groups-rings-fields

[viscid pewter]

then span(A) is a subset of C

[thorn delta]

you should see that
S ∪ T ⊂ span(span(S) ∪ span(T))
and then apply

if S \subset V for a vector space V, then span(S) \subset V

don't overthink the chain thing. Its just to make truth of S ∪ T ⊂ span(span(S) ∪ span(T)) more clear

[odd dirge]

ye, i already understood that, but how come
span(S ∪ T) ⊂ span(span(S) ∪ span(T)) is deduced from that,
I just don't get how span(S ∪ T) is connected to this
sorry if this should be pretty obvious

this i understood

[viscid pewter]

if C is closed

[thorn delta]

np, let V = span(span(S) ∪ span(T)) and X = S ∪ T.
We have X \subset V so span(X) \subset V

[viscid pewter]

ie. you can't combine anything in C to get anything not in C

then for any subset A of C

span(A) is in C

because you can't get out of C

[odd dirge]

alright, until now i got it

[thorn delta]

wait does that mean you understand now, or you don't understand anymore?

[odd dirge]

I do understand

[thorn delta]

alright cool, so that's one direction of the equality you want to prove

[odd dirge]

I meant that I understood this part

this span(S ∪ T) ⊂ span(span(S) ∪ span(T)) is still a mystery to me

[thorn delta]

span(S ∪ T) = span(X) ⊂ V = span(span(S) ∪ span(T))

[odd dirge]

hold on V = span(span(S) ∪ span(T)), this I didn't know

[viscid pewter]

S ∪ T ⊂ span(S) ∪ span(T)
and span(S) ∪ span(T) ⊂ span(span(S) ∪ span(T))

ie. A ⊂ B and B ⊂ C
so A ⊂ C

C is a span of something, so it is closed, so span(A) ⊂ C

ie. span(S ∪ T) ⊂ span(span(S) ∪ span(T)) which is what we want

[thorn delta]

that's what I wrote here. Indeed, span(....) is always a vector space

[odd dirge]

but is this always the case if S and T are subsets of V?

[thorn delta]

yeah. S ∪ T is just some bigger set, still contained V, so the span of it should be contained in V

[odd dirge]

but why is the span = V?

[thorn delta]

well so we have span(S ∪ T) ⊂ span(span(S) ∪ span(T)), and now we have to show span(span(S) ∪ span(T)) ⊂ span(S ∪ T)

[thorn delta]

i.e. its a separate argument

[odd dirge]

what do you mean by that?

[viscid pewter]

the span isn't V

there's no reason it should be

wait

yeah so

span(span(S) ∪ span(T)) is a different vector space, a subspace of V

[thorn delta]

span(S ∪ T) = span(span(S) ∪ span(T)). We just showed span(S ∪ T) ⊂ span(span(S) ∪ span(T)). You asked why span = V, i.e. why span(S ∪ T) = span(span(S) ∪ span(T)). The answer is that we have to prove that by showing span(span(S) ∪ span(T)) ⊂ span(S ∪ T)

^ RE: "what do you mean by that?"

[thorn delta]

I let V = span(span(S) ∪ span(T)) to try to make something more clear earlier oof the confusion

[viscid pewter]

yep

[odd dirge]

okay, I think I slowly start to understand what you are trying todo

[thorn delta]

in any case, S ∪ T ⊂ span(span(S) ∪ span(T)) implies that span(S ∪ T) ⊂ span(span(S) ∪ span(T)), which is one direction of the equality we're trying to prove. Now, we have to prove the other direction: that span(span(S) ∪ span(T)) ⊂ span(S ∪ T)

so, any idea on how to get started on showing span(span(S) ∪ span(T)) ⊂ span(S ∪ T)?
You shouldn't need any new ideas here

so you're saying that in your class, you proved that a finite sum of noetherian submodules is noetherian?

if that's the case, then yea you should be able to apply that fact to show that finite direct sums of notherian R-modules are noetherian
This is because each summand of a direct sum of modules is isomorphic to a submodule of the direct sum, and the sum of all of those submodules is the whole direct sum

[odd dirge]

I don't think that's the correct way,
but as span(M)= span(span(M)), so
span(span(span(S) ∪ span(T))) = span(S) ∪ span(T)
is that until now fine?

[viscid pewter]

no

span(span(span(S) ∪ span(T))) = span(span(S) ∪ span(T))

but you can't go any further

[odd dirge]

oh wait you're right

so we've still got V = span(span(S) ∪ span(T)) ?
as this was the assumption

[thorn delta]

I just did the V = span(span(S) ∪ span(T)) thing so it would be more clear how the lemma

if S is a set, and V is a vector space, and S \subset V, then span(S) \subset V
applies in showing that S ∪ T ⊂ span(span(S) ∪ span(T)) implies that span(S ∪ T) ⊂ span(span(S) ∪ span(T))

[wind locust]

much appreciated, thank you!

