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I've heard that in the study of finite fields, and other concepts related to finite fields, mathematicians have found a sort of gap: there are various results and things that seem like they correspond to a field with one element, $\mathbf{F}_1$, even though there is no such field.

Of course, there is an object which seems similar to a field with one element: the zero ring, $\mathbf{Z}_1$, defined as having one element, $0$, obeying the equations

$$0 + 0 = 0 - 0 = 0 \cdot 0 = 1 = 0.$$

The zero ring seems to behave like a field in every respect, except for the fact that it fails to satisfy $0 \ne 1$. (But the importance of the axiom $0 \ne 1$ isn't clear to me.)

However, in any case, there seems to be consensus that the zero ring is not the field with one element. People say that the zero ring "does not behave like a finite field" (Wikipedia) or that it "does not have the features that mathematicians need" (this Stack Exchange answer by a deleted user).

I'm not familiar enough with algebraic geometry to understand the properties that $\mathbf{F}_1$ is expected to have. Is there an elementary, undergraduate-level explanation of

  • why it seems like a field with one element "ought to exist" in the first place,
  • what properties we expect it to have and why, and
  • how the zero ring fails to satisfy these properties?

Or do I need to study algebraic geometry if I want to have an understanding of any of this?

(A side question: do we have a good answer to the question of whether the single element of $\mathbf{F}_1$ ought to be $0$, $1$, both, or neither?)

Sophie Swett
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  • As to your side question, a field must have a $0$ and a $1$, so if you only have one element, it is both the $0$ and the $1$ of your field. This, of course, hinges on whether or not $F_1$ is a field -- which is an interesting question that I am not qualified to address. – JonathanZ Jul 14 '20 at 16:39
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    A trivial Field with only one element were $1_F$, the multiplicative identity, is the same thing as $0_F$ then additive identity and there are no other elements is so trivial and useless that there is no point. We usually include in the definition of field as an axiom that $1 \ne 0$. So talking about a "field" where $1=0$ is like talking about a "field" where addition isn't commutative or any other axiom is violated. It's a legitimate algebraic structure but by definition, it isn't a field. – fleablood Jul 14 '20 at 16:42
  • The characteristic of fields, if non-zero, is a prime number. You wouldn't argue that $1$ is a prime number, would you? :-) – user804886 Jul 14 '20 at 16:44
  • "(But the importance of the axiom 0≠1 isn't clear to me.)" Hah!!! The reason for the axiom is ENTIRELY so this structure WON"T be a field! The thing is if we have the word "ring" and we have word "field" higher in the heirarchy there should be something distinguishing them. If $0=1$ then every thing that makes a field a field is vacuous..... Or... it's arbitrary. Let's just go with that.... It's an arbitrary distinction. – fleablood Jul 14 '20 at 16:47
  • "You wouldn't argue that 1 is a prime number, would you? :-)" Why not? A prime number is an integer whose only positive integer factors are $1$ and itself. And the only positive whole factor of $1$ is $1$ and itself, $1$... That got me through my first nine years of education and then I stubbornly went through the next three claiming "prime factorization ignoring the infinite occurances of $1$..."...... Okay... the point GOOD NEWS-- the OP is correct, the "trivial field" satisfies all axioms but one and the OP is correct. BAD NEWS-- we hamfistedly refuse to drop that axiom. – fleablood Jul 14 '20 at 16:52
  • It's a convention, and different authors make different conventions. The world doesn't fall apart if you let this be a "field." – Randall Jul 14 '20 at 16:53
  • "The world doesn't fall apart if you let this be a "field." Speak for yourself! My dog is going to leave me and I'm going to face the world with a sense of existential dread if the OP calls it a field.... Okay, I joke. I'm actually fine and have frequently have called this a field and, honestly, think it should be call "the trivial field" and think that a proposition should be prove that "the trivial field is the only field where $1=0$". – fleablood Jul 14 '20 at 16:58
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    From the linked article: "However, one of the key motivations of F1 is the description of sets as "F1-vector spaces"—if finite sets were modules over the zero ring, then every finite set would be the same size, which is not the case. " And as user804886 points out "The characteristic of fields, if non-zero, is a prime number" which, all my joking aside, is a problem if we consider $1$ to be prime (unique factorization fails). I think there are no basic issues with a trivial field but there are several advanced applied issues... but I suppose we could deal with those when they arise... – fleablood Jul 14 '20 at 17:09
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    One thing in particular that I'd like to find out is why it is that we would expect $F_1$-vector spaces to be pointed sets. – Sophie Swett Jul 14 '20 at 17:39
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    a standard result -- a ring is a field iff it has exactly 2 ideals. But the zero ring doesn't have 2 ideals. If you want to redo all the definitions, I guess you can try. – user8675309 Jul 14 '20 at 18:38
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    There are several definitions of varieties or schemes over $F_1$ by several authors, some of them very different. That does not mean that the is a field with one element. "Variety over $F_1$" is only a name, suggested by some analogies with varieties over $F_p$. – A.G Jul 14 '20 at 21:31

