Talk:Field with one element

What is "p.26" in the lead? --Vaughan Pratt (talk) 15:53, 2 October 2009 (UTC)

Probably refers to a page number in the cited reference. AxelBoldt (talk) 17:10, 23 October 2009 (UTC)

---

The introductory sentence refers to the field with one element as an object that "should" exist. Then it goes on to say that it does exist, namely as "the free algebraic monad generated by one constant". So why not just define it as that and be done with it? Or is that definition not universally accepted?

When people write F₁, do they have a concrete object in mind, or do they refer to a nebulous concept characterized by the properties in the Philosophy section? The article should make this clearer. AxelBoldt (talk) 17:10, 23 October 2009 (UTC)

Dangerous notation

The notation

$[n]_{q}!:=[1]_{q}[2]_{q}\dots [n]_{q}$

seems very dangerous, since $[n]_{q}!$ already has a different definition, namely as the factorial of the integer $[n]_{q}$. I would prefer $n!_{q}$ as less ambiguous. AxelBoldt (talk) 17:44, 23 October 2009 (UTC)

Non-existence of F₁

I think I disagree with the viewpoint that User:David Eppstein has taken in his rewrite of my rewrite of the lede, but I'm not sure I understand it. My issue is that the existence or non-existence of F₁ depends on what you mean by "F₁-algebraic geometry". In all the approaches I listed, there is an object which is given the name F₁. In Deitmar's, it's the trivial monoid; in Durov's, it's the free algebraic monad with one constant; in Borger's, it's the ring Z with the trivial Λ-ring structure; and so on. In none of these is F₁ a field, of course, but this is not disputed, because everyone agrees F₁ can't be a field. And nobody agrees that any of these is F₁ for certain. It may turn out that none or all of them have all the properties that F₁ ought to have. But to claim that F₁ is non-existent says to me that not only do none of the proposed constructions have all the desired properties, but that no construction of any kind of object can ever have all the desired properties. So I think I'm misunderstanding the edit; I don't see how it can be correct. Ozob (talk) 21:49, 21 November 2009 (UTC)

My understanding of the point of this subject (and I am far from an expert in it) is that it less about finding a mathematical object that can stand itself in the place of the missing field, and more about rearranging which algebraic objects we consider fundamental and which other ones are derived from these fundamental objects. So, classically, we start with a very concrete object, a field F, and derive from it other objects such as individual vector spaces F[x], F[x,y], etc or a category of F-vector spaces or a free algebraic monad or whatever. In this classical view, a field is a collection of elements and operations satisfying the usual axioms, and I think we can all agree that in this classical view F₁ doesn't exist. But instead of giving up when we discover that it doesn't exist, we are looking for a way of starting from something else as the fundamental object of our field theory and deriving most of the same things we could derive if we started from a genuine field but in a way that still works in characteristic one. For instance (though this is too naive to really work), we may take as our fundamental objects the categories that "look like" the categories of F-vector spaces defined from fields, according to some axiomatization of the properties of these categories, and we may then discover that our axiomatization describes not just the categories really coming from fields but also the category of pointed sets. If this replacement of fields by axiomatized categories allows us to rederive a large enough chunk of field theory, we can say that F₁ "is" the category of pointed sets, but it's still not true that this category is a field, any more than it's true that the other categories of F-vector spaces are fields. They're stand-ins for fields, but not themselves fields. —David Eppstein (talk) 23:45, 21 November 2009 (UTC)

PS a relevant quote from more of an expert, strongly critiqueing the way it was written here prior to my change: The field F₁ does not exist, so don’t try to make sense of sentences such as “The ‘field with one element’ is the free algebraic monad generated by one constant. It may be poetic (or more specifically, metonymous) to say that F₁ is something specific, but I think it's misleading, because whatever you define it as will not be a field with one element (in the standard sense of a collection of elements and operations satisfying the field axioms). Rather than being poetic, I think we should accurately describe what is really happening: we are defining a theory that is similar enough to field theory that many of our favorite theorems about fields are still true, but different enough from field theory that it works even in characteristic one even though true honest-to-God field theory doesn't. —David Eppstein (talk) 04:58, 22 November 2009 (UTC)

