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The ring is semiprime if $x\in R, xyx=0$ for all $y\in R$ implies $x=0$ or equivalently for $x\neq0$ exists $y_{0}\in R$ such that $xy_{0}x\neq0$.

I found an example of semiprime ring. However, I am not sure that if I understand properly.

$R=\left\{ \begin{pmatrix}a & 0\\ 0 & b \end{pmatrix},\; a,\, b\in\mathbb{Z}_{2}\right\} =\left\{ \begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix},\begin{pmatrix}0 & 0\\ 0 & 1 \end{pmatrix},\begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix},\begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\right\} $

This ring is a semiprime because

$\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix}y\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix}=\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix}$

Am I right? Could you give me some other example of semiprime rings?

Thank you.

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You need to check for any other $x\in R$, there exists a $y_0=I$ satisfying $xy_0 x=x^2$ is not $0$. –  asatzhh Aug 3 '13 at 16:59
    
@asatzhh: Why does it need to satisfy this condition which you wrote? –  Monika Aug 3 '13 at 17:02
    
Because you need to verify the "implies" but not the trivial result $0y0=0$. –  asatzhh Aug 3 '13 at 17:10
    
@asatzhh: Why your condition looks like you wrote? and what do you mean by $I$? –  Monika Aug 3 '13 at 17:15
    
Here $I$ means identity matrix, as is always in matrix group. –  asatzhh Aug 3 '13 at 17:20

1 Answer 1

The example you gave is indeed semiprime, but it is a complicated way to look at the ring $F_2^2$.

Given $(a,b)$ nonzero, one of $(a,b)(1,0)(a,b)$ or $(a,b)(0,1)(a,b)$ is nonzero.

Any semisimple or Von Neumann regular ring or prime ring is going to be semiprime.

I recommend that you try to show that the ring of linear transformations of any vector space is semiprime.

Rings that don't have nilpotent elements are also semiprime.

Another easy source is to take any semiprime ideal J in a ring R and use R/J. Semi prime ideals are easy to find: they're just the intersections of prime ideals.

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