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Stream: learning: questions

Topic: Field with one element

Keith Elliott Peterson (Dec 20 2024 at 00:33):

I'm just getting my toes wet in alg. geo. here, but the reason set theory is a kind of geometry over $\mathbb{F}_1$ is because $\mathbb{F}_1$ only has a single prime ideal (if such a field were to exist), correct?

Is a set just a variety in this case, and the points are the elements of that set?

John Baez (Dec 20 2024 at 00:48):

Getting your toes wet in algebraic geometry by thinking about the field with one element is kind of like starting to study music by listening to Cage's 4'33''. How can we answer a question like "if $\mathbb{F}_1$ existed, it would have a single prime ideal?" It's like a koan.

The idea is that if you have some variety systematically defined over all fields, you can count the number of points over $\mathbb{F}_q$ where $q$ is a prime power, and sometimes you get a polynomial function of $q$. Then if you set $q = 1$ you get the number of points of a very interesting set, which we pretend is that variety defined over $\mathbb{F}_1$, because it's actually defined in a way that closely resembles the variety were talking about. You have to look at examples, like the projective line or projective plane. I do some here.

By the way there's a performance of 4'33' on YouTube, but they obviously play it much too fast!

Keith Elliott Peterson (Dec 20 2024 at 01:00):

Mm. I was actually reading week 205, and it just struck me that if such a field existed, the geometry above it would consist of things with $k$-colored points for $k=1$. Obviously, such a field doesn't exist (not in any classical sense at least), but if it did, it would only have one ideal. Perhaps it might not be proper to call it prime, since it's not technically a proper ideal.

I agree, that performance is a tad too fast. It's supposed to be a slow burner.

John Baez (Dec 20 2024 at 01:18):

I just can't believe someone put a performance of 4'33' on YouTube where the video lasts for only 3 minutes and 43 seconds!

"You had one job...."

Keith Elliott Peterson (Dec 20 2024 at 01:19):

Moore's paradox moment.

Clearly, it's an uptempo remix to engage the youth.

Damiano Mazza (Dec 20 2024 at 13:05):

Yes $\mathbb F_1$ morally has to have only one prime ideal, but this is because all fields have exactly one prime ideal (namely the zero ideal, which is a prime ideal in any integral domain). So the relationship between sets and $\mathbb F_1$ is not due to it having only one prime ideal, otherwise this would be true for all fields. The reason is more related to what John explained (and I agree with his analogy $\mathbb F_1$/John Cage's 4'33'' :big_smile:)

For a bit more about geometry over $\mathbb F_1$, you can start by thinking that, if $k$ is a field, then $\mathrm{Spec}\,k$ is a one-point space, but "colored" with $k$. So in algebraic geometry there are many "one-point spaces", but none of them is the terminal object (like the usual one-point space), because no field is initial. Relativizing over any of them gives you geometry over the corresponding field: the category of $k$-schemes is literally the slice over $\mathrm{Spec}\,k$, in which $\mathrm{Spec}\,k$ is terminal, that is, it really behaves like the one-point space.

Since there is no initial field, there is no "absolute point", that is, a scheme $\ast$ whose underlying space is a point and such that there is a unique arrow $S\to\ast$ for every scheme $S$. The idea is that $\mathrm{Spec}\,\mathbb F_1$ should play the role of such an "absolute point". In fact, as far as I know (but I have an extremely limited understanding of these things), much of $\mathbb F_1$-geometry is about seeing $\mathrm{Spec}\,\mathbb Z$ (the spectrum of the integers) as a "curve" over $\mathrm{Spec}\,\mathbb F_1$, which you can't do in usual algebraic geometry because $\mathbb Z$ is initial in rings, so there's nothing "below" $\mathrm{Spec}\,\mathbb Z$.

This blog post tells you more about how the points of the underlying topological space of $\mathrm{Spec}\,\mathbb Z[X]$, which is the affine integer line, actually look more like the points of a product space, that is, more like a plane, and that this hints to the fact that it should be seen as the product of $\mathrm{Spec}\,\mathbb Z$ and $\mathrm{Spec}\,\mathbb F_1[X]$. But I am not sure you are going to learn much algebraic geometry by looking at this stuff!

Keith Elliott Peterson (Dec 20 2024 at 18:41):

Damiano Mazza said:

For a bit more about geometry over $\mathbb F_1$, you can start by thinking that, if $k$ is a field, then $\mathrm{Spec}\,k$ is a one-point space, but "colored" with $k$. So in algebraic geometry there are many "one-point spaces", but none of them is the terminal object (like the usual one-point space), because no field is initial. Relativizing over any of them gives you geometry over the corresponding field: the category of $k$-schemes is literally the slice over $\mathrm{Spec}\,k$, in which $\mathrm{Spec}\,k$ is terminal, that is, it really behaves like the one-point space.

Is this analogous to how the set $X$ is terminal in $\mathrm{Set}/X$?

David Michael Roberts (Dec 20 2024 at 20:58):

Yes

Keith Elliott Peterson (Dec 20 2024 at 22:10):

Okay, then this isn't as strange as I thought it would be.

John Baez (Dec 20 2024 at 23:17):

Well, but that's if we stick to working with schemes over a fixed field $k$: $\mathrm{Spec}(k)$ is terminal in the category of schemes over $k$, and since we've chosen a field, there's no pressure to get interested in the field with one element yet. The problem starts when we go beyond this, e.g. when thinking about the Riemann Hypothesis. There's a Riemann hypothesis for smooth varieties over a finite field $k$, called the Weil Conjectures, and those have been proven, but the actual Riemann Hypothesis is about $\mathbb{Z}$, and then we get very stuck, and people start desperately searching for the field with one element.

