Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: theory: mathematics

Topic: schemes over Q and their models over Z

John Baez (Apr 14 2024 at 15:54):

Chris Grossack (they/them) said:

are dramatically different over $\mathbb{F}_2$, but they're isomorphic over $\mathbb{F}_p$ for all primes $\neq 2$, since in those fields $2$ is invertible.

This was kind of my point earlier! Here's a proof theoretic angle, which might convince you,

By the way, it's not that I was unconvinced by what you wrote earlier: I just failed to understand it, mainly because I wanted to see a simple example of some nonisomorphic schemes over $\mathbb{Z}$ that become isomorphic over $\mathbb{Q}$, and you seemed to instead be giving me a suggestion for how to prove that such examples must exist. I didn't want to think about that: I wanted to just snatch an example out of thin air, because it seemed so obvious that they must exist. I was like those students who say, when you're telling them how to work a problem, "but what's the answer?"

If $X$ has two presentations over $\mathbb{Q}$, then there should be a proof of this fact (by which I mean something like a sequence of Tietze transformations but for ring presentations). But any such proof is finitely long, so one only has to divide finitely many times. But then the same proof will work over any $\mathbb{F}_p$ with $p$ large enough for all those divisions to be allowed!

That's nice. This sounds like it can become a proof of some nice general theorem.

John Baez (Apr 14 2024 at 16:07):

By the way, remember the stuff about "models" - that is, ways to choose a scheme over $\mathbb{Z}$ that gives some chosen scheme over $\mathbb{Q}$? And the stuff about "Neron models" - that is, initial models?

I'm guessing that while this affine scheme over $\mathbb{Q}$

$\langle u,v \vert u^2 + v^2 = 1 \rangle$

has lots of models

$X_{a,b} = \langle x, y \vert a^2 x^2 + b^2 y^2 = 1 \rangle$

namely one for each $a,b \in \mathbb{Z}$, the model $X_{1,1}$ is a Neron model. The point is that we have a map

$X_{1,1} \to X_{a,b}$

sending $x$ to $ax$ and $y$ to $by$, but usually this is not invertible since we're not allowed to divide by $a$ and $b$.

If $X_{1,1}$ is really a Neron model, it would be nice because it says the circle $x^2 + y^2 = 1$ is "better" than the other ellipses $X_{a,b}$ in some objective sense. But I'm actually not at all sure it's correct.

John Baez (Apr 14 2024 at 16:14):

Chris Grossack (they/them) said:

As an aside, I can't actually find anything on "Tietze transformations for algebra presentations"... Surely someone has studied that, though! Does it have a different name?

I don't know. I think there should be a general theory of Tietze moves going between all different finite presentations of any finitely presented algebra in any variety (in the universal algebra sense of 'variety'). It should look a lot like the familiar case of Tietze moves for group presentations. It should be in some book on universal algebra.

John Baez (Apr 14 2024 at 16:32):

Now let me try to work out the Hasse-Weil zeta functions of these schemes over $\mathbb{Z}$:

$A = \langle x,y \vert x^2 + y^2 = 1 \rangle$

and

$B = \langle x, y \vert x^2 + 4 y^2 = 1 \rangle$

I've never seen anyone do this, probably because it's too elementary. But it will help me understand the more fancy stuff in Calegari's paper.

John Baez (Apr 14 2024 at 16:33):

In each case the zeta function will be a product of 'Euler factors', one for each prime.

John Baez (Apr 14 2024 at 16:36):

Let's do $A$ first. The Euler factor for the prime $p$ is

$\displaystyle{ Z_p(A, s) = \exp \sum_{n > 0} \frac{\vert A(\mathbb{F}_{p^n})\vert }{n} p^{-ns} }$

where $A(\mathbb{F_{p^n}})$ is the set of points of our scheme over the field with $p^n$ elements.

John Baez (Apr 14 2024 at 16:40):

This formula probably looks a bit horrible (at least to category theorists) but there's a fairly elegant conceptual explanation, which I will only give on request.

John Baez (Apr 14 2024 at 16:42):

So what's $\vert A(\mathbb{F}_{p^n})\vert$? It's the number of solutions of

$x^2 + y^2 = 1$

where $x,y \in \mathbb{F}_{p^n}$.

John Baez (Apr 14 2024 at 16:47):

How many solutions are there? I'm guessing that - at least for most primes $p$ - there are $p^n + 1$.

John Baez (Apr 14 2024 at 16:54):

The reason I'm guessing this is that at least over $\mathbb{R}$ the 'quadric' $x^2 + y^2 = 1$ is isomorphic to the projective line, which is an ordinary line and a point at infinity. We can see this using stereographic projection.

John Baez (Apr 14 2024 at 16:54):

But over $\mathbb{F}_{p^n}$ the projective line has $p^n + 1$ elements.

John Baez (Apr 14 2024 at 16:55):

Let me check this guess in the case where it's most likely to be wrong: $p = 2$. (The prime 2 is bad news in general, and especially for polynomials of degree 2, e.g. the derivative of $x^2$ is zero in a field of characteristic 2.)

John Baez (Apr 14 2024 at 16:57):

So, in $\mathbb{F}_2$ the solutions of $x^2 + y^2 = 1$ are $(0,1)$ and $(1,0)$, so there are just two, not 3 as I was predicting.

