Showing that the characteristic of a commutative ring R without zero divisors is 0 or prime

Question

Question: Suppose that $R$ is a commutative ring without zero-divisors. Show that the characteristic of $R$ is either $0$ or prime.

I have established that every element in a commutative ring $R$ without zero divisors have the same additive order $n$.

Now, if no such additive order n exists, then the characteristic of $R$ is $0$.

Obviously, if a finite additive order exists, Char of $R$ is finite. How do I show that Char of $R$ is prime? It probably involves lagrange's theorem and the order of the element in $R$.

Hint is appreciated.

Thanks in advance.

Answer 1

Hint:

Consider the ring homomorphism $\;\begin{aligned}[t]\varphi\colon\mathbf Z&\longrightarrow R\\n&\longmapsto n\cdot 1_R\end{aligned}$

The characteristic of $R$ is the positive generator of $\ker\varphi$, i.e. of the ideal $\;\varphi^{-1}(0)$. Observe that, by definition, in an integral domain, $(0)$ is a prime ideal.

Answer 2

For rings not necessarily with identity, you can define the characteristic as the non negative generator of the ideal in $\mathbb{Z}$ formed by the integers $n$ such that $nr=0$, for all $r\in R$.

Suppose this ideal is not $\{0\}$; then it is $k\mathbb{Z}$, with $k>0$. Suppose $k$ is not prime, so $k=ab$, with $0<a<k$ and $0<b<k$.

By definition of $k$, there are $r\in R$ and $s\in R$ with $$ ar\ne0,\qquad bs\ne0 $$

Then $(ar)(bs)=(ab)(rs)=k(rs)=0$ so $R$ has zero divisors.