groups-rings-fields

[terse idol]

(problem statement being If $K$ is a subgroup of $G$ and $N$ is a normal subgroup of $G$, prove that $K/(K\cap N) \cong KN/N$.)

[cloud walrus]

ChiliLion

[tender wharf]

Trying defining a homomorphism out of KN into K/(K cap N)

actually this holds in general for order p, p >2

p prime

im sure it can be weakened to nonprime as well

but its basically due to the fact that you can embed the cyclic group

[dapper sluice]

I'm looking for some intuition for the BCH formula. I don't need to know how to prove it exactly, but I am looking for an argument for why the correction terms only involve nested commutators. Does anyone have a good argument for why this is?

Here is the BCH formula https://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula

[rose prism]

it is just from taylor expanding the exponential

not sure what would be a satisfying argument for why the commutator shows up

[dapper sluice]

hello again!

[dapper sluice]

yeah ok i think im seeing how that works now

[rose prism]

https://www.math.ru.nl/~mueger/PDF/BCHD.pdf

i guess your question was about why there isnt extra stuff

beyond the commutators

but idk commutators measure the degree to which things commute

and BCH is a reflection of the failure of the binomial formula when computing the taylor expansion of the log

i am sure there are more sophisticated interpretations of what is going on here, probably many in the pdf i posted

but if your aim is an explanation that doesnt require reading a proof of the formula, those are probably a bit out of reach

[dapper sluice]

yes exactly

[dapper sluice]

ok yeah i will look at this

[rose prism]

look at the first few pgs of that pdf

they do the calculation

[dapper sluice]

ok thanks

by binomial theorem, you mean expanding this, yes?

[rose prism]

also prop 2.7 and remark 2.8

[rose prism]

yes

the usual one for numbers (x+y)^n = sum_k n choose k … implicitly uses that xy=yx

[dapper sluice]

yeah ofc

so basically I expand that out, but use yx = xy + [y, x] to get everything into the form of (elementary school expression) + (commutators)

[rose prism]

i think so

[dapper sluice]

ok great, thanks

[rose prism]

and then if you write down the entire taylor series

and expand everything

and then rearrange it

i guess you get bch

[rose prism]

look at these though

maybe it will be illuminating idk

[dapper sluice]

yeah these are nice

[rose prism]

there should be a diffeq way

and maybe a physics way

[dapper sluice]

similar to e^A B e^-A = exp(Ad_A) B

[dapper sluice]

yes that would be nice haha

[rose prism]

maybe those references were not the best

but a few pages later they prove bch

[dapper sluice]

i mean i think your comment about the binomial theorem basically does it, so im happy

[rose prism]

ok

[dapper sluice]

maybe there's a slick argument in terms of thinking of commutators as derivatives, but idk

[rose prism]

well

i am sure there is

taylor series can be viewed in many ways

they allow to turn the operation of differentiation into an algebraic thing

[dapper sluice]

i found a kinda cool argument, though it is not for bch but just for arguing that the exponent is just made of commutators

https://mcgreevy.physics.ucsd.edu/f20/2020F-220-lectures.pdf

[rose prism]

what pg

[dapper sluice]

look around equation 3.4 on page 80

kind of a purely formal formula because it involves a time-ordered exponential, but it only involves commutators explicitly

[rose prism]

i see

is the time ordering even important

for this thing being a sum of integrals of e^{t ad_A}

ok i guess it is

seems like a nice argument that id appreciate better with more physics experience

[dapper sluice]

start at eq 20, if you are interested

https://scipp.ucsc.edu/~haber/ph215/TimeOrderedExp.pdf

[dapper sluice]

one thing though, regarding your original comment on binomial theorem, is that it is highly non-obvious (at least to me), that \emph{all} you get are nested commutators, and not terms like $[A, B]B + [B, A]A$

[cloud walrus]

nay

[dapper sluice]

@rose prism

[rose prism]

you’re right

i guess the fact that exp(ad) = Ad_exp is playing a big role here

i think that is allowing to convert everything into powers of ad

[south sinew]

Does anyone know how to draw this using latex?

[mighty kiln]

\includegraphics

[south sinew]

thank u
but, it is not just a graphic
I mean, the content of the graphic is written by latex

[minor fulcrum]

I have seen two perspectives on representations:

1. A representation of G on V is a group homomorphism G->GL(V)
2. A representation of G on V is a functor from the one object category corresponding to G to to the category Vect that takes the object to V, and this lets us describe the category of representations where objects are such functors and morphisms are natural transformations

For some reason, translating back and forth between these feels hard for me. Is there a way to take a one object groupoid and "find" the object in Grp which it corresponds to, and vice versa?

obviously the connection between the two can be described, but can it be described categorically?

[topaz solar]

Well, mapping • in G (category) to Hom(G, G) in Set, and considering the morphisms acting on it appropriately (which is also sorta G < Sym(G) after all) should suit this?

Something something yoneda idc

And G |-> category of G is easy by just specifying the object as •, morphisms by elements of G, and composition law from G

[minor fulcrum]

what are the elements of G

and what is their composition law

[topaz solar]

This relies on this being a small category but if your category isn’t small you should feel bad

[topaz solar]

??? It’s a group

[minor fulcrum]

it's an object in the category Grp, is what I mean

[topaz solar]

The composition law is like, the point?

Compose it with the forgetful then idc

[minor fulcrum]

this gives you elements, how do you get composition

[topaz solar]

Refusing to acknowledge a group as having a composition law built in is insanity I’m gonna be real

Don’t reinvent the wheel

[topaz solar]

This map I’ve described is functorial

And whether I can define it in terms of more elementary functors or operations on Grp is entirely up to what you take as elementary

But it’s a functor nonetheless

[minor fulcrum]

fair enough ig

[topaz solar]

And further, Grp is a subcategory of Set

It is a set

Alternatively, take the isomorphism with Grp and Grp(Set) and go that way

If the idea of elements and sets is heretical to you

Do note that this category stuff doesn’t really “give” you information as-is here

It’s a different language carrying it

[minor fulcrum]

I understand, the information is already in the categories. Was just curious if there's a purely categorical way to recover the one object groupoid from an element in Grp, if there isn't that is fine

[topaz solar]

You can get new information with category tools, like sheaves and such, but this one is just a reformulation

[minor fulcrum]

yeah

[topaz solar]

“Purely categorical” is a bit silly imo, since it’s a lot more effort for not much payoff

But you probably could

Should is a different question

[topaz solar]

This is probably the closest I can think up

But it’s not my purview, so to speak

[coral spindle]

@delicate orchid where the fuk in James & Kerber do they talk about Sylows and reps of normalisers???

Like surely they prove McKay for the symmetric groups

[delicate orchid]

Gonna be real with you boss I read 1.5 chapters of that book and then never looked back

Also McKay is proven for S_n???? Based

[coral spindle]

Wew…. McKay is a theorem

Britta Späth announced she had proven the inductive conditions

No paper is out tho

[delicate orchid]

Am I thinking of Alperin-McKay

Yeah I was

[delicate orchid]

Who? 😹

[coral spindle]

Wtf wew

Britta is a super prominent rep theorist

Recently she was involved in proving a reduction for Navarro-McKay?

[delicate orchid]

Navarro
Who? 😹

[coral spindle]

Can someone inform wew I shan’t be talking to him anymore

[delicate orchid]

I literally have his block theory book 2.5cm away from my hand as we speak

[shell pilot]

Let H be a normal subgroup of a group G and K be any subgroup of G. $HK={hk | h\in H, k\in K}$ is a subgroup of G. Show this using a subgroup test.

I actually went through and proved it was a subgroup by showing closure, identity element, and inverse. But I want to prove it using a subgroup test. Namely, if $ab^{-1}\in HK$ then HK is a subgroup.

What I have so far is just the basic idea of $ab^{-1}$.

$Let a=hk, b=h_1k_1$, where $h\in H and k\in K$ Then $ab^{-1}=hkk_1^{-1}h_1^{-1}$

My idea is just that since $k and k_1^{-1}$ are both in K, then their product is also in K. But then what other manipulations do I need to do to show that this is an element in HK?

[cloud walrus]

Soap_Opera

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

[tough raven]

Evidently this is of the form (something in H)(something in K)(something in H). If you can change the (K)(H) term to a (H)(K) term you would be done (and by setting h = k_1 = 1, you can see that this is necessary; in other words, for any subgroups H and K, HK is a subgroup iff KH \subseteq HK).
So given kh with k in K, h in H, how might you write it as (something in H)(something in K), given that you know that H is normal?

