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Rings are supposed to be associative and unital, but not necessarily commutative. Some definitions:

  • (Bass) A ring $R$ is said to have stable range $1$ if for all $a,b \in R$, whenever $Ra+Rb=R$, then there exists $x\in R$ with $a+xb$ being a unit.
  • (Warfield) A ring $R$ is said to be an exchange ring if it has the finite exchange property or equivalently (Nicholson) if for all $a\in R$ there exists an idempotent $e\in R$ with $e\in Ra$ and $1-e \in R(1-a)$.

Examples: Semiperfect rings have stable range $1$ and are exchange.

My question is whether someone has an example of a ring $R$ with stable range $1$ and Jacobson radical zero, that is not exchange?

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I know at least two:

  1. The integral closure of $\mathbb Z$ in $\mathbb C$
  2. The ring of holomorphic functions on $\mathbb C$.

Each nontrivial ideal of an exchange ring with Jacobson radical zero must contain a nonzero idempotent, but since these are both domains, this is clearly not the case for them.

I found these using this DaRT query.

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  • $\begingroup$ Nice! Thanks a lot. Yes, for an exchange ring that is a domain, an element x is either invertible or 1-x is invertible. $\endgroup$ – Hugo Jan 16 '20 at 21:13

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