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12:10 AM
What do these nutters call $efg = n$?
The Fundamental Theorem?
I was gonna write it off as Riemann Hurwitz
 
@BalarkaSen $f$ doesn't exist for Riemann surfaces right
every point is $\Bbb C$
 
I'm not sure it has a name
 
Yeah yeah
Do Riemann Hurwitz with basechange lol
 
And e and f are the same for all primes in Galois extensions
 
isn't g simply the number of prime factors?
 
12:14 AM
yeah
 
for non-Galois extensions, the formula is $\sum_{i=1}^g e_i f_i = n$
 
you mean e_i's are all equal in Galois
right
efg = n is my shorthand
 
where ef is understood to mean $\vec e \cdot \vec f$
 
Just pretend non-Galois extensions don't exist
 
i don't know what g means
 
12:15 AM
My textbook assignment one-shots all problems lmfao
It's literally one problem now
 
aha, ef means $\frac{\vec e \cdot \vec f}g$
 
Lol
 
what the hell are $\vec e$ and $\vec f$ ?
 
btw to supplement, $|\mathcal O_L/\mathfrak P^n| = |\mathcal O_L/\mathfrak P|^n$ because $\mathcal O_L/\mathfrak P^n = \left( \mathcal O_L \right)_\mathfrak P /\mathfrak P^n \left( \mathcal O_L \right)_\mathfrak P$ and $\left( \mathcal O_L \right)_\mathfrak P$ is a DVR because $\mathcal O_L$ is a Dedekind domain
 
You can also use a filtration right?
 
12:19 AM
what is a filtration
 
He means O_L/p^n = O_L/p + p/p^2 + ... + p^(n-1)/p^n
 
that
 
and then why is p/p^2 = OL/p?
 
Naka
 
what does naka do
 
12:22 AM
Ah maybe you need to localize, which is what you said
I should go to sleep
 
No man O_L/p^n is already local
You just shuffle
 
@BalarkaSen suppose $I=P$ and $P \mathcal O_L = \prod_{i=1}^g \mathfrak P_i^{e_i}$ so $|\mathcal O_L/P\mathcal O_L| = \prod_{i=1}^g |\mathcal O_L/\mathfrak P_i|^{e_i} = \prod_{i=1}^g |\mathcal O_K/P|^{e_i f_i} = |\mathcal O_K/P|^n$ so... $|\mathcal O_L/I\mathcal O_L| = |\mathcal O_K/I|^n$ but I'm sure you already knew
 
Yeah figured it out
CoolZz
 
i wonder if there's some cheap way to do it
I conjecture that for any nonzero ideal $I$ there is an order $C \subseteq \mathcal O_L$ such that $C$ is free over $\mathcal O_K$ and $C/IC = \mathcal O_L/I\mathcal O_L$
I mean this sounds like the opposite of cheap but it is what it is
 
lol
 
12:26 AM
fun fact: CIC is the name of a french bank
 
yeah it should follow from the classification of
 
cool fact
 
modules over Dedekind domains
i think those are all free + (something 2-generated)
quotients of
but these are the subfrees
WHO CARES
 
it's ok, OL is subfree, right
 
yeah
 
12:27 AM
subfree ?
 
submodule of free
 
sub of free
 
Ah ok
 
sous-libre
 
12:28 AM
also i'm 50% sure OL is locally free
 
lmao
this guy is a legend
 
0/10 he didn't say that in french
 
lol
the journalist was a brit who told him he cant respond in french back in england
 
subfree is equivalent to torsion-free, no?
at least over integral domains
ah no, obv not
only true with a finiteness hypothesis
 
Nobody cares bro
Torsion free is boring
$(x, y)$ is torsion free as a $\Bbb C[x, y]$-module
So what
 
12:31 AM
@BalarkaSen is the equivalent for curves true?
 