Jorge Soto-Andrade (Dec 20 2024 at 23:25):

John Baez said:

Getting your toes wet in algebraic geometry by thinking about the field with one element is kind of like starting to study music by listening to Cage's 4'33''. How can we answer a question like "if $\mathbb{F}_1$ existed, it would have a single prime ideal?" It's like a koan.

The idea is that if you have some variety systematically defined over all fields, you can count the number of points over $\mathbb{F}_q$ where $q$ is a prime power, and sometimes you get a polynomial function of $q$. Then if you set $q = 1$ you get the number of points of a very interesting set, which we pretend is that variety defined over $\mathbb{F}_1$, because it's actually defined in a way that closely resembles the variety were talking about. You have to look at examples, like the projective line or projective plane. I do some here.

By the way there's a performance of 4'33' on YouTube, but they obviously play it much too fast!

Hi! Sorry for jumping in a bit late. You have a nice "Grasmannian example" to motive the field with one element. I often ask my undergrads whether they can count how many k-dimensional subspaces they have in an n-dim vector space over the field with q elements (an easy linear algebre exercise). They get some nice polynomials in q. Then I say: look at those polynomials, what would you like to do if "nobody is looking". Quite often a (female student) says "evaluate at q = 1". Another (female) studen replies: "are you crazy, q was the number of elements of our finite field, so you can't take q = 1". A harsh dispute ensues, and then a (male) student intervenes as a mediator... (the genders are real) Of course, they have just discovered the Gaussian q-analogues of the combinatorial numbers n over k.
I guess the field with one element is quite popular in the American Bible Belt...

Jorge Soto-Andrade (Dec 20 2024 at 23:27):

Why? Because the Holy Bible says: "By their fruits you shall know them".

John Baez (Dec 20 2024 at 23:29):

How much fruit can you grow in a field with one element? :smirk:

Jorge Soto-Andrade (Dec 20 2024 at 23:30):

PS Pierre Cartier (see https://www.researchgate.net/publication/385750025_Pierre_Cartier_A_Visionary_Mathematician)
has some nice notes on the field with one element (available at IHES, I guess). Idem, Cristophe Soulé. One underlying idea is that a finite field is a cyclic group plus 0 with a clever addition.

Jorge Soto-Andrade (Dec 20 2024 at 23:35):

John Baez said:

How much fruit can you grow in a field with one element? :smirk:

More than you would have expected ... :smile:

Madeleine Birchfield (Dec 21 2024 at 03:55):

What happens if you set $q$ to a composite natural number instead of $1$?

John Baez (Dec 21 2024 at 22:16):

Good question! For prime powers we know what $\mathbb{F}_q$ - the unique field of that cardinality - and that's what we need to get all the combinatorics of various linear algebra puzzles to give formulas that reduce to familiar set-theoretic combinatorics when $q = 1$.

So what if $q$ is not a prime power? Is there a commutative ring that works in this case? Or do we need something weirder? That's the question I think you're really asking. And that's a great question.

There is more than one finite semisimple commutative ring (= products of finite fields) with $q$ elements when $q$ is composite. E.g. for $q = 24$ we have three

$\mathbb{F}_8 \times \mathbb{F}_3, \mathbb{F}_2 \times \mathbb{F}_4 \times \mathbb{F}_3, \mathbb{F}_2 \times \mathbb{F}_2 \times \mathbb{F}_2 \times \mathbb{F}_3$

There's a canonical "best" one that's a product of just one field for each prime in the prime factorization of $q$. Here it's $\mathbb{F}_8 \times \mathbb{F}_3$. Is that the commutative ring we should use to get the combinatorial formulas to work just like when $q$ is a prime power?

Sometime I'll check with a simple example.

There's another canonical choice of semisimple commutative ring with $q$ elements that's the product of as many field as possible, but that wouldn't give the usual field $\mathbb{F}_q$ when $q$ is a prime power, so that doesn't sound good.

John Baez (Dec 21 2024 at 22:21):

This stuff about semisimple commutative rings as generalizations of fields may sound weird, but it does show up naturally when you think about $L$-functions. For example James Dolan noticed that the Riemann zeta function shows up naturally when you decategorify this species (i.e. functor from the groupoid of finite sets to $\mathsf{Set}$):

The Riemann species assigns to any finite set the collection of ways of making that set into a product of finite fields.

Later we generalized the heck out of this fact.

Jorge Soto-Andrade (Dec 22 2024 at 04:52):

Madeleine Birchfield said:

What happens if you set $q$ to a composite natural number instead of $1$?

Nice question! Say $q = 6$ for instance. Then $q + 1$ = # of 1-dim subspaces over a "field with 6 elements" = 7. Seven what? Lines through the origin in the plane over the integers mod 6 are a bit weird, I am afraid. They do not make up seven. What are you counting, and where, to get seven? I first thought of Langrangians (I like to see symplectic geometry as just the geometry of the vectorial plane over the involutive ring = full matrix ring over a field, so vector lines = Lagrangians, so the two coordinates
$\star$ -commute ($\star$ = transposition), an analogue of Manin's plane, where the two coordinates $q$-commute) But also $7 = 2^2 + 2 + 1$, which suggests a connection with a finite projective plane .... What category is behind the vector geometry over the "field of 6 elements", as the category Set is behind the vector geometry over the field with one element?

Jorge Soto-Andrade (Dec 22 2024 at 04:55):

I would see ss involutive rings (rings with an anti-involution $\ast$ ) as a fruitful generalisation of fields (so you get hold of $\mathbb Z$ with $\ast = Id$ as well as of the full matrix ring over a field, with $\ast = transpose$ , which are both $\ast$ - euclidean :smile: )