Mathematics - 2017-07-24 (page 2 of 4)

Typhon

4:00 AM

yup

well.... b|a showing that x'|x and x|x' is trivial

for me anyways

but yeah what you just said

@AkivaWeinberger anyways, because x' and x divide each other we can say the following

Akiva Weinberger

So $\frac x{x'}$ is a unit (an element of a ring, in this case $\Bbb Z[x]$, whose inverse is also in the ring)

Typhon

oh

yeah, lol

suppose that we have the set of equivalency classes for mod x and the equivalency classes for mod x'. The conditions I just prescribed are the only instances where the two modular arithmeticae are one and the same. In other words, they are the only quadratic rings where it makes sense to say "Z[x]/x"

Akiva Weinberger

Why do we want the equivalence classes for x to be the same as for x'?

Typhon

@AkivaWeinberger of course, I can pick specific elements to do stuff, but I was curious about that particular one

@AkivaWeinberger Let Z[x] be a quadratic ring where x^2 + ax + b = 0. Let Z[x]/x be the equivalency classes for all elements of Z[x] mod x.

note: I rarely ever distinguish the two solutions except for the setup of the conjugate properties

Akiva Weinberger

That seems like it would be isomorphic to $\Bbb Z/N(x)$, actually

$\Bbb Z_b$

because $mx+n\equiv n\equiv(n\text{ mod }b)\pmod x$

Typhon

4:13 AM

let's not go too deep

I'll see what occurs.

though I will note that when b divides a all conjugates are equivalent mod x

i don't even know the term isomorphic

beyond some thing meaning "set equivalent involving some handwavy transformation"

i know in geometry it refers to surfaces being stretched or moved versions of each other

Akiva Weinberger

@Typhon No that's "homeomorphic"

Typhon

oh

see?

I don't know that stuff yet!

XD

regardless this is for an introductory essay in it

Akiva Weinberger

Homeomorphisms are not to be confused with homomorphisms, which are group theory maps that preserve the operation (and don't need to be bijective)

Typhon

basically, I've challenged myself to write a series of little mini-books on number theory in a minecraft server

their books are basically 50 pages long with 300 characters a page

Akiva Weinberger

Yeah topology has a bunch of weird things like that: homeomorphism, homotopy equivalence, and isotopy equivalence

@Typhon Cool

Typhon

4:18 AM

apparently the people on the server love buying copies in the player store

(no clue if they understand it though. Most are between 12-16 years old)

Akiva Weinberger

Huh weird

Typhon

but hey? maybe they'll grow to like number theory

well tbf anyone who understand polynomials can learn this

it's just a matter of mathematical maturity and logical thinking

not to trivialize it of course

I just mean that anyone can read about it with decent understanding.

granted who knows

maybe written books factor into some kind of OP armor recipe I'm not aware of. XD

and the amount of text written boosts the stats

so they buy it to use for making stuff.

shrugs

idc really. It is a fun challenge.

especially since the books do not allow you to move the cursor back on a page.

if you need to fix a typo on a page... you must rewrite that page

anyways

i have to go

to the toilet

XD

Akiva Weinberger

4:34 AM

@SinglePackAbs The parametric equation (sin(t)+2/3*sin(3t),cos(t)-2/3*cos(3t)) seems to be fairly close

Secret

Ok, here's a weird way to think about matrices:

Picture a world where people A and B are in. To each of them, they both see a normal looking, square grid. However, when they look at the other person's grid, they will see it squashed or distorted in the way given by the matrix

Therefore, we can then have a scenario where when A is walking north in A's view, to B, he will be walking east west

and likewise if B walks to the north in B's view, then to A he saw B is walking in the (insert suitable direction)

So in the end of the day, just like how we tend to understood linear maps by how they transforms the uni square (or more abstractly a basis set), into some stretched shape, the invariant property that is the generalisation of relative direction of two vectors is how the 2 views are distorted relative to each other, and that must be constant

So basically, if real life is like vector space and their dual space, then each of us have space relatively distorted when we look at other people's but looks normal when we look at ourselves. Thus it is relativity but beefed up such that you don't need to be traveling at some velocity to see other things distorted

Under this view, the adjoint thus describe how A will look like in B's view

(I will make this more rigorous later. This should agree to my guess about adjoints back in 2 years ago)

skullpatrol

4:51 AM

Thanks for sharing.

