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In a non-commutative ring (with identity), is it possible for an element which does not possess left or right inverses to generate the entire ring? i.e. $(r)=R$, where (r) is the two-sided ideal generated by $r\in R$ ?

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closed as too localized by Andrés E. Caicedo, Yemon Choi, Martin Brandenburg, Hailong Dao, S. Carnahan Nov 28 '10 at 5:53

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This question belongs on – S. Carnahan Nov 28 '10 at 5:54
The two answers below are pointing at a more general class of examples: any simple ring has the property that the two-sided ideal generated by any non-zero element is the whole ring. Building on this, it is easy to make examples in finite products of simple rings as well. – Emerton Nov 28 '10 at 6:59
up vote 7 down vote accepted

Yes, for example $r=\begin{pmatrix}1&0\cr0&0\end{pmatrix}$ generates the whole 2x2 matrix ring, $$ \begin{pmatrix}1&0\cr0&0\end{pmatrix} + \begin{pmatrix}0&0\cr1&0\end{pmatrix} \cdot \begin{pmatrix}1&0\cr0&0\end{pmatrix} \cdot \begin{pmatrix}0&1\cr0&0\end{pmatrix} = \begin{pmatrix}1&0\cr0&1\end{pmatrix}. $$

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Take for example the endomorphism ring of a finite dimensional vector space. There are only trivial two sided ideals. So the ideal generated by a non zero element is the whole ring.

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