Can integral domain be trivial?

I am studying Ring Theory from Herstein's book "Topics in algebra" and I would like to discuss one moment.

Definition 1. Let $R$ be a ring. The element $a\neq 0\in R$ is called zero-divisor if there exists $b\neq 0\in R$ such that $ab=0_R$.

Definition 2. An integral domain is the commutative ring which has not zero-divisors.

Also it is easy prove the following fact: If $D$ is an integral domain of finite characteristics then characteristics is a prime number.

Can trivial ring be an integral domain? If yes then it's characteristics is not prime which contradicts to the above fact!

But Herstein does not say nothing about triviality of integral domain.

Can anyone clarify this question, please?

• I would guess that this is an edge case that Herstein either didn't consider or didn't bother making note of (or possibly that the author doesn't consider the zero ring to be a ring - I had a professor who took that position once, although I don't think it's a great position to take). – Milo Brandt May 2 '18 at 19:58
• Yea, generally in the definition of integral domain the ring is required to be a ring with unity $1 \neq 0$. – wgrenard May 2 '18 at 19:59
• @wgrenard, Herstein does not require this condition. In his definition it need not to contain unity $1\neq 0$! – ZFR May 2 '18 at 20:01
• Yea, I see what you mean. I'm just saying I agree with the comment above mine. Because every definition I've ever read requires this. I just double checked in Lang's Algebra and it specifies there that $1 \neq 0$. – wgrenard May 2 '18 at 20:04
• You should check to see if Herstein allows the trivial ring to be included in his definition of a ring. – wgrenard May 2 '18 at 20:09

Eg. Ring $\mathbb Z$ has $0$ characteristic.
• Indeed, Herstein gives two definitions of characteristics for integral domain: the first one when it's zero and the other one when it's finite. In other words, you mean that if integral domain $D$ has finite characterics then it should not be trivial, right? – ZFR May 2 '18 at 20:13
• I know that Ring of Integers has $0$ characteristic. But I can't get why you mention this? – ZFR May 2 '18 at 20:21