Secret

If all of this holds, then change of basis of a matrix can be done geometrically, by distorting B's grid along with A's such that B's become the usual square grid and A's will be effectively multiplied by the inverse

Single Pack Abs

@AkivaWeinberger Thank you so much. Great!!!

Secret

This will be analogous to changing reference frames in physics

Typhon

5:24 AM

@AkivaWeinberger I'm back, but I kind of feel like going now. I'm going to head off for the night. Thanks for helping with that. It was a lot of fun. :-)

I like doing number theory with you.

skullpatrol

cya

Typhon

@AkivaWeinberger As a challenge, prove that there exists either a prime or a semiprime between any two consecutive squares. Apparently it has been proven before. Try to do it without external resources. Just... think about it.

skullpatrol

IBL?

Typhon

@skullpatrol dont know that acronym?

skullpatrol

Inquiry Based Learning

Typhon

5:27 AM

nah

more like...

"you've been posting fun riddles in chat. Here is one from me for a change."

skullpatrol

I was referring to the "Try to do it without external resources. Just... think about it." part :-)

Typhon

aah

ok

skullpatrol

That's what IBL is all about.

Typhon

oh ok

skullpatrol

It should be independent IBL.

Typhon

5:33 AM

ok

skullpatrol

Also known as The Moore Method.

Typhon

sure that's the moore method?

I thought the moore method was when you taught by having people prove a series of propositions?

skullpatrol

Yup, no textbooks, no asking for help etc.

Typhon

yeah

i had that when taking my basic abstract math class

except cause I'm an honors I have to do so many of these special projects (or take extra classes)

and basically I proved stuff about advanced stuff while also doing stuff in Z[sqrt{3}]

tbh, it is kind of funny when I say "i dont know what that is" and then they describe it and I'm like "oh, yeah I knew what that was. We never used any terminology."

Akiva Weinberger

5:48 AM

Yeah I find it's nice if you somehow have a chance to play with the ideas before learning the terminology for them

Like, it feels nice. You know the motivation for each definition and why it's defined like that.

Daminark

Hey guys

Typhon

@AkivaWeinberger no actually it is because the professor only had me proving basic stuff in the extra project, like.... stuff about the golden ratio

I decided on a whim that for the final paper I would prove stuff about the quadratics in general

the quadratic rings are 100% my conception

i figured "well, if im going to say something meaningful, why not say it about all of these appendings"

tbh, I'm not sure the professor expected that.

@AkivaWeinberger technically the uniqueness of the coefficients of quadratic integers is by this point a well established theorem. so technically you can also just call upon that.

Akiva Weinberger

6:19 AM

Ah fair

A poem on the Loch Ness monster:

> Ness Lake,
has snake

2

skullpatrol

Hi @Daminark

Sorry if I cutoff your poem @AkivaWeinberger

Akiva Weinberger

No I finished it

Daminark

How's it going?

skullpatrol

How do you get that indentation?

in chat @AkivaWeinberger

Akiva Weinberger

Start a line with "> "

skullpatrol

6:32 AM

>this

Daminark

4chan intensifies

Akiva Weinberger

With a space after the >

> this

Yeah

> thanks :-)

6:59 AM

Hi chat

Secret

> that

$ↄ^{ↄ}ↄ^{ↄ}_{ↄ_ↄ}$

Secret

7:29 AM

All open discussions are closed

There is no such thing as an open discussion

what about a clopen discussion?

Well since it is both open and closed, it is also a closed discussion

what about a neither discussion?

I have no idea

Actually, whether discussion is topological is an open question...

Alessandro Codenotti

8:29 AM

hi @Astyx

Astyx

Hi

The Raiders of Las Vegas

> Hi

Secret

9:06 AM

[Random]

Nonlinear integrals: Generalising integral operators such that they lose linearity. To be investigated later...

