Find the degree of the splitting field of the polynomial $x^6-7$ over $F_{3}$ (the field with $3$ elements).
Here, $x^6-7$|${x^6}^2-7^2 = x^{12}-1$ where $7^2 = 1\mod 3$ After this step, I couldn't make anything. How can I continue for solution? or if this step is wrong, what is the way to solve this question?
Thank you
On
The splitting field over $\Bbb{F}_3$ is a field $\Bbb{F}_{3^q}$. Let's check; $P(X)=X^6-7$ has $\pm 1$ as roots in $\Bbb{F}_3$ because $(\pm 1)^6=1\equiv 7\pmod{3}$
Now we have in $\Bbb{F}_3$ (remember that $a^3+b^3=(a+b)^3$)
$$(X-1)^3(X+1)^3=(X^2-1)^3=X^6-1=X^6-7$$
And so our polynomial have two distinct roots of multiplicity $3$ in $\Bbb{F}_3$ and the degree is $1$
Hint: $x^6-7=(x+1)^3(x+2)^3$ over $\mathbb{F}_3$. So the degree is $1$.