Mathematics - 2021-01-28 (page 3 of 4)

Vrouvrou

12:05

hello

i have a question about topology product

i'm searching an example of product of open sets do not make a topology

Sayan Chattopadhyay

So my notes say that if $\mathcal{h}$ is a lie subalgebra of a lie algebra $\mathcal{g}$ corresponding to a lie group $G$, then $\operatorname{exp}\mathcal{h}$ is a submanifold in a neighborhood of $1$. Is $\operatorname{exp}\mathcal{h}$ the integral manifold generated by the left invariant vector fields for a basis of $\mathcal{h}$ ?

user2103480

@MikeMiller topologist proof

geometrically obvious

for what it's worth, here's a more complete argument math.stackexchange.com/questions/1499978/…

Thorgott

12:23

@Vrouvrou is the union of two products of open sets a product of open sets?

robjohn

@politeproofs look here

Vrouvrou

12:39

@Thorgott i try $\tau_1=\{\emptyset, E_1,\{a\}\}$ where $E_1=\{a,b\}$ and $\tau_2=\{\emptyset, E_2,\{c\}\}$ with $E_2=\{b,c\}$

$\{a\}\times E_2\cup E_1\times \{c\}=E_1\times E_2$

so it can be true

Thorgott

that equality is not correct

there's an element in the RHS which is not in the LHS

Balarka Sen

@SayanChattopadhyay Yes, correct

Thorgott

13:00

hey Balarka

do you know an easy example of a super-exponentially distorted subgroup

Balarka Sen

dunno what distortion means

Vrouvrou

@Thorgott I I understand $(b,b)$ is in $E_1\times E_2$ but it is not in $\{a\}\times E_2\cup E_1\times \{c\}$

Alessandro Codenotti

@Thorgott there's a nice blog post about crazy distortion, let me see if I can first remember the author and then find it

Balarka Sen

whats the definition of distortion

Thorgott

Say, $H\subset G$ groups, both finitely generated. Pick generating sets $T,S$ respectively. Distortion of $H$ in $G$ is the function $n\mapsto\max\{\vert h\vert_T\colon h\in H,\vert h\vert_S\le n\}$, where the norms are the word norms. So distortion measures how far the inclusion is from being a (quasi-)isometry. Growth rate of this is independent of choice of generating sets, of course.

Balarka Sen

13:06

not divided by $n$?

Thorgott

nah

Balarka Sen

doesnt matter anyway

Thorgott

but, I mean, that doesn't change the amount of information you get

polite proofs

@robjohn I see, so I could have done that.

I wanted to do that...

But wasn't sure if I could do it.

Thorgott

this is denoted $\Delta^H_G$ and the inclusion being a quasi-isometric embedding is equivalent to $\Delta^H_G$ having affine linear growth rate

Balarka Sen

13:09

yeah thats why i wanted to divide by $n$

got it

Thorgott

examples: center of Heisenberg group is quadratically distorted

Balarka Sen

yeah makes sense

nbro

If someone is interested in reinforcement learning, here you have a potentially useful post.

3

Q: What are the biggest barriers to get RL in production?

I am studying the state of the art of Reinforcement Learning, and my point is that we see so many applications in the real world using Supervised and Unsupervised learning algorithms in production, but I don't see the same thing with Reinforcement Learning algorithms. What are the biggest barrier...

Thorgott

further example: subgroups of nilpotent groups are polynomially distorted (this was established by you can guess who)

Balarka Sen

yeah clear from Gromov's theorem

Thorgott

13:12

it's used in proving the theorem, apparently

other example: $\mathbb{Z}^n$ is exponentially distorted in $\mathbb{Z}^n\rtimes_M\mathbb{Z}$ if $M$ has an eigenvalue with absolute value different from $1$

Balarka Sen

how distorted are subgroups of (Free) x (Free)?

