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Proving that if $R$ is an integral domain, then $\mathrm{char}(R) = 0$ or $\mathrm{char}(R)$ is prime, has been discussed in MSE already; but all arguments consider that if $\mathrm{char}(R)$ is not a composite then definitely it is a prime. Why?! Yes if an integer $\ge 2$ is not a composite then it is a prime but this is irrelevant to $\mathrm{char}(R)$, as $\mathrm{char}(R)$ is not just a an integer $\ge 2$ as it must fulfill $\mathrm{char}(R).1=0$ as well; since a prime and 1 are both nonzero then in integral domain they can't satisfy $\mathrm{char}(R).1=0$ also. So $\mathrm{char}(R)=0$ is the only option and not "or $\mathrm{char}(R)$ is prime".

Why am I wrong?

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You are wrong because in a ring of characteristic $p$, we have $p = 0$. Note that $p$ is just short for $$ \underbrace{1 + 1 + \cdots + 1,}_{p\text{ times}} $$ and this may be 0 -- in fact, it becomes 0 precisely when a ring has finite characteristic.

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  • $\begingroup$ I think the point of misunderstanding is that $char(R) \notin R$? $char(R)$ is just a number. And multiplication of $char(R)$ in identity is not multiplication of the ring? (i.e. it just defines times of ring-addition) $\endgroup$ – Liebe Green Oct 24 '16 at 20:39
  • $\begingroup$ $\mathrm{char}(R) \notin R$ in a literal sense, but the characteristic of a ring is an integer, and integers make sense in any unital ring, by the definition in my answer. In fact, this gives a homomorphism $\mathbb Z \to R$ for any ring; that is, the multiplication and addition in $\mathbb Z$ agree with those of the "integers" in $R$. Which is why the argument in your question work. $\endgroup$ – Mees de Vries Oct 24 '16 at 20:58

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