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Let $R$ be a commutative ring with identity such that the characteristic of $R$ is $n$, char$R=n$. Prove that is $n>0$ then $R$ contains a subring isomorphic to $\mathbb{Z}$$_n$, the additive group of integers modulo n.

Attempt at proof: Consider an isomorphism $\phi: R\rightarrow\ \mathbb{Z}$$_n$ defined as $\phi(x)=[r]$$_n$ where $r$ is the remainder when x divides n. Showing then that $\phi$ is well defined and an isomorphism would conclude the proof.

Is this going in the right direction? Or am I totally off?

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    $\begingroup$ It is literally going in the wrong direction. Try $\mathbb Z_n \to R$. ;) $\endgroup$ Dec 4, 2013 at 1:34
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    $\begingroup$ Another comment is, what do you mean when $x$ divides $n?$. Your elements in the ring need not be integers, or multiples of 1. $\endgroup$
    – LASV
    Dec 4, 2013 at 1:42
  • $\begingroup$ There's no reason for $R$ to be isomorphic to ${\bf Z}_n$. Consider ${\bf Z}_n^2$. $\endgroup$
    – tomasz
    Dec 13, 2013 at 0:31

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Think of a map $\psi:\mathbb{Z}\to R$, where $\psi$ is the map you think it should be =]. Further, what is the kernel of this map?

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  • $\begingroup$ If the ring is commutative but without unity and $char R=p$ , a prime number then what will be the mapping? Please help me. $\endgroup$
    – Mini_me
    Aug 2, 2017 at 7:39
  • $\begingroup$ Here I guess you have taken the map $m \to m.1_R$ for m $\in \mathbb Z$ $\endgroup$
    – Mini_me
    Aug 2, 2017 at 7:41

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