Let $S$ be a ring with identity (but not necessarily commutative) and $f:M_{2}(\mathbb R)→S$ a non zero ring homomorphism ($M_{2}(\mathbb R)$ is the ring of all $2\times 2$ matrices). Has $S$ infinitely many nilpotent elements?
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1Do you mean "ifinitely many nilpotent elements"? If not, what does "infinitely nilpotent" mean here? – Geoff Robinson Jan 06 '13 at 15:09
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We have that $M_2(\mathbb R)/\ker f\simeq\operatorname{Im}f$. But $\ker f$ is a bilateral ideal of $M_2(\mathbb R)$, so $\ker f=(0)$. (It is well known that $M_2(\mathbb R)$ is simple.) This shows that $f$ is injective and sends nilpotents to nilpotents, so $S$ have infinitely many nilpotents. (It's easy to check that the matrix ring $M_2(\mathbb R)$ has infinitely many nilpotents.)