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Let $R$ a left artinian ring. Show that if $a \in R$ is not a right zero divisor, then $a$ is a unit in $R$.

Comments: I am tryed to do so: See the R-module $_{R}R$ and consider the function $f: _{R}R \longrightarrow _{R}R$ defined by $f(r)= ra$. As $R$ is artinian then $_{R}R$ has a composition series. I do not know if it's out there the way.

Croos
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1 Answers1

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Hint: consider the descending chain of left ideals $Ra \supseteq Ra^2 \supseteq Ra^3 \supseteq \dots $ and use the left artinian hypothesis.

lokodiz
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  • If $R$ is artinian then we have that exist $n$ such that $Ra^n = Ra^{n+1}=...$ then $a^n = ra^{n+1}$ to some $r \in R$ then $(ra-1)a^{n}=0$, but I can not see how to use the fact that $a$ is not a right zero divisor. – Croos May 10 '15 at 17:19
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    Since $a$ is not a right zero divisor, neither is $a^n$. More explicitly (at least for the case at hand), $0 = (ra-1)a^n = \big( (ra-1)a^{n-1}\big)a$, and since $a$ is not a right zero divisor, we must have $(ra-1)a^{n-1} = 0$. Now induct. – lokodiz May 10 '15 at 17:23
  • I confirmed you the correct answer, but as I see that $a$ is left-invertible? – Croos May 10 '15 at 18:11
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    One way to see this is as follows: since $ra=1$, $r$ is also a right non-zero-divisor (if $xr = 0$, then $0 = 0\cdot a = xra = x$). By the same argument as above, $r$ has a left inverse $s$, so $sr = 1$. But then $s = s \cdot 1 = s(ra) = (sr)a = 1 \cdot a = a$, so $s=a$ and we therefore have both $ra=1$ and $ar = 1$. – lokodiz May 10 '15 at 18:56