groups-rings-fields

[chilly ocean]

got nothing out of it

[hidden haven]

lol

They are like discrete models of topological spaces in a way

[chilly ocean]

whenever i went to shitter i opened random nlab articles and it was good waste of time

[chilly ocean]

oh thats another thing

model theory seems important

but idk where it fits im with my education

seems like ill need to get primed on it if i have srs interest in AT

[hidden haven]

Model theory is not that prominent in AT

You might be thinking of model categories

[chilly ocean]

yes lol

[hidden haven]

Those don't have anything to do with model theory

😵‍💫

[chilly ocean]

i have been forsaken

is model theory just super logic stuff

idk what model categories are

[hidden haven]

It is a way to apply logic to math to get results

[chilly ocean]

what the hell

[hidden haven]

Like groups are defined by the group axioms

[chilly ocean]

i read model category wiki intro

[hidden haven]

You call those axioms "the theory of groups"

[chilly ocean]

its algebraic topology abstracting away topologies 🤔

[hidden haven]

And then you can find out a lot about the category of groups from what the theory of groups looks like, for example if the theory doesn't use "there exists" anywhere then you can say some very nice stuff etc

[chilly ocean]

oh what

this sounds cool as hell

[hidden haven]

It is

Model categories are a way to treat certain arrows as isomorphisms by formally inverting them

[chilly ocean]

so its like learning the about a category without set theoretic axioms

or maybe still with them

not necessarily i should say

[hidden haven]

I have just started reading about model cats so don't trust my words too much but what I have understood is that they are like derived cats but with better behaved morphisms because of the extra classes of fibrations and cofibrations

You usually assume set theory when you do model theory

[chilly ocean]

because as it stands i only have naive understanding of construction of concrete categories

[hidden haven]

Unless you are doing inner model theory which is more a subfield of set theory than of model theory

[chilly ocean]

“inner”

[hidden haven]

[chilly ocean]

im scared of that word tbh

[hidden haven]

lol fair

[chilly ocean]

because it never is self apparent what it means

[hidden haven]

Right ye

No clue why inner automorphisms or inner models are called inner

[chilly ocean]

no yea

i never had it explained

[rustic crown]

I thought inner automorphisms were called inner because they come from inside the group

And 'rest' of them are called outer

Out(G) = Aut(G)/Inn(G)

[hidden haven]

Conjugation automorphisms

[next obsidian]

I feel like they’re the only automorphisms you can define internally, without any extra info

Maybe this is not true, but I don’t see any other automorphism you can define on an arbitrary group G just in terms of G itself

[chilly ocean]

Do you have any online resources to learn about that topic ?

[wooden ember]

Is inversion an inner automorphism 🤔

Cause it’s certainly internal

[next obsidian]

It isn’t a homomorphism in general

[wooden ember]

Fair enough, in an abelian group

Then it’s definitely not inner

But it is internal

[next obsidian]

Sure, but then we’ve exited being able to define it for arbitrary groups

We used extra structure of G

[wooden ember]

True

[hidden haven]

Here's a quick intro from #point-set-topology message

For more details you can refer to Beck's PhD thesis called triples, algebras and cohomology

[chilly ocean]

I'll read that, thank you !

[pastel cliff]

all elements of a finite group have finite order - does this imply that each element generates a cyclic subgroup?

[delicate orchid]

prove it to see

[next obsidian]

All elements always form a cyclic subgroup my man

[delicate orchid]

but yes, almost by definiition, that's true

[next obsidian]

Cyclic = has a generator

<g> has a generator

[delicate orchid]

emphasis on a generator

[next obsidian]

Hey Wew are you going for that

Like last chapter golden strawby

Whichever the incredi-hard one is

[pastel cliff]

i just made it to farewell for the first time

[pastel cliff]

noted

[delicate orchid]

8b first

[next obsidian]

But eventually?

[delicate orchid]

I got to 2 rooms before cassette the other day so it's just one lucky run away

and yes

obviously I'm gonna do fwg

[next obsidian]

[delicate orchid]

then maybe, finally, I can say I'm at least competent at something

[pastel cliff]

meanwhile i barely survived chapter 8

[delicate orchid]

tbh the base game aint' shit unless you're speedrunning or doing goldens, I'll dm you some REAL fuckin maps

[lavish nexus]

how is the weight defined here

given the action $diag(t_1, t_2, t_3) f (z_1, z_2, z_3) = f(t_1^{-1}z_1, t_2^{-1}z_2, t_3^{-1}z_3)$

[cloud walrus]

Iteribus

[lavish nexus]

given a monomial $z_1^{d_1}z_2^{d_2}z_3^{d_3}$ with $d_1+d_2+d_3=d$, $diag(t_1, t_2, t_3) f (z_1, z_2, z_3) = t_2^{d_1-d_2}t_3^{d_1-d_3}z_1^{d_1}z_2^{d_2}z_3^{d_3}$

[cloud walrus]

Iteribus

[lavish nexus]

is the weight $(d_1-d_2, d_1-d_3)$?

[cloud walrus]

Iteribus

[maiden heath]

Question about internal direct products
I have A and B which are subgroups of H, where A and B are isomorphic to S3, and i'm trying to show that the internal direct product of A and B is H, because i'm fairly certain that H is isomorphic to S6. I'm hoping that the internal direct product of S3 and S3 is S6.
Would someone be able to quickly guide me through how to find the internal direct product of S3 x S3?
Much appreciated

[next obsidian]

S3 x S3 is not even isomorphic to S6

The orders don’t match up

[maiden heath]

oh well that answers my question

of course

[next obsidian]

And the internal direct product of two subgroups is just isomorphic to their product as groups

If you did have H as the internal direct product of A and B, then H is isomorphic to S3 x S3

So I’m not sure what it means to find the internal direct product of S3 x S3

[maiden heath]

well I know that A and B are isomorphic to S3, and i'm trying to show that internal direct product of A x B = H, by using that fact, I must have the wrong isomorphism for H then

[next obsidian]

This doesn’t make sense

The internal direct product isn’t of A x B

It’s of A and B, and it ends up being isomorphic to the group A x B. The internal direct product is a way when you have two subgroups of a group, and they intersect trivially, their product is the group, and both are normal

You end up being able to write the supergroup as being isomorphic to the direct product of the two subgroups

The internal direct product isn’t an operation you can do to two groups G and H

[maiden heath]

ohhh right, I misunderstood that , thanks for clearing that up, I thought it was some form of operation

[next obsidian]

No

The direct product is

But the internal direct product is like when you can write a group G as the direct product of two of its subgroups in a really natural way

You can sort of relate the two by taking two groups G and H, then forming G x H

You then have subgroups of G x H which are isomorphic to G and H, namely G x {e} and {e} x H

It then follows that G x H is the internal direct product of those two subgroups

But this is different than “taking an internal direct product of G and H”

Internal direct product is a relative notion, it requires specific subgroups of a group. It’s kind of like normality, it doesn’t make sense to say that G is a normal group or something

Because normality is about G as it lives inside another group

[maiden heath]

Ok I see, my god you've just cleared up so much. Can't thank you enough

[next obsidian]

[gilded gull]

Is the symmetry group $C_{2}$

[cloud walrus]

Rishi

[gilded gull]

I just want to makes ure

and is this one $\mathbb{Z}$

[cloud walrus]

Rishi

[spice whale]

the second one is C_2

why would the next one be Z

it's not symmetrical

rotationally or reflectionally

[gilded gull]

wait the second one is C_2?

