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| Title of object: |
Baer ring |
| Canonical Name: |
BaerRing |
| Type: |
Definition |
| Created on: |
2006-04-23 23:39:13 |
| Modified on: |
2006-04-23 23:39:13 |
| Classification: |
msc:16U99, msc:47C15, msc:47C10 |
| Defines: |
Baer *-ring |
Preamble:
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Content:
Let $R$ be a ring with multiplicative identity $1$. Then $R$ is called a left \emph{Baer ring} if, for any subset $S$ of $R$, the left annihilator of $S$ is left principal, generated by an idempotent:
$$\operatorname{l.ann}(S):=\lbrace r\in R\mid rS=0\rbrace=Re.$$
A right Baer ring is defined similarly, by replacing the word left with right above. It turns out that a left Baer ring is a right Baer ring, and vice versa, so we may drop the word left or right in the name.
Clearly, a domain is a Baer ring. And, by Wedderburn-Artin theorem, a semisimple ring is also a Baer ring. Another example, found in operator theory, is the ring of bounded linear operators on a Hilbert space.
A closely related concept is that of a left (right) \emph{Baer *-ring}: it is a ring with involution $*$ such that the left (right) annihilator of any subset is left (right) principal, generated by a projection (an idempotent that is in addition a self-adjoint element).
A left Baer *-ring is a left Baer ring, a right Baer *-ring is a right Baer ring. And, interestingly, the notion of left and right is also redundent for Baer *-rings. For example, let's show left means right. Since a left Baer * ring $R$ is left, and consequently right Baer, the right annihilator of a subset $S$ has the form $\operatorname{r.ann}(S)=eR$ for some idempotent $e\in R$. Since $Se=0$, $e^*S^*=0$ or that $\operatorname{l.ann}(S^*)=Re^*$. But $R$ is left Baer *, $\operatorname{l.ann}(S^*)=Rf$ for some projection $f\in R$. So $fR=f^*R^*=(Rf)^*=(Re^*)^*=e^{**}R^*=eR=\operatorname{r.ann}(S)$. |
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