Mathematics - 2017-10-12 (page 2 of 9)

Leaky Nun

02:01

@Faust rip

Faust

@LeakyNun its just there no point in studying at all nothing we did in class or assignments was remotely close to a single question they were all completely unique and there was technically 7 of them to be answered in 50 minutes

i got it done in the nick of time

but lots of people didnt finish

Leaky Nun

I'm staying up at 3AM reading Galois theory

Faust

@LeakyNun is it 3 am now cause that statement makes nonsense...

Leaky Nun

yes, it is 3 AM now

and I'm reading Galois theory

Faust

you must like it

Leaky Nun

02:05

I do

Faust

i know nothing of the nonsense you speak

Brody

Galois theory, only 22 chapters away...

Leaky Nun

@TedShifrin @MatheiBoulomenos my book (by Ian Stewart) uses . for multiplication and wrote 2.4=8... I can officially burn the book now

Faust

is it a third year class or a 4th year class?

Leaky Nun

@Faust are you talking to me?

Faust

02:07

yeah

Leaky Nun

I have no idea

Faust

when do u leanr Galois theory

Leaky Nun

I just read it for myself

Faust

just wondering when im going to learn it

Leaky Nun

no idea

Faust

02:08

you can't see the future?

Leaky Nun

it isn't written on my syllabus

@Brody should I introduce it to you?

Brody

@Faust Typically, undergrad has a two-semester course introducing abstract/modern algebra. Galois theory comes into play the second semester

Leaky Nun

@Brody but which year?

Brody

Depends on the student no? @LeakyNun

Leaky Nun

@TedShifrin @MatheiBoulomenos the same book included a geometrical interpretation of the $\Bbb D_8$ Galois group of $x^4-2$... I can unburn it now.

Faust

02:10

@Brody thanks so ill learn it this semester cause im taking my second AA class now and my third next semester

Leaky Nun

@Faust should I introduce it to you?

I'm constantly amazed by it the more I know

Brody

@LeakyNun I don't have enough prerequisite knowledge to understand the basics

Faust

i might understand it

Leaky Nun

@Brody how do you know?

Brody

Because I haven't even started rings, fields, etc.

Leaky Nun

02:12

I can adjust my explanations

@Faust so I need to do two explanations at the same time lol

Faust

rings are just groups with 2 binary operations

Leaky Nun

you know what fields are right

Faust

yeah

Brody

With what axioms?

Faust

times and addition on the same group\

Leaky Nun

02:13

@Faust so $\Bbb Q$ is a field, and do you know what $\Bbb Q(\sqrt2)$ means?

Brody

I can just look this up

Leaky Nun

@Brody sure you can

@Faust you're missing the most important axiom that links the two binary operations together

Faust

its the field of rational adjioned with root 2

Brody

off to proofwiki...

Leaky Nun

@Faust and a homomorphism is a mapping that preserves addition and multiplication

a monomorphism is an injective homomorphism; an isomorphism is a bijective homomorphism; an automorphism is an isomorphism to itself, ok?

Faust

02:14

never seen a ring homomorphism b4 can we call it a homomorphism between vector spaces?

Leaky Nun

@Faust yes

Faust

ok rest i have seen before

Leaky Nun

everything has homomorphisms, groups, rings, fields, vector spaces

Brody

Ok like an abelian group under $+$ and a distributive semigroup under $\cdot$

Faust

only seen em for groups and vector spaces

Leaky Nun

02:15

@Faust prove that any automorphism of $\Bbb Q(\sqrt2)$ fixes $\Bbb Q$ (I can skip exercises if you want, and I'll just give you those results)

@Brody yes

Faust

what do you mean by "fixes"

Leaky Nun

maps elements of $\Bbb Q$ back to themselves

Faust

ok

i can belive that

Leaky Nun

ok

so every element in $\Bbb Q(\sqrt2)$ is in the form $a+b\sqrt2$

oh you know vector spaces

Faust

yeah from dif geo

Leaky Nun

02:17

so you must know that the degree of extension $\Bbb Q(\sqrt2):\Bbb Q$ is $2$?

