Mathematics - 2018-07-13 (page 1 of 2)

Daminark

12:57 AM

Yo @Mathein!

EnjoysMath

2:22 AM

Please check out my diagram chaser video. Taking functor image works!!!!

Introducing Zoomspace, a Category Theory Diagram Editor / Chaser

►

What do you guys think? Should I try to make a question out of it to promote it more

I have lots of questions

Daminark

If you have actual questions do ask but questions which are secretly just promotion, even if it's for something that's completely free, are frowned upon

EnjoysMath

2:40 AM

@Daminark did you see video?

Daminark

Haven't quite gotten the chance to watch it yet, I'm kind of in the middle of working, but I'll do so soon :)

EnjoysMath

I took the gal dang functor image !

:D

Maybe it can be a teaching tool

The bad news is that I have no idea how to do this in a general way, so every feature has to be coded. So there's no graph searching involved, I actually hand coded the take functor image feature

Prototank

any low dimensional topologists here?

Xander Henderson

3:05 AM

I'm pretty good at zero dimensional topology

is that low dimensional enough?

Prototank

Maybe too low :)

MatheinBoulomenos

@XanderHenderson I thought as a fractal geometer you're good at $\log_2(3)$-dimensional topology and stuff like that

Xander Henderson

well, sure

but that ain't topology no more

that's metric spaces!

Prototank

what is $\pi_1$ of the sierpinski gasket

Xander Henderson

(far superior to topology)

@Prototank at a guess, a free group with a bunch of generators

Prototank

3:08 AM

yeah that's my guess

Xander Henderson

but I don't actually know

(by "a bunch", I mean countably many)

Prototank

for instance, if we just look at limiting skeletons

Xander Henderson

sciencedirect.com/science/article/pii/S0166864109000224

Prototank

we'll find new generators in each one

hey sweet

my ideas were approaching the abstract

Xander Henderson

yeah, that seems to be the obvious way to do it

Prototank

3:11 AM

I'll take obvious

Daminark

Finally have a replacement computer

MatheinBoulomenos

@XanderHenderson random thoughts on the Sierpinski gasket: if we draw the center for each empty triangle and connect the centers in the obvious ways, we get a 3-valent tree with countably infinite layers. Infinite paths in such a tree that always go down one layer in each step (so in the Sierpinski gasket, always go to the center for smaller triangle) can be identified with 3-adic integers, so there seems to be some relation between $\Bbb Z_3$ and the Sierpinski gasket?

Xander Henderson

hrm...

it is possible

that being said, $\mathbb{Q}_p$ is homeomorphic to a Cantor set

for any $p$

MatheinBoulomenos

I think you mean $\Bbb Z_p$, as $\Bbb Q_p$ is not compact

Xander Henderson

$\mathbb{Q}_2$ is homeomorphic to the usual ternary cantor set

err..

yes

sorry... I've been working on a project that essentially "covers" $\mathbb{R}$ with Cantor sets

it ain't compact no more ;)

MatheinBoulomenos

3:20 AM

that sounds crazy

Xander Henderson

not particularly

but the Sierpinski gasket can be realized as the attractor of an iterated function system of three contracting similitudes

so each point in the set can be seen as the limit point of a sequence of maps

(yay symbolic dynamics)

so there may be an obvious identification with $\mathbb{Z}_3$

though I am not sure how well the metric structure aligns

MatheinBoulomenos

no wait, that was nonsense

I'm identifying $\Bbb Z_3$ with certain piecewise linear paths that play nicely with the Sierpinski gasket as each linear piece goes from a triangle to an adjacent triangle on a lower level

as I said, just some random thought

probably not interesting

or we can also map each 3-adic integer to a sequence

I just drew the elements in Z/3Z and Z/9Z

Xander Henderson

3:38 AM

yeah, that is essentially a version of the Bruhat-Tits tree, in terms of the symbolic dynamics of the generating iterated function system, no?