[spice whale]

no

[gilded gull]

You mean the first is C_2

[spice whale]

yes

sorry

[gilded gull]

oh ok good

im being stupid about the second one

[spice whale]

yep

the second one can't be rotated

[gilded gull]

So is it just the identity then

[spice whale]

the trivial group yes

[gilded gull]

O ok

thanks

[gilded gull]

does this work in showing that the composition of g w/ f is a group morphism

also I just realized I flipped the order at the end but ill fix it

[boreal lion]

assuming f: G -> H and g: H -> G are homomorphisms (group morphisms), then yeah, it works

[maiden heath]

Can someone please explain what how $H \tau$ works?

[cloud walrus]

ArtyLeAardvark

[maiden heath]

I understand it is a right coset, but i've never seen one of an operator before

[delicate orchid]

I imagine it's just tau applied to H but I'm not 100%

[coral shale]

That's on your book to hopefully define it

someplace before

[maiden heath]

I've figured it out, was just being stupid. Cheers

[gilded gull]

great thanks

[boreal lion]

Np

[devout crow]

you never heard of translational?

[spice whale]

if we include that then it becomes the additive group of R²

[devout crow]

how so

[gilded gull]

So I've done some examples w this just to see how it works out

I get how (0 1 1 0) is basically acting like h

and then the other matrix is going to end up adding powers since we do have exp

But what am I exactly trying to show?

Like am I trying to show that each element composed of matrix multiplication maps to one of the elements of $D_{k}$

[cloud walrus]

Rishi

[gilded gull]

So $h^{2}$ goes to the first matrix squared so thats the identity and then each power of the second matrix would be the same as each power of g?

[cloud walrus]

Rishi

[gilded gull]

And I guess I can show for $g^{k}h$ elements that they have the same order as the elements generated by the first matrix w powers of the second

[cloud walrus]

Rishi

[gilded gull]

But I don't really know how to put it all together cohesively

[lavish nexus]

Dk is symmetry of regular polygons

[gilded gull]

right

[lavish nexus]

you have rotation and reflection

[gilded gull]

yes

Like I conceptually understand that these two are the same

Its just I don't know how to show it lol

[lavish nexus]

you just need a rotation matrix

and a reflection matrix

[gilded gull]

Right

the first one is reflection

and the second is rotation

I would need more to show that they're isomorphic tho

[lavish nexus]

There’s a generator relation sr =r^(-1)s

[gilded gull]

o

[lavish nexus]

once this is satisfied you’re done

[gilded gull]

So am I trying to show they have the same order and they're cyclic ---> isomorphic?

Or not exactly cyclic tho

[lavish nexus]

the definition of Dk = (r, s) with relation r^k=1, s^2=1, sr =r^(-1)s

prove the two generators of this group satisfy these relations
then they must generate Dk

[gilded gull]

wait thats it

cool

So I just show the second matrix has order k

and obviously the first one has order 2

And then its just the last relation

[lavish nexus]

yeah

[gilded gull]

what does the last one mean again

It seems familiar

[lavish nexus]

try it yourself for D4 the symmetry of a square

[gilded gull]

Ok

thanks

sorry ab that i didn't get it right away lol

[lavish nexus]

np

[gilded gull]

oml im stupid

Lmfao thanks again

[fallow plume]

we're being asked to prove that K[x] being a PID implies K is a field. I've proved it's a division ring, but is this true even for non-commutative division rings? i looked around and it seems like it isn't?
(i can't accept someone elses proof so please don't post a proof, but this does seem sketchy and Idk if this is actually valid)

[thorn delta]

yea, this is false

oh wait nvm the counterexample i had in mind doesn't apply oops

[fallow plume]

mm yea I'm really confused cause I googled a bit and they always assume K is commutative

and I can't find results about division rings

[thorn delta]

@fallow plume wait

lmao

PIDs are commutative

K is a subring of K[x] so its automatically commutative as well

[fallow plume]

how have I not seen this before?!?!?

:rooDerp:

thanks lmao

[thorn delta]

np

[fallow plume]

wait are you sure? https://math.stackexchange.com/a/724547/715922

[pastel cliff]

this is probably stupid but D_8 isn't cyclic right??

cuz two generators

[fallow plume]

no, r and s

yee

[pastel cliff]

kk

[fallow plume]

???

whack

[thorn delta]

I think the common definition of "principal ideal domain" is "integral domain where every ideal is generated by a single element"

[pastel cliff]

is this valid

[fallow plume]

as opposed to?

[pastel cliff]

D_8 should count right

[fallow plume]

I'd say so.

it might just be me as a student tho, but I don't think it's common knowledge that D_8 is in S_4 tho

but idk

[pastel cliff]

i mean it's hw lol

[thorn delta]

as opposed to some alternative like every ideal is principal but the ring might not be an integral domain

[pastel cliff]

i'd hope the prof knows

[fallow plume]

hah well the problem is does he know that you know :P

oo int domain gotcha

i'll just hope that's how my prof defines it but idk

[thorn delta]

wdym D8 should count? Isn't that what they wrote

[thorn delta]

D_8 is a group of symmetries of a cube. Every element of D_8 can be thought of a map which maps the vertices of a cube to their new positions under the action of some symmetry-preserving transformation. That's why D_8 is a subgroup of S_4

like rotation by 90 degrees is a cycle (1 2 3 4) for example

[hidden haven]

They're often commutative by definition

[fallow plume]

oh cool okay

[fallow plume]

yee fully agree, just Idk what stage they're at

my prof would've marked me down in a heart beat

[thorn delta]

ohh wait i see thats what nitzeba wrote

for their proof

[chilly ocean]

are there equivalent ones

[cyan raft]

isnt the symmetry group of a cube S4?

every symmetry permutes the 4 diagonals

[thorn delta]

yes but not every permutation of four vertices preserves the cube.

