# Characteristics of a pretty ring

This is a problem from a test I took today.

Definition: A pretty ring $$R$$ is a ring with unity 1, not a field, and each nonzero element can be written uniquely as a sum of a unit and a nonunit element of $$R$$.

The first problem, which is much easier, is to find such a ring. A got difficulty at first but came out with $$R = \left\{\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}\mid a,b \in \mathbb{Z}_2\right\}$$ with usual operation done in modulo $$2$$.

The latter problem, which I can't solve yet is to find all possible characteristic of pretty ring.

From some experiment with almost the same ring, I came to conclusion that a pretty ring can only has characteristic 2. But, I can't prove it for the general case.

• You can do this construction with square matrices of any size, not just $2$. Jun 15 '19 at 10:37

We show that a pretty ring has exactly one unit. Indeed, if $$0 \neq u$$ is not a unit and $$e$$ is a unit, then $$(u+e) + 0 = e + u$$ tells us that $$u+e$$ is not a unit (otherwise we get the contradiction $$u=0$$). Nowe take two units $$e, \tilde{e}$$ then $$e + (u+ \tilde{e}) = \tilde{e} + (u+ e)$$ implies $$e=\tilde{e}$$ and hence $$1$$ is the only unit.

On the other hand every ring with only one unit is a pretty ring as we can write every $$x\neq 0$$ as $$x = 1 + (x-1).$$

Thus, we have for a unital ring $$R\neq \mathbb{Z}/2 \mathbb{Z}$$: $$R \text{ is a pretty ring} \quad \Leftrightarrow \quad \vert R^\times \vert =1$$ In particular we have $$-1=1$$ and thus a pretty ring has either characteristic equal to $$1$$ or $$2$$. Note that both cases are possible as the zero ring is a pretty ring.

Shorter proof: Assume that that the characteristic of the pretty ring is not $$2$$. Then we get from $$1+0=-1+2$$ must be a unit. Let $$u\neq 0$$ be a non-unit (exists as a pretty ring is not a field), then $$1+0=(u+1) - u$$ implies that $$u+1$$ is a non-unit. Then we get from $$(u+1)+1=u +2$$ that $$0=1$$, ie. our pretty ring is the zero ring. Therefore, a pretty ring has characteristic $$1$$ or $$2$$.

• @rschwieb I assume by contradiction the characteristic is not $2$ and then in particular $2$ is a unit. Jun 16 '19 at 16:09
• that alone is not sufficient to make 2 a unit. The characteristic of the integers is also not 2. Jun 16 '19 at 16:43
• @rschwieb Thanks for pointing out my stupid mistake. I changed the proof and showed something more general. Jun 16 '19 at 16:59
• @rschwieb One could also have left it like it was by noting $$-1 + 2 = 1 + 0.$$ This shows that either $0=2$ or $2$ is a unit. Jun 16 '19 at 17:14
• Yes, I like that latter version Jun 16 '19 at 18:02

Only a partial answer. Suppose $$n \geq 3$$ be the characteristic of ring. Then $$n.1 = 0$$ implies that $$1 = -(n-1).1 = -(n-2).1 + (-1).1$$, since $$-1$$ is unit. If $$-(n-2).1$$ is not unit then by the uniqueness we get $$-(n-2).1 = 0$$ which contradicts that $$n$$ is the characteristic. So $$-(n-2).1$$ must be unit. Since $$(n-2).1 + 2.1 = 0$$, which implies that $$2.1$$ is a unit. Hence characteristic can not be $$2$$, even can not be even number. So we have proved that such $$n$$ must be odd.

• $2$ cannot be the characteristic, since you assumed already that the characteristic $n$ is at least $3$. Jun 15 '19 at 10:05
• @GreginGre: Azlif wants the possibility of characteristic. Did I understood correctly? Jun 15 '19 at 10:09
• You are assuming that the characteristic is bigger than $2$. Of course then we get that the characteristic is not $2$ under this assumption... Jun 15 '19 at 15:02
• I realised that. Jun 15 '19 at 16:18