I think we're almost completely in agreement, and really our differences come down to a matter of style. My own viewpoint when I was revising the lede initially was that it was pretty obvious that there was no field with one element, and so the real focus of the lede should be on what one does to get around that. That is, the real focus should be on proposed theories of F₁-geometry and why they're interesting. But after thinking about it for a while longer, I think that's not quite right—the article is entitled "Field with one element", after all, so it needs to discuss up front the fact that there is no field with one element, there cannot be a field with one element, and whenever someone mentions a field with one element, they don't mean an actual field with one element. This is what your lede does. But I wasn't entirely happy with your lede, either. As far as I can tell, in F₁-geometry, one isn't so much interested in generalizing fields specifically as one is in generalizing all of abstract algebra, or at least all of commutative algebra. Furthermore, in these organized theories, one doesn't define things like polynomial rings and vector spaces over F₁ in an ad hoc way, one tries to define them in an obvious way and then prove that when these are considered over F₁ (because there's always some object which one calls F₁ in these theories), they turn out to be finite sets, or pointed sets, or whatever.

I've rewritten the lede again. Go ahead and edit it if you like; I like this version, but mostly I just want the lede to be clear about this rather muddy subject. Ozob (talk) 22:00, 22 November 2009 (UTC)

This article lacks an explanation as to why the generally agreed definition of a field requires 1 \neq 0

The trivial ring is not a field because by definition in a field 0 may not equal 1. But that does not tell us why the definition of a field includes this seemingly arbitrary condition. There must be good reasons, and I feel that this article is incomplete if it does not give those reasons.sephia karta | dimmi 14:55, 22 November 2010 (UTC)

Essentially, the problem is that the zero ring doesn't work like other fields. For instance, if A is a commutative ring, then A[x], the polynomials in one variable over A, is a commutative ring of Krull dimension one greater. Except in one case: If A is zero, then A[x] is still zero. Or, if you consider maps from a ring to the zero ring, there's always exactly one of them; so from the viewpoint of algebraic geometry, every space has a generic point given by the prime spectrum of the zero ring, and hence all spaces are connected. It's annoyances like this that make the zero ring not a field. It's kind of like why one is not a prime number; the standard definition is made so that the theory holds together well, and that may be unsatisfying, but that's just life.

I don't think this is really an appropriate topic for the present article. It's a better topic for field (mathematics) or integral domain. Ozob (talk) 02:23, 23 November 2010 (UTC)

Thank you for you answer!

I posit that a reader who would want to know why fields with one element are proscribed would come to this article first to find out. At the present moment, this article lists a couple of 'substitute' fields with one element that can perform its role in several contexts, without giving even a hint as to why what surely is the intuitively most obvious candidate for a field with one element falls short of how we want fields to behave.

I would add this to the article myself, but my knowledge of field theory is too limited. I'm afraid that I don't quite understand your second argument, could you elaborate? (The prime spectrum of the zero ring is empty, right? Do you mean the ideal (0)? How does this affect the trivial ring if it were a field?)

As for your first argument, you are talking about rings, not fields, and there are other good reasons why the trivial ring is generally considered a ring. For any type of structure, there are definitions, lemmata and constructions that are ill-defined or do not hold for the trivial case, yet still the empty set, the empty topological space, the trivial ring or the empty category are in principle permitted. What makes the 'trivial field' special? Is the trivial field that much worse (why?) or is it just convention? sephia karta | dimmi 16:50, 23 November 2010 (UTC)

Regarding my first reason: One expects particularly nice behavior from fields (as compared to other rings), not unusual and exceptional behavior. In this respect the zero ring is a failure, as it fails to have many properties shared by other rings. Regarding my second reason: If the zero ring were a field, then one would have to define the ideal (1) to be prime; otherwise one would lose the fact that a ring modulo a ideal is an integral domain if and only if the ideal is prime (because fields are integral domains). So the prime spectrum of the zero ring would become non-empty; the prime spectrum of a field would contain two points, not one; the prime spectrum of a discrete valuation ring would contain three points, not two; etc.; and one would also gain the strange generic point that I mentioned above. Any of these problems has a workaround, and it turns out that the simplest workaround is to declare that the zero ring is not a field.

