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Is the zero ring usually considered a domain or not? Wikipedia says:

  • The zero ring is not an integral domain; this agrees with the fact that its zero ideal is not prime. Whether the zero ring is considered to be a domain at all is a matter of convention, but there are two advantages to considering it not to be a domain. First, this agrees with the definition that a domain is a ring in which $0$ is the only zero divisor (in particular, $0$ is required to be a zero divisor, which fails in the zero ring). Second, this way, for a positive integer $n$, the ring $\Bbb Z/n\Bbb Z$ is a domain if and only if n is prime.

What are the arguments for/against this convention (besides the ones listed above)? What does the literature (i.e. your favorite textbooks) say on this matter? Note that I am specifically talking about domains, not integral domains; the only difference between them aside from the non-triviality assumption mentioned here is that domains are not required to be commutative.

Mario Carneiro
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  • I can't think of any argument for it. On the other hand, excluding the $0$ ring in the definition means to not have to consider it as a special case or exception in your proofs. – A.P. Jun 15 '15 at 17:57
  • People often exclude the zero ring even as a ring (i.e. By prefacing a statement with "Let $R$ be a ring with $0\ne 1$..."). It's pretty convenient for an ring to have a quotient which is a field. If someone wants to exclude/include the zero ring into integral domains, I'd like to know why they'd include it into rings. – PVAL-inactive Jun 15 '15 at 18:03
  • @A.P. There's no free lunch. If you make extra assumptions in your proof, that just means that it won't be as widely applicable, and users of the theorem will need to work harder to prove the extra assumptions, even if they are not necessary. I find it decidedly unnatural to go so far as to exclude the zero ring as a ring, because it has all the properties you would expect of a ring. The issue comes in when you start dividing things and $0$ is invertible, but domains are on the border of acceptability here. – Mario Carneiro Jun 15 '15 at 18:32
  • Sure, someone would need to prove the result for the $0$ ring. The point is that most results are either trivial or inapplicable for the $0$ ring, so having to mention it every time is just an annoyance. – A.P. Jun 15 '15 at 18:49

2 Answers2

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Is the zero ring usually considered a domain or not?

Nearly every ring theory text says it is not. I bet guessing that 90% of texts agree with this is a conservative guess.

The main thing is what the first sentence you cited tries to say (but does not do a good job with): we want to say that $R/P$ is a prime ring if and only if $P$ is a prime ideal. (For noncommutative rings, this becomes $R/P$ is a domain if and only if $P$ is a completely prime ideal, that is, one satisfying the commutative definition of "prime." This does not matter much for the following discussion.)

What does the literature (i.e. your favorite textbooks) say on this matter?

If $\{0\}$ were considered a domain, then $R/R$ would be a domain for any ring $R$, so that $R$ is a prime ideal of itself. This is (in every reference I know) explicitly ruled out by the definition of a prime ideal. You could of course object that this is just an equivalent question ("why shouldn't the whole ring be considered a prime ideal?")

A very similar discussion took place here where someone asked why the zero ring wasn't considered a field. The argument that I gave there applies pretty well in this situation too. Our intuition of primes being "indecomposable" leads us to the idea that a prime ring shouldn't decompose into other rings. Certainly decomposing a prime ring into a product of multiple prime rings would be unattractive. But if you let $R=\{0\}$ be a prime ring, then $R\cong\prod_{i\in I} R$ or any nonempty index set that you like.

In particular if we let $\{0\}$ be called a domain, then we'd have a domain that is isomorphic to infinitely many copies of itself, which certainly seems like poor behavior for a domain.

Community
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rschwieb
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  • Regarding "Why shouldn't the whole ring be a prime ideal?", for some reason I am much more comfortable with this definition. Similar to maximal ideals or filters in topology ($\emptyset\notin F$), without this assumption many statements tend to trivialize without adding the condition back in anyway. – Mario Carneiro Jun 15 '15 at 19:31
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I would like domains to have fields of fractions. Fractions are equivalence classes of ordered pairs $(p,q)$ where $q \neq 0$. Since all elements of the zero ring are zero, there are no such pairs and one deduces that the "field of fractions" should be the empty set. But the empty set is not a field (does not contain $0$ or $1$.)

