Mathematics - 2018-06-23 (page 2 of 3)

Abdelrhman Fawzy

11:05 AM

how he get this result?-start with bottom page-

mick

A closed thread is still visible , but answering or voting is not possible ? Right ?

Jo Be

What do you do when a calculation physically hurts you?

ÍgjøgnumMeg

@Mathein there's a line here saying that if $a, b \in \Bbb Z$ have the same p-adic expansion then $p^n\mid a-b$ for every $n$ so $a = b$; this is just because the only integer that is divisible by every power of $p$ is $0$ right? lol

MatheinBoulomenos

11:27 AM

@ÍgjøgnumMeg exactly

In ideal language, $\bigcap_{n \in \Bbb{N}} (p)^n = 0$. This is precisely the intersection of all kernels $\Bbb Z \to \Bbb Z/p^n\Bbb Z$ and by some general nonsense this intersection is the kernel of the homomorphism $\Bbb Z \to \Bbb Z_p=\varprojlim \Bbb{Z}/p^n\Bbb{Z}$

ÍgjøgnumMeg

Right, I think this is talked about in the next little section

the point was to establish an identification $\Bbb Q \hookrightarrow \Bbb Q_p$ whose restriction to $\Bbb Z$ gives $\Bbb Z \hookrightarrow \Bbb Z_p$

in fact this is on the next page lol

MatheinBoulomenos

it's really not much different from working with p-adic expansions, although that's not obvious when you see it for the first time.
But doing the elementary approach first is the right way to do it pedagogically

@ÍgjøgnumMeg lol

ÍgjøgnumMeg

Fair, thanks :)

MatheinBoulomenos

@ÍgjøgnumMeg if you want something to complement Neukirch, I can recommend the 2. chapter in Kato/Kurokawa/Saito "Number Theory 1: Fermat's Dream" that's just about 30 pages in a small book, but it's really engaging and insightful. They show how p-adics connect to some two-variable quadratic equations and quadratic reciprocity and have quite a unique style

ÍgjøgnumMeg

11:48 AM

@Mathein Nice! Thanks, I'll look it up

moteutsch

Can someone justify to me why the free group is worth studying?

ÍgjøgnumMeg

@Mathein what's the problem with defining addition in $\Bbb Z_p$ in the obvious way (adding the coefficients of the p-adic expansions)? Is it because these expansions are infinite sums so you have weird convergence issues? Neukirch hints towards "problems with defining addition/multiplication in the same way as one would when constructing the reals"

MatheinBoulomenos

@ÍgjøgnumMeg you want to have some carrying

You don't want to end up with formal power series $\Bbb{F}_p[[X]]$

ÍgjøgnumMeg

I see

this is his motivation for introducing $\varprojlim \Bbb Z/p^n \Bbb Z$ lol

err

what's the tex for that

MatheinBoulomenos

\varprojlim

@moteutsch from your profile, you seem to have a programming background, so allow me to answer a different, but closely related question first: why is the free monoid worth studying?
(In case you're not familiar, a monoid is like a group that doesn't necessarily have inverses)
The free monoid on a set is just the finite words if we take this set as an alphabet, or to put it differently all finite strings (including the empty string) where the letters are from our chosen set. For example, if you take a two element set {a,b}, then elements in the free monoid over that set look like "aba", "",…

geocalc33

12:05 PM

@MatheinBoulomenos to extend a finite group structure of functions from 2d to 3d can you rotate all the functions about an axis of rotation

MatheinBoulomenos

@geocalc33 I don't understand the question

geocalc33

So you have a family of functions and you spin them around an axis of rotation

Is this like the SO(3) group sort of

For one special rotation

Out of the many you could do

But I should add that the rotation is discrete in that it doesn't create a surface of revolution

But finitely many curves

MatheinBoulomenos

@geocalc33 sorry, I don't think I can help you

@ÍgjøgnumMeg that book by Kato et al. has the best poetry you'll find a math book, probably, lol

geocalc33

You don't understand?