[shell pilot]

Create a conjugation of H by writing kh = khk^-1k?

This is the solution but not sure how they create the last equality

$ab^{-1}=h_1k_1k_2^{-1}h_2^{-1}=h_1(k_1k_2^{-1})h_2^{-1}=(h_1h')(k_1k_2^{-1})$ for some $h'\in H$

[cloud walrus]

Soap_Opera

[coral spindle]

k_1k_2^-1 is in K

use the same commutator trick to 'move past' h_2^-1

[shell pilot]

Ok, so since H is normal then (k_1k_2^-1)h_2^-1 = h'k_1k_2^-1 for some h' in H because xh=h'x by normal definition

[tough raven]

(Incidentally, you don't necessarily need that H is normal (i.e. that gh = h'g for h in H, g in G) only that it's normalised by K (i.e. kh = h'k for k in G). The set of k such that k normalises H is always a subgroup N_G(H) of G containing H, and it suffices that K is contained in N_G(H) for HK to be a subgroup.)

[shell pilot]

Awesome, thank you @tough raven

[hidden cairn]

I need a hint for the first question

,rotate

[hidden cairn]

My idea is to find a homomorphism from GL_n(F) to {1, x, …, x^(q-2) }

With its kernel equal to SL_n(F)

Where (x, q) = 1

I tried phi(M) = x^(detM) but it’s not a homomorphism and doesn’t give 1

[tough raven]

I suppose that the latter refers to a cyclic group of order q-1 generated by x?

Although this is technically possible, note that you don't need a cyclic group, any group of order q-1 will do.

[tough raven]

(and the homomorphism should be surjective).

[hidden cairn]

Yes

[hidden cairn]

Yeah the cyclic group is the easiest example of such a group I could think of

So I used that but I don’t see other more convenient groups of order q-1 to work with

[tough raven]

It's the easiest to think of, but not the easiest to actually construct the homomorphism for.

[tough raven]

Well, since the group clearly will depend on F (even its size depends on F), try to come up with some group related to F.

[hidden cairn]

F - 0 under multiplication?

F^x elements of F with an inverse

Since q is prime

Only 0 would be excluded so order would be q-1 right

Oh I think phi(M) = detM works

Okay this was way more natural than my first thought lol

Thanks

[tough raven]

(As it turns out, F^x is actually cyclic of order q-1, so your first approach could have worked as well, but that works much less naturally - you have to go and find a generating element of F^x somewhat arbitrarily.)

[sage lodge]

does anyone have a further hint for this problem? not sure how to proceed

[rain grove]

In group (Z_n, +) what order is 0?

[delicate orchid]

I'd go with 1

[rain grove]

Why not 0?

[delicate orchid]

because it generates a subgroup of order 1 not 0

o(g) = |<g>| is a nice result and I wouldn't want to ruin it

also the definition of order is "smallest positive integer"

[rain grove]

Ohh really? In my book is just smallest natural number

[delicate orchid]

does you book include 0 in N

cause it shouldn't

[rain grove]

Ohh right yeah its not, I keep having my logic course in mind cause we there we said 0 \in N

[delicate orchid]

alas... it is not to be...

[tardy hedge]

Hey Wew what happened to you ? You were gone for a bit there

You missed my module theory grind

[delicate orchid]

a grave blunder

[tough raven]

Which part are you having a problem with?

[hidden wind]

anyone in here that might be interested in helping me trying to make sense of the contents of this book?:

[delicate orchid]

the entire book?

[hidden wind]

i’ve done some scattered reading in it but there’s lots of seemingly unifying things i don’t understand

[delicate orchid]

hmm... I do like coexter groups...

[hidden wind]

more like anything in the book tbh

also i’m not sure how representative this book is of the literature

[tardy hedge]

😂

[hidden wind]

some of the notation is surprisingly sumilar to something as old as burnside’s 1911 theory of groups

[delicate orchid]

perhaps posting the contents page would allow me to make more astute quips irregards to the co0peatnmi0mrbeuy[hto'pinm

could you imagine if burnside was real? that would be so cool...

[hidden wind]

whar

[delicate orchid]

ty

[hidden wind]

oh is that readable

no it’s not one sec

[delicate orchid]

yeah it is

dw dw

this is all fine and matches modern stuff

[delicate orchid]

except maybe the term "linear fractional group"? never heard of that in my life

oh they're like mobius transformations. Sure

[delicate orchid]

I cuold help with pretty much any chapter other than 7+8, but I'd struggle with some parts of 5 and 9

[hidden wind]

whole book is so beautifully typeset

[hidden wind]

amazing then i’ll definitely pester you mwahaha

[delicate orchid]

Mecejide would be good to ask about this stuff as well I think

[sage lodge]

I guess both and I'm not sure whether the given hint applies to both directions or only one

maybe struggling more to show that given G -> Z such a presentation exists

[delicate orchid]

my approach would be to observe that the map F_n -> Z (or even G -> Z) factors through the abelianisation, so it's enough to consider maps G_{ab} -> Z and Z^n -> Z. But I can't get any further atm

[tough raven]

Yes, that's more difficult. Since in the other direction, you map all generators to 1, an idea is take for generators elements mapping to 1 in Z: if you do this, any relation will have to have exponents summing to 0. So it suffices if G is generated by elements mapping to 1.

So given an arbitrary g in G, can you write g as a product of elements mapping to 1 (or their inverses)?

[sage lodge]

we're trying to show that a presentation exists given a surjective homomorphism phi : G -> Z right? and you are saying we can try taking the generators to be {x in G | phi(x) = 1}, and then if x1 x2 ... xn = e, we would have phi(x1 x2 ... xn) = phi(e) = 0, so the exponents in x1 x2 ... xn would sum to 0.

am I understanding correctly

[tough raven]

Yes. All you need to make sure this works is that the supposed generators actually generate G.

[sage lodge]

right

hmm

I think if phi was injective I could do it

[tough raven]

Why?

[sage lodge]

in trying to write g as a product of elements that map to 1, my idea was to look at the image of g, which would be some integer, and since Z is generated by 1 I could find some product of our generators that has the same image of g, but I can't guarantee that it equals g itself (without injectivity)

maybe I am missing something

[rain grove]

Why is Z_2 not a subgroup of Z

I meant obv. its not because 1 + 1 = 0 and in Z 1+1 = 2

But why is it like that? Is + defined differently or is 1 in Z_2 not the same element as in Z or smth else

[delicate orchid]

Z is torsion free and abelian so it can't have a torsion subgroup

[rain grove]

torsion?

[vapid vale]

if it were a subgroup, you should be able to name two elements that act like Z_2 in Z, under Z's group operation

[delicate orchid]

suppose there was a map f : Z_2 = <g | g^2 = 1> -> Z
this is defined by the image of g, so say f(g) = n in Z, then f(g^2) = f(1) = 2n = 0

implying that n is 0, so f was the zero map all along

this map isn't injective, and any embedding of Z_2 into Z clearly must be. So if there was a subgroup of Z isomorphic to Z_2 we would have a contradiction

[vapid vale]

you're right to say that 1 in Z_2 is not the same element as in Z

Z_2 is better named Z/2Z, the quotient of Z with 2Z if you're familiar with quotients

but essentially you take every multiple of 2 and set it to zero, and in Z_2 (the quotient group), 1 is a representative of every number with remainder 1 when divided by 2

you could very well call 1 "3" or "-11187281742585"

[rain grove]

ohh yeah in Z_2 elements are equivalent classes and in Z they are just whole numbers

[delicate orchid]

that's not particuarly important

[tribal moss]

You could define Z_2 such that its elements were literally the integers 0 and 1. Then it would be a sub_set_ of Z, but still not a subgroup because the group operation wouldn't be a restriction of the group operation of Z.

[delicate orchid]

exactly

I interpreted the question to mean "why are there no copies of Z_2 in Z" because not thinking about things up to isomorphism is an alien concept to me by this point

[rain grove]

Hmm ok i get it

[sly frost]

a simple point is that in Z every non-zero element has infinite order, while in Z/2Z every element has finite order

[delicate orchid]

That is a better way of phrasing this

[tough raven]

Do you need abelian?

[delicate orchid]

I clarified that simply because I didn't want to run the risk of being "ermmm acktully"ed

it was a protective measure

[quartz wind]

thought torsion was a modules thing, not group

specifically, shouldnt a torsion element be a non zero divisor? if it applies to groups how do you define zero divisor for a group

[fading field]

abelian groups are Z-modules, if you want to think of it that way

[sly crescent]

ermmm acktully you don’t need Abelianness

[quartz wind]

oh is that why you need abelian

youre still adding extra structure that isnt given rigjt

[fading field]

one could define torsion in a non-abelian group, it just won’t be a subgroup and also no one cares about that case (?)

