0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ It is literally going in the wrong direction. Try $\mathbb Z_n \to R$. ;) $\endgroup$ – Dustan Levenstein Dec 4 '13 at 1:34
  • 1
    $\begingroup$ 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 '13 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 '13 at 0:31
5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ If the ring is commutative but without unity and $char R=p$ , a prime number then what will be the mapping? Please help me. $\endgroup$ – nurun nesha Aug 2 '17 at 7:39
  • $\begingroup$ Here I guess you have taken the map $m \to m.1_R$ for m $\in \mathbb Z$ $\endgroup$ – nurun nesha Aug 2 '17 at 7:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.