Alessandro Codenotti

berstein.wordpress.com/2011/03/02/the-rips-construction-ii here you go, distortion not bounded above by any computable function, is this distorted enough for your taste? @Thorgott

Thorgott

oof

Balarka Sen

how distorted can copies of Z be?

Alessandro Codenotti

I think Paul Plummer showed this post to me a couple of years ago, but he hasn't been in chat for too long and I can't ping him, he surely knows more about this topic

Thorgott

13:20

@BalarkaSen idk, but apparently this has been investigated

Balarka Sen

random shit man its impossible to learn ggt

Thorgott

"Further, Olshanskii and Sapir show that the set of classes of distortion functions of finitely generated subgroups of a product of two free groups coincides with the set of classes of all Dehn functions of finitely presented groups."

from the blog Alessandro sent

Mike Miller

@user2103480 Dude I cited Hahn Banach you can't call that a topologist proof

user 147593

@AlessandroCodenotti his profile does not exist

Mike Miller

A topologist proof points out that if it's not a dense embedding then the image has closure a proper subset and you can write down a functional by first collapsing that

Balarka Sen

13:26

@Thorgott thats cool

Alessandro Codenotti

@BalarkaSen You just need to be Gromov

user 147593

and he hasn't chatted for 4 months

Balarka Sen

F_2 x F_2 has "flats" where curvature is 0, so I was wondering if you can move in and out of those to gain distortion

because if you move out of the flats to negative curvature by Morse lemma you're walking exponentially more

so it seems you can definitely gain exponential distortion

exponentially distorted cyclic subgroups in fact

but i dont know beyond this

i could be wrong tho

Thorgott

definitely seems like exponentially distorted Zs should be a thing

Balarka Sen

yeah but if anyone can tell me concretely thatd be you

Thorgott

13:42

I learned the concept of distortion like 3 days ago, I can't tell you anything

Thorgott

13:54

can't I just do like $\langle 1,s,t\vert t^nst^{-n}=2^n\rangle$?

Balarka Sen

yeah

Thorgott

hmm, is that non-trivial

user2103480

14:15

@MikeMiller that would have been very much enlightening since it's basically the statement cited there

That F is dense in E iff functionals on E are zero if they are zero on F

Mike Miller

as a topologists proof should be

porridgemathematics

15:00

is $D^2 / \sim$ where $f : \partial D^2 \rightarrow \partial D^2$ is $f(z) = z^2$ (viewing $D^2 \subset \mathbb{C}$), and $\sim$ is the equivalence relation generated by $a \sim f(a)$ homeomorphic to some well known object?

if $\sim$ just identified antipodal points then I could say $\mathbb{R} \mathbb{P}^2$, but given the third point in the equivalence class, it doesn't seem like this is $ \mathbb{R} \mathbb{P}^2$

BigSocks

15:24

just to make sure I am not going completely crazy:

$f(x) = e^{-1/x^2}$, $f^{(n)}(0) \neq 0,$ $\forall n \in \Bbb N$ right?

Thorgott

yes

very important fact

BigSocks

oh ok I looked up the errata of this book and it fixes this

thanks

Thorgott

no wait, I misread

copper.hat

huh?

Thorgott

they're all equal to $0$, not unequal

copper.hat

15:28

its part of the usual bump function

Thorgott

yeah, precisely

BigSocks

17

Q: How do you show that $e^{-1/x^2}$ is differentiable at $0$?

Define the real valued function $f(x)$ by $f(x) = e^{-1/x^2}$ if $x \ne 0$ and $0$ if $x = 0$. How do you show that $e^{-1/x^2}$ is differentiable at $0$? How do you show that this function is infinitely differentiable?

calculus real-analysis

@copper.hat answering 7 years ago

every day I remember calculus is absolutely nuts

Thorgott

haha nice

copper.hat

scary, i forgot

BigSocks

sorry to spook you. I too am afraid of this phenomenon occurring to me. For this reason I absolutely never write intelligent, correct answers on the internet for future me to get taken aback by

copper.hat

15:33

:-)

Sha Vuklia

For $k$ a field (algebraically closed if you wish) can we put a bound on the (minimal) number of generators for any ideal of $k[x_1,\dots,x_n]$?

