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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.

2 Answers2

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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 $$\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} +\begin{bmatrix} 0 & 0 \\ 0 & 01\end{bmatrix} \in I$$

N. S.
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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. .

Robert Israel
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  • You assume the field is closed, while it doesn't have to be! The answer linked in the comments explains how to do it – Patrick Da Silva Aug 12 '15 at 15:35
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    The field doesn't need to be algebraically closed. If one eigenvalue is $0$, the other is rational (since the trace is rational). – Robert Israel Aug 12 '15 at 17:01
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    Right. But this should be added to the answer, it's not hard to understand but worth mentioning! I didn't think that the fact that the matrices were $2 \times 2$ would be relevant (you can't use your trace argument if we work with $n \times n$ matrices, but the result is also true for $n \times n$ matrices). – Patrick Da Silva Aug 12 '15 at 17:03