[sly crescent]

Torsion just means non-identity elements with finite order

[quartz wind]

like you only have Z and +, but you need other operations for it to be a module

[tribal moss]

"Torsion" can be used about non-abelian groups -- but with care because the torsion elements are not necessarily a subgroup in that case.

[quartz wind]

i see

thanks

[sly crescent]

Some people care about that case

Whether all Artin groups are torsion-free is an open problem

[fading field]

okay i see

[delicate orchid]

they're isomorphic as categories, there's no extra structure

Z-mod and Ab, that is

[quartz wind]

idk what that means but you need Z to have multiplication right

[delicate orchid]

ok and what does that have to do with our abelian group

that's just saying that (a^x)^y = a^(xy)

which is counting

[quartz wind]

hrm okay i see

thanks

[delicate orchid]

np np

[quartz wind]

so is repeated addition the only multiplication that makes Z a ring with standard addition?

[delicate orchid]

are we really doing this

[quartz wind]

sorry its still not entirely clear how we can treat Z and Z2 as modules without loss of generality here. im sure youre correct its just not clear to me

[delicate orchid]

I said Z

I didn't say Z with a wacky multiplication

a different multiplication structure gives a different ring, and hence a different class of modules

[quartz wind]

okay i will revisit this later when im better at algebra thanks for the explanation

[shell pilot]

Is the principal ideal generated by an element of a ring just the product of all elements of a ring with other elements of the ring?

[quartz wind]

there isnt necessarily just one principal ideal

[shell pilot]

I'm just confused as to what it means since a is in R and r is in R

[quartz wind]

a ring generated by S is ideal iff S has only one element

so its the set of products of one particular element in R with all the other elements

[shell pilot]

Ok, so then what about this one

Let R be a commutative ring with unity and let $a_1,a_2,...,a_n$ belong to R. Then $I=<a_1,a_2,...,a_n>={r_1a_1+r_2a_2+...+r_na_n | r_i\in R}$ is an ideal of R called the ideal generated by $<a_1,a_2,...,a_n>$

[cloud walrus]

Soap_Opera

[shell pilot]

Is it like the cross product of the set with itself?

[quartz wind]

this is an ideal but not principal

[wraith cargo]

wdym cross product?

[quartz wind]

not sure what that means

[shell pilot]

R x R

[wraith cargo]

no?

[quartz wind]

thats cartesian product, but no its not

[wraith cargo]

I don't fully understand where your confusion is arising from

[quartz wind]

elements of a cartesian product are tuples, elements of an ideal are still elements of the ring

for example the set of even numbers $2\mathbb{Z} = {\dots, -2, 0, 2, 4, 6, \dots}$ is an ideal of $\mathbb{Z}$

[cloud walrus]

esca (@ with reply)

[quartz wind]

and this is the ideal generated by {2}, written $\langle 2\rangle$

[cloud walrus]

esca (@ with reply)

[coral spindle]

It should be mentioned that (bizarrely) ideals generated by some elements are also written with normal parentheses, e.g. (2)

[shell pilot]

But what about the ideal generated by all the elements in the Ring

As in the example I posted

Or obviously, tell me if I'm not reading it the right way!

Well, not "all" the elements in the ring, but multiple elements

[south patrol]

Smh

[coral spindle]

The element 1 is a generator of that ideal. What ideals contain 1?

Welp this is irrelevant anyway because that wasn't the question

Yeah it's literally explained right there idk what the question is

[shell pilot]

Like, if someone said, 1, 2, 3, 4, 5, 6, 7, 8, 9 all belong to a commutative ring with unity. Then the ideal <1, 2, 3, 4, 5, 6, 7, 8, 9> = what

[coral spindle]

described above

[shell pilot]

But would it be like {1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 4, 6, 8, 10, ...}

Or would it be a set of the squares of each element

[coral spindle]

It would consist of all sums of multiples of 1, 2, 3, 4, 5, 6, 7, 8, and 9

Like

I still don't know what the confusion is

[coral spindle]

The description is in the screenshot here

[shell pilot]

I'm just trying to turn it into an example

[coral spindle]

I don't know what that means

[shell pilot]

An example with a discrete finite set of integers

Not a definition with arbitrary elements

[coral spindle]

You wanna know what the ideal <1, 2, ..., 9> of Z is?

[coral spindle]

.

[shell pilot]

Precisely

Or even just a smaller generator like <1, 2, 3>

[coral spindle]

Have you thought about what I am pointing out?

[coral spindle]

I didn't say this for no reason

[delicate orchid]

you really are just plugging things into a definition here

[shell pilot]

Ok, so using 1 you get <1> = {1a_1 + 1a_2 + ... + 1a_n}

[delicate orchid]

you get <1> = {a1 : a in R}

[shell pilot]

Why the definition is confusing is because when you have <1, 2> is it {1a_1 + 2a_1 + 1a_2 + 2a_2 + ... + 1a_n + 2a_n}?

[delicate orchid]

why are you putting n things in there?

read the definition

n is the number of generators

[coral spindle]

read this definition again.

[coral spindle]

And you don't need the definition of <a, b, c...> for this question. Use the definition of an ideal.

What are the two properties that ideals have.

Three if you count non-emptiness, who cares

[shell pilot]

Ah, ok, so <1, 2> is just {1a_1 + 2a_2}? Not {1a_1 + 1a_2 + 2a_1 + 2a_2}

[delicate orchid]

non-emptiness is automatic based on the first property

[coral spindle]

Indeed.

[coral spindle]

||depends on if you're saying closure under addition or being a subgroup||

||but yeh||

[delicate orchid]

subgroup u lil nerd

[coral spindle]

um more elegant 🤓

My question wasn't rhetorical

[tardy hedge]

No fighting!!! NO fighting allowed !!!!

I can't have my seniors fighting like this 😭

[coral spindle]

Let $R$ be a ring and $I \unlhd R$ be an ideal. Suppose $1 \in I$. Describe the elements of $I$.

[cloud walrus]

Boytjie

[shell pilot]

Whoa whoa whoa

I already answered that

[coral spindle]

Wew and I are like brothers.... we hate each other and fight all the time

[coral spindle]

So what's your answer?

[tardy hedge]

Love that

[shell pilot]

I answered it wrong, but the right answer would be <1> = {1a | a is in R}?

[coral spindle]

OK that wasn't the question

(1) I wrote the ideal I, not <1>

Well yes that's it for now actually there's no (2).

What are the elements of I

[delicate orchid]

bro copied my homework 😭

[coral spindle]

Mhm

[shell pilot]

I = {1a | a is in I}?

[coral spindle]

You have just told me that I = {i | i in I}

Give me a description that isn't trivial.

Remember: what properties define ideals

[shell pilot]

Hmmm

[coral spindle]

You literally have two things you can do

[shell pilot]

A subring is an ideal if its left and right cosets are subsets of itself?

[coral spindle]

Subring?

[delicate orchid]

what the heeelllllllll

I mean true

[coral spindle]

Soap, do you know the definition of an ideal?

[shell pilot]

Better be true

[delicate orchid]

get this man a true

[shell pilot]

Gallian said it

[delicate orchid]

idk who that is I'm pretty sure that's an element

[coral spindle]

Like you have a question mark on the end

OK fine I will say that much of the mathematical world does not consider ideals to be subrings, but we'll move on from that because it's technical.

[delicate orchid]

you can weaken subring to additive subgroup, which is the normal definition

[coral spindle]

What I care about is the "absorption" property.

This is the strong property that ideals have.

Use it.

[coral spindle]

.

[tender wharf]

i think gallian's rings are not assumed to have 1

[coral spindle]

That is incredibly cringe

Gallian needs to be arrested

[tender wharf]

thats why he writes subring

[coral spindle]

I still think this is a poor way to define and introduce ideals, but hey ho.

It's redundant, makes it seem like you have to prove more than you actually do.

[delicate orchid]

they're the subsets of R that are equivalence classes under a relation invariant under multiplication and addition

[coral spindle]

incredibly based

But this is immaterial to the question at hand

[shell pilot]

Ok, let me think about your question again

Describe the elements of I?

[coral spindle]

Everything is in the question I wrote

[shell pilot]

Well, if it contains 1....then when the elements of the Ring are multiplied with it, you just get the elements of the Ring

So I = R?

