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The title pretty much says it all.

Of course, one answer (IMO unsatisfactory) to such questions goes something like "a definition is a definition, period." In my experience, mathematical definitions are rarely completely arbitrary. Therefore I figure there must be a good reason for insisting that a field be a non-trivial ring (among other things), but it's not obvious to me. I realize that a trivial ring would make for a very boring field, but it is also a rather boring ring, and yet it is not disallowed as such.

EDIT: I found a hint of a rationale in the statement that "the zero ring ... does not behave like a finite field," but I could not find in the source for this assertion exactly how the zero ring fails to behave like a finite field.

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Do you think the linear algebra over the zero ring looks like the linear algebra over a field? – Pierre-Yves Gaillard Jan 11 '14 at 16:22
@kjo the field with one element is not about the zero field. It is just a name, and it is not about having one element as in $\{0\}$. – user119256 Jan 11 '14 at 17:09
@kjo, it is rather unacceptable to react to an emminently sensible suggestion like that of Pierre-Yves in that way. It should be obvious that one of the things to consider when contemplating a change in the definition of fields is what change that would require in related contexts. Now linear algebra is a particularly important context where fields are useful, so thinking about consequences there is probably the very first thing to do. If you have not thought about this, then you definitely should, and Pierre-Yves suggestion of a particular useful statement of linear – Mariano Suárez-Alvarez Jan 11 '14 at 19:17
algebra to think about is something which is quite remote from condescension but actually quite helpful. – Mariano Suárez-Alvarez Jan 11 '14 at 19:18
-1 for asking a question and then acting uninterested in and offended by the useful comments (which were in content, at the level of a good answer) that someone took the time to leave. – Pete L. Clark Jan 12 '14 at 20:56

7 Answers 7

up vote 6 down vote accepted

Good question, and I don't have a complete answer. But here's a partial answer.

It is a general principle that if we can get away with using just identities (i.e. universally quantified equations), that's great, because it means we end up with a variety. The axioms of ring-theory fall into this category, which is why the category of rings is so well-behaved.

Failing that, if we can get away with using just quasi-identities, that's still pretty great, because we'll end up with a quasi-variety. By a quasi-identity, I mean a universally quantified axiom of the following form, where all the Greek letters represent equations. $$\varphi_0 \wedge \cdots \wedge \varphi_{n-1} \rightarrow \psi$$

Cancellative semigroups fall into this category, which is why the category of cancellative semigroups is also quite nice.

However, suppose we really, really need either logical OR $(\vee)$ or logical negation $(\neg)$ for one of our axioms. This happens, for example, with integral domains; we typically assume either of the following.

  1. $xy=0 \rightarrow x=0 \vee y=0$
  2. $a=0 \vee (ax = ay \rightarrow x=y)$

It also happens with fields:

$$\forall x(x=0 \vee \exists y(xy=1))$$

Anyway, the point is this. If we need either $\vee$ or $\neg$ to axiomatize a class of algebraic structures, then we may as well include some non-degeneracy axioms, like $0 \neq 1,$ because frankly, our category of models already sucks, and few more non-triviality axioms aren't going to make it any suckier.

I think that is why fields (integral domains, etc.) are typically given non-triviality axioms, while rings (groups, etc.) are not.

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Nice. I was thinking about how the category of rings was quite nice and you need the trivial ring for this nicety; the category of fields sucks [your words :-)] anyway and having or not having the trivial field doesn't do anything about it. So you might as well not have it, because you'd have to exclude it from most theorems anyway. But, one (maybe the) reason that the category of rings is so nice, it that you can describe rings with equations. So, I like the insight in this answer a lot. – Magdiragdag Jan 11 '14 at 18:50

And Bourbaki said (Algebra I, Chapter 1, §9, definition 1)

"A ring $K$ is called a field if it does not consist only of $0$ and every non-zero element of $K$ is invertible"

and fields did not consist only of $0$.

Bourbaki saw that the rings were good, and he separated the fields from the other rings.
He called the fields with commutative multiplication "commutative fields" and non-commutative fields he called "skew fields" .
And there were domains and there were division rings-the first day.

Edit: For miscreants, heretics, apostates, schismatics, infidels and other iconoclasts
Since your ilk might not know: this answer was shamefully plagiarized from lines 3,4,5 of this text.

