groups-rings-fields

[timber bay]

let me think about it for a second

[hollow comet]

Which notation?

[timber bay]

well, all the series notation for subfields

we started this monday, so i havent had much practice

[hollow comet]

Please

They are not subfields

They are subgroups

I think you had the same confusion the other day with Galois theory xd

[timber bay]

yeah I use them interchangeably apparently

i know the difference but i just dont ever think to clarify. we've went from field extensions to this

[hollow comet]

Ping me if you have further problems

[timber bay]

V_i (Wj \cap V{i+1} )

im not exactly sure what this means

[hollow comet]

Well

You have three subgroups of G there, right?

[timber bay]

treating it like a function?

yeah

[hollow comet]

So you have the intersection of two subgroups

[timber bay]

dont you mean subfields /s

[hollow comet]

Which is again a subgroup

And the product of two normal subgroups

Which is again a subgroup

[timber bay]

oh okay so the parentheses is just product

[hollow comet]

Oh, sure

[timber bay]

ah okay

[hollow comet]

It is there to be able to distinguish between A(B cap C) and (AB) cap C

[timber bay]

how does the result give us V with two indeces?

im not exactly sure what two indices would mean in this.

[hollow comet]

V_i, j is a new subgroup we are defining

We want a refinement

[hollow comet]

@timber bay did you manage to do it?

[timber bay]

i did not @hollow comet

i had to go afk for an hour or so

[hollow comet]

Oh, okay

Are you still going at it?

[haughty prism]

Is this right?

[timber bay]

i started on another problem because i needed to work on more stuff

but I can go back to it if you have tips

@hollow comet

[hollow comet]

I told you exactly how to do it xD

Just construct the series I told you

[timber bay]

okay yeah let me write this out

are the identities considered V_i's?

or do you just start with a non-identity element?

is this kind of the process? sorry for bad picture, camera is bad.

@hollow comet

do you want me to ping you still? i could see that getting annoying

[hollow comet]

Yes, that's kind of the thing

You have to do that with every i and every j

And all of those can be simplified

Because as you know, any ideal of Z is of the form nZ

So it will be clearer if you write them like that

[timber bay]

yeah the simplification will be a different step, which i dont really know how to do yet

[hollow comet]

Good luck

Hahaha

[timber bay]

so there will be i*j elements in the refinement?

[hollow comet]

More like n*m

[timber bay]

what are n and m?

[hollow comet]

The lengths of the original series

Serieses?

I don't know if it has a plural xd

[timber bay]

oh okay

series'

is what you do if there's an s on the end

usually

english is weird

how do we know what order to put these products in?\

[timber bay]

anybody have any advice for this?

[marble wagon]

how do you find the sum of roots?

in the polynomial

[timber bay]

im not sure what you mean

[marble wagon]

there is a very easy way to find the product and sum of roots

just by looking at the polynomial

[timber bay]

will it be related to 2?

somehow

[marble wagon]

the poly is (x-a1)(x-a2)(x-a3)

can you see where the sum and product will show up?

[timber bay]

oh i thought you were talking about the polynomial given

[marble wagon]

it is the polynomial given

x^3 - 4x^2 + 6x - 2 = (x-a1)(x-a2)(x-a3)

[timber bay]

wouldnt that mean that this poly is the minpoly?

oh nvm

[marble wagon]

from that expression you can quickly find the sum and product of roots

can you see how?

[timber bay]

not right off

[marble wagon]

expand it

[timber bay]

oh

seemed to have find that 4 = a1+a2+a3

well i got -4x^2 +6x-2 = -(a1+a2+a3)x^2+(a1a2+a1a3+a2a3)x -a1a2a3

[marble wagon]

yes

4 is the sum

2 is the product

[timber bay]

oh okay

[marble wagon]

so what's the minpoly?

[timber bay]

i didnt realize that about the 2

x-4 for sum?

[marble wagon]

yes

[timber bay]

x-2 for product?

[marble wagon]

well the product isnt asked

but sure

now for b you wanna find a poly with those roots

[timber bay]

oh okay so same process

[marble wagon]

so you could think of (x-a1^2)(x - a2^2)(x - a3^2)

why is this in Q?

[timber bay]

except with squares

yeah

[marble wagon]

can you see why it's guaranteed to be in Q?

[timber bay]

i dont

[marble wagon]

an element is in Q if the action of the galois group on it is trivial

in particular, symmetric polynomials in the roots will always be in Q

because the roots are permuted

so for example

consider an element f in the galois group of the fraction field over Q

f( (x-a1^2)(x - a2^2)(x - a3^2) ) = (x-f(a1)^2)(x - f(a2)^2)(x - f(a3)^2)

f is just going to permute the roots

so it's gonna fix the polynomial

therefore it is in Q

[timber bay]

hm okay

that makes sense

[marble wagon]

therefore it just remains to calculate the polynomial (x-a1^2)(x - a2^2)(x - a3^2)

by using the information you have on the roots

given by the initial polynomial

[timber bay]

thats what im doing now

[marble wagon]

its kind of a hassle

[timber bay]

thank you wolfram

[marble wagon]

lol

[timber bay]

x-4 seems to also be the root of this one

wait thats not right

oh wait im overcomplicating it

wait why is the answer not just (x-a1^2)(x - a2^2)(x - a3^2)

or will it just be that product, but replacing 4 with the sum of the three squares and replacing 2 with a1a2a3

@marble wagon

[marble wagon]

that is the answer

but you have to express it as a poly in Q

so you have to use some stuff

the 4 and 2 arent enough

you also need to use the 6

from the original poly

basically expand it out

and start eliminating the a_i

the sum of three squares isn't 4

but you can express in terms of (a1 + a2 + a3)^2

and your other information

[timber bay]

i got that when you expand (x-a1^2)(x - a2^2)(x - a3^2) the coeff of the x^2 is the sum of the squares

which gives that the coeff is 4 yeah?

[marble wagon]

no

why would it be

(a1+a2+a3) = 4

but you need (a1^2 + a2^2 + a3^2)

[timber bay]

don't you set the product equal to the original poly?

