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: deprecated: mathematics

Topic: etale maps of rings

John Baez (Aug 26 2023 at 20:58):

I'm trying to understand etale maps of (commutative) rings in various elementary ways, preferably with a minimum of stuff about schemes (other than affine schemes). The intuition is that an etale map of commutative rings

$f: A \to B$

is supposed to be one that gives rise to a covering

$\mathrm{Spec}(f) : \mathrm{Spec}(B) \to \mathrm{Spec}(A)$

with finitely many sheets. So there should be two conditions in the definition of "etale", one that captures the "covering" idea and the other being a finiteness condition.

John Baez (Aug 26 2023 at 21:01):

The answers to this Math Stackexchange question are a bit frustrating, but after reading them multiple times I can extract some useful information:

• Definition of etale for rings.

John Baez (Aug 26 2023 at 21:04):

The first answer uses a somewhat nonstandard definition, as the comments below it reveal.

John Baez (Aug 26 2023 at 21:09):

With that wrinkle removed, it seems to say a morphism of rings $f: A \to B$ is etale if:

A) it is formally etale: if $R' \to R$ is a square-zero extension (meaning its kernel $I$ has $I^2 = 0$) then $\mathrm{hom}_A(R,B) \to \mathrm{hom}_A(R',B)$ is a bijection.

B) $B$ is finitely presented as an $A$-algebra.

John Baez (Aug 26 2023 at 21:13):

The second condition is the finiteness condition and that's just fine.

John Baez (Aug 26 2023 at 21:16):

The first is the more important one. Saying that $R' \to R$ is square-zero extension should be a way of saying that the map $\mathrm{Spec}(R) \to \mathrm{Spec}(R')$ is a "first-order infinitesimal thickening".

John Baez (Aug 26 2023 at 21:17):

The easiest example is $k[x]/\langle x^2 \rangle \to k$ where $x$ is sent to zero.

John Baez (Aug 26 2023 at 21:18):

Here $\mathrm{Spec}(k)$ is a point, and $\mathrm{Spec}(k[x]/\langle x^2 \rangle)$ is a "point with first-order infinitesimal fuzz", or "infinitesimal arrow" for short.

John Baez (Aug 26 2023 at 21:21):

So, the "formally etale" condition on $A \to B$ seems to imply that any way of lifting a point along $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$ extends uniquely to a way of lifting a first-order infinitesimal arrow.

John Baez (Aug 26 2023 at 21:23):

This should be a way of saying that the map $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$ gives a bijective map from each tangent space of $\mathrm{Spec}(B)$ to the corresponding tangent space of $\mathrm{Spec}(A)$.

John Baez (Aug 26 2023 at 21:24):

But by letting $R \to R'$ be an arbitrary square-free extension it seems we are generalizing this considerably.

John Baez (Aug 26 2023 at 21:28):

The Stack Exchange answer says that $A \to B$ is formally étale (resp. formally smooth, formally unramified) if for every square-zero extension of $A$-algebras $R' \to R$ the natural map $\mathrm{Hom}_A(B, R') \to \mathrm{Hom}_{A}(B, R)$ is bijective (resp. surjective, injective), and

Differential-geometrically, unramifiedness, smoothness and étaleness correspond to the tangent map of $\mathrm{Spec} (f)$ being injective, surjective and bijective, respectively. In particular, étale is the generalization to the algebraic case of the concept of local isomorphism.

John Baez (Aug 26 2023 at 21:37):

Another answer says that a morphism of rings $f: A \to B$ is etale iff:

1) The module of Kaehler differentials $\Omega^1_{B/A}$ vanishes.

2) $B$ is flat as an $A$-module.

3) $B$ is finitely presented as an $A$-algebra.

John Baez (Aug 26 2023 at 21:46):

I don't see why conditions 1) and 2) here are equivalent to condition A) in the other answer (if that's how it works).

John Baez (Aug 26 2023 at 21:51):

I do know some stuff relating Kaehler differentials to square-zero stuff, but it would be really nice if

1) is equivalent to $\mathrm{Hom}_A(B, R') \to \mathrm{Hom}_{A}(B, R)$ being injective (i.e. $f: A \to B$ being formally unramified)

and

2) is equivalent to $\mathrm{Hom}_A(B, R') \to \mathrm{Hom}_{A}(B, R)$ being surjective (i.e. $f: A\to B$ being formally smooth).

John Baez (Aug 26 2023 at 21:52):

Aha - Prop. 3.5 in Milne's Etale Cohomology seems to say that if $B$ is finitely presented as an $A$-algebra, then 1) is indeed equivalent to $\mathrm{Hom}_A(B, R') \to \mathrm{Hom}_{A}(B, R)$ being injective (i.e. $f: A \to B$ being formally unramified). :tada:

John Baez (Aug 26 2023 at 21:59):

So the question is about 2).

