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I apologize in advance, as this is not a particularly precise question, but I was thinking a bit in my spare time and the following occurred to me:

There exists no nonassociative ring with a cyclic additive group generated by the multiplicative identity.

One can prove this pretty simply by an application of the distributive property (for arbitrary elements a, b, c, and multiplicative identity e):

(ab)c = (a(e+e...+e))c = (a+a...+a)c = ac+ac...+ac = a(c+c...+c) = a((e+e...+e)c) = a(bc)

But what about nonassociative rings with cyclic additive groups that are not generated by the multiplicative identity? Such a ring clearly must be infinite, but I see no obvious reason why one could not exist.

Moreover, it seems to me that such structures, while not truly associative, seem to be "almost associative" in some sense.

In particular, while the multiplicative identity does not generate the additive group, one can express any element as a product of the generator of the additive group and some element generated by the multiplicative identity. For example, say our generator is g. Then, for arbitrary element a and multiplicative identity e,

a = g+g...+g = g(e+e...+e)

By similar reasoning to what was done above, then, for arbitrary elements a, b, and c,

(ab)c = (ag)((e+e...+e)c)

In other words, any element that is generated by the multiplicative identity will associate, and any element which is not can be split up into a product of the generator and such an element (with the latter part, thus, associating).

So, here's my (admittedly hopelessly broad) question: Is this at all a useful notion that leads to anything interesting (maybe a classification of such rings)? Do such nonassociative rings even exist? If so, can anyone provide an example of one?

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Let $R$ be a unital, possibly nonassociative ring, and assume that $R$ is generated as an additive group by $g \in R$.

Note that the unique unital ring homomorphism $\mathbb Z \to R$ is well-defined - all that is required is the unit and distributivity. In particular for any $n \in \mathbb Z$ and $r \in R$, the expression $n \cdot r \in R$ is well-defined.

  • If $R$ has characteristic $k>0$, then $k \cdot g = 0$. It follows that $g$ has additive order dividing $k$, so the number of elements of $R$ is at most $k$. Since it contains the $k$ elements generated by the unit, that must be all of them, so $R \simeq \mathbb Z/k\mathbb Z$.

  • If $R$ has characteristic $0$, then the unit $e \in R$ satisfies $e = k \cdot g$ for some $k \in \mathbb Z$. But then $g = e \cdot g = (k \cdot g) \cdot g$, which by distributivity after expanding $k$ as a sum of $e$'s can be rewritten as $k \cdot (g \cdot g)$. So whatever the element $g \cdot g \in R$ is, it must have the property that $g = k \cdot(g \cdot g)$ - it generates the additive generator of $R$. Since the only additive generators of $R$ are $\pm g$, it follows that $k= \pm 1$ and $R \simeq \mathbb Z$.

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  • $\begingroup$ I follow your argument for the case of characteristic 0, but not for characteristic k>0; I do not see how it follows that e is a generator. $\endgroup$ – user3716267 Sep 23 '16 at 20:55
  • $\begingroup$ @user3716267 I thought that was the case you said was "obvious"? I'm not rereading your question in any case; I edited to expand on that part. $\endgroup$ – Dustan Levenstein Sep 23 '16 at 21:00
  • $\begingroup$ I had a mild brain fart and didn't realize that there's no way to have an infinite ring with finite characteristic if the additive group is cyclic. $\endgroup$ – user3716267 Sep 23 '16 at 21:03
  • $\begingroup$ @user3716267 I see. $\endgroup$ – Dustan Levenstein Sep 23 '16 at 21:05

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