[cyan raft]

i meant the diagonals?

[thorn delta]

wait, this whole time ive been saying "cube" when I meant "square"

[subtle ivy]

"a separable splitting field is the splitting field of a separable polynomial"

not sure if i just havent learned enough theory about splitting fields yet but im finding this kind of hard.

[hidden haven]

Are you asking why that is true?

[subtle ivy]

yeah

[hidden haven]

What definition of separable extension do you have?

Every element has separable minimal polynomial?

@subtle ivy

If that is the case, then you have to show that if an extension is generated by separable elements, then it is separable

You can use a lemma for this which says that F[a]/F is separable iff any homomorphism F → cl(F) has exactly (deg a) many extensions to homomorphisms of the form F[a] → cl(F)

Then you count extensions of such homomorphisms for the given field extension and you are done

[coral shale]

To construct an algebraic number in C whose argument is an irrational multiple of 2pi, I can consider a polynomial like

$$f(x) = x^2+x+1$$

Right?

[cloud walrus]

Shuri2060

[coral shale]

in particular

$x^n \not\equiv 0 \pmod{x^2+x+1}$

[cloud walrus]

Shuri2060

[coral shale]

For all n

Is my thought process.

[lethal dune]

roots of f(x) are w, w^2 which has argument 2pi/3 and 4pi/3 which are rational multiples

[coral shale]

eh?

[lethal dune]

or I understood it wrong?

[coral shale]

where does my thinking go wrong

[coral shale]

are they???

that isnt the equation of unity

i chose

[lethal dune]

I don't think that'll even be an algebraic extension

[coral shale]

???

now im confused

wut

[lethal dune]

[coral shale]

x^2+x+1 is irreducible

in Q[x]

[lethal dune]

but you require the roots to have argument which is irrational multiple of 2pi

in C

[coral shale]

and i claim the roots of this polynomial

satisfy this requirement

(not totally sure which is why i wanna ask)

[lethal dune]

,w argument of (-1+sqrt(3)i)/2

[cloud walrus]

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[coral shale]

eh?

hmm where the heck did my thinking go wrong

[lethal dune]

idk if that's what you are asking

[coral shale]

Since that is a rational multiple

$x^n \equiv 0 \pmod{x^2+x+1}$

[cloud walrus]

Shuri2060

[coral shale]

This must be true for some n right?

[lethal dune]

yes

[coral shale]

ohhhh waiittt

[lethal dune]

n=2 works

nah

[coral shale]

huh?

[lethal dune]

ignore that

[coral shale]

no no i think i wrote wrong

[chilly ocean]

I dont think such n mighht exist

[coral shale]

$x^n \equiv k \pmod{x^2+x+1}$

[cloud walrus]

Shuri2060

[coral shale]

We want this to be true for k in R

and indeed we can get this by uh

[chilly ocean]

,w factor x^3 - 1

[cloud walrus]

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Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by catfan1337#9812

[coral shale]

wow i chose a terrible example

But anyways so yeah I was wondering if the above is possible

[lethal dune]

if argument is irr multiple of 2pi then the extension will not even be algebraic

[coral shale]

Ok, and I wanted to convince myself why/why not this was true

it seems if you keep multiplying a by itself

[lethal dune]

idk

[coral shale]

you will eventually get some real multiple of a

[lethal dune]

i was only thinking of roots of unity

[coral shale]

$x^n \equiv k \pmod{x^2+x+1}$

[chilly ocean]

sry wahhts the actual problem or question?

[cloud walrus]

Shuri2060

[coral shale]

Me thinking

[chilly ocean]

You're the actual problem

[coral shale]

I am thinking about stuff like uh

L : K finite algebraic extension

and uh stuff idk

For variety of reasons I want to convince myself of the above fact being true or not

[chilly ocean]

wat fact

[coral shale]

Can we find an algebraic number with irrational multiple of 2pi

for its arg

[chilly ocean]

like k = 2pi * irr?

[coral shale]

yh

[chilly ocean]

hhmm probably hard to answer

[coral shale]

Which is equivalent to seeing if

[chilly ocean]

because like idk 2pi * e we dont know if its algebraic

[coral shale]

$x^n \equiv k \pmod{f}$

[cloud walrus]

Shuri2060

[coral shale]

Does this always have a solution for some integer n

L:K alg extension

[chilly ocean]

if you can choose any k then yeah

[coral shale]

k in K

f in K[x]

[coral shale]

why ? 🤔

[chilly ocean]

my intuition says so

[coral shale]

🔫

Let $L:K$ be an algebraic extension. Let $f\in K[x]$.

Is there always a solution to

$$x^n \equiv k \pmod{f}$$

for some $n\in\bZ, k\in K$?

[cloud walrus]

Shuri2060

[coral shale]

So I think this is the full statement

[coral shale]

which should be the generalisation of this question

If not give an example 👀

wait n = 0 is a solution

i mean n > 0

🤔

[lethal dune]

is $L \simeq K[x]/(f(x))$?

[coral shale]

not necessarily

Do we want this?

🤔

[chilly ocean]

wait Im thinking this may be true for finite L/K?

idk

[coral shale]

god sorry

yes I need that to be a finite extension

Let $L:K$ be a finite algebraic extension. Let $f\in K[x]$.

Is there always a solution to

$$x^n \equiv k \pmod{f}$$

for some $n\in\bZ_{>0}, k\in K$?

[cloud walrus]

Shuri2060

[coral shale]

And f not constant maybe

I initially felt this might not be true

But I keep thinking of examples involving roots of unity

[lethal dune]

I'd guess no

[cloud abyss]

How to solve this problem?

[coral shale]

write down what <8, 14> is

according to the defn

[lethal dune]

$x^n-k \equiv 0 (\mod f) => f|x^n-k$ for some k means f(x) must have roots roots as x^n-k

[cloud walrus]

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

[south patrol]

Do you know what the group generated by some elements corresponds to in Z

[coral shale]

right and i keep thinking examples involving this

so what wouldnt....

[cloud abyss]

Can solve that for me and explain it ti me

[coral shale]

lol?

Can you write down

what the definition is

from your notes, instead

what does this thing < > mean?

x^n + x + 3 I think this works

Hope it does

for my Q

[hidden haven]

Probably take x² + 2x + 2 or something similar

[cloud abyss]

{8^n(14^m)|n,m is an element of Z}

[coral shale]

good

[hidden haven]

Some irreducible that has terms other than the leading one and the constant term

[coral shale]

wait not good

[south patrol]

Surely it's not that

Ye

It's additive

[coral shale]

Z is a group under addition

not multiplication

[coral shale]

i tried x^2+x+1 cus thinking this and epicly failed earlier

[cloud abyss]

Ow addition since 0 has no inverse

[hidden haven]

oh how does it fail?

