groups-rings-fields

[lethal dune]

now I can finally try to understand this

[gusty halo]

btw

what is this about

[hidden haven]

Type theory probably

Only mniip and clerk would know

Oh it is called Lawvere

Yeah I think he did some type theory

And other category stuff

But that doesn't look like the other category stuff

It's something in categorical logic for sure lmao

[delicate orchid]

ah I recognise this yes yes it is the diagram constructing the trivial abelian grape from just the empty set and abelian varieties

[lethal dune]

[rustic crown]

[chilly ocean]

Can anyone please suggest some good playlists on group theory

[delicate orchid]

I quite like this one https://www.youtube.com/watch?v=IP7nW_hKB7I&list=PLi01XoE8jYoi3SgnnGorR_XOW3IcK-TP6

[chilly ocean]

Thanks wew lads

[lethal dune]

https://youtube.com/playlist?list=PLwV-9DG53NDxU337smpTwm6sef4x-SCLv
this is also a playlist I like

[hidden haven]

rubrix cube

[delicate orchid]

roobick cub

[lethal dune]

no that's not a rubicks cube, that's a category

[hidden haven]

stop doing catgegory theory

[lethal dune]

shouldn't have taught me

[hidden haven]

[lethal dune]

lol classic evil apprentice arc

[delicate orchid]

the entire maths server when they see a triangle with a G and a H in the corners:

[hidden haven]

me when shape

[broken stirrup]

what's the example of a family of finite groups whose direct product has an element of infinite order?

[delicate orchid]

me when I see D_3 and I go "Omg first iso theorem"

[hidden haven]

Z/nZ product over n probably

take (1,1,...)

[delicate orchid]

I don't think so moldi

that has order n

[hidden haven]

what

[delicate orchid]

if you're taking Z/nZ product with itself n times

[hidden haven]

product over n

is different from product n times

lul

[delicate orchid]

rip

my example is just

$\bigoplus_{p = prime} C_p$

[broken stirrup]

he's right Z_n will do

[cloud walrus]

Moldilocks ✓

[hidden haven]

Same idea

[cloud walrus]

Wew Lads Tbh

[hidden haven]

What do you mean direct sum

lmao

noob

[lethal dune]

lol

[delicate orchid]

yeah I'm direct summing them come at me

[hidden haven]

That doesn't work

lol

[lethal dune]

yes lol

[hidden haven]

Every element has finitely many non zero entries

[lethal dune]

direct sum \neq direct product for infinite case

[delicate orchid]

ah yeah ok maybe I just wanted to use the fancy symbo lthen?!?!

[lethal dune]

the initial object of ....

[hidden haven]

Observe

[delicate orchid]

anyway my idea is:
C_2 is generated by one element of order 2, C_3 is generated by one of order 3, so on
these are all coprime, take this product then you get (1, 1, 1, 1, ...) which can never be the identity

[hidden haven]

Copy my hw but make it not obvious

[delicate orchid]

you think I have the FOGGIEST idea what "product over n" means

[hidden haven]

[lethal dune]

btw the product of countable groups is the direct sum right? and not the direct product?

[delicate orchid]

[hidden haven]

You are asking if product is product?

[lethal dune]

product in the category grp,

🙈

[hidden haven]

It is the product group

That is why it is called the product group

🙈

[lethal dune]

[hidden haven]

🐱 theorists, I swear

[lethal dune]

which one then?

[delicate orchid]

anyway moldi what does product over n mean

[hidden haven]

this

[hidden haven]

product is product not sum

[delicate orchid]

ah I see

[lethal dune]

why does it say sum then

[hidden haven]

Because that is the coproduct

Also this is the abelian case which I don't think you mentioned

There is no direct sum of non abelian groups

Only free product

That universal property is the universal property of the coproduct

[broken stirrup]

Moldilocks can i ask for your current degree?

[hidden haven]

MSc first year

[lethal dune]

ohhhh I reversed the arrows

[delicate orchid]

WTF

[lethal dune]

I know right

[hidden haven]

[delicate orchid]

wow way to make me feel (deservedly) like a hack fraud

[lethal dune]

moldi should be banned from the server

[delicate orchid]

I legit thought you were a PhD student at minimum moldi

[hidden haven]

Have I become what I hate

[broken stirrup]

me too

[hidden haven]

[broken stirrup]

dude has a great sense of understanding math

[delicate orchid]

if you hate people who actually seem to know what's going on, then I might have some bad news

[lethal dune]

doesn't waste 5hr watching cat videos online

[hidden haven]

High school prodigies

Like moth

[final oasis]

I feel bad that cat talk is weird to me

[hidden haven]

Lmao

Me

[lethal dune]

you watch CATegory videos not cat videos

[hidden haven]

Have you seen how much time I spend in #cats

I check it more frequently than my DMs

🙈

[dusty sapphire]

and wbu ryu i mean may i ask you current degree?

[lethal dune]

msc-2

[dusty sapphire]

cool

i think you are also PhD scholar

[hidden haven]

You're older than me

I thought you were like first year at best, given the anime pfp

[lethal dune]

age probably

[hidden haven]

Is there another measure of oldness

[lethal dune]

[delicate orchid]

waittt MSc students in the US are older than I am I think

[lethal dune]

ya i'm not that good lol

[delicate orchid]

you have 4 years undergrad right?

[hidden haven]

3

I had 3

[delicate orchid]

of course you did

[hidden haven]

I am 22

[delicate orchid]

ah ha I'm younger

[lethal dune]

ok then same age

[hidden haven]

Nice

How old wew

[unreal portal]

wikipedia says that any field is a local ring, which makes sense because the only maximal ideal is (0). But it also says a local ring satisfies $1\neq 0$ and the sum of any non-units in $R$ is a non-unit. But in the field $\mathbb Q$, for example, wouldn't $-2 + 3 = 1$ violate that condition?

[delicate orchid]

21

[hidden haven]

Nice, better learn all that category theory in the next year

[lethal dune]

ok so you have an excuse, not me

[dusty sapphire]

Where are you from?

[delicate orchid]

considering my group theory course is currently being taught by a cat theorist, maybe

[lethal dune]

you mean country?

[dusty sapphire]

yep

[cloud walrus]

cgodfrey

[lethal dune]

India

[dusty sapphire]

same😁

[lethal dune]

ya IK

[hidden haven]

-2 and 3 are both units in Q

[delicate orchid]

-2 and 3 are units, every element other than 0 is a unit in a field

and 0+0 = 0

[hidden haven]

copy my hw but make it subtle

[dusty sapphire]

college?

[unreal portal]

wow idk how I got that wrong lmao

thanks

[delicate orchid]

listen here just because you didn't misspell field twice doesn't mean

(felid and feild in case you were wondering)

[dusty sapphire]

WB?