[coral spindle]

Yes

[shell pilot]

The sum of the elements of R

[coral spindle]

So let us now go back to our question

[coral spindle]

bro

No sums

[shell pilot]

If it's in the example we were looking at, sorry

[coral spindle]

Nothing was said about sums

OK

[shell pilot]

Go ahead

Keep going

[coral spindle]

The ideal <1, 2, ..., 9> of Z

Do you agree that it contains 1, 2, ..., and 9?

This should be very clear from the definition

[shell pilot]

Absolutely

[coral spindle]

So 1 is in the ideal <1, 2, ..., 9>

So what is the ideal <1, 2, ..., 9>

[delicate orchid]

open problem

[coral spindle]

a real thinker

[delicate orchid]

equivalent to CFSG

[coral spindle]

Might be independent of ZFC

[coral spindle]

me 30 minutes ago was so naive, so young, so innocent...

[delicate orchid]

me 30 minutes ago was playing a lil boogie woogie rag time jazz

[coral spindle]

Yo that's cool

[delicate orchid]

now I've hopped on the 'craft (2011)

[shell pilot]

Sorry, power outage

So 1 is in the ideal generated by 1, 2, ..., 9. Is that what you are saying? Just making sure before I try to answer it using sums haha

<1, 2, 3, 4, 5, 6, 7, 8, 9> = {r_1(1)+r_2(1)+...+r_9(1)+r_1(2)+r_2(2)+...+r_9(2)+...} or is it {r_1(1)+r_2(2)+...+r_9(9)}?

[topaz solar]

Or nilpotent

[barren sierra]

1: what does your textbook say
2: Since r_1, ..., r_9 are arbitrary, you can just combine r_1(1) + r_2(1) + ... r_9(1) = (r_1 + r_2 + ... + r_9)(1) so the two sets you wrote are equivalent, can you see?

[topaz solar]

Why are you doing insane sums

Literally what are you doing

[shell pilot]

Here is the book definition again

[barren sierra]

(please go look at your textbook definition)

oh good

[topaz solar]

You need to actually look at it

[shell pilot]

Ok, so it says that a_1, ..., a_n belong to R

But those aren't necessarily the same elements as the r_i's in the ideal generated by the a_i's

[topaz solar]

The r_i is not in the ideal

It’s literally just sums of multiples of the a_i

[barren sierra]

For example lets consider a polynomial ring

R[x], polynomials with real coefficients

What is (x^2), the ideal generated by x^2?

just apply the definition

what sorts of polynomials live in that ring

[shell pilot]

Well

According to that definition, it would be r_1(x^2) + r_2(x^2) + ... right?

[south patrol]

I am stuck on smth I think is easy sad.

[barren sierra]

sure, but you can simplify this considerably

[topaz solar]

Stop doing multiple coefficients on each generator

[barren sierra]

also notice in the definition the number of terms in the sum is n = the number of generators

which hints how to simplify

[shell pilot]

The definition says that the r_i's are elements of the Ring

[barren sierra]

also + ... is not well defined, infinite sums are not well defined

[south patrol]

The issue being that like okay say A/I has (A/I)[p^n] = (A/I)[p^oo], then if p^k a is in I^m, it is in I lol so in fact p^n a is in I. But then I don't see how to do anything without multiplying by a again lol

Like ideally you'd want to show p^mn a is in I^m or smth

[barren sierra]

where does the p^oo come from?

[south patrol]

Well as in like

All p-power torsion = p^n torsion

By hypothesis

[barren sierra]

Oh I see I've never seen that notation before (although idk of any notation in the first place ig)

hmmmm

[south patrol]

(This is standard notation for p-power torsion)

Hmm

[coral spindle]

Come on man. We just covered an EXTREMELY relevant result regarding this

We proved something extremely relevant

[south patrol]

I mean one thing that should be possible is using the SES 0 -> I/I^n-> A/I^n -> A/I or smth

[coral spindle]

I didn't have you prove that thing for nothing!

[south patrol]

The hint I have is 'devissage' lol

[barren sierra]

am I dumb or can you not like just apply the hypothesis of A/I after you say "so in fact p^n a is in I"

never heard that term before

[south patrol]

Idk what you mean lol, that is what I did to say p^n a is in I

Ph

[barren sierra]

idk I'm just spitballing here idk the answer here either

[south patrol]

Maybe I should ask in advanced alg lol idk

[barren sierra]

maybe cause other help is going on in here

[shell pilot]

I'm just confused...

[nimble folio]

Does anyone know any good resources to learn about tensor products (of modules)

[coral spindle]

OK

[nimble folio]

Chapter 10.4 of dummit and foote covers them but its a bit difficult for me

[coral spindle]

I liked the coverage in Rotman's Introduction to Homological Algebra. I'm not sure if its easier than dummit and foote, you may just have to work with the definition and get used to it a bit

[south patrol]

I learnt via atiyah-macdonald's commutative algebra which was fine

[barren sierra]

Rotman is good yea

[nimble folio]

Dummit and foote doesn't really use definitions LOL

[coral spindle]

Tensor products are just kind of difficult no matter what

OK that's fair

[nimble folio]

they are being very verbose

[coral spindle]

I haven't read D&F

[barren sierra]

D&F is very verbose yea (I don't like it lol)

[coral spindle]

ig the benefit is that it has 'everything' in it?

Is that what people think?

[barren sierra]

it's a good reference for UG algebra after you already know it and need to look at it again for refresher

[nimble folio]

i've heard the opposite since a lot of stuff is worked through in the exercises

so far i've enjoyed it though chapter 10 has been great

[south patrol]

I havw never rly used D+F 😌

[shell pilot]

@coral spindle I found another definition

[barren sierra]

this is the same as your first definition

[shell pilot]

So I think why I'm confused is because which r_i goes with which s_i? Is there some sort of order imposed on these sets?

[barren sierra]

?

it's arbitrary

[fossil shore]

Luckily for you, most alg top books hand-wave the point-set stuff, so you basically only need the basic notions of topologies, connectedness and compactness. U might like Hatcher - whom I hate lol

[shell pilot]

e.g. If Z[x] is our ring and 4Z is our subset then <4z> = <..., -4, 0, 4, ...> = a sum of products of elements, but that sum could be different depending on which elements you multiply together

[south patrol]

I would more just say you don't really need much point set stuff regardless of what you're doing

(For most things in algebraic topology)

[barren sierra]

can you elaborate what you mean?

[shell pilot]

Like if you multiply 100 from 4Z with a polynomial vs multiplying 1000 with that same polynomial

[barren sierra]

there's no polynomials going on here

so idk what you mean?

nvm

[shell pilot]

Isn't Z[x] a ring of polynomials with integer coefficients?

[barren sierra]

I missed the [x] lol

ok yea so that's fine

[shell pilot]

Oh, gotcha

[barren sierra]

take 2x

[shell pilot]

Ok

[barren sierra]

100 and 1000 are both in 4Z

and 100*2x and 1000*2x are both in <4Z>

[shell pilot]

What about 2x^2? It's in there too, so does it also get multiplied by 100 and 1000?

[barren sierra]

Is 2x^2 of the form r * s for some r in Z[x] and s in <4Z>?

100*2x^2 and 1000*2x^2 are both of that form, so those elements are in <4Z> but what about just 2x^2?

[shell pilot]

Here is an answer that shows what I"m talking about

Why is r paired with x and s paired with y?

Won't that sum of products be different depending on what you choose for each variable?

[barren sierra]

can you answer my question first?

[shell pilot]

2x^2 is not in <4Z>

[barren sierra]

yea!

since it's not of that form

[barren sierra]

to answer this, ok lets make this more concrete

uhhhh lets say we're in Z

and lets say r = 2 and s = 5

so then (2, 5) = {2x + 5y | x, y in Z}

it's the set of all possible sums of this form

I could also write {2y + 5x | x, y in Z}

ordering is arbitrary, I just need to give those variables a name

[shell pilot]

Oh

[barren sierra]

does that clear things up?

[shell pilot]

set of all possible sums of that form

[barren sierra]

yea!

[shell pilot]

That is terrible notation

[barren sierra]

that's exactly what this definition says

how is it terrible notation 😭

it's just basic set notation

[shell pilot]

It should be a set of sets

Well

What I mean is

[south patrol]

Wdym

Why should it be a set of sets

I mean okay it isn't 100% formal but it is unambiguous what it means and something worth getting used to

[shell pilot]

It should be { {r_i_1a_1 + r_i_2_a_2 + ... + r_i_na_n} }

[south patrol]

Why?