MatheinBoulomenos

yeah I don't understand and I also don't understand the motivation behind this

ÍgjøgnumMeg

12:17 PM

@Mathein hahah nice, that's what I look for in a text on number theory

geocalc33

Sometimes the best motivation is curiosity @MatheinBoulomenos

MatheinBoulomenos

@ÍgjøgnumMeg you'll see what I mean when you read the comparison of reals and p-adics to the sun and the stars

ÍgjøgnumMeg

jesus hahaha

MatheinBoulomenos

I wasn't joking with "they have a unique style"

ÍgjøgnumMeg

lol

@Mathein I guess in general if you take two p-adic integers then their sum mod $p$ would not be the same as the $p$th power of their sum mod $p$, due to carrying issues? I think this is what I'm reading lol

MatheinBoulomenos

12:32 PM

mod $p$ it's alright when you just mean the 0-th coefficient. But the addition mod $p$ has consequences via carrying to the result for the next coefficient

ÍgjøgnumMeg

I see

MatheinBoulomenos

but for all $n$, adding and multiplying p-adic integers is the same as adding and multiplying in $\Bbb Z/p^n \Bbb Z$ if you just look at the first n coefficients

Hence this inverse limit stuff

ÍgjøgnumMeg

yeah and the coefficients retain their value modulo higher powers of $p$

makes sense

If I try and write down $12, 14$ in $\Bbb Z_p$ and add their $2$-adic expansions (where the coeffs are in $\Bbb F_2$) coefficient-wise then I just get $2$ lol

fair enough

MatheinBoulomenos

yeah, exactly

Shaun

1:01 PM

Hello :)

MatheinBoulomenos

Hello! @Shaun

ÍgjøgnumMeg

hi @Shaun

Shaun

I have a system of equations that need solving. I don't have any tools to do so, however. Shall I share them here?

They are as follows.

2XY+X^2+Y^2+X^2Y+Y+XY^2+X=0, for X\neq Y and X, Y\in\{J, K, L\}.

Please help me get started in solving them. They occurred in my PhD research.

MatheinBoulomenos

What values are allowed for X and Y?

Shaun

The values for X and Y are pairs of J, K, L that are not equal; I want to solve for J, K, L.

Oh, and X and Y are nth roots of unity for some n.

MatheinBoulomenos

1:13 PM

I'd say maybe try Gröbner bases, but I don't know the details on that

just heard that they can solve nonlinear systems of equations

Shaun

1:24 PM

Okay. Thank you, @MatheinBoulomenos.

Shaun

1:34 PM

I've just realised that J=K=L=0 or -1 by symmetry.

@Semiclassical, does that seem right to you? Are they the only solutions?

(Hello by the way. I'm sorry: I didn't mean to be rude.)

ÍgjøgnumMeg

1:55 PM

@Mathein Neukirch says we take $a \in \Bbb Z$ to be $(a \bmod p, a \bmod p^2, \dots ) \in \prod_{n=1}^\infty \Bbb Z/p^n\Bbb Z$ and that $\Bbb Q \subseteq \Bbb Q_p$ in the same way; that doesn't make sense to me :o

Sorry for the question spam lol

MatheinBoulomenos

yeah, that doesn't make sense

ÍgjøgnumMeg

"We obtain $\Bbb Q$ as a subfield of the field of p-adic numbers $\Bbb Q_p$ in the same way" lol

oh well

$\Bbb Z_p$ is an integral domain anyway right? So I can just localise

MatheinBoulomenos

@ÍgjøgnumMeg right

and we only need to localize at one element: $p$

ÍgjøgnumMeg

nice

MatheinBoulomenos

Everything else is already invertible (you can use a geometric-series argument for that, but it will also follow from Hensel's lemma which you will see later)

In fact, not only $\Bbb Z$ embeds into $\Bbb{Z}_p$, but also $\Bbb{Z}_{(p)}$, the localization at the complement of $(p)$

ÍgjøgnumMeg

1:59 PM

Cool, that's given as a bit of motivation at the start

MatheinBoulomenos

now notation is really confusing, lol. We have $\Bbb{Z}[1/p]$ the localization at $p$ which would also be denoted $\Bbb{Z}_p$ if it wasn't for the the conflict of notation, we have $\Bbb{Z}_{(p)}$ and we have $\Bbb{Z}_p$, the p-adic integers. And then there are also $\Bbb{Z}_p[1/p]$ which is just $\Bbb{Q}_p$ and some madmen even use $\Bbb{Z}_p$ for $\Bbb{Z}/(p)$

