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



1 Answer 1


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 $$(m1_R) (n1_R) =mn1_R$$

  • $\begingroup$ Hey, thanks I stand corrected. How to write the proof of this statement ? $\endgroup$
    – T.D
    Mar 16, 2020 at 14:44
  • $\begingroup$ @T.D. I expanded a bit. $\endgroup$ Mar 16, 2020 at 15:02

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