Computing Galois group of polynomial over finite field

Question

What is the Galois group of $(x^2-1)(x^2-2)...(x^2-p+1)$ over $\mathbb Z_p$ for an odd prime, $p$? There are exactly $\frac{p-1}{2}$ squares in $\mathbb Z_p$ but my guess is the group is $\mathbb Z_2$.

Answer 1

By the uniqueness of finite fields of a given degree, there is one and only one field of order $p^2$. Since not all elements of $\Bbb F_p$ are squares, you know that if $a$ is a non-square then

$$\Bbb F_p[\sqrt{a}]=\Bbb F_p[x]/(x^2-a)=\Bbb F_{p^2}$$

is an extension of degree $2$. But then since another non-square $b$ would also generate an extension $\Bbb F_p[b]$ of degree $2$, it must be that these two are the same field, i.e. $\Bbb F_p[\sqrt{b}]=\Bbb F_{p^2}$ as well. So the splitting field for that polynomial is $\Bbb F_{p^2}$ since that field has all the square roots of all elements of $\Bbb F_p$ in it. The Galois group of this extension has order $2$, hence must be $\Bbb Z/2\Bbb Z$, as you surmise.

Answer 2

Also one can use the fact that $\mathbb{F}_{p^2}/\mathbb{F}_p$ is Galois, in other words is normal, so one can write $\prod (x^2-i)$ actually splits in $\mathbb{F}_{p^2}$. So, ans is $\mathbb{Z}_2$