Mathematics - 2017-12-27 (page 1 of 5)

GFauxPas

00:00

resistance is futile, you will be taken mod your commutator

Antonios-Alexandros Robotis

2

Q: A question about the assassinator (={associated primes}) and the support of a module.

This question is motivated by a proof in Bruns' and Herzog's book on "Cohen Macaulay Rings". Let $$R,\mathfrak{m}$$ be a Noetherian local ring, $M \neq 0$ a finitely generated $R$-module. Suppose further that $\mathfrak{p} \in Ass(M)$ and that $x \in \mathfrak{m}$ and also that $x$ is a nonzer...

ac.commutative-algebra

assassinator

Daminark

I often refer to terms as "this guy"

So sometimes I say "So this guy dies"

Antonios-Alexandros Robotis

Yeah, a few of my professors this semester referred to various objects as "gadgets" which I found funny.

Daminark

One of mine referred to objects of a category as "widgets"

anon

also "technology," "machine for turning X into Y"

there are a couple MO questions about what the word "yoga" means

Semiclassical

00:11

Physicists like to use creation and annihilation operators

Leaky Nun

You know that: $\alpha_1 v_1+\dots \alpha_n v_n=0\RIghtarrow \alpha_1=\dots\alpha_n=0$. Now, you want to prove that: if $\beta_1(v_1+v_2)+\dots\beta_n(v_n+v_1)=0$ then $\beta_1=\dots \beta_n=0$. If this is not possible, then you will can to write somw element of $C$ as linear combination of the others. Try to play with that. — yemino 46 secs ago

hmm

RIP english

> you will can to write

Semiclassical

(Aka ladder operators aka raising/lowering operators. But that’s not nearly so colorful)

Akiva Weinberger

00:26

@LeakyNun You want to prove that if $\{v_i\}$ is linearly independent then $\{v_i+v_{i+1}\}$ is?

Leaky Nun

@AkivaWeinberger no, I'm just quoting an instance of RIP english

Akiva Weinberger

That's not actually true

Leaky Nun

@AkivaWeinberger it depends on the parity of the dimension

Akiva Weinberger

I mean I think it's true if $n$ is odd but not otherwise

Leaky Nun

exactly

Akiva Weinberger

00:27

I said that without seeing what you said

Leaky Nun

great minds think alike

@AkivaWeinberger AKA "I am sniped"

Akiva Weinberger

/snʌɪpt/

Leaky Nun

btw feliz navidad

Akiva Weinberger

You too

Mike Miller

01:18

@Daminark The only character I have ever played as is Lil Mac

mac tonite

Daminark

01:52

@Mike Little Mac is p good but the recovery...

Eric Silva

the punch out kid is in smash?

Daminark

Yeah

Vinícius Machado Vogt

Too hot for studying math: climatempo.com.br/previsao-do-tempo/15-dias/cidade/363/…

72 ºF at 00:00.

I need lower temperatures... :(

It is 22 ºC...

I would like it to be winter all year round...

I need a good book recommendation...

I am studying Ring Theory...

Any suggestions?

Very basic Ring Theory. I am a newbie at it.

Daminark

02:13

I've been told to look at Atiyah and MacDonald by at least 2 people, though some have said it's not good for beginners and instead prefer looking at general algebra books.

Dummit and Foote, Hungerford, Herstein, etc

Xander Henderson

02:29

I second the nomination of Dummit and Foote

I would not recommend either Hungerford

Hungerford is a good reference, but is not very pedagogical

I was also recently told that I should have a look at Aluffi's book

Algebra: Chapter 0

which takes a more category-theoretic view of introductory algebra

Vinícius Machado Vogt

@Daminark and @XanderHenderson: Thank you very much!

