Arrow

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$Mathematics$ Mathematics

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Arrow

Oct 21, 2023 21:29

@LukasHeger actually, I am not sure about the formula $\operatorname{lcm}(f,g)\gcd(f,g)=\alpha fg$ when $A=\Bbbk [x_1,\dots ,x_n]$, so I'm not convinced by the idea I sketched before.

Arrow

Oct 21, 2023 20:19

@LukasHeger I don't follow. 1) Is my argument above alright for the UFD case? 2) What is your intuition for suspecting the assertion holds for domains?

Arrow

Oct 21, 2023 20:16

Ah, ok. I think I got it. If $uf+vg=0$ then there's a low-degree common multiple. By the remarks above this will contradict $\operatorname{lcm}(f,g)\gcd(f,g)=\alpha fg$ by taking degrees...

Arrow

Oct 21, 2023 20:14

Oh, great! Why?

Arrow

Oct 21, 2023 20:12

I'll try to think about it, but I'm very slow. Just for context, here's my motivation:
I know that polynomials with a common factor have zero resultant. I am wondering whether the converse holds for polynomials over $A$ when $A$ itself a multivariate polynomial ring over a field.

Arrow

Oct 21, 2023 20:08

@TedShifrin I don't see a reason for them to be associates in general

Arrow

Oct 21, 2023 20:06

@LukasHeger I agree. I'm trying to reverse engineer some conditions for when a minimal degree element of $\langle f,g\rangle$ will be the gcd. Am I wasting my time?

Arrow

Oct 21, 2023 20:03

Ah, actually the problem disappears again if I restrict $A$ to be a mulivariate polynomial ring, since rank is defined over UFDs or something like that...

Arrow

Oct 21, 2023 20:02

Ah, right! Great. I don't have motivation to generalize that far for now :)
Now a bit of a different question. I'm thinking of the extended Euclidean algorithm for finding a gcd of $f,g$ in terms of row-reducing their Sylvester matrix. For example I computed the polynomials $x^2+7,x^3+3x+6\in \mathbb Z[x]$ admit $74$ as a gcd with respective coefficients $−2x^2−3x+8,2x+3$.

Morally, I'd like to say the degree of the gcd should equal "the" rank of the Sylvester matrix. But I recall the rank of matrices over domains to be undefined (though I don't know why). Does it make sense to claim the de…

Arrow

Oct 21, 2023 19:57

@LukasHeger Actually, I'm over generalizing. I'm just interested in the case of multivariate polynomials over a field

Arrow

Oct 21, 2023 19:55

Well played :) I wonder if an equally sneaky trick works if I assume f,g to be monic?

Arrow

Oct 21, 2023 19:52

I think your ring is not a polynomial ring in the sense that it's not isomorphic to $A[x]$ for some commutative ring $A$.

Arrow

Oct 21, 2023 19:51

@LukasHeger I'm okay with assuming $A$ is a domain, but I'm not sure that resolves the problem

Arrow

Oct 21, 2023 19:49

@LukasHeger I can't think of one :\

Arrow

Oct 21, 2023 19:48

Let $A$ be a commutative ring. Suppose $f,g\in A[x]$ admit a gcd. Is $fg/\gcd(f,g)$ an lcm of $f,g$?

Arrow

Mar 5, 2023 21:03

@Thorgott ah, jolly good. The iso "straightens the parabola" and translates between the projection from the parabola to the y-axis that @TedShifrin mentioned and the 2-cover given by the squaring map.

Arrow

Mar 5, 2023 20:56

If $A[x]\overset{x\mapsto x^2}{\to}A[x]$ is the branched 2-cover, what's the inclusion $A[x^2]\hookrightarrow A[x]$?

Arrow

Mar 5, 2023 20:52

@Thorgott the target should be $A[x]$ instead of $A[x^2]$

Arrow

Mar 5, 2023 20:47

Agreed, my name was poorly chosen.

Arrow

Mar 5, 2023 20:45

I don't understand why the latter map is not a ring morphism - I am defining the image of the generator $x$ as the polynomial $x^2$.

Arrow

Mar 5, 2023 20:43

Noob question: what is the geometric interpretation of the inclusion $\Bbbk[x^2]\hookrightarrow \Bbbk[x]$? Is it the branched cover $z\mapsto z^2$? If so, what is the geometric interpretation of the squaring map $A[x]\overset{x\mapsto x^2}{\to}A[x^2]$?

Arrow

Sep 26, 2021 22:29

@Thorgott jolly good. Thanks!

Arrow

Sep 26, 2021 21:17

@Thorgott since this monoid need not itself be saturated, perhaps most natural is to ask the question for the saturation $(1+I)_\mathrm{sat}$. That is, when are $I,(1+I)_\mathrm{sat}$ set-theoretic complements?

Arrow

Sep 26, 2021 21:10

@Thorgott, I see. I should have asked my question for the "more saturated" multiplicative set $A^\times +I$.

Arrow

Sep 26, 2021 18:23

Given an ideal $I\vartriangleleft A$ of a commutative ring, when is $1+I$ its set theoretic complement in $A$?

Arrow

Sep 24, 2021 16:21

How should one think of the localization of $A[x]$ at $1+\langle x\rangle$?