John Baez (Aug 26 2023 at 22:00):

Here's the connection between Kaehler differentials and square-zero stuff. If we have a ring homomorphism $f: A \to B$, here is one way to define the module of relative Kaehler differentials $\Omega^1_{B/A}$.

John Baez (Aug 26 2023 at 22:01):

We use $f$ to think of $B$ as an $A$-module and let $I$ be the kernel of the multiplication map

$B \otimes_A B \to B$

John Baez (Aug 26 2023 at 22:02):

Then we let

$\Omega^1_{A/B} = I/I^2$.

David Michael Roberts (Aug 27 2023 at 05:39):

John Baez said:

So, the "formally etale" condition on $A \to B$ seems to imply that any way of lifting a point along $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$ extends uniquely to a way of lifting a first-order infinitesimal arrow.

A more formal way if saying this might be via some kind "valuative criterion", or something analogous to one. Perhaps using a Weil algebra, as in the way people build models of synthetic differential geometry

John Baez (Aug 27 2023 at 06:40):

I don't know what a "valuative criterion" is, but on good days I remember what a Weil algebra is. If I remember correctly, the infinitesimal arrow $k[x]/\langle x^2 \rangle$ is a nice example of a Weil algebra over the (commutative) ring $k$. But it's also a square-zero extension of $k$ since its an extension of rings whose kernel squares to zero: $x^2 = 0$. The condition for a map to be formally etale involves an arbitrary square-zero extension of an arbitrary ring. So you're making me wonder: what's the difference between a Weil algebra over a ring and a square-zero extension of a commutative ring?

One huge difference seems to be that the Weil algebra is geometrically over the ring, i.e. equipped with a ring homomorphism from it, while the square-zero extension is geometrically 'under' that ring, i.e. equipped with a ring homomorphism to it.

I'm pretty confused since $k[x]/\langle x^2 \rangle$ is both.

John Baez (Aug 27 2023 at 06:43):

Oh, I guess this happens for any Weil algebra over any ring: we get maps both ways.

Hypatia du Bois-Marie (Aug 27 2023 at 06:49):

John Baez said:

I don't know what a "valuative criterion" is

https://stacks.math.columbia.edu/tag/0BX4 it's basically formalizing exactly what you said above (e.g. the generic and specific point of the DVR, or the decapitated tangent vector, or (?) the Weil algebra).

There are other "elementary" ways of thinking of étale-ness too, e.g. with arithmetics (Hilbert: decomposing ideals, Artin: extending valuations), the points in that topos (the "henselian traits"), Jacobian criterion and inverse "function" theorem

John Baez (Aug 27 2023 at 06:53):

Thanks! I'm scared that if I try to learn all those things I'll never finish. :upside_down: I'm really trying to understand etaleness in terms of separability and Kaehler differentials. So what I really want to know is a couple of things like:

Given a ring homomorphism $A \to B$, is $B$ being flat as an $A$-module equivalent to $\mathrm{Hom}_A(B, R') \to \mathrm{Hom}_{A}(B, R)$ being surjective (i.e. $f: A\to B$ being formally smooth)?

This is the main question buried in my earlier text.

John Baez (Aug 27 2023 at 07:00):

I was trying to see why this definition of $A \to B$ being etale:

A) it is formally etale: if $R' \to R$ is a square-zero extension (meaning its kernel $I$ has $I^2 = 0$) then $\mathrm{hom}_A(R,B) \to \mathrm{hom}_A(R',B)$ is a bijection.

B) $B$ is finitely presented as an $A$-algebra.

is equivalent to this one:

1) The module of Kaehler differentials $\Omega^1_{B/A}$ vanishes.

2) $B$ is flat as an $A$-module.

3) $B$ is finitely presented as an $A$-algebra.

3) is the same as B), and given either of these Milne says 1) is equivalent to half of A): that is, 1) is equivalent to $\mathrm{hom}_A(R,B) \to \mathrm{hom}_A(R',B)$ being a injection.

So we'd be done if we could show that 2) is equivalent to $\mathrm{hom}_A(R,B) \to \mathrm{hom}_A(R',B)$ being an injection, at least if the other conditions hold.

Hypatia du Bois-Marie (Aug 27 2023 at 07:14):

John Baez said:

Given a ring homomorphism $A \to B$, is $B$ being flat as an $A$-module equivalent to $\mathrm{Hom}_A(B, R') \to \mathrm{Hom}_{A}(B, R)$ being surjective (i.e. $f: A\to B$ being formally smooth)?.

Maybe by manipulating flatness (def: tensoring with a flat module is an exact functor) with the adjunction with hom?

