Mathematics - 2017-10-10 (page 5 of 5)

CausingUnderflowsEverywhere

23:00

Math is so easy when you just use a lined sheet of paper instead of the space $they$ provide.

Leaky Nun

@MatheiBoulomenos the hell

@CausingUnderflowsEverywhere use a lined sheet of paper to find the Galois group of $x^5-x+1$

CausingUnderflowsEverywhere

define Galois

Leaky Nun

Galois group

In mathematics, more specifically in the area of abstract algebra known as Galois theory, the Galois group of a certain type of field extension is a specific group associated with the field extension. The study of field extensions and their relationship to the polynomials that give rise to them via Galois groups is called Galois theory, so named in honor of Évariste Galois who first discovered them. For a more elementary discussion of Galois groups in terms of permutation groups, see the article on Galois theory. == Definition == Suppose that E is an extension of the field F (written as E/F and...

Mary Star

Could someone of you take a look at my question (it's about change of variables): math.stackexchange.com/questions/2466703/… ?

Ted Shifrin

@Mambo: It seems that Hadamard wrote a paper on this too, and it's necessary and sufficient that $\sum (b_{n+1}-b_n)$ converges. Oddly, it seems that this multiplying factors can be chosen independent of the original convergent series.

CausingUnderflowsEverywhere

23:01

define group

Leaky Nun

Group (mathematics)

In mathematics, a group is an algebraic structure consisting of a set of elements equipped with an operation that combines any two elements to form a third element and that satisfies four conditions called the group axioms, namely closure, associativity, identity and invertibility. One of the most familiar examples of a group is the set of integers together with the addition operation, but the abstract formalization of the group axioms, detached as it is from the concrete nature of any particular group and its operation, applies much more widely. It allows entities with highly diverse mathematical...

CausingUnderflowsEverywhere

oh okay I have both a lined sheet of paper and a rubik's cube. I think I can do this

Ted Shifrin

@Mambo: We should have checked MSE first!!

Leaky Nun

good

Mambo

@TedShifrin That's very weird.

Ted Shifrin

23:03

It looks like Borel's lecture is also constructing the terms without paying attention to the particular convergent series.

CausingUnderflowsEverywhere

let me just make sure my lined sheet of paper is mathematical.

define lined sheet of paper

Leaky Nun

@CausingUnderflowsEverywhere hucking fell

I got back-trolled

CausingUnderflowsEverywhere

poor hucking I hope they didn't hurt themselves

Leaky Nun

lol

Ted Shifrin

@Mambo: I suggest you hunt on MSE. Here's another.

MatheiBoulomenos

23:05

@LeakyNun because you asked, $\operatorname{Aut}(\mathbb{Q}(X)/\mathbb{Q}) \cong \operatorname{PGL}(2,\mathbb{Q})$

Leaky Nun

@MatheiBoulomenos right, someone told me that earlier

MatheiBoulomenos

you can replace $\mathbb Q$ by any field, too

Leaky Nun

I see

but what about $\operatorname{Aut}(\overline{\Bbb Q})$ lol

Faust

morning

Leaky Nun

:o my book says "Galois group $\Gamma(L:K)$"

@Faust top of the morning

Mambo

23:07

@TedShifrin The very original proof of Riesz brother's theorem : If $\mu$(Borel measure on $[-pi,pi]$) is of analytic type, then it is absolutely continuous wrt Lebesgue measure, uses such construction.

MatheiBoulomenos

As I said, the absolute Galois group of $\mathbb Q$ is a mystery

Leaky Nun

@MatheiBoulomenos :c

Ted Shifrin

OK, @Mambo.

G'night, @Faust.

MatheiBoulomenos

Think about it, the inverse Galois problem (which is still open, but solved for many groups) asks if every finite group is quotient of that group

Mambo

I will follow up on those MSE links. Thanks again

Leaky Nun

23:09

@MatheiBoulomenos :O

Faust

Let $G = \mathbb{Z^3}$ and consider
$N = {(i, j, k) ∈ \mathbb{Z^3}| i + 2j = 3i − k = 0}.$ where N is a normal subgroup. i want to define a homomorphism to $ \mathbb{Z^2} $ what do i use?