[marble wagon]

no

this is a different poly

you have to calculate it

[timber bay]

oh okay so you use the 2, 4, and 6 to express the product of the (x-ai^2)?

[marble wagon]

yes

[timber bay]

ah

okay cool

[timber bay]

this is hard to compute

[marble wagon]

really?

it's not that difficult if you do it in an organized way

[timber bay]

there are elements like (a1^a2^2+a1^2a3^2+a2^2a3^2) that i dont know how to deal with

[marble wagon]

the idea is the following

you have to set an ordering of which elements are worse

and deal with those first

for example we say a1 > a2 > a3 in terms of being worse

so we want to get rid of higher powers of a1 first

so we have

(a1^2 a2^2+a1^2 a3^2+a2^2 a3^2)

the highest power is a1^2

so we subtract the element a1^2 + a2^2 + a3^2 multiplied by an appropriate constant

my explanation isnt amazing

let me find it for you

[timber bay]

the constants wont have a's in them at all, right?

[marble wagon]

here

p.622 of dummit & foote

has a good description

the elementary symmetric functions are the terms of your original polynomial

s1 = a1 + a2 + a3

s2 = a1a2 + a1a3 + a2a3

s3 = a1 a2 a3

[timber bay]

yeah

[marble wagon]

so your problem is exactly that: decomposing symmetric polynomials in terms of s1, s2, s3

read the procedure in exercise 38 and carry it out

[timber bay]

what does lexicographic mean

[marble wagon]

it just means the ordering where a1 > a2 > a3

[timber bay]

oh

[marble wagon]

so to compare two monomials

you look at the powers of a1 first

if they are the same you go to a2

etc

[timber bay]

so A is a constant yes?

[marble wagon]

yeah

a constant in Q

i mean you dont need to prove exercise 38

you just need to carry the procedure out

[timber bay]

yeah im kind of confused by it

[marble wagon]

right so let's see

i'll do one example

and you can try the other ones

[timber bay]

okay

[marble wagon]

let's do a1^2 + a2^2 + a3^2

[timber bay]

i would think to start with s1^2

[marble wagon]

so the highest monomial is a1^2

the coefficient is 1

[timber bay]

do you want the expanded poly?

where a = a1 b=a2 and c = a3

[marble wagon]

you have to do it for each of the 3 coefficients

im doing it for the coefficient of x^2

which is a1^2 + a2^2 + a3^2

well negative that

anyway the highest monomial is a1^2, coefficient 1, and the powers are 1 s1^(2-0) s2^(0-0) s3^(0-0) = s1^2

that means that the first step is subtracting s1^2

[timber bay]

what do you mean by highest monomial?

[marble wagon]

you have to take the monomial which is highest in the ordering

[timber bay]

oh i see

[marble wagon]

so we subtract (a1^2 + a2^2 + a3^2) - s1^2 = -a1a2 - a2a3 - a1a3

now the highest monomial is a1a2

coefficient -1

and the new term is

-1 s1^(1-1) s2^(1-0) s3^(0-0) = -s2

we subtract and we get 0

so the polynomial is s1^2 - s2

[timber bay]

how have you determined the powers, like (1-1) and (1-0)?

[marble wagon]

they are the difference of the powers

this is annoying because the notation the book uses for the powers is a1 a2 a3

[timber bay]

yeah haha

[marble wagon]

and that's what we are using for the monomials

but the powers for the monomial -a1 a2 are (1,1,0)

[timber bay]

oh okay

so this was all for the coeff of x^2

[marble wagon]

it's basically an organized way of chopping it off part by part

yeah I intentionally did the one that wasn't so hard

[timber bay]

how shoule i repeat this to get the whole poly?

[marble wagon]

you have to do the coefficient of x and of 1 as well

[timber bay]

and add them all together?

[marble wagon]

i mean, what we noticed is that the coefficient of x^2 is -(s1^2 - s2) = - (4^2 - 6) = -10

[timber bay]

all the s's?

[marble wagon]

you know the value of the s's

from the original poly

remember?

[timber bay]

yeah

so add all the s's in terms of a's?

[marble wagon]

so we know right now the start is x^3 - 10 x^2 + ...

[timber bay]

oh wait

nvm

oh i missed something of yours

oih okay

ill try it out

thanks

-a1^2a2^2a3^2 will just be s3^2, so 4?

am i understanding that correctly>

okay let me try this here. so looking at coeff of x, its (a1^2a2^2 + a1^2a3^2 + a2^2a3^2) so that will be s1^(2)s2^(2)s3^0?

which is a huge polynomial.

so im pretty sure my exponents are off

[marble wagon]

yes

and uhh

no

the biggeest monomial is a1^2 a2^2

so the exponents are (2,2,0)

the exponents of the s are the differences

which are 2-2, 2-0, and 0

so it's s2^2

@timber bay

[timber bay]

ohh okay

2-2

which is zero

[marble wagon]

yeah

[magic owl]

What conditions do you need for G/K=H

To imply G=HxK

For G,H,K groups

Is it just that K be normal

[woven delta]

No

Look up projective modules

And the splitting lemma

Also K must be normal to do a quotient

@magic owl

[magic owl]

Ah all right, I doubt my proof strategy will work in that case

[woven delta]

What do you mean?

[magic owl]

I was hoping to solve a larger problem using that

[woven delta]

There are some very simple counterexamples lol

[magic owl]

Yeah I’m certain, I was just wondering if there were sufficient conditions that this problem satisfied

[woven delta]

If your quotient group is free abelian then you're good

Also G should be abelian

[magic owl]

Unfortunately neither are

A priori anyway

[woven delta]

Tbh this is usually not true, so I wouldn't count on it

[magic owl]

Yeah, I basically have groups X/R=A

And the goal is AxR=X

[woven delta]

Is that actually the question?

[magic owl]

But I probably have to use more facts about the groups

It’s regarding fundamental groups and retractions

[woven delta]

What is the question exactly?

