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00:00 - 18:0018:00 - 00:00

12:23 AM
@user2103480 oh I thought you were a probabilist for a moment
 
@geocalc33 why not? just the matrix of the transforms of the individual elements.
 
@JoeShmo well, I'm still a master's student, so technically I'm nothing of that professionally - but probabilist was essentially right, since I'm mostly doing probability theory at the moment (and what's necessary to do it, like a bit of fourier analysis, pde and such)
 
12:57 AM
lmao my lecturer's handwriting is so good, automatic latex copying software can read it
 
nice
 
@user2103480 more acurate description is nerd
 
@EdwardEvans says the one in the math chat at 2am
wait
 
wait
 
damn
 
1:06 AM
can't sleep so I'm listening to my own shitty midi music
 
uhh like separate tones?
what do you mean by that
video game style?
 
you know guitar pro?
 
Hi @SayanChattopadhyay, not much, nearing the end of the semester.
 
@EdwardEvans I have now searched for the stuff on youtube and am horrified
 
1:09 AM
eerie
 
I cannot make out the lyrics
 
tfw you wake up in an abandoned 18th century house and you hear this playing in the next room
 
rofl
what's with the sudden nerd influx
 
tfw 2am and still TeXing
 
1:11 AM
@EdwardEvans listen to it again, after hearing the original at like 1:30
it'll get even more creepy
 
jetzt neu 28 black hanf
listening again
 
want to listen to some way better music, very original from germany?
 
if it's david hasselhoff then no
 
it's something I'd call "eastern german hardbass"
at least the second part of it
 
jesus christ I can hear the lyrics
@user2103480 as in, Kalkbrenner style?
 
1:16 AM
@EdwardEvans no, like in russian hardbass style
from 8:48
 
standard Alpha Industries shirt
makes me wanna bomb a gram and chew my cheeks out
 
yeh. the whole video lol
 
Let $M$ be a module over a PID $R$ such that $M_{(p)} \neq 0$ or infinitely many primes $p$. Show that $M$ is not f.g. Proof: If $M$ is finitely generated, then $M = M_t \oplus F$, where $M_t$ is the torsion part and $F$ is free. Since $M$ is finitely generated, $M_t$ is finitely generated (this is b/c we're working in a PID). Then $M_t = \oplus_{p} M_{(p)}$, where the summation is over finitely many primes $p$ for which $M_{(p)} := \{m \mid \exists \mbox{ s.t. } p^n m = 0 \}$.
 
let's see when berlin clubs open again haha
 
Question: Is there anything else to prove?
 
1:18 AM
We have Ausgangssperre in Mannheim
10pm - 5am
 
That's harsh. Gonna be like that most of the winter I'd assume
 
bleh
 
numbers aren't going down for some reason
 
Hmm...I feel like there should be more to prove...but I don't see it...
 
Reichsbürger obviously
 
1:19 AM
definitely not helping
 
lol this is cool
 
@user193319 won't $M=R$ have $M_{(p)}=R_{(p)} \neq 0$ for all primes $p$?
wait, so $M_{(p)}$ is not the localization?
that's nonstandard notation I think
$M[p^\infty]$ is more standard I think
 
$M_t := \varnothing$ should become standard notation
 
wat
 
The M T set
weeeeyyy
 
1:23 AM
...
 
hahahaha
I keep proving stuff for p-adic Hodge theory by just smashing tensor products together but I don't feel like I'm getting any feel for why stuff is happening
I'm just formally tensorplaying
I mean, I got $\mathcal{O}_{\widehat{K^{ur}}} \otimes_{\mathcal{O}_K} T^G \cong T$ for $T$ a f.g. $p^n$-torsion $\mathcal{O}_{\widehat{K^{ur}}}$-module and then extended that to all $T$ by STFGMPID but I don't feel like I actually know what's going on, I just algebra'd
just expects Lukas to remember wtf I'm talking about
 
what is $\mathcal{O}_{\widehat{K^{ur}}}$
 
the integers of the completion of the maximal unramified extension of a local field K
 
1:50 AM
I don't know much about tensor products
 
@LukasHeger No, it is defined as $M_{(p)} := \{m \mid \exists \mbox{ s.t. } p^n m = 0 \}$.
Does the proof seem correct? Something about it seems off.
 