@TedShifrin its not night time!

Leaky Nun

@Faust you use a pen and a paper to solve the equations parametrically

Ted Shifrin

What kind of homomorphism to you want to define, @Faust? BTW, every subgroup of an abelian group is normal, of course.

Faust

n(1,-2,-6)

Ted Shifrin

Answer my question. What does the homomorphism have to do with $N$?

MatheiBoulomenos

23:10

Just any homomorphism? Send everything to zero

Leaky Nun

@MatheiBoulomenos lol

Ted Shifrin

I'm happy with that, @Mathei. Then I can quit.

Leaky Nun

and (1,-2,-6) isn't even right

sorry, it is 1-dimensional.

Ted Shifrin

getting fed up

Leaky Nun

@TedShifrin sorry :c

Faust

23:12

@TedShifrin G/N isomorphic to $\mathbb{Z^2}$ ?

Ted Shifrin

So you want a homomorphism $\phi\colon G\to\Bbb Z^2$ such that ... ?

Faust

ker of the homo morphism is N?

Ted Shifrin

Bingo. Now do it.

Faust

well i got n(-2,1,-6) or w.e as a generator?

Ted Shifrin

Forget that.

How do you know what $N$ is in the first place?

Faust

23:14

it was given in the question?\

Ted Shifrin

How was it given?

Faust

You may assume that N is a normal subgroup of G. Find a familiar
group H such that G/N ∼= H

"?

or am i misunderstanding the question

Ted Shifrin

So we don't even know that $H = \Bbb Z^2$ is correct, by the way.

Faust

no that was my assumption

Ted Shifrin

But you're missing the main point. How was $N$ given to you?

Faust

23:17

cause the ker phi

is cyclic and infinite

nvm

thats not correct

N looks like its two planes that intersect in a line

Ted Shifrin

But $N$ is given to you by integer-coefficient equations.

Faust

but the lines all busted so points that lie ona line

Ted Shifrin

What does kernel mean?

Faust

kernel of phi is all the elements that phi maps to the identity i have no idea what kernel means

Ted Shifrin

What's the identity in the groups you're working with?

What's the identity in $\Bbb Z$, $\Bbb Z^2$, $\Bbb Z^3$, etc.

Faust

23:21

well in N the identity is n(-2,1,-6) where n is in the intergers?

0

(0,0)

(0,0,0)

Ted Shifrin

Right. So can you write down a homomorphism $\phi\colon\Bbb Z^3\to\Bbb Z^2$ so that $\phi(i,j,k)=(0,0) \iff (i,j,k)\in N$?

Faust

send i+2j to 0 and 3i-k to zero

oh

Ted Shifrin

So tell me what $\phi(i,j,k)$ is.

Faust

nope

hmm

not a nontrivial one

Ted Shifrin

Hint: This is totally immediate and you should have written it down in one second.

You're obviously over-thinking ridiculously.

Faust

23:30

well i mean let $ \phi(i,j,k) =(0,0) \forall (i,j,k) \in N $

Ted Shifrin

That's obviously crap.

It's for all $(i,j,k)\in\Bbb Z^3$.

Faust

lol

i dont understand

ones in three dimensions

the others in 2

Ted Shifrin

So?

Faust

i mean i can just drop off the last one?

Ted Shifrin

No.

Faust

23:31

send it to (i,j)

Ted Shifrin

Then does precisely $N$ get sent to $(0,0)$?

No, all things with $i=0$ and $j=0$ get sent to $(0,0)$.

Faust

i mean everything of the form (-2n,n,-6n) gets maped to (0,0)

Ted Shifrin

Not with your $\phi$ you just made up.

It really is not helpful having that formula. It is distracting you completely.

If I ask you to give me a linear map $T\colon \Bbb R^2 \to\Bbb R$ so that $T(x,y)=0 \iff x-y=0$, can you write one down?

I'll take either a formula $T(x,y) = $ ... or a matrix.