[hidden haven]

damn

[lethal dune]

are you tracking me

[dusty sapphire]

no just ig

[hidden haven]

I mean statistically that is what anyone should guess

Given indian into math

[lethal dune]

first guess should be UP not WB

[lethal dune]

yeah all those western fakers

[hidden haven]

Do you really not know how dominant WB is in math in India

Half of CMI is Bengali

[dusty sapphire]

No you are from WB😂

[lethal dune]

yeah lol

half of my college is also bengali

even faculties

[hidden haven]

[lethal dune]

[dusty sapphire]

hope so because you are here so,

[lethal dune]

that makes 0.01% probably

[dusty sapphire]

IISER?

[lethal dune]

no

[dusty sapphire]

IITs

[lethal dune]

IIT

[hidden haven]

uh oh

[dusty sapphire]

JAM

cool

[lethal dune]

jam sucks

try other institutes

[dusty sapphire]

IITB

wb moldi

[lethal dune]

what about you? what degree

[hidden haven]

CMI

[dusty sapphire]

Post graduated😁

[lethal dune]

oh preparing for entrance?

[dusty sapphire]

yep

[lethal dune]

cool

[final oasis]

How should I think of projective limits

In the case of rings in particular

[molten viper]

Could someone help me with the following problem
I need to show the polynomial $x^4 - x^2 + 1$ is in the ideal $(x^2 + y^2 - 1, xy-1)$

[cloud walrus]

Phil With Flex Tape

[molten viper]

Whoops, menat to post this in #algebraic-geometry, apologies

[prisma shuttle]

what ring is this over

[prisma shuttle]

what why

this is abstract alg

[clever rampart]

I should read Ideals, Varieties and Algorithms again

[molten viper]

Funny enough, that book is sitting right next to me

[clever rampart]

Indeed it was your problem which reminded me of that

[molten viper]

It's unspecified, so I figure it's R[x,y]

The hint for the problem was "multiply the second equation by xy + 1

[delicate bloom]

what were the equations

[molten viper]

$x^2 + y^2 - 1, xy-1$

[cloud walrus]

Phil With Flex Tape

[prisma shuttle]

In that case, just note that $x^2(x^2+y^2-1) - (xy+1)(xy-1) = x^4-x^2+1$

[cloud walrus]

JustKeepRunning

[delicate bloom]

these are expressions lol

I'll assume they're =0 in the ideal I guess haha

[molten viper]

oh well thanks @prisma shuttle

[sharp sonnet]

sage can do it

[molten viper]

good to know!

[prisma shuttle]

in general if u are given this type of question

just subtract (or equivalently add because everything has an additive inverse) and multiply by elements of the ring repeatedly

until u can produce the desired polynomial

[clever rampart]

I am not sure how useful that is unless you have some "progress measure" in place

[molten viper]

It's the first time I've done this sort of problem so

[hidden haven]

Universal property, or just keep an example in mind (ring of power series, p-adics etc)

It's defined by how it projects into smaller things

[molten viper]

and now the fun part, finding polynomial gcds 😭

wait nope that's the next problem

[clever rampart]

If you haven't already seen it, there is an algorithm for it in the book (unless I am misremembering). You can use sage to implement it and relieve yourself of hand calculation.

[hidden haven]

So the structure maps for the power series are the "projections" that quotient higher order terms. The projective limit is just a way to glue together all the data of the polynomials in a way that this works out

@final oasis

[molten viper]

Well thanks all

[delicate bloom]

I don't know if this is actually true but I think of projective limits as kind of like removing zero divisors from a sequence of rings and giving you an integral domain

[final oasis]

I was looking p-adic case in particular

[hidden haven]

Definitely not in general because if you take projective limit of a constant diagram you'd get back that same ring so nothing happens in that case

[delicate bloom]

I'm willing to reclaim that with a disclaimer "without trivial cases" or "edge cases" or something

[next obsidian]

@molten viper

[final oasis]

Should the projective limit satisfy some commuting diagram property

[next obsidian]

A little hack for this

When you mod out by xy - 1

This is actually saying y = x^-1

So when you look at x^4 - x^2 + 1 and x^2 + y^2 - 1, just pretend y = 1/x

That should help you see how it’s in the ideal

In this case

Just multiply by x^2

And you end up with x^4 - 1 + x^2

If you replace y with 1/x

And multiplied x^2 + y^2 - 1 with x^2

[prisma shuttle]

yo guys

can someone explain wut the notation in purple means

[next obsidian]

Lmfao

What the hell

Maybe this is like

Polynomials over C

Except you allow any sort of positive rational exponent?

That sounds right to me

Yeah yeah yeah

Then nothings irreducible

[prisma shuttle]

these papers are like impossible

[next obsidian]

That seems right

[prisma shuttle]

they don't explain anything

[next obsidian]

Just intuit

[prisma shuttle]

at least none of their notation

[next obsidian]

Channel your inner Chmonkey

[prisma shuttle]

how do u tell

[next obsidian]

Cuz u can just keep dividing the exponent by 2

x^b = (x^b/2)^2

That’s my intuit at least

Maybe there’s an issue with like x^2 + 1

But I dunno

Maybe this isn’t what it is

https://arxiv.org/pdf/1905.07168.pdf

Maybe this has some info you are looking for

[rustic crown]

(x+i)(x-i) 🙈

so fundamental theorem of algebra would reduce this to showing that x^r - a aren't irred. as ajy element of C[X; Q+] will be in C[X^r] for some r.

which we can just factor once more... by difference of squares?

[hidden haven]

Yes google it, there is a universal property

It's also called inverse limit

But either should give you results

[next obsidian]

Oh god Hurb

The EXPONENTS are in Q^+

Not the friggin coefficients

[next obsidian]

Yeh yeh yeh yeh smart det

[rustic crown]

[rustic crown]

lol i didn't know the notation, you were the one who explained it just now >.<

okie that's probably not a complete proof. we would need to show that those aren't units. but we can define the degree like usual which shows that pretty easily.

[chilly ocean]

guys what the steps we need to prove when we have to show that a subset is a subgroup

[rustic crown]

it's closed under all operations

which means it should contain the identity, is closed under inverses and products of elements.

[chilly ocean]

hmmm

did my professor do all three steps here ?

so when you are asked these type of questions i should just grab two arbritary elements always right?

[rustic crown]

that proof is still incomplete

[chilly ocean]

bruhhh my professor posted these solutions for midterm review LOL

[rustic crown]

so the criterion they're using is that if H is a non-empty subset of G such that for every a, b in H we have ab^{-1} in H then H is a subgroup.

non-empty is important, but no one cares lol.

[chilly ocean]

hmm okay

so basically each element has an inverse?

this is called the subgroup criterion?

[rustic crown]

yea, everything in G has an inverse by the group axioms. but if we start with an element of H, we would somehow need to say that inverse is not only an element of G, but also it lies in H.

