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?

  • 1
    $\begingroup$ 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). $\endgroup$ – Milo Brandt May 2 '18 at 19:58
  • $\begingroup$ Yea, generally in the definition of integral domain the ring is required to be a ring with unity $1 \neq 0$. $\endgroup$ – wgrenard May 2 '18 at 19:59
  • $\begingroup$ @wgrenard, Herstein does not require this condition. In his definition it need not to contain unity $1\neq 0$! $\endgroup$ – ZFR May 2 '18 at 20:01
  • $\begingroup$ 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 $1 \neq 0$. $\endgroup$ – wgrenard May 2 '18 at 20:04
  • $\begingroup$ You should check to see if Herstein allows the trivial ring to be included in his definition of a ring. $\endgroup$ – wgrenard May 2 '18 at 20:09

Generally when one says characteristic is finite it means characteristic is non-zero number.

Because if you consider zero characteristic as finite case then there many non trivial rings with zero characteristic.

Eg. Ring $\mathbb Z$ has $0$ characteristic.

  • $\begingroup$ 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? $\endgroup$ – ZFR May 2 '18 at 20:13
  • $\begingroup$ Yes! That’s what above example shows. $\endgroup$ – Mayuresh L May 2 '18 at 20:17
  • $\begingroup$ I know that Ring of Integers has $0$ characteristic. But I can't get why you mention this? $\endgroup$ – ZFR May 2 '18 at 20:21
  • $\begingroup$ I mentioned this only to explain when we say finite characteristic it means its non zero finite number. $\endgroup$ – Mayuresh L May 2 '18 at 20:23
  • $\begingroup$ See also math.stackexchange.com/questions/98605/… $\endgroup$ – lhf May 2 '18 at 23:46

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