Actually, I'd already be happy to see an example of an ideal generated by at least $n+1$ elements

oh, I found the answer

BigSocks

shouldn't the minimal number of generators for an ideal there be $1$? just pick any $x_i$ and $(x_i)$ is an ideal?

Sha Vuklia

No

Mike Miller

For an arbitrary ideal

Sha Vuklia

(x1,x2) isn't generated by 1 element

Thorgott

15:41

doesn't exist by Krull's principal ideal theorem

BigSocks

oh an arbitrary ideal

Mike Miller

@Thorgott ?

Thorgott

uh, not the principal version, of course

the generalization, whatever it's called

Mike Miller

I try to understand every time Sha comes in with these questions and leave clueless lol

That's on me tho

Thorgott

@Mike I'm saying there is no ideal that cannot be generated with $n$ elements

Mike Miller

15:42

Hmmmmmm

BigSocks

maybe a consequence of Nullstellansatz?

spitballin

Sha Vuklia

@MikeMiller Wait what xD I'm confused

@Thorgott is that a solution to my question, or what?

Mike Miller

I thought there were varieties in A^2 or A^3 which required more generators than you would expect

Sha Vuklia

I found an ez induction argument actually

Mike Miller

@ShaVuklia I just mean I am very very bad at commutative algebra / algebraic geometry

Thorgott

15:45

oh yeah, I guess it's just a dévissage argument

BigSocks

gesundheit

Thorgott

no wait, I was thinking modules

I don't see the induction argument

how does it go?

Sha Vuklia

Ok I said it was easy, but this argument actually involves the generalised Nullstellensatz (which I wouldn't have thought of myself). I'm still going through some details, but this is the post: math.stackexchange.com/questions/422182/…

BigSocks

@ShaVuklia yooo I guessed right

Thorgott

ah, that's a nice argument, but it's only for maximal ideals

(on the upside, it also gives the exact number of generators for those)

I don't think you can avoid something like Krull's theorem for arbitrary ideals

porridgemathematics

15:52

anyone know what my space is? :D

Thorgott

ugly

porridgemathematics

hehe

Thorgott

don't know the actual answer, sorry

porridgemathematics

but definitely not $\mathbb{R} \mathbb{P}^2$ right?

is there an easy way to say it isnt?

Mike Miller

@porridgemathematics I don't think your space is Hausdorff

Let $\alpha = e^{2\pi i \beta}$ where $\beta$ is irrational

Then the orbit of $\alpha$ under your squaring map (forwards and backwards) is a dense countable subset of the boundary circle

And you're collapsing that all to a point

Any neighborhood containing that will intersect any neighborhood of 1

porridgemathematics

16:01

oh wow

Thorgott

yeah nvm, my Krull argument doesn't work directly

I had the wrong statement in my head

porridgemathematics

@MikeMiller thanks!

Thorgott

this is annoying

ah, that's nice

I suspected the space isn't Hausdorff, but cba to think of an argument

Mike Miller

Whenever your equivalence classes are non-closed you should be very skeptical that things will work out

In fact I think if the quotient is Hausdorff then the equivalence classes are closed

porridgemathematics

I guess that makes sense, because $\{ [x] \}$ must be closed, so its preimage must be closed?

thats a good condition to check, yeah

Thorgott

16:09

yeah, so equivalence classes closed is equivalent to the quotient being T1

porridgemathematics

right

and the (forward + backward) orbit is not closed, because its dense, so there are rational points on the circle that are limit points of the dense orbit but certainly not in the dense orbit?

is that the idea

Mike Miller

16:22

Right

It is not closed because it is dense but countable

There's some condition about the quotient being Hausdorff when the relation is closed a subset of X x X but there are additional requirements and I always forget it

