# Show splitting field of $t^5 + 1$ over $\mathbb{F}_3$ is degree 4

I would like to show the splitting field of $t^5 + 1$ over $\mathbb{F}_3$ is degree 4.

I see we can factor this as $(t+1)(t^4 - t^3 + t^2 -t + 1)$. I would then like to show the factor $g(t) = t^4 - t^3 + t^2 -t + 1$ is irreducible over $\mathbb{F}_{3}$.

But I can not see a way of doing this without going for a brute-force contradiction. It has no linear factors, but I do not know how to rule out it being a product of two irreducible quadratics.

I would like to then conclude we are done since the extension by adjoining just one of the roots of $g(t)$ gives a degree 4 extension, but since the multiplicative group of a finite field is cyclic, we can get to all other roots so this extension is simple (and hence, the splitting field). Is this correct?

The degree of the splitting field of $\Phi_{10}(x)$ over $\mathbb{F}_3$ is given by the least $k\in\mathbb{N}^+$ such that $3^k\equiv 1\pmod{10}$, hence $k=4$ by direct inspection.
We see that our polynomial has no roots from $\{1,-1,0\}$ and easy to see that $x^2+1,$ $x^2+x-1$ and $x^2-x-1$ they are unique irreducible polynomials with degree $2$.
We have $$t^5+1=(t^4 + 2t^3 + t^2 + 2t + 1)(t+1),$$ and the first factor $t^4 + 2t^3 + t^2 + 2t + 1$ has no root and no quadratic factor, since $x^2+1,$ $x^2+x+2$ and $x^2+2x+2$ are the unique irreducible polynomials of degree $2$, and it is obvious that they are not dividing it.
• Yes, you are right:) I suppose there was a similar question on $x^5-x-1$ not too long ago... – Dietrich Burde Jan 9 '18 at 14:02