[thorn delta]

npnp

[odd dirge]

i am sorry even though you guided me this far, but I've still got no clue

[thorn delta]

no problem. Are you stuck on span(S ∪ T) ⊂ span(span(S) ∪ span(T)) or span(span(S) ∪ span(T)) ⊂ span(S ∪ T) now?

[odd dirge]

span(span(S) ∪ span(T)) ⊂ span(S ∪ T)

[thorn delta]

alright yea, this one is a little tougher tbf

so, span(S ∪ T) is a vector space, right? So if we can show
span(S) ∪ span(T) ⊂ span(S ∪ T), then by our lemma,
span(span(S) ∪ span(T)) ⊂ span(S ∪ T)
which is what we want to prove

therefore, we need to show span(S) ∪ span(T) ⊂ span(S ∪ T)

[odd dirge]

span(S) ⊂ span(S ∪ T) and span(T) ⊂ span(S ∪ T) so span(S) ∪ span(T) ⊂ span(S ∪ T)?

[thorn delta]

yep

[odd dirge]

and we're done just with that?

[thorn delta]

yep

[odd dirge]

alright, I am gonna go over it again after a good sleep!
thanks for keeping me company until now! and kaisheng aswell

[thorn delta]

npnp

[chilly ocean]

Is there anyone available to tutor Galois theory with me for the next 5-6 weeks? I’m really committed to learning with someone, even if they are wanting just to refresh themselves on group and field theory. Please message me if you are interested 🙂

[lethal cipher]

If I knew a single thing about Galois theory, then I'd definitely be interested. But that is just not the case

[median pawn]

Is there a good way to check if R is a field for different choices of F?

or is this just another painful problem?

[cloud walrus]

Hausdorff

[median pawn]

So if the matrix on the left has an inverse, ac - bd = 1 and ad = -bc

R is already a commutative ring with identity, so I think we don't have to check any other conditions for R - {0} to be an abelian group

[rustic crown]

there is uwu

if you think in terms of matrcies, you want the determinant to be non-zero whenever at least one of a and b is non-zero

det = a^2 + b^2

[median pawn]

yeah true

so we want a^2 + b^2 to be non-zero if and only if both a and b are not zero

[rustic crown]

yep

without loss of generality, you can assume that b is not 0

so we get (a/b)^2 = -1

does this equation have any solutions?

[median pawn]

this makes sense in Q,R,C but in F_p can you divide like that?

[rustic crown]

yep, that's what a field is... i can divide by non-zero things

equivalently i'm multiplying by the multiplicative inverse

which exists because all these are fields

[median pawn]

yes makes sense

[rustic crown]

if x^2 = -1 has no solutions, the condition will hold and we'll get a field

conversely, if it has a solution say a, then the matrix
[a -1]
[1 a]
is non-zero and not invertible

there is an equivalent way of doing it btw

which is more ring theoretic

[cloud walrus]

Hausdorff

[rustic crown]

yep!

[median pawn]

for F_p is it more like writing down the entire multiplication table

p = 5,7

[rustic crown]

nah you don't have to write it out entirely

just plug in x = 0, 1, ... p-1 and see if there is a root or not

you can save a lot of trouble with some simple group theory

it turns out x^2 = -1 has a solution in F_p for odd prime p if and only if p = 1 mod 4

[median pawn]

x^2 = -1 in Z_p means x^2 = p-1 right?

[rustic crown]

yep

[median pawn]

very cool, thanks a lot!!

[rustic crown]

the other way of doing this is this

we have a map F --> R

which sends a to the corresponding scalar matrix a * I

we can extend this to a map F[x] --> R sending x to the matrix
[0 -1]
[1 0]

this matrix squared is -1 * I

so the kernel of that map is x^2+1

this gives us an isomorphism

R = F[x]/(x^2 + 1)

so R is field if and only if x^2+1 is irreducible

[median pawn]

you're too good, @rustic crown!

[rustic crown]

extending follows from the universal property of polynomial rings

you just need to specify where x goes

rest everything is forced upon you

[median pawn]

yep i figured

[rustic crown]

so R is like a way of adjoining sqrt(-1) to your field

if it already contains a sqrt(-1) then you'll get something weird

[median pawn]

yes like in C

[rustic crown]

C[x]/(x^2+1) = C[x]/(x+i)(x-i) = C * C

same thing for F_5

[median pawn]

just confirming. for F_5, R is not a field, but for F_7, it is

[rustic crown]

yep!

[median pawn]

Just another question;

I only want a hint for this one

[rustic crown]

what if it isn't a prime number?

[chilly ocean]

what if

the world would end

[rustic crown]

[median pawn]

nice, i got a contradiction to the minimality of n!

[rustic crown]

yay!!