And that criterion is basically never useful imo

Alessandro Codenotti

@MikeMiller open quotient map? (not sure if that's the additional requirement you had in mind, but it's one that works)

Thorgott

I think this is a very useful criterion

quotient maps by group actions are always open, so these abound in nature

this is what I'd use in testing when, say, the quotient of a manifold by a Lie group action is Hausdorff

Mike Miller

Nerd

Just do it by hand

Nah I agree this is probably a straightforward route to getting that fact for proper actions on l.c. spaces

For simple cases though I just work by hand

Thorgott

yeah, fair

like doing a hyperplane argument for projective spaces

but dunno, this is probably already more convenient when you're doing Lens spaces

Mary Star

16:37

What transformation is the composition of a glide reflection and a rotation?

BigSocks

16:52

Ok so I think bump functions are supposed to be zero divisors in the ring of smooth functions, but what do you multiply them by to get the zero function?

Fargle

@MaryStar Well, because a glide reflection is orientation-reversing and a rotation is orientation-preserving, you know that, whatever your composite is, it must be orientation-reversing. Therefore it's either a reflection or a glide reflection. Then think about whether it can have any fixed points---that will answer it.

(Depending on where your center of rotation is, I think both are possible.)

Thorgott

@BigSocks what does it mean for the product of two functions to be the zero function

Ted Shifrin

@Fargle Yup. Howdy :)

DiscoLemonade

Question I had before bed:. How does a computer compute a number raised to a power of a rational number, let's say 3^2.1 ??

Ted Shifrin

Howdy, @BigSocks and @Thor.

BigSocks

17:07

Hi @TedShifrin

DiscoLemonade

How did they do this before computers too?

Thorgott

hey Ted

BigSocks

@Thorgott $ f(x)*g(x) = h(x)$ and $h(x) = 0 \forall x \in \Bbb R$

Ted Shifrin

If you haven't heard of it before, @DiscoLemonade, calculators use the CORDIC algorithm.

Fascinating stuff. Read about it.

When I was in high school, we used logarithm tables (and slide rules) to do computations. Believe it or not.

Thorgott

I don't know what computers do, but one way this could be done by hand is by using something like the Binomial series

BigSocks

17:09

but here we have to have $f(x) \neq 0$ and $g(x) \neq 0$. I guess they can be zero for some $x$? Maybe if one was zero for all the $x$ such that the other was not...

Ted Shifrin

So you would do your problem by taking the log and using the log table to find log and multiply and then "unlog."

Thorgott

@BigSocks you're on the right track

Ted Shifrin

@Thor: I always thought calculators/computers used Taylor series to do trig, exponentials, etc. They don't.

Thorgott

yeah, I remember we had this conversation for trig functions before

Ted Shifrin

Damn, you're getting old :)

Thorgott

17:11

lol

BigSocks

So the bump function is zero outside of what I am guessing is an interval of radius $1$, but I can't remember. We just make it so the other function is zero there and not zero everywhere else. like a $bump^{op}$ type of guy

Ted Shifrin

Would that we had bump functions in the analytic/algebraic category. It would make life so simple. :D

DiscoLemonade

Just spent 5m reading. A) I really suck at math above a 1st year uni level, B) I wish I was working in the 70s where you'd have to be intelligent to get stuff done like this. I wish there were more constraints in engineering.

Ted Shifrin

I remember the 70s :)

DiscoLemonade

I was born in 85, but I repair some equipment that was build in early 90s. They have schematics you are supposed to interpret yourself, and they give a theory of operation, and then they say "go". Now each section of the manual is hand-holding, and you can't work on a third of the things you used to be able to.

Ted Shifrin

17:22

Think about car engines. It's all computerized now. My first car (71 Saab) I could take apart the automatic choke, clean it out, put it back together. I could adjust the carburetor. Now it's all a big black box.

