Prove /disprove $M_2=\left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix}; a, b , c , d \in \mathbb{Q}\right\}$ is simple ring 
Show that $$M_2=\left\{\begin{pmatrix} a & b   \\ c & d \end{pmatrix}; a, b , c , d  \in \mathbb{Q}\right\}$$ is  simple  ring

My proof :
we know that  $M_2$ is  a ring  under adition  and matrix  multiplication and  has unity \begin{pmatrix} 1& 0  \\ 0 & 1 \end{pmatrix}. we can find  two element  $A $ and $B $ in $M_2$  such that $AB  \neq \begin{pmatrix} 0& 0  \\ 0 & 0 \end{pmatrix}$. For  example $ A=\begin{pmatrix} 1& 0  \\ 1 & 0 \end{pmatrix}$ and $ B= \begin{pmatrix} 1& 1 \\ 0 & 0 \end{pmatrix}$
If we show that $ M_2$ has no non trivial ideal , then $ M_2$ become a simple ring
Let $A$ be any ideal of $M_2$ . If $A= \{0\}$ , $0 $ being a $2 \times 2$ null matrix, then there is nothing to prove . let $A \neq\{0\}$. Then there exist a nonzero matrix  $X \in A$ of the form  $ X=\begin{pmatrix} a_{11}& a_{12}  \\ a_{21} &  a_{22} \end{pmatrix}$
Since  $X$ is a nonzero matrix , atleast one the $4 $ entries in $X$ is nonzero . let $a_{12} \ne 0 \in \mathbb{Q}$
We choose four matrix in  $M_2$  as follow
let $ P=\begin{pmatrix} 1& 0  \\ 0 & 0 \end{pmatrix}$,
$ Q=\begin{pmatrix} 0& 0  \\ 1 & 0 \end{pmatrix}$,
$ S=\begin{pmatrix} 0& 1  \\ 0 & 0 \end{pmatrix}$,$ T=\begin{pmatrix} 0& 0  \\ 0 & 1 \end{pmatrix}$
Now  by doing  multiplication  $ PXQ=\begin{pmatrix} a_{12}& 0  \\ 0 & 0 \end{pmatrix}$ and $ SXT=\begin{pmatrix} 0& 0  \\ 0 & a_{12} \end{pmatrix}$
Since   $X\in A$  and $A $ is an ideal of $M_2 $, therefore $PXQ + SXT \in A$
now $ \begin{pmatrix} a_{12}& 0  \\ 0 & 0 \end{pmatrix} +\begin{pmatrix} 0& 0  \\ 0 & a_{12} \end{pmatrix}=\begin{pmatrix} a_{12}& 0  \\ 0 & a_{12}   \end{pmatrix} =K\in A$
since $a_{12} \neq  0 \in \mathbb{Q}$, $a_{12}^{-1} \in \mathbb{Q}$ that is  $K^{-1}  \in M_2$
Since $A$ is an ideal of $M_2$  that is $ KK^{-1} = I  \in A$
Thus $A$ is an ideal of $M_2 $ containing the unity $ I  $ it  implies $A = M_2$
Hence prove  that $M_2$  is  simple ring
Is my proof  is correct ?? yes/No

$SXT=\begin{pmatrix} 0& 1 \\ 0 &  0 \end{pmatrix}\begin{pmatrix} a_{11}& a_{12}  \\ a_{21} &  a_{22} \end{pmatrix}\begin{pmatrix} 0& 0  \\ 0 &  1 \end{pmatrix}=\begin{pmatrix} a_{21}& a_{22}  \\ 0 &  0 \end{pmatrix}\begin{pmatrix} 0& 0  \\ 0 &  1\end{pmatrix}=\begin{pmatrix}0& a_{22}  \\ 0 &  0 \end{pmatrix}$.
If you want the $a_{12}$ as the last entry of the matrix you can do $QXT$.
The rest is correct.