Semiclassical

given the symbolics dynamics connection, I'm tempted to replace those with their base-3 forms

MatheinBoulomenos

ah, I'm not very familiar with tha stuff, but probably yes

Xander Henderson

yup, that is the usual trick

since the metric really isn't playing a role

Semiclassical

the connection of the cantor set to base-3 stuff is about the closest I've ever been to actually understanding p-adics :P

Xander Henderson

ugh... bedtime

Semiclassical

3:44 AM

dangit Numerical Recipes

why must you sound so good and yet omit a crucial detail

shakes fist in impotent fury

Failed to be a Mathematician

5:03 AM

0

Q: Wedge product of a 2-form with a 1-form.

(Wedge product of a 2-form with a 1-form).* Let $\omega$ be a $2-$form and $\tau$ a $1-$ form on $\mathbb R^3$. If $X, Y, Z$ are vector fields on $M$, find an explicit formula for $(\omega ∧\tau )(X,Y,Z)$ in terms of the values of $\omega$ and $\tau$ on the vector fields $X,Y,Z.$ Let ...

Am I correct?

Fargle

6:16 AM

Idempotents other than $1$ are always zero-divisors, right? $e(1 - e) = 0$ but $1 - e \neq 0$ unless $e = 1$.

Daminark

Yup @Fargle

Anyway so I realized that the paper my mentor sent uses a different elliptic curve for its Kronecker-Weber type theorem than Silverman so I guess I'll just compute torsion for both

Right now doing $E[3]$ for $E:y^2 = x^3 + x$

x coordinates are given as solutions to $3x^4 + 6x^2 - 1$

Good lord this is gonna be fun

loch

6:35 AM

if you feel like cheating you can learn how to compute them using SAGE :p

Balarka Sen

@Daminark Algebra is fun yes

Meanwhile, I'm writing an answer on the Thom class ;)

Daminark

I mean I did fully do out one case by hand so I guess I might not feel quite as guilty

@Balarka I mean this is the computational bit, it's in service of a theorem which is dank af

Balarka Sen

You know what's dank though? The Thom class

Daminark

Abelian extensions of $\mathbb{Q}(i)$ are contained in those given by adjoining the coordinates of torsion points of $y^2 = x^3 (\pm ?) x$

Balarka Sen

Oh wow.

Daminark

6:38 AM

I remember someone tried explaining the Thom class to me not that long after I learned what the dihedral group was

As you can guess, my use of the word "tried" was rather deliberate

Balarka Sen

lmao

Daminark

Err, Thom isomorphism

I assumed that since both were named Thom they're probably related somehow

Then again if Thom is anything like Cauchy I'm very wrong

Balarka Sen

They are :)

LOL

well, the thom class is very simple. if you have a vector bundle it's a cohomology class that goes slicesloosh through the zero section

Daminark

But yeah somehow I know that's supposed to give you Poincare duality

Balarka Sen

I can link you my answer after I'm done writing if you want

Daminark

6:41 AM

Go for it

Balarka Sen

I was going to parse it in a super pretentious way but since you'll be reading I'll parse it in an even more pretentious term

Daminark

I'd be offended if you didn't tbh

Alessandro Codenotti

6:53 AM

Hi @Dami @Balarka

Yo

Hey @Alessandro

What's up gang

@CupFever hi

hi pal

7:06 AM

under single elimination, I don't think it means much for a team to enter semifinal

if the teams are linearly ordered, what can the actual rank of a team be, provided that it is in the semi-finals?

Cup Fever

yup, and the shoot-out is all luck

Leaky Nun

I asked you a math question lol

Alessandro Codenotti

I guess if you defeat $x$ teams to get there you can only say that you are at least the $x+1$ best team, starting from the bottom, but not much more?

Leaky Nun

I don't think so

not every team can enter the second stage to be defeated by you, say

and I think you mean the x+1 worst team?