DiscoLemonade

Yep.

Wait, a guy/gal that knows math AND can fix a car?

A colleague of mine has a B-I-L who is a math prof, but needed help to fix his bicycle... Simple turning of screws/wrenches.

Ted Shifrin

Well, never could seriously fix, no. But I dabbled. Even the morning before one of my qualifying exams, I remember.

Thorgott

5

Q: Size of a linear image of a cube in $\mathbb{Z}^d$

Suppose that we have an element $v = (v_1, \dots, v_d) \in \mathbb{Z}^d$ such that $\gcd(v_1, \dots, v_d) = 1$. Then $v$ is contained in some base of $\mathbb{Z}^d$ (seen as a free-abelian group or a free module over $\mathbb{Z}$). In particular, there exists a regular integer matrix $A \in \math...

linear-algebra group-theory modules abelian-groups

this is interesting

DiscoLemonade

Qualifier exams... "Come spend a year of your life at our institution and THEN we'll tell you if you're good enough"

BigSocks

Ok what if you had like a little "bump donut" around $(-2,-1) \cup (1,2)$ that was also somehow smooth. that times the bump function would surely be $0$ everywhere right?

Mary Star

17:32

@Fargle We have the gliede reflection $\kappa \begin{pmatrix}x\\ y\end{pmatrix}= \begin{pmatrix}x\\ -y+2\end{pmatrix}+\begin{pmatrix}2 \\ 0\end{pmatrix}=\begin{pmatrix}x+2\\ -y+2\end{pmatrix}$ and the rotation $\delta \begin{pmatrix} x\\ y\end{pmatrix}=\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix} x -1\\ y-1\end{pmatrix}+\begin{pmatrix} 1\\ 1\end{pmatrix}$. Both have as fixed line the $y=1$.

To check what we have do we take an arbitrary point and see what the image of the composition is?

Thorgott

@BigSocks yes

Alessandro Codenotti

@TedShifrin he wouldn't remember if he were getting old

Thorgott

more generally, you want disjoint supports

BigSocks

right yeah. that I think is the key. But for some reason I think you also need it to be compact, so it couldn't just be something like "1-(the bump function)" on the entire $\Bbb R$

Thorgott

nothing to do with compactness

the issue with 1-bump is that there are points at which 1-bump and bump both are non-zero

BigSocks

17:42

yeah that's true. basically everything at the slanty bits I guess

speaking very formally of course

BigSocks

17:58

ok so taking a subring of the field of rational functions $k(x,y)$ that is generated by $k, x, $ and $y/x^i$ for $i \in \Bbb Z_{\geq 0}$, how do I see that the ideal of the 2 latter guys is actually just the ideal generated by $x$?

Like I can't divide by $x$ so how do I build any of the $y/x^i$?

furthermore how do I even get $y$. Seems like a hoax to me

Astyx

What do you mean "the ideal of the 2 latter guys"

in $k[x, y/x^i]$?

BigSocks

$(x, y/x^i)$

Karl Kroningfeld

Essentially this, I think: mathoverflow.net/questions/82052/… Swap x and y

Astyx

That's all of $k[x, y/x^i]$

BigSocks

@Astyx ok yes

but somehow there is a claim saying that ideal is the same as just $(x)$

maybe I am misunderstanding

Karl Kroningfeld

18:05

Yeah

BigSocks

I'll write out exactly what it says

Karl Kroningfeld

@Astyx How?

Astyx

I'm tempted to say by definition

Karl Kroningfeld

You're saying the ideal = k[x,y/x^i]?

Astyx

yes

Karl Kroningfeld

18:07

No way

Astyx

Well it depends in which ambient space you're taking the ideal

Which is what I was asking in the first place

Karl Kroningfeld

It's an ideal of a ring.

The only other thing is a field, when you get the whole field.