(It is also subfree, for what it's worth... totally not flat also)
 
for every divisor $D$ of a curve $X$ there is an open subset $S$ containing $D$ with trivial class group
 
Sounds off man what about an elliptic curve
 
what about it
 
Take $D$ to be a torsion point or something
IDK maybe not
 
12:34 AM
if you remove enough points do you always get an open subset of A1?
 
That's not even topologically true rofl
 
idk ag man
 
me neither
its a good q
ill think when i am awake
i have N assignments fml
 
look this is what I think
project to P1 or something
remove all the ramification points
and then remove one fibre to make it subset of A1
then you get n times open subset of A1, profit
 
then you get some cover of an open subset of A^1
is that good enough
 
12:37 AM
does that have trivial class group
 
no way right
surely not
thatd be stupid
 
but open subset of A1 has trivial class group
 
i agree
 
but it's a cover so it's trivial on some open subset right
like i want it to look like disjoint union of n copies of some open subset of A1
surely that has trivial class group
 
but the cover will look pretty nontrivial in general, like it'll be elliptic curve minus 4 points
thats a 2-sheeted cover of A^1 minus 3 points
anyway idk how to compute class group
run that stupid Cl(X) -> Cl(X - Z)
what was it
 
12:40 AM
but doesn't cover mean it's a disjoint union on some open subset
 
topologically you can refine but zariski opens are massive
If Cl(X - Z) was trivial then $\Bbb Z$ -> Cl(X) would be surj
but this $\Bbb Z$ is cyclic generated by the line bundle corresponding to Z
 
so how can this be surjective my dude
Cl(X) = X
if X is elliptic curve
:3
ie rank 2
rekt yes? admit it
 
maybe being free has nothing to do with having trivial class group
lol
 
lol
 
12:44 AM
I was going to write specifics but Leaky had the presence of mind of screenshotting it instead :p
 
just have to know where all the theorems are in hartshorne
thats basically how algebraic geometry functions
 
you mean stacks project
 
also
 
also what does Cl(X) = X mean? isn't X a torus?
 
Pic_0 X = X
Cl is just Pic_0 right
or what
 
12:46 AM
Cl is closure
 
isn't Cl(X) a discrete group
 
That moment when Thorgott is the voice of reason
 
i think the class group is just the elliptic curve group
change my mind
 
I am reason itself
 
Cl is Pic_0 is E man
 
12:49 AM
nice letters bro
 
or is it
look up the definition lmao
 
> Definition. Let X satisfy(*). Two divisors D and D' are said to be linearly
equivalent, written D ~ D', if D - D' is a principal divisor. The group
Div X of all divisors divided by the subgroup of principal divisors is
called the divisor class group of X, and is denoted by Cl X.
 
principal divisors are trivial line bundles right
and divisor sum is tensor product of line bundles
 
CaCl = Pic when integral. Cl = CaCl when noetherian, normal, regular
 
so its just Pic isnt it
 
12:53 AM
@Astyx Ganz verboten!
 
CaCl lmao
suddenly chemistry
yeah man Cl = Pic
 
You can ask for NaCl.
 
CaCl isn't even a legit molecule structure
Ca is 2+ and Cl is just -
 
big brain
who remembers that
CaCl_2 yeah
 
I certainly remembered that when I was y'all's age.
 
12:55 AM
i think leakys question is why it would make sense to give that a structure of a variety (scheme)
 
oh that was just meme
it makes sense but i cant be arsed to explain
Picard scheme
 
then what is Cl(X) as a discrete group?
 
it is the discrete torus lol
 
Every time you say meme I wonder how Hippa is doing.
 
@TedShifrin i can't say that per hipaa
what is a discrete torus?
 
12:56 AM
it is just X as a group man
 
growls
 
what do you mean discrete groups
this is the world of algebraic geometry
 
I thought Cl(X) is finitely generated
 
everything is discrete
why is it so
 
because Cl(P^n)=Z and i extrapolated lol
 
12:57 AM
lol
 
Um.
 
the point is if you randomly pick points on the torus they wont be equivalent as divisors
the equivalence is exactly the same as elliptic curve addition or whatever
 
That's not right.
 
which bit
 
Equivalence is exactly the same as addition.
 