Mar 23 '15 at 14:15, by Mats Granvik

In engineering there is a way to handle many dimensions. It is called "kvotekvationer" in Swedish.
http://web.abo.fi/fak/tkf/vt/Eng/education_VTG.htm

Cannot read (insert language)

Tobias Kildetoft

@Secret It is Swedish, as indicated

(not that there seems to be a link to the actual course notes, just a detailed list of contents)

skullpatrol

How many students are going to be in your algebra course? @TobiasKildetoft

Tobias Kildetoft

9:35 AM

@skullpatrol I don't know yet, but my guess would be something like 50-70

skullpatrol

I see.

Secret

10:16 AM

Random fact: There are no analogues of zero divisors in the + substructure of an algebra structure (unless an additive absorber exists, which then the treatment will be similar to the case for zero divisors). This is because a zero sum $a+b=0, a,b\neq 0$ is really saying: $a,b$ are additive inverses of each other

However, it is unsure if $a+b=1$ are any special. One example of this is $2+3=1 \mod 4$

37

Q: What is Abstract Algebra essentially?

In the most basic sense, what is abstract algebra about? Wolfram Mathworld has the following definition: "Abstract algebra is the set of advanced topics of algebra that deal with abstract algebraic structures rather than the usual number systems. The most important of these structures are groups...

> The properties of how elements interact under operations is a more general, abstract notion of what we do with numbers when we do algebra.

and this is why it is quite visual to me

Anything that reminds me of space is visual

Having said that, I suck at topology

Danu

@AkivaWeinberger lol

Secret

11:34 AM

"Flow diagram" illustration of the proof many posts up

yesterday, by Alessandro Codenotti

The book I'm reading defines "An algebra $A\neq 0$ is said to be a division algebra if for all $a,b\in A$ with $a\neq 0$ the two equations $ax=b$ and $ay=b$ have unique solutions in $A$" (here algebra means algebra over $\Bbb R$ even though it shouldn't make a difference in this definition)

It appears Alessandro's question have accidentally opened new territory in the Repository of Unnatural Algebraic Structures

Johan Larsson

Is there a name for a shape like this?

The section of the ring with parallel lines

Dattier

11:55 AM

Hi, $$\text{Give an approximate value to } 10 ^{-6} \text{ of }\sin(2^{1000}).$$

Astyx

Wild guess : -0.159202

Tim Davids

hi all

opisthofulax

you mean the section of the toroid

?

Secret

Visual calculus

Visual calculus by Mamikon Mnatsakanian (known as Mamikon) is an approach to solving a variety of integral calculus problems. Many problems that would otherwise seem quite difficult yield to the method with hardly a line of calculation, often reminiscent of what Martin Gardner calls "aha! solutions" or Roger Nelsen a proof without words. == Description == Mamikon devised his method in 1959 while an undergraduate, first applying it to a well-known geometry problem: Find the area of a ring (annulus), given the length of a chord tangent to the inner circumference. (Perhaps surprisingly, no additional...

this one so strongly reminds of those "vectorizoids" I dreamt about some days ago in terms of concept (but obviously not in terms of appearance nor working principle)

Waiting

@robjohn Hey! How is it going?

Secret

12:10 PM

I wonder, if an area preserving transformation can be found so that the area corresponds to complicated integrands can be morphed into easier shapes, and then have the area computed. That might help compute a lot of definite integrals...

Tim Davids

12:22 PM

Can anyone see why $$\int_{-\infty}^{\infty}A^{\dagger}_{y}A_{y}dy = 1$$ where
$$\hat{A}_{y} = \int_{-\infty}^{\infty}\frac{e^{-(x'-y)^2/(4V)}}{(2\pi V)^{\frac{1}{4}}} |x ' \rangle \langle x' | dx'$$

Akiva Weinberger

@Secret Oh I love that

Mamikon's great

Secret

hmm, how to even compute the adjoint of this integral operator (not seeing anything complex here...)