Leaky Nun

I wouldn't take a matrix

Dragneel

Sup guys
I'm uncertain if I got the contrapositive correctly.
Here's the original statement https://i.gyazo.com/d02ad718d212705def039b68ab704135.png

and I got $\exists n \in \mathbb{N}, K > \sqrt{n} \wedge [n \neq K*L \vee K > L]$

Faust

23:36

@TedShifrin not in any kind of math words. you want to take everything and send it to the diffrence of its first and second cordiante

Ted Shifrin

So write down the formula $T(x,y) = ...$ that says that!

MatheiBoulomenos

I would show that the space of $(x,y) \in \mathbb{R}^2$ such that $x-y=0$ is one-dimensional subspace, consider the quotient $\mathbb{R}^2/\langle(1,1)\rangle$, use rank-nullity to show that it is one-dimensional and compose the quotient map with an isomorphism $\mathbb{R}^2/\langle(1,1)\rangle \cong \mathbb{R}$ :P

Leaky Nun

@Dragneel yes

Ted Shifrin

smacks @Mathei

Leaky Nun

@MatheiBoulomenos let them do it

Dragneel

23:37

@LeakyNun Great! Thanks :)

Leaky Nun

and it's \langle \rangle you LaTeX illiterate (just kidding)

@MatheiBoulomenos are you going to claim next that it is a one-form?

MatheiBoulomenos

Okay, that smack was deserved

Leaky Nun

@MatheiBoulomenos give me an inverse Galois theory problem :P

MatheiBoulomenos

@LeakyNun Show that, if the ground field is allowed to be anything, any finite group is the Galois group of some field extension

Faust

if y=x take (x,x) to 0 if y=x+1 take (x,x+1) to -1 etc?

Leaky Nun

23:40

@MatheiBoulomenos :o what

MatheiBoulomenos

Yup

Faust

youd be at it for a long time cause there unaccountably infinite unique y

Leaky Nun

how at all

Faust

i dunno i really dont understand this crap O.o

Leaky Nun

no, I was talking to Mathei.

Faust

23:42

Ah

MatheiBoulomenos

I'll give you a lemma, with which you may not be familiar with: If $K$ is any field and $G$ is a finite group of automorphisms of $K$, then $K/K^G$ is Galois with Galois group $G$

this is due to Artin iirc

Ted Shifrin

@Faust: This is nuts. You told me in words to take the difference of the first and second coordinate. You can't write the formula?

Leaky Nun

@MatheiBoulomenos that's just the Galois correspondence innit

@Ted is getting fed up woo

Faust

@TedShifrin but it says you want to take it to 0

that only works when y=x

Ted Shifrin

Precisely.

What's the damn formula?

MatheiBoulomenos

23:43

@LeakyNun not quite. It's used in the modern proofs of the Galois correspondence, but it's an imporant lemma in itself

Faust

x-x=0?

Ted Shifrin

I need $x$ and $y$.

Write down what you said when you said "take the difference of the first and second coordinate."

Leaky Nun

@MatheiBoulomenos but I can't just adjoin any element ;_;

MatheiBoulomenos

I think with that lemma you can do it

Leaky Nun

say what

Faust

23:44

@TedShifrin x=y?

Leaky Nun

wait, I misread the lemma the first time

Faust

err y=x*

Ted Shifrin

Where is the difference of the coordinates?

Leaky Nun

@MatheiBoulomenos so you need to construct $K$ first lol

Faust

like x-y =0 ?

MatheiBoulomenos

23:45

@LeakyNun right

Faust

like x-y = a

Ted Shifrin

Sigh. $T(x,y)=x-y$. Then $T(x,y)=0 \iff x-y=0$.

Leaky Nun

@MatheiBoulomenos is it completely arbitrary or based on pre-existing fields?

Ted Shifrin

Now go off in a quiet corner and write down $\phi$ in the obvious, correct way.