[coral shale]

cus cube roots of unity

are its roots

2 of them

😂

[south patrol]

Nothing except 1 or - 1 has a multiolicative inverse in Z

[coral shale]

im guessing this works but unsure

i will wolfram

forgot quadratic formula exists

[hidden haven]

Oh ok

[coral shale]

,w arg((-2+sqrt(4-8))/2)

[cloud walrus]

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Requested by Shuri2060#9525

[coral shale]

uh

[south patrol]

lol

[coral shale]

fail again somehow

[hidden haven]

[coral shale]

i get the sinking feeling this is true

[chilly ocean]

I dont know how weird the polynomial should be for it to be irr * pi

[lethal dune]

[coral shale]

but uh

[hidden haven]

Quadratics cannot work

[coral shale]

hmmm

[chilly ocean]

take K=Q and x^2+3x+2

[south patrol]

to clarify, are we saying 'do there exist n,x,k such that ...'

like nothing is fixed right

[coral shale]

x is irrelevant

its uhhh

a variable

[chilly ocean]

all roots of x^n-k lie on a circle

any trans element will do

[south patrol]

isn't it an algebraic extension

[coral shale]

bruh what, this has real roots

[south patrol]

(x+2)(x+1) lol

[subtle ivy]

every element of the extension is the root of some separable polynomial over the original field

[chilly ocean]

Uh

[hidden haven]

Ye then that hint works

[chilly ocean]

You want (x^n-k)/f(x) in K[x] right?

For some n>0 and k in K

[coral shale]

uhhhhhhh

[chilly ocean]

[coral shale]

im not totally sure what i want --- lets go with

Complex root to integer polynomial with irrational multiple of 2pi argument

for now

before thinking about general case D:

Moldi thinks no quadratic will work for some reason 🤔

[chilly ocean]

I was answering the question in front of me

|| ||

[chilly ocean]

[hidden haven]

Adjoining a root of a quadratic adjoins its discriminant whose square is in K

[coral shale]

So uh... you are convinced I should definitely be able to find

one with irrational 2pi arg

[hidden haven]

[coral shale]

If yes I will hunt rather than trying to prove its true 😂

[chilly ocean]

Id be super surprised if that was true

[coral shale]

rlly?

[hidden haven]

This seems annoying and false

Lol

[coral shale]

hmmmmmm

this is annoying to verify

cubic formula wtf

[hidden haven]

Quadratics can't be counterexamples and x^n - a can't be counterexample, so

[coral shale]

I was hoping our answer would be obvious D:

Why was I thinking about this... something about basis something something

[chilly ocean]

I'd look at degree 5 at least, weird stuff happens there

[hidden haven]

Hopefully false for cubics

But

[proud bear]

What about ||(x-(2+i))(x-(2-i))||

[coral shale]

uh

,w arg(2+i)

[cloud walrus]

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[coral shale]

ah

[hidden haven]

Why are you looking at argument

[coral shale]

and this can be made into a cubic

the angle?

cus my thought process relates to

exponentiating the root

until we get some multiple of it

like uh... i want to confirm whether a^n eventually is a multiple of a or not

[hidden haven]

Yeah but there can be an element other than the root whose power is in K right

[coral shale]

if that makes sense

[hidden haven]

oh

You are only looking at the root

Ye makes sense

[south patrol]

this question has too many variables

[coral shale]

(x-(2+i))(x-(2-i)) = x^2 - 4x + 5

wait did i get scammed by moldi lul

,w simplify (x-(2+i))(x-(2-i))

[cloud walrus]

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[hidden haven]

Yes

[coral shale]

xd

[subtle ivy]

uhmm not sure this works... im not trying to prove the extension is separable, im trying to prove it is the splitting field of a seperable polynomial.

"Let L/M be the splitting field of a polynomial f over M, where f may or may not be seperable. show that if L is seperable, it is the splitting field of some seperable polynomial g over M."

[coral shale]

Back to what I was doing..... I think I am using euclidean algorithm to get this. But the remainder won't always be a constant.

x^n = gf + k
Man why was it so hard to find a counter

[hidden haven]

Ah then choose some finitely many generators of L over M, and take the LCM of their minimal polynomials to be g

[olive mirage]

Any representation theorists here? I have a group algebra Z[G] and a subgroup H of a finite group G, and a character chi of H and I'm looking at the quotient of Z[G] by the ideal I generated by the set of 1 - chi(h) h. I assume this is a standard construction that seems to be a slight variant of the augmentation ideal. Does it have a name? Do you believe the claim that Z[G] / I has rank [G:H]?

[pastel cliff]

in S_2 the elements are {id, (1,2)} - can i say that the (1,2) is the same in S_2 as S_3 and S_4

[barren sierra]

You mean (1 2)

(1, 2) is a tuple, not a permutation

But uh yes

(1 2) is the same in all those

It permutes the elements 1 and 2

[pastel cliff]

yeah sorry if notation is off

i just mean like

permuting two elements is the same whether the overall total is 2 or 3 or 4

[barren sierra]

Yes

Well permuting 1 and 2 is the same

In S_4

[pastel cliff]

ahhhhh that was my next question

[barren sierra]

(3 4) also permutes 2 elements

[pastel cliff]

but is different i presume

[barren sierra]

But (3 4) is not in S_3 or S_4

[wraith obsidian]

(3 4) is in S₄

[barren sierra]

However you can take the group {id, (3 4)} and use an isomorphism to rewrite it as S_2

So they're the same in a sense

But different objects

[wraith obsidian]

in the end isn't this discussion largely the same as „is ℝ a subset of ℂ or just isomorphic to a subfield of ℂ“?

[barren sierra]

Yea lol

[pastel cliff]

im just trying to rationalize whether or not S_3 and S_2 are subgroups of S_4

[wraith obsidian]

and dependent on the question whether you consider a permutation of X with support in Y to be the same as the restricted permutation

[pastel cliff]

so yeah p much lol

[barren sierra]

They are essentially

If you wrote that

No one would misunderstand

[wraith obsidian]

Well the constructivist viewpoint would be that you have a pretty canonical embedding of Sym(Y) into Sym(X) so you can identify them (whenever Y\subseteq X)

was that too much abstract nonsense or why the reaction @pastel cliff ?