Tim Davids

@Secret Yeah I don't think there is anything complex...quite tricky.

0celoñe7

@TimDavids So just multiply, what do you get?

You will get a delta $\delta(x'-x'')$

Tim Davids

12:37 PM

@0celoñe7 Huh, how would you get that?? Let me try writing the working of what I have...

Emilio Pisanty

@TimDavids How is that tricky? If there's nothing complex, then everything is real, and the conjugate of a real number is itself.

@TimDavids because $\langle x'|x''\rangle = \delta(x'-x'')$.

0celoñe7

@TimDavids $\langle x'|x''\rangle$

Secret

The integral operator does seemed naively like a "continuum sized diagonal matrix" which means its transpose is probably going to be the same as itself... Not sure how to justify this heuristic, though, given the complications of unbounded operators...

Emilio Pisanty

@Secret There's absolutely no mystery here.

Tim Davids

@0celoñe7 Let me try to write the working.

Emilio Pisanty

12:42 PM

$\hat A_y$ is the operator that takes $f(x)$ and returns $\hat A_y f$ where $(\hat A_y f)(x) ={(2\pi V)^{-1/4}}e^{-(x-y)^2/(4V)}f(x)$.

it's essentially trivial to show that it's self-adjoint and bounded

Secret

Ah yes, that gaussian function keep it bound

Johan Larsson

@opisthofulax yeah

GFauxPas

good morning mathematician friends

Semiclassical

Memo to self: If I ever have to bike 2.5 miles to catch a 7am bus, make sure to bring water

GFauxPas

good idea

Tim Davids

12:50 PM

@EmilioPisanty I get $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{‌(x'-y)^2/(4V)}}{(2 \pi V)^{\frac{1}{4}}}\frac{e^{(x''-y)^2/(4V)}}{(2 \pi V)^{\frac{1}{4}}}\delta(x'-x'')|x' \rangle \langle x'' | dx' dx'' dy$$ is that right so far?

Semiclassical

It's 7:48 now, and I still have another 15 minutes before I'll be able to get water

Emilio Pisanty

@TimDavids yes

Semiclassical

Welcome to our side of the chat :) @EmilioPisanty

0celoñe7

@Semiclassical What?

@Semiclassical Ah.

Semiclassical

(Probably superfluous but I haven't seen you over here before)

0celoñe7

12:52 PM

@Semiclassical No welcome for me :(

Astyx

The dark side

Semiclassical

Lol, welcome back @0celoñe7

Emilio Pisanty

@TimDavids Seriously, this is a trivial calculation. All it takes is put the symbols in and then crank the symbol-manipulation handle. But it's not going to do itself ─ you actually do need to crank the handle.

@Semiclassical hiya

Tim Davids

@EmilioPisanty Okay, I don't have a great deal of experience working with dirac delta functions but is the idea to use $$\int_{a}^{b}f(x')\delta(x-x')dx' = f(x)?$$

0celoñe7

What else could you possibly use?

Emilio Pisanty

12:59 PM

^ that.

0celoñe7

@TimDavids It is generally preferable to actually try to use it for a few minutes instead of asking people if you should use it.

Semiclassical

Here's an old dilemma for delta functions, whose resolution I'm blanking on

Vrouvrou

hello, please what is a cut-off function ?

0celoñe7

@Vrouvrou Something that is =1 on some set and =0 outside of a slighly larger one.

Emilio Pisanty

@0celoñe7 @TimDavids Otherwise known as: it's better to ask for forgiveness than for permission

Semiclassical

1:01 PM

If you integrate $\delta(x)$ over the real line, you get 1 (duh).

0celoñe7

cringe

Semiclassical

What about over $[0,\infty)$?

0celoñe7

That's not a dilemma, that's physicists being, well, physicists :P

Semiclassical

Lol

0celoñe7

@Semiclassical If you view the delta as a measure, then that should be 1. Is this a contradiction?