[tight flare]

i guess they're just saying non empty is obvious there

which is reasonable on that example

[delicate bloom]

I feel like their way of proving it is weird

[chilly ocean]

i feel like that proof is too simple

[delicate bloom]

just do ab and show ab*ab=aabb=1 cause it commutes

[chilly ocean]

he did do that

[delicate bloom]

he did a bunch of inverse junk

[chilly ocean]

idk why

[rustic crown]

well you only prove that it's closed under *

but inverse is obvious ig?

cause ever element is inverse of itself there >.<

[tight flare]

ye was wondering where the inverse part would be

[chilly ocean]

so what is the goal for these type of problems?

[tight flare]

to show that it's a subgroup

[chilly ocean]

like do i start by getting 2 random elements?

[tight flare]

[delicate bloom]

it shows the inverse because it's squaring (ab) itself

[chilly ocean]

and then run it through the function to see if it works for any elements?

[rustic crown]

just check the 3 properties any way you want. it's not like one is wayyyy too much work than the other. they are all pretty much the same thing.

[tight flare]

yeah you'll just be checking axioms i guess

[delicate bloom]

at the end of the day the proof is so easy the prof left out some step cause it's not too serious lol all discussion at this point is way overthinking haha

[tight flare]

take arbitrary elements, use given properties

It's one of the rare algebra discussions i can participate in

i enjoyed it.

[chilly ocean]

so

[tight flare]

[chilly ocean]

do i check for the 3 axioms

or no

or do i do what he did

and just do algebra until i get 1

?

[tight flare]

I mean if you get that kinda task ig it's more safe to check all 3 axioms especially when it just got introduced but that's kinda on you to assess what your prof wants you to do

[chilly ocean]

what are the 3 axioms

inverse

identity

associativity?

[rustic crown]

no closure

associativity would be inherited from G itself

[chilly ocean]

just closure

hmm

isnt closure gonna be knocked out when i prove inversees

[rustic crown]

nope

[tight flare]

why would it

[chilly ocean]

okay so

if you were to do it the classic 3 step way

how would u begin ?

[rustic crown]

see it's either 3 step or 2 step. no big difference

the easiest way to show a set is non-empty is by handing over an element of it... and identity is going to be easiest one to hand over

now either you check a*b and a^{-1} separately or just do both in a single shot by checking a*b^{-1}

[tight flare]

pew

[chilly ocean]

okay so if i would to do it the 3 step way

a^2 = B^2 = 1

closure: a^2 x b^2 = 1 right?

[rustic crown]

call that subset H
(i) 1^2 = 1 so 1 in H
(ii) if a, b in H, then a^2 = b^2 = 1, so (ab)^2 = a^2b^2 = 1, which means ab in H
(iii) if a in H, then (a^-1)^2 = (a^2)^-1 = 1 which means a^-1 in H

[chilly ocean]

ahhh

okay

what is (i) specifically doing?

[rustic crown]

showing that identity of G also lies in H

after you restrict the operations of G to H, these are then showing that H forms a group on its own

a group needs to have a 1

you go into the jungle, find the 1, and show to me that it lies in H

[chilly ocean]

oh okay so basically we grab a 1 from G and show that it works in the H world using the given function

[rustic crown]

yea

[chilly ocean]

fuck man

this class is so ass

and im assuming (ii) is closure

[rustic crown]

yep

(technically, closure under *. because all the three are closure under some operation >.<)

identity is a 0-ary operation, it's just a function 1 --> G
inverse is a unary or 1-ary operation. i : G --> G
and product is a binary or 2-ary operation * : G x G --> G

[chilly ocean]

isnt (i) actually showing its nonempty?

[rustic crown]

yep, it's also doing that

[chilly ocean]

okay things in my brain are making connections now

finally

[rustic crown]

[tight flare]

[quick harness]

If polynomial f has an even sum of coefficients, does it tell you anything else about it's properties ?

Divisibility or something like that

[rustic crown]

it tells you that 1 is a root of f mod 2

f(1) is just the sum of coefficients

[quick harness]

Thanks a lot

[delicate bloom]

sorta related for fun, if the sum of coefficients is odd and the constant term is odd, then f has no integer roots, might be fun to try to prove that

[quick harness]

My majors isn't mathematics, so coming up with proofs isn't really my idea of fun

[rustic crown]

.<

[chilly ocean]

whats your major

[delicate orchid]

we will never know

[quick harness]

CS

And for some reason, I have to take Abstract Algebra

[chilly ocean]

how do you know if a group is isomorphic

find an isomorphism

like how would i know if something is not isomorphic

for example my prof said D4 isnt isomorphic

not isomorphic to...?

"isomorphic" is a property of a pair of groups

not of a single group

oh

well

the claim was that all groups of order 8 are isomorphic

but thats false

but idk why

[delicate orchid]

you can pick your favourite groups of order 8 and show that if there's an isomorphism between them we get a contradiction

for example, C8 has an element of order 8 in it, D4 doesn't, so they can't be isomorphic

[chilly ocean]

idk how to create my own groups of order 8

[delicate orchid]

you can always take the cyclic group of order n

[broken stirrup]

There many nice properties of isomorphism. You could start checking order of groups, order of elements a.e. if G contains an element of order n but not H then G and H can't be isomorphic and etc

[chilly ocean]

so are u saying if group G has an element with order 4

and Group H doesnt have any element with order 4 they cant be isomoprhic?

[delicate orchid]

yes

[broken stirrup]

yes because isomorphism preservers orders

[delicate orchid]

^

I can explain why if you want

[broken stirrup]

it preserves structure basically

being isomorphic is bein "same" in some sense

[delicate bloom]

you can think of isomorphism as relabelling the group elements but not really changing anything

[chilly ocean]

oh okay

they act the same but can be of different elements

like each element in both groups behave the same right?

since the order of everything is basically the same

[broken stirrup]

yes, you have the 'same' group with elements named differently

[chilly ocean]

are there some groups that i should know?

like that would be helpful for an exam

[delicate orchid]

cyclic groups, Z is a good one

D_n

[chilly ocean]

when you mean by cyclic group of order 8 its basically a group with 8 elements that are constructed by one element right?

[delicate bloom]

cyclic groups C_n, symmetric groups S_n, dihedral groups D_n at a minimum imo

[delicate orchid]

S_n, A_n

I always remember A_n with n > 4 if I need a big simple group

[chilly ocean]

A_n is?

[delicate bloom]

along with their orders and generators

[delicate orchid]

the group of even permutations

[delicate bloom]

klein 4 group and various other specific ones that you see pop up can't hurt either

[chilly ocean]

and Z is the modulo groups right?

