TITLE: Splitting field of a polynomial over Z_5
QUESTION [6 upvotes]: Find the splitting field of the polynomial $$x^3+x+1$$ over $\mathbb{Z}_5$.
I can see that the given polynomial is irreducible over $\mathbb{Z}_5$ So, $$\dfrac{\mathbb{Z}_5 [x]}{(x^3+x+1)}$$ is a field and isomorphic to $\mathbb{Z}_5(\alpha)$ where $\alpha$ is the image of $x$ under the canonical map $$\pi \colon \mathbb{Z}_5[x]\to \dfrac{\mathbb{Z}_5 [x]}{(x^3+x+1)}.$$
So, $\alpha$ is a root of the polynomial over $\mathbb{Z}_5(\alpha)$ and it factorizes as $$x^3+x+1=(x-\alpha)(x^2+\alpha x+\alpha^2+1)$$ over $\mathbb{Z}_5(\alpha)$.
But now I can't conclude anything about $x^2+\alpha x+\alpha^2+1$.
Any help will be highly appreciated.
Thank you!

REPLY [5 votes]: Hint: If $\alpha$ is a zero of $f(x)= x^3+x+1$, then by the Frobenius automorphism $a\mapsto a^5$, $\alpha^5$ and $\alpha^{25}$ are also zeros.
Hence, $f(x) = (x-\alpha)(x-\alpha^5)(x-\alpha^{25})$.
Remark:
Since $\alpha$ is a zero, $\alpha^3+\alpha+1=0$ and so $\alpha^3=-\alpha - 1$.
Thus $\alpha^5 = -\alpha^3-\alpha^2 = -(-\alpha-1)-\alpha^2 = -\alpha^2+\alpha+1$.
Moreover, by Frobenius, $\alpha^{25} = (\alpha^5)^5 = (-\alpha^2+\alpha+1)^5 = (-\alpha^2)^5 + \alpha^5 +1^5 = -\alpha^{10} +\alpha^5 +1 =\ldots$, where the computation of $\alpha^{10}$ needs to be considered.