Vrouvrou

1:04 PM

@0celoñe7 thank you ^_^

Semiclassical

it is if you think of the delta function as a limit of symmetric functions

Emilio Pisanty

@Semiclassical that's ill-defined. You can specifically require a symmetric delta function, in which case $\int_0^\infty \delta(x)\mathrm dx=\frac12$, but unless you make that explicit then it's undefined.

0celoñe7

@Semiclassical That's because that limit is in the topology of $\mathscr D'$ or $\mathscr S'$ (duals of the $C_0^\infty$ functiosn or Schwarz functions).

GFauxPas

Semi I have a book that offers a definition of dirac delta by the limit of a sequence of sequences and he leaves the measure definition in the appendix, want me to go get it?

Semiclassical

Yeah. Which it really points to is the extent I tend to be sloppy about what a delta funcltion really is

Nah.

0celoñe7

1:06 PM

The resulting integral is not defined for such a definition.

GFauxPas

he says the measure definition is more rigorous but the limit of a sequence of sequences works

Semiclassical

Right.

And if the two ever disagree it's because the limit way isn't actually well-defined

Secret

I
Never
Understood
Limits

0celoñe7

Actually the limit is in the weak* topology.

@Semiclassical Oh, it's perfectly well defined for certain situations.

Tim Davids

So then you get $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{e^{(x''-y)^2/2V}}{(2 \pi V)^{1/2}}dy|x'' \rangle \langle x'' | dx'' = 1?$$

GFauxPas

1:08 PM

I think he uses limit in distributions sense where the sequences in the sequence have compact support

0celoñe7

It's just that the function $\chi_{[0,\infty)}$ is not a smooth function.

GFauxPas

something like that

0celoñe7

@GFauxPas Exactly.

Semiclassical

@0celoñe7 oh, sure. I meant that it doesn't work in every scenario

GFauxPas

do you have a counter example on hand semi?

Semiclassical

1:11 PM

The limit approach doesn't, I mean

GFauxPas

in my risk class our definition was something like

Emilio Pisanty

@TimDavids that's still correct

0celoñe7

@Semiclassical If you care about such things, the pages 23-25 of "Weakly Differentiable Functions" by Ziemer explains in what sense the convergence you're talking about is.

Semiclassical

I think we just argued the function @0celoñe7 stated is an example where the limit approach gives different examples depending on what approximation of identity you choose

Tim Davids

@EmilioPisanty "Still correct" ...Thanks

0celoñe7

1:14 PM

@Semiclassical I think the limit definition is really only there to make sense of numerical approximations and the like. The best definition is of course $\langle \delta,f\rangle=f(0)$.

Semiclassical

Whereas the measure approach always gives 1 since 0 is in the support of $\chi_{[0,\infty)}$

0celoñe7

That's a perfectly valid definition.

Emilio Pisanty

@TimDavids I'm not sure why you need someone to hold your hand, but I guess if that's all you need then I can do that.

0celoñe7

@Semiclassical The two viewpoints are (i) measure theory (ii) operator on certain functions.

abenthy

Why is $\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$ not a linear algebraic group?

Secret

1:15 PM

When s.harp and fargle get on , I need to discuss something with them. It turns out the zero divisors in $\Bbb{Z}/n$ for composite $n$ actually follows a pattern

$\textbf{Conjecture}:$ The zero divisors of $\Bbb{Z}/n$ are precisely those which is resulted when $n$ is divided by those primes $p$ such that $p|n$

0celoñe7

The sequence definition is tertiary and comes from trying to "explain" it to engineers.

Tim Davids

@EmilioPisanty That's quite nice of you, thanks.

Tobias Kildetoft

@Secret That is not correct unless you allow more primes

@Secret Actually, not even then.

Emilio Pisanty

@TimDavids No, seriously, this is as routine as it gets. It does take some time to learn how the symbol-pushing goes, but it's ultimately just symbol-pushing. All it takes is not being shy when it comes to cranking the handle.

Secret

But e.g. for $\Bbb{Z}/6$, the zero divisors are $2 = 6/3, 3 = 6/2$?