[waxen hedge]

A way to do this question is to look at the (unique) morphism of rings from $\mathbb{Z}$ to $D$, which send 1 to $1_D$ (and so $k\in\mathbb{Z}$ to $ k\cdot 1_D$)
It's kernel is a prime ideal of $\mathbb{Z}$ since $D$ is an integral domain, so it's either $0\mathbb{Z}$ or $p\mathbb{Z}$ with a prime $p$

[cloud walrus]

Adrien

[waxen hedge]

This explains why we say "characteristic 0" instead of "characteristic ∞"

[median pawn]

Isn't this too direct? It's the last exercise in a section so I thought it might be somewhat challenging

[cloud walrus]

Hausdorff

[hidden haven]

yes.

[median pawn]

and since F_j is a field, we're done lol

[hidden haven]

😌

[median pawn]

I think this will hold for other stuff also

Rings in general? Integral domains?

[hidden haven]

Yes, most algebraic structures

For model theoretic reasons you could say

[median pawn]

nice! it really wasn't worth being the last exercise was it

[hidden haven]

This is a special case of a construction called a direct limit

[median pawn]

oh cool

[serene radish]

Hi, I'm trying to prove that a permutation in S_n is a power of a cycle iff it has a cyclic decomposition with cycles of the same length. What I've done so far, I just considered the cases when we have a m-cycle, and gcd(m, k) = 1 or gcd(m, k) is not 1, for k being some power of that m-cycle. In the case of gcd(m, k) = 1, it's clear that there can be only one cycle with order m, but I have trouble when gcd(m, k) != 1. Then, I've supposed having less than gcd(m, k) cycles in decomposition, but that's not possible, since we'd have a cycle with order greater than m/gcd(m, k). But the issue is, what happens the when number of cycles is greater than gcd(m, k), then all cycles have order less or equal than m/gcd(m, k), since otherwise we'd get the previous case?

[hidden haven]

This generalizes to uncountable chains + directed sets (special kinds of preorders rather than totally ordered)

[median pawn]

very cool, thanks!

p.s. i saw directed sets in analysis, for nets, lol

[hidden haven]

oh nice lol

[hidden haven]

I think you can explicitly construct the cycle that gives this permutation when raised to some power

See what happens when you exponentiate a cycle, then work backwards

This construction won't be unique

I don't think you have to worry too much about gcd's anywhere

[serene radish]

I realize what I get when I raise some cycle to some power, but I'm not sure how to use that

[hidden haven]

Let the cycle decomposition of the permutation be into c_i, i=1,...,m. Let each cycle have length k, with jth entry of c_i being c_ij (this is after you fix a way of writing the cycle)

Then consider the cycle

c_11 c_21 ... c_m1 c_12 ... c_m2 ... c_1k ... c_mk

You should just be able to read off the mth power of this

Just noticed that I might be using different variable names than you were, so be a bit careful with that

[serene radish]

Ah, I was proving the other implication

But it seems that other implication would be useful here

[hidden haven]

Wait you want to prove that mth power of a cycle decomposes into cycles of the same length?

[serene radish]

Yeah, and also if a permutation has cycles of the same length then it is a power of a cycle

[hidden haven]

Yeah I've given a proof of this

[median pawn]

I think it's extremely cool how x² + 1 = 0 has infinite solutions in quaternions (over ℝ) but only two solutions in ℂ.

[lethal dune]

what the sols?

[median pawn]

x = ai + bj + ck where a² + b² + c² = 1

[lethal dune]

"a finite group cannot be written as a union to 2 proper subgroups". how do I approach it?

[woeful flint]

I've seen the notation $\mathcal{O}_K$ in my commutative algebra class where $K$ is a finite field extension of $\mathbb{Q}$. It seems to be related to the ring of integers but my lecturer makes no attempt to explain it. I've looked online and can't fine anything either, anyone have any explanation?

[cloud walrus]

Kraft Macaroni

[waxen hedge]

This is usually the notation for the ring of integers !

[woeful flint]

I guess the thing that confuses me is just the fact that $\mathbb{Z}$ is usually used

[cloud walrus]

Kraft Macaroni

[woeful flint]

I guess what I'm trying to get at is how does $\mathcal{O}_K$ differ from $\mathbb{Z}$

[waxen hedge]

Ok I inderstood
It's not the classic ring of integers composed of integers
It's the normal closure of $\mathbb{Z} in your field extension

[cloud walrus]

Kraft Macaroni

[hidden haven]

Use the fact that cosets of a subgroup are either equal to it or disjoint

[waxen hedge]

It's the set of $x\in K$ (a finite extension of $\mathbb{Q}$) such that there exists integers $a_1,...,a_{n-1}\in\mathbb{Z}$ such that $x^n+a_{n-1}x^{n-1}+...+a_1x+a_0=0$

[hidden haven]

You just have to decide which coset you want to consider

[cloud walrus]

Adrien

[woeful flint]

ohhhhh

Tysm

[waxen hedge]