[barren sierra]

remember these ideals are subsets of the original ring

[shell pilot]

Ok, I think what I mean is they shouldn't use numbered indices in that definition

[barren sierra]

so each element of the ideal should be in the ring

oh

I mean for arbitrarily many generators you need some way to label them

how else would you label them

[shell pilot]

I guess so, it's just that Gallian's definition makes it look like r_1 can only be multiplied by the same indexed a element

Almost like the sets are ordered and then each element has a specific product element in the other set

[glad osprey]

Could you say the ideal generated by S is kind of a linear combination of S with coefficients in R?

[topaz solar]

…it is

[barren sierra]

that's exactly what it is

[topaz solar]

It’s just sums of multiples. You get one coefficient for each of your generators

[shell pilot]

per set

There are sets within the set

well

[south patrol]

What

[topaz solar]

Tf you mean, no there aren’t

[shell pilot]

not sets within the set, but sums within the set

So it becomes a set with many different sums of different combinations of coefficients

All possible linear combinations

[topaz solar]

For every object in the ideal, there are coefficients that give it as a sum of generators

[south patrol]

It is the set of elements which can be written in the form r_1 a_1 + ... r_n a_n foor somoe n>=0 and r_1,...,r_n in R

The elements don't care what you call them when you're quantifying over all of them

[shell pilot]

Right, but just like in the example we looked at <r, s> = {rx + sy, ry + sx | x, y \in R}

So it's the set of all possible linear combinations

[south patrol]

Yes

[shell pilot]

x doesn't HAVE to go with r

It can go with s as well

[south patrol]

r is just a letter being quntified over

[shell pilot]

So the indices should not be assigned strictly

These definitions with numeric indices make it seem like r can only go with x and s can only go with y because the order is (r, s) (x, y)

[south patrol]

it dooesn't matter if u call it r or s

or r_1

or whatever

[shell pilot]

Definitely answered my question though

[delicate orchid]

I was literally gone for an hour and mfs still can't substitute things into a definition

[nimble folio]

Wow I really like Rotmans explanation of tensor products

[delicate orchid]

tensor products are when you compose bimodules

[nimble folio]

Havn't gotten there yet but I like how (specifically for tensor products) Rotmans book gives a definition and immediately follows up with concrete examples

[delicate orchid]

that's ok. You shouldn't get there

lemme take a look at this book

[rotund aurora]

wasn't Poincaré a hater of point set topology

[south patrol]

Lol idk

[rotund aurora]

https://mathshistory.st-andrews.ac.uk/Biographies/Poincare/quotations/

Point set topology is a disease from which the human race will soon recover.

[south patrol]

Based

[rotund aurora]

but idk more

[south patrol]

Prophecy of simplicial sets

😌

[delicate orchid]

I must be looking at the wrong book cause this ain't "immediate"

[topaz solar]

Real

[coral spindle]

I age greatly when I use this server

[delicate orchid]

fret not........ for the reaper will come for us all

[barren sierra]

Ok so the map F(u) is just u

I've got left exactness

So all that's left is to show if A/pA = 0 then F(v) is surjective

Let c in C[p], so pc = 0. Then c in C means there is some b in B, v(b) = c. I would like to show that pb = 0. We have that 0 = pc = pv(b) = v(pb) which means pb is in the image of of u. Thus there is some a in A, u(a) = pb

But A = pA so there is an a' such that a = pa'. This implies that u(a') = b

aaaaaand then I'm stuck

[rocky cloak]

This only implies that
p u(a') = pb

Which means that p(b - u(a')) = 0.

Now think about v(b - u(a'))

[barren sierra]

ah

yea you right

thanks

[rocky cloak]

Note you can also prove this with the snake lemma, as F is the kernel of multiplication by p and G is the cokernel

Or note that F = Hom(Z/p, -) and show that G = Ext(Z/p, -)

[barren sierra]

Hmmm I need to understand Ext more so I may try to work that out myself

[tough raven]

This, but also

[tough raven]

The only other possibility is m * n := -mn (which is isomorphic to Z with the standard multiplication by n |---> -n). For any multiplication, if 1 * 1 = x then m * n = xmn by repeated addition. This is valid for any x if you just want a rng, but if you want a multiplicative identity e, then e * n = xen = n => xe = 1 => x = +-1.

[quartz wind]

i see, thanks

i think my main question is that if someone talks about $\mathbb{Z}$, can i assume we have it as a ring with all its normal properties? "subgroup of Z" makes me think of it as only an additive group, with no preconceived notion of multiplication, is that wrong interpretation?

[cloud walrus]

esca (@ with reply)

[tough raven]

Yes. Absolutely. Without thinking twice or even once. No-one ever refers to ℤ as a ring and means another ring structure unless they very clearly say so first and if someone violates this, you can probably raise a mob to lynch them.

[quartz wind]

okay tysm

[tough raven]

And subgroup of ℤ probably means an additive subgroup, i.e. subgroup of the additive group (ℤ, +).

[hidden cairn]

I wonder if there’s a way to justify G = {1H, …., y^{p-1}H } without using a homomorphism

Can someone check my proof for the third problem?

[proud kraken]

Hi, is there a correct channel for me to discuss the Unified Field Theory of Everything? My model maintained dimensional consistency across quantum scale - celestial scale objects and I now need to model and predict model with my Unified Field Theory of Everything Theory. Thanks in advance

[mighty kiln]

Have you tried vixra

[proud kraken]

Hi, no I am not familiar. I am searching for others to work with, in all honesty I can code the system environment needed, but I’m sort of so excited about resolving the UFToE I’d like to find someone who can ensure me that I’m not going insane.

[coral spindle]

What children think mathematicians do: "I just proved the theorem that says that we're living in a simulation... checkmate elon musk"
What mathematicians actually do: "Can someone please help me work out the 69th cohomology of the piss ring, a ring that I defined for no reason and that nobody but me cares about"

[languid trellis]

Speak for yourself. I just proved that the universe has the geometry of Spec(pissring)

[delicate orchid]

what mathematicians actually do: "what if I took this construction...... and made it into a category...."

[dull ginkgo]

Dude

Shut up

Stop trying to advertise

You are going to get advanced access removed

[coral spindle]

Sweetie, he's making fun of you. It was a joke.

[faint forge]

Getting laughed out of the room is not even close to being persecuted for your beliefs; this is an incredibly offensive comparison

It's this kind of takes that get you laughed out of the room

[dull ginkgo]

<@&268886789983436800>

This is getting ridiculous

[mighty kiln]

The reason you're not being taken seriously is that every two weeks or so someone comes into this server and claims they have a proof of the Riemann hypothesis or some other long-standing math or physics problem, and then proceed to demonstrate zero actual knowledge of math or physics above high school

[dull ginkgo]

This is why we need applications for advanced math channels.

[deft spade]

Thanks lol

I was going to timeout but I looked at their short post history and figured it's not worth it

[tough onyx]

isn't the theory of everything about like.... physics?

[faint forge]

Such a promising career so prematurely cut short

[coral spindle]

Bold to assume it was meaningful at all

[tough onyx]

so it might make more sense in #geometry-and-manifolds or something

[delicate orchid]

so pissed I missed whatever messages just got deleted

mods show me the logs

[mighty kiln]

[faint forge]

Just another crank with a persecution complex.

[deft spade]

90 messages btw

[coral spindle]

They just started having a fit and compared themselves to someone who got burned alive

[delicate orchid]

lol

[coral spindle]

This is literally pissring cohomology

[faint forge]

I miss that high schooler who invented the best prime-generating function one day and solved RH the next, and compared himself to Galileo.

Similar vibes but more entertaining.

[next obsidian]

I miss Jesse. Rip. He didn’t die, he just never is around

[tough raven]

It's about everything, duh.

[rain grove]

Z_n has a subgroup of order k <=> k | n

Both => and <= follow from lagrange theorem right?

[coral spindle]

No, only => is from Lagrange

Lagrange does not guarantee the existence of any subgroups

[rain grove]

But couldn't I also say: if k does not divide n then there is no subgroup or order k because every order of subgroup (of a finite group) divides the order of group

[coral spindle]

Yes, that is called contraposition.

This has nothing to do with algebra. The proposition A => B is logically equivalent to the proposition not B => not A.

[rain grove]

Ohh wait so im proving => again

[coral spindle]

Yes.

You will need to use properties of Z_n specifically to prove this, because this is false for groups in general.

You cannot make some slick argument using Lagrange here.