[tranquil parcel]

isn't A5 the smallest nonabelian simple group

[delicate orchid]

I wrote this whole thing up and it would be a shame not to post it

[delicate orchid]

no it's just the group of integers under addition

it's a nice infinite group if you ever need one

[chilly ocean]

Oh

okay

and all abelian groups are isomorphic right

[delicate orchid]

absolutely not

[tranquil parcel]

Another good one is Q/Z, it's an example of an infinite torsion group under addition

[chilly ocean]

or there was a theorem

[delicate orchid]

truly horrifying

[tranquil parcel]

No, consider C_2 and C_3, both abelian, but different cardinalities

[chilly ocean]

all abelian groups are...

ill get there lol

[delicate orchid]

all finite abelian groups are products of cyclic groups, that might be it

[oblique leaf]

Can someone tell me what this question even means? It seems to me that $$id\otimes\phi$$ is ill-defined as $$L\otimes M$$ contains finite sums of tensors but the $$\phi$$ is only defined for singleton tensors

[cloud walrus]

alyosha

[gusty halo]

you extend it linearly

so $\sum_i \ell_i \otimes m_i$ is mapped to $\sum_i \ell_i \otimes \phi(m_i)$

[cloud walrus]

Phil

[oblique leaf]

@gusty halo that's what i thought that probably means but i wasn't sure because in that case isn't this question trivial?

[gusty halo]

yes pretty much

[oblique leaf]

because we are supposed to show that idxphi is a homomorphism

ok

thanks!

[hidden haven]

Not entirely right? How do you know that this is well defined?

[oblique leaf]

@hidden haven could i do this to prove it is well defined? I wasnt sure because this “proof” uses the fact that phi is a homomorphism before i have proven it is an actual map

I was wondering if i could do that

[delicate orchid]

you've given that phi is a homomorphism right?

[oblique leaf]

@delicate orchid i am supposed to show it is a homomorphism

[oblique leaf]

but i first proved the properties of homomorphism and then use those properties to show that the map is well defined

oh wait sorry

those phi is in fact id tensor phi

i wrote it wrong

[hidden haven]

You're applying phi to tensors

[hidden haven]

Oh

[hidden haven]

Yeah you can't do this lol

Use the universal property of tensors

[delicate orchid]

"use the universal property" is basically just code for "I sentence you to death"

[hidden haven]

[delicate orchid]

for well definedness I'd just show that if x' \otimes y' = x \otimes y then their images are the same

[hidden haven]

But that is not easy

Tensors are extremely hard to work with

Maybe it's easy in this case idk

[delicate orchid]

[oblique leaf]

@hidden haven wait so you don't know how to do it either?

[hidden haven]

Usually you can't get far with tensors without using their universal property

[delicate orchid]

no moldi will know

[hidden haven]

I do

Use universal property

Lol

[delicate orchid]

I can kind of see how the universal property would work here

basically just shove the whole problem into LxM -> LxN

[hidden haven]

Yep

Then the universal map L × N to L ⊗ N

Here's something I wrote about this a while ago

This is what we get from the universal property. When you have some construction defined by a universal property, and it can be proven to exist for all objects (for every module M, M ⊗ N exists) then there is a canonical way to turn it into a functor. There are 2 kinds of universal properties:

(universal construction for M) is the initial map: M → (some functor applied to universal object)

(universal construction for M) is the terminal map: (some functor applied to universal object) → M

"Tensor with N" can be viewed as a construction of the first type. It is the object M ⊗ N along with a map M → Hom(N, M ⊗ N) that is initial among maps M → Hom(N, P). Here the "some functor" is Hom(N, -). You might not have seen the universal property stated in this form before, but you can think about what this universal map is (don't spend too much time on it though). This is a more categorical way of stating universal properties, and from this it's not even clear that tensoring is symmetric, but this tells you about the induced functor.

Given a map
M → P
You have a diagram
M → P
↓ ↓
Hom(N, M ⊗ N) Hom(N, P ⊗ N)

Notice the composite map from M to Hom(N, P ⊗ N). The universal property above says the the left vertical map is initial among such maps, so there exists a unique map M ⊗ N → P ⊗ N such that when you apply Hom(N, -) to it, it completes the above square in a commutative way. We define this to be the image of the original map under - ⊗ N. You could check that this is exactly the map you described. This is the general construction of a left adjoint to a given functor (Hom(N, -) in this case), assuming it exists. If we chose a universal property of the second kind, we'd have a right adjoint to a given functor by a dual construction.

[delicate orchid]

yeah so you just have to show id × \phi a homomorphism from L × M -> L × N and then map it back by composing it with the universal map?

[hidden haven]

Might be helpful for general understanding because this is a very common construction

[hidden haven]

Didn't get that

You have a map L × M to L ⊗ N

By composing the previous 2 maps

Use the universal property to induce a factoring through L ⊗ M

[delicate orchid]

good lord

[hidden haven]

This is really useful to know it's how we construct functors from universal properties, like the free group functor from the universal property of free groups etc

[delicate orchid]

I mean sure

but I have no clue how or why this is relevant

[gusty halo]

i guess otherwise you'd have to look at the construction of the tensorproduct to check things

and you dont want that

[delicate orchid]

I dunno I think I kinda want that

[hidden haven]

The question is about how L ⊗ - is a functor

[gusty halo]

have you seen that definition

[delicate orchid]

yes

[gusty halo]

wew lad

[hidden haven]

And if you know the general construction this is trivial

Wew have you tried working with quotients like proving that the standard generators and relations representation of D_n gives a group isomorphic to the symmetry group of polygons?

That is an ugly proof and relies very highly on the group being finite

[delicate orchid]

not via quotients

I imagine it's nice using the universal product of the free group though

[hidden haven]

How else do you work with generators and relations

Generators and relations = quotient of free group

[delicate orchid]

I sick them in the langle rangle

[hidden haven]

[delicate orchid]

I know that they are quotients of the free group

but I learnt that literally 3 days ago

[hidden haven]

Well my point is that even this is very hard to prove. The only proof I know is by constructing a homomorphism from the presentation to the concrete group using universal properties of free and quotient groups, prove that it is surjective, then use finiteness hacks and count cardinalities

And in general it's hard to even answer whether < generators | relations > gives a trivial group

Like how do you know whether 2 elements here are distinct?

