# R contains a subring isomorphic to Z/car(R)

I am currently working on rings, and the following theorem in my course is stated without proof :

Let R be a (unital) ring. Then there exists a subring of R ( denoted S) that is isomorphic to $$\mathbb{Z}_\mathrm{char(A)}$$

The characteristic is defined as follows :

We say that R has characteristic $$c$$ if $$c$$ is the smallest integer such that $$1_R^c=0_R$$. If no such integer exists, we say that the ring has characteristic 0.

I have proven the following lemma :

If $$R'$$ is a subring of $$R$$, then $$\mathrm{char(R)}=\mathrm{char(R') }$$ which is easy as $$1_R=1_{R^\prime}$$

Now I can understand why a char of zero means there exists a subring is infinite but why is it isomorphic to the integers ??

Same for rings of prime characteristic...

A bit of help would be welcome

Thanks

Nobody is saying that a ring of characteristic zero is isomorphic to the integers, but rather that it contains some subring isomorphic to the integers. The subring isomorphic to $$\mathbb{Z}_{\mathrm{char}(A)}$$ is precisely the subring consisting of all integer multiples of $$1_R$$. There is a homomorphism $$\mathbb{Z}_{\mathrm{char}(A)}\to R$$ obtained by sending $$1\mapsto 1_R$$, and this homomorphism is injective, so the image is a subring isomorphic to $$\mathbb{Z}_{\mathrm{char}(A)}$$.
To see that the homomorphism exists and is injective, note that if we send $$1\mapsto 1_R$$ this is an injective homomorphism of abelian groups because $$\mathrm{char} (R) 1_R$$ is the identity and this is by definition the smallest multiple that is the identity. It is a homomorphism of rings because $$(m1_R) (n1_R) =mn1_R$$