# $R$ a ring with 1 and charac. $n>1$ (resp. 0) $\implies R$ contains a subring $\simeq \Bbb{Z}_n$ (resp. $\Bbb{Z}$). What happens if $R$ has no 1?!

Definition: Let $R$ be a ring. We say that $n\in \Bbb{Z}_+$ is the characteristic of $R$ if it is the least positive integer such that $n r=0$, for all $r\in R$ (here $nr$ denotes $r+r+\dots+r$, "$n$ times"). If such $n\in \Bbb{Z}_+$ does not exist, we say that $R$ has characteristic zero.

If the ring has identity $\mathbf{1}$, one can easily show that the above definition of characteristic of $R$ coincides with the least positive integer such that $n\mathbf{1}=0$, in the case of positive characteristic; and that there is no positive integer such that $n\mathbf{1}=0$, in the case of null characteristic.

One can show also:

$(*)$ If $R$ is a ring with identity $\mathbf{1}$ with characteristic $n$, then $R$ contains a subring isomorphic to:

1. $\Bbb{Z}$, if $n=0$;
2. $\Bbb{Z}_n$, if $n> 1$.

So, are there any examples of rings $R$ and $S$, both without identity, such that $R$ contains a subring as in $(*)$ and $S$ does not (also in both cases: $n=0$, $n>1$)?

• Examples of $S$ are fairly straightforward. For 1 consider the (non-unital) ring of even integers, and for 2 perhaps the (non-unital) rings $n\Bbb{Z}/n^2\Bbb{Z}$ – sharding4 Jul 1 '17 at 18:01
• Hmm. The examples of $2\Bbb{Z}$ and $n\Bbb{Z}/n^2\Bbb{Z}$ may actually go the other way if you allow ring homomorphisms to send $1$ to a non-identity element. – sharding4 Jul 1 '17 at 18:14
• @Anderson Felipe Viveiros That's exactly what I mean by the category of rings you work out with in fact. The usual definition of a ring homomorphism doesn't allow what sharding4 suggests. – user321268 Jul 1 '17 at 18:16
• @mayer_vietoris I've got your point! Thank you. I was really working with homomorphisms sending $1$ anywhere. I was considering homomorphisms even between rings without identity: for me, they just need to preserve operations. :) – Derso Jul 1 '17 at 18:20
• @AndersonFelipeViveiros: Generally, for people whose default interpretation of "ring" is a ring with unit, $1$ is considered to be one of the operations (a "nullary operation", often called a "constant") to be preserved. Compare with the notion of a monoid homomorphism. – Hurkyl Jul 1 '17 at 18:21

## 3 Answers

I think it's a matter of convention in principle. How do you define the characteristic? For me its the positive integer $n$ occurring in front of the ideal $n\mathbb{Z}$, which is the kernel of the unique map $\mathbb{Z} \rightarrow R$, that makes $R$ a $\mathbb{Z}$-algebra. This map necessarily (in the category of rings I'm working) gets $1$ to $1_R$, hence in that case it's kind of mandatory to have a unital element to define the characteristic.

However, although I haven't seen an algebraist in practise working with rings with non-unital element, probably you can define the characteristic to be given if you think the embedding of it (which always exists) into a ring with a unit element and work out the characteristic in that case.

• As I wrote in the post, the given definition generalizes the definition of characteristic of rings with identity. – Derso Jul 1 '17 at 18:07
• Sorry, misunderstanding. You're right of course, I read your question before giving my reply, it's a kind of rhetorical question in fact. What I mean is that your definition is correlated with the one I wrote out above, hence unit element is necessary in the sense of the usual definition I'm using. – user321268 Jul 1 '17 at 18:14

If the ring $R$ contains an idempotent $e$ (i.e. an element such that $e^2=e$) then the subring ${\Bbb Z}e$ gives your example.

Examples for characteristic $0$: Consider $\mathbb Z \times 2\mathbb Z$ clearly this is a ring without identity, containing $\mathbb Z$ as a subring.

For a ring which does not contain a subring isomorphic to $\mathbb Z$ consider the ring with additive group $\mathbb Q / \mathbb Z$ and trivial multiplication (i.e. $ab = 0$ for every $a,b$), then this ring has characteristic $0$, but it cannot contain a subring (not even an additive subgroup) isomorphic to $\mathbb Z$, as every element in $\mathbb Q / \mathbb Z$ has finite order. We could have also used $\mathbb Z$ with trivial multiplication, since the only element of $Z$ with the usual multiplication that satisfies $a^2=0$ is $0$.

To give examples in characteristic $n$, let us denote the additive group of $\mathbb Z / n \mathbb Z$ considered as a ring with trivial multiplication by $R_n$. Then $\mathbb Z / n \mathbb Z \times R_n$ is a ring of characteristic $n$ that contains $\mathbb Z / n \mathbb Z$ as a subring and $R_n$ is a ring of characteristic $n$ that does not contain $\mathbb Z / n \mathbb Z$ as a subring. ($\mathbb Z / n \mathbb Z$ does contain an element that does not square to $0$, but $R_n$ does not.)