Sym(X) is defined as the group of permutations of a set X, and Sₙ := Sym({1, …, n}) is the shorthand notation you're aware of

[wraith obsidian]

*identify Sym(Y) with the corresponding subgroup of Sym(X), sorry

[pastel cliff]

yeah some conversations in this channel make me dizzy lol

[wraith obsidian]

The main point is that S₂ and the subgroup of things in S₃ leaving {3} fixed are „structurally the same“ and therefore we can justify calling them „the same“ when we're doing group theory

so one never even bothers to distinguish which Sₙ the permutation (1 2) belongs to

because it doesn't matter

[barren sierra]

^

[tender kestrel]

anyone know how to solve this?

i was trying to form a mapping that represents reflections and rotations, and (13)(24) is clearly a reflection but (12) isn't a rotation so im sorta stuck

[wraith obsidian]

Have you tried multiplying these elements and see which other permutations you can get?

[tender kestrel]

I have, but I get permutations like (12) * (13)(24) = (1324) which doesnt make sense as a transformation on squares

[wraith obsidian]

Well you have successfully found a 4-cycle

I mean if you label the corners 1→2→3→4 clockwise this is not a rotation, sure

But it is if you label them as 1→3→2→4

[tender kestrel]

and then (13)(24) is still a reflection

[wraith obsidian]

„being isomorphic to“ means „being the same after relabeling“, so there's a good hint regarding your desired relabeling :>

Well doesn't the dihedral group contain a reflection?

[tender kestrel]

yes, i was just stating an observation

[wraith obsidian]

Ah, alright

thinks whether I'm missing a double l in relabeling somewhere

[tender kestrel]

HAHA

[wraith obsidian]

Ah, both is okay. https://en.wiktionary.org/wiki/relabel#English

[tender kestrel]

okay i think it makes more sense now

in terms of showing the isomorphism

if we define (12) * (13)(24) as a rotation and (13)(24) as a reflection then surjectivity and injectivity are fairly trivial

so we could say that phi((12)) = r, phi((13)(24) = s, but then how do we show that phi((12) * (13)(24)) = phi(12) * phi((13)(24))

oops sorry replace the (12) w/ (12) * (13)(24)

[stone fulcrum]

Calculate both sides

What's phi((12) • (13)(24))?

[tender kestrel]

shouldn't we be calculating phi((1324) * (13)(24))

[stone fulcrum]

Technically we need phi(x•y)

For any x,y

But it might be enough to be exhaustive

[tender kestrel]

by exhaustive you mean i should take every pair of elements in <(12), (13)(24)>?

[stone fulcrum]

Yeah I can't think of an easier way other than to define f on every single element and show that f splits over all •

[tender kestrel]

yeah

[stone fulcrum]

Unless you can say with certainty that D8 is the only 8 element group with a certain property, then show your group has it

[tender kestrel]

i tried finding all 8 values generated by <(12), (13)(24)> but im stuck @ 6

[stone fulcrum]

So I calculated a little bit. I'll call the two elements a, and b for ease.

Worth noticing is that both a and b have order 2. Neither of them can possibly be a rotation

ab and ba are inverses and order 4

[stone fulcrum]

And done. ab works as the rotation, a works as the reflection. Probably not the only choices.

Oh haha you got that already

[runic hemlock]

I've heard that the category of O_X modules for some ringed space (X, O_X) does not necessarily have enough projectives. Does anyone know an explicit example?

[runic hemlock]

Not sure if this fits this channel

[bold egret]

According to Wikipedia, a binary relation is a generalization of functions and has less restrictions than a function

what's something that you can do with binary relations that you can't do with functions?

(kinda don't know if this is the right channel tbh)

[next obsidian]

A function is a subset S of X x Y such that for each x in X, there is a unique pair (x,f(x)) in S

This is basically the graph, you think of the function f assigning x to that y value f(x)

So this is why you need a unique pair (x,f(x)), this is saying for each x, you have a single value f(x)

A binary relation is just a subset of X x Y

[bold egret]

Thank you

So you could have a fixed element a in X and have the set (a, y) where y is in Y and that is a binary relation?

But not a function

[next obsidian]

Sure if you like

Or you could just say that xRy for any x,y in S

xRy being shorthand for like saying (x,y) is in S x S

[gilded gull]

Find a subgroup of S7 of order 10

Does D5 work

[gilded gull]

and is this (1 2 4)(3 6 7) with each cycle being even so the permutation is even

and it has order 3

?

[lapis trail]

Yes to your first question

The others I'd need to get up to check. Sorry

[gilded gull]

oh thats ok

thanks

But Ig moreso what I need to check with that one

Both cycles have order 3

so is the lcm(3,3) just three?

[lapis trail]

Are you're asking about what the order of the full permutation is?

Given the orders of its disjoint cycles?

[gilded gull]

yea

I'm psure its just 3

I just want to make sure im not being stupid lol

[lapis trail]

Oh yeah. The order is the LCM of the orders of disjoint cycles. That makes sense to me

[gilded gull]

great

[lapis trail]

It's been a minute since I've studied these

[gilded gull]

Oh 100% when I finish this class I'm going to forget like half of it

[next obsidian]

disjoint cycles commute

[gilded gull]

But I'm just going to review everything from calc 1 to analysis before I go to college so

[next obsidian]

so yeah, it's the lcm

[gilded gull]

thanks

[next obsidian]

I said something false

It doesn't follow only from the fact that they commute, but also that they sorta can't cancel each other as well

[gilded gull]

yeah because they're disjoint

[next obsidian]

yup

[next obsidian]

[gilded gull]

Are these the only symmetric permutations?

or are there some im missing

[next obsidian]

for that polynomial?

[gilded gull]

yea

[next obsidian]

Uhhh

[gilded gull]

srry if my writing is bad its just 1, 2, 3, 4

[next obsidian]

looks right to me?

[gilded gull]

cool

[next obsidian]

wait

[gilded gull]

Oh shoot

[next obsidian]

no

[gilded gull]

identity

[next obsidian]

the second one isn't

symmetric

I think it multiplies by -1

[gilded gull]

Oh because 4 can't go to one

[next obsidian]

no that's okay

it's just you went from x3 - x4 to x4 - x3

right?

[gilded gull]

yes

so 4 has to go to 2

it can't go to 1

[next obsidian]

right but then you end up with the first one

[gilded gull]

What if they both flip then it could work?

[next obsidian]

yeah

[gilded gull]

So instead of that one i could do 1 to 4 and 2 to 3

[next obsidian]

so that's the missing one

[gilded gull]

and then the identity

[next obsidian]

(12)(34)

[gilded gull]

O shoot thats another

Ur right

[next obsidian]

no that's the only 3 I think

(12)(34), (13)(24)

the identity

[gilded gull]

what about (1 4) (2 3)

[next obsidian]

oh yeah

[gilded gull]

wouldn't the negatives then cancel

o

ok great

thanks

[next obsidian]

swag

[gilded gull]

I just realized something

Can we only have even permutations because we need the value to be the same and not negative?