12:59 AM
$\text{Pic}_0(E) \cong E$ as groups if $E$ is an elliptic curve
This is what I meant
 
right, points cant be equivalent at all because that would be an isomorphism with P1 right
 
Addition is defined using equivalence and an inflection point.
 
@LeakyNun something like this
 
sth sth Abel-Jacobi
 
yeah but you dont have to go so far to understand the isomorphism above, it is explained by what Ted said
A + B + C = O implies there's a function vanishing on A, B and C
a linear function
 
1:02 AM
this is huge Riemann Roch moment
nvm yeah they're just on a line
 
yeah lol
 
so what's Pic(E)
 
Pic_0(E) + Z
 
where's the Z from
 
Z keeps track of degree
 
1:04 AM
great
 
Pic(E) -> Z is degree
Ok
This stuff sucks
I am going to do 80's topology
 
so in general Pic0(X) is a curve if X is a curve?
 
How's Pic0 defined ?
 
its a torus
Pic_0 of a genus g curve is a g-dimensional torus iirc
@Astyx kernel of Pic(E) -> Z, degree of divisor
I gotta sleep
Bye
It's 7 AM
 
g'night
 
2:06 AM
So I'm given that the Partial Sum $S_N$ for $\sum_{n \geq 2} a_n$ is given by $S_n=\frac{n-2}{n+6}$.
Then the terms $a_n = S_N - S_{N-1} = ... = \frac{8}{(n+5)(n+6)}$.
And $a_2 = \frac{8}{(2+5)(2+6)} = \frac 17$.
Since the series starts at 2, we should have that $S_2=a_2$.
But $S_2 = \frac{2-2}{2+6}=0$.
What am I doing wrong?
 
2:55 AM
@Jeff think about why $a_n = S_N - S_{N-1}$
 
because $S_{N-1} + a_n = S_N$.
so what should I get from that @leaky?
 
this is a domain issue
you're applying (a formula that only works for some values $n$) for all values of $n$
in particular, $S_n = \frac{n-2}{n+6}$ should only be valid for $n \ge 2$ (get a new textbook if this textbook doesn't specify this)
are you sure $n \ge 2$ wasn't specified?
 
It's not a textbook question. It's from the teacher (who offered a problem last week that @TedShifrin said had a mistake in it).
 
brilliant
 
Anyway, I think it should be for $n>=3$ and I also think it should still be consistent for a_2.
 
3:02 AM
if you want $n=2$ you would need to have $S_1$
but clearly $S_1$ needs to be $0$
 
@LeakyNun, OK. I'm not sure what you're getting at. What I don't get is that shouldn't the equation for a_2 be consistent with S_2?
 
the formula $a_n = S_n - S_{n-1}$ (why do you keep using $n$ and $N$ like they're different?) only works for $n > 2$ I think
I assume you mean $S_N = \sum_{n=2}^N a_n$?
 
$n$ is the index, $N$ is what you go up to (like the top of the summation symbol in your most recent message), @LeakyNun
 
yeah but $a_3 = S_3 - S_2$ right
 
yes
 
3:09 AM
so the two indices should be the same?
$a_n = S_n - S_{n-1}$
 
understood. hold on for a minute please. got a situation over here.
 
3:34 AM
OK. I want to make sure I get this right

The question says summation from n=2 $\sum_{n=2}^{\infty} a_n$.
It says $S_n = (n-2)/(n+6)$.
Then it says a) Find the sum of $\sum_{n=1}^{\infty} a_n$ (but I think that's just a typo).
It says "What is $a_1$?", which is also probably a typo.
It says "Find $a_n$ for $n \geq 2$.
So the question thinks that the formula for $a_n$ should match $S_n$ for $n=2$, right?
 
I think you should ask your teacher lol
 
3:47 AM
@LeakyNun She hasn't responded (she's not known for responding to her email... and I don't mean just 'quickly').