[sharp sonnet]

additional exercise: show that the id tensor phi defined here is the same as id tensor phi as an element of Hom(L, L) tensor Hom(M, N)

also is well definedness even an issue in above

[hidden haven]

It's really difficult symbolically, and the easiest way is to use the univ prop of quotients

[sharp sonnet]

isnt it part of the construction/definition of tensor product that defining maps on pure tensors works

[delicate orchid]

yeah I know it's hard to get properties from presentations in general if you're doing it directly

[sharp sonnet]

so you do it once and then never worry again

[hidden haven]

This is a special case of that so I assume you wouldn't be allowed to do this

[sharp sonnet]

i mean well definedness of the map L x M into L tensor N

[hidden haven]

Like if you have proven that tensoring is a bifunctor

Then you wouldn't be doing this exercise

[sharp sonnet]

well, you observe the map is bilinear

[hidden haven]

Yes

[sharp sonnet]

and then you are done

[hidden haven]

That's what I gave as a hint

Because the solution they gave was "define phi by extending it linearly, and it is a homomorphism because it's extended linearly"

If I understood it correctly

They didn't mention going to the product

[sharp sonnet]

ah ok but

i dunno wouldnt you show this in a more general way

kinda weird exercise overall

[sharp sonnet]

this is still nice

[hidden haven]

Should ask them to directly show that it's a bifunctor

[tranquil parcel]

Why exactly is it that Q under addition is not isomorphic to Q_+ under multiplication?

cant i do like 2^x as my homomorphism

[hidden haven]

You can't half everything

Under multi

2 doesn't have a sqrt

But every element has a half

[tranquil parcel]

i see so 2^x needs irrational solutions

[hidden haven]

Yes

[sharp sonnet]

you can also look at torsion

and other stuff

[tranquil parcel]

i tried, it seems neither group has torsion

[sharp sonnet]

i dont think there is an "exact" reason

[hidden haven]

Cardinality of ℚ_+ is half that of ℚ

Half |ℚ| - 0.5

Because 0 is also removed

[sharp sonnet]

additive group has torsion group {0} and multiplicative {\pm 1}

[hidden haven]

[sharp sonnet]

or am i brainfart

[hidden haven]

ℚ_+

[oblique leaf]

@hidden haven Thank you! I just found the commutative diagram

[hidden haven]

Only considering positive things here

[sharp sonnet]

i am indeed brainfart

or rather i dont read

i eject myself from this conversation again

[tranquil parcel]

lol

[hidden haven]

[sharp sonnet]

i am go learn about multifunctors

[tranquil parcel]

i don't see any algebraic difference between these two groups

[hidden haven]

Is there anything to learn about multifunctors past the definition

[sharp sonnet]

i dont know, i didnt know what a bifunctor is until you just mentioned it

[hidden haven]

What do you mean by algebraic difference

[sharp sonnet]

i mean i did, but i didnt know the word

[hidden haven]

Oh

Lol

Just a functor from a product of 2 categories

[tranquil parcel]

The problem posits they aren't isomorphic but I can't see why

[hidden haven]

I have you 2 reasons

Every element of ℚ can be halved

But you can't halve every element of ℚ_+

[sharp sonnet]

i am check nlab for examples of multifunctors

[hidden haven]

Because halving under multiplication means taking a sqrt

[delicate orchid]

ahhhhh that's kinda what I was thinking but expressed way better

[hidden haven]

[tranquil parcel]

oh i see, so what we're saying is there is this equation 2x=q, and for every q under addition we can find a rational solution, but every q under multiplication this isn't the case

[hidden haven]

Yep

[delicate orchid]

I was thinking there was something weird with 2^-1+2^-1 = 1 which means there's some nonsense that might be happening, was this along the right lines?

[hidden haven]

Bruh

Idk

[delicate orchid]

sorry

[sharp sonnet]

i have one more thing that works (?)

• is free on primes, + is not

[lilac trench]

can anyone tell me why euclidean domain

[hidden haven]

Seems legit

[lilac trench]

just why

[hidden haven]

Long division

Factorising stuff

Nom theory

[lilac trench]

no comprendo

[hidden haven]

https://tenor.com/view/cat-bite-nom-attack-slo-mo-gif-12221453

Comprendo now?

[lilac trench]

hmmmm

i guess what im having trouble understanding is the what the norm function has to do with anything

aka euclidean function

[hidden haven]

We want to be able to write things as quotient*divisor + remainder

And we want remainder to be smaller than divisor in some sense

Think of natural numbers or polynomials

The norm is a size function that makes this happen

Then you can run the Euclidean algorithm to find GCDs etc

[lilac trench]

ok i'll let this simmer 🧑‍🍳

[hidden haven]

Which is useful in nom theory and polynomial study which is much of ring/field theory

[hidden haven]

We want this so that the Euclidean algorithm terminates

[lilac trench]

jeez how long did it take to get this knowledgeable

[hidden haven]

Technically my entire life

[barren sierra]

lol I feel that

I barely know anything compared to these people 💀

[hidden haven]

Damn this channel is being really nice to me today

[tranquil parcel]

the knowledge is in us all along

[lilac trench]

ok so I was thinking about the polynomial ring, and to even be able to use the Euclidean algorithm, there has to be some notion of ordering, so intuitively this is what the norm is; a method of ordering the ring

[tranquil parcel]

to a degree yes, the analogy kind of breaks down for complex numbers

[lilac trench]

oh ok

[broken stirrup]

seriously when did you start learning abstract algebra?

[tranquil parcel]

but that intuition is exact for Z

[delicate orchid]

2nd time moldi has been asked something to this effect today

[hidden haven]

It was a joke

UG 1st year is when I learned linear alg and some group theory

[delicate orchid]

ah so, what

4 years?

[hidden haven]

Kinda works still, but yeah I'd say it's better to think of it as assigning a size to each element rather than ordering maybe

[hidden haven]

3.5

[delicate orchid]

I really have some catching up to do

[lilac trench]

ahh yes, then it makes more sense in C if its size rather than ordering

[hidden haven]

What norm are you thinking of for ℂ btw?

The standard norm doesn't actually work since these ring theory norms are only natural numbers

You can use the trivial norm here I think which just gives everything a 0

Or maybe it was that 0 has norm 0, everything else has norm 1

But this works slightly differently from magnitude since it's just about factoring

Wait we don't even assign a norm to 0 do we

I don't rember 💀

[lilac trench]

im lost

[tranquil parcel]

i think the norm of 0 is just defined to be 0 in every case

[lilac trench]

thats what i have in the defn

[hidden haven]

Lol what I'm saying is that this dividend norm is different from the standard magnitude

[delicate orchid]

I think we do, it just has to be 0

[hidden haven]

Interesting

[delicate orchid]

the trivial metric generates a norm where everything is 0 so I assume that's the one you're referring to

[hidden haven]

Because when we talk about degrees of polynomials, we either define deg 0 = -∞ or undefined

For the additivity of degree to work out

[hidden haven]

No lol

This is norm for Euclidean domain

Not analysis stuff

[delicate orchid]

how have I covered euclidean domains but never even heard of this wtf

[sharp sonnet]

the assessor in my abstract algebra exam told me i will never become a mathematician if i keep making careless mistakes because i forgot to exclude 0 from the definition of euclidean function

[hidden haven]

Might have been called Euclidean function

[hidden haven]

Bruh

[sharp sonnet]

i actually noticed later too

because of a suggestive question i guess

anyways, this is how i remember

through pain

[delicate orchid]

oh ok I have seen this function before

I just don't think we ever named it anything

like I've seen multiple examples of it but never the general case if that makes sense

cool to know it's a thing

[barren sierra]

