# Finding irreducible factors of $f=t^6-3$ for the fields $F = \mathbb{ Q, Z/5Z, Z/7Z}$

Let $f = t^6-3 \in F [t]$. Construct a splitting field $K$ of $f$ over $F$ and determine $[K : F]$ for each of the cases: $F = \mathbb{Q, Z/5Z, Z/7Z}$. Do the same thing if $f$ is replaced by $g=t^6+3$.

I am just lost on how to do this one. One way to construct auch a field would be to find irreducible factors of $f$ in each of the three cases and just mod $F[t]$ out by that irreducible polynomial, and then modding the resulting field by another remaining irreducible polynomial, and so on, until it stops. However, the difficulty lies in figurimg out the irreducible factors of $f$ or $g$. It is clear (by pluggin in numbers) that $f$ and $g$ have no linear factors in any of the fields. A friend told me that I can show that$f$ is itreducible by a hard contradiction, but I am not sure.

I realize that there are lots of parts to my question; any help, however partial, would be great and may be selected as an answer.

• I'm sure I saw a very similar question recently, even down to the $\mathbb 5$ and $\mathbb 7$. Was that yours? Did you delete that and repost it? – Lord Shark the Unknown May 13 '18 at 2:27
• @LordSharktheUnknown Approach0 did find several questions about this polynomial over $\Bbb{F}_7$: 1 and 2 among them. But, I couldn't find this precise question. Of course, if it was very recent, then it's not in that database yet. – Jyrki Lahtonen May 13 '18 at 5:59

In each case the splitting field is $K(\alpha,\zeta)$ where $\alpha$ is a sixth root of $\pm3$ and $\zeta$ is a primitive sixth root of unity, so a zero of $X^2-X+1$.
I would work out $K(\zeta)$ first of all, then ask what the degree of $\alpha$ is over that. For instance $\Bbb Q(\zeta)=\Bbb Q(\sqrt{-3})$. If $\alpha^6=-3$ then $\alpha^3\in\Bbb Q(\sqrt{-3})$ so $|\Bbb Q(\alpha,\sqrt{-3}):\Bbb Q(\sqrt{-3})|\le3$.
On the other hand, $\Bbb F_7(\zeta)=\Bbb F_7$ and $\Bbb F_5(\zeta)=\Bbb F_{25}$ as $6\mid(7-1)$ and $6\nmid(5-1)$. In $\Bbb F_7$, $3$ is neither a square nor a cube, so $\alpha$ has degree $6$ over $\Bbb F_7$ etc.