I would to like to know which product of cyclic groups the group $A^\times$ of units of the quotient ring $$ A = \mathbb F_5[X] / ((X^2-2)^2) $$ is isomorphic to.
I know that $A$ is not an integral domain and has characteristic $5$. As the polynomial $f= X^2-2$ is irreducible over $\mathbb F_5[X]$, I think that $\pi(g) \in A^\times$ if and only if $g$ is coprime with $f$. In particular, I've obtained $|A^\times|=5^4-5^2 = 600$.
But I don't know how to continue. Any held would be appreciated !
Because $600=2^3\times3\times5^2$ it suffices to find an element of order $2^3$, and to show that there is no element of order $5^2$.
For the first, note that taking the quotient $A/(X^2-2)$ yields a surjection $$\pi:\ \Bbb{F}_5[X]/((X^2-2)^2)\ \longrightarrow\ \Bbb{F}_5[X]/(X^2-2),$$ and hence a surjection of the unit groups. The latter has a unit of order $2^3$, hence so does $A$.
If the order of $a\in A^{\times}$ is a power of $5$ then $\pi(x)=1$, so there exist $a,b\in\Bbb{F}_5$ such that $$x=1+(aX+b)(X^2-2),$$ from which it follows that $x^5=1$. So there is no element of order $25$ in $A^{\times}$, hence $$A^{\times}\cong\Bbb{Z}/2^3\Bbb{Z}\times\Bbb{Z}/3\Bbb{Z}\times(\Bbb{Z}/5\Bbb{Z})^2.$$