Let $p$ be an odd prime. How do I compute the Galois group of $F/Z_p$ where $F$ is the splitting field of $\prod_{i=1}^{p-1} (X^2-i)$ over $Z_p$?
Since exactly $\frac{p-1}{2}$ integers from $1\leq i\leq p-1$ are the roots of $X^2-i$ in $Z_p$, let's write $F=Z_p(\sqrt{i_1},...,\sqrt{i_k})$ where $k=\frac{p-1}{2}$ and $\sqrt{i_j} \notin Z_p$.
This leads me to guess that the Galois group is $(Z_2)^k$, but is it really true? How do I prove this?
You should note that for any non-squares $x$ and $y$ of $Z_p$, there is some $a\in Z_p$ such that $x=a^2y$. In this way, once you adjoin to the field the square root of one of such elements, you have adjoin the square root of all the other roots. In other words, $$Z_p(\sqrt{i_1},\ldots,\sqrt{i_k})=Z_p(\sqrt{i_l})$$ for any $l$. Hence $F/Z_p$ is cuadratic and the Galois group is just $Z_2$.