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Apr 1, 2013 at 10:08 history edited Dag Oskar Madsen CC BY-SA 3.0
changed the first sentence
Apr 1, 2013 at 10:01 comment added Dag Oskar Madsen I think I am kind of cheating with this example and not giving you the sort of answer you are after. Maybe there should be more conditions on the ring $R$?
Apr 1, 2013 at 9:32 comment added Mariano Suárez-Álvarez @yeshengkui: there is a canonical isomorphism $M_n(A\times B)\cong M_n(A)\times M_n(B)$, and the subspace $M_n(A)$ is an ideal of $M_n(A\times B)$ under this identification.
Apr 1, 2013 at 7:38 comment added yeshengkui @Dag:Thanks for your answer. But why is $M_2(M_2(R))$ an ideal of $M_2(R)$? Have you assumed that $M_2(R)$ is a direct product of $M_2(M_2(R))$ and $M_2(\bb{C})$ to define the map $M_2(R)\rightarrow M_1(R)$? Why can we have such a product?
Apr 1, 2013 at 3:41 history answered Dag Oskar Madsen CC BY-SA 3.0