[magic owl]

Don’t want too much help, but it’s given a retraction X onto A by r

Show that pi(X)=Ker r x i*pi(A)

Where i is the homomorphism induced by the inclusion A into X

i is the inclusion i* is induced*

[woven delta]

Yeah, it definitely depends on the specific case

Good luck

[magic owl]

It’s a totally general retrzction though

I’ll think on it I guess

[uncut girder]

The determinant is the unique elementary alternating n-tensor on R^n.

I love this definition. Its so much better than the axiomatic or the inductive definition.

[marble wagon]

is it?

[fickle brook]

define elementary tensor?

[marble wagon]

norm preserving probably

[bleak abyss]

Nah it's what some call pure tensor

Instead of a sum of them

Or actually hmm now I'm not sure

Okay so the way I'm used to thinking about it is, the space of alternating n-tensors is 1-dimensional

[uncut girder]

yes

[bleak abyss]

So you want to send the identity to 1, that characterizes the determinant

[uncut girder]

i was trying to define elementary alternating k tensor from scratch

in one sentance

[bleak abyss]

I mean the main confusion was the definition of elementary, I had thought this word was synonymous for "pure tensor" but if our space is 1-d then that's kinda irrelevant, what's more important is scaling

[marble wagon]

right so it probably means norm preserving

[uncut girder]

???

[somber bramble]

We defined it as the unique multilinear and alternating map in the dual space of square matrices fulfilling det(I) = 1

[marble wagon]

what do you mean by the dual space of square matrices?

that sounds like R^(n x n)

which isn't really the case

[somber bramble]

space of linear functions from Mat(nxn,F) -> F where F is the base field

that's what dual space of finite dim. VS means

(for infinite dim I think you need to also explicitly state continuity or sth)

oh, hm, though, det isn't actually linear so I did misspeak

so just pretend I said "function to the base field" rather than "element of dual space"

[fickle brook]

$(\bbR^n)^{\otimes n}$

[cloud walrus]

Ann:

[fickle brook]

wait no

wrong space

[marble wagon]

the point is that's misleading sascha, det isn't the top form in R^(nxn) (there is no such form unless n = 1)

it's the top form in (R^n)^n

so it's like, the (norm preserving) top form in the matrix row/colspace

[somber bramble]

ah, fair enough, it's multilinear in the rows/columns, not the entires; is that what you mean?

[marble wagon]

yeah

[chilly ocean]

aye so what does it mean to left multiply by G?
in the context of isomorphisms with permutations

o nevermind im dumb

[bleak finch]

Let at^2 + bt + c be an irreducible polynomial over some field, and consider F[x]/(at^2 + bt + c). Is it true that for any r in F[x]/(at^2 + bt + c), r = x - ca^{-1} + (y - ba^{-1})t for some x, y in F?

[woven delta]

@bleak finch let x be t^2

And multiply both sides by a

Note that the quotient space is a field so things work nicely

[heavy jasper]

This is in section 4.5 of dummit and foote. Im having trouble seeing how 52 and 53 work together - if your group G in 52 is non abelian, then it would act as a counterexample to 53? Then in 52, since its abelian and simple, G must be cyclic? I know that since 53 is supposed to use 52 in its proof its (maybe?) cyclic reasoning, but I feel like it makes sense regardless? basically im confused about whether 53 implies more information about the group G in 52/is the problem statement in 52 purposely being more general than it actually is in practice, as in not telling reader properties of G that must be true as part of the problem statement. I hope that made sense, thanks in advance

[woven delta]

The first part is trivial

Look at the normalizer of their intersection

@heavy jasper

[bleak finch]

How do I show a ring is a reduced local ring?

[fickle brook]

what's the defn of a reduced local ring

[bleak finch]

That's what I would like to know

[bleak finch]

How do I show a ring is a reduced local ring?

Better question.

How do I show that the ideal F[t]/((t - a)^2) has a maximal ideal?

[fickle brook]

you mean the ring F[t]/( (t-a)^2 ) has a maximal ideal?

doesn't every ring have one, by Zorn's?

[bleak finch]

@fickle brook I need to show that F[t]/( (t-a)^2 ) is local, and that involves showing that there is only one maximal ideal.

[fickle brook]

oh, a unique maximal ideal

you should've specified

[bleak finch]

Sorry

[fickle brook]

hmm

well i mean the quotient by that ideal must be a field

so like i'd try to find an ideal, show it's maximal, and then try to show no other ideal is maximal? or sth?

[bleak finch]

What is sth

Maybe I'll post the whole problem

Is there a trick to show that F[t]/( (t-a)^2 ) is a local, irreducible ring?

[fickle brook]

what's a reducible ring?

[elder valley]

The correspondence theorem will give you local

[bleak finch]

reducible: I think it means a ring with an element such that e^2 = 0, so irreducible means no such element

[elder valley]

Well t-a is reducible under that definition

[fickle brook]

irreducible = no nilpotent of index 2?

[bleak finch]

@elder valley yes, but recall that (t-a) is modded out in F[t]/( (t-a)^2 )

[elder valley]

The square of it is

[bleak finch]

Whoops, I got it reversed. Reduced = no square nilpotent and non-reduced = has such an element. I needed to show that it is non-reduced.

[elder valley]

Ah. Then I gave away the answer lol

[bleak finch]

Now for the correspondence theorem...I'm not sure how that can be used to show that it is local.

[elder valley]

What definition of local are you using?

[bleak finch]

unique maximal ideal

[elder valley]

The correspondence theorem tells you exactly what the ideals of the quotient ring are in terms of the ideals of F[x]. And F[x] is a PID so it's ideal structure is very simple

[bleak finch]

My guess is that (t - a) is an ideal. My guess is that the proper sub-ideals of ((t-a)^2) are the ideals of F[t]/( (t-a)^2 ), and there is only one sub-ideal so...

[elder valley]

Yeah that's exactly right

The fact that the exponent is 2 doesn't really matter, it will work the same for any power. So locality isn't necessarily coming from there being only one sub-ideal, it's that the ideals contained in (t-a)^n form a containment chain

Actually it's not sub-ideals, it's super-ideals. Ideals that contain ((t-a)^2)

[bleak finch]

If this town

Is just an apple

Then let me taaaake a biiiiite....