John Baez (Apr 14 2024 at 16:58):

(It's possible that I'm getting in trouble because I'm working with an affine quadric instead of a projective one, or it could be that the prime 2 is doing its usual evil things.)

John Baez (Apr 14 2024 at 16:59):

In $\mathbb{F}_3$ we get solutions $(0,\pm 1)$ and $(\pm 1, 0)$ and no others (right?). So we are getting 4 solutions, as I predicted.

John Baez (Apr 14 2024 at 17:03):

(You may wonder why in this category theory discussion group I'm taking every question and exploring it by means of the most mundane calculations possible, rather than simply hitting it with a well-chosen functor or using a short exact sequence or something. It's because I want to become familiar with algebraic geometry from the ground up: if I don't even understand the solutions of $x^2 + y^2 = 1$ over a finite field how can I claim to know anything about the subject?)

John Baez (Apr 14 2024 at 17:07):

In $\mathbb{F}_5$ we have numbers $0, \pm 1, \pm 2$, with $(\pm 2)^2 = -1$, so the solutions of $x^2 + y^2 = 1$ are $(0,\pm 1), (\pm 1, 0)$ and... that's all??? Just 4 solutions again???

John Baez (Apr 14 2024 at 17:09):

Okay, so either I'm making computational mistakes or I don't understand the pattern at all. Now I'm guessing the pattern may be related to $p \bmod 4$, since $-1$ has a square root in $\mathbb{F}_p$ iff $p = 1 \bmod 4$ (except for the annoying case $p = 2$).

Chris Grossack (they/them) (Apr 14 2024 at 17:27):

I'm sure you're interested in doing this computation yourself, but I actually did it a few years ago for exactly the same reason as you! I find it impossible to learn things like this without computing. Though I used sage to try every possible solution over finite fields both to speed things up and to make sure I didn't miss any!

Chris Grossack (they/them) (Apr 14 2024 at 17:28):

If you want to cheat at all, you can find it in this pdf: https://grossack.site/assets/docs/weil-conjectures-talk/paper.pdf

Chris Grossack (they/them) (Apr 14 2024 at 17:29):

I was going to write a blog post going into these kinds of computations in more detail (and explaining how the Weil conjectures themselves are REALLY concrete numerical predictions! It's only the proofs that are fearsome)... But it looks like I never got around to writing it, and by now I've forgotten a lot of things

John Baez (Apr 14 2024 at 17:51):

Chris Grossack (they/them) said:

If you want to cheat at all, you can find it in this pdf: https://grossack.site/assets/docs/weil-conjectures-talk/paper.pdf

I was considering cheating, because I was getting a bit tired and my predictions weren't panning out very well. But now I'm remembering some more theoretical stuff that may help me guess the answers, or at least motivate me to persevere. It's in Gilles and Szamuely's book Central Simple Algebras and Galois Cohomology, which is a great introduction to Galois descent through some concrete examples - namely, the classification of central simple algebras, which is actually how Noether, Hasse and Brauer stumbled into the cohomology of groups!

John Baez (Apr 14 2024 at 17:53):

Among the easiest of the central simple algebras are the 'quaternion algebras', which are precisely the 4-dimensional simple algebras over a field whose center is that field - or more concretely, algebras over a field with generators $i, j$ and relations

$i^2 = a, j^2 = b, ij = -ji$

John Baez (Apr 14 2024 at 17:54):

For example all quaternion algebras over $\mathbb{R}$ are isomorphic to the quaternions (take $a = b = -1$ for example) or the algebra of 2 $\times$ 2 real matrices (take $a = b = 1$ for example).

John Baez (Apr 14 2024 at 18:00):

And Brauer and Severi noticed that any quaternion algebra gives a projective curve called a 'conic', which is defined by quadratic equations.

John Baez (Apr 14 2024 at 18:00):

So my rather feeble attempt to count points on what I was calling 'quadrics' is actually connected to the classification of conics and thence to quaternion algebras and thence to Galois descent!

John Baez (Apr 14 2024 at 18:01):

Let me try to say a bit more about this - by now I'm fired up and don't mind talking to myself.

John Baez (Apr 14 2024 at 18:02):

We can write elements of any quaternion algebra as

$w 1 + x i + y j + z k$

where $w,x,y,z$ are elements of the field it's over, $1$ is the unit of the quaternion algebra, and $k = i j$.

John Baez (Apr 14 2024 at 18:03):

So, our quaternion algebra has a 3d space of imaginary elements:

$\{ x i + y j + z k \}$

John Baez (Apr 14 2024 at 18:09):

The square of an imaginary element is always real - I think it equals this:

$(x i + y j + z k)^2 = ax^2 + by^2 - ab z^2$

John Baez (Apr 14 2024 at 18:10):

And this is a homogeneous quadratic polynomial on $\mathbb{F}^3$, where $\mathbb{F}$ is our (hitherto nameless) field.

John Baez (Apr 14 2024 at 18:11):

So it defines a curve in the projective plane over $\mathbb{F}$, called a conic.

John Baez (Apr 14 2024 at 18:12):

Concretely this curve consists of triples $(x,y,z)$ with

$ax^2 + by^2 - abz^2 = 0$

modulo rescaling.

John Baez (Apr 14 2024 at 18:13):

We can define conics over any field to be curves in the projective plane defined by nontrivial homogeneous quadratic equations... I hope that's the right definition....

