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Let $(R,+,\times)$ be a finite ring. $R^\times$ denotes the set of all invertible elements, i.e., units in $(R,\times)$.

Find finite rings $(R,+,\times)$ such that for every unit $r\in R^\times\setminus\{1\}$, $r-1$ is a unit.

I know that finite fields $\mathbb{F}_q$ are such rings. Then I try to prove that they must be finite field.

My Attempt:

Assume that $r$ and $r-1$ are units. Then there exist $x,y\in R^\times$ such that $$rx=xr=1$$ and $$(r-1)y=y(r-1)=1.$$ Now we have $$(r-rx)y=(r-1)y=1=rx$$ and thus $$(1-x)y=x.$$

I do not know how to contiue...

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    $\begingroup$ It might be useful to keep in mind that every finite integral domain is a field. $\endgroup$ – Gerry Myerson Aug 6 '19 at 9:54
  • $\begingroup$ @GerryMyerson How to say that such rings are integral domain? It is obvious if it is for every non-zero element $r$ rather than for every non-identity unit $r$. $\endgroup$ – Zongxiang Yi Aug 6 '19 at 10:00
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    $\begingroup$ If it's not an integral domain, it has a zero divisor, call it $r$. So $r$ is not a unit, so $r+1$ can't be a unit, so $r+2$ can't be a unit, and so on. Maybe you can make some progress, knowing there have to be all those non-units. $\endgroup$ – Gerry Myerson Aug 6 '19 at 10:06
  • $\begingroup$ @GerryMyerson if the characteristic is $p$, then $r+p=r$ for all $r\in R$. So we can only say that if there is one non-unit, then there are at least $p$ non-units. $\endgroup$ – Zongxiang Yi Aug 6 '19 at 10:10
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    $\begingroup$ @WlodAA Note that $(r-1)y=1$ and $rx=1$. So $(r-rx)y=(r-1)y$ and it follows. $\endgroup$ – Zongxiang Yi Aug 7 '19 at 2:02
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Boolean Algebras B are "such rings".

Indeed, they have only one unit, $\ B^x=\{1\}.$

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    $\begingroup$ I guess you mean Boolean rings. This shows the guess in the comments is not true.. $\endgroup$ – Berci Aug 6 '19 at 21:01
  • $\begingroup$ Thank you. (I have a weak connection between the above and below my nose). $\endgroup$ – Wlod AA Aug 6 '19 at 21:34
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    $\begingroup$ Logically, Boolean Algebras $R=B$ are "such rings". However it is trival where $R^\times\setminus\{1\}=\emptyset$. $\endgroup$ – Zongxiang Yi Aug 7 '19 at 2:05
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    $\begingroup$ Zhong Xiang Yi, true. It still contributes to a better and complete understanding of the problem (as simple as it is). $\endgroup$ – Wlod AA Aug 7 '19 at 2:08

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