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Today I answered this question Characteristics of a pretty ring and wondered whether one could characterize these rings.

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 $R$.

Note 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 (different from $\mathbb{Z}/ 2 \mathbb{Z})$ 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 $R\neq \mathbb{Z}/2 \mathbb{Z}$: $$ R \text{ is a pretty ring} \quad \Leftrightarrow \quad \vert R^\times \vert =1 $$ Now, my question is the following:

Are all pretty rings of the form $$ R_I = \Pi_{i\in I} (\mathbb{Z}/2\mathbb{Z})$$ where $I$ is a set with cardinality different from $1$?

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The answer is no : indeed the notion of pretty ring can be axiomatized as a first order theory in the language of rings, and it has an infinite model, therefore it has a model of cardinality $\aleph_0$, whereas your examples are finite or uncountable.

If you don't know any model theory, that's not an issue : just note that since the only unit is $1$, any subring of a pretty ring is itself pretty, and surely an infinite ring has a countable subring.

It is however true that any pretty ring has characteristic $2$ : if $2\neq 0$, write $2$ as $1+$ a nonunit to get a contradiction.

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  • $\begingroup$ Honestly I don't see how one gets a countable infinite subring. Do you have a reference where I could look up the model theory I'd need to understand your statement? $\endgroup$ – Severin Schraven Jun 16 at 12:42
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    $\begingroup$ @SeverinSchraven: That is the downward Löwenheim-Skolem theorem. $\endgroup$ – Henning Makholm Jun 16 at 12:44
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    $\begingroup$ Just look at a subring generated by countably many elements : it is countable $\endgroup$ – Max Jun 16 at 12:45
  • $\begingroup$ @HenningMakholm Thanks for the key word $\endgroup$ – Severin Schraven Jun 16 at 12:49
  • $\begingroup$ @Max Of course, that was a stupid question. $\endgroup$ – Severin Schraven Jun 16 at 12:50
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$\mathbb F_2[X]$ is pretty but does not have the form you conjecture.

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  • $\begingroup$ Indeed, you are right. Nice example! $\endgroup$ – Severin Schraven Jun 16 at 16:13

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