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 $1$?


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @SeverinSchraven: That is the downward Löwenheim-Skolem theorem. $\endgroup$ – Henning Makholm Jun 16 at 12:44
  • 2
    $\begingroup$ 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

$\mathbb F_2[X]$ is pretty but does not have the form you conjecture.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.