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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...

Zongxiang Yi
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    It might be useful to keep in mind that every finite integral domain is a field. – Gerry Myerson Aug 06 '19 at 09:54
  • @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 $r$. – Zongxiang Yi Aug 06 '19 at 10:00
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    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. – Gerry Myerson Aug 06 '19 at 10:06
  • @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. – Zongxiang Yi Aug 06 '19 at 10:10
  • Well, applying the hypothesis to the subring generated by $1$, we easily get that the characteristic is prime, so $R$ must be an algebra over $\Bbb F_p$. Playing around with (minimal) polynomials, I guess, it could lead to the desired result, that is, $R$ actually is a field. – Berci Aug 06 '19 at 11:26
  • @GerryMyerson, $\ 0\ $ is not a unit but $\ 0+1\ $ is a unit. – Wlod AA Aug 06 '19 at 11:28
  • @Wlod, I'm not sure what your point is. We already know that when $r$ is the unit $1$, then $r-1$ is not a unit; we're wondering what happens if for every unit $r$ other than $1$, $r-1$ is a unit. – Gerry Myerson Aug 06 '19 at 12:00
  • @GerryMyerson, I guess, I didn't follow the thread of the discussion. Sorry. – Wlod AA Aug 06 '19 at 18:21
  • Zhong Xiang Ti, I don't see where did (−) come from? (after words "Now we have"). – Wlod AA Aug 06 '19 at 18:29
  • @Berci Yes, it is true if the ring is generated by $1$. But how can I say such rings must be generated by $1$ ? – Zongxiang Yi Aug 07 '19 at 01:59
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    @WlodAA Note that $(r-1)y=1$ and $rx=1$. So $(r-rx)y=(r-1)y$ and it follows. – Zongxiang Yi Aug 07 '19 at 02:02
  • Zhong Xiang Yi, thank you (blind me, so embarrassing :) ). – Wlod AA Aug 07 '19 at 02:06

1 Answers1

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Boolean Algebras B are "such rings".

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

Wlod AA
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