Let $R$ a conmutative ring, $K\subset R$ a field.

The characterist of $R$ is $n\in\mathbb{N} \:\mid\; \text{Ker}(\varphi)=(n)$, with $\varphi:\mathbb{Z}\longrightarrow R$ the canonical homomorphism.

I have proved that for any given field $F$, its characteristic is either $0$ or a prime number using:

Let $ab\in\text{Ker}(\varphi)\Longrightarrow 0=\varphi(ab)=\varphi(a)\varphi(b)$. Because $F$ is a field, then $\varphi(a)=0$ or $\varphi(b)=0$. So $\text{Ker}(\varphi)$ is a prime ideal in $\mathbb{Z}$, so $n$ is prime.

However, I think I can not apply it to any ring because that ring does not have to be a domain.

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    $\begingroup$ Why would you prove this with $\;ab\in\ker\phi\;$ instead of simply $$a\in\ker\phi\implies 0=\phi(a)=a\phi(1)=a\cdot 1_K\;??$$ The same can be done with a commutative ring. Your deduction that $\;\ker\phi\;$ is a prime ideal in the integers because $\;\phi(a)=0\;\;or\;\;\phi(b)=0\;$ seems to be unjustified. $\endgroup$ – Timbuc Jan 5 '15 at 13:32
  • $\begingroup$ If $\varphi(a)=0$ then $a\in\text{Ker}(\varphi)$. The same for $b$. It is the definition of a prime ideal on any ring. How do I deduce that $\text{Ker}(\varphi)$ is a prime ideal? $\endgroup$ – user203327 Jan 5 '15 at 13:38
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    $\begingroup$ Oh, I see your point now! You wanted to prove the kernel of the homomorphism is prime. Yes, if you don't have a domain then you may have problems to apply what you did with fields as the "characteristic" may not be prime. For example, in the ring $\;\Bbb Z/4\Bbb Z=\Bbb Z_4$ = the rings modulo $\;4\;$ , the "characteristic" would be $\;4\;$ as is the minimal natural number of times for which $\;1_4+1_4+...=0\pmod 4\;$. $\endgroup$ – Timbuc Jan 5 '15 at 13:43

There are two ways to interpret what you write:

  1. You do not insist that $K$ and $R$ have the same multiplicative-neutral element. Consider the ring $R=\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p^2\mathbb{Z}$ (with coordinatewise operations). Then $K= \{(a,0) \colon a \in \mathbb{Z}/p\mathbb{Z} \}$ is a field contained in $R$. The characteristics are different.

  2. You do insist that $K$ and $R$ have the same multiplicative-neutral element. The characteristics are (of course) the same essentially just by applying the defitions; note $1_K$ and $1_R$ that are the same element.

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    $\begingroup$ Clear and complete. I was going to argue that if the identity elements are the same, the image of $\varphi$ still lies in $K$, so the kernel is the same. $\endgroup$ – Lubin Jan 5 '15 at 13:46
  • $\begingroup$ Thank you @Lubin for the additional information and the positive feedback $\endgroup$ – quid Jan 5 '15 at 13:47

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