[bleak finch]

a is an algebraic integer iff it is a root of a minimal polynomial with coefficients in Z.

What is a "minimal polynomial"?

Nvm wikipedia has informed me

[elder valley]

Monic polynomial*

[bleak finch]

Let D \equiv 1 mod 4. Why is Q[\sqrt{D}] of the form x + ((1 + \sqrt{2})/2)y?

[marble wagon]

you probably mean the ring of integers

not Q[\sqrt{D}]

but the integral closure of Z inside it

it's a computation that you have to carry out

computing the integral elements over Z

in the 1 mod 4 case there's a cancellation that happens and so the halves can also get in

it doesnt happen in the 3 mod 4 case

[bleak finch]

Let D = 1 + 4k and D be square-free.

[marble wagon]

being square free doesn't actually matter, it's just a simplyfing assumption

because extending by \sqrt{D} and \sqrt{D k^2} is the same

[bleak finch]

(x - (a + b√(1 + 4k)))(x - (a - b√(1 + 4k))) = x^2 -2x(2a) + a^2 - b^2 (1 + 4k)

I see no cancellation that is happening.

[marble wagon]

what is this

you just showed these elements are integral

[bleak finch]

I'm trying to find the "cancellation that happens" in the 1 mod 4 case

[marble wagon]

so that Z[\sqrt{D}] works

now you have to show that the others also work

the ones with 1/2

[bleak finch]

Okay I get it.

I think

[marble wagon]

put a 1/2 in there

and you'll see what happens

[bleak finch]

So how do I know there isn't a fractional case when D equiv 2 mod 4?

[marble wagon]

i dont exactly remember but you basically just try in general

you write the equation and it doesnt work

just an element (a/b) + (c/d)D

[bleak finch]

makes sense

And you were right: there WAS a cancellation that happened!

[modest pecan]

Let F be a finite field of characteristic p. Prove that lFI = p^n for some positive integer n.

[somber bramble]

nah

[somber bramble]

again, nah. don’t feel like doing more algebra today

(if you’re hoping to get help, maybe say so, and also say what you need help with specifically. people are not going to just solve the problems for you)

[modest pecan]

I'm thinking about these problems. Seeing if I could maybe think with other people.

[marble wagon]

for 7 you just do what it says

use (sqrt(2) + sqrt(3))^2 to get an element of the form a sqrt(2) + b sqrt(3)

and with that one and sqrt(2) + sqrt(3) you get both

[modest pecan]

a = sqrt(2) + sqrt(3)

(a^3 - 9a)/2 = sqrt(2)

(11a - a^3)/2 = sqrt(3)

[fickle brook]

use (sqrt(2) + sqrt(3))^2 to get an element of the form a sqrt(2) + b sqrt(3)

not exactly of that form

[modest pecan]

this would prove the other inclusion, but this doesn't follow the problem's "guideline"

[somber bramble]

what ann said, {√2, √3} is not a basis of ℚ(√2, √3)

(for the obvious reason of the latter being a 4-dimensional vector space over ℚ)

you’re missing two basis elements

[fickle brook]

one is obvious

||pun intended||

[somber bramble]

btw, random insight I had a few days ago:
I was annoyed for the longest time that there was the notation F(x) for rational functions over F, as well as F(a) for F adjoint a

they’re actually just the same thing

[modest pecan]

how do I know [ℚ(√2, √3) : ℚ] = 4?

[somber bramble]

consider intermediate extensions

you can find a tower of (two) extensions which you can both show to be of degree 2 rather easily

[modest pecan]

so ℚ(√2, √3) = ℚ(√2)

[somber bramble]

and degrees multiply

yes, and ℚ(√2)/ℚ is degree 2 (that one’s easy to see)
ℚ(√3,√2)/ℚ(√2) can be shown to also be of degree 2 with the same argument, but you have to show that √3 does not actually lie in ℚ(√2)

(to see how, find the irreducible polynomials)

[modest pecan]

x^2 - 3 = 0

would I have to show this is irreducible in ℚ(√2)

[somber bramble]

yes. which means (as its degree is 2) you have to show it has no roots in ℚ(√2)

luckily you know its roots

[modest pecan]

any tips for the other problem?

(Let F be a finite field of characteristic p. Prove that lFI = p^n for some positive integer n.)

I don't think I have a complete enough idea for the structure of fields

[fickle brook]

isn't that a linear algebra problem lmao

[modest pecan]

it was in dummit and foote

[fickle brook]

try to show F is a vector space over Z/p

you'll automatically get that it's findim

[bleak abyss]

Not the right channel, maybe in #precalculus or something

[marble wagon]

what do you mean middle school isn't advanced mathematics?

[somber bramble]

Normal math is linear equations
Once you get to squares it's advanced as shit

And don't even talk about the quintic

[random crag]

god forbid fractions in the exponent

[short oxide]

[lyric basin]

[timber bay]

im not sure how to start this one. im not good at composition series. i was thinking like s3 as a subgroup of s4 but im not sure if that's the right process.

[bleak abyss]

@timber bay you need a normal subgroup in particular, not just a subgroup

But in order to do a composition series, you need to have simple quotients, and if you know the group is solvable (since that's what the question is asking), you should be aiming for these to be abelian.

Thus, what you wanna do is find a surjection from S_4 to Z/p where p is prime, and find the kernel

What's the obvious one?

[timber bay]

well i know its not z/24

[bleak abyss]

24 isn't the primest of numbers, I'll give you that :P

[fickle brook]

[timber bay]

is there an obvious one?

[bleak abyss]

Think back in terms of subgroups, what's the immediate roflsubgroup of S_n, for any n?

I'm not sure if calling it "roflsubgroup" communicated what I want to, but like, the first thing that comes to mind

[timber bay]

S_(n-1)

[bleak abyss]

Not quite

Like, S_(n-1) doesn't sit cleanly in S_n, there are n ways at least to do it

[timber bay]

well <e> is trivial

[bleak abyss]

And it's not normal

Give me a normal subgroup

Or a hom to Z/p for some (hint: small) prime

[timber bay]

hmm

im not sure

[bleak abyss]

A_n

[timber bay]

oh of course

we just started talking about the alternating group

is it abelian?

