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There's a question in Herstein:

Prove that the ring $R$ of $2\times 2$ matrices defined over $\Bbb{Q}$ contains only two ideals: $(0)$ and $R$.

This seems to say that if I take any non-zero element, say $a=\begin{pmatrix} 1&1\\1&1\end{pmatrix}$, then $aR=R$.

This implies that for any $\begin{pmatrix} p&q\\r&s\end{pmatrix}\in R$, there exists $\begin{pmatrix} f&g\\k&l\end{pmatrix}\in R$ such that $\begin{pmatrix} f&g\\k&l\end{pmatrix}\begin{pmatrix} 1&1\\1&1\end{pmatrix}=\begin{pmatrix} p&q\\r&s\end{pmatrix}$.

It is clear that such a matrix $\begin{pmatrix} f&g\\k&l\end{pmatrix}$ need not exist in $R$, as the system of equations $f+g=p$ and $f+g=q$ does not have a solution unless $p=q$.

Is th question wrong then?

Thanks in advance!

  • For reference, this is Section 3.4, question 12. –  Dec 01 '13 at 11:45
  • You are misinterpreting the meaning of $aR = R$, it does not mean that you apply $a$ and obtain the same matrix as before. – Mark Fantini Dec 01 '13 at 11:45
  • I am not doing that at all. What I understand from $aR=R$ is that for every $r\in R$, there exists $r'\in R$ such that $r'a=r$. –  Dec 01 '13 at 11:47
  • Long story short, the ideal generated by $a$ is not $aR$. – Hagen von Eitzen Dec 01 '13 at 15:41
  • The more fundamental error is forgetting that this is a non-commutative ring, and that the word "ideal" should therefore be qualified by "left", "right" or "two-sided". The two-sided ideals are the only kind of which only two exist (the other types have infinitely many of them); the ideal $aR$ is only a right ideal. – Marc van Leeuwen Dec 01 '13 at 16:57
  • Just to confirm, the ideal generated by $a$ is $RaR$ then? Thanks –  Dec 02 '13 at 08:15

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