Alessandro Codenotti

Well the best starting from the bottom is what happens when I forget that worst is a word

@LeakyNun ah, right, the other teams also win or lpse according to that order

Well so if you defeat someone you're also better than everyone they defeated

If you win at the first stage you're surely better than one team, at the second stage than three team and so on, someone who knows combinatorics can work out a recurrence relation or something

Leaky Nun

7:20 AM

or 2^n-1

oh so it isn't actually so bad

if you enter the semi-finals then you are better than 1/4 of the teams, i think

ok so it isn't actually so good also

Alessandro Codenotti

I think a more interesting question is the expected value of your position in the total order, knowing you're at the semifinals, and randomizing the pairings at the beginning of the tournament

Leaky Nun

right

Alessandro Codenotti

Which might be best solved by having a computer run it a million times :P

Leaky Nun

the old monte-carlo

AncientSwordRage

7:38 AM

hello

Leaky Nun

@loch hi

Cup Fever

8:27 AM

hello @AncientSwordRage and welcome :-)

Secret

8:51 AM

sciencedaily.com/releases/2018/07/180710185359.htm

> Ultimately, Sadler said, the study's findings don't suggest that students should drop high school calculus altogether, but rather shows that success in the subject -- whether in high school or college -- comes more from having a strong foundation. That foundation starts early and every year of great math teaching, even as far back as Algebra I in eighth grade, contributes to math proficiency that pays off in college.

"The one thing the paper says is if your background is strong, if you really know your algebra, geometry and pre-calculus, you're going to do well in college calculus," Sadl…

link.springer.com/article/10.1140%2Fepjd%2Fe2018-80684-y

I wonder if quantum optics will provide a experimental mathematics way to have deeper understanding of Meiger G functons and hypergeometric functions

agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/2017WR021864

Persistent homology

See homology for an introduction to the notation. Persistent homology is a method for computing topological features of a space at different spatial resolutions. More persistent features are detected over a wide range of spatial scales and are deemed more likely to represent true features of the underlying space rather than artifacts of sampling, noise, or particular choice of parameters. To find the persistent homology of a space, the space must first be represented as a simplicial complex. A distance function on the underlying space corresponds to a filtration of the simplicial complex, that...

Porous set

In mathematics, a porous set is a concept in the study of metric spaces. Like the concepts of meagre and measure zero sets, a porous set can be considered "sparse" or "lacking bulk"; however, porous sets are not equivalent to either meagre sets or measure zero sets, as shown below. == Definition == Let (X, d) be a complete metric space and let E be a subset of X. Let B(x, r) denote the closed ball in (X, d) with centre x ∈ X and radius r > 0. E is said to be porous if there exist constants 0 < α < 1 and r0 > 0 such that, for every 0 < r ≤ r0 and every x ∈ X, there is some point y ∈ X with ...

Now I wonder, whether we can use an infinite version of this model to investigate the pore topologies of porous sets whoose metric is difficult to compute directly

sciencedaily.com/releases/2018/06/180606120408.htm

> Chakrabarty, whose doctorate is in chemical physics, said that physicists normally solve problems by mathematically describing a cause and effect and marrying the two for a solution. But this new tool cares nothing about the cause, only about mathematically capturing the effect.

Perhaps we should go one step higher: A model that regardless of how unpredictable the inputs and outcome is, still produce some desired outcome

Call it noise control, or more generally, unpredictability forcing

If anyone has the ability to impose a constraint on completely unpredictable phenomenon such that regardless on how unpredictable it is, the outcome will fall within some prescribed threshold, then you control unpredictability itself

sciencedaily.com/releases/2018/06/180605112115.htm

Would this work in politics?

VP.

9:23 AM

Hi, I'm looking for a book about the history of mathematical discoveries

like how the distance between sun and earth was discovered, archimedes and etc

Cup Fever

VP.

so in one book, history with mathematical details about it

nice thank you!

Cup Fever

np

:-)

Secret

Directed acyclic graph

In mathematics and computer science, a directed acyclic graph (DAG ( listen)), is a finite directed graph with no directed cycles. That is, it consists of finitely many vertices and edges, with each edge directed from one vertex to another, such that there is no way to start at any vertex v and follow a consistently-directed sequence of edges that eventually loops back to v again. Equivalently, a DAG is a directed graph that has a topological ordering, a sequence of the vertices such that every edge is directed from earlier to later in the sequence. DAGs can model many different kinds of information...

