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Let $f:\mathbb{Z}\to R$ be given by $f(n)=n1$ for each $n\in\mathbb{Z}$, where $R$ is a ring with identity. $f$ is clearly a homomorphism. Now $\ker f$ is an ideal of $\mathbb{Z}$. But $\mathbb{Z}$ is a principal ideal ring, whence $\ker f=\{n\in\mathbb{Z}:n1=0\}=(p)$ for some integer $p$.

Claim: $p=\text{Char}R$.

Proof of claim: $f(p)=p1=0\implies 0=0x=p1(x)=p(1x)=px$ for each $x\in R$. Thus $\text{Char}R\mid p$. Now $\text{Char R}=pq+r$ for some integers $q,r$ where $0\leq r<p$, by division algorithm. Therefore $f(\text{Char}R)=f(pq+r)$ from which it follows that $r=0$ as $(\text{Char}R)\ 1=0, pq\in\ker f$ and $f$ is a homomorphism. Thus $p\mid \text{Char}R$. Hence $\text{Char}R=p$.

Is my proof alright?

Moreover the text says that each ring with identity and characteristic zero contains a subring isomorphic to $\mathbb{Z}$. As I see it this subring should be $f(\mathbb{Z})$. Am I correct? Please help. Thanks.

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    $\begingroup$ Yes, your proof is good. If you assume $R$ has characteristic $0$, then based on your argument, ker$(f)$ would have to be $(0)$, hence $f$ is an isomorphism of $\mathbb{Z}$ onto $f(\mathbb{Z})$. $\endgroup$ – quasi May 20 '17 at 7:31

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