John Baez (Apr 14 2024 at 18:14):

And there was a witty guy named Witte, who showed this:

Theorem. Isomorphism classes of quaternion algebras correspond bijectively to isomorphism classes of conics.

John Baez (Apr 14 2024 at 18:14):

(This should really be an equivalence of categories.)

John Baez (Apr 14 2024 at 18:15):

There's more: we can classify both these things using Galois cohomology.

John Baez (Apr 14 2024 at 18:15):

But now I have to go hear Scottish folk music at my local pub!

John Baez (Apr 15 2024 at 09:13):

Hmm, this is project is sprawling outwards in some ways that will be frustrating to anyone reading this. Let me summarize where I am right now.

John Baez (Apr 15 2024 at 09:16):

I was trying to compute the Hasse-Weil zeta functions of these affine schemes over $\mathbb{Z}$:

$A = \langle x,y \vert x^2 + y^2 = 1 \rangle$

and

$B = \langle x, y \vert x^2 + 4 y^2 = 1 \rangle$

Don't be intimidated: this simply amounts to counting the number of solutions of these equations over all finite fields $\mathbb{F}_q$. (There's a unique finite field of size $q$ for each prime power $q = p^n$.)

Since these are quadratic equations, this should not be extremely hard. People have done it for similar-looking cubic equations, as part of "computing the Hasse-Weil zeta function of an elliptic curve". This is supposed to be much easier.

John Baez (Apr 15 2024 at 09:20):

However, it's better to work with projective varieties than affine varieties - that'll let us use more theorems. So now, for each finite field $\mathbb{F}_q$, I want to count the number of nonzero solutions of

$x^2 + y^2 = z^2$

modulo rescaling, i.e. counting the solutions $(x,y,z)$ and $(ax,ay,az)$ as the same for any $a \ne 0$.

This new improved equation is homogeneous: the rescaling of any solution is another solution. The variable $z$ is called a homogenizing variable - it replaces the constant $1$ I had before, and makes the equation homogeneous.

John Baez (Apr 15 2024 at 09:21):

The solutions of this new improved equation, mod rescaling, are points in a projective variety I'll call $A'$.

Similarly I want to count the points in a projective variety I'll call $B'$ coming from the equation

$x^2 + 4y^2 = z^2$

I'll call this variety B'.

John Baez (Apr 15 2024 at 09:28):

Projective varieties of this sort, given by a single homogeneous quadratic equation in 3 variables, are extremely famous: they're called conics. They've been studied intensively since 200 BC by a host of people including Apollonius of Perga, Kepler, Descartes, Fermat, and many more modern algebraic geometers, so I'm not trying to do anything new.

John Baez (Apr 15 2024 at 09:30):

Why switch to the projective setting? In their book Central Simple Algebras and Galois Cohomology, Gille and Szamuely mention a result:

It is a well-known fact from algebraic geometry that a smooth projective conic defined over a field k is isomorphic to the projective line $\mathrm{P}^1$ over k if and only if it has a k-rational point.

John Baez (Apr 15 2024 at 09:32):

I think "has a k-rational point" simply means the homogeneous quadratic equation describing our conic has a nonzero solution in the field k. (If I'm wrong someone please correct me.)

John Baez (Apr 15 2024 at 09:35):

If I'm right about this, the conic A', coming from

$x^2 + y^2 = z^2$

has a k-rational point when $k = \mathbb{F}_2$, e.g. x = 1, y = 0, z = 1. So, it must be isomorphic to the projective line over $\mathbb{F}_2$. So, it should have 3 points.

John Baez (Apr 15 2024 at 09:37):

And it does have 3 points! The nonzero solutions of $x^2 + y^2 = z^2$ are

$(x,y,z) = (1,0,1), (0,1,1), (1,1,0)$

and rescaling does nothing in $\mathbb{F}_2$, so we get 3 points.

John Baez (Apr 15 2024 at 09:37):

The weird one is the so-called "point at infinity", where the homogenenizing variable $z$ is zero.

John Baez (Apr 15 2024 at 09:41):

If we were working over the real numbers, the equation $x^2 + y^2 = z^2$ wouldn't have a nonzero solution with $z = 0$. The corresponding projective variety (nonzero solutions mod rescaling) would be simply the circle! So I'm just saying the ordinary circle doesn't have a point at infinity.

John Baez (Apr 15 2024 at 09:42):

But the hyperbola does have a point at infinity, as you can see by looking at $x^2 - y^2 = z^2$: it has exactly one nonzero real solution mod rescaling with $z = 0$.

John Baez (Apr 15 2024 at 09:48):

Over $\mathbb{F}_2$ we have 1 = -1, so the circle and the hyperbola are the same! So it might have been touch-and-go whether the circle over $\mathbb{F}_2$ has a point at infinity or not... but it does.

John Baez (Apr 15 2024 at 09:50):

Anyway, using the sledgehammer provided by Gille and Szamuely, we instantly see the conic A' has $q + 1$ points over the finite field $\mathbb{F}_q$ if it has a "rational k-point" for k = $\mathbb{F}_q$.

John Baez (Apr 15 2024 at 09:51):

And I'm hoping this means A' has $q+1$ points if $x^2 + y^2 = z^2$ has any nonzero solutions over $\mathbb{F}_q$.

John Baez (Apr 15 2024 at 09:51):

But it always does, namely x = 1, y = 0, z = 1.