AncientSwordRage

9:51 AM

someone removed a tag from my question (set-theory) and I was wondering why one tag was relevant (graph-theory) and the other one wasn't?

4

Q: Ways to construct sets of magic the gathering factions by colour combinations?

Magic: the Gathering is a 25 year old card game where each card can have a 'colour identity'. There are five such identities ([White (W), Blue (U), Black (B), Red (R), Green (G)] ignoring colourless), and when a new collection of cards are released they are often grouped into factions. Factions...

combinatorics graph-theory

Secret

10:09 AM

My guess is it's not set theory as defined by the tag. You are not asking things like thoerems, proofs, or constructing sets using set theoric axioms. The set theory tag is often reserved for questions related to e.g. ZFC and other foundational issues

This question is more on asking how to get some set given requirements, thus it is more a combinitorics question

AncientSwordRage

11:00 AM

@Secret so is it not graph theory either?

Secret

There might be a graph theoretic solution, but it does not sounds like graph theory at first glance (you are trying to form the n factions satisfying some criteria, I don't see how you can invoke edges here). However, leave that tag here for now, and let the community to decide whether it stays, cause its presence may inspired the answerers to give a graph theoretic solution

Leaky Nun

hi @loch

LegionMammal978

11:35 AM

Something small that's stumped me: using only a straightedge and a compass, where the compass's circle must go through another point, then given a segment AB and an arbitrary point C, how does one produce a point D such that AB = CD?

I tried a solution, but it only worked if C didn't get too close to AB.

chris

12:23 PM

hey chat, silly question but I can't answer: does the projective closure of the tangent space of an affine variety always increase the dimension by 1 (assuming the point is smooth)? I mean, if $X\subset\mathbb{A}^n$, $\dim(T_p X)=n$, does $\dim(\tilde{T}_p X)=n+1$ (with $\tilde{T}$ the projective closure)?

loch

hi @LeakyNun

@chris i dont quite follow - why would the dimension increase by $1$? the projective closure of $\mathbb{A}^n$ is $\mathbb{P^n}$

chris

hey @loch.... sorry but I'm quite confused about the projective tangent space: I was reading Harris' book, in which the projective tangent space is defined as the projective closure of the affine tangent space

but in some lecture notes (which i don't link because they're not in english), I read this

Leaky Nun

@loch so if a prime ideal is a subset of a union of prime ideals then it is a subset of one of them?

chris

"in the porjective case, assuming $X\subset \mathbb{P}^n$ be a projective variety (locally isomorphic to $k^n$), and $[p]\in X$, the basis of $T_{[p]}(X)$ is given by $\{\frac{\partial }{\partial x_0},\ldots, \frac{\partial}{\partial x_n} \}$, so
$$ \dim T_{[p]}(X)=n+1=\dim(X)+1$$

there are just these few lines

loch

@LeakyNun ah this is prime avoidance right - and you can replace prime ideal is a subset with any ideal

chris

12:35 PM

we have $\dim(X)=\dim(k^n)=n$, but why $\dim T_{[p]}(X)=n+1$?

Leaky Nun

@loch prime avoidance ?

loch

@chris this seems confusing to me too. also, a projective variety isn't locally isomorphic to $k^n$.. $\mathbb{P}^n$ is though (and perhaps that's what you meant)

@LeakyNun oh the fancy name of your statement

Leaky Nun

what is the name of the intersection version?

chris

so if $\dim(X)=n$, the identity $\dim T_{[p]} (X)=n+1$ does not hold?

(assuminng $[p]$ is smooth)

loch

@chris it could hold if say $[p]$ is a singular point, but if it's a smooth point then the dimension of the tangent space is equal to the dimension of your variety.