Tobias Kildetoft

1:17 PM

@Secret $4$ is also a zero-divisor

Secret

ah, nvm then...

Alessandro Codenotti

every multiple of a zero divisor is a zero divisor

Tobias Kildetoft

The zero-divisors are precisely the non-invertible elements in this case

Tim Davids

@EmilioPisanty Yeah I know I'm not joking. I also know it is simple stuff for most everyone here. But yeah I get your point, need to learn to do the donkey work...

Alessandro Codenotti

As in every finite ring with unity

Secret

1:22 PM

then how do I explain this weird pattern when I multiplied the zero divisors to all elements in the ring:

e.g.
2*{0,1,2,3}={0,2,0,2}

2*{0,1,2,3,4,5} = {0,2,4,0,2,4}
3*{0,1,2,3,4,5} = {0,3,0,3,0,3}

2*{0,1,2,3,4,5,6,7}={0,2,4,6,0,2,4,6}
4*{0,1,2,3,4,5,6,7}={0,4,0,4,0,4,0,4}
6*{0,1,2,3,4,5,6,7}={0,6,4,2,0,6,4,2}

What are these repeating periods I am seeing here whenever the whole ring is being multiplied by the zero divisors?

Tobias Kildetoft

@Secret I don't see anything to explain there

@AlessandroCodenotti Hmm, is it obvious that all non-invertible elements are zero divisors in any finite ring with unity?

user84215

if $X=\cup_\alpha A_\alpha$ and $f:X \to Y$ such that $f|A_\alpha$ is continuous for each $\alpha$, then $f$ is continuous?

Tobias Kildetoft

@aminliverpool Only if the subsets are open

Alessandro Codenotti

pick an element $a$ and consider the map $x\mapsto ax$, if this is injective then it's also surjective since the ring is finite and there is an $x$ with $ax=1$, if it's not injective then there are distinct $b,c$ mapped to the same element and $a(b-c)=0$

user84215

Suppose $A_\alpha$ is closed for each $\alpha$.

Tobias Kildetoft

1:28 PM

@AlessandroCodenotti Ahh, of course

@aminliverpool Then definitely not in general (take any Hausdorff space and cover the space by singletons)

Semiclassical

yay water

user84215

Consider countable collection.

Tobias Kildetoft

@aminliverpool I don't see why that would help

0celoñe7

@aminliverpool You need them to agree on overlaps.

Tobias Kildetoft

@0celoñe7 That does not make sense. The function is already defined on the entire space

user84215

1:32 PM

Suppose $\{A_\alpha\}$ is a countable collection.

Secret

$2*\Bbb{Z}/4 = \text{something that looks like two sets of}\Bbb{Z}/2$

$2*\Bbb{Z}/6 = \text{something that looks like two sets of}\Bbb{Z}/3$
$3*\Bbb{Z}/6 = \text{something that looks like three sets of}\Bbb{Z}/2$

$2*\Bbb{Z}/8 = \text{something that looks like two sets of}\Bbb{Z}/4$
$3*\Bbb{Z}/8 = \text{something that looks like four sets of}\Bbb{Z}/2$
$4*\Bbb{Z}/8 = \text{something that looks like two sets of}\Bbb{Z}/4$

$3*\Bbb{Z}/9 = \text{something that looks like three sets of}\Bbb{Z}/3$
$6*\Bbb{Z}/9 = \text{something that looks like three sets of}\Bbb{Z}/3$

There's this "partition" thing going on

0celoñe7

@TobiasKildetoft Ah, well.

Tobias Kildetoft

@Secret Yes, you are partitioning the ring into cosets with respect to the ideal generated by the element

0celoñe7

@aminliverpool The gluing lemma works for a finite number of closed sets.

user84215

yes

user84215

1:33 PM

but for countable ?

Secret

so the action of multiplying the zero divisors of each ring to the whole ring generates the ideals of that ring?