Astyx

If it's in $k(x,y)$ that's the whole space, yes

Which was my first question

BigSocks

Let $R$ denote the subring of the field $k(x,y)$ of rational functions in two variables generated by $k,x,$ and$y/x^i$, for $i \in \Bbb Z_{\geq 0}$. Let $M_0$ denote the ideal $(x, y/x^i, i \in \Bbb Z_{\geq 0})$. Clearly, $M_0 = (x)$, and it is easy to check that $M_0$ is maximal. Moreover, $\cap^\infty_{i=1} M^i_0 \supset (y/x^i , i \in \Bbb Z_{\geq 0}$). Let $A := R_{M_0}$, and denote by $M = (x)$ its maximal ideal.

Astyx

If it's in $k[x, y/x^i]$, again it's the whole ring

Karl Kroningfeld

18:09

c'mon

dc3rd

hold on @Thorgott and @TedShifrin, I was under the impression Taylor Series are what's used to approximate the trig functions on calculators.....

Astyx

@BigSocks You're missing an (

BigSocks

Thank you

He puts the $i \in \Bbb Z_{\geq 0}$ in the parenthesis, so I left it there, but it doesn't seem to actually mean that the natural numbers are generators of this ideal and are also somehow generated by $x$

that would be an even bigger sell idk

Astyx

Is this in Hartshorne?

BigSocks

nah an errata from Lorenzini

alpha.math.uga.edu/~lorenz/corrections.html

Karl Kroningfeld

18:12

Implicitly in Hartshorne (see the link I gave above).

BigSocks

pages 69-70

@KarlKroningfeld and yeah this is true

I looked at that but am waiting for the muse to speak

Karl Kroningfeld

Part of what you want to show is that $y\in (x)$, right?

BigSocks

Yea

Karl Kroningfeld

So you need $xf=y$ with $f$ an element of the ring.

BigSocks

Aha. And $f = y/x$ because it’s in the ring

Mmmmm

Karl Kroningfeld

18:17

Here's something cool that Hartshorne mentions in his chapter on cohomology that is actually more elementary. First, background: if $A$ is a Noetherian ring, then for any injective $A$-module $I$ and any $x\in A$, the homomorphism $I\to I_x$ is surjective. Here is an example showing that you need the hypothesis $A$ is Noetherian: let $A=k[x_0,x_1,\dots,x_n,\dots]/(x_0x_1,x_0x_2^2,\dots,x_0x_n^n,\dots)$. If $I$ is any injective $A$-module that contains $A$, then $I\to I_{x_0}$ is not surjective.

Astyx

I was talking about Bigsocks' way, but I got why it was interesting

BigSocks

Wait what about my way

@KarlKroningfeld gunna have to stare at this more. Not immediately clear to me why or how it’s related

But thanks nonetheless

DiscoLemonade

Ok, is there a way for my mobile browser to resolve Tex $ formatting? It is getting tough.

Karl Kroningfeld

@BigSocks Unrelated

BigSocks

Ooo ok got worried for a second

Karl Kroningfeld

18:23

@BigSocks yeah, you can show that any generator of the ideal in question belongs to (x) in a similar way

user2103480

@dc3rd no that's mostly way too slow

Astyx

@KarlKroningfeld Why is it not surjective?

Karl Kroningfeld

@Astyx Because it got in a bad mood.

It's kinda counterintuitive.

But, one can show that $1/x_0$ does not belong to $I$.

(If I'm thinking about it right...)

dc3rd

@user2103480 then what procedure is used?

robjohn

@DiscoLemonade Did you look at the link given in the room description for LaTeX in chat?