It's the ring of integers of $K$

[cloud walrus]

Adrien

[woeful flint]

that's way easier to work with

my lecturer literally just plucked it out of thin air

[waxen hedge]

With that definition, you could try to show that $\mathbb{Z}$ indeed is the ring of integers of $\mathbb{Q}$

[cloud walrus]

Adrien

[woeful flint]

hmm

i'll give it a go

[waxen hedge]

(with $\mathbb{Q}$ considered as a extension over himself)

[cloud walrus]

Adrien

[waxen hedge]

(of dimension 1)

The ring of integers of a finite extension share many properties with the usual ring of integers (the most important being that there are both Dedekind's rings)

It's a "natural" generalization
Be careful : most of the time there are not UFD or ideal principal domain

(see the "ideal class group")

[lethal dune]

how do you know Q is a male

[waxen hedge]

I don't

[hidden haven]

I asked him, I can confirm

[woeful flint]

Thanks for all your help mate

[upbeat juniper]

you can play around with orders. let n,a,b,c be the orders of the large group, the two proper subgroups and the intersection respectively. then n = a+b-c. since a,b <= n/2, the only possibility is a=b=n/2 and c=0, which is impossible since the intersection contains the identity

[woeful flint]

@waxen hedge Hey I'm here to bother you again. Lets say $K = \mathbb{Q}(\sqrt{-5})$ then in this case $\mathcal{O}_K = \mathbb{Q}(\sqrt{-5})$ right?

[cloud walrus]

Kraft Macaroni

[chilly ocean]

are there algebras in which operatiosn distribute over themselves?

[waxen hedge]

No ! Actually it depends of the residue of -5 mod 4.
It can be either $\mathbb{Z}[\sqrt{-5}]$ or $\mathbb{Z}[\frac{1+\sqrt{-5}}{2}]$
I never remember which one it is, if I'm not wrong here we have -5=3 mod 4 so it will be the first one
You can search "ring of integers of quadratic extensions", it's pretty standard

[cloud walrus]

Adrien

[lethal dune]

it was a putnum problem, what about 3 subsets?

[upbeat juniper]

V_4={1,a,b,c} is the union of {1,a}, {1,b} and {1,c}

[lethal dune]

uu

nice

[sinful mirage]

can someone give an explicit example, which is in H_1 otimes H_2,but not the set tey give?

linear combinations of those?

because arbitrary element in H_1 otimes H_2 is a linear comb of $\sum_[i,j} a_{ij} e_i \otimes f_j$, where e_i ,f_j are basis of H_1 and H_2 respecitely

[cloud walrus]

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

[coarse storm]

I think so, yea.

[sinful mirage]

and now,coordinate-free

how could one argue this?

without choosing any basis a priori

[chilly ocean]

Does anyone know anything about group cohomology

group homology*

[sturdy marsh]

@sinful mirage coordinate-free doesnt do anything for you here. To produce such an element, you would need to choose a bunch of linearly independent vectors.

[sturdy marsh]

yes, what about it

[chilly ocean]

Ive looked online and cant find a textbook or nice article about it

everything is group cohomology

[sturdy marsh]

weibel has homology

[chilly ocean]

i just need reading resources

[sturdy marsh]

and brown's book

[chilly ocean]

group homology though

[sturdy marsh]

brown's book is very good

yes group homology

[chilly ocean]

ok browns ty

i have to present in one day

one the topic

[sturdy marsh]

[chilly ocean]

and idk anything rn besides an overview of algebrsic topology

[sturdy marsh]

what is your presentation on

like historical motivation?

[chilly ocean]

its just presenting on the idea i think

im assuming motivation,definition,examples

its only 20min

i have all 12 hours to prepare for now

i thought that is enough, you dont think so?

[sturdy marsh]

what is enough

[chilly ocean]

enough to prepare 20min presentation

[sturdy marsh]

sure

[chilly ocean]

im assuming i only need around 2-3 hours to make beamer presentation with all information once im done reading

[sturdy marsh]

depends on how well you know the material

[chilly ocean]

lol

i looked up computation on wikipedia

[sturdy marsh]

if you already know the basic idea, then it shouldnt take you too long yea

[chilly ocean]

idk what its for lol

i just think i know computation

and idk how to make projective resolutions, but I can make free ones

[sturdy marsh]

historically, people wanted to compute the (co)homology of spaces which are completely determined by their homotopy groups

for example, eilenberg maclane spaces, classifying spaces, etc.

and they discovered that there's a purely algebraic way to do it

[chilly ocean]

algebraic way to classify spaces without talking about the spaces?

[sturdy marsh]

algebraic way to compute the cohomology

[chilly ocean]

oh

[sturdy marsh]

i.e. from only the data of the group

not the topology of the space

[chilly ocean]

just looking at the R coefficients or something?

