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Stream: learning: questions

Topic: From the functor of points to the points of a scheme

Damiano Mazza (Jan 21 2024 at 16:58):

In the "functor of points" approach one sees a scheme as a functor

$X:\mathbf{CRing}\longrightarrow\mathbf{Set}.$

In the introduction to the second edition of EGA I, Grothendieck shows how, in the affine case, one may recover the points of the underlying scheme. The idea is that the points of $X$ should morally be given by

$\bigcup_{k\text{ field}} X(k)$

(see also this Mathoverflow question). However, the above is a proper class, so one considers the following equivalence relation: given $f\in X(k)$ and $f'\in X(k')$, we set $f\sim f'$ iff there exists a common field extension $i:k\to K$, $i':k'\to K$ such that $X(i)(f)=X(i')(f')$. The intuition is that $X(k)$ is the set of solutions of some kind of system of polynomial equations in the field $k$, and we shouldn't distinguish, say, the solution $3$ as a rational number from the solution $3$ as a real or complex number (but we should distinguish it from the solution we would usually denote by $3$ in the field $\mathbb F_7$, for example).

Damiano Mazza (Jan 21 2024 at 16:59):

Now, if $X=\mathbf{CRing}(R,-)$, i.e., if it is the functor of points of $\mathrm{Spec} R$ for some commutative ring $R$, then $f\in X(k)$ means that $f$ is a homomorphism $R\to k$, and it turns out that

Lemma. $f\sim f'$ iff $\ker f=\ker f'.$

It is immediate to check that $\ker f$ is a prime ideal of $R$ because the target is a field. It is a bit less immediate (at least for me) but it is also true that any prime ideal of $R$ is of the form $\ker f$ for some $f:R\to k$ with target a field. So we get exactly the points of the underlying topological space of $X$ qua locally ringed space. Cool!

Damiano Mazza (Jan 21 2024 at 17:04):

My problem, in all this, is that I can't prove the lemma. The left-to-right direction is immediate. The converse escapes my (very limited) commutative algebra skills. Here is how I tried to go about it.

Damiano Mazza (Jan 21 2024 at 17:07):

We have a commutative ring $R$ and two homomorphisms $f:R\to k$, $f':R\to k'$ with $k,k'$ fields such that $\ker f=\ker f'$. We need to find a field $K$ which is a common extension of $k$ and $k'$ and which makes $f$ and $f'$ equal by postcomposition. An obvious possibility is to compute the pushout $k\otimes_R k'$ in $\mathbf{CRing}$, which is a commutative ring, then we take an arbitrary maximal ideal $\mathfrak m$ in it, and $k\otimes_R k'/\mathfrak m$ is the field we're looking for.

Damiano Mazza (Jan 21 2024 at 17:07):

There's problem though: we never used the assumption $\ker f=\ker f'$! But there's a catch: by taking a maximal ideal, we implicitly supposed $k\otimes_R k'$ to be non-zero. So, I would like to prove that $\ker f=\ker f'$ implies $k\otimes_R k'\neq 0$, but this is where I get stuck. Does anyone know how to do this?

Damiano Mazza (Jan 21 2024 at 17:08):

Another, "bonus" question: does the above recipe for reconstructing the points of the underlying topological space work for any scheme $X$, or just for the affine ones?

Todd Trimble (Jan 21 2024 at 18:03):

This may be killing a gnat with a bazooka, but one idea is that first of all $R/\mathfrak{p} \otimes_R R/\mathfrak{p} \cong R/\mathfrak{p} \otimes_{R/\mathfrak{p}} R/\mathfrak{p} \cong R/\mathfrak{p}$, and also that the field of fractions $K$ of $R/\mathfrak{p}$ is a flat module over $R/\mathfrak{p}$ (because it is a filtered colimit of free modules $\frac{1}{x} R/\mathfrak{p}$ where $x$ ranges over nonzero elements. So when you apply $K \otimes_{R/\mathfrak{p}} -$ to the inclusion $R/\mathfrak{p} \hookrightarrow K'$, you again get a monomorphism (exploiting flatness).

(This may be harder than necessary, but I think something along these lines will work.)

John Baez (Jan 21 2024 at 18:32):

I obviously need to up my commutative algebra game since I wouldn't have been able to think of Todd's argument, nor do I see a more elementary proof. I feel there should be something more elementary.

Damiano Mazza said:

Another, "bonus" question: does the above recipe for reconstructing the points of the underlying topological space work for any scheme $X$, or just for the affine ones?

Lemma 26.13.3. of the Stacks Project may help here:

Let X be a scheme. Points of X correspond bijectively to equivalence classes of morphisms from spectra of fields into X.

and also the surrounding material.