MatheiBoulomenos

@LeakyNun I guess you need to know that some field exists

Leaky Nun

23:47

@MatheiBoulomenos I only know the obvious fields that everyone knows lol

MatheiBoulomenos

@LeakyNun I meant "any" field

Leaky Nun

finite fields, rational, extended rational, algebraic completion of rational, rational rational functions, real, complex

MatheiBoulomenos

you can take any field and start from there

Leaky Nun

and I don't know how to construct "any" field in general, unlike groups

you just let letters in groups

e a b c whatever

MatheiBoulomenos

I meant an arbitrary field

I can assure you, you know enough fields to solve this problem

Leaky Nun

23:49

every finite field is contained in $S_n$ (Cayley)

Faust

@TedShifrin i finally understand what the hell that says your talking about the kernel of T(x,y) you know you want it to be 0 so you want T(x,y) to be defined in such a way so that T(x,y)=0 iff x-y=0

MatheiBoulomenos

I guess you mean finite group

Leaky Nun

@MatheiBoulomenos yes

MatheiBoulomenos

Cayley's theorem is the right approach

Faust

so in my original thing i want $\phi (i,j,k) = (i+2j,3i-k) $

Ted Shifrin

23:52

Yup. Finally! ... Now you can apply the Fundamental Homomorphism Theorem provided you check that $\phi$ is what?

Faust

subjective homomorphism

Leaky Nun

@Faust be objective

MatheiBoulomenos

I think we can show that it is a homomorphisms objectively

Leaky Nun

@MatheiBoulomenos and Aut(Z2^n) = S_(2^n-1) right

not sure if this helps

MatheiBoulomenos

that doesn't really help

Z2^n is not a field

Leaky Nun

23:53

hmm

Faust

the homomorphism part isn't bad the surjective part will take me a minute

Leaky Nun

@MatheiBoulomenos any hint?

MatheiBoulomenos

Suppose you have a field $k$. Do you know some way to make it bigger?

Faust

@TedShifrin anyway thank you very much i have been staring at that for like an hour before i asked you.

Leaky Nun

@MatheiBoulomenos adjoin elements lol

Ted Shifrin

23:55

Right. You do need to check surjective. So remember that all you have to do is find something that hits an arbitrary element. You don't have to necessarily find everything.

@Faust: I hope you learned a lesson here. Go back to basics!!

MatheiBoulomenos

@LeakyNun Right, but what kind of elements can you adjoin if you know nothing about the field?

Leaky Nun

@MatheiBoulomenos a transcendental element

MatheiBoulomenos

Okay

Leaky Nun

that doesn't make $S_n$ embed in the automorphism...

MatheiBoulomenos

Well, one transcendental element doesn't

Leaky Nun

23:57

....

you win

you can't distinguish between the transcendental elements.

Faust

@TedShifrin i think the problem was that i didnt really understand what was being said exactly and my book only has exercises not examples =\

MatheiBoulomenos

Sure you can, just give them different names

Leaky Nun

@MatheiBoulomenos I mean

if you have n transcendental elements, you can't distinguish them, so the Aut group is S_n

I mean, disregarding the PGL thing

MatheiBoulomenos

the Automorphism group is certainly not S_n

Leaky Nun

I mean, S_n is a subgroup of the Aut group

MatheiBoulomenos

23:58

right, that's the important part

Ted Shifrin

@Faust: Don't forget to study your lecture notes and make sure you understand them.

Leaky Nun

an explicit construction would have $\Bbb Q(e,e^e,e^{e^e},\cdots)$ (Lindemann-Weierstrass)

Ted Shifrin

I'm glad Leaky and Mathei have become algebra friends. I can escape :)

Leaky Nun

@TedShifrin we're still doing Galois theory :)

Faust

@TedShifrin my lecture notes are so messy that i cant read them i think ima have to borrow notes from someone going forward cause he does a buncha the excerises in class

MatheiBoulomenos

23:59

well, you don't need to be so explicit, just take $F(X_1,X_2, \dots ,X_n)$, where $X_i$ are trancentendal by fiat

Leaky Nun

@MatheiBoulomenos I want to be explicit

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$Mathematics$ Mathematics