[next obsidian]

I was going to say

I think that might be true, but I wasn't 100% sure

[gilded gull]

I mean it isn't for all cases

[next obsidian]

since the polynomial you look at to evaluate parity of a permutation is a different one

[gilded gull]

like one of them (x1+x2)(x3+x4)

that wouldn't matter

[next obsidian]

right but that doesn't have differences

[gilded gull]

and also if its squared

x1-x2

[next obsidian]

I think is the difference

[gilded gull]

^2

[next obsidian]

yeah, I really don't know

I think it's true that only even permutations preserve this, but I'm not sure why

[gilded gull]

I mean for this for specifically the reasoning is fine

Idk how to generalize it any further

[next obsidian]

Well for this specific one I think you prove it by just enumerating them all

[gilded gull]

maybe for groups of binomials w even order?

[next obsidian]

and you observe that it's only even permutations

[gilded gull]

even power* not order

[next obsidian]

I just mean I'm not sure if there's a deeper reason

[gilded gull]

^^

[next obsidian]

I think there is, but I don't know it

[gilded gull]

O

I mean in the text there's something ab even preserving the sign of the determinant

and determinant is defined this way

Oh s*** thats probably it lol

[next obsidian]

¯_(ツ)_/¯

[gilded gull]

not determinant

discriminant

oops

[next obsidian]

¯_(ツ)_/¯

[gilded gull]

\

[next obsidian]

oh yeahso that's the definition yeah

but this isn't the discriminant

you're missing like x2 - x3 for example

[gilded gull]

Oh true

[next obsidian]

So yeah, idk I think there's a relation here

[next obsidian]

This was what I was referring to here

[gilded gull]

but it could be like $D_{1, 2/3}$

[cloud walrus]

Rishi

[gilded gull]

idk

wait nvm

[next obsidian]

yeah idk what that means lol

[gilded gull]

its just like

excluding certain elements

Idk i need to finish the problems lol

thanks again

[gilded gull]

Express (123) (456) as the power of a single cycle in S6. Can you generalize
this result?

I have no idea what to do for this

actually could it be (1 4 2 5 3 6)^2

and then to generalize its just interlacing the elements and then squared

[coral shale]

(123)(456) has order 3.

If it is the power of a single cycle, then that single cycle must have order a multiple of 3

let's say 3k.

ie. it must be a 3-cycle or 6-cycle.

Obviously 3-cycles don't work, so it must be some 6-cycle, which you managed to find.

You can generalise this result to (1...n)(n+1...2n) in S2n, I suppose.

[gilded gull]

so what I found works?

and if so when generalized is it always squared?

[white jackal]

for part a, is the largest possible order of an element is a^mn?

assuming "a" is an element of that product

[coral shale]

I can think of some m and n where that isn't true.

[hidden haven]

Order would just be mn in that case, not a^mn. And that is wrong

[gilded gull]

Prove that Sn is generated by the elements (12), (23), (34), . . . , (n − 1 n)

Would I want to do this by induction?

Like for n = 2 we can just transpose once since theres only two elements

but like

whats the inductive step here

Or would it follow from any permutation can be writen as a product of disjoint cycles

So I would just have to show that the disjoint cycles are formed by the transposition

but then how do I do that

[hidden haven]

Bubble sort

[toxic zephyr]

when proving that something is a group/ring/field/etc., specifically in showing the existence of an identity or inverse, is it more standard to do something like

1. take a=[some arbitrary element]. consider x=[the identity element]. then show that a*x=x*a=a. then finally say thus x=the identity and we have shown the existence of an identity element

or

2. take a=[some arbitrary element]. consider 1=[the identity element]. then verify that a*1=1*a=a. thus it has been shown that 1 is indeed the identity element and we have shown that it exists

or is there another sort of proof structure that is more conventional?

[next obsidian]

aren't these the same thing?

this is indeed how you show the existence of an identity

but I really don't see how these are different

[coral shale]

1 is just a symbol

x is another symbol

[toxic zephyr]

yeah im sure they're both "fine"/sufficient proofs id just rather do it in the way considered most conventional.

[next obsidian]

I literally don't see how these are different

[toxic zephyr]

basically which symbol is more common

ik its nitpicky but im kinda anal about stuff like that

[coral shale]

take a=[some arbitrary element]. consider 🍌=[the identity element]. then verify that a🍌=🍌a=a. thus it has been shown that 🍌 is indeed the identity element and we have shown that it exists

[toxic zephyr]

lol

i think im just a bit iffy on calling something 1 before ive shown its the identity element

[coral shale]

I don't understand why the symbol 1 needs to even be mentioned

You are proving ax = xa = a for all a

for some specific x you have chosen

hence showing 1 := x

the symbol x or 1 won't actually appear in the proof

[toxic zephyr]

my book writes the axiom as "there exists an element denoted 1 such that blah blah". so i take proving that axiom very literally, as showing there exists an element called 1 which satifies that property.

[coral shale]

Let x in whatever

3x = x3 = x

Therefore 3 is the identity, since 3 in whatever

You have shown exactly that

[toxic zephyr]

thanks i hate it /s

idk these proofs just seem too easy so im assuming its all about the very rigorous and explicitly listed steps, and proving axioms to the letter/symbol

[coral shale]

Above, I identify 3 as the witness to the existential statement

Hence I show it.

What I did was plenty explicit

[toxic zephyr]

so then is abstract algebra really this easy? at least these kind of proofs

[coral shale]

Yes, the first bits are easy

You are learning basic arithmetic again, but abstractly

[toxic zephyr]

fair. so far it just seems like "we define it exactly how you would think to define it in generality".

[subtle ivy]

in my experience algebra is blissfully simple up until it is not, and then it can become very, very hard.

it becomes apparent that the simplicity and "boringness" of algebra is a very hard-won battle, the farther along you go

at least in my view

[coral shale]

Just came across compositum. If L is a field, and A, B fields in L, then

AB is the smallest subfield of L containing A and B.

In set notation, is this correct, or could I somehow write it differently/more intuitively?

$$AB=\left\{\frac{\sum^k_{i=0}a_ib_i}{\sum^l_{j=0}a_jb_j} : k, l\in\bN, (\forall i, j)(a_i, a_j\in A, b_i, b_j\in B)\right\}$$

(Or alternatively)
$$AB=\left\{\frac{\sum^k_{i=0}a_ib_i}{\sum^l_{j=0}a_jb_j}\right\}$$```

[cloud walrus]

Shuri2060

[coral shale]

I can also write A(B) and B(A) (A adjoin B, and vice versa) I suppose?