Kinda stuck in seeing how #3 applies to #4

3 tells us that $|\mathcal{C}| = [S_n : A_n C_{S_n}(\sigma)] * [A_n : Cl_{A_n}(\sigma)]$

[cloud walrus]

Spamakin🎷

[barren sierra]

and we know $|Cl_{A_n}(\sigma)| = [A_n : C_{A_n}(\sigma)]$ by orbit-stabilizer

[cloud walrus]

Spamakin🎷

[barren sierra]

but idk where to go from there

I've seen a proof for this that doesn't use problem 3

however since I am kinda on the struggle bus in this course

I would like to figure out the intended solution

[frank lake]

ok I'm back

and I'm feeling a bit confident

I'm going to show that for groups with finite order, multiplication in <a> is done by addition mod n.

let's say I have i and j

if I say that $(i + j) \mod n = k$, I want to prove from this that $a^i a^j = a^k$

[cloud walrus]

♡ChubbyMuffins♡

[frank lake]

according to the divison algorithm, in the equation (i + j) mod n = k, a = (i + j), b = n, q is unknown and r = k

therefore (i + j) = qn + k

$a^ia^j = a^{i + j}$

[cloud walrus]

♡ChubbyMuffins♡

[frank lake]

and $a^{i + j} = a^{qn + k}$

[cloud walrus]

♡ChubbyMuffins♡

[frank lake]

$a^{qn + k} = (a^n)^q a^k = e^q a^k = a^k$

[cloud walrus]

♡ChubbyMuffins♡

[frank lake]

therefore, a^i a^j = a^k

does this work as a valid proof?

and for groups with infinite order, since $a^ia^j = a^{i + j}$ you can just figure out from there that multiplication in <a> is simply done by normal addition

[cloud walrus]

♡ChubbyMuffins♡

[frank lake]

😭 pls guys judge my proof

[barren sierra]

looks good to me

[barren sierra]

Can anyone help with this?

[frank lake]

thanks 🙂

[barren sierra]

here's my proof that doesn't use the prior problem #3

[frank lake]

😰 what book is that?

[barren sierra]

but like I want to know how to use that for this problem

[barren sierra]

my HW lol

[frank lake]

damn

[barren sierra]

tbf this is a grad level algebra course so you'll get here soon enough

[frank lake]

I see

that's pretty cool

[proud bear]

this works right? if $\sigma$ commutes with an odd permutation, then $A_nC_{S_n}(\sigma)$ is a subgroup that strictly contains $A_n$ because $C_{S_n}(\sigma)$ contains an element not in $A_n$. the only subgroup containing $A_n$ is $S_n$ so $k=\lvert S_n:A_nC_{S_n}(\sigma)\rvert=1$, meaning $\mathcal C$ is the union of one conjugacy class of $A_n$. that conjugacy class contains $\sigma$, so it is $\mathrm{Cl}_{A_n}(\sigma)$.

if $\sigma$ does not commute with any odd permutation, then $C_{S_n}(\sigma)\subseteq A_n$ so $A_nC_{S_n}(\sigma)=A_n$. thus $k=2$ meaning that $\mathcal C$ is not also a conjugacy class of $A_n$

[barren sierra]

Hmmm

I think that makes sense but I'll have to look at that again later

Also I'm working on this

and I think I can take this for granted but

How do I show that a permutation commutes with the elements of it's cycle decomposition

(it's part of the hint)

I'm trying to prove it myself and I'm not quite sure

oh wait I'm dumb NVM

the elements of the decomposition are disjoint

wait no that doesn't make sense

or does it? idk

[hot hill]

@subtle ivy Are you around?

I'm confused about a part of this preliminary proof of Zorn's Lemma: https://i.uguu.se/qhdYVcVo.png . In particular I'm confused about showing that for a non-empty linearly ordered subset of D, the supremum of that set is in D.

I'll ask there, thank you.

[upper cape]

Hi just checking that my understanding of transitive group actions is correct. Let G be a group, of order 4, and suppose that G acts transitively on the set E={a,b,c,d}. My understanding is that implies that for any x,y in E, there is a group element g so that gx = y. Surely then this would imply that if ga=a, then gx=x for all x in E

Does that make sense? The argument is that there are only four elements, and we must have g_i with g_1a=a, g_2a=b, g_3a=c, g_4a=d, thus only g_1 can be the identity, and since |G|=4, then g_1 must be the identity

[hot hill]

@upper cape Wish I could help you 🙂

answered my question, thanks.

[lavish nexus]

Yes

[pastel cliff]

i need sum quik dumb help :P

here algebra is just some set closed over some operation, not over a field or anything, just an algebraic structure

my claim is wrong tho bc the smallest "subalgebra" w those properties is {1} and idk what property Q\ {0} might have that {1} doesnt

[rustic crown]

Q \ {0} contains Z \ {0}

[pastel cliff]

quoi

[rustic crown]

what if you say (Q \ {0}, *) is the smallest subalgebra containing Q \ {0}?

[pastel cliff]

i cant cheese it that hard

fr tho any ideas?

like what property of Q sans 0 would separate it from {1}

cuz no addition

[rustic crown]

this should do...?

non-zero rationals are quotients of two non-zero integers.

[pastel cliff]

oh

hm

[rustic crown]

you would need to specify infinitely many elements, finite won't do

[pastel cliff]

i was thinking that could cheese it too lol

[rustic crown]

well you can say containing primes and -1 (and closed under multiplication, inverses)?

[pastel cliff]

okok i'll brb

[rustic crown]

(Q \ {0}, *) as an abelian group is isomorphic to (Z/2Z) ⊕ (infinite direct sum of Zs)
so it has infinite rank, specifying finitely many elements won't be enough to get the whole thing

[pastel cliff]

i mean just adding "smallest with mmore than one element" feels like cheese

or that

[rustic crown]

we can try adding another closure statement?

[pastel cliff]

this is very early in the course we havent technically defined isomorphisms so i cant use that

like what lol

i learned manifolds today too

[rustic crown]

i still haven't

[pastel cliff]

crumpled up pieces of paper glued together

15 minutes left

fuck it

im cheesing it

[rustic crown]

another bad way is by saying smallest subalgebra A of R such that A u {0} is a field >.<

[pastel cliff]

i would love to but he hasnt defined fields

like it's week two dawg teach me something

[rustic crown]

i mean just add to your claim in the end that "such that A u {0} is also closed under +"

but i'm sure that they just want you to say some sort of containment of Z \ {0}

[pastel cliff]

i chose to cheese with "more than one element"

fits easily wiht a proof i had inn mind

dumb onne

[rustic crown]

they say a "set of numbers"

[pastel cliff]

but proof nonetheless

[rustic crown]

instead of containing "a few numbers"

[pastel cliff]

horseshit proof lets fucking go

[rustic crown]

wait that won't do the job

consider the set {2^n | n in Z}

inverse of 2^{n} is 2^{-n}

[pastel cliff]

welp

[rustic crown]

@pastel cliff

[pastel cliff]

i had to submit lol

i figured it wasnt enough but alas

[rustic crown]

i told you na, just finitely many elements won't do the job

[pastel cliff]

[rustic crown]

pick a prime that doesn't appear in any of those finitely many elements, that prime would never belong to your A

[pastel cliff]

this was probably the way to go

[rustic crown]

yea >.<

[pastel cliff]

i just didnt get how it was a "property"

[chilly ocean]

Futurama theorem example on wolfram lol

[pastel cliff]

wait

there might be a grace period

[rustic crown]

well the question was smallest something which contains a "set of numbers" and satisfies "certain properties"

[pastel cliff]

yeah i was fixatinng on the properties part

assuming i fix the hypothesis did the proof look ok?