ÍgjøgnumMeg

I mean, the comparison where every $f \in \Bbb C(z)$ can be written as a quotient of elements of $\Bbb C[z]$ and that it has a Taylor series expansion around $z = a$ as long as $z - a$ doesn't divide the denominator

Leaky Nun

@MatheinBoulomenos hey

ÍgjøgnumMeg

and this has a comparison with $\Bbb Q$ and $\Bbb Z_{(p)}$

MatheinBoulomenos

@ÍgjøgnumMeg that's a nice analogy

@LeakyNun hi

Leaky Nun

2:02 PM

my presentation went brilliant

ÍgjøgnumMeg

@Leaky congrats man

MatheinBoulomenos

@ÍgjøgnumMeg yeah, elements in $\Bbb{Z}_{(p)}$ are functions on $\operatorname{Spec}(\Bbb{Z})$ which don't have a "pole" at $(p)$

ÍgjøgnumMeg

ha nice

MatheinBoulomenos

@LeakyNun nice! congratulations

Leaky Nun

ok to be fair, $\Bbb Q_p$ is the completion of $\Bbb Z_{(p)}$ right

so they're related

MatheinBoulomenos

2:05 PM

@ÍgjøgnumMeg the p-adic evaluation gives you the order of the pole or zero at the point $(p)$ for the rational number which is basically a meromorphic function on $\operatorname{Spec}(\Bbb{Z})$

@LeakyNun the completion of $\Bbb{Z}_{(p)}$ is $\Bbb{Z}_p$

Leaky Nun

isn't localization at $p$ denoted by $\Bbb Z_{(p)}$

oops

MatheinBoulomenos

@LeakyNun no, I mean the localization at a single element

Leaky Nun

away from $p$!

anyway, I mean, make $p$ invertible

and you get $\Bbb Z_{(p)}$

MatheinBoulomenos

No, $p$ is not invertible in $\Bbb{Z}_{(p)}$

Leaky Nun

$p$ is invertible in $\Bbb Z_p$ then

MatheinBoulomenos

2:07 PM

no

Leaky Nun

I mean localization

MatheinBoulomenos

it's not invertible in p-adic integers

then yeah

Leaky Nun

one of the localizations makes $p$ invertible

ÍgjøgnumMeg

hahaha

MatheinBoulomenos

sure

ÍgjøgnumMeg

2:07 PM

#notation

Leaky Nun

and then the completion of that localization is $\Bbb Q_p$

MatheinBoulomenos

yeah, that's correct

but you have to distinguish between completion for a valuation and for an ideal

for the latter, you don't want your ideal to be the unit ideal

Leaky Nun

what is your favourite ring-theroetic construction of $\Bbb Q$?

as opposed to, you know, the analytic construction

MatheinBoulomenos

So "the completion of $\Bbb{Z}[1/p]$ is a different thing than the completion of $\Bbb{Z}$, because for the latter we can take an inverse limit over the quotient"

ÍgjøgnumMeg

$\operatorname{Quot}(\Bbb Z)$

MatheinBoulomenos

2:09 PM

analytic construction for $\Bbb{Q}$?

Leaky Nun

oops

$\Bbb Q_p$

ÍgjøgnumMeg

lol

Leaky Nun

i wanna kill myself

ÍgjøgnumMeg

I'm just learning about it now

usukidoll

WHAAAAAAAAAAAAAAAAAAAAAAAAA

MatheinBoulomenos

2:09 PM

take the quotient field of Witt vectors for $\Bbb{Z}/p\Bbb{Z}$ :P

ÍgjøgnumMeg

so my favourite construction is by default the only construction I know

MatheinBoulomenos

It took a whole 90 minute seminar talk to show that Witt vectors form a commutative ring in full generality

usukidoll

Anyway...are there texts similar to Probability: An Introduction Second Edition? I need more ebooks

MatheinBoulomenos

more seriously, I like the inverse limit thing. It's how I think about p-adic numbers

Leaky Nun

@MatheinBoulomenos ok that's nice

wait no that isn't

that's not elegant enough, you used two steps

@MatheinBoulomenos or maybe use cohomology

MatheinBoulomenos

2:11 PM

I think of $\Bbb{Z}_p$ as the more fundamental object anyway

ÍgjøgnumMeg

@Mathein I definitely find that more attractive than just writing them as infinite series of powers of $p$ lol

Leaky Nun

@ÍgjøgnumMeg oh no, in Witt vectors it's still infinite series of powers of $p$

you're just choosing different coefficients

MatheinBoulomenos

@LeakyNun did you see my blog post? I started a series on rep theory

ÍgjøgnumMeg

Teichmüller bois

according to wikipedia

Leaky Nun

@MatheinBoulomenos how is that different from any old group rep course?