David Michael Roberts (Apr 15 2024 at 10:14):

John Baez said:

The solutions of this new improved equation, mod rescaling, are points in a projective variety I'll call $A'$.

Similarly I want to count the points in a projective variety I'll call $B'$ coming from the equation

$x^2 + 4x^2 = z^2$

I'll call this variety B'.

4x^2 is a typo

John Baez (Apr 15 2024 at 11:37):

Thanks - fixed!

John Baez (Apr 15 2024 at 13:23):

So, it looks like our projective conic A', given by

$x^2 + y^2 = z^2$

is maximally boring: over every finite field it's isomorphic to the projective line! All the 'trickiness' I was bumping into earlier comes from working with the affine conic

$x^2 + y^2 = 1$

which leave out points at infinity (namely nonzero solutions with $z = 0$).

So here's what I was actually seeing:

• over $\mathbb{F}_2$ our projective conic A' does contain a point at infinity, namely $(1,1,0)$ so the affine version was missing a point, so $x^2 + y^2 = 1$ had just 2 solutions ($x = 1, y = 0$ and $x = 0, y = 1$) instead of the full 3 points we expect from the projective line over $\mathbb{F}_3$.

• over $\mathbb{F}_3$ our projective conic A' does not contain a point at infinity, so all the points lie in affine space, so $x^2 + y^2 = 1$ has 4 solutions ($x = \pm 1, y = \pm 1$) which is just the right number to be all the points in the projective line over $\mathbb{F}_3$.

and so on.

John Baez (Apr 15 2024 at 18:21):

Let me do just one more, to show they get a bit weirder:

• Over $\mathbb{F}_5$ our projective conic $A'$ conic contains two points at infinity. There are plenty of solutions of $x^2 + y^2 = z^2$ with $z = 0$, each of which gives a point at infinity in our conic, namely $(\pm 2, \pm 1, 0)$ and $(\pm 1, \pm 2, 0)$. That's 8 solutions! But points in our conic are solutions mod rescaling and $\mathbb{F}_4$ has 4 nonzero elements so we really just get 8/4 = 2 points at infinity in our conic. The are more points in our conic that are not points at infinity, coming from these other 16 solutions of $x^2 + y^2 = z^2$: $(\pm 1, 0, \pm 1),$ $(\pm 2, 0, \pm 2), (0, \pm 1, \pm 1),$ $(0, \pm 2, \pm 2)$. These give 16/4 = 4 points in our conic that are not points at infinity. So we get 6 points in our conic, which is just the right number to be all the points in the projective line over $\mathbb{F}_5$.

John Baez (Apr 15 2024 at 19:22):

This funny business confirms what they've been telling us at least since 1822, namely that projective varieties are better behaved than affine varieties! Using the sledgehammer of Gille and Szamuely's result, we know the projective conic $x^2 + y^2 = z^2$ over some field $\mathbb{F}_q$ is isomorphic to the projective line over that field if it has a point, which it always does (namely (1,0,1)). So computing its Hasse-Weil zeta function is a snap!

John Baez (Apr 15 2024 at 19:26):

First we calculate the local zeta function at each prime $p$. Since we're calling this conic $A'$ and it has $p^k + 1$ points over $\mathbb{F}_{p^k}$, we have

$Z_p(A',s) = \exp \left( \sum_{k = 1}^\infty \frac{|A'(\mathbb{F}_{p^k})|}{k} p^{-k s} \right)$

$= \exp \left( \sum_{k = 1}^\infty \frac{p^k + 1}{k} p^{-k s} \right)$

John Baez (Apr 15 2024 at 20:31):

$= \exp \left( \sum_{k = 1}^\infty \frac{p^k}{k} p^{-k s} \right) \exp \left( \sum_{k = 1}^\infty \frac{1}{k} p^{-k s} \right)$

John Baez (Apr 15 2024 at 20:34):

What's going on? $A'$ is the projective line, and it's a disjoint union of an affine line with $p^k$ points over $\mathbb{F}_{p^k}$ and a single point. Local zeta functions multiply under disjoint unions, and the two factors here are the local zeta function of the affine line and the local zeta function of a point!

John Baez (Apr 15 2024 at 20:36):

Let's work out the second factor, which is the local zeta function of a point:

$Z_p(\mathrm{point}, s) = \exp \left( \sum_{k = 1}^\infty \frac{1}{k} p^{-k s} \right)$

John Baez (Apr 15 2024 at 21:57):

Remember

$\sum_{k=1}^\infty \frac{x^k}{k} = -\ln(1-x)$

so

$\sum_{k=1}^\infty \frac{1}{k} p^{-ks} = -\ln(1-p^{-s})$

so

$Z_p(\mathrm{point}, s) = \exp \left( \sum_{k = 1}^\infty \frac{1}{k} p^{-k s} \right) = \exp(-\ln(1-p^{-s}))$

so

$\displaystyle{ Z_p(\mathrm{point}, s) = \frac{1}{1-p^{-s}} }$

John Baez (Apr 15 2024 at 22:01):

The Weil Conjectures (now proved) say the poles and zeros of the local zeta function of a smooth projective variety over $\mathbb{F}_p$ are determined by its etale cohomology groups. This local zeta function only has pole at 0, no zeros, and I think that's because the point has only 0th cohomology.

