Splitting field of $x^9-x$ over $\mathbb{Z}_3$.

Question

Let $F$ be a splitting field of $x^9-x$ over $\mathbb{Z}_3$.

$1$. Show that there is an $\alpha \in F - \mathbb{Z}_3$ such that $\alpha^2=2$.

$2$. Show that $\mathbb{Z}_3(\alpha)$ is a splitting field of $x^9-x$.

$3$. Show that $\mathbb{Z}_3(\alpha)$ is a cyclic group and find its generators expressed in terms of $\alpha$.

I found that the factorization of $x^9-x$ in $\mathbb{Z}_3[x]$ is $$x(x+1)(x+2)(x^2+1)(x^2+x+2)(x^2+2x+2).$$ Should it be $\mathbb{Z}_3[x]/\left<x^2+2x+2\right> = \{0,1,2,\alpha,\alpha+1,\alpha+2,2\alpha,2\alpha+1,2\alpha+2\}$? And what's next? I spent hours on this problem but I still couldn't figure it out how to solve it. Can someone help me please? Thanks.

Answer 1

For question $1$:

From factorisation of $x^9-x$ in $\mathbf Z_3[x]$, there is a element $\alpha\in F$ such that $$+1=0\iff\alpha^2=-1=2.$$

For question $2$:

$\alpha+1$ is a root of $x^2+x+2=x^2+x-1$ since $$(\alpha+1)^2+(\alpha+1)-1=\alpha^2-\alpha+1+(\alpha+1)-1=\alpha^2+1=0.$$ The other root is $-\alpha+1$. Similarly, $\pm\alpha-1$ are roots of $x^2-x-1=x^2+2x+2$.

Answer 2

The factorization you got is the key. In fact there is a more general phenomenon at play here. In general,

For any finite field of order $q$, $F_q$, and $n \in \mathbb{N}$, the product of all monic irreducible polynomials over $F_q$ whose degrees divide $n$ is equal to $x^{q^n}-x$.

In this problem, $q=3$ and $n=2$. So to factor $x^9-x$, all you had to do was find the monic irreducibles of degree 1 and 2 over $F_3 = \mathbb{Z}_3$.

Anyway, notice that $x^2+1$ shows up as a factor of $x^9-x$. So since $F$ is a splitting field of $x^9-x$, it contains the roots of $x^2+1$. Let $\alpha$ be either one of the roots of $x^2+1$. Then $\alpha^2 -2 = \alpha^2+1 = 0$. So $\alpha^2 = 2$.

Now $\mathbb{Z}_3(\alpha) \cong \frac{\mathbb{Z}_3[x]}{x^2+1} \cong \{a_0 + a_1 \alpha \ | \ \alpha^2 = 2\} = \{0,1,2,\alpha,1+\alpha,2+\alpha, 2\alpha, 1+2\alpha,2+2\alpha\}.$

To show that $\mathbb{Z}_3(\alpha)$ is a splitting field of $x^9-x$, just show that the other factors of $x^9-x$ split in $\mathbb{Z}_3(\alpha)$. That is just plug in things from $\{0,1,2,\alpha,1+\alpha,2+\alpha,2\alpha, 1+2\alpha,2+2\alpha\}$ to those other quadratics until you find the roots. It will all work out. It isn't elegant, but it works.

Now for part 3. My guess is that the group you are talking about it $Gal(\mathbb{Z}_3(\alpha)/\mathbb{Z}_3)$.

Let $f(x) = x^9-x$. Then taking the derivative, $f'(x) = 9x^8-1 = -1$. A polynomial has $b$ as a repeated root if and only if $f(b)=0$ and $f'(b) =0$. Since $f'(x) = -1 \neq 0$ for all $x$, $f(x)$ has no repeated roots.

Thus $f(x)$ is a separable polynomial. One of the characterizations of a Galois extension is that is it the splitting field of a separable polynomial. Thus we have just shown that $\mathbb{Z}_3(\alpha)$ is a Galois extension of $\mathbb{Z}_3$. Therefore $|Gal(\mathbb{Z}_3(\alpha)/\mathbb{Z}_3)| = [\mathbb{Z}_3(\alpha):\mathbb{Z}_3] = 2$.

So our Galois group is order 2 and we only have to find the non-identity element. This element will take $\alpha$ to the other root of $x^2+1$. The other root of $x^2+1$ is $2 \alpha$. Let $\sigma:\mathbb{Z}_3(\alpha) \to \mathbb{Z}_3(\alpha)$ be the automorphism fixing $\mathbb{Z}_3$ such that $\sigma(\alpha) = 2 \alpha$. Then $Gal(\mathbb{Z}_3(\alpha)/\mathbb{Z}_3) = \{e, \sigma\}$.