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Statistically, at least in the U.S., atheists are actually more likely to recognize the plagiarism :) – nbubis Jan 12 '14 at 23:46
Dear @nbubis: how is that? Do you mean atheists in the U.S. know Genesis better than believers? – Georges Elencwajg Jan 13 '14 at 0:30
@nbubis: thanks for the link. The results of the survey are rather counterintuitive and thus interesting. – Georges Elencwajg Jan 13 '14 at 8:24

I think of the requirement $0\neq1$ for fields as a consequence of the desires for (1) fields to be integral domains and (2) integral domains to satisfy $0\neq1$. Of course that just shifts the question to: Why do I want integral domains to have $0\neq1$? That desire comes from my (quite general) inclination to treat the empty set just like other finite sets. Here's how that's relevant: The key clause in the definition of integral domains (besides the axioms for commutative rings with unit) is that if $xy=0$ then $x=0$ or $y=0$. It follows immediately by induction that, if a product of $n\geq 2$ factors is $0$, then there must be a $0$ among those factors. The same holds trivially for $n=1$ under the obvious convention that the product of a single factor is that factor. So I'd like it to hold also for $n=0$. Now the product of no factors is $1$, so what I want is that, if $1=0$, then there is a $0$ in the (empty) set of factors. As there is no $0$ (or anything else) in the empty set, I infer that $1\neq0$.

(The same underlying idea explains why I don't regard the integer $1$ as prime and why I want lattices to have top and bottom elements.)

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The zero ring not only cannot be a field, it cannot even be either a local ring or an integral domain. If it were either of those things, the ideal $R\subset R$ would need to be prime for any unital commutative ring $R$, which is awkward and pointless, creating unnecessary difficulty everywhere.

This is pretty much the same reason that $1$ is not considered a prime number.

There are hypothetical advantages to a field that includes into every other field, but these are not realized in practice by the zero ring. In a typical examination of the "Field with one element", one looks at something that isn't a ring at all, such as the monoid $(\{0,1\},\times)$.

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+1 for the comparison with $1$ not being prime, although it would be nice to be able to say exactly why this is analogous (do certain things factor into fields like numbers factor into primes?). However, the zero (unital) ring does not include into every field, or into any other ring for that matter. It is a terminal object in the category of unital rings, not an initial object. (And yes, saying $1\neq0$ implies one is dealing with unital rings.) – Marc van Leeuwen Jan 11 '14 at 17:15
@MarcvanLeeuwen Yes, I should have put that better. This is an excellent justification for thinking of the "field of one element" as a monoid. – Slade Jan 11 '14 at 17:19
@MarcvanLeeuwen As for $1$ not being prime, the picture in my head is really this one: the zero ring cannot be field because its spectrum is the empty set, and the empty set cannot contain a point. The analogy of a "factorization" is just the spectrum as a list of points, which is non-unique if we can include the empty set. – Slade Jan 11 '14 at 17:25
@MarcvanLeeuwen Also, literally, $(1)$ would be a prime ideal in $\mathbb{Z}$. – Slade Jan 11 '14 at 17:26

At some level, this is just a extension of vadim123's answer. A good definition has three properties

  1. it covers a good class of "useful cases"
  2. it can be used to prove interesting and non-trivial theorems
  3. it (hopefully) has a high-level motivation.

Consider the notion of a field. It certainly satisfies 1 and 2 ($\bar{\mathbf Q}$, $\mathbf R$, $\mathbf C$, $\mathbf F_q\ldots$ being good examples of 1, and Galois theory being a good example of 2). Now consider 3. One would like a field to be a "simple commutative ring," i.e. a ring for which any module is a direct sum of copies of irreducibles, and such that there is (up to isomorphism) only one irreducible. The trivial ring satisfies this, but consider the definitions of "simple group" and "simple Lie algebra." Simple Lie algebras are not allowed to be trivial, and simple groups are tacitly assumed to be nontrivial. In other words, if you're going to study a field $k$, you want to do things like:

  • Consider varieties over $k$
  • Look at Galois extensions of $k$
  • Do linear algebra over $k$

For all of these, allowing the trivial ring to be a field makes most interesting theorems have an additional assumption "$k$ is nontrivial."

For example:

  • we want there to actually be nontrivial varieties over $k$
  • either $k$ should be algebraically closed (meaning one can do geometry over it) or there is interesting arithmetic coming e.g. from abelian varieties over $k$
  • we want a theorem like: "the $K$-theory of the category of vector spaces over $k$ is $\mathbf Z$"

You could say that adding the nontriviality hypothesis to the definition of a field is arbitrary and ad-hoc, but I respond that all definitions are (to some degree) arbitrary and ad-hoc. Mathematicians are interested in what interesting or useful theorems you can prove, and are generally only interested in "better" definitions if they a) make theorems cleaner or b) make proofs nicer. Allowing the trivial ring to be a field accomplishes neither of these.

Finally, note that there is a (not entirely rigorous) theory of a "field with one element," but it is much more sophisticated than simply saying "the field with one element is the trivial ring." For exampe:

  • A vector space over $\mathbf F_1$ is pointed set
  • $\operatorname{GL}(n,\mathbf F_1)=S_n$
  • The Riemann hypothesis "should" follow from considering $\operatorname{Spec}(\mathbf Z)$ as a curve over $\mathbf F_1$.

None of these work if we do the naive thing and set $\mathbf F_1=0$.

