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Let $R$ be a ring such that $R$ has no non-trivial right ideals. If there exists a nonzero element $a \in R$ with $aR=0$, prove that $|R|= p$ where $p$ is prime.

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Set $I=\{b\in R:bR=0\}$. This is a right ideal and then $I=0$ or $I=R$. By hypothesis $I\ne 0$, so $I=R$.

Thus we have $xy=0$ for all $x,y\in R$. Then $(R,+)$ is an abelian group with no nontrivial subgroups, so $|R|=p$ a prime number.

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