# Error in Herstein's "Topics in Algebra", zero ring has characteristic $1$ which is not prime

I was trying to prove the following statement:

"If an integral domain has a finite characteristic then the characteristic of the integral domain is a prime number"

This made me look at the definition of an integral domain, as given in the book, "Topics in Algebra " by I.N Herstein (2nd Edition) in Chapter -3 (Ring Theory) page number-126. The definition is as follows:

If $$R$$ is a commutative ring, then $$a\neq 0\in R$$ is said to be a zero-divisor if there exists a $$b\neq 0\in R,$$ such that $$ab=0.$$

A commutative ring is an integral domain if it has no zero-divisors.

Later in the book, on page number-129, the definitions of the characteristic of an integral domain are given as:

Definition: 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 $$m = 0.$$

The ring of integers is thus of characteristic $$0,$$ as are other familiar rings such as the even integers or the rationals.

Definition: An integral domain $$D$$ is said to be of finite characteristic if there exists a positive integer $$m$$ such that $$ma = 0$$ for all $$a\in D.$$

If $$D$$ is of finite characteristic, then we define the characteristic of $$D$$ to be the smallest positive integer $$p$$ such that $$pa = 0$$ for all $$a\in D.$$

The problem is with a remark that followed these definitions above. The remark was:

It is not too hard to prove that if $$D$$ is of finite characteristic, then its characteristic is a prime number.

Acccording to the definition of an integral domain as given in the book, the ring $$R=\{0\}$$ i.e containing only the zero element is also an integral domain.

Now, we note that, $$m.0=0$$ for any integer $$m.$$ So, $$R=\{ 0 \}$$ is not of characteristic zero. But, $$R$$ has a finite characteristic. The smallest positive integer that $$m$$ can be, in this case is $$1.$$ But $$1$$ is not a prime number. So, the thing given in the remark is false, because, even $$R$$ is an integral domain with a finite characteristic, the characteristic of $$R$$ is not a prime, and neither is it a composite number.

But nearly every where, I have found the statement:

"If an integral domain has a finite characteristic then the characteristic of the integral domain is a prime number"

true in general. This is why, I think that some definition in the book might be misleading or better say incorrect such that it contradicts such well- established statement.

Any help regarding this issue will be highly appreciated.

• Some books require that rings have a distinct 1, so that there is no $0$ ring. This is just a matter of convention. Check where your text defines rings. Sep 28 at 3:23
• @CharlesHudgins Ok, but in this book, the author does not require that the ring must have a unit element. They even mention it explicitly. So, they are allowing zero rings. Sep 28 at 3:24
• It's also common to say that integral domains must have a distinct $1$. Check your book's definition of integral domains. In any case, if this is an error, it's not a serious one. This is the one and only exception to the oft-cited fact that finite characteristics are prime. Sep 28 at 3:27
• @CharlesHudgins Even in the context of unital rings, there is still a zero ring: you would need an additional axiom $0\not=1$ to rule this out. (More generally, any purely equational theory - such as the theories of groups, unital rings, and non-unital rings, but not the theory of fields - has a one-element model.) Bringing things back to the OP, in my experience this axiom (+ the existence of 1 in general, if we're working non-unitally) is part of the definition of integral domains. Sep 28 at 3:27
• @NoahSchweber that's why I edited my comment to say distinct 1. I probably could have been clearer. Sep 28 at 3:28

Typically, an integral domain is defined to additionally require that $$0 \neq 1$$.

Another way to express this requirement is as follows. For all $$n \in \mathbb{N}$$, for all sequences $$a_1, \ldots, a_n$$, if the product $$a_1 \cdots a_n = 0$$ then there is some $$i$$ such that $$a_i = 0$$. The case $$n = 0$$ is precisely saying that $$0 \neq 1$$. For if we have $$1 = 0$$, that’s the same as saying the empty product is $$0$$. But there is no $$a_i$$ at all that could equal $$0$$.

If you drop the requirement $$0 \neq 1$$, the above Lemma about finite products only applies when $$n \geq 1$$.

• Oh, this is nice! Sep 28 at 4:38
• @QiaochuYuan This is actually the same sort of logic for why local rings require that $0 \neq 1$. It turns out that the local ring condition is equivalent to the following: for all finite sequences $a_1, \ldots, a_n$, if $a_1 + \cdots + a_n$ is a unit, then some $a_i$ is a unit. There are a lot of similar things across mathematics - for instance, why the whole space is open in topology. Sep 28 at 4:43

As has been discussed in the comments, there is this minor annoyance that one way or the other you need to exclude the zero ring from the definition of an integral domain. One way to say this is that you want to define integral domains in such a way that the integral domains should be exactly the subrings of fields (so that integral domains always have fields of fractions), which the zero ring isn't. (One way or the other you also need to exclude the zero ring from the definition of a field.)

• Should I define that the zero ring is not an integral domain? This is because as we see if we consider zero ring as an integral domain it creates some problems like the one in the OP. Sep 28 at 4:29
• @Thomas: yes, that's what I just said. Sep 28 at 4:37