@user193319 I see nothing wrong but I guess you could be more explicit about why there are only finitely many primes for which $M_{(p)} \neq 0$
 
@LukasHeger Okay. Thanks! For that part I was just citing theorem, but I'll definitely go back to the proof to see why that's the case.
 
you can reduce to the case of a cyclic torsion module since $M \to M_{(p)}$ plays nicely with direct sums
 
2:34 AM
If $M$ is a f.g. torsion module over a PID such that there exists $m \in M$ with $ann(m) = (r)$, where $r = order(M)$, what exactly is $order(M)$? Is it that element that annihilates everything in $M$? I.e., $rM = 0$?
 
Been working on this problem for a couple of hours. Any hints would be appreciated.
I tried to use law of cosines/sines to write a system of equations but it's not working so well.
 
3:08 AM
Did you try law of sines? I have not even drawn a picture, but it seems more promising.
 
I did but am not getting anywhere
 
@EdwardEvans thank you very much for this gift.
I loved it. Truly loved it :)
 
Anyone can help me with my problem?
 
3:29 AM
@TedShifrin Sorry to badger you, but just wondering if you drew it or thought of how to do it.
 
0
Q: Order of a Module

user193319Here is the problem I am working on: If $M$ is a finitely generated torsion module over a PID $R$ such that there exists $m \in M$ with $ann(m) = (r)$, where $R \ni r = order(M)$, what can you say about the invariant factors? What exactly is $order(M)$?

 
 
1 hour later…
4:39 AM
@AfronPie Did you try following my hunch? Work on it. It is exactly right.
One last hint: You have two triangles with a common side.
 
zeta(x) and pi(x)
delicious!
 
EM4
4:58 AM
what book is this from @LeonhardEuler
Hello @TedShifrin
 
@EM4 titchmarsh's theory of the Riemann zeta function
really good book
but sometimes his reasoning is hard to understand
 
EM4
I am not ready for that book.
 
@EM4 why?
 
EM4
what you believe the preq to study for that book.
 
analytic NT, complex and real analysis
 
EM4
5:00 AM
what is NT?
I am taking complex right now.
Real next semester.
 
NT=number theory
 
EM4
ohhhh
not ready for yet it, just yet.
I have this little problem in AA.
is driving me nuts. LOL
 
ah, I know only a few (two or three) people who are interested in analytic NT
 
Howdy, EM4
 
EM4
I would learn it until i can't do any more STEM.
How are you Ted?
 
5:07 AM
Still kicking, and what are you up to?
 
here are more:
this is what I call "attractive math"
lool
 
EM4
you are crazy (in a good way) my friend @LeonhardEuler.
 
ask titchmarsh to prove RH and he will promptly do it
 
EM4
I am dying @TedShifrin
 
I sorta warned you, EM4
 
5:10 AM
@EM4 why am I crazy?
just wanted to know why you called me crazy
 
EM4
crazy in good way....this is cool math and you find it attractive.
 
want more?
 
EM4
you did warned me, is my mistake that professor is nuts in teaching.
please do :)
Ted, I have Abstract problem...which is driving nuts.
 
What is it?
 
he literally gave 7 proofs of this
seven
 
EM4
5:14 AM
number 8.
I know those two cycles are disjoint.
is it better to make those cycles into permutations and do the product?
 
@SayanChattopadhyay w h a t.
 
I don’t know what they want, EM4.
 
There's no way any of those are core courses (except probably Riemannian geometry). They all look like electives. Why are there 7 electives in one semester?!?
 
You can write the formula telling where each number goes.
 
That's insane.
 
5:18 AM
He's not doing them all, feynhat.
 
@SayanChattopadhyay Katz's book is supposed to be elementary. He doesn't even assume knowledge of manifolds and bundles.
 
EM4
that was I thinking about.
 
Just give the mapping element by element. I find this exercise silly.
Ordinarily we aim for a product of disjoint cycles and say we're done.
 
EM4
yes

i find this problem waste of time.
 
5:34 AM
Agreed.
But you could write the answer in 30 seconds and we’ve wasted 10 minutes.
 
EM4
yes! I did by turning into permutation and do it that way.
which took seconds to do.
 
OK, move on.
 
EM4
yes, sir!
what you think about Laurent Series?
 
0
Q: A special constant arising from upper bounds for the $n$th prime number with a conjectured value of $\frac{3}{\log 4}$

epic_mathIt is well known that $$p_n>n\log n$$ for all positive integers $n$. I wanted to flip this inequality: i.e., I wanted the inequality $$p_n<A\,n\log n\quad\forall n\ge2$$ where $A$ is some constant. I wanted $A$ to be as small as possible. I don't think much is known about this constant. According...