[rain grove]

hm ok

[coral spindle]

Hint: n = kx for some x. Can you use x to produce a subgroup of Z_n?

[rain grove]

kx = n * 1 = 1 in Z_n, so x is an element of Z_n that has order k. And then <x> is a subgroup of order k

[coral spindle]

kx = n * 1 = 1
?

that has order k.
Remember that the order of an element is the least n for which it goes to the identity. You've not shown that your k is the least such number yet.

[glad osprey]

Remember the identity in Z_n is 0, not 1

[rain grove]

ahh yeah damn wait

[rain grove]

Can I just say: let k'<k, then k'x = 0 and kx = 0 and so k'x = kx => k' = k

[coral spindle]

No this is confusing Z and Z_n

k'x = kx => k' = k
This is false in Z_n

k'<k
This is meaningless in Z_n

k'x = 0 and kx = 0
This is true in Z_n, but if you want to use k' < k you need to work in Z.

Hint: use the theorem on the division algorithm

[rain grove]

Idk

How

[coral spindle]

Let's work in Z.

Suppose that there were some integer 0 <= k' < k such that k'.x = 0 in Z_n

Then k'x < kx = n

So the remainder of k'x upon division by n is k'x

(by the theorem on the division algorithm)

But k'x = 0 mod n, so the remainder is 0.

So k'x = 0 and k' = 0

Edited to add positivity assumption

[rain grove]

In this step you say n = a * k'x + k'x for some a \in Z?

[coral spindle]

No this is backwards

k'x = nq + r for a unique pair of integers (q, r) such that 0 <= r < n.

But k'x = n0 + k'x so r = k'x is the remainder.

[rain grove]

Ok but then we conclude k' = 0 and not k' = k

oh but 0 is not integer

[coral spindle]

??

[rain grove]

or natural number

[coral spindle]

0 is an integer

Arguably zero is a natural number

[rain grove]

OK but 0 can't be order of the group?

[coral spindle]

It cannot be the order of x because the order is always positive, yes.

So in fact there is no positive integer below k that kills off x in Z_n, and we are done.

[rain grove]

nicee

[hidden wind]

i’m a bit surprised… i can’t find any intuitionistically valid proof that over a field xy = 0 implies x = 0 or y = 0

at least after staring at it for 15 min

the proof by contradiction is clear though

[tough raven]

Intuitionistically, it probably depends on your definition of field.

E.g. if "for all x, x = 0 or there exists y, xy = 1" holds, then apply that to x. In the first case, you're done; in the second case, you can multiply by the inverse of x to conclude that y = 0.

[delicate orchid]

@coral spindle just had to use the Zassenhaus lemma dawg I'm cooked

[coral spindle]

DAS SCHMETTERLING?????

U are truly cooked. Why you using it?

[delicate orchid]

you don't even wanna know...

[coral spindle]

ok I trust u

[delicate orchid]

I'm not sure if it even works because it's fuckin Zassenhaus

I just have the Zassenhaus set up so like...

it's like when u have a quotient by an intersection u always raise a lil eyebrow ykwim

[coral spindle]

like mr dwayne 'the rock' johnson

[delicate orchid]

yeah it didn't really do much

[coral spindle]

many such cases

Isn't it basically just used for Jordan–Hoelder and nothing else

[delicate orchid]

we may never know

[south patrol]

This is what I thought lol

[hidden wind]

oh

[junior badge]

Let P = (x - 1) in R = K[x, y]/(xy) with K = the reals. I'd like to show that the localization of R at the multiplicatively closed set R\P is K[x]_{(x-1)}.

Intuitively, K[x, y]/(xy) consists of polynomials with no mixed terms, i.e. polynomials of the form f(x, y) = f_x(x) + f_y(y). Elements in the localization R_P look like f(x, y)/g, where g ∉ (x - 1), while elements in K[x]_{(x - 1)} look like f(x)/g, where g ∉ (x - 1). I don't quite see how to construct a homomorphism... Hints? (Also tried using the universal property -- but it was kinda messy, so I think constructing homos is the way to go.)

[chilly radish]

Or even the physics server

[rain grove]

Continuing from here, I need to prove that is the only group of order k

[coral spindle]

You should try this on your own first.

[rain grove]

let <x> <x'> both be subgroups of Z_n of order k

im just dancing around this formulas for some time now and idk what it means:

kx = 0 and kx' = 0.
there exists r and l natural numbers that for g in Z_n: rg = x and lg = x'
krg = 0 and klg = 0 so n divides both so then...

[hidden wind]

ok so i've probably done a silly here and need some help... the exercise is: from a field F, form E = F x F with addition given coordinatewise and with product (a,b)(c,d) = (ac - bd, ad + bc) and determine conditions under which E is a field

that (E,+) is an abelian group is obvious, and i checked that the product is commutative though i forgot to check inverses exist oops, but what i'm wondering about is distributivity

i expanded (a,b)((c,d) + (e,f)) = (a,b)(c+e,d+f) and compared it with (a,b)(c,d) + (a,b)(e,f) but if my calculations check out (which i would never take for granted) they're the same without any additional conditions

[coral spindle]

What's the product?

Oh wait that is the product

What's the sum?

[hidden wind]

coordinatewise

[coral spindle]

OK

[chilly radish]

This is just Cayley-Dickson right

[hidden wind]

is perhaps the inverses where the missing condition is?

[coral spindle]

That is certainly the most restrictive condition on a field

[hidden wind]

because i haven't found any ehehe, but i assume it may be may brain is just so used to working over a fields that i just used one without realizing

oh ok

[coral spindle]

I believe so

[hidden wind]

so this may actually be one of those rare cases in which my algebraic manipulations do check out (i just forgot the most important one) mwahaha

also i've waited way too long before having like a proper look at these things

i've just taken my fields for granted for years

[chilly radish]

I think you can just define the norm and use that to form inverses

Oh actually

This only works if you have a notion of conjugation on F (which might be trivial). I think if F=C then E would have elements with 0 norm

Not sure if that means they're non invertible but i'd imagine so

[hidden wind]

ooh say no more sounds fun lemme figure the rest out myself

[chilly radish]

Sure thing

Spoiler ||I think (i,1), (i,-1) multiplies to 0||

[coral spindle]

||If the field F is R then you're looking at C, right? 1 = (1, 0) and i = (0, 1).||

||So right, you need some condition. I will say no more.||

[hidden wind]

yeah now i only have to see whether or not i have my basic linear algebra in order, i have two equations and two unknowns... Mnemosyne be with me

[glad osprey]

Would you ever use a CAS to carry out some of those calculations? I would hate doing those by hand

[hidden wind]

ok now i have explicitly found that every nonzero (a,b) has an inverse and i... still don't see the condition, other than like a^2 + b^2 ≠ 0

[hidden wind]

i quite enjoy doing these things by hand :S

[hidden wind]

the inverse is (a, -b)/(a^2 + b^2), but i'm not sure where the condition on F comes in

[hidden wind]

oh

[glad osprey]

I get so nervous I'm making a mistake, and the nervousness increases as the calculations get longer, so I love having the computer just tell me the answer

[hidden wind]

(i,1) very suspiciously does not satisfy i^2 + 1^2 ≠ 0

sorry for the double negation

[hidden wind]

well i find it very instructive to keep track of the symmetries in your expressions as you go... then it's somehow intuitively clear when something is off

[hidden wind]

conclusion: silly brain currently too reals-minded and forgetting a^2+b^2 may be 0 even if (a,b) ≠ 0 in arbitrary fields

thanks for the help Boytjie and Shin

[glad osprey]

yeah, that's true I guess, maybe I should get better at doing things by hand

did you check associativity of the product btw?

[coral spindle]

You get associativity for free if ||you see this as F[x]/(x^2+1)|| so we can alleviate jens' suffering a bit

But you are right in principle

[glad osprey]

phew 🙏 associativity would've been the most tedious I think

[coral spindle]

For sure

oopsie this actually just solves the question

[delicate orchid]

oopsie woopsie! ;3

[coral spindle]

I didn't realise lol

Algebra is so wonderful

[delicate orchid]

no it isn't

[coral spindle]

shut up i hate you

[hidden wind]

lemme see if i can type out a verification of assoc with my left hand as i eat walking omw home…

[coral spindle]

n o

[delicate orchid]

a brave move

a foolish move

[hidden wind]

foolish is my middle name

mwahaha

[delicate orchid]

this skulduggery cannot persist.... verify the associativity or PERISH

[hidden wind]

actually nvm i’ll just take in the scenery as i walk

[hidden wind]

i suppose one could also represent E* as a matrix group, from which assoc would follow immediately

if it can be represented as such, which i admit i do not know

[chilly radish]

I meant in the case F=C but F isn't considered with its involution

So E=C^2

But yea if you start with R you just get C

[chilly radish]

Anime protagonist activities

[glad osprey]

I think we can represent E as ||[a -b; b a]|| because of how ||E = C when F = R||

[hidden wind]

yeah i think that checks out but i can’t be bothered to actually check rn

[shell pilot]

When you have a factor group G/H, then do you consider H to be the "zero element" of the factor group G/H?

