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?


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.

  • $\begingroup$ 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

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.