John Baez (Apr 15 2024 at 22:04):

We can multiply all these local zeta functions, one for each prime, to get the Hasse-Weil zeta function of a point:

$\displaystyle{ Z(\mathrm{point}, s) = \prod_p \frac{1}{1 - p^{-s}} = \zeta(s) }$

Yes, the Riemann zeta function! We're using the Euler product formula here.

John Baez (Apr 15 2024 at 22:08):

Next we can work out the Hasse-Weil zeta function of the affine line:

$Z_p(\textrm{affine line}, s) = \exp \left( \sum_{k = 1}^\infty \frac{p^k}{k} p^{-k s} \right)$
$= \exp \left( \sum_{k = 1}^\infty \frac{1}{k} p^{-k (s+1)} \right)$

This is just what we got for the point, but with $s+1$ replacing $s$, so

$\displaystyle{ Z_p(\textrm{affine line}, s) = \frac{1}{1 - p^{-(s+1)}} }$

and multiplying all these local zeta functions we get the Hasse-Weil zeta function of the affine line:

$\displaystyle{ Z(\textrm{affine line}, s) = \prod_{p} \frac{1}{1 - p^{-(s+1)}} = \zeta(s+1) }$

John Baez (Apr 15 2024 at 22:18):

Remember we're trying to compute the local zeta functions of our conic $A'$, which is just projective line, and we know we can get these by multiplying those for the point and the affine line. Here's what we get for the local zeta functions:

$\displaystyle{ Z_p(A', s) = Z_p(\mathrm{point},s) Z_p(\textrm{affine line}, s) = \frac{1}{(1 - p^{-s})(1 - p^{-s+1})} }$

These have poles at $s = 0$ and $s = 1$, which the Weil Conjectures explain as coming from the two nonvanishing etale cohomology groups of the projective line.

Multiplying all these we get the Hasse-Weil zeta function:

$\displaystyle{ Z(A',s) = \zeta(s) \zeta(s+1)}$

John Baez (Apr 15 2024 at 22:35):

But the point of this thread was supposed to be this. Two schemes $X, Y$ over $\mathbb{Z}$ can give isomorphic schemes over $\mathbb{Q}$. There's no reason to expect $Z(X,s) = Z(Y,s)$. But some of you have argued convincingly that these two zeta functions are the same except for finitely many Euler factors, because $X$ and $Y$ must agree "except over finitely many primes".

At this stage I'm just trying to see that happening in some simple examples. My plan is to compare the projective conic $A'$

$x^2 + y^2 = z^2$

(which I have by now beaten to death) with another projective conic that's isomorphic over $\mathbb{Q}$. I originally wanted to use the projective conic $B'$ given by

$x^2 + 4y^2 = z^2$

This seemed likely to be nonisomorphic only at the prime 2.

John Baez (Apr 15 2024 at 22:41):

But thanks to that killer theorem mentioned by Gille and Szamuely, we know that over $\mathbb{F}_2$ this conic $B'$ is also isomorphic to the projective line, just like $A'$, merely because it has a point! I'd like to check this directly, or at least see some evidence for it, because it's a bit surprising.

John Baez (Apr 15 2024 at 22:48):

But we can also look at the projective conic $C'$ given by

$4x^2 + 4y^2 = z^2$

This is again isomorphic over $\mathbb{Q}$, but it clearly has no points over $\mathbb{F}_2$. That should be enough to ensure its local zeta function at $p = 2$ is different, and its Hasse-Weil zeta function is different.

John Baez (Apr 15 2024 at 22:52):

In fact $C'$ has no points over any finite field $\mathbb{F}_{2^k}$, since in this field $4 = 0$. So its local zeta function at $p = 2$ is the zeta function of the empty set! It's

$Z_p(\textrm{empty set}, s) = \exp \left( \sum_{k = 1}^\infty \frac{0}{k} p^{-k s} \right) = 1$

John Baez (Apr 15 2024 at 22:57):

Otherwise its local zeta functions should be the same, so its Hasse-Weil zeta function matches that of our friend $A'$ except that it's missing the factor of $1/(1 - 2^{-s})$. So we get

$Z(C',s) = \zeta(s) \zeta(s+1) (1 - 2^{-s})$

Kevin Carlson (Arlin) (Apr 16 2024 at 01:53):

For me the evidence that a projective conic is isomorphic to the projective line if it has a rational point goes like this: draw a line through the rational point of every slope in the projective line ($k\cup\{\infty\}$). Since these lines hit the conic once, they hit it twice—at the very least we’re sure about this in the simplest cases, because it amounts to the fact that a quadratic with one rational root must have two!

Kevin Carlson (Arlin) (Apr 16 2024 at 01:53):

I’m definitely just picturing the affine case when I imagine this.

John Baez (Apr 16 2024 at 09:08):

Yes, this is a lot like the argument that Gille and Szamuely sketch out. The way you're saying it works best for algebraically closed fields, or at least 'quadratically closed fields', where every quadratic polynomial can be factored into 2 factors. But they're claiming it works for all fields, including the finite fields I'm dealing with, which are never algebraically closed and - I guess - never even quadratically closed. It's already rather amazing that this works over the rational numbers!

So let me see how they organize the argument to avoid this.

John Baez (Apr 16 2024 at 09:11):

It's just a sketch but it goes like this:

It is a well-known fact from algebraic geometry that a smooth projective conic defined over a field k is isomorphic to the projective line $P^1$ over k if and only if it has a k-rational point. The isomorphism is given by taking the line joining a point P of the conic to some fixed k-rational point O and then taking the intersection of this line with $P^1$ embedded as, say, some coordinate axis in $P^2$.