@ÍgjøgnumMeg right

MatheinBoulomenos

2:14 PM

@LeakyNun I'm taking a leisurely approach, I'm not teaching a course and I'm not writing a textbook

Leaky Nun

the latex in example 1.7...

MatheinBoulomenos

so I can give many examples and incidental remarks

@LeakyNun :P

ÍgjøgnumMeg

there's an external examiners' panel meeting next week to discuss the dissertation grades this year and my supervisor said there's a chance some of the grades might be revised downwards :(

MatheinBoulomenos

@LeakyNun I'm taking a different approach than some standard textbooks, e.g. Serre

that's not visible yet, though

Leaky Nun

@MatheinBoulomenos are you free to teach me the 6 sheaves induced by a map?

provided that you want to

MatheinBoulomenos

2:17 PM

I don't really want to, sorry

Leaky Nun

ok, that's fine

1

Q: Is there an algebraic extension $K / \Bbb Q$ such that $\text{Aut}_{\Bbb Q}(K) \cong \Bbb Z$?

Is there an algebraic field extension $K / \Bbb Q$ such that $\text{Aut}_{\Bbb Q}(K) \cong \Bbb Z$? Here I mean the field automorphisms (which are necessarily $\Bbb Q$-algebras automorphisms) of course. According to this answer, one can find some extension of $\Bbb Q$ whose automorphism gr...

@MatheinBoulomenos am I bs-ing?

MatheinBoulomenos

no

maybe you can add why $\Bbb Z$ is not a Galois group

No wait, how do you know that the Galois group of the fixed field is $\Bbb Z$?

Leaky Nun

the OP added it himself

@MatheinBoulomenos because that's just the automorphisms?

MatheinBoulomenos

it doesn't quite work for that for infinite extensions

Leaky Nun

you aren't having less or more automorphisms

you have exactly what you started with

MatheinBoulomenos

2:22 PM

take any non-closed subgroup in an infinite Galois extension

that works for finite groups, yeah

Leaky Nun

ok, take $\Bbb Z$ in $[\overline{\Bbb F_2} : \Bbb F_2]$

MatheinBoulomenos

yeah, fixed field is $\Bbb F_2$

Leaky Nun

right

but that isn't what the question is asking

MatheinBoulomenos

Galois group is $\widehat{\Bbb{Z}}$

Leaky Nun

it isn't that you choose a subgroup

you're given that those are all the automorphisms

MatheinBoulomenos

2:23 PM

but that's what you're using

Leaky Nun

not really

I'm saying, if the fixed field of Aut(K/F) is L, then Aut(K/F) = Gal(K/L)

because they're equal set-theoretically

MatheinBoulomenos

ah, okay

yeah that works

sorry for my incompetence

Leaky Nun

hardly incompetence

MatheinBoulomenos

I was thinking of a theorem of Artin that only works for finite groups of automorphisms

Leaky Nun

what is it

0

Q: Constructive predicative Hausdorffification without Choice

Can the left adjoint to the inclusion functor $i : \mathbf{Haus} \to \mathbf{Top}$ be constructed (1) constructively, (2) predicatively and (3) in ZF? If all three conditions (i.e., (1), (2) and (3)) are not possible, why and what is the best we can do?

general-topology axiom-of-choice

oh and maybe you could have a look

MatheinBoulomenos

2:26 PM

If $L$ is a field $G$ is a finite subgroup of the automorphisms of $L$, then $L/L^G$ is Galois with Galois group $G$

remember the exercise I gave you some time ago that every finite group is Galois group?