[bleak abyss]

The quotient is

[timber bay]

oh

[bleak abyss]

S_n/A_n is Z/2

That's the point

[timber bay]

oh of course

[bleak abyss]

If you're trying to prove a group is solvable you want a chain of normal subgroups so that the quotients are just Z/p for various primes

[timber bay]

since 2* |A_n| = |S_n|

[bleak abyss]

So now we have A_4 ≤ S_4

[timber bay]

left cosets equal right cosets

[bleak abyss]

Now even I don't quite recall offhand what the next guy looks like... So let's see

I know A_4 is the rotation group of the tetrahedron

[timber bay]

A_3?

oh interesting

there is the double transposition subgroup, A_3, and V_4 whatever that is

[bleak abyss]

Do we have a unique subgroup of any given order?

Aside from the stupid cases

[timber bay]

no?

[bleak abyss]

Like idk if there's an order 6 subgroup at all

[timber bay]

is there an order 2 subgroup?

like i know theres got to be

like (0), (12) or something

[bleak abyss]

That's not even

[timber bay]

oh true

[bleak abyss]

You'll need a double transposition

[timber bay]

yeah

[bleak abyss]

Oh wait

Wait okay so

{0, (12)(34), (13)(24), (14)(23)}

I'm pretty sure that's the only subgroup of order 4

Yeah like

S_4 surjects to S_3 by acting by conjugation on the double transpositions

And the point is that the image of A_4 under that hom is gonna have order 12/4 = 3

So okay double transpositions are normal in A_4

With quotient Z/3

And that guy is the Klein 4 group

So it's trivial from there

So here's the series

[timber bay]

oh dope

i got it

how do you denote the double transposition subgroup?

[bleak abyss]

{e} ≤ {e, (12)(34)} ≤ {e, (12)(34), (13)(24), (14)(23)} ≤ A_4 ≤ S_4

Replace ≤ with \lhd because I'm on phone and lazy to TeX

[timber bay]

im just doing \leqslant

[bleak abyss]

Anyway I have an exercise for you now

I'm pretty sure there's another composition series for it

[timber bay]

so for proving b, well show theres an isomorphism from S_4 to F[x] for any f(x) deg less than or equal to 4, and subgroups of S_4 have orders 1,2,4. so we can then kind of conclude that F[x] is solvable?

is that the right reasoning

[bleak abyss]

Oh lemme see real quick

[timber bay]

yeah cant we just choose another subgroup of the order 4 subgroup? and it should also be a normal subgroup

[plain sequoia]

@timber bay Without any more context, if H is normal in G and K is a subgroup of H then K is also normal in G

so I dunno if that answered your question or not

[timber bay]

yeah

what question were you trying to answer?

[timber bay]

should i just try to find a refinement in order to prove a?

would that be the best method?

our F is not a field of characteristic 0, right, because its mod2

which is what we've been working with for the past week

[oblique river]

i'm just joining in, but: if you want to do part a, you should first figure out what group G is

@timber bay

[timber bay]

yeah I'm not having a good grasp of it.

it contains all the roots of the poly but mod 2?

[oblique river]

you're still trying to think in characteristic 0

[timber bay]

I know 0 and 1 aren't roots of f(x) obv

[oblique river]

correct

the roots aren't in Z/2Z

in the same way that in characteristic 0, the roots of that polynomial aren't in Q

[timber bay]

so it'll be something like Z/2Z(i) or something?

[oblique river]

"or something", yes

meaning it's not literally Z/2Z(i)

but it is an extension of Z/2Z

[timber bay]

so like z/2z adjoined something

adjoined the descriminant?

[oblique river]

yes, and that something is "the roots of that polynomial"

[timber bay]

so sqrt(-3)

[oblique river]

no

[timber bay]

oh wait

yeah it's not Q

[oblique river]

the roots of that polynomial

[timber bay]

it has to be the whole thing yeah

the whole root

[oblique river]

what is the degree of that extension

[timber bay]

Z/2Z(-1+/-sqrt(-3)/2)

2

[oblique river]

uh oh

you just divided by 2

but 2 = 0

so you jsut divided by 0

[timber bay]

oh

hmm

well

[oblique river]

you dont need a formula

at all

the thing you're adjoining is literally just the root of that polynomial

[timber bay]

so just call it alpha or something

[oblique river]

sure

does F(alpha) have the other root of that poly?

[timber bay]

no

or are they equal mod 2 or something

the two roots

[oblique river]

you're trying to think about the roots in characteristic 0 again

everything is living in characteristic 2

but in any case, the answer was yes. over any field, once you add one root of the quadratic, you get the other one for free

always

if you have a polynomial f(x) = x^2 + bx + c

then c is the product of the roots

[timber bay]

oh okay

[oblique river]

so once you have one root alpha, the other is c/alpha

which you have because it's a field

[timber bay]

hmm

[oblique river]

ok so we've established that the splitting field is K = F(alpha)

[timber bay]

c/alpha? why that?

[oblique river]

because the product of the two roots is c

if one of the roots is alpha, then the other is c-alpha

[timber bay]

why does the +bx thing not impact it?

[oblique river]

what? -b is the sum of the roots

literally multiply out (x-alpha)(x-beta)

[timber bay]

oh okay

oh I see what you're saying.

[oblique river]

the point of that was just to say that for any field and any quadratic, oncey ou have one root you automatically have the other

so now we have established that K = F(alpha)

what is the degree of K over F

[timber bay]

2?

I wouldn't be 0

[oblique river]

wait why would it ever be 0?

what? degree is always a positive integer

[timber bay]

well 2 is 0 mod 2z

[oblique river]

no

[timber bay]

yeah

yeah I know

[oblique river]

degree is not a thing that happens mod 2

[timber bay]

yeah

[oblique river]

degree is a nonnegative integer. it's the dimension of K as a vector space over F

but yes, the answer is 2

is K/F galois?