@LeakyNun i personally always referred it also by prime avoidance - but you might have to google to see if that's the right name lol

chris

12:44 PM

thanks @loch

Leaky Nun

1:04 PM

@loch

you call both of them prime avoidance?

loch

uh the reason i do that is because ambrus pal calls both of them that when i took his calss in commutative algebra

lol

user193319

Does anyone know how to find the online group permutation calculator? I can't seem to find it anymore...All I find are permutation/combination calculators.

Leaky Nun

@loch lol

Balarka Sen

m in m + k

(m+k) - m = k

Leaky Nun

@BalarkaSen ?

Balarka Sen

1:08 PM

(m+k) - (k + k)

= m - k

k + (m - k) = m

k.

Leaky Nun

are you alright

Balarka Sen

Not quite, figuring out dimension of transverse things lmao

codim X + codim Y = codim X \cap Y

Leaky Nun

dimension doesn't exist

math is a construct

Balarka Sen

ur face is a construct

Leaky Nun

no u

Balarka Sen

1:10 PM

no no u

no no no u

im slain

r e k t

gottem

gottem good

user193319

To show normality of a subgroup, does it suffice to conjugate by the generators of the group?

Leaky Nun

1:12 PM

yes

user193319

1:33 PM

Do any two left cosets of a normal subgroup have the same cardinality?

Vrouvrou

hello what is the traduction in French of "evenly covered"

?

Leaky Nun

@user193319 yes

user193319

Thanks!

Vrouvrou

hello, someone understand French here?

Semiclassical

hi chat

rschwieb

1:42 PM

@Semiclassical Are you saying "hi cat" in French?

Semiclassical

yes. yes, I totally am

there is no other possible interpretation

philmcole

Hello, I have two norms and a sequence which converges to different limits under the two norms. Does that imply that the norms are NOT equivalent?

Secret

2:04 PM

24

Q: Example of sequences with different limits for two norms

I was explaining to my students that if there is an inequality between two norms, then there is an inclusion between their spaces of convergent sequences, with matching limits. I then proceeded to show examples of such inequalities on the normed spaces they knew, and counterexamples of sequences ...

gn.general-topology

Semiclassical

equivalence of norms is by definition a matter of their ratio being a bounded function for all possible inputs

and it's not hard to find sequences whose norms converge to different limits but for which the ratio remains bounded.

so convergence of sequences to distinct limits is not sufficient for norm inequivalence.

philmcole

Okay thanks Semi!

Semiclassical

(of course, there are examples in the other direction as well. as a particular case, consider the scenario where one norm converges to a finite value but the other converges to zero.)

philmcole

The norms I'm looking at is the supremum norm and the 1-norm on the space of continuous functions from $[a,b]$ to $\Bbb R^d$ (or $\Bbb R$). Do you know if they are (in)equivalent?

Semiclassical

no clue.

philmcole

2:10 PM

Okay :P

Secret

supremum norm is basically the $||||_{\infty}$ norm, right?

philmcole

yes

Semiclassical

this looks pertinent: math.stackexchange.com/q/1645878/137524

philmcole

@Semiclassical Oh interestingly they provide an example of a sequence which converges to one for the supremums norm and to zero for the 1-norm. This would mean that they are not equivalent since the ratio is unbounded right?

Semiclassical

should, yeah

philmcole

2:14 PM

What you linked to suggests the result is even stronger. Nice :)

Semiclassical

namely, consider a localized function $f\in L^1$ and consider $\lambda f(\lambda x)$ in the limit $\lambda\to 0$

seems like a nice example

philmcole

nice thanks a lot

Semiclassical

And since all you need is a counterexample, you can pick any appropriate $f$ you want

oh, hah, i didn't even see that second answer

philmcole

but both use the fact that under one norm the sequence converges to zero so the ratio of the norms explodes right?