John Baez (Apr 16 2024 at 09:21):

Minhyong Kim once pointed out that over the rational numbers this has consequences for Pythagorean triples.

All mistakes here are mine, not his! We work affinely and consider the rational circle

$x^2 + y^2 = 1$

Points in here are pairs of rational numbers whose squares sum to one; clearing denominators these give Pythagorean triples!

Now we claim this rational circle is isomorphic to the rational projective line. We do this by considering the line

$y = 0$

We claim that for each point $(x,0)$ on this line with $x$ rational, the line from $(x,0)$ to $(0,1)$ intersects the rational circle! I.e., it intersects the unit circle in a point with rational coordinates. I'd need to do some algebra to check that.

John Baez (Apr 16 2024 at 09:23):

If this is true, we get a bijection between the rational affine line and the rational circle minus the point $(0,1)$. With extra work this extends to an isomorphism between the rational projective line and the rational circle.

John Baez (Apr 16 2024 at 09:23):

But this proves the existence of lots and lots of Pythagorean triples!

Morgan Rogers (he/him) (Apr 16 2024 at 09:35):

John Baez said:

Yes, this is a lot like the argument that Gille and Szamuely sketch out. The way you're saying it works best for algebraically closed fields, or at least 'quadratically closed fields', where every quadratic polynomial can be factored into 2 factors.

Is it not the case that a quadratic polynomial with a known linear factor can be factorized in any field? I'm pretty sure you can extract the other factor from the coefficients of the polynomial even without fancy machinery, as long as it's a finite point (it doesn't have to be "rational" in a setting where that doesn't make sense).

David Michael Roberts (Apr 16 2024 at 11:57):

The line from (x,0) to (0,1) does indeed intersect at a point with rational coordinates! At least in characteristic not 2. You have to calculate in an extension field a priori, because you use the quadratic formula (and here I'm using the fact 2 is invertible), but then the thing you want to take the square root of turns out to be a perfect square :-)

I happened to have to do this with my student earlier this year, to prove that every conic in the projective plane with a rational point has a parametrisation in the way that Pythagorean triples do.

John Baez (Apr 16 2024 at 12:01):

Oh, nice! Thanks, @Morgan Rogers (he/him). So that's the "magic" behind what's going on here. On the one hand it's very simple, but on the other hand it really does magically show that there are enough Pythagorean triples $m^2 + n^2 = k^2$ for the points $(m/k, n/k)$ to be dense in the unit circle.

Here are my thought processes in response to hearing what you said. They're a bit amusing:

At first I thought this: "Oh! Yes, you write down the quadratic formula for the two roots of your quadratic. It may not make sense in a field where not every number has a square root. But if one of the roots makes sense, so does the other, because the only difference between them is that $\pm$ sign."

Then I thought: "But wait, the quadratic formula also involves dividing by $2 a$, so this argument seems not to work in characteristic 2".

Then I thought: "But wait, in characteristic 2 the $\pm$ also doesn't do anything. So something funny is going on in characteristic 2. Is there some better way to write the quadratic formula so it makes sense in characteristic 2? Or what?"

Then I thought: "But wait, he's saying that if a quadratic has one factor in some field then it has two. Isn't this practically the definition of 'factor', that we're writing $ax^2 + bx + c$ as $(x-r_1)$ times some other polynomial... which can only be linear, and thus the other factor we're looking for?"

David Michael Roberts (Apr 16 2024 at 12:07):

For characteristic 2, I found this comment https://mathoverflow.net/questions/46505/how-to-solve-a-quadratic-equation-in-characteristic-2#comment112047_46505 that shows that for a monic quadratic x^2 + bx+c = 0 over F with char(F)=2, you either have b=0, or you can change variables to get something like x^2 + x + c=0.

David Michael Roberts (Apr 16 2024 at 12:10):

So you can work with a field extension with sqrt(-c), in the first case, or else, if you are in the second case and over a finite even characteristic field F, you check that the trace map tr: F -> F_2 applied to c gives 0, and then you know you have a root in F (this is from section 3 of http://eudml.org/doc/41332).

John Baez (Apr 16 2024 at 12:13):

Thanks so much, David! It will take me a while to absorb all that. (For example, why should the trace map applied to c give zero? Are you saying the trace map applied to everything in F is zero, or that c happens to have some special property, or can be assumed to have it?) However, I am retired so it's time for me to finally learn how to solve quadratic equations.

David Michael Roberts (Apr 16 2024 at 12:16):

In a more pedestrian viewpoint, @John Baez , you consider x^2 + bx + c in F[x] as lying in Fbar[x], where F -> Fbar is some algebraic closure. Then you factorise it there into linear factors. If one of the factors lies in F[x], then the other factor has to be, too. It's like that elementary exercise whereby you get students to show that if the real number ab is rational, and a is rational, then b has to be rational :-)

John Baez (Apr 16 2024 at 12:19):

Yes, this is sort of like my final burst of enlightenment:

"But wait, he's saying that if a quadratic has one factor in some field then it has two. Isn't this practically the definition of 'factor', that we're writing $ax^2+bx+c$ as $(x−r_1​)$ times some other polynomial... which can only be linear, and thus the other factor we're looking for?"

David Michael Roberts (Apr 16 2024 at 12:19):

I have no idea about the trace thing, it was in the paper I linked!

