Let $R$ be a given ring assigned by $R=\left\{\begin{pmatrix}a&b\\0&c\end{pmatrix}:a,b,c \in \mathbb{Z}\right\}$. Let us consider the ideal $I$ of $R$ which is given by $I=\left\{\begin{pmatrix}0&b\\0&c\end{pmatrix}:b,c \in \mathbb{Z}\right\}$. Prove that $I$ is a prime ideal as well as a maximal ideal of $R$.
I am trying to show it by observing that $R/I$ is an integral domain. Let us take any elements $M_1$ and $M_2$ from $R/I$. Then if $M_1M_2$ belongs to $I$ under matrix multiplication, then it is easily observed that one of $M_1$, $M_2$ must be in $I$, which gives us the integral domain prove of $R/I$. Is my way of proceeding is in correct way? Please help me to solve it.
Consider the map
$$R \to \mathbb Z, \begin{pmatrix}a&b\\0&c\end{pmatrix} \mapsto a.$$
Check that this is a ring homomorphism (basically you have to check that if you multiply two matrices of $R$, then their diagonal elements are multiplied).
It is clearly surjective and its kernel is $I$, hence $R/I \cong \mathbb Z$.
In particular $I$ is not maximal, because the kernel of $R \to \mathbb Z \to \mathbb Z/2\mathbb Z$ is a larger ideal.