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

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.

• Hint: If $m1 = 0$ with $m$ minimal and $m$ is not a prime, how can you find two non-zero elements which multiply to $0$? – Tobias Kildetoft Aug 2 '16 at 8:03
• It is not mentioned that this is a ring with unity. I had though about something along your comment. – Mathematicing Aug 2 '16 at 8:04
• Then replace $1$ by any element from the ring in my hint. – Tobias Kildetoft Aug 2 '16 at 8:05
• @TobiasKildetoft It is not possible to find any two non-zero element such that their product is zero. This follows from the fact that R contains no zero-divisors. – Mathematicing Aug 2 '16 at 8:08
• Yes, I know that, and hence you would reach a contradiction based on the assumption that $m$ was not a prime. – Tobias Kildetoft Aug 2 '16 at 8:09

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.

• What about the Zero Ring, $R=\{0\}$, this ring is a commutative ring with a unit and has characteristic 1 – Eduardo Jun 17 '17 at 4:19
• @Eduardo Usually a ring with no zero divisors is assumed to be not trivial. – egreg Jun 17 '17 at 6:26
• Could you provide me some reference with this kind of observation? – Eduardo Jun 17 '17 at 18:41
• @Eduardo Not at the moment; anyway, it is implied by the question, isn't it? Your remark is valuable anyhow. – egreg Jun 17 '17 at 19:16

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.

• This only works for a ring with identity. – Jacob Wakem Aug 2 '16 at 21:41
• In commutative algebra, we generally consider only rings with identity, unless otherwise specified. – Bernard Aug 2 '16 at 21:48
• Ah yes that is a well known standard. I didn't tell you you were wrong. – Jacob Wakem Aug 2 '16 at 22:07
• What about the Zero Ring, $R=\{0\}$, this ring is a commutative ring with a unit and has characteristic 1 – Eduardo Jun 17 '17 at 4:18
• I have the convention that in a ring, $0\ne 1$. – Bernard Jun 17 '17 at 8:53