Ring of rational 2x2 matrices has no proper ideals I'm trying to prove the ring $R$ of all rational 2x2 matrices has no (two-sided) ideals other than $(0)$ and $R$. 
My attempt: Let $I$ be an ideal other than $(0)$. Then there exists some nonzero 2x2 matrix $a \in I$. The proof is easy if $a$ is invertible, because then $aa^{-1} = id \in I$ so $I =R $. I can't prove the result for noninvertible $a$. In order to show $I =R$ when all we know is there exists some noninvertible $a \in I$, we must fine some matrix $b$ s.t. $ba = id$ or $ab = id$ but that is impossible. For instance, take $a = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$. No 2x2 matrix multiplies $a$ to give the identity. 
Alternatively, we could prove that any nonzero ideal in $R$ must contain some invertible matrix. 
I'm not sure how to proceed in either case. 

Hint If $a \neq 0$, then my multiplying $a$ to the left and right by elementary matrices (i.e. by row and column reduction) you can get 
$$\begin{bmatrix} 1 & 0 \\
0 & 0 \end{bmatrix} =EaF \in I$$
Then 
$$\begin{bmatrix} 0 & 0 \\
0 & 1 \end{bmatrix} =\begin{bmatrix} 0 & 1 \\
1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\
0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\
1 & 0 \end{bmatrix} \in I$$
Thus
$$\begin{bmatrix}1 & 0 \\
0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \\
0 & 0 \end{bmatrix} +\begin{bmatrix} 0 & 0 \\
0 & 01\end{bmatrix} \in I$$