[timber bay]

yes

[oblique river]

why

[timber bay]

well there's a galois group of k/F at least

of automorphisms of the roots

brb

[oblique river]

ok well what is the size of that group

you know it divides the size of the extension

[timber bay]

back

2 roots so the size is 2?

identity and transpositoin

or

@oblique river

[oblique river]

yes

[timber bay]

oh okay

was beginning to doubt myself

[oblique river]

you should verify that the numbers aren't the same I guess, so that that actually defines a nontrivial automorphism

but okay, so the group has order 2

[timber bay]

yeah

[oblique river]

how many groups of order 2 are there

[timber bay]

well 1 essentially

[oblique river]

what is it

[timber bay]

z/2z

[oblique river]

is that group solvable

[timber bay]

S_2

lol

yEAH

it is

it has abelian quotients

(z/2z)/(e) = z/2z

[oblique river]

I mean "it has abelian quotients" is not a good enough reason to be solvable

every group has an abelian quotient

but not every group is solvable

[timber bay]

well the only subgroup is trivial

[oblique river]

so what is your reason

for Z/2Z being solvable

[timber bay]

the series expansion has abelian quotients i meant

[oblique river]

ok great

yes, our group is solvable

[timber bay]

since the only subgroup is the identity

[oblique river]

Z/pZ is solvable for any p

uhhhhhh Z/2Z is also a subgroup of Z/2Z lol

[timber bay]

ah

[oblique river]

in general any abelian group is solvable

[timber bay]

nontrivial

oh

that makes sense

an abelian group mod an abelian group is another abelian group?

we know that for sure?

[oblique river]

that is a super basic group theory fact, you should know that

[timber bay]

uh oh

honestly algebra's been a pretty big blur

[oblique river]

or at the very leas you should be able to prove it in like 30 seconds

[timber bay]

im trying to grasp the ideas

yeah probably

[oblique river]

but the ideas are useless if you don't know what to do with them or how to use them in examples!

[timber bay]

youre right

okay so is the proof of the extension by radicals trivial?

like dont we know that for each root, the product is 1, and the sum is -1

well the sum doesnt matter

[oblique river]

what is a radical extension?

it's an extension of the form "adjoining the nth root of some element"

so you need to show that alpha to some power is in F

cuz then alpha will be an nth root of that number

[timber bay]

i can do that brute force right?

like directly showing the roots of the polynomial squared or something is in F

[oblique river]

is alpha^2 in F?

[timber bay]

yeah it should be shouldnt it?

[oblique river]

nope

alpha satisfies alpha^2 + alpha + 1 = 0

so that means alpha^2 = alpha + 1

which is not in F

what about alpha^3

[timber bay]

oh okay

= alpha^2 + alpha?

is that right?

[oblique river]

yes

and what is alpha^2

[timber bay]

oh okay so alpha^3 =2alpha +1

[oblique river]

yes

is that in F?

[timber bay]

nope

[oblique river]

why not?

[timber bay]

oh

it is

2alpha is 0?

so 2alpha+1 is in F?

[oblique river]

yes because 2 = 0!

2alpha + 1 = 1

because 2 = 0!

[timber bay]

yeah

well 2z = 0

we dont know alpha is an integer?

[oblique river]

what?

[timber bay]

or does this not apply?

[oblique river]

it's certainly not an integer!

[timber bay]

yeah

[oblique river]

alpha is a root of a polynomial over a field of characteristic 2

the integers aren't even present in this problem

[timber bay]

Z/2Z is what im talking about

[oblique river]

yeah

2 = 0

so 2a = 0a

for any element a

of any characteristic 2 field

[timber bay]

even for non integer a?

[oblique river]

start with 2 = 0

multiply both sides by a

[timber bay]

which i thought alpha was

[oblique river]

that's how math works haha

multiply both sides of an equality by the same thing and you still have equality

[timber bay]

yeah im just thinking the fact that alpha is non-integer makes 2alpha not necessarily 0

but if it is i can just accept that

[oblique river]

no, stop

it's not something you have to accept on faith

[timber bay]

sure

[oblique river]

it's literally like the most basic property of algebra from like 5th grade

if you have an equality

and you multiply both sides by the same thing

then you still have equality

[timber bay]

yeah we didnt talk about modular arithmetic in 5th grade

i understand that

[oblique river]

this isn't about modular arithmetic!

[timber bay]

were dealing with z/2z

i can just accept it

its okay

[oblique river]

but it's not a Z/2Z fact

or a Z/7Z fact

it's a general fact about any algebraic structure

[timber bay]

like imagine if a = 1/2.

[oblique river]

NO!

1/2 is not an element in Z/2Z

or any characteristic 2 field

[timber bay]

neither is alpha

[oblique river]

because you can't divide by 2!

alpha actually exists somewhere

1/2 doesn't exist

because it's just 1/0

and you can't divide by 0

[timber bay]

ok

i see

[oblique river]

like, if a = b

that means they are literally equal

literally the same

[timber bay]

yes

[oblique river]

completely interchangable

so a*c must equal b*c

because a is literally equal to b

[timber bay]

i understand the basic algebra. but i was expecting the fact that alpha is not an integer to mess things up

[oblique river]

nothing here is an integer

don't think of 0 or 1 as integers

they are not integers

they are elements of the field Z/2Z

sure, that field happens to be a quotient of the integers

but that's irrelevant

[timber bay]

man, this is a buncho bananas

[oblique river]

no i'm a buncho bananas, this is just field theory

[timber bay]

oh dang

okay so alpha ^3 is in F

which is enough to prove K is an extension by radicals

[oblique river]

yes

[timber bay]

dude got a week or so left and you won't have to deal with me anymore

lol

until grad school

im not that good at this stuff but i enjoy doing it. i can see myself not giving up

[oblique river]

dont give up! :)

[wide oxide]

@timber bay thats how I feel

I'm not that good but I enjoy it

At least you've made it to grad school

I've been out of under for a few years, and I'll be lucky if I can get into grad within the next 2

I never learned abstract algebra, but it's one of those fundamental things I wish I had

As a math post grad I'm somewhat ashamed

[timber bay]

I'm not in grad school yet

but hopefully I get there

[topaz solar]

grad is oof

[solar wyvern]

@oblique river what do covering spaces have to do with galois theory/groupsis

Hatcher give some copout answer

[oblique river]

@solar wyvern it's not so much that there is a literal connection as just a lot of similarities in how they behave

for example, there is a notion of a regular covering space, which is similar to a galois extension (as opposed to just any old field extension)

and the group of deck transformations behaves like the galois group in the sense that there is a "galois correspondence" between subgroups and subcovers

that's order-reversing, and normal subgroups H correspond to regular subcovers with deck transformation group G/H

this is one of the motivations for ideas of "covering spaces" in algebraic geometry. i put them in quotation marks because they're usually not gonna be manifolds or anything nice

which are more connected to galois theory

[solar wyvern]

so etale stuff?

