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I'm exploring rings that have the following property:

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For all $x \in R$, and for all $n \in \mathbb{N}$, there exists $y \in R$ such that $\sum_{i=1}^{n} x = y \cdot x$. $\rule{10cm}{0.4pt}$

Any ring with unity satisfies this, $y = \sum_{i=1}^{n} 1$. Any Boolean ring also satisfies this, $y = \sum_{i=1}^{n} x$.

My general questions are:

(i) Are there non-Boolean rings without unity that have this property? Or better: are there any rings with elements of infinite order without unity that have this property? (ii) Is this property interesting? (iii) What if we require that $y$ be unique?

Thank you

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1) yes, any ring with local right identities has this property. That is, if for every $x$, there exists $e$ such that $xe=x$. Such an example is $\prod _{i=1}^\infty \mathbb Z$.

Actually if you think about it, this is equivalent to your condition, since $\sum_{i=1}^n=y$ does the trick.

2) rings with local identities are interesting.

3) Let me share a bit about what I puzzled out about what happens if you additionally required right identities to be unique.

The first thing you notice is that all the right identities are idempotent:

$x=xe=xe^2\implies e=e^2$

So each idempotent is its own unique right identity.

But $0\cdot e=0$ for every idempotent, so $e=0$, if $0$ has a unique right identity. Unfortunately this means that $R$ is the ring $\{0\}$. So with the uniqueness assumption on all right identities, everything collapses.

Perhaps it can be saved by supposing that just nonzero elements have a unique right identity?

Supposing that, let $e$ be a nonzero idempotent and $f$ be an idempotent such that $ef=e$, it must be that $f=e$, because $e$ is its own unique right identity.

For any element $r$ and idempotent $e$, $e+re-ere$ is an idempotent, and in particular if $r=x$ and $xe=x$, you have $f=e+x-ex$ is an idempotent. Furthermore, $ef=e$. If $e$ is nonzero, it must be that $e=e+x-ex$ and then we see that $x=ex$ also, so the right identity is actually a two-sided identity.

Another thing, suppose that $e$ and $f$ are two idempotents that commute. Then $ef$ is an idempotent such that $(ef)e=ef$ and $(ef)f=ef$, therefore either $ef=0$ or else $e=ef=f$. The upshot is that if two different idempotents commute, they have to be orthogonal. In a commutative ring with identity, it would mean there are no nontrivial idempotents.

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  • $\begingroup$ Very clear, thank you. $\endgroup$ – Jonathan Hebert Jan 11 '18 at 3:11
  • $\begingroup$ @JonathanHebert Added a few of the thoughts that I had which I couldn't enter with my phone. $\endgroup$ – rschwieb Jan 11 '18 at 15:01

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