Anyway, why aren't I right that a_2 should equal S_2
Because I think I'm still right about that. Even if we fix the typos.
 
can you define Sn?
 
$S_n=(n-2)/(n+6)$.
 
4:26 AM
@Jeff Luckily, Ted's memory isn't what it once was and he doesn't remember the problem in question.
 
I'd be happy to remind you, but I don't have pleasant memories of it, @TedShifrin.
 
You're probably better off just not speaking to me altogether :D
 
:D
 
Oh, the conditionally convergent integral. Your teacher wasn't wrong. I think you misinterpreted or misquoted her.
Search is such a wonderful thing. ? not ?
Anyhow, lucky for you, I haven't read the problem you and Leaky were discussing. So I'm innocent this time.
By the way, the very first equation you wrote down, @Jeff, makes no sense with the letters here. What is the partial sum, explicitly? $S_n = \sum_{k=2}^n a_k$ ? Nowhere do I see this correctly written.
 
4:46 AM
@TedShifrin, right. I was misusing capital $N$ Leaky explained that issue.
And yes, that is what $S_n$ was supposed to be.
 
So, if $S_n\to L$ as $n\to\infty$, that tells you what $\sum_{k=2}^\infty a_k$ will be.
 
Right @Ted. My question is why $S_2$ does not equal $a_2$.
 
It should.
If we have the correct definition of "partial sum."
So $a_2=0$.
 
5:01 AM
@Ted We are given that we are summing from 2, $\sum_{n=2}^{\infty}$ and that $S_n=(n-2)/(n+6)$.
I've calculated $a_n = 8/((x+5)(x+6))$ (and checked it several times, including on wolfram).
Since we're starting at 2, it seems $a_2$ should equal $S_2$.
However, $S_2= (2-2)/(2+6)=0$, but $a_2 = 8/(7 \cdot 8) = 1/7$.
 
There's just so much wrong here.
You agreed that what I typed was the correct definition of $S_n$. Where did you get the formula for $a_n$? And it should not have $x$'s in it.
I think Leaky already explained your mistake to you.
 
Where does what I typed conflict with anything you said?
And I used $x$ accidentally where i meant $n$ on the second line, but otherwise i don't see anything wrong.
 
So we agree that $S_n = \sum_{k=2}^n a_k = (n-2)/(n+6)$ for $n\ge 2$.
You should not use the same letter for $S_n$ and then in the summation, as $n$ is the upper limit, not the dummy variable.
So I agree that for $n\ge 2$, we have $a_{n+1} = S_{n+1}-S_n$.
 
@TedShifrin Yes, we agree. I'm using the letters the prof. gave. I agree we shouldn't use the same letter (which was why I originally, and also incorrectly, used capital $N$).
@TedShifrin For $n \geq 2$. But I don't see why it wouldn't work for $a_2$, too. Every sequence/series I've seen till now always did.
 
I agree that $a_{n+1} = \dfrac 8{(n+6)(n+7)}$ for $n\ge 2$.
 
5:13 AM
@TedShifrin OK.
 
So $a_n = \dfrac 8{(n+5)(n+6)}$ for $n\ge 3$. We know nothing about $a_2$ from this.
We learn what $a_2$ is from $a_2=S_2 = 0$, as we agreed.
 
The question says (exactly), "Find a formula for the general term $a_n$ when $n \geq 2$....". Is that maybe a typo?
 
Well, yes, although you know what $a_2$ is by different reasoning.
I wouldn't call $n=2$ a general term, though.
 
OK. So next, I'm (personally) curious why the formula wouldn't work for $a_2$? Every sequence/series I've seen till now it has.
 
When before have you had a problem just like this, starting sums at $k=2$?
You have to be careful about exact hypotheses.
 
5:17 AM
Well, I don't remember precisely. But is that the reason, because it starts at 2?
Because that is my suspicion.
 