John Baez (Aug 27 2023 at 07:16):

I should try - thanks.

It's interesting how you managed to change all the double dollar signs in my text to single dollar signs while quoting it, thus making the LaTeX no longer work. That would seem to require effort.

Hypatia du Bois-Marie (Aug 27 2023 at 07:20):

John Baez said:

I should try - thanks.

It's interesting how you managed to change all the double dollar signs in my text to single dollar signs while quoting it, thus making the LaTeX no longer work. That would seem to require effort.

Probably zulip android has a bug, I didn't change anything (x

John Baez (Aug 27 2023 at 07:24):

I figured something like that must be happening - I can't imagine you'd go to so much work to make sure none of my math symbols work. :upside_down:

Simon Burton (Aug 27 2023 at 09:55):

John Baez said:

The easiest example is $k[x]/\langle x \rangle \to k$ where $x$ is sent to zero.

The infinitesimal arrow is $k[x]/\langle x^2\rangle$ , yes?

David Michael Roberts (Aug 27 2023 at 10:10):

@Simon Burton yes, I believe so.

@John Baez
I was thinking something along the lines as being explored in section 2 of these notes: https://websites.umich.edu/~viktorb/etale_summer2020/Notes/etalemaps.pdf Is it enough to ask for specific square-zero extensions, i.e. the sort you get from Weil algebras (I'm kinda guessing here). The definition of formally etale looks to me like a unique lifting property filling some kind of square, no?

David Michael Roberts (Aug 27 2023 at 10:21):

One think that is true is that a map of rings is etale precisely when it is "standard etale" "at every prime". In scheme language, one can characterise etale maps of affince schemes if for a point in the domain and its image in the codomain, there is are affine neighbourhoods of both of them so that the maps restricts to a map of these neighbourhoods, and this map is "standard affine" https://stacks.math.columbia.edu/tag/02GI (see also https://stacks.math.columbia.edu/tag/0G1A)

Thinking about standard affine maps should be helpful in gaining intuition about how the algebra is supposed to reflect the topology, since they are built up in some very specific series of steps. And maybe one needs to consider specific special cases to built up intuition for even these! Say working over a field, to start, the second Stacks Project link above mentions generalising the example of a separable finite field extension.

John Baez (Aug 27 2023 at 10:21):

Simon Burton said:

The infinitesimal arrow is $k[x]/\langle x^2\rangle$ , yes?

Yeah - that typo was all over, and I'll fix it. $k[x]/\langle x \rangle$ is just $k$! :rolling_on_the_floor_laughing:

John Baez (Aug 27 2023 at 10:28):

David Michael Roberts said:

Simon Burton yes, I believe so.

John Baez
I was thinking something along the lines as being explored in section 2 of these notes: https://websites.umich.edu/~viktorb/etale_summer2020/Notes/etalemaps.pdf Is it enough to ask for specific square-zero extensions, i.e. the sort you get from Weil algebras (I'm kinda guessing here).

That would be cool - I don't know.

Note that the general square-zero extensions allow us to talk about lifting not just for points with first-order infinitesimal thickenings but arbitrary affine schemes with first-order infinitesimal thickenings. So I can map a big fat variety into the base, lift it to the total space, extend the map to the base to a first-order thickening, and etaleness will assure me that thickening has a unique lifting to the total space! It seems hard to achieve this level of generality by demanding a lifting property just for a limited supply of Weil algebras.

John Baez (Aug 27 2023 at 10:29):

But maybe you can do a reduction of this sort:

David Michael Roberts said:

One think that is true is that a map of rings is etale precisely when it is "standard etale" "at every prime". In scheme language, one can characterise etale maps of affince schemes if for a point in the domain and its image in the codomain, there is are affine neighbourhoods of both of them so that the maps restricts to a map of these neighbourhoods, and this map is "standard affine" https://stacks.math.columbia.edu/tag/02GI (see also https://stacks.math.columbia.edu/tag/0G1A)

John Baez (Aug 27 2023 at 10:33):

I guess one question is what sort of "point", or "prime", we're talking about here! If a "point" is any prime ideal, the points can be quite big and spread out. Anyway, I'll have to read about this stuff.

John Baez (Aug 27 2023 at 10:33):

David Michael Roberts said:

The definition of formally etale looks to me like a unique lifting property filling some kind of square, no?

Yes, exactly. The only reason I didn't say it that way is that it's so annoying to draw diagrams here.

John Baez (Aug 27 2023 at 10:51):

It's the usual sort of thing you expect from a covering space, like a unique path lifting property:

You've got your would-be etale map of schemes $p: E \to B$. We say $p$ is formally etale if whenever I choose any map $i : A \to X$ coming from a square-zero extension, and any maps $f: A \to E$ and $g: X \to B$ making the resulting square commute, there always exists a unique $\tilde{g} : X \to E$ making the whole resulting diagram commute.