[oblique river]

yep

things like "If L is a field extension of K then Spec L is an etale cover of Spec K with "deck trasnformation group" equal to Gal(L/K)"

[solar wyvern]

dumb but isn't spec of a field a point space

[oblique river]

as a topological space, yes

but Spec isn't just a topological space, it's a scheme

and as schemes they are different

[solar wyvern]

how

[oblique river]

well just in the same way that you can have two different group structures on the same set, you can have two different scheme structures on the same topological space

the idea is that the scheme structure tells you what the functions on your space are

Spec K is a point, so a function on Spec K should just spit out a single number

in this case, an element of K

similarly, {functions on Spec L} = L

the object "Spec K" remembers that it came from K

[solar wyvern]

ah, so you can distinguish fields

[oblique river]

yep

one sec

so schemes can also detect more subtle things

for example consider the ring R = C[x]/x^2

here C is the complex numbers for concreteness

this ring has a unique prime ideal

so once again Spec R is just a single point as a topological space

but functions on Spec R are just elements of R, i.e. things of the form a + a'x

where x^2 = 0

let's try to think about what this should mean. let's take two functions a + a'x and b + b'x and multiply them

(a+a'x)(b+b'x) = ab + (ab' + a'b)x

no x^2 term because x^2 = 0

does "ab' + a'b" ring a bell to you at all? like does that expression look familiar?

( @solar wyvern tell me if you want me to stop lol)

[solar wyvern]

sorry was jus getting food

[oblique river]

np, do you want me to spoil the answer to my last question

[solar wyvern]

gimme a minute

[oblique river]

k

just @ me

[solar wyvern]

tempted to say leibniz (sic?) rule

[oblique river]

yes!

we should think of the a' and b' as like derivatives

so a function on Spec R looks like a + bx

a is the value of the function, and b is the value of its first derivative

[solar wyvern]

seems...linear

[oblique river]

think of it as the linear approximation of a function

if you start with Spec C[x], that's just C

via the corerspondence (x-a) <--> a

and functinos on C are just polynomials, i.e. elements of C[x]

there is a quotient map C[x] --> C[x]/x^2

well, let's start with the quotient map C[x] --> C[x]/x = C

quotienting by x is like saying "just take the constant term", which is the value of the function at 0

so Spec C is really just like the point 0 sitting inside Spec C[x]

[solar wyvern]

so the 0th order approx sorta?

[oblique river]

yep

and then the quotient map C[x] --> C[x]/x^2 is like "take the linear approximation at 0"

because you're discarding all the higher order terms

all you're left with is f(0) and f'(0)

the map is f(x) --> f(0) + f'(0)*x

so that's why we think of Spec C[x]/x^2 as not just the point 0

but the point 0 with some fuzz around it

because a function on this scheme isn't just a number, it's like a function and its derivative

[solar wyvern]

hmm, so would C[x] --> C[x]/(x-α)^n be the n-th order approx around α？

[oblique river]

yep

so (just taking alpha = 0), C[x]/x^3 would be quadratic approximation

So Spec C[x]/x^3 is still a 1-point topological space

but functions on it comprise of "the value of the function, the value of its derivative, and the value of its second derivative"

(at x = 0)

so there's some kind of "second order fuzz"

[solar wyvern]

I "know" there's more info than the top space but it takes a bit of getting used to 😄

[oblique river]

yep, the trick is to think algebraically

Spec R is "the thing whose functions are elements of R"

the whole issue with topological spaces is that they can't capture that often

here is a reddit post I made about schemes a while ago: https://www.reddit.com/r/math/comments/a0dd6j/mathematicians_what_is_your_favourite_algebraic/eah33w3/?context=3

[solar wyvern]

so umm, can we talk about this in sheafy terms?

[oblique river]

here is the main idea:

Let's step back a second. Recall that if R,R' are two k-algebras with corresponding varieties V,V', then a map R -> R' gives you a map V' -> V, and vice versa. For example, the map C[x] --> C[x]/(x) corresponds to the map on varieties "include the point {0} into C". Thinking in terms of functions, the map C[x] --> C[x]/(x) is like "take a function on C, and only remember its value at 0".

From this point of view, there is a map from Spec(C[x]/(x^2)) (the "variety" (really a scheme) associated to C[x]/(x^2)) into C = Spec(C[x]) corresponding to the map C[x] --> C[x]/(x^2). The inclusion is like "include {0, the tangent space of 0 in C} to C" and the map on rings is like "take a function on C and remember both its value at 0 and the value of the first derivative".

yeah, we can talk about sheaves

[solar wyvern]

I read the lovering expository thing on sheaves and sorta seems like it'd be handy

[oblique river]

yeah, a scheme is just a topological space with a sheaf of rings

[solar wyvern]

what's a variety

[oblique river]

a subset of C^n cut out by polynomial equations

[solar wyvern]

oh ya, is it necessarily over base field C?

[oblique river]

traditionally yes

but not necessarily I suppose

[solar wyvern]

some algebraically closed field?