I love Arnol'd: quora.com/…

quora.com/How-can-we-do-calculus-without-the-concept-of-a-limit

xD

Daminark

@Xander what's wrong with Hungerford? I used it briefly and it seemed alright. I recommend D&F on general principle as being pretty decent but for the most part I find it to be a bit slow

Xander Henderson

@Daminark As I said, it is encyclopedic, but not very pedagogical

it is also a mixed bag of proofs, proof sketches, and "leave it to the reader"s

he often belabours very simple ideas for several paragraphs

then quickly breezes through difficult proofs, leaving the difficult heavy lifting to the exercises at the end of the chapter

Daminark

I see

Xander Henderson

also, it doesn't consistently use the same font throughout the book

that, actually, pisses me off more than almost anything else

Daminark

02:42

Lmao, I wonder what the reason for that is

Xander Henderson

I. Don't. Know.

but it makes me angry

Daminark

I want to caption that somehow but I couldn't think of one

Xander Henderson

heh

Daminark

"When you can't even apply dominated convergence"

Antonios-Alexandros Robotis

Lang's algebra is the best :thonking:

Mike Miller

02:57

@Daminark who needs recovery when u have side+B

Akiva Weinberger

@BalarkaSen I don't get it, halp

Daminark

@Mike

Antonios-Alexandros Robotis

lol

Xander Henderson

03:14

Baez taught the intro grad real analysis a year or two ago; many students applied DCT without justifying the application all over the place

Baez eventually sent the class an email with the phrase "DCT is not a magical spell! You have to check the hypotheses!"

which prompted this:

Daminark

Xander: amazing

anon

03:30

poor convergence theorem

Leaky Nun

03:59

@KasmirKhaan a normal subgroup is closed under conjugation, i.e. inner automorphisms; a characteristic subgroup is closed under every automorphism

the statements above are both criteria, not definitions

Let H be a subgroup of G.
H is normal if $gHg^{-1}=H$ for all $g$ (definition)
H is characteristic if $\varphi(H)=H$ for all $\varphi \in \operatorname{Aut}(G)$ (definition)
Conveniently, H is normal if $gHg^{-1} \subseteq H$ for all $g$ (property)
Conveniently, H is characteristic if $\varphi(H) \subseteq H$ for all $\varphi \in \operatorname{Aut}(G)$ (property)

also, "$gHg^{-1}=H$ for all $g$" and "$gHg^{-1} \subseteq H$ for all $g$" are equivalent

Adeek

04:31

hahaha

hi everyone

Leaky Nun

@Adeek what is funny about that

Adeek

@LeakyNun I like hadoken

I am big fan of street fighter in general

Leaky Nun

I see

> To see that the topograph is connected, take the particular positive
defmite quadratic fonn that has a well with α = β = γ = 1 at
some chosen superbase. Then, by climbing down, we see that any
component of the topograph must contain a well, the three vectors of
which must be those that yield the three smallest primitive values of
the form. This well can only be "our" well. Therefore the component
must be "our" component, and so the topograph is indeed connected.

this is one of the best proofs I've seen

Leaky Nun

04:44

@MatheinBoulomenos froehliche weihnachten

MatheinBoulomenos

danke!

merry christmas to you, too!

Leaky Nun

sounds like

happy wine night

Adeek

Guten Tag @MatheinBoulomenos

MatheinBoulomenos

Guten Tag @Adeek Merry Christmas!

Adeek

Thanks dude :) Merry Christmas to you too :)

I am really enjoying learning about algebraic cycles

very fun

I was watching some lecture of spencer bloch and he was mentioning category of motives and how it is related to algebraic cycles

and that if some of conjectures were solved then we would build category of motives somehow which solves a lot of interesting problems

Daminark

04:56

Hey @Adeek and @Mathein!

Adeek

hi @Daminark

MatheinBoulomenos

Hey @Daminark Merry Christmas

Daminark

How's everything going?

MatheinBoulomenos

Pretty good, thanks

Leaky Nun

@MatheinBoulomenos all products exist -> all equalizers exist -> all limits exist?

where A -> B -> C means A and B imply C

MatheinBoulomenos

04:59

no

Leaky Nun

why not

MatheinBoulomenos

Consider the category of torsion-free abelian groups

no wait, that doesn't work

consider instead the category of faithful abelian groups

Leaky Nun

what is a faithful abelian group?

does it pray 5 times each day?