Leaky Nun

hmm

Sir Cumference

Quick question: Suppose the second derivative of a function $v(t)$ can be written as a function of $v$ (and not $t$). Does that imply that the first derivative of $v$ can be written as a function of $v$ alone?

mercio

probably not

for example look at $v(t)=t^2$

Sir Cumference

@mercio It's second derivative is constant

mercio

so ?

Sir Cumference

2:30 PM

How could you construct a second order autonomous ODE with that?

mercio

$v''(t) = 2$

Sir Cumference

Actually, nvm, you're right

K thx. The other question I had was

12 hours ago, by Sir Cumference

May sound dumb, but quick question: suppose a function $\vec{v}(t)$ parameterizes a Jordan arc for all $t \in I$. Then can we express its tangent vector as a function of $\vec{v}$ alone, which is defined over $I$?

jcora

hi all I'm trying to understand the last part of this maclaurin series for ln(1+x+x^2). it's not in english but I think it's basic enough to follow:

I don't understand how these coefficients are derived

MatheinBoulomenos

@LeakyNun I don't know much about foundational issues

Leaky Nun

ok

@MatheinBoulomenos do you have a categorical description of fields?

MatheinBoulomenos

2:48 PM

no

you can translate the definition to categorical language but I don't think that's useful in any way

Leaky Nun

@MatheinBoulomenos example 1.14 is wrong: you need the algebra to be a central $K$-algebra

MatheinBoulomenos

No you don't

a $K$-algebra always means that $K$ is contained in the center of the algebra

that's just the definition of an algebra

Leaky Nun

then what is a central $K$-algebra?

MatheinBoulomenos

that the center is not larger than $K$

Leaky Nun

interesting

what is the Lie algebra of an algebraic group?

MatheinBoulomenos

2:51 PM

Tangent space at the identity

Leaky Nun

what is an algebraic group?

MatheinBoulomenos

a group object in the category of varieties over a field

Leaky Nun

oh, we're in the classical pov

mercio

a representable functor from the category of k-algebras into the catgory of groups

Leaky Nun

now that's the non-classical pov

MatheinBoulomenos

2:53 PM

@LeakyNun not that classical, I haven't said that the field is algebraically closed

Leaky Nun

hmm

oh and what was the third question?

MatheinBoulomenos

there was no third question, the first one I gave you was split into two questions

Leaky Nun

:o

like a split torus?

MatheinBoulomenos

yeah, sure, the important part of the exercise was $\Bbb{F}_{p^2}^\times$ which is a split torus over $\Bbb{F}_{p^2}$

The exercise was to compute the kernel of the homomorphism between tori $\Bbb{F}_{p^2}^\times \to \Bbb{F}_p^\times$ given by the norm map

Leaky Nun

right

ÍgjøgnumMeg

2:58 PM

hmm the wikipedia article for $\Bbb Z_p$ says $\Bbb Q_p \cong (p^{\Bbb N})^{-1} \Bbb Z_p^{\times}$; why $\Bbb Z_p^{\times}$?

Leaky Nun

link?

ÍgjøgnumMeg

P-adic number

In mathematics, the p-adic number system for any prime number p extends the ordinary arithmetic of the rational numbers in a different way from the extension of the rational number system to the real and complex number systems. The extension is achieved by an alternative interpretation of the concept of "closeness" or absolute value. In particular, p-adic numbers are considered to be close when their difference is divisible by a high power of p: the higher the power, the closer they are. This property enables p-adic numbers to encode congruence information in a way that turns out to have powerful...

mercio

why the -1 ?

MatheinBoulomenos

either you or wikipedia is confusing two things $\Bbb{Q}_p^\times = p^{\Bbb Z} \times \Bbb{Z}_p^\times$ and $\Bbb{Q}_p=\Bbb{Z}_p[1/p]$

ÍgjøgnumMeg

@mercio localising at $p^{\Bbb N} = \lbrace p^n : n \in \Bbb N\rbrace$

Leaky Nun

3:01 PM

@ÍgjøgnumMeg I think that's just bs

MatheinBoulomenos

you're right, wikipedia is wrong

Leaky Nun

the earlier sentence is also bs

ÍgjøgnumMeg

Lol

love wikipedia

Leaky Nun

"uniquely written as $p^{-n} u $ with $n$ natural number and $u$ unit in the integers"

would have been right if $n$ is integer instead

MatheinBoulomenos

true

ÍgjøgnumMeg

3:02 PM

also

actually dw

I tend to use $\Bbb N = \lbrace 1, 2, ...\rbrace$ which makes the notation $p^{\Bbb N}$ wrong for me

lol

MatheinBoulomenos

it doesn't really matter of you're multiplicative subset contains $1$ or not

ÍgjøgnumMeg

how come?

