Is $x^6-3$ irreducible over $\mathbb{F}_7$?

Question

I know that $\mathbb{F}_7=\mathbb{Z}_7$, and the all possible solutions of $x^6-1=0$ over $\mathbb{Z}_7$ are 1~6, so if we let the root of equation $x^6-3$ as $t $ then the solutions of $x^6-3=0$ is $t$~$6t$, which shows that all the solutions are not in $\mathbb{Z}_7$. However I can't say that $x^6-3$ is irreducible by this fact(since there could exist 2 or more degree polynomial divides $x^6-3$.) Is it irreducible over $\mathbb{Z_7}$? Why?

Answer 1

Assume that you have an irreducible polynomial $P$ of degree $d$ dividing your polynomial. Then taking $r$ to be the class of $X$ in the field :

$$\mathbb{F}_{7^d}:=\frac{\mathbb{F}_7[X]}{(P)} $$

We know that $r^6=3$ in this new field (since $P$ divides $X^6-3$). On the other (because you know Lagrange's theorem) you have that :

$$r^{7^d-1}=1 $$

Now I claim that both equations are not compatible when $d=2$ or $3$. Let us do the case $d=2$. In that case we have $r^{48}=1$ and $r^6=3$ but $48=6.8$ so : $r^{48}=3^8=3$ mod $7$ which is not $1$.

And the case $d=3$ $r^{342}=1$ and $r^6=3$ but $342=6.57$ so that $r^{342}=3^{57}=3$ mod $7$ which is not $1$.

Hence you cannot have an irreducible polynomial of degree $1$, $2$ or $3$ dividing $X^6-3$, this means that your polynomial is irreducible.

Answer 2

Same question here. I am new on this site. I dont know how to share same question link. (as a comment or answer). But I wanna inform every one that there is same question.

and this was my answer in the other question: 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.

Answer 3

Since you can get $7 | x^6-1$ for all x by Fermat Little Theorem and $(x^6 -1 , x^6 -3) \leq 2$; hence, 7 doesn't divide $ x^6 -3$. Hence, it is irreducible.