Yes, of course.
You can only use $a_{n+1} = S_{n+1}-S_n$ for the values of $n$ for which you know a formula for $S_n$. But note that this tells you $a_{n+1}$.
 
Got it. It's a matter of noting the index carefully.
I like using $n+1$ for the next term, too, instead of $n$.
 
You're less likely to mess up your index range the way I did it.
Knowing $S_n$ for $n\ge 2$ tells you $a_k$ for $k\ge 3$.
 
@TedShifrin But also, $a_k$ works out right for sums that start at 1 and/or 0.
(I'm guessing)
 
Don't guess. Sit and work it out. :)
 
5:22 AM
I will. Thought not at the moment.
 
5:42 AM
@robjohn $p$ is the partition function and $\phi$ is Euler's totient function
$\frac{p(4)}{\phi(0)}$ equals 5
ah yes 5
btw I had already solved it last night
I tried it another way and found that $$c = 3 - 2\alpha-2\alpha^2--2\alpha^3-2\alpha^4$$
here $\alpha$ is the 5th root of unity
 
6:00 AM
sets are like stackexchange questions
both can be open or closed
a question can be both open and closed if it is closed as duplicate and the original question is open
and can be neither open nor closed if the the user is writing the question and hasn't finished writing
 
6:40 AM
 
sequence defined by regurgitation
 
 
4 hours later…
10:51 AM
What is the meaning of 'closed discrete subspace of R^n'?
discrete points?
 
@love_sodam I was confused as first as well, but I think it just means closed + discrete, and subspace just means subset
 
the subspace is discrete in the induced topology
 
or equivalently, every point is isolated
 
So really just discrete points
 
i guess
 
10:58 AM
Thanks!
 
11:34 AM
@EdwardEvans it just cracks me up how the queen pronounces 'prosperity' here at 1:25 youtube.com/watch?v=SSi4CTN2Jfs
prosperiteh
so noble
the rolled r's make such a difference
 
lmao
They used to pronounce house like hice
and then it was deemed TOO royal and out of touch
imagine that
 
12:03 PM
CARM ON INGERLAND
 
 
2 hours later…
1:34 PM
Hello, I have a problem regarding first hitting time for irreducible Markov chains. Someone has given a solution, but I'd like to know other approaches. Thank you.
1
Q: Show that the probability that first passage time is finite satisfies this BVP

MikeI am doing some exercises in E-Li-Vanden-Eijnden's Applied Stochastic Analysis and I meet this problem: (Exercise 3.19) Consider an irreducible Markov chain $\{X_n\}$ on a finite state space $S$. Let $H\subset S$ and define the first passage time $T_H=\inf\{n:X_n\in H\}$, $h_i=P_i(T_H<\infty)$. ...

 
2:01 PM
just found like 50% of our exercises in some topology inline repository from a french dude
 
2:32 PM
Why is the $T_1$ axiom called so
 
Trennungsaxiom 1
 
@TedShifrin have you seen the documentary on Chern ("Taking the Long View: The Life of Shiing-shen Chern"), and if you have, is it recommended?
 
3:32 PM
Hi there, I'm confused by the following proof math.stackexchange.com/a/2784145
My confusion may solely be in the line "and since I is finitely generated, one can localize at an element h such that $I_h = 0$", but I'd have to see after someone explains that line
 
the point should be that localization at $\mathfrak{p}$ being $0$ is equivalent to every element being $R\setminus\mathfrak{p}$-torsion, but that + f.g. implies there's a single element of $R\setminus\mathfrak{p}$ annihilating $I$ and then localizing at that gives $0$ too, I think
 
Let me think about that, although it's not clear to me why $I$ is finitely generated
Wouldn't every element need to be torsion for 'some' element of $R\backslash \mathfrak{p}$ rather than being $R\backslash\mathfrak{p}$-torsion (which atleast by name seems to suggest that every element in $R\backslash\mathfrak{p}$-kills each element)
(Just by the definition of localisation I mean)
Namely having m/1 = 0/1 just means that there exists some s\in S such that sm = 0, rather than needing all elements of S to kill m
 
