I am searching for a ring $R$ (preferably semisimple) and a nonzero primitive idempotent $e\in R$ such that the corner ring $eRe$ is not local. I am inspired by the theorems on semiperfect rings in which there are idempotents $e_i$ with each corner ring $e_iRe_i$ a local ring. I have also scanned the ERT site (the encyclopedia of ring theory), but to no avail. Thanks for any cooperation!
A non-local corner ring
1
$\begingroup$
abstract-algebra
ring-theory
noncommutative-algebra
idempotents
local-rings
1 Answers
2
You seek a primitive idempotent which is not a local idempotent.
This is not possible in a semisimple ring, since these two notions coincide for semiperfect rings (Proposition 23.5 in Lam's First course in noncommutative rings)
Without additional conditions, it is fairly easy to find such idempotents: for example, $1\in \mathbb Z$ is primitive but not local.
-
0Thanks for the answer! Would you please give me a "nontrivial" non-local idempotent which is primitive? Could the ring $R$ be Boolean in this regard? – 2017-02-27
-
0Sorry! I meant by ERT the database of Ring Theory! – 2017-02-27
-
1$(0,1)\in\mathbb Z\times \mathbb Z$ is a "nontrivial" example. For boolean rings: obviously $eRe$ being an indecomposable Boolean ring implies $eRe=F_2$. So it is not possible in Boolean rings. – 2017-02-27
-
0Why $eRe$ is boolean when $R$ is so? – 2017-02-28
-
1@karparvar Because a subset of a boolean ring will satisfy the same identity? – 2017-02-28