# Find finite rings $(R,+,\times)$ such that for every unit $r$, $r-1$ is a unit except $r=1$.

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...

• It might be useful to keep in mind that every finite integral domain is a field. – Gerry Myerson Aug 6 '19 at 9:54
• @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 6 '19 at 10:00
• 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 6 '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 6 '19 at 10:10
• @WlodAA Note that $(r-1)y=1$ and $rx=1$. So $(r-rx)y=(r-1)y$ and it follows. – Zongxiang Yi Aug 7 '19 at 2:02

Indeed, they have only one unit, $$\ B^x=\{1\}.$$
• Logically, Boolean Algebras $R=B$ are "such rings". However it is trival where $R^\times\setminus\{1\}=\emptyset$. – Zongxiang Yi Aug 7 '19 at 2:05