Give an example of a quotient ring R/I such that R/I have unity but R does not have unity. In a book we are given that $R=$ ring of matrices of the type $$\begin{pmatrix}
a&b\\0&0\\
\end{pmatrix}$$and $I$ of the type$$\begin{pmatrix}
0&x\\0&0\\ 
\end{pmatrix}$$
over the integers. I am not able to find the identity of $R/I$. I tried using the definition but I failed. Can someone tell me about the identity of $R/I$ and that $R$ has no identity. Also it is given that $R/I$ is commutative but $R$ is not?  
 A: $R/I$ can be seen as matrices of the type $$\begin{pmatrix}
a&*\\0&0\\
\end{pmatrix}$$
where $*$ means "don't care". Multiplication in this ring is by $$\begin{pmatrix}
a&*\\0&0\\
\end{pmatrix}\begin{pmatrix}
b&*\\0&0\\
\end{pmatrix}=\begin{pmatrix}
ab&*\\0&0\\
\end{pmatrix}$$ where I've just totally ignored whatever is going on in the $*$ squares. So it's clear that $R/I\cong\mathbb{Z}$ and the identity is $$\begin{pmatrix}
1&*\\0&0\\
\end{pmatrix}$$
A: To see that $R$ has no (multiplicative) identity, let's first take a look at how multiplication works in $R$:$$\begin{pmatrix}a & b\\0&0\end{pmatrix}\begin{pmatrix}c&d\\0&0\end{pmatrix}=\begin{pmatrix}ac&ad\\0&0\end{pmatrix}$$ Now, suppose $R$ has an identity element. Recall that an identity must be idempotent. Determine the form of an $x$ satisfying this property and prove that nothing of this form can be the identity--for this, consider $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $\begin{pmatrix}1&1\\0&0\end{pmatrix}$.
A: Note that $\begin{pmatrix}a&b\\0&0\end{pmatrix}\mapsto a$ is a ring homomorphism(!) $R\to \mathbb Z$ with $I$ as kernel.