[sturdy marsh]

any coeffecients

[chilly ocean]

so i should definetly start with talking about group cohomology first

[sturdy marsh]

cohomology is a bit more intuitive, yes

as youre taking invariants

instead of coinvariants

[chilly ocean]

i can guess that homology is just the dual version

but im assuming people do group homology to make some calculations easier?

[sturdy marsh]

they do group homology for the same reason they do homology

but when the group is finite, something special happens

[chilly ocean]

ill have to read browns book, cohomology of groups?

[sturdy marsh]

there's a norm map for finite groups, which allows you to combine group cohomology and homology to get something called tate cohomology

[chilly ocean]

woah

[sturdy marsh]

you get an extra long long exact sequence

[chilly ocean]

lol wtf is that

[sturdy marsh]

combining cohomology and homology

[chilly ocean]

6 terms long LES?

[sturdy marsh]

no infinite

[chilly ocean]

wtf

[sturdy marsh]

even the usual LES is infinite

[chilly ocean]

no

i mean the parts where the homology group same index

[sturdy marsh]

0 --> H^0 ---> H^0 ---> H^0 ---> H^1 ---->....

associated to an SES of G-modules

[chilly ocean]

G-modules are just group modules?

[sturdy marsh]

yes

[chilly ocean]

are they group actions on groups

[sturdy marsh]

modules over Z[G]

[chilly ocean]

or is there more structure

is that group ring?

[sturdy marsh]

linear group actions on modules

[sturdy marsh]

yes

[chilly ocean]

ok

so my goal is to introduce group homology and maybe at 15min mark i can mention tate cohomology

[sturdy marsh]

for tate cohomology you get an LES of the form ---> H^-1(A) ---> H^-1(B) ----> H^-1(C) ---> H^0(A) --->H^0(B)--->H^0(C) ---> H^1(A) ---> H^1(B) ---> H^1(C) --->...

so infinite in both directions

and this streamlines a lot of the "dimension shifting" arguments that you see in group cohomology

[chilly ocean]

oh so by extra long you mean it goes into negatives

its only been mentioned but idk what negative homology groups supposed to be

[sturdy marsh]

you dont need to consider H^0 separately

negative cohomology groups are homology groups

except H^-1 is a bit different to make the LES work

there's a duality theorem for this, and you can use it to define a cup product

[chilly ocean]

oh i forgot about those

[sturdy marsh]

anyway, brown's book is a great resource

have fun!

[chilly ocean]

yy

ty

[sturdy marsh]

cup products in brown's book are a bit hard to understand if you're seeing it for the first time tho

[chilly ocean]

ive seen them before

[sturdy marsh]

the cartan-eilenberg approach is easier

they just use dimension shifting a bunch to define cup products

[chilly ocean]

as i understand it, its giving group structure throigh addition of different cohomology groups

[sturdy marsh]

ring

the cup product is the multiplication

[chilly ocean]

oh that makes sense for anticommutivity thingy

i understand how the wedge product is a sort of cup product

thats my concrete intution

idk if i have to go that far though

im only presenting on group homology and i plan to compute one example

[sturdy marsh]

there's a super easy abstract definition of cup products using the yoneda product

[chilly ocean]

no clue what that is

[sturdy marsh]

but you cant compute anything using that description

anyway, good luck!

[chilly ocean]

ty

[woeful flint]

I'm trying to show the ideals $(xy),(3x-3y-1)$ in $\mathbb{Z}_{(3)}[x,y]$ are coprime

[cloud walrus]

Kraft Macaroni

[woeful flint]

how can I go about this?

[sinful mirage]

why can every element of the tensor product $V_1 \otimes V_2$ be written as $\sum_{i} \phi_i \otimes \psi_i$, where $\phi_i \in V_1, \phi_i \in V_2$?

[cloud walrus]

ProphetX

[coarse storm]

I would try to write x and y as a sum of elements from those ideals.

[coarse storm]

What definition of tensor product are you using?

[sinful mirage]

universal property

or equivalently,V otimes W=F[VxW]/~

[chilly ocean]

think of small examples

like R tensor R2

[chilly ocean]

or you can think of the opposite

what do you suppose every tensor product can be written as

[sinful mirage]

$\sum_{ij} c_{ij}\phi_i \otimes \psi_j$

[cloud walrus]

ProphetX

[chilly ocean]

thats the same thing

[sinful mirage]

why?

[chilly ocean]

cij are arbitrary scalars

if you have a vector av for a a scalar and b a vector you can just let av=w which is also a vector

so scalars are kind of arbitrary

[sinful mirage]

hm

[chilly ocean]

like think of R tensor R2

3(4) tensor (2,1) is the same as (12) tensor (2,1)

[sinful mirage]

but c_{ij} not equal c_{i} d_{j}

[chilly ocean]

the scalars may not be equal but you can use c_ij to make it so that the indexing works out if that makes sense

[sturdy marsh]

it's just V x W/~

[sinful mirage]

how?