Tobias Kildetoft

@Secret This is just a general thing about any (commutative) ring and any element in that ring

Alessandro Codenotti

@aminliverpool Cover $\Bbb Q$ with singletons, it has the same problem @Tobias was pointing out earlier

Semiclassical

Hmmmm

Is there a simple way in Mathematica to create a contour plot consisting only of contours which pass through specific points?

user84215

then what is $f$ ?

Tobias Kildetoft

1:37 PM

@aminliverpool Any non-continuous function will do

Semiclassical

I mean, if you tell it to do a contour plot of $f(x,y)=f(x_1,y_1)$ then the resulting level set will include the point $(x_1,y_1)$. But it may contain other disjoint sets as well.

user84215

$Q$ with which topology?

Tobias Kildetoft

@aminliverpool Order

Alessandro Codenotti

the one inherited from $\Bbb R$

user84215

Under which condition is $f$ continuous?

Secret

1:46 PM

another interesting thing is how the cosets looks e.g.

$2*\Bbb{Z}/8 = \{0,2,4,6,0,2,4,6\}$
$4*\Bbb{Z}/8 = \{0,6,4,2,0,6,4,2\}$

Barring those zeros, the sequences are the reverse of one another

and there is always a zero divisor that gives a coset which has the form {0,x,0,x,0,x,0,x...}

Tobias Kildetoft

@Secret That is because you keep picking even numbers

Felix.C

Does the residuum of a function at a given point tell us something about the series of convergence of the laurent series of that point? I have an example here where it seems to have some relation between those two parameters but at first glance I don't see the relation between them...

Secret

ah yes, sorry, Z/9 does not obey that rule

Semiclassical

@Felix.C It really shouldn't. The rate of convergence of $c+cx+cx^2+\cdots=c(1-x)^{-1}$ doesn't depend on $c$.

Secret

and I suspect $Z/121, Z/169, Z/p^2$ etc. should have even weird cosets

Semiclassical

1:50 PM

Though if you mean the Laurent series at $x=1$, then one certainly has that $-c(x-1)^{-1}$ has a residue of $-c$ at $x=1$.

Tobias Kildetoft

@Secret Sure, the unique non-trivial proper ideal of those will have index (and size) $p$

Semiclassical

@Felix.C What example did you have in mind?

Alessandro Codenotti

@aminliverpool preimage of open sets is open. The standard topology on $\Bbb Q$ has a basis of "intervals" $(a,b)\cap\Bbb Q$

user84215

Ok. I got it.

user84215

16 mins ago, by aminliverpool

Under which condition is $f$ continuous?

Semiclassical

2:02 PM

lol, this is pretty good:

dieselsweeties.com/ics/355

abenthy

2:29 PM

Can you prove that $SL(n,\mathbb{Z})$ is not of finite index in $SL(n,\mathbb{Q})$?

Tobias Kildetoft

@abenthy Yes, essentially the same way you would do it for $n=1$

arctic tern

@TobiasKildetoft not sure what you mean

SteamyRoot

If $n = 1$, don't you just have $1$ in both cases? o.O

arctic tern

a finite set of coset representatives would have finitely many primes in their denominators, and any SL(n,Q) matrix involving other primes in the denominator would not be in the group generated by the coset reps and SL(n,Z)

Secret

hmm... that's an interesting circular pattern....

I need better ideals skills to appreciate this

abenthy

2:37 PM

The usual thing I would like to do is get an exact sequence $0 \to SL(n,Z) \to SL(n,Q) \to SL(n,Q/Z) \to 0$, but $Q/Z$ has no (interesting) ring structure.

Tobias Kildetoft

@abenthy It is not a normal subgroup, so there is no way to get such a sequence

SteamyRoot

@arctictern Hmmm, my first idea was to take the cosets generated by $\operatorname{diag}(p,1/p,1, \dots, 1)\operatorname{SL}(n,\mathbb{Z})$.

arctic tern

why say "generated by" when you gave a whole coset? :P

Tobias Kildetoft

@SteamyRoot Right, I was thinking of the $GL$ version

abenthy

The quotient $SL(n,\mathbb{Q})/SL(n,\mathbb{Z})$ is not a group, right?