MatheinBoulomenos

as long as it's non-empty you can still localize and get the same ring as if you added $1$

ÍgjøgnumMeg

fair

mercio

I don't think p-adic numbers are localisations

ÍgjøgnumMeg

3:05 PM

I'm not qualified to make any comment on that

MatheinBoulomenos

@mercio but in the notation $\Bbb Q_p \cong (p^{\Bbb N})^{-1} \Bbb Z_p$ there's a localization

mercio

aaah

ÍgjøgnumMeg

but $\Bbb Z_p$ is an integral domain and $\Bbb Q_p$ is the field of fractions of $\Bbb Z_p$

mercio

that's a strange notation

Leaky Nun

@ÍgjøgnumMeg theorem: for any multiplicatively closed subset $S$ of ring $A$, $S^{-1} A \equiv {\overline S}^{-1} A$, where $\overline S$ is the saturation of $S$ defined by the property that it is the smallest superset of $S$ with the property that $xy \in \overline S \implies (x \in \overline S \land y \in \overline S)$

so if $S$ is nonempty then $\overline S \ni 1$

@mercio any fraction field is a localization

MatheinBoulomenos

3:07 PM

I think he knows that ...

mercio

yeah but it's simpler to write Zp[1/p]

Leaky Nun

@MatheinBoulomenos 0 is dense in R/Q :o

mercio

lol

MatheinBoulomenos

@LeakyNun everything non-empty is dense in that

Leaky Nun

:o

mercio

3:11 PM

actually I thought your wikipedia quote was a typo for $p^{\Bbb Z} \times (\Bbb Z_p^*)$

Leaky Nun

that would have been an accurate description for $\Bbb Q_p^\times$

unfortunately it still doesn't include 0

mercio

oh right

geocalc33

Hey math folks

ÍgjøgnumMeg

that was my main confusion

"pretty sure $0 \in \Bbb Q_p$"

mercio

x)

geocalc33

3:14 PM

Hey guys

ÍgjøgnumMeg

@geocalc33 yo

geocalc33

I wany Germany to win the world cup

mercio

what's a world cup ?

geocalc33

Soccer

ÍgjøgnumMeg

Öschtriiiich

are they even in the world cup

I don't follow it

geocalc33

3:18 PM

Germany is one of the best team s

MatheinBoulomenos

Iceland should win

Alessandro Codenotti

Pfff obviously Italy will win as usual

MatheinBoulomenos

Sorry, that was mean

ÍgjøgnumMeg

there was an English pub here offering free pints for every goal scored by Scotland, Wales, or Northern Ireland in the world cup

Alessandro Codenotti

I didn't read the second message but I didn't find the first mean at all, rather accurate instead

geocalc33

3:33 PM

Is there geodesics of a circle

Great circles

Alessandro Codenotti

You want geodesics on a circle or on a sphere?

geocalc33

Say you generate $n$ equally spaced geodesics on a sphere

Sphere

Alessandro Codenotti

Then the geodesics are all and only the great circles

geocalc33

Infinitely many would completely cover the sphere correct?

ÍgjøgnumMeg

@Mathein it's actually an exercise here to show that $\Bbb Z_p^\times$ is all of the $p$-adic integers that are non-zero mod $p$, so it makes sense that you only need to localise at $p$

bleh#

Alessandro Codenotti

3:37 PM

@geocalc33 yep

(If you choose them well, of course you can also pick infinitely many geodesics without covering the whole sphere)

geocalc33

Is there any point in generating a finite number of them ?

Just to generate a basic idea of a sphere

Like if you wanted to approximate the shape given the geodesics

Obviously a circle rotated around discreetly would give you an idea of a sphere

But for more complicated objects

I wonder what topic that would fall under within mathematics