[sturdy marsh]

that's a definition

[chilly ocean]

the field is implicit

[sinful mirage]

it should be the free vector space

no

[chilly ocean]

oh you mean free

[sturdy marsh]

vector spaces are always free

[sinful mirage]

F[VxW] why [VxW]?

[sturdy marsh]

youre not taking the free vector space on VxW

VxW is already a vector space

[sinful mirage]

i'm so confused

I did the construction with the free vector space

and it seemed to be fine

let X be a set. then F[X] is a vector space

choose X=VxW

[chilly ocean]

oh

[sturdy marsh]

F[VxW] is an infinite-dimensional thing

if F is an infinite field

[sinful mirage]

do we understand the same thing under F?

F[VxW]:=free vector space on VxW

@thorn delta we did the construction with F[VxW]

why can we ignore F?

[sturdy marsh]

oh wait I think I might have messed up

gimme a sec

[chilly ocean]

i recognize it as V x W/~

[thorn delta]

wdym F?

the field?

[sinful mirage]

no

F[VxW]:=free vector space on VxW

[thorn delta]

Yea, but wdym by "ignore F." F is just a symbol

[chilly ocean]

is this not redundant

[sinful mirage]

well,people say it's not

[sturdy marsh]

sorry youre right

[sinful mirage]

why is F[VxW]/~ same as VxW/~?

[sturdy marsh]

no

it is not

[thorn delta]

VxW is just a set. F[VxW] is the free vector space on VxW, which is what u want

[sinful mirage]

yes

but if V,W are vector spaces

then so is VxW

why is VxW iso to F[VxW]

[hot lake]

because F(S) is the set of finite linear combinations of the i(s)

[thorn delta]

this is only if we make VxW a direct product

[sinful mirage]

yes

[thorn delta]

here, we just have VxW the set, and we make VxW a vector space in a different way: by taking the free vector space on the set V x W

[sinful mirage]

yes

[chilly ocean]

oh ok

[sinful mirage]

this is what I thought too

[chilly ocean]

this is silly tho

[hot lake]

F(VxW) is not VxW though

[thorn delta]

shit oops

[sinful mirage]

I don't really see what you mean

[chilly ocean]

why not just let V x W be a vector apace?

and then mod out by usual equivalence relations

wtf is point of making V and W sets and then looking at its free construction

[hot lake]

how do you add (v1,w1) + (v2,w2)

[sturdy marsh]

you dont take the free vector space on VxW

[hot lake]

you need extra elements to do that

[thorn delta]

F[VxW] is not VxW as a set prophet (hopefully i didn't confuse you). Its like the set of "all formal linear combinations" of elements of VxW

[hot lake]

you can't keep everything inside VxW

[sinful mirage]

yes

but

[chilly ocean]

i thought u just dont do that lol

ok

i see why free part needed

[sinful mirage]

???

[sturdy marsh]

it's the free Z-module on V x W

modulo relations

[chilly ocean]

so silly

[sinful mirage]

how?

this states

there exists only one vector space up to iso. that is the free vector space

wtf

[chilly ocean]

?

[sinful mirage]

how can R be a free vector space?

[chilly ocean]

ok im working on presentation hope this clarifies

[sinful mirage]

free vector space of what?

[hot lake]

of {1}

[sinful mirage]

lol

[sinful mirage]

how would I prove this?

in generality

it seems so shocking

[hot lake]

also that definition is a little weird unless they include that B is linearly indpendent

[sturdy marsh]

zorn's lemma

[hot lake]

also when they say "is a free vector space" they mean "is isomorphic to a free vector space"

[sinful mirage]

yes

lol

do they actually mean every vector space has a basis

and the basis is given by B?

[hot lake]

if you are being petty and looking at who exactly the elements are, they will be different because x is not i(x)

[summer cradle]

chad

[hot lake]

zorn's lemma implies that every vector space has a basis yes

[thorn delta]

(um what is hamel basis)

[hot lake]

a basis over Q ? idk

[sturdy marsh]

it's just basis in the usual sense

[thorn delta]

ah okay strange

[hot lake]

also, is that a different book ? the one from before only has 13 chapters

[sinful mirage]

yes

this is a mathphys book

[sinful mirage]

the book from yesterday is ok

I went through all chap of tensor prod

but now trying to see different things

[thorn delta]

the sketch goes like this:

1. every linearly independent set is contained in a maximal linearly independent set (requires zorns lemma)
2. a maximal linearly independent set is a basis
3. the empty set is linearly independent
   so every vector space has a basis

[sinful mirage]

yes but

why is the statement of every vector space having a basis equivalent to every vector spcae being free

this is my confusion

I can accept that every vector space has a basis by zorn

[thorn delta]

oh well a space is free on its basis

[hot lake]

if B is a basis of V, then V is isomorphic to F(B)

[sinful mirage]

why?

let B be a set. then F(B) si a vector space

true

Let V be a vector space and B its basis

why F(B) iso V?