MatheinBoulomenos

i.e. abelian groups $A$ with the property $\forall n \in \Bbb Z: (\forall a \in A: na= 0) \Rightarrow (n=0)$

Daminark

@LeakyNun kek

Leaky Nun

05:04

so the action of Z, i.e. as a Z module, is free

what happened to a=0

MatheinBoulomenos

I had the wrong definition

Leaky Nun

RIP "a=0" 2017-2017

MatheinBoulomenos

I confused torsionfree and faithful

This category has arbitrary coproducts, just take direct sums of abelian groups

but if you consider the map $\Bbb Z \to \Bbb Z$ multiplication with $2$, then this doesn't have a cokernel.

Leaky Nun

you obviously misinterpreted my statement?

MatheinBoulomenos

what was your statement supposed to say?

Leaky Nun

05:08

if all products exist and all equalizers exist, then all limits exist

MatheinBoulomenos

yes, this is true

why did you write this in such a horrible and confusing way?

Leaky Nun

how do you construct a pushout?

@MatheinBoulomenos because I'm a horrible and confusing person

by pushout I mean pullback

MatheinBoulomenos

(A->B->C) is not even well-defined, because conditionals are not associative

Leaky Nun

so we take it to be right-associative

that's the convention

in mathematical logic, not in general maths

Adeek

@LeakyNun why doing abstract non-sense ?

just kidding

Leaky Nun

05:10

@Adeek because I'm an abstract and non-sensical person

Adeek

what is the question ?

Leaky Nun

the limit-existence theorem

3 mins ago, by Leaky Nun

if all products exist and all equalizers exist, then all limits exist

and my question is how you construct a pullback from products and equalizers

MatheinBoulomenos

If your category has products and equalizers, then you construct the pullback from two morphisms $f:A\to C$ and $g:B\to C$ like this. Compose the projection $A\times B \to A$ with $f$ and the projection $A\times B \to B$ with $g$, then take the equalizer of these two compositions

Adeek

I don't think product exists --> all equalizers exist

Leaky Nun

@Adeek facepalm

Adeek

05:12

oh

@LeakyNun sorry I have a habit of skipping words while reading.

Leaky Nun

@Adeek sorry have habit skipping while

Adeek

:P

Adeek

05:46

@Daminark want to check something with you

to make sure I am not insane

here ?

Daminark

I can try to help

Adeek

I figured it out :P

GFauxPas

06:05

so are you insane

Daminark

Kek

Typhon

06:25

i am about to finalky answer this question

0

Q: Verifying an integration method for step function integrals.

I should note that this is a symbolic integration algorithm and therefore intermediate steps DO appear to be unjustified. The purpose of this question is to justify or reject this algorithm as always giving correct solutions. Suppose we have a function $f(x)$ that has no vertical asymptotes and ...

calculus integration algorithms floor-function

it only took me two years to do so

Typhon

07:03

@LeakyNun havent done every subtheorem yet but the core structure is thre. Ive started nodding off so ita best i hit the hay lest i botch my proof.

Leaky Nun

08:01

The easiest way to get to the real problem is usually asking Why five times. — Gordon Oct 21 '12 at 17:25

Typhon

08:18

@LeakyNun i dont want to start a huge conversation but am i wrong in saying that guy was pretty darn well trying to tick me off?

it really felt like it

Leaky Nun

no comment

Typhon

fair enough :)

Leaky Nun

I'd only say that it isn't the first time you got worked up over nothing

Typhon

@LeakyNun the guys first post to me was that it has been a year and i still need to learn logic with really no justification other than that integrals can lack closed form and my algorithm was wrong (both of which i was well aware of).

he consistently claims that if one doeant know something then they need to learn logic. Im not the first hes berated and kicked out. He once kicked a guy out for what appeared to be someone saying sets were defined (not in the mathematical sense but rhether the sense that the word set has a lexigraphicak definition in the dictionary).

