# Compute the Galois group for the root field of the polynomial $x^3$+$2x$+2 over $Z_3$

Compute the Galois group for the root field of the polynomial $x^3$+$2x$+2 over $Z_3$

Choose a root $\alpha$ in the root field of the polynomial.

We found three roots in this root field, $\alpha$, $\alpha$+1, $\alpha$+2.

What are the automorphisms then? Is the Galois group just these three? Or is it just the first 2 since you can generate the third one with the second one?

The elements of the Galois group are not elements of the field. You may take one element of the group to be the one that sends $\alpha$ to $\alpha+1$. This is often designated $\alpha\mapsto\alpha+1$. When you do it twice, the $1$ remains fixed, as it must, and the $\alpha$ of the image goes to $\alpha+1$, so the upshot is $\alpha\mapsto\alpha+2$. Do it once more and $\alpha\mapsto\alpha$, the identity. So you have $3$ things in the Galois group, as you must, for a normal cubic extension.
Let $f(x) = x^{3}+2x+2$ over $\mathbb{Z}_{3}$. $f$ is clearly irreducible over $\mathbb{Z}_{3}$, as it has no roots in $\mathbb{Z}_{3}$ - this can be checked easily. Let $\alpha$ be a root of $f(x)$, and consider the extension $\mathbb{Z}_{3}(\alpha)$. As you correctly noted, the other roots of $f(x)$ are given by $\alpha+1$ and $\alpha+2$. Hence, $\mathbb{Z}_{3}(\alpha)$ is the splitting field of $f(x)$.
From here, if you like, you can use the fact that $[\mathbb{Z}_{3}(\alpha):\mathbb{Z}_{3}] = 3$ (since $\deg(f(x)) = 3$) to conclude immediately that $\mathrm{Gal}(\mathbb{Z}_{3}(\alpha)/\mathbb{Z}_{3}) \cong \mathbb{Z}_{3}$.
If you want to more concretely answer the question, consider the possible automorphisms of your field extension. You have to map $\alpha$ to another root of $f$, so what are your options? There's clearly the identity map. Consider the map $\sigma$ which takes $\alpha$ to $\alpha+1$. Is this an automorphism? Consider now $\sigma^{2}$. Where does it take $\alpha$? $\alpha + 1$? See if you can show that $\langle \sigma \rangle = \{1, \sigma, \sigma^{2}\}$ is the complete automorphism group of this extension. Then, clearly, $\mathrm{Gal}(\mathbb{Z}_{3}(\alpha)/\mathbb{Z}_{3}) \cong \mathbb{Z}_{3}$. I'm happy to say more if you need clarification.