[sturdy marsh]

write down an isomorphism

[sinful mirage]

in finite dimensions say, I can choose basis and write down iso

in inf dim why?

[sturdy marsh]

same argument

and you already chose a basis

[hot lake]

well B is a subset of V so that gives you a map from F(B) into V

it sends sum ai i(bi) to sum ai bi

if that's not surjective then you immediately deduce that B was not a generating subset ?

if that's not injective then you immediately deduce that B was not linearly independent

I feel this is just a lot of definitions pushing :/

also all those sums are FINITE sums

even if B is infinite

[sinful mirage]

yep

[hot lake]

iirc you defined F(B) as the set of functions from B to the field with FINITE support

[sinful mirage]

yes

[thorn delta]

(i can't stand this def)

[hot lake]

me neither because they forgot about B needing to be linearly independent

but it's kinda expected that physicists do everything wrong

[thorn delta]

the "based space" definition that prophet has captures the notion of "free" much better

[sinful mirage]

he's a mathematician

who tries to teach physicists math

[hot lake]

I am revoking their mathematician license

but yeah then uuuh

I will put the blame on the fact that writing books is very hard

[thorn delta]

in general we think of a vector space "V being free on X" if a set map f : X -- > W induces a unique linear map T : V --> W which restricts to f.
Clearly this is the case when X is a basis for V (i think you proved that V is free on its basis by this definition before)

[hidden haven]

🥅 ⤴️

[hot lake]

universal properties to the rescue !

[languid moss]

up-to what exponent does rational root theorem hold for

[sinful mirage]

yse this I proved

we did it together here

[thorn delta]

and the upshot is that this definition is the same for groups, modules, algebras, etc... in case you encounter those

[languid moss]

anyone 😦

[hidden haven]

I have never seen a condition on the exponent. What is the context?

[limpid edge]

is = necessarily an equivalence relation here

reflexivity is easy: if you have x < x then you also have x > x but (i) says only one of these should be true (contradiction)

[hot lake]

= is equality

[limpid edge]

here isn't = defined wrt <

[languid moss]

I got it

Thanks

[limpid edge]

< is given as a set B \subset ( S \times S )

[hot lake]

that looks circular

[limpid edge]

and then x=y iff neither (x, y) nor (y, x) are in B

[limpid edge]

I meant

[hot lake]

< is just a relation ?

[limpid edge]

yea

[hot lake]

well = is the relation {(x,x) | x in S}

[limpid edge]

oh wait really

[limpid edge]

it doesn't directly say this tho in the definition

that's why i wanted to check

[hot lake]

the definition of an order ?

[limpid edge]

yeah

[hot lake]

I think it assumes you know what "=" means

[limpid edge]

i thought that's not the point

everyone knows what < means

but it's introducing it abstractly

[hot lake]

yeah but they introduce the symbol < in the preceding sentence

telling you that it's going to be the subject of a definition

[limpid edge]

hm I guess

[limpid edge]

but say we continued down this path

could we similarly show = is symmetric and transitive given that = is just a binary relation and nothing else

i just got symmetry: suppose x = y but y > x, or x < y, which contradicts (i)

i can't get transitivity

fuck

[inland otter]

Hi, Im studying some introductory notes on field theory.
Kind of blocked on this question: if L//K is algebraic are L//F and F//K algebraic.

Seems like they shouldnt be if L//K is of infinite degree but I cant find a counterexmple

[limpid edge]

@hot lake this is what i meant originally

[limpid edge]

ok i get more why they would assume this now

but it's still kind of peculiar that you can prove symmetry and reflexivity of = without the assumption (and probably transitivity too but I can't figure it out)

[long obsidian]

Does $R/Z$ have any zero divisors?

[cloud walrus]

fajitas

[next obsidian]

That isn’t a ring

[long obsidian]

Is $\mathbb{R}$ a free $\mathbb{Z}$- module?

[cloud walrus]

fajitas

[next obsidian]

No

[long obsidian]

Why is that?

[next obsidian]

There’s only a single map R -> Z

The zero map

[long obsidian]

Thank you

Is my reasoning about this correct, the torsion elements of the $\mathbb{Z}$-module $\mathbb{R} / \mathbb{Z}$ is $\mathbb{Q}$ the rationals

For instance if I have $q=m/n$
Then $n*(q+Z)=m+Z=Z$ since m is an integer. No other torsion elements exists outside of $\mathbb{Q}$

[cloud walrus]

fajitas

[hidden haven]

Yes

[long obsidian]

Thank you!

[verbal harness]

Why does a group's cayley table have to form a latin square? I thought I had why it was worked out, but then I found that I only had sufficient evidence that the identity element would appear exactly once per row and per column .. but no clear guarantees for the other elements.