[oblique river]

yeah, you can do it over R too but it gets harder

C is nice because it has a topology whereas other fields don't really

[solar wyvern]

yeah, was wondering if the top was relevant

[oblique river]

traditionally the study of varieties was about their topology

[solar wyvern]

or if people cared about like

[oblique river]

things like cohomology

[solar wyvern]

\bar{Q_p}

or positive char

[oblique river]

which relies on topology. so algebraic geometry is trying to like generalize that to other fields using etale stuff

maybe C_p so that it's complete

but yeah

[solar wyvern]

ah, so you might care about completeness

(makes sense if you care about top props)

[oblique river]

probably, idk though I dont think about that kinda stuff

[solar wyvern]

sorry if this is sorta 🤤

[oblique river]

its fine haha

you're not 🤤

[scenic harness]

(is it ok if i leave a reminder here to read this?) (If not it's too late sorry)

[solar wyvern]

that work?

[scenic harness]

Wew, thanks flim

[solar wyvern]

np

thank buncho

[oblique river]

👍

[solar wyvern]

just reading the thing you posted

[elder valley]

Cool discussion! 👍

[stone fulcrum]

🤔

[brisk granite]

Do you guys know any good abstract algebra textbooks

I'm new...

[chilly ocean]

Abstract Algebra by Dummit and Foote I guess

[woven delta]

Topics in algebra by Herstein is nice

[crimson mango]

I just wrote the weakest proof ever about how alpha and beta along with there inverses will always be an even permutation

[woven delta]

What does that even mean?

[crimson mango]

its a proof about how a^-1b^-1ab will always be even

[woven delta]

Isn't that obvious though

[crimson mango]

ya

but my proof is weak i think

[woven delta]

You can decompose it into an even number of 2 cycles

That's a proof

What did you say?

That's a really cute fact actually

It tells you that the commutator subgroup is A_n for n>4

[crimson mango]

i just used that if beta is even then so is beta^-1

and i split it up into three cases

[woven delta]

Why though

[crimson mango]

where one could be even and odd, both even, and both odd

[woven delta]

Just decompose alpha and beta into 2 cycles

Then alpha and beta inverse will have the same decomposition but backwards

And just count how many 2 cycles there are

[crimson mango]

ya but you are not given any cycles

[woven delta]

The point is you can always decompose them

[crimson mango]

yes i know

but there is nothing to decompose?

[woven delta]

So given a decomposition

Of alpha and beta

[crimson mango]

I would have to use word but then i would still have to break it down into three cases

[woven delta]

You also have a decomposition of aba^-1b^-1

Which is composed of 2(#a + #b) 2 cycles

Where #a refers to the number of 2 cycles in your decomposition

[crimson mango]

I understand that everything can be broken down into 2-cycles

[woven delta]

So given a decomposition

We construct a decomposition of aba^-1 b^-1

What cases did you do?

Like if one was even and one was odd of something?

I guess that works also tbh

I guess the easiest proof is just knowing that sgn is a homomorphism from S_n to {-1, 1}

And that sgn(a^-1)=sgn(a)

So sgn(aba^-1b^-1)=sgn(a)^2 sgn(b)^2

That's a 2 second proof

[crimson mango]

Ya i just split it into three cases

Anyways im go watch game of thrones

[topaz solar]

Alright, so I have 3 objects, w, v, and x

assigned to each object is a set (I, J, and K, respectively)
Is there any way to "detangle" wI + vJ + xK such that there exists no overlap $(wa + wb = w(a\cup b)$ as well)?
I've been trying at this for a while

[cloud walrus]

Darkrifts:

[topaz solar]

I have no idea what to do at this point since I'm p convinced you cant by operating only pairwise to send $aI + bJ = aI\setminus J + bJ\setminus I + (a+b)*(I\cap J)$

[cloud walrus]

Darkrifts:

[topaz solar]

Every method I try (which, granted, is only a small amount) ends with a duplicate section which was the first separated

[plain sequoia]

If $G \simeq H$ and $K \simeq L$ then $G \times K \simeq H \times L$ right?

[cloud walrus]

markus:

[fickle brook]

well why don't you verify that yourself 😛

[solar wyvern]

what's simeq

[plain sequoia]

Isomorphic

[marble wagon]

there's an obvious candidate for your isomorphism

check that it works

[solar wyvern]

@oblique river what does it mean to have an open cover by affine schemes, and is this equivalent/relatex to your scheme as a sheaf being representable?

not sure if this belongs in #geometry-and-manifolds

[topaz solar]

Update: figured out my problem

[oblique river]

@solar wyvern are you talking about the definition that's like "a scheme is something that has an open cover by affine schemes + some gluing condition"?

also, representable by what? normally I feel like people talk about functors being representable meaning "are of the form Hom(X, -) for some scheme X"

[solar wyvern]

@oblique river for the second I was looking at this def
An affine scheme is a scheme that as a sheaf on the opposite category CRingop of commutative rings (or equivalently as a sheaf on the subcategory of finitely presented rings) is representable. In a ringed space picture an affine scheme is a locally ringed space which is locally isomorphic to the prime spectrum of a commutative ring. Affine schemes form a full subcategory Aff↪Scheme of the category of schemes.
https://ncatlab.org/nlab/show/affine+scheme

yeah for the first

[oblique river]

yeah so like a nonaffine scheme wont be representable in that category

CRingop

I think

like P^1

[solar wyvern]

ummm, so then for that nonaffine scheme how do we cover it by affine schemes

how is it "locally" affine

[oblique river]

like P^1

think of P^1(C) being covered by two copies of A^1(C)

that's a very geometric picture

and the scheme story is the same idea

[solar wyvern]

tbh not quite entirely comfortable with projective space especially geometrically

[oblique river]

the sphere S^2

puncture at the north pole and you have a plane R^2 left

puncture at the south pole and you have an R^2 left

so S^2 is covered by two copies of R^2

(or C, really)

[solar wyvern]

and if you identify antipodal points S^2 is RP2?

[oblique river]

i mean yes but not relevant here

[solar wyvern]

oop

[oblique river]

we're talking about covering S^2 with two copies of C

each C is open

and "affine"

[solar wyvern]

affine because it's literally C right?

[oblique river]

yes

C = Spec C[t]

which is affine

but P^1(C) (which is topologically S^2) is not Spec R for any ring R

[solar wyvern]

How would you go about proving that

here and maybe in general

here at least