(field extension is just the opposite of subfield)

Faust

dont even know what that notation says

oh

Leaky Nun

$\Bbb Q(\sqrt2)$ as an extension field of $\Bbb Q$

Faust

i understand

Brody

spectates while doing HW and listening to a talk

Leaky Nun

@Brody look, I said I can adjust to you if you want

Faust

02:18

what does the 2 mean

Leaky Nun

@Faust dimension of the vector space

Faust

ok

Leaky Nun

2 mins ago, by Leaky Nun

so every element in $\Bbb Q(\sqrt2)$ is in the form $a+b\sqrt2$

Faust

yeah

Brody

@LeakyNun it's fine. I don't want to impede Faust. a bit busy regardless

Leaky Nun

02:19

and if $r$ is an automorphism of $\Bbb Q(\sqrt2)$, then $r(a+b\sqrt2) = r(a)+r(b)r(\sqrt2) = a+br(\sqrt2)$.

Faust

ok

Leaky Nun

so an automorphism is completely determined by the image of the generators

Faust

ok

Leaky Nun

we know that $2 = r(2) = r(\sqrt2^2) = r(\sqrt2)^2$

so $r(\sqrt2) = \pm \sqrt2$

so we have exactly two automorphisms of $\Bbb Q(\sqrt2)$

Faust

ok

you have 2 generators

Leaky Nun

02:21

we know that automorphisms form a group under composition (exercise in group theory)

@Faust I only have 1

as a vector space it does have 2 generators

but as a field its only generator is $\sqrt2$

($1$ is a given)

Faust

intresting ok

Leaky Nun

49 secs ago, by Leaky Nun

we know that automorphisms form a group under composition (exercise in group theory)

ok?

Faust

i can take that (dont actually know it to be true but i know wha tit says)

Brody

why need mention an operation be closed if we already define $\cdot\, :\, S\times S\to S$?

Leaky Nun

lol, I thought you know groups @Faust

@Brody some don't mention that

but just for the sake of clarity

Brody

02:23

but it's superfluous, yes?

Leaky Nun

yes it is

Brody

oh ok

Faust

i have never seen an automorphism map in a class i just know what it is

Leaky Nun

but many people forget that

@Faust you haven't studied group automorphisms?

Brody

ty

Faust

02:24

@LeakyNun i know what it is but i have never learned it in a class

Leaky Nun

so one automorphism is the identity here and the other automorphism is kind of conjugate, ok?

what group do they form?

lol there's only one group of order 2

Faust

$S_2$ :p

Leaky Nun

good

So $\newcommand{Gal}{\operatorname{Gal}}\Gal(\Bbb Q(\sqrt2):\Bbb Q) = S_2$

($\Gal(L:K)$ means the group of automorphisms of $L$ that fixes $K$, here it's superfluous as every automorphism fixes $\Bbb Q$ but we write it anyway.)

@Daminark hi

Faust

ok

intresting

you really dont have to use $S_2$ i was just being cheeky :p

Leaky Nun

@Faust why not?

Faust

02:29

$\mathbb{Z_2} $ is more normal way to write it

Leaky Nun

Can you find $\Gal(\Bbb Q(\sqrt2,\sqrt3):\Bbb Q)$?

1. note that an automorphism is completely determined by the image of the generators
2. note what equations the generator have to satisfy
3. identify the valid automorphisms
4. identify the common name of the group

Faust

$\mathbb{Z_6}$ ?

Leaky Nun

why?

Faust

intuitive guess

Leaky Nun

come on lol

Faust

02:32

i have no justification lol

Leaky Nun

alright, let's work it out together

let $r \in \Gal(\Bbb Q(\sqrt2,\sqrt3):\Bbb Q)$.

Then, $r(\sqrt2)^2 = r(\sqrt2^2) = r(2) = 2$ and $r(\sqrt3)^2 = r(\sqrt3^2) = r(3) = 3$.

Faust

ok

Leaky Nun

Therefore, $r(\sqrt2) = \pm \sqrt2$ and $r(\sqrt3) = \pm\sqrt3$.

So there are $4$ automorphisms.

What group do they form?