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If it's not part of the definition, then almost every proof that uses that definition will need a special case to handle the trivial ring. To eliminate that wasteful and annoying step, many people write the definition to exclude the trivial ring.

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If that's the rationale, it'd be the weakest rationale for a mathematical definition that I've ever seen. For one thing, the same thing could be said of the trivial ring, but this ring is not ruled out by the definition. Moreover, in mathematical writing it is commonplace to obviate such annoyances with stipulations like "we henceforth assume that all fields under consideration have $1\neq 0$." – kjo Jan 11 '14 at 16:24
@kjo: That it makes things more efficient is the best reason to choose a convention. Regarding the trivial ring when studying ring theory, I can think of several a variety of basic things that would be made very awkward if rings were required to be nontrivial... but nothing springs to mind that is awkward because of the existence of the trivial ring, except possibly in the case of restricting attention to integral domains (which is closely related to the case of studying fields anyways). – Hurkyl Jan 11 '14 at 18:28
@kjo, your comment is quite strange.I cannot imagine what possible criterion you could come up for the goodness of definitions if it is not their convenience! Moreover, it is precisely doing a stipulation like «we shall assume that all fields under consideration have 0 and 1 distinct» what the mathematical community has done when restricting fields to be nonzero! – Mariano Suárez-Alvarez Jan 11 '14 at 19:11

It is merely a matter of convenience that fields (and sometimes domains) are required by definition to satisfy $1\ne 0,\,$ i.e. the trivial ring $\{0\}$ is excluded. There are various motivations for such. Without this convention many definition and theorems would require cumbersome exceptions to handle trivial degenerate cases. Also in domains and fields it often proves very convenient to assume that one has a nonzero element available. This permits proofs by contradiction to conclude by deducing $\,1 = 0,\:$. or that $\,0\,$ is invertible, or equivalent field "absurdities" More importantly, it implies that the unit group of a domain is nonempty, so unit groups always exist. It would be very inconvenient to have to always add the proviso (except if $\;\rm R = 0)$ to the ubiquitous arguments involving units and unit groups. More generally it is worth emphasizing that the usual rules for equational logic are not complete for empty structures. That is why groups and other algebraic structures are always axiomatized to prevent nonempty structures (see this sci.math thread for further details).

Remark $ $ Perhaps worth mentioning is that there are some examples on MSE of just how confusing things can get if one starts reasoning in the zero ring, esp. in a proof by contradiction. For one example of such, see the long discussion in comments to my answer here, where it took a surprisingly long time to convince some readers that a reinterpretation of one of Rudin's proofs (by contradiction) was actually valid, and had a natural interpretation in the trivial ring (where $0/0 = 1$). See also the closely related discussion in this comment thread. If those discussions aren't enought to convince one of the pedagogical difficulties with such, then I suspect nothing will! $ $ Note: when reading those threads, be sure to click on "show more comments" so you can read the entire discussion (the top voted comments are skewed by the misunderstandings in the earlier part of the discussions).

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Algebraic structures aren't always axiomatized to prevent empty structures. While I can see why such a thing would be preferable to people who focus on the connections to logic and prefer to use the version of first-order-logic in which $\exists x: \top$ is a tautology, the convention becomes very undesirable without such preferences -- e.g. it means $\mathbf{Set}$ is not a variety, and in some varieties of universal algebra, small limits aren't guaranteed to exist. – Hurkyl Jan 11 '14 at 18:37
If one permits empty structures then one must modify many of the commonly used logical systems to restore completeness - see the linked sci.math discussion. Rather than doing that, for single-sorted logics it is usually more convenient to exclude empty structure. But the same is not true for many-sorted structures, since the structure can still prove very interesting if one of the sorts is empty. So one finds more discussion of these matters in applications using many-sorted structures, e.g. constructive/computational algebra and programming languages and logics for such. – Bill Dubuque Jan 11 '14 at 18:45
@BillDubuque: thanks; those links you posted are instructive. BTW, one reason why I've been somewhat resistant to convenience as a rationale is recalling a professor of algebra who was constantly compelled to add the caveat "assuming the ring is non-trivial" in his lectures. I'm trying to understand why this inconvenience is not enough to warrant disallowing zero rings by definition. Of course, in the end these arguments on the basis of "convenience" devolve into very subjective accountings of how much inconvenience is acceptable, etc. The answer given by... – kjo Jan 11 '14 at 21:13
...user18921 begins to explain this. – kjo Jan 11 '14 at 21:15
The trivial ring is a ring -- we need it, e.g. because we want the quotient of any ring by any ideal to be a ring, and it is in practice highly nonobvious whether an ideal $I$ is equal to $R$ or not. Moreover we want a functor from rings to groups given by taking units. So this motivates the following comment: the answer suggests that the trivial ring has "empty unit group". According to my convention (which I thought was the standard one), in the trivial ring $0 = 1$ is a unit, so the unit group has one element, not none. – Pete L. Clark Jan 12 '14 at 20:46

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