Semiclassical

ya

philmcole

2:19 PM

so your intuition was spot on

Semiclassical

another way to say it is that the condition for norm equivalence is that for any v there exist c,C>0 such that c p(v)<= q(v) <= Cp(v) for norms p,q

philmcole

yeah this is what I know

Semiclassical

but if p(v) = 0, then you'd need 0 <= q(v) <= 0

which is a bit of a problem if q(v) !=0

philmcole

Or you divide by p(v) and then you get your boundedness condition

Semiclassical

ya

just different ways of saying the same thing

but I like the statement that q(v) would need to be both nonnegative and nonpositive :)

aaaanyways

philmcole

2:22 PM

yeah :)

Btw how do I integrate something like this:

$$f_n:[0,1] \to [0,1], \; f_n(x)=\begin{cases} 2nx & \in [0,\frac{1}{2n}] \\ 2 - 2nx & x \in [\frac{1}{2n},\frac{1}{n}] \\ 0 & x \in [\frac{1}{n},1] \end{cases}$$

Can I split the intervall up and integrate the pieces?

Semiclassical

you could, sure. but you can also just draw it and see that it's a triangle :)

2

philmcole

Yeah but there is another function looking similar which has an integral which is non zero ;)

Wait I'm looking for it

Vrouvrou

@LeakyNun hello

philmcole

$$\begin{aligned}[]f_n(x) = \left \lbrace \begin{array}{ll} n^2x & \text {für } x \in \big [0,\frac {1}{2n}\big ] \\ n^2\left (\frac {1}{n}-x\right ) & \text {für } x \in \big [\frac {1}{2n},\frac {1}{n}\big ] \\ 0 & \text {für } x \in \big [\frac {1}{n},1\big ]\end{array} \right .\end{aligned} $$

has integral $1/4$ apparently

Semiclassical

okay? this one shouldn't have zero integral either

i mean, $f_n(x)\geq 0$ on $[0,1]$

philmcole

2:31 PM

I thought it was zero because I split it up and what I was left with was $-1/n$ which goes to $0$

Semiclassical

and since it's strictly positive on [0,1/n], it'd better not give zero integral

philmcole

for $n \to \infty$

Semiclassical

then i'm afraid you've done something wrong. the integrand is never negative on [0,1], so it'd better not have a negative integral

philmcole

okay lemme check again

right, I think it should be $1/2n$

which still goes to zero though

(talking about the first function)

Semiclassical

the geometric way to do it is to note that the graph consists of a line segment from (0,0) to (1/2n,1), a line segment from (1/2n,1) to (1/n,0), and zero thereafterr

so that's a triangle with height 1 and base 1/n, so the area should be 1/2(1)(1/n)=1/2n

same idea works for the other integral: you've got line segments connecting (0,0) -> (1/2n,n/2), (1/n,0)

so that's a triangle with base 1/n and height n/2, hence area 1/2(1/n)(n/2) = 1/4

philmcole

2:39 PM

yeah that's probably the fastest way to see it

okay ty

Semiclassical

also, writing your second function as $g_n(x)$, you have $g_n(x) = n f_n(x)/2$

so the two are really saying the same thing up to an overall constant

philmcole

nice I didn't spot this one

Semiclassical

it's more obvious once you realize that it's the same triangle graph in both

philmcole

yeah

Leaky Nun

3:13 PM

@MatheinBoulomenos @loch what do we know about rings where for each x and y either x divides y or y divides x?

am I just talking about valuation rings?

geocalc33

4:02 PM

can you tell what kind of curvature a network has just by inspection?

Semiclassical

for small ones, maybe? but typically one detects curvature by computing some measure of it

for instance, i've seen entropy-based measures of network curvature

see for instance: youtube.com/watch?v=z8xSKIlmQf8

loch

@LeakyNun for integral domains i think so

Balarka Sen

56

Q: Independent evidence for the classification of topological 4-manifolds?

Is there any evidence for the classification of topological 4-manifolds, aside from Freedman's 1982 paper "The topology of four-dimensional manifolds", Journal of Differential Geometry 17(3) 357–453? The argument there is extraordinarily complicated and a simpler proof would be desirable. Is the...

gn.general-topology gt.geometric-topology

Is it really true that Freedman's theorem is that much folk? I thought it was well-understood and studied.

Semiclassical

4:21 PM

I really do tire of knee-jerk flags.