I'm learning about factor rings and this proof says to suppose R/A is an integral domain and ab \in A. Then (a + A)(b + A) = ab + A = A, the zero element of the ring R/A.

So when you have a factor ring R/A, then do you consider A to be the "zero element" just like in a factor group?

[topaz solar]

Yes

Intuitively, it’s like “mod A”

[topaz solar]

Z mod n is just Z/(n) = Z/nZ

You probably should have seen factor groups before this though

[shell pilot]

Oh, yes, definitely. I'm just trying to compare factor rings to factor groups

[topaz solar]

Well, as far as addition goes, it’s exactly a group quotient

Because an ideal is an additive subgroup

[rotund aurora]

If L, M are Galois extensions of K and L cap M=K then G(LM/K)=G(L/K) x G(M/K) right? (even when the extensions are infinite)

[shell pilot]

You say that because a ring is a group under addition?

[topaz solar]

Yes

[shell pilot]

Thank you!

[topaz solar]

It should certainly contain it, right? So it would be if we can’t have any extra weird automorphism yeah?

I think

Infinite extensions are a bit spooky though, so, take my “I believe yes” with a grain of salt

But if I’m correctly thinking about orbits & types, then uh

[rotund aurora]

well I think the proof is the same as in the finite case, you don't need to use finiteness or anything

send sigma in G(LM/K) to (sigma|_L, sigma|_M). This is obviously injective, so it is enough to show it is surjective. But given sigma in G(L/K), since every x in LM can be written uniquely as x=lm with l in L, m in M, the map lm-->sigma(l)m is well defined and is an automorphism

[shell pilot]

Working on this first part of the proof that R/A is an integral domain if and only if A is prime.

Suppose that ab \in A. We know that ab + A = A by closure under addition. ab + A = (a + A)(b + A). So why does either (a + A) or (b + A) have to be equal to A?

Is it because A is the zero element of the factor ring which means that since it's an integral domain, either a + A = 0 or a + A = A (or b + A = 0 or b + A = A)?

[tender wharf]

yes

that is correct

[shell pilot]

Well, I need to fix the "= 0" part because A is the "zero element"

[tender wharf]

=0 is fine

[shell pilot]

Ok, gotcha

[tender wharf]

call it abusive notation

[shell pilot]

haha

Thank you!

[tender wharf]

I think you don't even need the quotation marks here

the element A is precisely the additive identity of the quotient ring R/A

No worries

[topaz solar]

Should be fine yeah?

[shell pilot]

In the second half of the proof, why are they saying (a + A)(b + A) = 0 + A? Wouldn't that mean that ab = 0?

[mighty kiln]

They're proving by contradiction that R/A is an ID
So assume two nonzero elements of R/A multiply to 0, and show a contradiction

[rocky cloak]

ab is in A, so ab+A=A which = 0 in R/A

[shell pilot]

And ab \in A since 0 is the additive identity?

[rocky cloak]

(a+A)(b+A) = A is exactly equivalent to ab in A.
And A is the 0 in R/A. So this (a+A)(b+A) = A would be two elements multiplying to 0

[shell pilot]

Ok, I was just thinking about the definition of (a+A)(b+A) = ab + A so if ab + A = 0 + A then ab = 0

[rocky cloak]

No, ab need not equal 0.

You only have ab in A

[shell pilot]

Oh ok, I'm just wondering then what the 0 + A means and why they didn't write just A

[rocky cloak]

Just to emphasize that this is the zero element in R/A

[toxic sapphire]

I'm not sure if I'm just missing an important part of the text but I'm not sure what this means:
"write down the elements of the group generated by (123456)..."

(123456) here is a permutation in cycle notation

I'm not sure what "the group generated by (123456)" would be

I don't see how that function generates a group (or what exactly "generates" means here), but I doubt the author would use the word group in a loose way (since it's a book on groups)

[knotty badger]

So are we viewing the cycle as an element of S_6?

[toxic sapphire]

yes

oh, is the group S_6 itself then?

[knotty badger]

Nope!

So, the subgroup generated by the cycle would be the smallest subgroup of S_6 which contains the cycle

[toxic sapphire]

I'm not sure what a subgroup is yet although that is the next section of the book

[knotty badger]

Ah, I see

[toxic sapphire]

is there an informal notion of it that I could reasonably use for now, or maybe it was a misplaced problem?

[knotty badger]

I think…

Suppose you wanted to include that cycle in a group

What would be a good group operation to try?

[toxic sapphire]

I mean, a subgroup seems pretty intuitive, and I can work with this probably, but I figure I'm not intended to be using that ideology quite yet

ok so if this cycle were an element in a group...
since it's a function, function composition seems like a reasonable group operation

[knotty badger]

Yep!

And - what other group elements can you deduce exist?

[toxic sapphire]

I'd say the other permutations of length 6 if you didn't already tell me it wasn't S_6. maybe cycles that can be combined to produce (123456)?

[knotty badger]

Well

What are the axioms of a group?

[toxic sapphire]

that's a good place to start yeah. loosely:
closure - if a and b are in S then ab is in S
associativity
identity
inverses

[knotty badger]

Mhm - and, at the moment, those aren’t satisfied with (123456) alone, right?

What I would suggest is - using these axioms alone, see what other group elements you can deduce must exist

[toxic sapphire]

I think that's a great idea. thanks! I'll come back if I'm at a loss again

hmm. (for functions a and b) should the existence of ab (a compose b) in the group imply the existence of a and b?

hm nevermind this was a premature thought. I'll stew on this some more

[lime badge]

let k = k' k", where k' is odd and k" is a power of two.

Why then does CRT imply F_2[x-1]/(x^k - 1) = F_2[x]/(x^k'-1) (+) F_2[x-1]/((x-1)^k")?

[delicate bloom]

what have you tried so far

[lime badge]

Well F_2[x-1] = F_2[x], so I can rewrite this as

F_2[x]/(x^k - 1) = F_2[x]/(x^k'-1) (+) F_2[x]/(x^k")

So I guess it's just a matter of showing x^k'-1 and x^k" are relatively prime and that they multiply to x^k -1, unless I screwed up everything above

I feel like relatively prime should be obvious but there's probably some result about how the degrees of the polynomials are related in this situation that I'm forgetting. But these polys don't multiply to the original poly

[rocky cloak]

You mean rewrite as
F2[x]/((x+1)^k - 1) = ...

So you need to fiddle with when binomial coefficient are even or odd yeah

[lime badge]

Oh whoops, yes

[cloud lynx]

Give an example of a non-abelian group of order 120 and an element in it that has
order 6. Has anyone an idea? or how to approach this exercise?

[tribal moss]

The first thing to do is to notice that 120 is a factorial...

[cloud lynx]

S5

lol

oh bruh

thank you

[tribal moss]

Then you just need an element of order 6.

[cloud lynx]

thats more tricky

[tribal moss]

Do you know some ways to determine the order of a permutation?

[cloud lynx]

no

i dont

[tribal moss]

Does "cycles" ring a bell?

[cloud lynx]

yes

by just looking how many permutations we need in order to get the identity?

for example (a b) has order 2 because we need to change a b then change it again

so 2

[tribal moss]

Right. And how about, say, (1 2)(3 4)?

[cloud lynx]

https://tenor.com/view/ponke-ponkesol-monke-monkey-cry-gif-11763130847124576621

i think i have to refresh permutations

maybe 4?

ord((1 2)) *ord((3 4)) = 2 * 2 = 4

would be my first thought

[tribal moss]

A reasonable first guess -- but try to write out the powers of (1 2)(3 4) explicitly.

[cloud lynx]

what do u mean by powers of ?

[tribal moss]

I mean if you have a group element x, then by definition its order is how far in the list x, x², x³, .... you need to go before you meet e. The entries in that list are "powers of" x.

[cloud lynx]

or by just determine the lcd(ord((1 2)), ord((3 4)))= 2*2/2=2?

[tribal moss]

lcm, but yes!