@LeakyNun any clue what he meant by past incidents thiugh? I know me and amwhy bicker now and then but he sounded likd he was talking about a sock account or something more serious than an argument.

very puzzled by that

Leaky Nun

@Typhon he probably mistook you for another use who caused much trouble here. I've already told him once. I'll tell him again.

Typhon

08:29

@LeakyNun Fair enough. I know a few users thought I was flag spamming and it got to the point where a staff member was contacted and flat out said it wasnt me.

Daminark

@Mathein wait so what exactly is the deal with the p-adics anyway?

Typhon

other than that, I cannot see what would fall into that category

Leaky Nun

@Daminark what do you mean?

Typhon

@LeakyNun why are people pushy with advanced subjects anyways? I really should be studying diff eq at this point, but i got bored. Do people think Im a professor or something?

Daminark

On one hand, how I get the vague idea of how they work insofar as closeness sorta flips to having highest powers of 10

But not much beyond that. Also I wonder what they're used in

Balarka Sen

08:33

Look up Hasse's local global principle

Typhon

@Daminark mind if i ask what the most advanced math you know is?

Leaky Nun

@Daminark "p-adic" "10"

Balarka Sen

You can define 10-adics, sure

Leaky Nun

but p stands for prime

and (to quote my professor) 10-adic isn't interesting because Z10 = Z2 * Z5

Daminark

@Typhon so, in terms of the math I'm actually competent at, I'm gonna say basics of analysis, group theory, and linear algebra

Leaky Nun

08:35

so everything about 10-adics can be inferrred from 2-adics and 5-adics

like the fact that there are 4 idempotents

Typhon

sounds like good stuff

Leaky Nun

this follows from the fact that there are 2 idempotents in Z2 and Z5

Typhon

so youre a sophomore in college, I presume?

Leaky Nun

@Daminark what happened to topology and category theory?

Balarka Sen

@LeakyNun $\widehat{\Bbb Z} = \varinjlim \Bbb Z/n\Bbb Z$ is a perfectly interesting object

Daminark

08:36

Also some basic number theory I guess. In terms of what I'm acquainted with but can't honestly say I'm good at, some algebraic topology

Balarka Sen

The 10-adics sit inside it

Leaky Nun

@BalarkaSen fair enough

the profinite integers

@BalarkaSen sure, as it's just the product of the p-adics

Typhon

what is topology?

i mean i know it has to with surfaces

but thats all i know

Daminark

That's part of it

Leaky Nun

Nov 2 at 14:01, by Leaky Nun

@GFauxPas topology is finding the invariant in the variant

Nov 2 at 14:02, by Balarka Sen

That's like the most pretentious garbage I have heard in some time

Nov 2 at 14:02, by Balarka Sen

It literally means nothing

Typhon

08:38

lol

Daminark

Topology is basically about continuous functions

MatheinBoulomenos

Topology is like geometry without a notion of length, volume, angles or proofs

Daminark

Or...

Leaky Nun

@Daminark rip manifolds

Daminark

@MatheinBoulomenos lol I see you read math facts

:P

Leaky Nun

08:38

@MatheinBoulomenos if you include proofs, you get projective geometry

Typhon

@MatheinBoulomenos so drawing on paper, then?

Leaky Nun

projective geometry is euclidean geometry done without compasses

so you only get to have straightedges

Daminark

But yeah so the idea is that in R^n/metric spaces, you have a notion of continuity. Turns out, it need not depend on having distance, you can define a certain system of sets on a space and talk about continuous functions on there, in some sense inspired by open balls in a metric space

Typhon

which is euclidean again?

Daminark

And that'll agree with what you get by delta-epsilon

Typhon

08:40

is that the one in 3D?

Balarka Sen

@AkivaWeinberger Your form should have an antiderivative.

You have to solve a small PDE :)

Typhon

nvmd

Balarka Sen

Then you can forget about the curve

Daminark

So topology is interested in those spaces

Typhon

my brain hurts

is this what you learn in college?

Daminark

08:42

@Leaky I think even the idea in differential topology is that when you can stack a smooth structure on something, it makes it easier to work with

Leaky Nun

30 secs ago, by Typhon

my brain hurts

rip brain 2017-2017

Typhon

...