As far as the topic of the article, I have to say that I am not sure what to do about your objection. The field with one element is not an elementary topic; a reader who is not already familiar with fields is not going to get anything out of reading about F₁, not here or anywhere else. I agree that the zero ring sounds like a plausible candidate at first glance, but the reason why it's not is because it doesn't have the expected properties of F₁. In order to explain why that's the case, the article has to explain the expected properties of F₁. If the reader understands those, then the failure of the zero ring is obvious. So I stand by my statement that this is not the right place to discuss why the trivial ring is not a field. Ozob (talk) 22:50, 23 November 2010 (UTC)

I'm sorry to say I still don't find your example persuasive - I would rather say that that the zero ring would still not be an integral domain. The statement that all fields are integral domains would gain one exception but we would not have any of the other complications you mention. That said, I'm happy to leave this discussion at this. sephia karta | dimmi 12:10, 24 November 2010 (UTC)

R is an integral domain iff ab = 0 implies a = 0 or b = 0; in the zero ring this is obvious. It's not the only characterization of an integral domain, though; there's also: R is an integral domain iff multiplication by a non-zero element is injective; in the zero ring this is obvious. Or: R is an integral domain iff it is reduced (has no nilpotent elements) and irreducible (has a unique minimal prime); in the zero ring these are obvious (zero is not a nilpotent element by definition, and the ring has only one ideal). So unless you impose the condition that zero is not one, the zero ring is an integral domain. And if you do let the zero ring be an integral domain, then it's also got to be a field: It has no non-zero prime ideals; every non-zero element is (vacuously) invertible; it is its own field of fractions; it has Krull dimension zero.

For me, the really compelling reasons are geometric. Spec 0 should be the empty set. But I don't know how to explain that unless you already believe it; it's a judgment to say that it should. Ozob (talk) 14:01, 24 November 2010 (UTC)

The morally correct form of the first definition is that R is an integral domain if any finite product of nonzero elements is nonzero. This implies 0 ≠ 1 by taking the empty product. The zero ring also fails your third definition: its unique ideal is not prime, as prime ideals are (for good reasons) by definition proper subsets of R.—Emil J. 19:46, 6 August 2014 (UTC)

What do the sources say? Deltahedron (talk) 21:29, 6 August 2014 (UTC)

The original question was not a question about sources; it was a question about why the sources were as they were. There's no dispute about the standard definition here (for that, see Talk:Integral domain#Variation of definition, a conversation which I found inexplicably frustrating).

I agree that having any finite product of nonzero elements be nonzero is a good property of integral domains. I also agree that prime ideals should be proper subsets of R. But if someone were to come along and ask, "Why not add an exception for the zero ring to those two statements?", my only answer is (still) that it just doesn't seem to work as well. And I feel like further discussion of that point is mostly a discussion of mathematical aesthetics. After all, logically speaking we could define the zero ring to be an integral domain and then put exceptions in our other definitions and theorems where necessary. Ozob (talk) 02:58, 7 August 2014 (UTC)

field with no elements?

Has anyone ever come up with something like that? Double sharp (talk) 04:22, 16 April 2016 (UTC)

Not that I'm aware of. Logically speaking, it's even more nonsensical than F₁. But F₁ was suggested by developments in other theories, while I don't think anyone has ever found evidence of a field with no elements. Ozob (talk) 14:25, 16 April 2016 (UTC)

On the "French-English pun"

The linguistic pun for $\mathbb {F} _{\text{un}}$ only works if the language is Catalan instead of French and if the English dialect is Irish English, due to the differing ways of pronouncing the vowel 'u' in the languages and dialects involved. 69.156.109.238 (talk) 20:49, 18 November 2019 (UTC)

As a written pun, rather than a spoken pun, it works in the standard dialects. —David Eppstein (talk) 21:11, 18 November 2019 (UTC)