Mike Miller

Yeah I see the message because I was pinged with it and it is wholly unreasonable

Hard to parse as anything but directed rules lawyering

Anyway, I don't think the tone in that thread is correct. I think people doing topological concordance seem to have an understanding of his work.

Semiclassical

i'm annoyed in two ways, really: one is the flagging itself

and then there's the fact that said flags were evidently validated

which...feh

Balarka Sen

Thanks @Loong, if you lifted the ban.

@MikeMiller Thanks for answering. Yeah I have seen texts and notes on Bing topology which seem to have a documentation of the ideas behind the proof.

So the whole thread seemed very off to me

(I haven't attempted to read Freedman's notes on Bing topology & 4-manifolds beyond the proof of PL Schoenflies theorem)

Mike Miller

Out of context it could have looked like he was yelling at me instead of expressing surprise about a link

In context it is clearly meh

Balarka Sen

No, that would be

@MikeMiller ??? WHAT THE SHIT??

Semiclassical

4:32 PM

the fact that I didn't see anyone (until Loong) enter the room makes me dubious that anyone actually checked to see the context (though they of course could have gone directly to the transcript)

Balarka Sen

@HDE (random numbers) entered and exited, but clearly did not think of the ban worthy to be lifted

Semiclassical

just "oh, someone used a profanity? okay, valid flag"

MatheinBoulomenos

@LeakyNun that's a generalization of valuation rings for non-domains yeah

Leaky Nun

I see

MatheinBoulomenos

It has a name, but I don't remember

You still get that it's equivalent to the poset of ideals being totally ordered

Leaky Nun

4:37 PM

ok

Fargle

Looks like they're called uniserial? @Mathein @Leaky

Leaky Nun

I see

rschwieb

@LeakyNun Rings for which x|y or y|x are equivalent to having ideals linearly ordered

uniserial, or sometimes chain rings

MatheinBoulomenos

Ah right

rschwieb

although I like uniserial best since it stays away from confusion

MatheinBoulomenos

4:39 PM

Uniserial modules are also important even over non-uniserial rings

The Prüfer p-groups are examples of uniserial Z-modules

Balarka Sen

Uniral rings: Rings which are pretty shitty

MatheinBoulomenos

I think it comes up in structure theory for Artinian modules

rschwieb

@MatheinBoulomenos Hmm, really... I'm not up to speed on it :)

There's a structure theory for artinian modules?

MatheinBoulomenos

Under some conditions in the base ring yeah

rschwieb

I know that uniform modules are a well-used class. They let you use uniform dimension

MatheinBoulomenos

4:41 PM

Saw a paper once, forgot the details

rschwieb

Let me know what it is if you find it

MatheinBoulomenos

Okay

rschwieb

There's a general theorem about which rings have modules, all of which are direct sums of cyclic modules

It turns out that they are exactly what they call (Artinian) serial rings

MatheinBoulomenos

I think you get that under some conditions, Artinian modules are direct sums of uniserial modules or something like that ...

rschwieb

I bet so, yeah ( serial just means direct sum of uniserials)

Daminark

5:08 PM

Hey there nerds

Alessandro Codenotti

Hi @Dami

Balarka Sen

SUBATOMIC PENETRATION RAPID FIRE THROUGH YOUR SKULL

loch

hi

rschwieb

5:24 PM

Now let's all get out our sliderules and dragon dice. We have nerding to do.

Leaky Nun

@loch hi

loch

@LeakyNun hi

Leaky Nun

> In an affine scheme if a finite number of points are contained in an open subset then they are contained in a smaller principal open subset.

stacks.math.columbia.edu/tag/00DS

mut chun is this?

loch

lol
it's saying if you have a finite number of points $\{x_i\}$ in an open subset $U\subset X$, then there exists a principal open $D(f) \subseteq U$ such that $x_i\in U$ for all $i$

if you don't have finiteness of course it's false, e.g. by taking all the points of $U$ (and assume $U$ is not principal)

geocalc33

5:51 PM

Thanks for the link @Semiclassical

Semiclassical

np. i saw a talk about that stuff a few years ago

there's a few other discussions of it online, so google around if you find it interesting

Alex K Chen

6:24 PM

is linear algebra needed to learn any of introductory (i.e the Murthy texts) algebraic number theory/analytic number theory ?