Daminark

Typhon: yeah basically. Also to answer your question from earlier, I'm actually a third year in college :P

Typhon

ah

Leaky Nun

7 mins ago, by Daminark

@Typhon so, in terms of the math I'm actually competent at, I'm gonna say basics of analysis, group theory, and linear algebra

Daminark

08:43

But yeah so let's take the degree of a map. I think you can interpret it via homology

Leaky Nun

basics????

Typhon

last year of high school for me

Leaky Nun

by which lens is that basics @Daminark

MatheinBoulomenos

p-adics are really useful in algebraic number theory. They have a lot of nice properties which makes them easier to deal with (for example, every finite extension has a solvable Galois group and a ring of integers generated by a single element) and some statements about rationals can be reduced to statements about p-adics. Balarka mentioned Hasse's local-global principle, but there is more.

Kasmir Khaan

Hi mathein :D

MatheinBoulomenos

08:43

In some sense, the behavior of the prime number p, say in an extension L/Q is reflected in the completions

Kasmir Khaan

Hi leaky :D

MatheinBoulomenos

Hi Kasmir

Leaky Nun

@KasmirKhaan did you read my stuff?

Kasmir Khaan

Mathein I want you to clear something for me

Balarka Sen

@Daminark Gonna post something in Washington

Kasmir Khaan

08:44

abotu normal subgroup

@LeakyNun let me see

Daminark

Because the top homology group of an orientable manifold is Z, so if you have a continuous map, the induced map on top homology is a homomorphism from Z to Z. Those are determined by f(1), that's the degree. Now I'm guessing that's gonna be rather hard to compute

But if your manifold is smooth, you can get that number by computing integrals, which is both an interesting connection as well as possibly being computationally helpful

MatheinBoulomenos

what question do you have about normal subgroups? @Kasmir

Typhon

@LeakyNun any basic logic videos for high schoolers? Apparently I dont know my stuff even though i do.

Daminark

And lol I said basics because I don't exactly want to go and say I know analysis, then someone asks me about spectral theory or something and I'm like derp

Kasmir Khaan

@LeakyNun yeah i read them now thanks leaky :)

Leaky Nun

08:46

@Daminark isn't spectral theory about linear algebra?

Kasmir Khaan

@MatheinBoulomenos well , the idea of H is normal, if gHg' =H for g in G or gHg' is contained in H for all g in G , they are not the same

so which is it

Leaky Nun

@KasmirKhaan that's just what I posted

Kasmir Khaan

I know leaky

MatheinBoulomenos

they are equivalent

Kasmir Khaan

I want to be extra sure

Leaky Nun

08:47

"gHg' =H for g in G" and "gHg' is contained in H for all g in G" are equivalent

Daminark

It's a thing in functional analysis

Kasmir Khaan

then why do we use both

Leaky Nun

but you can't pull out the "forall g in G" and claim that I have said that "gHg' = H and gHg' is contained in H" is equivalent for all g

@KasmirKhaan because the latter is easier to verify

Daminark

Like in finite dimensions the stuff is easy, you just have a finite set of eigenvalues

Kasmir Khaan

the contained part

?

Leaky Nun

08:47

yes

Balarka Sen

@Daminark Suppose $f : M \to N$ is a map between orientable manifolds. If $x \in N$ is a point such that it admits a neighborhood $U$ around it for which $f^{-1}(U) = V_1 \cup \cdots \cup V_n$ where $V_i$ are disjoint neighborhoods of points $y_i \in f^{-1}(x)$ and $f$ restricts to a homeomorphism $V_i \to U$, then degree of $f$ is the sum of the degrees of $f^* : H_n(V, V - y_i) \to H_n(U, U - x)$.

Leaky Nun

to check that two sets are equal, you need to check that one is contained in the other and that the other is contained in one

but it turns out that you only need to check one direction for all g

Balarka Sen

This is the local degree theorem in TOP

Kasmir Khaan

so it can happen that gHg' is contaiend in H , without being equal to H ?