Mike Shulman (Jan 21 2024 at 19:28):

It looks to me like it's the right-to-left direction that you're trying to prove. The way I would think of approaching it is to take the quotient $R/K$ where $K=\ker f = \ker f'$. Since $K$ Is a prime ideal, $R/K$ is an integral domain that maps injectively to $k$ and $k'$; thus its field of fractions ought also to inject into both of them. Now I feel like that should mean that $k$ and $k'$ should have a "union" over this field of fractions, but my field theory isn't strong enough to be positive.

Mike Shulman (Jan 21 2024 at 19:29):

Hmm, possibly the existence of such unions is already implicit in the transitivity of $\sim$.

Todd Trimble (Jan 21 2024 at 19:55):

Oh yes, Mike's right, my suggestion didn't say anything about a common field extension. I was really just focused on proving Damiano's conjecture that $k \otimes_R k' \neq 0$.

John Baez (Jan 21 2024 at 20:02):

I feel that this business of trying to find the smallest field that contains $k$ and $k'$ is connected to what people call the "composite" or "compositum" of fields:

• Wikipedia, Composite of fields.

John Baez (Jan 21 2024 at 20:03):

but in the study of "composites" they tend to assume $k$ and $k'$ are both subfields of some larger field, e.g. an algebraic closure of $F$ if both $k$ and $k'$ are algebraic extensions of $F$.

John Baez (Jan 21 2024 at 20:03):

I suspect that this larger subfield is a kind of crutch.

Todd Trimble (Jan 21 2024 at 20:20):

Oh, I see. So assuming $k \otimes_R k' \neq 0$ (which is what I had focused on), which is the coproduct of $k, k'$ in the category of commutative $R$-algebras, take a maximal ideal in this commutative ring, say $\mathfrak{m}$. Then
$(k \otimes_R k')/\mathfrak{m}$ is a field. But then we have a composition of maps

$k \to k \otimes_R k' \to (k \otimes_R k')/\mathfrak{m}$

where the first map is a coproduct coprojection. Since this is a map between fields, it is injective. So $k$, and $k'$ by the same argument, both inject into a common field, and we are done.

Damiano Mazza (Jan 21 2024 at 22:00):

Thanks to everyone for your answers/suggestions! @Mike Shulman, I did swap left and right, sorry! :blush: (I edited it now). I think that your idea is related to @Todd Trimble's argument. Mixing the two, here is a possible proof.

Damiano Mazza (Jan 21 2024 at 22:00):

Like I said in my original post (and as observed by Todd in his last comment), given $f:R\to k$ and $f':R\to k'$ with $k,k'$ fields, to prove that $\ker f=\ker f'$ implies $f\sim f'$ it is enough to find a non-zero ring $S$ and (necessarily injective) homomorphisms $g:k\to S$ and $g':k'\to S$ such that $gf=g'f'$, because then any $S/\mathfrak m$, where $\mathfrak m$ is a maximal ideal of $S$, is a field with the property we're looking for. My idea was to take $S:=k\otimes_R k'$, and Todd's argument (if I understand it correctly) shows that it works, but there's a simpler argument, not invoking flatness.

Damiano Mazza (Jan 21 2024 at 22:01):

We start by letting $\mathfrak p:=\ker f=\ker f'$, which is a prime ideal of $R$. As both Todd and Mike suggested, both $f$ and $f'$ factor through $R/\mathfrak p$, which is an integral domain, via monos $m$ and $m'$ into $k$ and $k'$, respectively. Let $K:=\mathrm{Frac}(R/\mathfrak p)$ be the field of fractions of $R/\mathfrak p$. By its universal property, we have that $m$ and $m'$ factor through $K$ via monos $K\to k$ and $K\to k'$.

Damiano Mazza (Jan 21 2024 at 22:01):

Now, these exhibit $k$ and $k'$ as $K$-algebras. In particular, they are non-zero $K$-vector spaces, and their tensor product $S:=k\otimes_K k'$ as $K$-algebras is the same as their tensor product as $K$-vector spaces. In particular, it is non-zero, so we found our $S$!

Damiano Mazza (Jan 21 2024 at 22:04):

Here's the proof summarized in a quiver commutative diagram.

Damiano Mazza (Jan 21 2024 at 22:11):

@John Baez, for the general question, I'll see what the Stacks Project has to say! (It's probably going to take me a while...).

Damiano Mazza (Jan 21 2024 at 22:20):

I guess the lesson I learned here is that "every span in the category of fields may be closed", in the sense that two extensions $k,k''$ of a field $K$ always have a common extension $K_0$. There's nothing canonical about $K_0$ because it depends on a choice of a maximal ideal of the algebra $k\otimes_K k'$, but it exists. This is the "union" Mike was hinting at. The comments in this Math Stack Exchange question seem to say that it is related to the notion of compositum of fields mentioned by John.

Damiano Mazza (Jan 21 2024 at 22:22):

The tl;dr version of the proof then is: since $\ker f=\ker f'=:\mathfrak p$, both $f:R\to k$ and $f':R\to k'$ factor through a common field, namely the field of fractions of $R/\mathfrak p$. Then we conclude by applying the above observation.