Related to my last question on MSE
 
5:50 AM
I like Laurent series.
 
I too like Laurent series because I like series
 
@feynhat Yeah man, the have something called an overload semester. It has 6 courses anyway, compounded with the fact that I want to if possible do my thesis outside the institute, so I also have to try to either shift the courses from my last year to this semester, or shift them to the last semester of my degree. I am contemplating which one to go with
 
EM4
Laurent Series is making me to like series Haha.
 
Because if my thesis extends a little, I would not be able to finish the degree in time because of a not finishing the credit requirements. And all the six courses are electives and I have anyway finished the number of humanities electives that I had to take. So it's all math
 
love series. love products. love special integrals. love special functions. love analysis.
the key to a good life
 
EM4
5:56 AM
what classes you taking @SayanChattopadhyay
this correct @LeonhardEuler
 
okay let me do smth that I do here daily
 
6:27 AM
Example of induction that cannot work: youtu.be/CEUSRxvsrS8
 
6:39 AM
how can I prove that $$p_n<\frac{3}{\log 4}n\log n\quad\forall n\ge2?$$
$p_n$ being the $n$th prime?
please help this is very important
any idea?
 
why is it very important?
 
@copper.hat this helps me in deriving a theorem for prime numbers
 
do you know it is true and are looking for a proof or are you trying to determine if it is true?
 
it is true
very probably
 
so you are asking for someone to find a proof for you?
true and very probably are quite different things.
 
6:46 AM
@copper.hat uhh.. yes
I dunno a proof but it is true
 
ty
this is what I wanted
but I have an idea
weird idea
use induction
I know what to do
there are many ridiculously hard theorems that were proved using induction
do you know any upper bound for prime gaps?
ah I found smth
but I don't want to assume the RH :(
@copper.hat I have shown that it is true for all n till 17. Now I use the principle of strong induction:
oh sorry it is too long can't mention here.
bye let me do smth. sorry for wasting your time (if I did so)
 
7:22 AM
@SayanChattopadhyay Yikes. That many courses is already too much, but doing them online would perhaps be even more burdensome.
I hope it works out for you.
 
123
Hi All...
 
7:41 AM
@SayanChattopadhyay Who is teaching EG, btw?
 
Alok Maharana, I would have preferred shane or Kapil, but Shane is taking Hom Cohom
Sardar taking GGT is the only incentive I have for GGT
 
Any chance I could get the zoom (or whatever) link for EG?
 
The course structure for EG is Kapil's initiative apparently, he designed the course.
@feynhat Sure as soon as I get it, but it would be great if you drop in an email to Alok
 
Sure. I will write to him. When does the course start?
 
I think sometime in January
 
8:04 AM
Just wanted to know
How to write the citations in an article?
 
it depends on the journal.
 
So, for example, how would I cite this:
 
123
Why a^0 = e in group theory. If there is no operation we are taking for order, why it is equal to identity???
 
3
Q: Does an element of a group to the 0th power equal the identity?

Bobby LeeMy textbook doesn't explain this well at all. I was thinking about how a group follows the axiom that $xx^{-1} = x^{-1}x = 1$, where $x$ is some element of the group, $1$ is the identity and $x^{-1}$ is $x$'s inverse. The book says that the powers of some $x$ work with the binary operation on it...

 
123
@LeonhardEuler Thanks. But it not clear the point $a^0$ means no operation is taking for identity. If there is no operation why we take identity.
 
8:17 AM
This is an operation
We are still raising it to a power
It's just the same as x^0=1 for a real number x
 
123
$xx^{-1} = e$ It is an operation. but we can not write it $a^0$ , here $0$ shows no operation take place
what about multiplication?
$0$ shows no operation. because power shows how many times we take operation to get identity.
 
Ok
I am not much good in algebra
 
123
@LeonhardEuler Okay No prob frnd.
 
 
4 hours later…
12:19 PM
When I am giving a system of equations representing the generators and relations of an abelian group, I have to put it into Smith normal form to get a direct sum description of the abelian group. When I'm putting it into Smith normal form, am I allowed to perform row reductions or am I only allowed to do column operations?
 
