# Doesn't an integral domain automatically imply that is it is of characteristic zero?

I was reading about Ring Theory from the book Topics in Algebra by I.N Herstein. I encountered the following definitions:

• Definition 1: If $$a\neq 0$$ is in a commutative ring $$R,$$ such that there exists a $$b(\neq 0)\in R$$ such that $$ab=0$$, (,where $$0$$ is the zero element of $$R$$), then $$a$$ is called a zero-divisor of $$a.$$

• Definition 2: A commutative ring is called an integral domain if it has no zero-divisor.

• Definition 3: An integral domain $$D$$ is said to be of characteristic $$0$$ if the relation $$ma=0,$$ where $$a\neq 0$$ is in $$D$$ and where $$m$$ is an integer, can hold only if $$m=0.$$

I feel that in fact all the integral domains $$D$$ are of characteristic $$0,$$ as since, $$D$$ is an integral domain it is never possible that $$ab=0$$ with $$a\neq 0 ,b\neq 0.$$ So, if say, $$pq=0$$ then, $$p=0$$ or $$q=0.$$ Again, if $$p\neq 0$$ then, surely $$q=0.$$

So, isn't it unnecessary to write Definition $$3$$ ? This is because Definition $$2$$ implies Definition $$3$$ we are just giving another name for integral domains, right?

Please correct me, if I am mistaken.

• @infinitezero Maybe. Commented Jul 24, 2023 at 10:22
• This would be true if any commutative ring contained the ring of integers, but it doesn't. And that's precisely the point of characteristic. Commented Jul 24, 2023 at 11:05
• There is already a perfect answer, but the integrity domain thing means that we cannot have $ab=0$ for members $a,b\in R$ unless one of $a$ and $b$ is zero in $R$. The characteristic zero thing means that we cannot have $ma=0$ where $m\in\mathbb{Z}$ and $a\in R$ unless either $m=0\in\mathbb{Z}$ or $a$ is zero in $R$. In writing $ma$ we are not using the ring multiplication, but are instead seeing $R$ as a module over $\mathbb{Z}$, so $m$ can be thought os as a sort of scalar that is multiplied by the element of the ring. Commented Jul 24, 2023 at 13:53

You are mistaken: for example, the ring $$\mathbb{Z}/2\mathbb{Z}$$ (or more generally, $$\mathbb{Z}/p\mathbb{Z}$$ for $$p$$ prime) is an integral domain but does not have characteristic zero.
The point is that we have to distinguish between multiplication by "true integers" (which is relevant to calculating the characteristic) and multiplication by ring elements (which is relevant to integral domain-ness); inside $$\mathbb{Z}/2\mathbb{Z}$$ the only nonzero element is $$1$$, so to check that it's an integral domain you just need to check that $$1\cdot 1\not=0$$.
It's a good exercise to show for integers $$n \geq 2$$ that $$\mathbb{Z}/n\mathbb{Z}$$ is an integral domain iff $$n$$ is prime.
• To slightly expand on this answer: the definition of the characteristic of a ring only uses the operation $+$ and the constants $0$ and $1$. On the other hand, the definition of an integral domain only uses the operation $\cdot$ and the constant $0$. Commented Jul 24, 2023 at 13:01