Leaky Nun

@AlexKChen yes

MatheinBoulomenos

@AlexKChen absolutely

you also want advanced linear algebra, i.e. the structure theory for finitely generated modules over a PID, at least for algebraic number theory

Alex K Chen

oh darn. Well then what are some good books ? The ones by Strang are the absolute crap I have ever read, and Axler is too abstract for a first-timer like me.

MatheinBoulomenos

I didn't learn LA by a book, just from the lectures I took, so I can't help with that

though after you picked up some basic LA, you can work with the LA section of an abstract algebra book (which you'll also need), a lot of general abstract algebra books have a section on that

Alex K Chen

@MatheinBoulomenos I will be working through Herstein Abstract Algebra (with ZERO lin algebra knowledge), so what other things I need to read for linear algebra ?

Also what do you mean by "some basic LA"?

MatheinBoulomenos

6:40 PM

you should know stuff like vector spaces, linear transformations, matrices, traces/determinants, eigenvalues/eigenvectors, dimension and the important results on these

Alex K Chen

that is basic :O

rschwieb

Sanity check: Can one construct an infinite field in which every element is a root of unity? Isn't it possible just to build an infinite ascending tower of fields of, say, characteristic $p$, and then their union is an infinite field with the property?

Alex K Chen

yea so what other things I need to read besides Herstein for LA (even the basic ones) ?

MatheinBoulomenos

@AlexKChen I don't know Herstein's book, he seems to have 11 pages on vector spaces, but that can hardly replace a more detailed treatment

you want to have abstract algebra up to Galois theory for finite extensions at least

@rschwieb a field has this property iff it has characteristic $p$ and is an infinte algebraic extension of the prime field

Alex K Chen

And for intro anal nt, do I need to know complex anal ?

MatheinBoulomenos

6:43 PM

@AlexKChen I should mention that I'm thinking of a "standard intro ANT" book/course

Alex K Chen

Also the princeton lectures in analysis are the easiest books on complex analysis, or are there some easire book ?

MatheinBoulomenos

there are some books which are more elementary and introduce some necessary algebra as needed along the way

I'm not that knowledgable in analytic number theory and I don't know about complex analysis books

rschwieb

@MatheinBoulomenos Ahh when you put it that way, it seems obvious. I guess the algebraic closure of $F_p$ would work just as well.

Alex K Chen

@MatheinBoulomenos OK and for reading complex analysis (and also anal nt), do I need to know real analysis ?

rschwieb

And I'm not crazy that the algebraic lcosure of $F_p$ is countable, right?

MatheinBoulomenos

6:46 PM

@rschwieb no you're not

any algebraic extension of an at most countable field is countable

rschwieb

whew brain still works

MatheinBoulomenos

@AlexKChen yes, you need to know stuff like convergence etc. rigorously

Alex K Chen

ok, thanks ! bye

MatheinBoulomenos

@AlexKChen you might want to have a look at "Algebraic Number Theory" by Frazer Jarvis

it's specifically targeted at undergrad students unike most other books which are graduate level

the prerequisites are really modest

@AlexKChen if you're interested in learning algebraic number theory early, I think Jarvis' book is really good

Hi @KasmirKhaan

Kasmir Khaan

@MatheinBoulomenos Hey :D

MatheinBoulomenos

6:56 PM

Hi @AlessandroCodenotti

Alessandro Codenotti

Jarvis is much better than Neukirch for a first exposition to the subject in my opinion, but Neukirch does stuff in greater generality and has more stuff

Hi @Mathei

MatheinBoulomenos

@AlessandroCodenotti it's a completely different level

Alessandro Codenotti

I know, I mentioned it because Neukirch pops up in all the "suggest me an ANT book" threads but I don't think it's appropriate as a first approach

MatheinBoulomenos

It depends on how much you know

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