Astyx

18:34

Dinnertime for me

Seeya

Hmm I'm going to have to think about this more

Seeya

Karl Kroningfeld

Seeya, my suggestion works though I was being imprecise (I meant 1/x_0 is not in the image of $I$)

robjohn

@dc3rd often the CORDIC algorithm is used for the transcendental functions on a calculator (or they were years ago)

Semiclassical

i'm having periodic internet outages while doing remote teaching work and it's driving me up a wall

dc3rd

I was just perusing the Wiki entry on the CORDIC algorithm, seems interesting

BigSocks

@KarlKroningfeld sure ok, that makes sense. Another question about this same thing is why do you get that infinite intersection to be nontrivial? There seems to be no $x^n$ for any $n \in \Bbb N$. what would you multiply to get any of the $y/x^i$, say $y/x$

robjohn

18:39

@dc3rd I used it for computing trig functions and inverse trig functions for QuickDraw GX (MacOS). Some of the code survives in the current Mac OS and some was ported to early Android code (I don't know if it is still used there, they may use an FPU now).

(and who knows; the FPU might use the CORDIC algorithm internally)

Karl Kroningfeld

@BigSocks Two things: 1. What do you mean when you say infinite intersection? 2. What do you get when you multiply $x$ with itself in $k[x,y/x^i]$?

dc3rd

@robjohn what is an FPU? and what kind of algorithm is in use now for calculators ?

robjohn

@dc3rd Floating Point Unit

@dc3rd If they don't use an FPU, it is most likely still the CORDIC algorithm

dc3rd

Ah...I do know gloating point numbers.

I mean floating....

what would a gloating point number look like I wonder......

BigSocks

@KarlKroningfeld For 1. I guess if you take a finite one, say $(x) \cap (x^2)$ you just get $(x^2)$, and that has all the $y/x^i$ because you can get $y/x$ by multiplying by $y/x^{2}$ to get $y$ and by large powers of $x$ in the denominator to get $y/x$. This I guess works for any finite case, but yeah I don't know what you get for the infinite one...

Sha Vuklia

18:47

Oh guys btw

it's not true that every ideal of k[x1,...,xn] is generated by =<n elements

BigSocks

@KarlKroningfeld and for 2. I suppose it's $x^2$

Sha Vuklia

so @Mike was right that there are indeed alg var that need more generators

BigSocks

A local noetherian domain I guess has maximal ideal that cannot be written down with $n$ or fewer generators @ShaVuklia

Sha Vuklia

what's n here then?

the dimension?

BigSocks

yea

Thorgott

18:50

tfw my TA just criticized me for calling the monomial basis the easiest basis for a space of polynomials, cause that's a subjective claim

Karl Kroningfeld

@BigSocks It's still unclear what you're intersecting. What are you trying to do with that intersection?

@BigSocks Righ', iterating you show that all $x^n$ belong to the ring.

BigSocks

the powers of the maximal ideal $M$, $M^i$ for $i$ natural. And what the author wants to do is show that the intersection is not trivial whenever you're working in a local domain with principal maximal ideal in order to show that it is not sufficient that all maximal ideals be principal to show that all ideals in general are principal

@KarlKroningfeld

@Thorgott math.stackexchange.com/questions/546152/…

show him this question- the accepted answer agrees with you

Karl Kroningfeld

@BigSocks Ah, nice.

Thorgott

MSE is the only true academic source

BigSocks

yep it's full of only facts which, by definition, could not possibly care about your TA's feelings

@KarlKroningfeld yeah but that works at each finite stage. idk why it would hold in the limit though

don't say induction, please, my heart won't be able to take it

Karl Kroningfeld

18:57

@BigSocks You have a family of sets $S_i$ and $z\in S_i$ for all $i$. Perhaps, write out what $\bigcap_i S_i$ is in set notation.

Thorgott

logicians owned

BigSocks

oh wait, but I can't get $x$ in $(x^2)$ can I

Karl Kroningfeld

That's not a problem.

BigSocks

@Thorgott sobbing

Semiclassical

@dc3rd the number of votes the last president received?

