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I have the following question.

Let $R$ be a ring with identity.

Prove or give a counter example:

If $a \neq 0$ is a non-unit element of $R$, then the ideal generated by $a$ is a proper ideal of $R$ where,

$\langle a \rangle=\{\sum_{i=1}^{n}r_{i}as_{i}:r_{i},s_{i}\in R,n \in \mathbb{N},i = 1,2,...,n\}$

Thanks in advance.

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    $\begingroup$ Let R = M_n(K) where K is a field then M_n(K) doesnot have any proper ideals except 0. $\endgroup$ – random123 Sep 8 '15 at 17:15
  • $\begingroup$ @random123 Please consider making that into a solution. $\endgroup$ – rschwieb Sep 8 '15 at 17:22
  • $\begingroup$ @random123 Thank you very much. $\endgroup$ – user28083 Sep 8 '15 at 17:29
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Let $R$ be a commutative ring with identity then it is a known fact that two-sided ideals in $M_n(R)$ are of the form $M_n(I)$ where $I$ is an ideal of $R$. It is proved and discussed here What are the left and right ideals of matrix ring? How about the two sided ideals?. So if you consider $R = K$ to be a field then only zero and $K$ are the ideals of $K$ and hence the only two sided ideals of $M_n(K)$ are $M_n(0) = 0$ and $M_n(K)$ which is the full ring.

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