So, as we're looking for an element of S5 with order 6, we need to divide 5 elements into some cycles, such that the lcm of their lengths is 6.

[cloud lynx]

so maybe it is useful to have one with order 3 and one with order 2? because both are coprime soo 3*2/1=6

but how does a element with order 3 looks like lol

(a b c)?

[tribal moss]

Yes.

[cloud lynx]

soo (a b c d) has order 4 and so forth?

[tender wharf]

yes

[cloud lynx]

soo ( a b c)( d e)

[tender wharf]

an n cycle has order n which makes sense

[cloud lynx]

ahh okk thank youu guys

sorry my bad english btw

[tender wharf]

your english is fine

[cloud lynx]

https://tenor.com/view/sad-gif-13261737358986789359

i have another question

soo

if we have a non cyclic group

and arbitrary many cyclic groups

[tender wharf]

sure

[cloud lynx]

and we do the cartesian product of them then the product is not cyclic right? because one of the entry is not cyclic

[delicate orchid]

yes this is true

but note that it's very rare for the product of only cyclic groups to be cyclic

[cloud lynx]

yes for example Z2xZ2

order of (0,0) is 0

[coral spindle]

No that's incorrect

The order of (0,0) is 1

[cloud lynx]

and order of (1,0),(1,1),(1,0) is 2

yes 1

[coral spindle]

This is very important

[cloud lynx]

sry

ok thank youu guys

[hidden wind]

x^0 is the identity e for any element x in any group… it is an instance of the more general identity that x^(k ord x) = e for all integers k

[lime badge]

So there is Lucas's theorem you can apply to determine the parity of the binomial coefficients in my case. Suppose k = 2^i(2j+1). Then all the even coeffs vanish and the smallest odd (k choose m) is when m = 2^i

Though there are also higher m's with the property that (k choose m) is odd I believe

[cloud lynx]

so we know that ZmxZn is cyclic only if m and n are coprime. Does is only work for the product of two finite cyclic groups? or does it also work for arbitrary many finite cyclic groups with the condition of coprime orders

[coral spindle]

It is more complicated for the general case, but it does not matter, because you can reduce to the case of two.

Well actually now that I think of it properly, what I say is misleading.

It simply reduces to the case of a product of two and generalises how you'd expect.

[cloud lynx]

sooo it works for arbitrary many?

[coral spindle]

Try proving it

[cloud lynx]

https://tenor.com/view/crying-crying-meme-gif-26327776

i gonne try wait

it works

[delicate orchid]

this really is a rn_image_picker_lib_temp_9575f55f-df3e-4c6f-affe-8 moment

[coral spindle]

I am not downloading that

[cloud lynx]

broo whyyy

thats not a virus or anything:((

[finite turtle]

Given how this proof of "showing every finite group G with a normal subgroup H has a composition series containing N" works:

It seems as long as you exhibit a non-trivial, proper normal subgroup H of finite group G, you can take any chain of normal subgroups with a property on its factors and prove it on such a group G by induction by piecing togther the chain and using the induction hypothesis on the small subgroups H, and G/H

i feel like my thinking must be wrong, could anyone correct me in pointing out on how the above mentioned proof must be limited to composition series, rather than any chain of normal subgroups? Or maybe the proof (from stack exchange) is not correct?

[barren sierra]

You need simplicity in the factors for it to be a composition series

Otherwise if you just want to show every group has a chain of normal subgroups there's a pretty trivial one: {e} ≤ G

or in this case a chain of normal subgroups containing H: {e} ≤ H ≤ G

but this has 0 guarantees of simplicity in the factors

[finite turtle]

yes i know this, but im saying that for example if you want to show every finite group has a series/chain where very quotient is abelian, you could use the same logic as long as you come up with a non-trivial normal subgroup of G

[coral spindle]

No you couldn't

Because your base case would fail

[delicate orchid]

I'm confused can't you just take 1 < G/[G,G] < G

or are we doing the "factors have to be simple" woke mind virus

[barren sierra]

why are we talking about abelian now

[finite turtle]

could you not use strong induction?

[coral spindle]

What's the base case?

I am aware we are using strong induction

[barren sierra]

How would weak induction even work lol

"hm yes the previous group"

[coral spindle]

Let me elaborate

[delicate orchid]

u can order the finite groups

[finite turtle]

haha how silly of me forgot you still need base case

[coral spindle]

Every group has a composition series because of the base case: if a group has no normal subgroups, it is by definition simple

[barren sierra]

fake news woke math

[coral spindle]

But you cannot do the same with Abelian groups because there is not the same thing

[delicate orchid]

I have a feeling that every group may have an abelian subgroup

just a hunch

[finite turtle]

well i'm assuming you can exhibit a non-tribial normal subgroup, but yes you still need to prove a base case if you impose pproperty on the quotient, correct?

[coral spindle]

Thank u for the trivial group xoxoxox

[chilly radish]

All horses are the same colour

[viral halo]

Is the following way for showing Q(sqrt(2),sqrt(3))=Q(sqrt(2)+sqrt(3)) valid? (From a book). We need to show that Q(sqrt(2)+sqrt(3)) is a subset of Q(sqrt(2),sqrt(3)) and that Q(sqrt(2),sqrt(3)) is a subset of Q(sqrt(2)+sqrt(3)). The first inclusion is obvious, for the second one let A=sqrt(2)+sqrt(3) equiv to A-sqrt(2)=sqrt(3), implies A^2-2Asqrt(2)+2=3, equiv to sqrt(2)=(A^2-1)/(2A). This implies that sqrt(2) is in Q(A), since sqrt(3)=A-sqrt(2) also sqrt(3) is in Q(A)=Q(sqrt(2)+sqrt(3)). Together this implies that Q(sqrt(2),sqrt(3)) is subset of Q(sqrt(2)+sqrt(3)). Combined with the inclusion Q(sqrt(2)+sqrt(3)) is a subset of Q(sqrt(2),sqrt(3)) it follows that these are equal. This method of introducing an "A" and defining it as A=sqrt(2)+sqrt(3), I'm not sure why it works, don't we have to show that sqrt(2) and sqrt(3) can be written as a linear combination of positive powers of (sqrt(2)+sqrt(3))? (am confused)

[lone niche]

We introduce A to more compactly write sqrt(2)+sqrt(3) AND to show in a clear way that sqrt(2) and sqrt(3) can be expressed as rational functions in A (=sqrt(2)+sqrt(3)). Remember that Q(A) can be thought of as the set of rational functions in A with coefficients in Q.

[viral halo]

Would it make sense to say that Q(sqrt(2)+sqrt(3)) is isomorphic to Q[x]/(m(x)) where m(x) is the minimal polynomial for sqrt(2)+sqrt(3), since m(x) is by definition irreducible it follows that Q[x]/(m(x)) is a field. So divison by powers of A=sqrt(2)+sqrt(3) is defined makes sense (I assume this is what you mean by rational functions in A)?

[lone niche]

Unfortunately I have to go. But the way to see it is that Q(A) is the field of fractions of Q[A], and Q[A] is polynomials in A.

[viral halo]

I have to read more about field of fractions then, not comfortable with them, ty anyways

[delicate orchid]

what you did works, uginp

[shell pilot]

Only trying to prove that A is an ideal of R[x]. Here is my proof so far.

Let $a \in A$ and let $b \in A$. Then using the ideal test, $a - b = (a_nx^n+...+a_1x)-(b_nx^n+...+b_1x)$. So simplifying this gives $(a_n-b_n)x^n+...+(a_1-b_1)x$ which is an element of A. Does that seem correct? Other answers use other ways of proving it, but this was my own version

[cloud walrus]

Soap_Opera

[shell pilot]

This is assuming that a_i != b_i...I guess...so maybe that proof doesn't work

[tender wharf]

if its equal its still fine

(why?)

[shell pilot]

Oh, wait, yeah I was thinking zero/nonzero constant term

Ok, so I guess I'm wondering if that proof works? I just need to also check that ra and ar are both in A when r \in R and a \in A

[tender wharf]

the proof shows that A is a subring (gallian subring)

i will mention that you havent said A is nonempty yet

[shell pilot]

What is a good way to say that A is nonempty besides saying A is nonempty haha. Is it ok to provide an example like x \in A so A is nonempty? (I'm also working on my proof writing skills)

[tender wharf]

produce an element in A

there is an obvious choice

[shell pilot]

0?

0 * r for any r in R is 0 which has a 0 constant term so 0 \in A

So A is nonempty

[tender wharf]

yes

[tender wharf]

this was not needed btw