I want to know if there is a simple form of the ring $$R=\mathbb{Z}_{9}[x]/(x^2-3,3x)$$
I tried to start with the equations $3x\equiv0$ and $x^2\equiv 3$. So, $3x^2\equiv 0$ and $3x^2\equiv 9$. The ideal $I=(x^2-3,3x)=((x-3)^2)$.
I don't know if is true. Also, the form I got is not simple enough. If my solution was true, what is the ring $$R=\mathbb{Z}_{9}[x]/((x-3)^2)$$
I don't think that this ring has a simpler representation.
However, it has a nice property: The only ideals are $(x^0) = R$, $(x^1) = (x)$, $(x^2) = (3)$, $(x^3) = \{0\}$. Therefore, $R$ is a chain ring as the lattice of ideals forms a chain.
Thus $R$ may be thought of as a brother of the ring $\mathbb{Z}/27\mathbb{Z}$, which is another chain ring of the same order.