[waxen hedge]

Yes

[waxen hedge]

You need to suppose that you're not dividing by 0, else it seems correct

[chilly radish]

Quick Q, if G,H are fin gen free abelian, G injects into H, then is rank(G)<=rank(H)?

[lethal dune]

yes

[chilly radish]

It sounds correct but I can't seem to come up with a satisfactory proof

[lethal dune]

if rank(G) < rank(H) then you should be able to show it's not injection

[chilly radish]

U mean rank(H)<rank(G)

Yea I get that's the idea but I'm not sure how to do that. Z should have IBN, but i'm not sure how exactly it follows that every independent set should have size <=rank(G)

[lethal dune]

ok how about, since G is embedded into G, it's isomorphic to a submodule of H

but rank(G) = rank(im(G)) <= rank(H)

[chilly radish]

That was my thought. Not sure why rank is monotonic tho w.r.t inclusion

Wait nvm

That's obvious isn't it

Yea ok I was just overthinking this

That's what I ended up writing I was just second guessing myself lmao

Thanks

[lethal dune]

ig there's something else you can try, say {a1, a2, ..., an} be generators of G and {b1, b2, ..., bm} be generators of H (with m<n) then if phi is our map then phi(ai) maps n objects to m objects so from PH there's at least one overlap i.e. phi(ai) = bk^s and phi(aj) = bk^t so you should be able to construct an element that gets mapped to 0 by phi

[chilly radish]

Wait why are you assuming that you're mapping into the basis

Or am I misunderstanding you

[lethal dune]

no you are right, I shouldn't assume that

but you should be able to find an non zero element that gets mapped to 0,

some kind of power manipulation

[lethal dune]

nah this is better

[chilly radish]

Yea the contradiction solution would be really annoying

[spice whale]

$\begin{tikzcd}
G \arrow[r, "f"] \arrow[d, "g"] & H \arrow[d, "\tilde{\varphi}", dashrightarrow] \
{\ker f} & {\ker g}
\end{tikzcd}$

[cloud walrus]

coalison (shyshu for honorable)

[spice whale]

can tilde phi be an isomorphism

[bleak abyss]

Shouldn't it be ker(f) includes into G

[next obsidian]

This diagram don’t make no sense

[spice whale]

ok I'm bad

I'll work out what i mean

[pastel cliff]

can a subgroup have the same order as a group?

not counting the trivial case of a group being its own subgroup

[spice whale]

the group is a subgroup of itself

oh

[pastel cliff]

id think not right

[spice whale]

actually yes

2Z is a subgroup of Z

[pastel cliff]

mmm but inf order

[spice whale]

yep

[pastel cliff]

i was thinking in finite order but i guess

[next obsidian]

Lol bro

If it’s finite order you can’t even be bijective to a proper subset

[pastel cliff]

[spice whale]

yeah I'm not sure how this would work

[pastel cliff]

mmm so if an element has the same order as the group, then G = cyclic group generated by that element

[spice whale]

yes

if it's finite

[pastel cliff]

uh oh

[delicate orchid]

pigeon hole principle moment

don't you uh oh me buster

[spice whale]

if it's infinite...

[pastel cliff]

ok

[delicate orchid]

I do have a question if any chatters on the line rn have any experience with computing Schreier generators let me know cause I've been at this for 3 hours and I'm not getting it

[pastel cliff]

#abstract-chill

[delicate orchid]

go on....

https://cdn.discordapp.com/attachments/359052581022203914/940656085893799936/unknown.png

[spice whale]

you can probably make a proper subgroup with equal cardinality you silly billy

[delicate orchid]

thinkin bout (R, +)... (I got no idea if this is an actual counterexample btw)

[spice whale]

it seems like it

[hexed panther]

Hi, is there a consensus on the "best" textbook for introductory Galois Theory? I'm an undergrad with a little background in group theory and I'm wondering if galois theory is accessible. (I'm in the "prépa" system in france and we have to do a presentation at the end of our second year outlining the significance of a mathematical (or physical) result or theory and how it can be applied. I'd like to dig into galois theory for this, hence why I'm looking for an introductory textbook)

Oops just realized there's a channel for this, sorry (I'll repost if I get no answers here)

[rigid cave]

I read a bit of Rotman and it was good. I think Artin is good as well

I once found great lecture notes, let me try to find them again

[cursive temple]

stewart was nice and had a lot of historical stuff aswell

[rigid cave]

oh ye

IMO this is good http://www-groups.mcs.st-andrews.ac.uk/~martyn/5836/MT5836lecturenotes.pdf

starts with rings and works it up

[hexed panther]

ah this looks wonderful

thanks

[delicate orchid]

reading time

[chilly ocean]

For infinite groups the answer of existence is in the positive if we're talking about exponent rather than order

Like for countably infinite direct product of Z/pZ, the exponent is p and so is the exponent for every proper subgroup

[maiden ocean]

Can you elaborate a little bit on what your exposure to algebra is? have you seen any ring or field theory?

oh eek im about to lose internet so i wont be able to respond for like an hour haha sorry

but I will when i get home

[hexed panther]

Tbh I don't really know... I know what rings and fields are, we've done exercises on them but I don't think it's anywhere near a full course on ring or field theory. So I don't think it counts as really having seen them

I don't think we have any of the big theorems of abstract algebra

[coral shale]

I think you need basic ring/field theory down

or you're going to be

In particular, the concept of the different types of rings/hierarchy (ID, UFD, PID, ED, Field) crops up

And (maximal) ideals

Because you are interested in quotients

===
For a field K, the idea of taking the quotient
K[x]/(f)
where f is a minimal polynomial (or (f) is a maximal ideal)
to create a new field is v. important

[hexed panther]

Yeah we have quotients and ideals

Like yeah we have the basics but we don't have stuff like the sylow theorems

[barren sierra]

I'd say Sylow theorems are part of the basics

bold of me to say since I am bad at this stuff still

[delicate orchid]

they are kinda basic

[hexed panther]

yeah hence why i'm not sure what counts as "the basics"

like we don't have the equivalent of a course in group theory but we have definitions and some properties

[barren sierra]

I'm still bad at using them

[hexed panther]

but anyways i'll look at the lecture notes that were sent they look relatively accessible ig?

