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

Question

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?

Answer 1

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.

Answer 2

The hint.

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 $2$.

Answer 3

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.