I'm having some difficulty with the following problem:
Define an isomorphism of rings $(\mathbb{Z}/5\mathbb{Z})[X]/(X^2) \cong A$ where $A = \Bigg\{ \begin{pmatrix} a & a-b\\ a-b & a \end{pmatrix} : a,b \in \mathbb{Z}_5\Bigg\}$ is a commutative subring with unit of $\mathcal{Mat}_{2\times 2}(\mathbb{Z}_5)$.
Knowing that a matrix of A is invertible iff $a+b \neq 0$ and that the set $N$ of non-invertible elements of A is a principal ideal and that $N$ is the only ideal of $A$ other than $(0)$ and A and which is prime.
I think there is an error in the exercise. Suppose that there were a ring isomorphism $\varphi: \frac{(\mathbb{Z}/5\mathbb{Z})[X]}{(X^2)} \overset{\sim}{\to} A$. Then $$ \varphi(X) = \begin{pmatrix} a & a-b\\ a-b & a \end{pmatrix} $$ for some $a,b \in \mathbb{Z}/5\mathbb{Z}$. Since $X^2 = 0$, we have \begin{align*} \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix} &= \varphi(0) = \varphi(X^2) = \varphi(X)^2 = \begin{pmatrix} a & a-b\\ a-b & a \end{pmatrix}^2\\ &= \begin{pmatrix} 2a^2 - 2ab + b^2 & 2a(a-b)\\ 2a(a-b) & 2a^2 - 2ab + b^2 \end{pmatrix} \, . \end{align*} This yields the system of equations \begin{align} 0 &= 2a^2 - 2ab + b^2 = 2a(a-b) + b^2\\ 0 &= 2a(a-b) \, . \end{align} Subtracting the second equation from the first, we have $b^2 = 0 \implies b = 0$, which then implies that $a = 0$, too. Thus $$ \varphi(X) = \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix} = \varphi(0) \, , $$ contradicting the hypothesis that $\varphi$ is injective.