[barren sierra]

I will say that the graduate level abstract algebra course I am doing is doing intro galois theory at the end of the course

so I'd say if you've taken a full undergrad course in abstract alg

and then are willing to study some graduate material

[hexed panther]

yeah that's what i was afraid of

[barren sierra]

then maybe go for learning galois theory

[hexed panther]

but i figured there's a huge amount of leeway with the use of the terms "intro" and "basic" in math haha

[hexed panther]

alright

[barren sierra]

but like ring theory does have alot of polynomials which are the subject of study of galois theory

so not having a strong understanding of those might be a bit rough

[hexed panther]

yeah understandable

i'll talk to my teacher about it

[barren sierra]

is $(f(c))_{c \in F}$ just notation for a product of all evaluations $f$ at all $c \in F$?

[cloud walrus]

Spamakin🎷

[barren sierra]

like as a (potentially infinite) tuple?

I've just never seen that notation lol

[coral shale]

I have no idea of them and I haven't needed them yet in my course 👀

[delicate orchid]

it's not infinite but I believe so yes

[barren sierra]

oh

yea $F$ is finite

[coral shale]

What is sylow needed for in galois

[cloud walrus]

Spamakin🎷

[barren sierra]

[barren sierra]

I meant basics of algebra

[coral shale]

Sylow is basic?

[delicate orchid]

yup

[barren sierra]

I'd say so

[coral shale]

It came at the end of my group theory course in yr 2 (and couldnt be bothered)

[barren sierra]

interesting

that seems late

[delicate orchid]

it was near the end of my first one iirc

[coral shale]

Dont think you need it specifically in galois do you?

[barren sierra]

not in Galois but like

if you don't know them and you try to learn Galois theory

what other holes do you have

ya know?

[delicate orchid]

it occasionally comes up in places

[coral shale]

I'm doing ok currently

[barren sierra]

I see

[coral shale]

The list of things I mentioned above has served me well enough

(and field extensions)

But I'm only halfway through my course.

[maiden ocean]

you absolutely do not need to know the sylow theorems to do galois theory

Especially if you're just doing a student project

You need to know like literally what groups and automorphisms are and then like a chapter of ring and field theory

You basically just need to understand what a normal subgroup and a quotient is lol

[coral shale]

God I missed that 'not' for a sec lmao

[hexed panther]

Alright that's reassuring

Although I'd need substantial understanding of it, it's not a small project

There's a 1 hour interview with a jury so it's not just hey we can't solve quintics

But alright thanks for all the advice :)

[chilly ocean]

we can solve quintics

[coral shale]

If K is a field, then do you know why K[x]/(f) where f is a minimal polynomial basically 'adjoins' a root of f to K?
This is probably one of the most important starting points imo

[maiden ocean]

The only situation in which the sylow theorems or whatever would be necessary is if you're literally like, doing a galois theory pset and proving that galois groups of order x are y or whatever

Theres really not a ton of group theory involved in the main theorems

[chilly ocean]

what's a pset?

[hexed panther]

Yeah I was just giving an example to try to pinpoint where we're at in group theory

[maiden ocean]

Proble mset

[chilly ocean]

whats a proble mset

[maiden ocean]

A problem set

[chilly ocean]

ah okey

[coral shale]

$\bZ^\times = {-1, 1}$

[cloud walrus]

Shuri2060

[coral shale]

Usual convention 👀

[chilly ocean]

wait no im stupid

kill me

[maiden ocean]

Well its not a group so

thats a start

[chilly ocean]

kill me

[maiden ocean]

Happens

[coral shale]

$\bZ^*:=\bZ - {0}$

[chilly ocean]

Anyway

[cloud walrus]

Shuri2060

[coral shale]

I think this is usual convention

for future reference

[chilly ocean]

it isn't

[coral shale]

R^x refers to the set of units if R is a ring is the usual convention

[chilly ocean]

Is $\mathbb Q^{\times}$ a semidirect product of two non-trivial groups?

[cloud walrus]

Carla_

[maiden ocean]

its abelian so it would have to be a direct sum of abelian groups

[chilly ocean]

true

[coral shale]

I think I have one but I may be wrong

Consider all elements of Q as integer quotients a/b written in simplest form

Consider the elements where there are no powers of 2 in a or b

I think this forms a (normal) subgroup?

[chilly ocean]

normalness is automatic

[coral shale]

Right, so quotient Q by this

[chilly ocean]

dont take away powers of 2 in a

and i think it works

[coral shale]

I believe this subgroup and the powers of 2 would be a semidirect product

[coral shale]

oh

wait we're considering Q under multiplication only right

[maiden ocean]

if Q^times is the direct sum of like A and B or whatever both nontrivial then we have a neq 0 in A and b neq 0 in B so that (a, 0) and (0, b) are nonzero in A oplus B = Q^\times. but (a, 0) times (0, b) is (a * 0, b * 0) = (0, 0). This cant happen lol

[chilly ocean]

yea under multiplication only so forget maybe

[maiden ocean]

Actually

Nvm

The identity in Q^times is not 0 what the fuck was i saying lol

[coral shale]

$$A := {2^n : n\in\bZ}$$
$$B := \left{\frac ab : a, b\in \bZ^*\land\gcd(a, 2) = \gcd(b, 2) = 1 \right}$$

I think this is what I was referring to formally

hopefully.

And the claim is A x B = Q

I probably can chuck the gcd(a, b) = 1 condition

[cloud walrus]

Shuri2060

[delicate orchid]

B is the localisation of Z at (2) so I imagine taking the direct sum with (2) just gives you Frac(Z) = Q

makes sense to me

[maiden ocean]

That would be kind of a neat result

If the direct sum of A_(p) and (p) as modules was isomorphic to Frac(A) for A an integral domain, as a module

[delicate orchid]

at least as multiplicative groups, I get why it might fail as additive

[maiden ocean]

Well

Q^times here is multiplicative

[chilly ocean]

whats a multiplicative group

and an additive

[delicate orchid]

$(Q^*, \times)$ vs $(Q, +)$

[coral shale]

what do you study 'Frac' in as well lul

[cloud walrus]

Wew Lads Tbh (200 🍇) ✓

[barren sierra]

Frac is some ring theory stuff

[delicate orchid]

ring theory

[barren sierra]

Frac(R) = the ring of fractions over R

[delicate orchid]

field, but yeah

[chilly ocean]

why is the former multiplicative and the latter additive?

[barren sierra]

right

[coral shale]

And I was thinking fractal, nvm me

[barren sierra]

field

[delicate orchid]

the left multiplies... the right adds...

I'm confused where the confusion is

[chilly ocean]

that makes no sense

[barren sierra]

CONVENTION (not a hard fast rule) says that addition is used for commutative stuff

[chilly ocean]

i can rename * to +

[barren sierra]

like people will write (G, +) to imply G is abelian

[delicate orchid]

ok cool but then that won't be the multiplicative group of the rationals