Does every ring $R$ have a subring $S$ with cyclic additive subgroup $(S,+)$?

I was wondering if the statement above is correct or not.

I think it is, but I'd like to know if I got it right:

For every ring $R$, $1\in R$. Hence, if we take the additive subgroup $\langle 1\rangle$, it's obviously cyclic.

That's very basic, and therefore I thought I've missed something.

Thank you!

  • 2
    $\begingroup$ You might want to prove that $<1>$ is a subring, so it is closed under the product too (that is easy in any case) $\endgroup$ Aug 11, 2021 at 19:24
  • $\begingroup$ @LorenzoPompili You are right, thank you very much! $\endgroup$
    – Ludwig
    Aug 11, 2021 at 19:26

1 Answer 1


Your are missing a step: You claim that every ring $R$ has a subring $S$ with cyclic additive group. Yet you only show (well, note) that the additive subgroup generated by the unit element $1_R\in R$ is cyclic. But is it a subring too?

The answer is, indeed, yes. In fact, this is tightly related to the special role $\mathbb Z$ plays among all (commutative) rings with unity. When you consider $\langle1_R\rangle\subseteq R$ you obtain the set


"embedded" into $R$ (well, up to characteristic; i.e. $n\cdot 1_R=0$ might occur for $n\ne0$ in $R$). To see that $\langle1_R\rangle$ makes a perfectly fine subring define $f\colon\mathbb Z\to R,\,1\mapsto1_R$, check that it is a ring homomorphism and convince yourself that $\operatorname{im}(f)=\langle 1_R\rangle$.

This ring homomorphism can be define always and $\mathbb Z$ is essentially the only ring with unity who does the job (this is -intentionally though- a bit vague; making it precise goes beyond the scope of this question). Moreover, we define the characteristic of a ring $R$ as the index of the kernel of this map if it is not an embedding (in the latter case we say that the ring has characteristic $0$).


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