Arrow

May 6, 2021 08:37

why does reducing to the identity mod a square-zero ideal imply being an isomorphism?

Arrow

May 6, 2021 08:36

Arrow

Jun 13, 2020 12:42

@Thorgott thanks

Arrow

Jun 13, 2020 10:40

@Thorgott why?

Arrow

Jun 13, 2020 10:01

Hi. Could anybody answer this question I have about an answer about splitting fields? Thanks!

Homotopy Theory

A room for anyone interested in homotopy theory, or any nearby...

Arrow

Jan 30, 2021 08:48

Hi, does anyone know where Grothendieck first defines connections as de Rham descent data?

$Mathematics$ Discussion between Roland and Arrow

Imported from a comment discussion on math.stackexchange.com/q...

Arrow

Apr 27, 2020 10:26

Have a great week!

Arrow

Apr 27, 2020 10:26

Dear @Roland, forgive my late reply, I only remembered to check this room now! Again, many thanks for your answer. I will have many more questions about these things, so I hope it's okay to ask here.

Arrow

Apr 21, 2020 15:41

(Regarding my (2) above, I understand a cocycle need not be a circle, but the winding number's finite so we can work with finitely many circles.)

Arrow

Apr 21, 2020 14:45

Ah, I see I have missed you giving an alternative to the pathological Hawaiian earring. You write to consider the pushforward along the inclusion into the plane. The inclusion of what? I'm a bit confused. Again, infinite thanks for your kindness!

Arrow

Apr 21, 2020 14:33

I guess the last comment is correct, and that the pushforward along $\mathbb C^\times \subset \mathbb C$ is still locally acyclic.

Arrow

Apr 21, 2020 14:33

Regarding your comment that $H^1(\mathbb{C},j_*\mathbb{Z})=0$, I confess I haven't really learned to calculate cohomology yet - I just stumbled on this comment which appears to suggests the cohomology is non-zero.

Arrow

Apr 21, 2020 14:22

Dear @Roland, I think I understand the algebra of your example, but I'm still worried about my geometric intuition. I'm trying to understand just the intuitive difference between being locally acyclic vs cohomology presheaves sheafifying to zero. Do the following interpretations seem reasonable to you?

1. Being locally acyclic means that a sheaf $F$ admits a cover on which its restrictions have no cocycles.
2. Zero sheafification means that given any cocycle of a sheaf $F$ admits an open cover on which its restrictions are zero.

Arrow

Apr 18, 2020 14:31

Exactly what you wrote.

Arrow

Apr 18, 2020 14:19

(I might take some time to reply)

Arrow

Apr 18, 2020 14:19

Right now I don't quite understand the distinction between being locally acyclic vs having vanishing sheafification for the cohomology presheaves.

Arrow

Apr 18, 2020 14:19

Right now I can't quite talk (I'm with people), but if it's okay, I'll use this chat room to ask you things.

Arrow

Apr 18, 2020 14:17

Hello @Roland! Honestly, I cannot thank you enough!

Arrow

Apr 18, 2020 14:14

@Roland I think I don't understand the gap between being locally acyclic vs the cohomology presheaves having zero sheafification. The later means any cocycle is locally zero, and the former means there are locally no cocycles... (P.S thanks for your patience. I always learn a lot from you.)

Arrow

Apr 18, 2020 14:14

@Roland but when you push along $\mathbb C^\times \subset \mathbb C$ you have the origin too. How can the generating cycle for the pushforward be trivialized on any open neighborhood of the origin?

Arrow

Apr 18, 2020 14:14

Dear @Roland, that is very instructive, thank you. The derived direct image $\mathrm Rf_\ast$ takes a sheaf $F$ to the sheafification of the presheaf $V\mapsto \mathrm H^\bullet(f^{-1}V,F)$. Proper base change identifies its stalks for nice $f$, but I'm not sure how to think of its sections in general. The sections of a sheafification are equivalence classes of compatible families that coincide eventually, so it seems $\mathrm Rf_\ast F(V)$ consists of equivalence classes of compatible families of cocycles locally on $f^{-1}V$?

Arrow

Apr 18, 2020 14:14

Dear @Roland actually I'm even struggling with the geometric content of the vanishing sheafification of the cohomology presheaves: if $U\mapsto \mathrm H^\bullet (U,F)$ has zero sheafification then every cocycle admits an open cover on which it restricts to zero, right? So every cocycle is locally zero. On the other hand, consider the pushforward of the constant sheaf $\underline{\mathbb Z}$ along $\mathbb C^\times \subset \mathbb C$: its first cohomology seems to also be $\mathbb Z$, but I don't see what open cover trivializes the cocycle. What am I missing?

Arrow

Apr 18, 2020 14:14

(I see formally why the sheafification is zero - it boils down to homology of an injective resolution whose higher cohomologies vanish. My problem is with the geometry and with the case where $f$ is a fiber bundle or submersion.)

Arrow

Apr 18, 2020 14:14

Dear @Roland, I'm struggling with geometric intuition here. (My question is motivated by trying to understand the sheafification of this presheaf, which is the derived pushforward.) If the sheafification is zero (in particular noninjective) then compatible cocycles can glue to distinct cocycles. What are some simple pictures illustrating this? (You say the higher cohomologies are never sheaves.)