Show that $R=\left\{\begin{pmatrix} 0 & a\\0 & b\end{pmatrix}:a,b\in\Bbb Q\right\}$ is a subring of $M_2(\Bbb Q)$ and $I=\{A\in R:A^2=0\}$ is an ideal of $R.$ Finally, show that $R/I\cong \Bbb Q.$
I was able to show that $R=\left\{\begin{pmatrix} 0 & a\\0 & b\end{pmatrix}:a,b\in\Bbb Q\right\}$, is a subring of $M_2(\Bbb Q)$ and $I=\{A\in R:A^2=0\}$ is an ideal of $R.$
However, I was not able to prove that $R/I\cong \Bbb Q.$
My strategy was to find an epimorphism from $R$ to $\Bbb Q$ with kernel $I.$ If I was able to find one such mapping, then from First Isomorphism Theorem I could've said, $R/I\cong \Bbb Q.$
I tried constructing several epimorphisms $\phi$ from $R$ to $\Bbb Q$ such as, $\phi(\begin{pmatrix} 0 & a\\0 & b\end{pmatrix})=a+b$ or $\phi(\begin{pmatrix} 0 & a\\0 & b\end{pmatrix})=ab$ or $\phi(\begin{pmatrix} 0 & a\\0 & b\end{pmatrix})=a+b-ab$ or $\phi(\begin{pmatrix} 0 & a\\0 & b\end{pmatrix})=a+b-1$ or $\phi(\begin{pmatrix} 0 & a\\0 & b\end{pmatrix})=(a-1)(b-1),$ for all $\begin{pmatrix} 0 & a\\0 & b\end{pmatrix}\in R.$
However, none of them turned to be a valid homomorphism in the first case.
Finally, I tried to find isomorphisms from $R/I$ to $\Bbb Q.$
But this made me take a closer observation at what the set $R/I$ is?
This made me realise that $R/I=R$. So, the question now asks us to show that $R\cong \Bbb Q.$ Now, this looks much confusing.
Any help regarding this issue will be greatly appreciated.
Note that if $A=\begin{bmatrix}0&a\\0&b\end{bmatrix} \in R$, then $A^2=0$ implies $\begin{bmatrix}0&ab\\0&b^2\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$, then $A$ must be of the form $\begin{bmatrix}0&a\\0&0\end{bmatrix}$. Thus $I=\left\{\begin{bmatrix}0&a\\0&0\end{bmatrix}\, | \, a \in \mathbb{Q}\right\}$.
Try the map $\phi:R \longrightarrow \mathbb{Q}$ defined by $$\phi\left(\begin{bmatrix}0 &a\\0&b\end{bmatrix}\right)=b.$$
Kernel of this map is $\ker(\phi)=\left\{\begin{bmatrix}0&a\\0&0\end{bmatrix}\, | \, a \in \mathbb{Q}\right\}.$