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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.

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

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  • $\begingroup$ @rschwieb I assume by contradiction the characteristic is not $2$ and then in particular $2$ is a unit. $\endgroup$ Jun 16 '19 at 16:09
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    $\begingroup$ that alone is not sufficient to make 2 a unit. The characteristic of the integers is also not 2. $\endgroup$
    – rschwieb
    Jun 16 '19 at 16:43
  • $\begingroup$ @rschwieb Thanks for pointing out my stupid mistake. I changed the proof and showed something more general. $\endgroup$ Jun 16 '19 at 16:59
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    $\begingroup$ @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. $\endgroup$ Jun 16 '19 at 17:14
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    $\begingroup$ Yes, I like that latter version $\endgroup$
    – rschwieb
    Jun 16 '19 at 18:02
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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.

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

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