3:47 PM
@tigre I mean the first one, yeah
 
Oh right, I see now that's normal language
I don't follow how you're using that + f.g. implies there's a single element [etc]
Oh wait
For each generator of $I$ you find an element in $R\backslash \mathfrak{p}$ killing it
And then multiply them all? (which is a finite product so well defined if $I$ is finitely generated)
 
Sounds about right
 
Okay, I think that does kill my confusion after all
Other than not knowing why $I$ is finitely generated
 
no wait, this isn't assuming ID, so that doesn't work
hmm, how did the argument go again
 
$I$ being fg should follow from Noetherian property
It's a sub of a finite rank free module.
Integral domain is not a big deal, you can always pass to irreducible components.
 
3:55 PM
ah no, there's no issue, the product is non-zero as we're in the complement of a prime ideal
 
I'd like to drop the assumption that A is noetherian if possible
But in that case I'm willing to add that M is finitely presented
 
Then it's alright
Finite presentation is precisely finite generation of the relator module
 
Okay, I think I'm universally happy
I was also wondering what the comment is meant to be telling the author to be careful about though
 
Your confusion was acting nilpotently, localization killed it.
 
Good call^
Where is the commenter intending for the author to reduce to that case?
 
4:00 PM
Is there any quick way to see the signs of intersections of products of disjoint circles in a torus?
E.g.
S^1 x S^1 x pt and pt x pt x S^1
 
Signs are only present when you orient everything
 
Hm? I assume the "standard" orientation on the torus but that is probably not what you mean? Also, since I honestly don't know this, what orientation (that is not dependent on the ambient space) does the outward pointing normal field correspond to?
 
OK, let's say you orient all the copies of S^1 in S^1 x S^1 x S^1 in the standard way; that gives a product orientation on all these various subtorii. Are we happy with that?
Then the intersection of S^1 x S^1 x pt and pt x pt x S^1 is a positively-oriented point.
If you negate the orientation of the last S^1 factor you get a negatively-oriented point
Locally the picture is the xy-plane intersection z-axis in $\Bbb R^3$.
 
@BalarkaSen ok, so one vector points towards me and the other to the right
 
If you have right hand rule, all those axes are positively oriented and $\Bbb R^3$ gets the product orientation, intersection is +. If you flip the orientation of the z-axis you are down to left-hand rule, so -
Right hand rule just means to me that x, y and z-axes are oriented as $-\infty \to \infty$
 
4:25 PM
0
Q: Find and classify critical points of $x^4-y^4-4xy^2-2x^2$, second derivative test is inconclusive.

KranAs in the title I found critical points to be $(1,0)$-saddle point, $(-1,0)$- local minimum and $(0,0)$ is the one I have problem with. Second derivative test is inconclusive in this case. If it was a saddle point I should find some curve in the graph that has inflection point at $(0,0)$ to prove...

why does Jyrki Lahtonen's comment suggest that (0,0) is a saddle point?
 
4:43 PM
Can you see why it's a critical point first ? @satan29
 
@Astyx yes
 
Ok, so it's either a minimum, a maximum, or a sadle point
 
yes.
 
The fact that it's positive along the curve $y^2+2x =0$ shows it's not a maximum at 0,0
The fact that it's negative along the curve x=0 shows it's not a minimum
 
@Astyx why
?
 
4:46 PM
The value at $(0,0)$ is 0
 
AH!
oh I see it now
Thanks a lot @Astyx
 
glad to help
 
1
Q: Is a triangle a $1$-dimensional manifold or a $2$-dimensional manifold?

Petros KLet me explain, in my course on differential geometry we define a 2-d manifold to be a normal surface. We usually take the zero function of that equation to look for the Jacobian to have the maximum rank and said that equation was an (m-p) manifold. But we consider a triangle on $ \mathbb{R}^n ...

 
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