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Let $R$ be a ring with descending chain condition on right ideals. Suppose $l(R)=0$ (the left annihilator of $R$) and $\exists c\in R$ with $r(c)=0$ (the right annihilator of $c$).

Show that $R$ has identity.

I think I can prove that there is a left identity, although I am not confident about it. I cannot seem to find a way to construct an identity, so if someone could point me down a path, that would be great!

Cheers

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  • $\begingroup$ How would you prove that there is a left identity ? $\endgroup$ Mar 23 '16 at 13:00
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That's a nice question. Thanks for sharing it.

For $a \in R,\,$ $aR := \{ar\mid r \in R\}$ is a right idal (note: it is not necessarily the ideal generated by $a$). Hence $$cR \supseteq c^2R\supseteq \cdots$$ is a decending chain of right ideals and by decending chain condition the chain stabilizes, i.e. $c^nR = c^{n+1}R$ for some $n > 0$. In particular, there is $e \in R$ s.t. $$c^n \cdot c = c^{n+1}\cdot e.$$ So $c^n(c-ce)=0$ and since the right annulator of $c$ is zero, we obtain $c-ce=0.$ Multiplying from the right yields for all $r \in R:\,\, 0=cr-cer=c(r-er)$ i.e. $r-er$ is a right annulator of $c$. Hence $r=er$, and $e$ a left identity.

Fix $r \in R$. Then for all $s \in R$: $$(r-re)s=rs - r(es)=rs - rs =0$$ Thus $r-re$ annulates $R$ from the left and since the left annulator of $R$ is zero we find $r=re$ and $e$ is also a right identity.

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  • $\begingroup$ This was my line of thought too, but I got $c^n=c^{n+1}e$ And didn't think of doing the same thing with $c^{n+1}=c^{n+1}e$ . Bravo $\endgroup$
    – rschwieb
    Mar 24 '16 at 1:37
  • $\begingroup$ Nice, I was getting convinced there had to be some modification to the left identity. Thanks a lot. $\endgroup$
    – Rhys Evans
    Mar 24 '16 at 10:44
  • $\begingroup$ @rschwieb: Right, if we know that $R$ has an identity, it follows that $c$ has a right inverse. $\endgroup$ Mar 24 '16 at 12:30

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