John Baez (Aug 27 2023 at 10:53):

If $\tilde g$ always exists we say $p$ is formally smooth. If $\tilde g$ is always unique we say $p$ is formally unramified. I'll admit I don't have a good geometrical sense for why these two concepts deserve to be called smooth and unramified.

John Baez (Aug 27 2023 at 10:54):

I should look at a map of schemes that has what I'd call ramification, i.e. branch points.

Tim Hosgood (Aug 27 2023 at 23:32):

if i'm not mistaken, one thing you might find useful in searching the literature is that your condition (A) is often expressed in terms of the functor of points (in this Zulip I guess people would say "Yoneda" instead)

Tim Hosgood (Aug 27 2023 at 23:35):

I don't have an immediate answer, but my first reference to check would be the CRing project: http://math.uchicago.edu/~amathew/chetale.pdf

Tim Hosgood (Aug 27 2023 at 23:44):

but I'm a bit lost with all the conversation above, so let me try distilling your question down: are you asking if "$B$ is flat and finite over $A$ implies that $A\to B$ is formally smooth"?

Tim Hosgood (Aug 27 2023 at 23:47):

if so, have a look at the proof of Theorem 3.9 in the CRing link above (which refers to Theorem 2.16 for the main argument)

Tim Hosgood (Aug 28 2023 at 00:16):

I just had a leaf through EGA IV, because if a statement about étale things is true and classical enough then it's probably there, and is sort of is, but not in a way that I think will be particularly enlightening. Here's a rough idea of what happens there (and consequently probably, I guess, how it's argued in other classical references):

1. étale = smooth + fibres are of dimension 0 (or = smooth + quasi-finite, from a more SGA point of view)
2. smooth "=" formally smooth + locally of finite presentation
3. formally smooth = flat + locally of finite presentation + $\Omega_{X/Y}$ is locally free of rank equal to the dimensions of the fibres (and thus 0 in the case of étale)

The proof of point (3), which is the one that you care about, is EGA IV.4 Proposition 17.15.15, but the proof of this is... arduous

Tim Hosgood (Aug 28 2023 at 00:18):

But hey, the good news is that you can ignore all of the above and just look at some notes by Bhargav Bhatt: the theorem you're looking for is Theorem 4.9 in these notes.

Edit: these notes are nice and concise and don't require going down a rabbit hole of proofs citing other proofs, but it just feels wrong to not give the "correct" reference, which is what these notes turned into: Tag 025K in the Stacks project. They also tell us that the fact that these two definitions are indeed equivalent is known as the functorial characterisation of being étale (this is why I said "functor of points" above).

John Baez (Aug 28 2023 at 07:24):

Tim Hosgood said:

but I'm a bit lost with all the conversation above, so let me try distilling your question down: are you asking if "$B$ is flat and finite over $A$ implies that $A\to B$ is formally smooth"?

I was explaining a bunch of stuff I've just been learning, hence the bloated nature of my remarks, but in the process I bumped into that question. Actually I was asking "If $B$ is finite over $A$, is it true that $B$ being flat is equivalent to $A \to B$ being formally smooth?" But an implication either way is great, so thanks a million!

John Baez (Aug 28 2023 at 07:25):

Even more actually...

John Baez said:

I was trying to see why this definition of $A \to B$ being etale:

A) it is formally etale: if $R' \to R$ is a square-zero extension (meaning its kernel $I$ has $I^2 = 0$) then $\mathrm{hom}_A(R,B) \to \mathrm{hom}_A(R',B)$ is a bijection.

B) $B$ is finitely presented as an $A$-algebra.

is equivalent to this one:

1) The module of Kaehler differentials $\Omega^1_{B/A}$ vanishes.

2) $B$ is flat as an $A$-module.

3) $B$ is finitely presented as an $A$-algebra.

3) is the same as B), and given either of these Milne says 1) is equivalent to half of A): that is, 1) is equivalent to $\mathrm{hom}_A(R,B) \to \mathrm{hom}_A(R',B)$ being a injection.

So we'd be done if we could show that 2) is equivalent to $\mathrm{hom}_A(R,B) \to \mathrm{hom}_A(R',B)$ being an injection, at least if the other conditions hold.

Tim Hosgood (Aug 28 2023 at 18:12):

ok, then I think the Theorem 4.9 that I cite above actually does tell you (with some work in converting your condition (A) to the condition there, which looks slightly different but is, I think, equivalent) that these two things are equivalent!

John Baez (Aug 28 2023 at 22:39):

GREAT! I will check it out. It may take a while, but I seem to be slowly getting pulled into trying to understand etale maps from various points of view.