David E Speyer
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  • I don't buy this argument. Of course you aren't going to get a field of fractions, because the multiplication isn't commutative. But I'm not even sure you will get a well-defined skew field / division ring with the usual construction. Even the total ring of fractions, which allows zero divisors, has to assume that the ring is commutative at the outset. – Mario Carneiro Jun 15 '15 at 18:09
  • See also: Ore condition, which details the conditions under which one can expect a noncommutative ring or domain to have an associated division ring of fractions. It would seem that zero ring cannot be an Ore domain for the results on that page to hold. – Mario Carneiro Jun 15 '15 at 18:22
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    @MarioCarneiro What do you mean the zero ring isn't commutative? $0 \times 0 = 0 = 0 \times 0$. – David E Speyer Jun 15 '15 at 19:26
  • I mean that domains in general aren't commutative, so that "I would like domains to have fields of fractions" is wishful thinking. What it seems you mean is "I would like integral domains to have fields of fractions", in which case there is no problem since the zero ring is not an integral domain. – Mario Carneiro Jun 15 '15 at 19:43
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    @Mario: It is common in many parts of mathematics to define a domain to be a commutative ring without zero divisors. In particular, many people who study commutative algebra do not like the term integral domain, because the word "integral" has a completely different meaning in that subject. (David: your proof of the commutativity of the zero ring can be simplified!) – Pete L. Clark Jun 15 '15 at 21:08
  • @PeteL.Clark If a domain is an (evidently nonzero) commutative ring without zero divisors, then what is an integral domain (as that is my exact definition of integral domain)? FYI, the context of this question is in defining these notions in a formal system, so the goal is to have a definition that everyone agrees on, even across different areas of study (to the extent that is possible). – Mario Carneiro Jun 15 '15 at 22:02
  • @Mario: I for instance do not use the term "integral domain" in my papers, though I may say something like "By a domain we mean a commutative nonzero ring without zero divisors". I didn't quite understand what you meant by defining these notions in a formal system: in a formal system, everything means exactly what you say it means. – Pete L. Clark Jun 15 '15 at 22:20
  • Not everyone agrees whether $A \subset B$ means containment or proper containment, so I think having a "universally understood terminology" is prohibitively difficult. It is better to be flexible, and to assume, when possible, that people are using terminology so as to make correct, meaningful statements (as in David's answer). – Pete L. Clark Jun 15 '15 at 22:21
  • @PeteL.Clark Of course, in a formal system (as in any math paper or book), everything means what you define it to mean. However, it is also desirable to make sure that one's definitions matches the "general usage", in order to ease understanding and minimize misconceptions. The constraint in a formal system, though, is that you must make one definition once and for all - you can't change the definition later in the development when you have moved on to another topic. (e.g. to me $A\subset B$ means proper containment and $A\subseteq B$ means containment, because a choice had to be made). – Mario Carneiro Jun 15 '15 at 23:51
  • @Mario: No, not all terms are defined in math papers; there's some amount of cultural understanding required, most often enough to make the paper difficult to read by the greater community. Anyway, what you say is desirable certainly is, but what I'm saying is that it turns out not to be possible in many cases. Therefore in practice it is more important to choose some reasonable terminology that people won't complain too loudly about, and make sure to explain it clearly. If you read math papers dogmatically expecting $A \subset B$ to mean proper containment, you'll get in trouble. – Pete L. Clark Jun 16 '15 at 00:01
  • (I'm also not completely sure that I understand what you mean by "formal system". A formal system is something like the first-order predicate calculus. One has a defined alphabet consisting of logical symbols, variable names, and so forth. I have never seen a formal system in which "domain" appeared: that would either be a strange name for a variable or a very confusing way to write a product of six things!) – Pete L. Clark Jun 16 '15 at 00:04
  • @PeteL.Clark This is the formal system I am talking about, a proof system and library of proofs called Metamath. I'm trying to develop abstract algebra in it, but if you can understand the linked statement you will know that the current definition of a domain does not include the non-zero assumption, which is why I opened this question. (What does "a very confusing way to write a product of six things" mean? What six things are you referring to?) – Mario Carneiro Jun 16 '15 at 01:30
  • The six things are d o m a i and n. My point was that a "domain" is usually not part of the alphabet of a formal system. But I see what you mean: you have not just a formal system but also a collective enterprise to find and exhibit formal proofs of mathematical theorems within that system. Certainly you will want to choose terminology carefully while doing such a project. But as above, you can't choose it so carefully that everyone will know what you mean w/o reading an attached documentation/glossary of terms, so I hope that the project is putting effort into that too. – Pete L. Clark Jun 16 '15 at 02:36
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    (I checked out the project. It looks like you folks are doing very nice work.) – Pete L. Clark Jun 16 '15 at 02:38
  • I wonder if perhaps a trick like https://mathoverflow.net/a/397075 could make the construction work. Instead of defining the field of fractions directly as the set of ordered pairs $(p, q)$ for nonzero $q$ (with the usual operations), define it as the ring generated by $(p, 1)$ and $(1, q)$ for nonzero $q$, under the equivalence relation $(pq, rq) \equiv (p, r)$. So for the trivial ring, its field of fractions is also trivial. – user3840170 Jan 19 '25 at 16:18
  • Yeah, I thought about this more and found out that the notion of a field of fractions can be generalized to non-domains without much trouble, and there is a construction of it that works well for the trivial ring: https://mathoverflow.net/a/486313/157594. – user3840170 Feb 09 '25 at 08:49