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Are these two definitions equivalent?

  1. A ring $R$ is called semiabelian by Yiqiang Zhou if its identity $1$ can be written as a finite sum $1 = e_1 + \cdots + e_n$ of mutually orthogonal idempotents $e_i$ such that each corner ring $e_iRe_i$ is abelian.

  2. A ring $R$ is called semiabelian by Weixing Chen if every idempotent $e$ of $R$ is either left semicentral or right semicentral, where an idempotent $e$ of $R$ with complementary idempotent $f = 1-e$ is called left semicentral if $fRe = 0$ and right semicentral if $eRf = 0$.

Both definitions are proper generalizations of the notion of an abelian ring, and the first definition includes all semiperfect rings.

Follow-up: The answer from leslie townes shows that these definitions are not equivalent. More specifically, it shows that $(1) \not\Rightarrow (2)$. Does $(2) \Rightarrow (1)$?

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By "abelian" do you mean commutative? – Qiaochu Yuan Jan 7 '12 at 7:40
By abelian, I mean all idempotents in the ring are central. – Anononym Jan 7 '12 at 17:18
up vote 3 down vote accepted

If you fix any commutative ring $R$ it seems to me that the ring $M$ of all $2 \times 2$ matrices with entries from $R$ (with the usual matrix operations: entrywise addition and multiplication according to the usual formulas) is semiabelian according to the first definition but not semiabelian according to the second.

Let $e_1 \in M$ denote the $2 \times 2$ matrix with a $1$ in the upper left corner and zeros everywhere else, and let $e_2 = I - e_1$ (where $I \in M$ denotes the usual $2 \times 2$ identity matrix). Then $e_1$ and $e_2$ are mutually orthogonal idempotents summing to $I$, and each of the rings $e_i M e_i$ is isomorphic to $R$ and hence commutative. So $M$ seems to verify the requirements of the first definition.

On the other hand, $e_1 M e_2$ is the set of strictly upper triangular matrices and $e_2 M e_1$ is the set of strictly lower triangular matrices; neither of these are $\{0\}$ so this seems to show that $M$ does not satisfy the requirements of the second definition.

Of course there is nothing special about $2 \times 2$ matrices here; it seems that essentially the same things could be said of the ring of $n \times n$ matrices over a commutative ring $R$ by adjusting the discussion only a little bit.

By all means, read this very critically. I could be making a mistake--- ring theory is not my specialty.

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Incidentally, this answer shows that the second definition does not include all semiperfect rings. – Anononym Jan 7 '12 at 17:17

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