1 Answers1

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I think approaching the "field with one element" from the perspective of the field axioms makes it seem a bit silly, when it really is (in my opinion) deep mathematics, so lets use the definition of a field as a ring all of whose (finitely generated) modules are free.

One can reconstruct a commutative ring $R$ from its category of $R$ modules, so this isn't losing anything, and in this sense we can look for categories that behave just like $R-Mod$ for $R$ a field. These categories are fantastic in many senses, but one important fact is that every object decomposes into a direct sum of simple objects, and there is a unique simple, up to isomorphism.

Then the observation is that the category $Set$ satisfies a lot of these desirata, in particular this last point. The category of sets carries a lot of the categorical structure of $Vect_k$, such as an internal $\otimes$ functor, and internal hom functor, satisfying the tensor Hom adjunction (these are $\times$ and functions from X to Y, respectively), but also carries more of the vector space specific functors, such as exterior and symmetric powers, mapping a set to its set of sub/multisets of size $m$. So in this sense, the category of sets strongly resembles the category of $k$ modules, aside from the slight issue that its not additive.

Hopefully this has convinced you that someone could draw parallels, but one would be right to be skeptical at this point if I asked you to believe that this is deep mathematics. So now I'll try give some examples of nontrivial parallels, that don't require much background.

  1. The automorphism group of the sum of $n$ copies of our simple object admits a non-obvious map to an abelian group (the determinant/sign map, respectively), and the kernel is in general, a simple group ($A_n$ and $SL_n(k)$).

  2. We can turn the set of $m$ dimensional subspaces of an $n$ dimensional vector space into a projective variety $Gr(m,n)$, and the number of points of $Gr(m,n)$ over the field with $q$ elements is a polynomial $P(q)$ in $q$. When we set $q=1$, we recover the number of subsets of an $n$ element set of size $m$. Furthermore, if we have a cyclic group of order $n$ acting transitively on our $n$ element set, it acts on the set of $m$ element subsets, and the value of that same point counting polynomial at a primitive $n$th root of unity, $P(\zeta_n)$ yields the number of $m$ element subsets that are invariant under this cyclic group action. This is the cyclic sieving phenomena, and has shown up in a variety of counting problems.

Finally, there is a machine (algebraic K-theory) that one can apply to the system of compatible groups $GL_n(R)$ for any ring $R$, and applying this to $S_n"="GL_n(\mathbb{F}_1)$ yields the stable homotopy groups of spheres, which are incredibly rich and hard to understand objects. This can be stated as the algebraic K groups of $\mathbb{F}_1$ "are" the stable homotopy groups of spheres.

These are just a few examples I know of, there are many more parallels, but hopefully this is enough to convince you that there is deep mathematics at play here. The degeneration of field-theoretic behaviour is why these analogies are termed the "field with one element", and it seems like our restrictive notion of a field (as defined by its axioms) is not sufficient to capture the full spectrum of "field theoretic behaviour".

Alekos Robotis
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Chris H
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  • You wrote "there is a unique simple". Is there a word missing? – MJD Apr 15 '24 at 11:16
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    I omitted the word module here, this is standard (though somewhat informal) when talking about modules in settings where simple modules are principally considered. – Chris H Apr 16 '24 at 07:04