Here's a better paper: https://www.academia.edu/29601746/On_the_solution_of_algebraic_equations_over_finite_fields

David Michael Roberts (Apr 16 2024 at 12:20):

In the case of x^2+c, you seem to get a repeated root, owing to the fact square root is respects addition (!) in this setting.

Morgan Rogers (he/him) (Apr 16 2024 at 12:21):

More concretely $(x-p)(x-q) = 0$ expands to $x^2 - (p+q)x +pq = 0$, so if I know $p$ and my equation is $x^2 + bx + c = 0$ with coefficients in my field I can deduce that $q = -p-b$.

Morgan Rogers (he/him) (Apr 16 2024 at 12:23):

(Note that I used monic polynomials here; since you're considering the general case @John Baez where "the" leading coefficient might reduce to 0 in a given field, there is still the earlier problem to examine)

David Michael Roberts (Apr 16 2024 at 12:25):

Anyway, when you do the algebra in odd-characteristic, it's quite evident when you go to extract the square root, as you get cancellation under it and then have something like sqrt(k^2), and the two square roots give you either the point you started with, or the new one.

I can imagine that in the two cases in a characteristic 2 finite field, it should be possible to check that the result turns out to not like in the extension needed to naively take the square root.

David Michael Roberts (Apr 16 2024 at 12:28):

Oh, this is fun. If you have a finite char 2 field and one solution A of x^2 + x + c = 0, then the other solution is 1+A (!).

John Baez (Apr 16 2024 at 12:51):

Thanks, D & M. I think Morgan's "more concrete" approach is very nice when it comes to the application to Pythagorean triples (which has nothing to do with characteristic 2: it's about $\mathbb{Q}$).

Spelling that out:

Take the unit circle and draw a straight line from its top point $(0,1)$ to any point $a \in \mathbb{Q}$ on the $x$ axis. Since this line hits the circle at one point with rational coordinates, it must hit the circle at another point $(x,y)$ with rational coordinates!

Why? Simply because a quadratic with rational coefficients and one rational root must have two rational roots.

So, points with rational coordinates are dense in the unit circle.

But Morgan's approach lets us find all these points. The algebra gets messy, but the principle is simple.

However, I seem to be getting bogged down in calculation mistakes. :crying_cat:

Morgan Rogers (he/him) (Apr 16 2024 at 13:11):

You're not the only one, I just corrected a sign error on my formula for $q$ from earlier...

John Baez (Apr 16 2024 at 13:12):

Maybe I can blame my problems on you then. :wink:

Morgan Rogers (he/him) (Apr 16 2024 at 13:14):

It might have been easier to use the other formula of $q = 1/p$, but of course that only works as long as $p$ is not $0$

John Baez (Apr 16 2024 at 13:27):

Okay, with Morgan's fixed formula everything works great.

Theorem. Points with rational coordinates are dense in the unit circle..

Slick Proof. Take the unit circle and draw a straight line from its top point $(0,1)$ to any point $a \in \mathbb{Q}$ on the $x$ axis. Since this line hits the circle at one point with rational coordinates, it must hit the circle at another point $(x,y)$ with rational coordinates! The reason is that if a quadratic equation with rational coefficients has one rational root, the other root must also be rational. Such lines hit the circle in a dense set of points. So points with rational coordinates are dense on the unit circle. $\qquad$ █

Explicit Proof. The line hits the circle where $x^2 + y^2 = 1$ and also $x = a(1 - y)$, so the points of intersection must have

$a^2(1-y)^2 + y^2 = 1$

$a^2 - 2a^2y + a^2y^2 + y^2 = 1$

$(a^2+1)y^2 - 2a^2y + (a^2-1) = 0$

$y^2 - (2a^2/(a^2+1)) y + (a^2-1)/(a^2+1) = 0$

We know $y = 1$ is a solution of this quadratic since our line crosses the circle at $(0,1)$ - we don't have to check this!

Thus, Rogers' Lemma says that $y = -1 - b$ is also a solution of the above quadratic, where $b$ is the second coefficient of the quadratic. That is,

$y = -1 + 2a^2/(a^2+1) = (a^2-1)/(a^2+1)$

is also a solution. This gives

$x = a(1-y) = 2a/(a^2 + 1)$

Thus, for any number $a$,

$(x,y) = \big( 2a^2/(a^2+1), (a^2-1)/(a^2+1) \big)$

is a point on the unit circle. And if $a$ is rational this point has rational coordinates. These points are dense in the circle since rationals are dense in the line and the map

$a \mapsto \big(2a^2/(a^2+1), (a^2-1)/(a^2+1) \big)$

from the line to the circle minus $(1,0)$ is a homeomorphism. $\qquad$ █

Morgan Rogers (he/him) (Apr 16 2024 at 13:30):

Please don't go around calling this Rogers' Lemma, I don't want the algebraists to hate me :joy:

John Baez (Apr 16 2024 at 13:31):

Don't worry, I won't. But at least I'm calling the corrected formula Rogers' Lemma.

John Baez (Apr 16 2024 at 13:40):

Later I'll return to thinking a bit more about quadratics in characteristic 2, based on @David Michael Roberts's comments. Now I have to do some actual work!