Faust

what about $ r (\sqrt2 \sqrt 3) $

Leaky Nun

$r(\sqrt6) = r(\sqrt2)r(\sqrt3)$, so its value is also completely determined by the image of the generators

Faust

02:34

ah

k now big question

what do you use it for?

Leaky Nun

good question

Now, among the determiners of the group structure, there is this thing called chain of subgroups

for example, $S_3 > A_3 > V_4 > \Bbb Z_2 > \{e\}$

Faust

wth is $ V_4 $

Leaky Nun

@Faust you really need to study more group theory :P

$V_4$ is Klein's four group

Faust

thats why i am taking GT classes :p

Leaky Nun

$\Bbb Z_2 \times \Bbb Z_2$

$\{e,a,b,c\}$ where $a^2=b^2=c^2=e$ and $ab=c$ and $bc=a$ and $ca=b$

Faust

02:37

oh i know it just never seen it written as $V_4$

Leaky Nun

oh ok

are we convinced that $V_4$ is a subgroup of $A_3$?

oh what, I meant $S_4 > A_4 > V_4 > \Bbb Z_2 > \{e\}$

Faust

well yes

Leaky Nun

@Faust what is the subgroup?

in terms of permutations

Faust

just take all the reflections

err

permutations

hmm

Leaky Nun

hmm?

Faust

02:41

hmm

Leaky Nun

lol

Faust

(132)

is one of the things i want

i think

i dunno i want the refelections O.o

no thats not quite right

that would be 3 of them

and one order 1 identity so 4

no thats right

Leaky Nun

which 4?

Faust

what ever the equivlant to a reflection ( order 2) theres 3 of them

and then the identity

Jasper

Hello @Faust and @LeakyNun!

Leaky Nun

02:46

so which?

@Jasper hi

Faust

(123) (132) (e) (234) ?

that doesnt look right

there even at least :P

Leaky Nun

facepalm

should I tell you the answer?

Faust

normally we would write it out in terms of r's and j's

Leaky Nun

what is r and j?

Faust

thats why i ddint say it

Leaky Nun

02:48

lol what

r for rotation, j for ??

Faust

flip

horizontially

Jasper

j for jerk, lol

Faust

@Jasper how you doing?

Leaky Nun

@Faust write it out in terms of r and j if they are more convenient for you

Jasper

@Faust Not good. What about you?

Faust

02:50

thats where im confused cause i get 5 elements.... e rj r^2j r^3 j and j

@Jasper had a really hard midterm today but not bad you try getting some help?

Leaky Nun

@Faust what is (rj)(rj)?

Jasper

@Faust Yes, I will continue taking my meds and also see a therapist in a couple of weeks.

Faust

e

@Jasper i wish you all the luck man

Leaky Nun

@Faust isn't that the notation for $\Bbb D_8$?

Faust

yeah im totally lost

what the hell are te four elements of order 2 in $A_4$

Leaky Nun

02:53

@Faust how many elements are there?

Faust

4!/2

12

alegedly

Leaky Nun

so just list them out :P

Brody

ill...so in the subtractive quasigroup on integers, right substraction is just addition?

Jasper

Never heard of quasigroup, lol.

Faust

(1234)(1324)(1423)(e)

Brody

02:54

@Jasper I'm trying to learn more words

Faust

no that isnt even in $ A_4 $

those are odd

Leaky Nun

right

@Brody I don't think so

Brody

hmm

Faust

@LeakyNun im an idiot

there are more than one set of options!

Leaky Nun

@Faust no you aren't

@Faust hmm?

Faust

02:57

(123)(132) (124)(e) is fine

Leaky Nun

@Brody yes, I think so

@Faust what is (124)(124)?

Faust

that subgroup

Leaky Nun

so what are the elements of the subgroup?

Faust

the identity

Leaky Nun

and?

Faust

02:58

the subgroup of (123)(132)(124) e is ismorephic to $ \mathbb{Z_2} x \mathbb{Z_2} $

theres a buncha choices which was confusing me

Leaky Nun

@Faust list the elements out

one by one

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$Mathematics$ Mathematics