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

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!

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

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
$$\{0,\pm1_R,\pm2\cdot1_R,\pm3\cdot1_R,\dots\}$$
"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 $$0$$).