BigSocks

18:58

@KarlKroningfeld bc at each step of the way I keep all the $y/x^i$ basically huh...

yeah that's gotta be it

dc3rd

@Semiclassical 🤣

Karl Kroningfeld

Or, it is what the intersection is... :P

BigSocks

man that one was kinda clear. Thanks for the patience @KarlKroningfeld

dc3rd

@Semiclassical "Big number of votes, so many votes it is impossible to count them all....we're talking huge"........something he would say or along those lines....

BigSocks

@KarlKroningfeld lol, I'm not owned, I'm not owned

Karl Kroningfeld

19:00

It shows that you don't have to represent an element of the intersection in the same way in each set.

Semiclassical

i guess the real 'gloating point number' is what he got in 2016

BigSocks

@KarlKroningfeld right bc they come from the products of different $x^n$ at each step I guess

Karl Kroningfeld

That might help explain why taking the intersection of a chain of sets and then taking its image using a function does not give the intersection of the images in general.

Ted Shifrin

@dc3rd I had been too, until we had a fascinating lecture in our Math Club at UGA fifteen years ago or so. Read up on the CORDIC algorithm.

Mike Miller

There are a lot of lies like that

I dislike lying of that form

dc3rd

19:08

Yes @TedShifrin, @robjohn was enlightening me on FPU

Ted Shifrin

Hmm, what's FPU?

robjohn

@TedShifrin: I mentioned the CORDIC algorithm to dc3rd and the Floating Point Unit

Ted Shifrin

Oh. floating point. duh.

Karl Kroningfeld

It is all a lie... This is a hologram... @MikeMiller

Ted Shifrin

It's amazing how I got through most of my math career not having been properly informed on this point. Very annoying. But I've proselytized ever since.

(Damn, that word is hard to spell.)

robjohn

19:12

@TedShifrin and easy to mistake for another...

Ted Shifrin

You're too witty for me this morning, @robjohn.

Balarka Sen

I read it as "prosthetized" on first glance and imagined a cyberpunk Ted

robjohn

@TedShifrin yeah, I'd really hate to be accused of prosecuting when only proselytizing... ;-p

Balarka Sen

Advent of calculators may be called prosthetization of mathematics

Ted Shifrin

Next, Balarka will be starting a prosthetization ring.

robjohn

19:16

@TedShifrin Most likely without identity

Semiclassical

my brain automatically starts playing either Blue Monday or Sweet Dreams whenever I hear the word cyberpunk

Ted Shifrin

I would assume it's non-associative, in fact.

robjohn

The idea is to replace the original with a prosthetic, so yeah, non-associative.

robjohn

19:29

@TedShifrin: amWhy got me thinking about holiday decorations for the room...

Maybe not

Ted Shifrin

LOL, talk about "easy to mistake"

robjohn

It's a very unfortunate acronym

Ted Shifrin

Even an acrid acronym.

copper.hat

19:45

vd does not seem to be the usual term in the usa. was it at some stage?

Ted Shifrin

Sure it is. For both things.

user2103480

@Thorgott postmodern reacts only

@Thorgott wtf he's a logician

copper.hat

20:02

when i grew up std was a term for direct dialing...

Ted Shifrin

and it's still an abbreviation for "standard."

Semiclassical

one better be careful with the phrase "standard diagnosis", then

Astyx

or "standard injection"

Semiclassical

"It's the community standard!"

Ted Shifrin

canonical, @Astyx, not "standard." :)

Thorgott

20:13

natural injection

functorial, even

Marcel

20:31

Is there some kind of construction that is equivalent to prime multiplication in that it is easy to compute n = p*q and also easy to check whether p | n, but where the equivalent of finding the greatest common denominator is hard? (i.e. there is no efficient algorithm for gcd(n_1, n_2))? Or is the existence of such a function a consequence of divisibility checking (or its equivalent) having an efficient algorithm?

dc3rd

20:56

Quick Linear algebra question: If I have a function: $T: V \to W$ and I have the basis for a domain. This doesn't necessarily mean that my map $T$ is linear?

Transcript for

$Mathematics$ Mathematics