[lethal dune]

oh shit

it's not chill

[pastel cliff]

#abstract-chill

pretty much becomes chill with moldi and wew lads around

i speak only facts

[dusty sapphire]

?

[rustic crown]

3 + 2sqrt(2) = (1+sqrt(2))^2

so K = Q(sqrt(2), w)

[chilly ocean]

Is it true that the existence of root-finding algorithms implies uncomputable numbers are transcendental over the field of computable reals? I am pretty sure since if we have an uncomputable number and suppose it is the root of some computable polynomial, then you can just find it with the algorithm (contradiction)

I think that is a really cool fact

[pastel cliff]

@rustic crown hi gonna bother u more

[chilly ocean]

Like how nuts is it that not only are we only ever playing around with a countable subfield of the reals in any particular example, but also that without invoking some kind of universality (like saying ok we're talking about ALL reals now) or just specifically setting something as uncomputable by hypothesis, you can't even reach uncomputable reals. Not even algebraically via any computable reals, even theoretically

[pastel cliff]

once i type my fuckin latex

computer gonna die

i'll be back det lol whoops

[chilly ocean]

Wait so I knew about this uncomputable stuff for a few months already and thought it was cool and made this connection with transcendental extensions but after some more googling I just learned about un-definable reals... my soul is crushed

and not only do definables strictly contain the computables, they're countable and also a subfield, so are undefinables transcendental over the definables?

I am not familiar so I can't put 2 and 2 together in terms of whether being a root of a definable polynomial would mean you can define a dedekind cut in logic or not

Oh god and you can keep getting stricter and stricter field extensions via using different models for definability, still all with countable base fields and probably transcendental

[delicate bloom]

I'd be interesting in playing with this later but I have to learn the definition of a computable number, and I'm about to go to sleep lol

[chilly ocean]

me too 😭 gn fam

[pastel cliff]

@rustic crown ok hi

can u help me the proof super quick pls

[pastel cliff]

i wanna rewrite this kinda proof but instead using the Z\0 statement but im having some trouble translating it

[rustic crown]

notice that elements of Q \ {0} are precisely quotients of Z\{0}

A contains Z \ {0}, so if a, b in Z\{0} are arbitrary, by the closure properties, we know A must contain 1/b and consequently a/b.

This shows that Q \ {0} is contained in A

[pastel cliff]

oh wait can i just say "since the desired set A is closed, and must have multiplicative inverses for all elements, and contain Z\ {0}, the set must contain all rationals"

i was typing as u sent

so i just finished typinng lol

but same idea i think?

[rustic crown]

yep

[pastel cliff]

yay

your wording seems a bit more rigorous tho

[rustic crown]

for the other direction use that Q \ {0} is a subalgebra of (R, *) satisfying the required properties. (as A is smallest, it must be contained in Q \ {0})

[pastel cliff]

mmmm

monke brain want use contradiction

[rustic crown]

🙈

[pastel cliff]

actually i thin i can...?

[rustic crown]

yea, what you did was fine.

if A had any more elements, it wouldn't be smallest...

[pastel cliff]

oh word

thank you det <3

im convinced the eevee-motes exist solely for det to communicate

[rustic crown]

[hidden haven]

Therefore eevee emotes must be removed

[pastel cliff]

MOLDI AWAKENS

[hidden haven]

Hello abstract-chill

[pastel cliff]

algebra hw done... time to do stats

,ti

[cloud walrus]

The current time for nitezba is 01:37 AM (EST) on Thu, 03/02/2022.

[pastel cliff]

god i LOVE being a dual degree, EVERYONE should do a dual degree

[hidden haven]

Back in my day, dual wasn't invented yet

[pastel cliff]

moldy moldi

[rustic crown]

didn't we have a dual for math and cs?

[hidden haven]

It doesn't count because I never took CS seriously

[pastel cliff]

yall go to same uni?

that's literally what im doing tho

[hidden haven]

yes

[pastel cliff]

oh damn that's cool

cs major is the only reason im forced to take stats

https://tenor.com/view/tn-j-gif-23950225

[lethal dune]

stats

[pastel cliff]

i'll see yall on the other side

[next obsidian]

Chmonkey

[chilly ocean]

catfan1337

[tough raven]

Interesting
In what context can subalgebras of R not have 0?
And need you to additionally specify that they are multiplicatively closed?

[hidden haven]

Algebraic structure

[dusty sapphire]

in Z_n, is every maximal ideal generated by prime element p such that p divide n?

[hidden haven]

Yes, by the correspondence theorem

[dusty sapphire]

yeah, similar for prime ideal right

[hidden haven]

Yes

[dusty sapphire]

prime ideal iff maximal in Zn

okay

non-zero

[molten viper]

Could y'all give me some advice on how to show a polynomial is in a specific ideal (just a moment while I type it out)

I wish to show $p = f(x^2 - 4) + g(y^2 - 1)$ is in the ideal
$(2x^2 - 3y^2 - 11, x^2 - y^2 - 3)$

[cloud walrus]

Phil With Flex Tape

[molten viper]

meh whatever you get the point

latex being annoying

[south patrol]

do you mean for the f and g to be there

[molten viper]

yes I do

[delicate orchid]

are f and g also polynomials?

[south patrol]

or is it a product

[molten viper]

correct

it's a product yeah

[south patrol]

Oh so all you mean is to show x^2-4 and y^2 - 1 are in the ideal then

Equivalently

Fair

[delicate orchid]

consider g = -1 and f =1
and f = 2, g = -3
something funny might occur

[molten viper]

The goal is showing 2 ideals are equal, and idk if using specific polynomials would help me

p being in the ideal $(x^2 - 4,y^2 - 1)$

[cloud walrus]

Phil With Flex Tape

[delicate orchid]

what I'm saying is that if you use those specific polynomials you'll get the two generating polynomials of the other ideal

also what polynomial ring is this in?