Tobias Kildetoft

2:40 PM

right

arctic tern

SL(n,Z) is normal in SL(n,Q) because it is normal in the supergroup diag(Q)*SL(n,Z), no?

abenthy

This fact is basically the statement that $SL(n,\mathbb{Q})$ is not an arithmetic subgroup of $SL(n,\mathbb{R})$. I would have expected this to be a common example :/

Tobias Kildetoft

@arctictern $SL(n,k)$ is close to being simple for any field, also for the rationals

Akiva Weinberger

@Secret I've seen versions of that where they colored each pixel according to its number

like on a scale from blue to red or something

Secret

They have the $\Bbb{Z}/15$ example

Secret

2:53 PM

Btw, this stuff I proved many hours earlier, There's a subtle catch

16 hours ago, by Secret

$\textbf{Revised version}$

$\textbf{Theorem}$: Let $A$ be a magma with the absorber $0$, and satisfying one of the following: $\forall a,b\in A,a\neq 0,\exists ! x: ax=b$ or $\forall a,b\in A,a\neq 0,\exists ! y: ya=b$. If $A$ contains a zero divisor, then the zero divisors can only have one sided inverses.

$\textbf{Proof}:$ Suppose we take the division rule to be $ax=b$. If $A$ has a zero divisor, that is $\exists c, ↄ\in A,c,ↄ \neq 0, cↄ=0$, then by definition, the division rule also holds for $c,ↄ$. Pick some $a \in A$ such that

It turns out you can only have those "invertible zero divisors" proved here if your structure is infinite

this is because in finite dimensions, any injective mophism is automatically surjective and vise versa

and thus attempt to invert zero divisors is the same is trying to invert a nontrivial kernal, which is forbidden in finite dimensions

(should really have checked my proofs more carefully next time to not waste 5 unnecessary hours on trying to construct a finite example)

On the flip side, I am becoming a bit more comfortable working with infinite structures now, though I need to learn more about the isomorphism theorems and etc. in order to further cut down the amount of time spent hitting walls

Tobias Kildetoft

@Secret I still think you should read at least some introductory text on abstract algebra. You keep hitting walls that would have been thoroughly examined in such a text

Secret

I am thinking about Ted's book, but I suspect there are more introductory ones that familarise with rings

though I am never good with book recommendations, let me see if MSE has good suggestions...

0celoñe7

@Secret Pinter...

Alessandro Codenotti

@Tobias since you're one of the algebra guys around here do you happen to know of classification results for finite dimensional associative division algebras over fields different from $\Bbb R$?

I can do the commutative case, but I have no idea about the general one

Tobias Kildetoft

@AlessandroCodenotti So for a finite field it is pretty trivial of course.

Alessandro Codenotti

3:01 PM

why is that?

Tobias Kildetoft

Because those are finite

hence finite fields

Alessandro Codenotti

oh, of course

SteamyRoot

Algebraically closed fields are fun as well :^)

Tobias Kildetoft

I think over something like $\mathbb{Q}$ it should be essentially the same as for the reals, but I never really studied this stuff

Alessandro Codenotti

in general I proved (modulo mistakes) that the commutative, associative, finite dimensional division algebras over a field are precisely the algebraic field extensions

Tobias Kildetoft

3:03 PM

Actually, scratch the comment on the rationals, as they are clearly way harder than the reals even in the commutative case

Alessandro Codenotti

yeah, there are a lot more algebras over $\Bbb Q$ than $\Bbb R$

Tobias Kildetoft

I think this might all be the sort of thing studied more generally as part of composition algebras, but those are as far as I recall mainly of interest when they are non-associative

But I think there is some result somewhere about which ones are associative

(on the other hand, this is probably specific to the reals)

Secret

@0celoñe7 Downloaded, luckily there's a whole book in the math.umd.edu website

Alessandro Codenotti

I see, thanks

Astyx

3:52 PM

Are there no superfields (not sure about terminology) of $\Bbb C$ ?

Tobias Kildetoft

@Astyx Sure, just not finite dimensional ones

Astyx