1:00 PM
Brainfart: If I have a PID $R$ and an f.g. $R$-module $M$ and an extension $S \to R$, is the following a thing

$$M \cong R^r \oplus (R \otimes_S \bigoplus_i R_i) \cong R \otimes_S (R^r \oplus \bigoplus_i R_i)$$

where the $R_i$ are the torsion part of $M$. Basically I'm brainfarting because i've swapped around the tensor product with the direct sum (cuz the CoMmUtE) but for some reason my brain doesn't like it
erm
it doesn't look like a thing
ergh there's information missing
 
no clue about your specifics, but $A\oplus(B\otimes C)\cong B\otimes(A\oplus C)$ is not what it means for direct sum and tensor product to commute
 
yeah
thanks
lmao
I can give slightly more context if ya want
minus arithmetic bullshit
 
so assume $M$ is torsion-free, then the iso just reads $R^r\cong R\otimes_SR^r$ and surely that's sometimes wrong?
 
hi chat
 
Hello Astyx
 
1:11 PM
I've proved inductively that an f.g. p^n-torsion R-module $T$ is isomorphic to $R \otimes_S T^G$ where $G$ is a group acting semilinearly on $T$. Now I'm trying to extend this to all $R$-modules via STFGMPID. Then I have

$$T \cong R^r \oplus \bigoplus_i T/(p^i) \cong R^r \oplus \bigoplus_i R \otimes_S (T/(p^i))^G \cong R^r \oplus R \otimes_S \bigoplus_i (T/(p^i))^G$$
woops
err
ffs
 
If I want to prove a formal power series $P\in k[[X]]$ such that $P(0) \ne 0$ is invertible, is it sufficient to say there's a valuation v on k[[X]] such that v(P) = 0 ? It feels too easy
 
In the context of $\Bbb Z_p$, your element $(a_n)_{n \in \Bbb N}$ is a unit iff $p\nmid a_0$
now translate that into the language of valuations and replace $\Bbb Z_p$ with your favourite power series ring over a field
I guess
 
I mean, that sounds reasonable to me modulo details
 
Right, now obviously I just pretended that that oplus and otimes could be swapped around because in that case I get what I want lmaooo
 
@Astyx you can also invert it by hand fwiw
 
1:19 PM
note that $k[[X]]\cong \varprojlim k[X]/(X^n)$
 
How do you do it by hand ? It sounds super technical
 
write down another power series and multiply them out, then equate that to 1
 
that's lame
 
you'll find that if the leading coefficient of the first one is invertible, this determines the coefficients recursively
it's one line of arithmetic, nothing scary
 
Oh yeah that makes sense
 
1:22 PM
@EdwardEvans what does that tell me about units
 
Tbf I'm representating an element as compatible sequences and then hitting it with the relevant valuation and wildly assuming this will work exactly the same as for Z_p
 
man who cares about Z_p
damn algebraists
 
I'm not gonna be offended because I agree
 
How the turns have tabled
 
hahaha
The thick plottens
Hmm the $(-)^G$ is zeroth Galois cohomology and the category of those reps is additive (I think abelian but idk) so I can at least change $\bigoplus_i (T/(p^i))^G \rightsquigarrow \left( \bigoplus_i T/(p^i)\right)^G$
err H^0 is additive* lmao
I suck at category theory
 
1:35 PM
what does that have to do with cohomology
just have the group act component-wise on the sum
and then that's obvious,no?
 
idk, in my group we had some problems because G acts semilinearly
but idk I'm lame at this
 
idek what semilinearly means, that should work for any group action
ok nvm not all
yeah ok youre probably right
 
But the fact that (-)^G is H^0 clears that up anyway
ergh this needs details to make any sense, G is a Galois group and the T/p^i are vector spaces over the fixed field of G
I'm gonna just ask in my group lol
err I mean (T/p^i)^G are vector spaces over the fixed field of G lmao
 
2:06 PM
Suppose I start with a system of equations representing the relations on generators of an abelian $A$, and then convert this into a matrix. If I find the Smith normal form of it and find that it is $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix},$$ does this mean that $A \cong \Bbb{Z}/(1) \oplus \Bbb{Z}/(7) \cong \Bbb{Z}_7$?
...because $(1) = \Bbb{Z}$.
 