Reid Barton (Apr 16 2024 at 19:18):

https://en.wikipedia.org/wiki/Pythagorean_triple#Stereographic_approach

Kevin Carlson (Arlin) (Apr 16 2024 at 21:44):

Now for Eisenstein triples! This whole argument was a thing we used to challenge students with at AoPS. Of course they were way better at solving equations than I am.

John Baez (Apr 17 2024 at 14:53):

I summarized some of the ideas in this conversation here:

• Pythagorean triples and the projective line, The n-Category Cafe.

I thanked @Morgan Rogers (he/him) for that lemma, but I resisted calling it Rogers' Lemma.

Reid Barton (Apr 17 2024 at 16:00):

By the way, I don't know if this was mentioned earlier, but over a field of characteristic 2, the projective variety defined by $x^2 + y^2 = z^2$ is not isomorphic to the projective line as an algebraic variety, because it's not smooth: after all, it's just a funny way of writing the equation $(x+y+z)^2 = 0$.

Reid Barton (Apr 17 2024 at 16:01):

If you try to do this stereographic projection trick, you'll find that every line (other than the line $x+y+z = 0$) meets this variety in a double point, and never two distinct points.

Morgan Rogers (he/him) (Apr 17 2024 at 16:01):

Thanks John! I do feel that Kevin should get a mention too though :wink:

Reid Barton (Apr 17 2024 at 16:05):

For the purpose of counting its points over a finite field, though, this doesn't really matter, because the projective variety defined by a linear equation like $x + y + z = 0$ is isomorphic to the projective line (as a variety).

Reid Barton (Apr 17 2024 at 16:11):

Reid Barton said:

the projective variety defined by $x^2 + y^2 = z^2$ is not isomorphic to the projective line as an algebraic variety, because it's not smooth:

I suppose this might be debatable depending on the definitions of "variety" and "defined by", but in any case, it's definitely not a "smooth projective conic".

John Baez (Apr 17 2024 at 18:02):

Okay, so that's how it wriggles out of the theorem I mentioned in the blog article. I should fix the article.

John Baez (Apr 18 2024 at 10:42):

Reid Barton said:

By the way, I don't know if this was mentioned earlier, but over a field of characteristic 2, the projective variety defined by $x^2 + y^2 = z^2$ is not isomorphic to the projective line as an algebraic variety, because it's not smooth: after all, it's just a funny way of writing the equation $(x+y+z)^2 = 0$.

I've been thinking about this more. It seems to turn on a lot of nuances of definitions.

If we think of $x^2 + y^2 = z^2$ as defining a scheme over some field of characteristic 2, it's not a [[reduced scheme]] because the stalks have nontrivial nilpotents. There's an abstract definition of variety saying it's an integral scheme with some extra properties, and to be integral a scheme needs to be reduced, so we're not getting one of those here.

But the most common definitions of projective variety seem to say they're subsets of projective space with some properties, and $x^2 + y^2 = z^2$ picks out the same subset of $k\mathrm{P}^2$ as $x+y+z = 0$ in characteristic 2, so the former is a projective variety iff the latter is.

John Baez (Apr 18 2024 at 10:45):

However, the usual definition of projective variety demands the field $k$ to be algebraically complete, so according to that definition shouldn't really be talking about projective varieties over $k = \mathbb{F}_2$.

John Baez (Apr 18 2024 at 10:46):

I find all these definitions a bit niggly, and different people use different variants, so I find them hard to remember.

John Baez (Apr 18 2024 at 10:49):

Anyway, I agree with your main point, which is that we shouldn't think of $x^2 + y^2 = z^2$ as defining a smooth projective conic over $\mathbb{F}_2$. Instead it defines a "double line", a degenerate limit of conics. This is more obvious if we write it as $(x+y+z)^2 = 0$.

John Baez (Apr 18 2024 at 10:51):

So at least in some hand-wavy way I'd say 2 is a "prime of bad reduction" for $x^2 + y^2 = z^2$. But amusingly this doesn't throw off the count of points for this thing over the fields $\mathbb{F}_{2^n}$: it has the same number of points as you'd expect from a smooth projective conic over a finite field $\mathbb{F}_q$, namely $q+1$. So I think my zeta function calculations are still correct.

David Michael Roberts (Apr 19 2024 at 00:42):

Hmm, are you supposed to count the points of the set, or the points with multiplicity? I presume the former, since that makes things work out, but it did make me pause.

John Baez (Apr 19 2024 at 09:03):

The version of the Hasse-Weil zeta function I understand counts points without multiplicity. But I've only compared it to textbook material in two cases:

• smooth projective varieties over finite fields (the Weil conjectures) where the multiplicity is just 1, so everything works fine

and

• elliptic curves over Q, where people often mutter portentous warnings about 'primes of bad reduction' but my calculations on the n-Category Cafe show that counting points without multiplicity of the Neron model over finite fields gives the usual textbook answer for the zeta function.

John Baez (Apr 19 2024 at 09:08):

I don't really understand Neron models very well at all, hence this thread where I've been trying to see how different schemes over Z giving the same scheme over Q can have different Hasse-Weil zeta functions according to the definition I understand.

John Baez (Apr 19 2024 at 09:11):

For the Artin L-function of Galois representations it's easier to find clear explanations about what to do about primes of bad reduction, i.e. 'ramified' primes. Ramification should be related to double points.

John Baez (Apr 19 2024 at 09:13):

So, I may turn to studying those a bit more. But I should also see what people say about zeta functions of schemes with double points over finite fields!