Leaky Nun

@KasmirKhaan yes

Kasmir Khaan

08:48

i mean what would fail ?

some elements dont conjugate to get inside H ?

Leaky Nun

@KasmirKhaan no

that's what contains means

Kasmir Khaan

that is what I dont get

Leaky Nun

it means that after conjugating, you always get inside H

MatheinBoulomenos

btw the top homology of a orientable manifold is not necessarily Z if the manifold is not compact

Leaky Nun

but not everything in H is a result of conjugating

Daminark

08:49

But let's say you have an infinite dimensional operator, the spectrum is trickier to deal with. The proof in C^n that operators have eigenvalues is just FTA. The proof in Banach spaces that you have a spectrum (not necessarily eigenvalues, could also fail surjectivity since rank-nullity is dead) uses Liouville

Leaky Nun

consider $\langle a,b\mid aba^{-1} = b^2 \rangle$

then $a \langle b \rangle a^{-1} = \langle b^2 \rangle $

I think this is a subgroup of a linear group

I forgot the representation

Balarka Sen

@MatheinBoulomenos In fact it's always 0 if it's noncompact

Daminark

Tru. Though the always being 0 business is interesting

Kasmir Khaan

hmm

let me Think about that

Typhon

@LeakyNun lol you never did check if my proof was sound?

Daminark

08:52

Re local degree theorem: sweet

Kasmir Khaan

there is still something off for me ,trying to distinct them

Daminark

Re p-adics: also sweet

Typhon

What are p adics?

people who drink urin?

Balarka Sen

if you're noncompact you can deformation retract the manifold to a codimension 1 subcomplex

Kasmir Khaan

so we have that gHg' is contained in H for all g in G, is equivalent to gHg' = H for all g in G @LeakyNun

Balarka Sen

08:53

Prolly takes work to prove it though

Kasmir Khaan

I rememeber some example where gHg' was contained in H but was not normal

Dont remember it fully

Leaky Nun

@KasmirKhaan I just gave you one

Kasmir Khaan

@LeakyNun but I dont see the distinction why would that fail to be normal, ( did not do the example yet )

i mean from the definiton i gave

gHg' is contained in H, but not equal to H

Leaky Nun

try to prove that thing you just stated first

Kasmir Khaan

what can go wrong

Leaky Nun

08:55

the equivalence

Daminark

@Typhon kek. Basically you define the distance between two numbers differently, and at mentioned above by Mathein, the stuff is useful in ANT

Leaky Nun

then you can know why it is not normal

Kasmir Khaan

okay ill do tha tnow

Typhon

What is ANT?

Kasmir Khaan

i need to show that H is contained in gHg' @LeakyNun

Leaky Nun

08:56

basically $ab = b^2a$, so $ab^n = b^{2n}a$ and $a^mb = b^{2^m}a^m$, so $a^m b^n = b^{2^{m} \cdot n} a^m$, so $(b^m a^n) (b^p a^q) = b^{m+2^p \cdot n} a^{n+q}$

@KasmirKhaan yes

MatheinBoulomenos

ants are small insects

3

Kasmir Khaan

algebraic number theory @Typhon

Typhon

I fell right into that one

Kasmir Khaan

@MatheinBoulomenos -.-

=p

Typhon

Number theory as in...?

MatheinBoulomenos

08:57

number theory as in number theory

We ask questions about integers, rationals and related objects

And in the process of answering some of those, we unleash a lot of theory

Leaky Nun

@MatheinBoulomenos do you have a representation of the group $\langle a,b \mid b^a = b^2 \rangle$?

Kasmir Khaan

okay leaky

that was easy

Typhon

Uuh ok

Leaky Nun

@KasmirKhaan show me

Kasmir Khaan

gHg' is contained in H for all g in G

conjugating by g'

Typhon

08:59

so whether or not 3 is even?

Kasmir Khaan

we get that H is conained in g' H (g') '

MatheinBoulomenos

I guess. That's not a very interesting question though

Leaky Nun

more like, the class group of the ring of integers of Q[sqrt5] @Typhon

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$Mathematics$ Mathematics