 
1 hour later…
3:10 PM
I think this is right. Now I asked to find generators $x,y,z$ of $M$ such that $ax=0$, $by=0$, and $bz = 0$, where $a,b,c \in \Bbb{Z}$ are given integers. But I don't think this is possible. If my calculations are right, the above matrix is the Smith normal for the matrix I was given, and so $A$ is indeed isomorphic to $\Bbb{Z}_7$.
Hence, question is equivalent to finding $[x],[y],[z] \in \Bbb{Z}_7$ which generate $\Bbb{Z}_7$ (the notation $[x]$ denotes the equivalence class of $x \in \Bbb{Z}$) satisfying the relations. But if $a=b=1$ and $c=2$, then $c[x]=0$ is equivalent to $0 = 2[z] = [2z] = [2][z]$, which means $\Bbb{Z}_7$ has zero divisors, which is impossible.
Btw the way, the matrix I started with is $$\begin{pmatrix} 2 & 1 & -3 \\ 3 & -2 & -1 \\ 1 & -3 & 2 \\ \end{pmatrix},$$ but I believe the above is the correct normal form of this matrix (I used a Smith normal form calculator to verify my answer).
Does my reasoning sound correct?
 
3:48 PM
Suppose $f,g,h \in \Bbb{Q}[x]$ such that $f | gh$ but $f$ doesn't divide $h$. Does it follow that $f | g$?
 
@Thorgott all of my above considerations are dumb and the result fell out of the ass of $T \cong \varprojlim T/p^nT$
 
duh
 
if Galois cohomology commutes with limits then I'm done lol
 
@user193319 ask yourself the same question for integers instead
 
@Thorgott Oh, wait, it works for prime elements, so no it doesn't hold in general, right?
That is, if $f$ is prime, then $f | gh$ implies $f | g$ or $f | h$.
 
3:57 PM
that alone doesn't yet mean it's wrong
come up with counter-examples
 
4:16 PM
Hi
Is anyone there?
I'm doing a very basic geometry problem
I get the answer A which is correct if I assume that ABD and DBC are similar triangles. The trouble is I can't see how they are similar triangles?
Any help would be greatly appreciated
 
4:31 PM
Okay, so I have a 4x4 matrix (I know it's entries) and I was able to find its minimal polynomial, which is $x^3 + x$. Is it possible to get the rational canonical form from this much information? I know the minimal polynomial is the largest invariant factor and the others divide it in a "sequential" manner.
 
@Turbo Can you find a relation between the angles ABD, DBC and ABC ? Can you find one between DCB and DBC ?
 
@Astyx Well DBC+DCB=90 degrees and ABD+DBC=ABC?
 
Yes, now what is the angle ABC ?
 
90 degrees
 
So can you find a relation between ABD and ACB ?
 
4:40 PM
I'm trying, but I can't :/
 
Subtract the equations you wrote above
 
DCB-ABD=0
so DCB=ABD
so that's one angle down
 
I claim that that's enough
 
Because they both have 90 degrees angle?
So all three angles are the same?
 
yes
 
4:48 PM
So it has nothing directly to do with the fact that they both share BD?
 
I'm not sure what you mean by that
 
They both share the length BD
but they're different sides of the respective triangles
 
right
 
I was trying to see if I could use that side to show that they were similar but I can't then right?
 
no
 
4:51 PM
So I can?
 
No, I meant you're right, you can't use that side to show this
 
Great, thank you so much!
:)
 
The rational canonical form is a direct sum of its companion matrices. The largest invariant factor $x^3+x$ is a 3rd degree polynomial, so the companion matrix is 3x3, which means there can only be one other invariant factor---namely, $1$. Does this seem right? The matrix is a direct sum of $[1]$ and the companion matrix associated to $x^3+x$, right?
There can only be one other invariant factor because the matrix is 4x4.
Is there any flaw in my reasoning?
 
5:16 PM
Wait...Isn't the product of all invariant factors equal to the characteristic polynomial? The characteristic polynomial is $x^4 + x^2$, which means that $x$ is also an invariant factor. So, what's the companion matrix of that? I think it's still $[1]$, right?
 
if $X,Y$ are homotopy-equivalent top spaces, are there points $x_0,y_0$ such that $(X,x_0),(Y,y_0)$ are pointed homotopy-equivalent?
 
Oh, wait...maybe the companion matrix of $x$ is $[0]$...Does that seem right?
 
5:57 PM
@Thorgott what does homotopy equivalence mean for pointed spaces exactly?
 
the same as regular, but basepoint gets fixed throughout
more precisely, pointed homotopy equivalence means you have pointed maps in both directions whose compositions are pointed homotopic to the respective identities. pointed map means preserving basepoint. pointed homotopy between $f,g\colon(X,x_0)\rightarrow(Y,y_0)$ is a map $H\colon X\times I\rightarrow Y$ such that $H(-,0)=f$, $H(,-1)=g$ and $H(x_0,t)=y_0$ for all $t$.
 
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