[molten viper]

Q[x,y]

[delicate orchid]

Q
nice, so all these are also units

[molten viper]

yeah you're way above my head with that one

And the problem I'm having is idk how precisely to show p can be written in terms of those generating polynomials

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

I've gtg now but hopefully this helps

[molten viper]

alrighty

thanks

[shadow coyote]

prove that the set of algebraic numbers is countable

i'm not looking for a full solution, but a hint perhaps

[sharp sonnet]

try to write them as the union of countable sets

[shadow coyote]

hmm

oh bruh

is this like: group polynomials by sum |a_i| if a_i are the coefficients

for each possible sum of absolute values, you have a finite number of polynomials

and you will hit each polynomial at some point

[north sand]

all polynomials x^n have the sum 1

[shadow coyote]

oh F

that sucks

[sharp sonnet]

you can add the degree to your sum i think

[shadow coyote]

oh yes that does it

gg

wait what am i saying

yeah ok that works bc if we want sum |a_i| + degree = n then degree ≤ n and thus there are finitely many a_i too

[pastel cliff]

gonna have to add this disclaimer to literally everything i ask in here aren't i

[hidden haven]

Just use good notation bro

[pastel cliff]

forwarding this message to my prof, surely he'll respond to His Eminence, Moldilocks

[hidden haven]

He better

[sharp sonnet]

tell them that everyone on the math server agrees

[pastel cliff]

surely he'll cower in fear of His Excellency, Lochverstärker

[chilly ocean]

I think I like Modern Algebra

[barren sierra]

Are outer automorphisms over G just representatives of the equivalence classes Aut(G)/Inn(G)? If so does that make every inner automorphism an outer automorphism since they represent one of those classes?

[hidden haven]

Outer automorphisms are automorphisms that are not the inner automorphisms

Or so I have always assumed

[barren sierra]

My prof said that but also said Out(G) = Aut(G) / Inn(G)

[hidden haven]

Yeah according to wikipedia the outer automorphism group refers to that

[barren sierra]

So am I right?

[hidden haven]

In mathematics, the outer automorphism group of a group, G, is the quotient, Aut(G) / Inn(G), where Aut(G) is the automorphism group of G and Inn(G) is the subgroup consisting of inner automorphisms.
An automorphism of a group which is not inner is called an outer automorphism.

On the wikipedia page for outer aut group

https://en.wikipedia.org/wiki/Outer_automorphism_group

[barren sierra]

So the outer automorphism group doesn't consist of outer automorphisms, it consists of sets of outer automorphisms.

But the inner automorphism group consists of inner automorphisms?

[hidden haven]

Yes

[barren sierra]

Wonderful

[inner needle]

Hey folks, how do I prove this simple result? Let G and H be finite groups. If G x H is cyclic then every subgroup of G x H is of the form A x B for some subgroups A and B of G and H, respectively.
Of course, G and H would both be cyclic in this case.

Should I proceed by counting the number of subgroups of the form A x B and show that it is equal to the number of divisors of |G|•|H|?

Oh, I got it. Let d(n) be the number of divisors for n. Then the number of subgroups of the form A x B is d(|G|)•d(|H) = d(|G|•|H|), since |G| and |H| are relatively prime. Since G x H is cyclic, it has d(|G|•|H|) subgroups.

Any other way to go about this?

Any constructive way in particular. Given a subgroup of G x H, how do we find such an A and B?

[quick harness]

What does subring of (R, +, •) generated by some number, for example √2, look like ?

[chilly ocean]

I supppse powers of sqrt2 and sums of those powers

[quick harness]

thanks

[barren sierra]

I'm confused about Sylow subgroups

suppose |G| = 60

actually uh lemme pick a better number

ok

suppose |G| = 180

180 = 2^2 * 45 and 2 does not divide 45
180 = 3^2 * 20 and 3 does not divide 20

does this mean there's a Sylow subgroup of order 4 and a Sylow subgroup of order 9?

[proud bear]

Yes

[barren sierra]

ok cool

but clearly Syl_2(G) != Syl_2(G)

got it

so we don't really care about the set of all sylow subgroups

just the set of all sylow subgroups for some specific prime

[proud bear]

Yeah thats what Syl_p(G) is. Set of sylow subgroups of order p^am

[nocturne cobalt]

Hi, I'm a bit stuck on this question: it says Let O(3,1) Be the Lorentz group. Show that O(3) is a subgroup of O(3,1).

Do I have to just take a matrix in O(3) and show that it preserves the minkowski inner product?

i.e. $\Lambda^{T} \eta \Lambda^{T}=\eta$

[cloud walrus]

Elena002001

[nocturne cobalt]

where \lambda is some O(3) matrix

[nocturne cobalt]

<@&286206848099549185>

[south patrol]

We have an embedding of O(3) in O(3,1) and we know it'll be closed under multiplication / have inverses etc so that ought to be sufficient

fortunately this should be quick from the definition

[nocturne cobalt]

yeah i thought this would be the general idea to answering this question - just apply subgroup properties to it

but it turns out that its not the answer

[hidden haven]

Same as that of the reals

Dimensions multiply when you tensor

Product of any 2 infinite cardinals is their max

[devout crow]

the following diagram is the answer to a homework problem, which was to `show $\bZ^n$ is not a quotient of $\bZ[x]$ for $n > 2$'. my tutor went through the answer today, but I don't follow it fully. the argument is by contradiction: suppose there is a surjective ring homomorphism $\varphi\colon \bZ[x] \twoheadrightarrow \bZ^n$. then the following diagram (below) can be made to commute. the image of the horizontal arrows is $\bZ_2^n$, but $\bZ_2 [x] / \langle x^2-x \rangle}$ has 4 elements, so the other route has image of cardinality at most 4. this is a contradiction if $n>2$.

[cloud walrus]

∧res
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[devout crow]

diagram:

I guess my question is really: why must the diagram be able to commute? (a priori, before we know this setup isn't possible)

[hidden haven]

Let me begin by saying

That is a very weird way of solving this when you can just say that Z^n has idempotent elements while the other ring doesn't

[hidden haven]

But the diagram commutes by repeated application of this

Or I guess you could view it as a single application

The claim is that the kernel of the composite at the top contains the ideal (2, x^2 - x)

And this follows from the structure of Z_2^n

Everything is idempotent and 2 is 0

[south patrol]

doesn't rly answer the question but e.g. my solution was just you can explicitly see what the image of the composition of the surjective ring homomorphisms would be

as in, $\sum a_i X^i \mapsto \sum_i \phi_i(a_i) \phi(x)^i = a_0 + \phi(x) \sum_i \phi_i(a_i)$ and there are only 4 options after reducing mod 2

well that also just shows that the kernel is as moldi says

[hidden haven]

Wasn't the question about why the diagram commutes

[cloud walrus]

potato

[hidden haven]

Oh you are saying that your response doesn't answer it?

[south patrol]

Well perhaps this also shows why the kernel is as we say aha

oh i meant my response is more geared towards the problem rather than commuting diagram altho perhaps it helps too anyway