\equiv 1$ for all $a\not\equiv0\pmod{7}$.
Since finite fields are uniquely determined by their order, we can follow the following idea: If $x^6$ had a factor of degree $2$, then a root of that factor would generate $\mathbb{F}_{7^2}$.  Since $-1$ is not a square in $\mathbb{F}_7$, we know that $x^2+1$ is irreducible in $\mathbb{F}_7$.  Therefore, let $\alpha$ be a root of $x^2+1$, then $\mathbb{F}_{7^2}=\mathbb{F}[\alpha]$.  We can check if there is a root of $x^6-3$ in this field.  More precisely, consider
$$
a_1\alpha+a_2.
$$
It seems natural to substitute $a_1\alpha+a_2$ for $x$ in $x^6-3$ and this would work, but it is easier to multiply the entire expression by $x$ to consider $x^7-3x$.  Substituting $a_1\alpha+a_2$ into this expression results in 
$$
(a_1\alpha+a_2)^7-3(a_1\alpha+a_2)=a_1\alpha^7+a_2-3(a_1\alpha+a_2).
$$
This follows from Fermat's little theorem (and observing that most binomial coefficients have a factor of $7$).  Moreover, we know that $\alpha^2=-1$, so we can substitute this to get $-4a_1\alpha-2 a_2$ for $a_1\alpha+a_2$ to be a root, it must be that both $-4a_1=0$ and $-2a_2=0$, this only happens when $a_1=a_2=0$, but this doesn't satisfy the original equation.
You can continue with this logic for the cubic case as well.
A: I wanna use an easy tool: 
1) it is not neccesary to check $3$ is square of not in $\mathbb F_7$. $x^6-1 \equiv 0 \mod 7$ for all $x \in \mathbb F_7^*$ which is well-known.
2) Use same trick again $3^6 \equiv 1 \mod 7$. That means $x^{36}\equiv 1 \mod 7$. Now $36\mid(7^6-1)$ trivially, it is $(7-1)(7^5+\cdots+1)$.  Moreover, $36 \not \mid (7^n-1)$ for $n <6$.  This means it is irreducible.
