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: using Jordan canonical form, any non-invertible $2 \times 2$ rational matrix is similar to either $\pmatrix{0 & 1\cr 0 & 0\cr}$ or $\pmatrix{\lambda & 0\cr 0 & 0\cr}$ (and the similarity can be done using a rational matrix).  Express $I$ as a linear combination of matrices obtained from either of these by multiplication with other matrices.
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