[next obsidian]

It’s because of cancellation

Or like, inverses plus associativity

So like fix g and h

The “equation” gx = h has at almost one solution

Because x has to be g^-1h

Then the existence portion is true because well… if you wanna solve gx = h, then x = g^-1h works

[verbal harness]

(g^-1)h or g^(-h)?

[lethal cipher]

Okay, so I get why phi(F) is a field, but I am unsure how to extend that to show S is also a field.

[next obsidian]

The latter doesn’t make sense

So the former

[next obsidian]

Onto means surjective my guy

[lethal cipher]

oh, guess who didn't read the question...

[verbal harness]

lol 🙂 Can't take powers of group elements gotcha 😉

[next obsidian]

Yeah

Does that make sense tho?

What I said

What i did was show that every column (or row idk) has exactly one of each element basically

[verbal harness]

Agreed, I'm looking it over and making sure I digest it properly though.

[next obsidian]

Yee

Like, at most 1 is the like “injective” part

The gx = h has at most 1 solution

The fact each element shows up is the like surjective part

I guess what I proved is technically a little bit off

If you can figure it out from what I said then that’s good, or I can make directly prove it

[verbal harness]

I'm trying to pick apart the associativity axiom because something feels funny about it (kind of like Euclid's fifth postulate funny), and also it's the only one difficult to visually confirm via a cayley table.

[next obsidian]

Oh don’t bother lmao

It’s an axiom just enforce it

Idk how you’re supposed to take a table and “see” that the operation is associative

[verbal harness]

injective: not too many. surjective: not too few. bijective: botha those 🙂

[next obsidian]

Right

Or okay

Here’s a much simpler proof

[verbal harness]

Well you can see if it's abelian for example, it just reflects across the primary diagonal (assuming normalized row/column order as I basically always do)

[next obsidian]

Each row / column having every element once is the same thing as saying the function f:G -> G given by f(g) = gx is bijective

Or f(g) = xg

Sorry, here we’ve fixed x now

This represents either the x-row or the x-column

But this function has an inverse, given by f^-1(g) = gx^-1

So it’s bijective

[next obsidian]

Well the issue with this comparison IMO is being abelian is a property which only requires pairs of elements which is easy to see on a table

[verbal harness]

OK that function isn't a group element itself though, so while I intuitively believe it would have an inverse I can't see a justification for that straight away. 🙂

[next obsidian]

For associativity you need triples

No like

As a function

Compose them and the function is

g -> gxx^-1

Or gx^-1x

But this is just identity

[verbal harness]

That's why one thought I've been toying with is a 3d cayley table for associative operations. :B

[next obsidian]

[viscid pewter]

aaaaaaaaaaaaaaaaa

[hidden haven]

Bruh in our first semester, saketh and I came back from class after seeing the definition of a group, and we heard someone say that all groups of order ≤ 4 are abelian, and then we basically brute forced every single possible multiplication table for order ≤ 4 and figured out all 5 such groups

Associativity was by far the most annoying

We didn't have product groups or cyclic groups or anything lmao

[next obsidian]

Yes associativity is god awful

[hidden haven]

We literally just brute forced multiplication tables

[next obsidian]

Because it requires triples

[hidden haven]

Ye

[next obsidian]

So the number of stuff you have to verify just explodes

[next obsidian]

Also meme

[hidden haven]

Yes, and now we would just say that we know that Klein 4 is a group

But we verified associativity

All of this by hand

[next obsidian]

Eww

[hidden haven]

[next obsidian]

I feel like it would have been more economic to just learn GAP

[hidden haven]

@gritty sparrow do you remember this

[next obsidian]

And program something to spit them out

[hidden haven]

Lol it actually didn't take too long

[next obsidian]

Also wtf Saketh goes to the same uni as you?

[hidden haven]

30 mins or so

Yes

We are roommates

[next obsidian]

Det too?

Wtf

[hidden haven]

When we are not home for lockdown

Yes det same uni 1 year junior

[next obsidian]

Why does Saketh know so much more commutative algebra than you

[hidden haven]

Because he took commie alg lol

I didn't

[next obsidian]

SAD!

[hidden haven]

I did logic instead 🤡

[next obsidian]

Saketh knows a lot a lot of commutative algebra

He helped Chmonkey a lot

[verbal harness]

I haven't tried brute forcing it yet, but the smallest loop I've found so far without associativity is of order 4

[hidden haven]

Ye he even took a topics course on it

[next obsidian]

Oh lol

[verbal harness]

When I think about group elements in their role as transformative functions over another domain, I have a difficult time visualizing what associativity really means in that perspective. A composition chain is just doing each of the functions to an input in the given order. So I think parentheses can be seen as replacing a portion of such a chain with whatever single transformation is isomorphic to it. And then that leaves associativity as roughly replacing overlapping parts of a chain not changing it's answer. But assuming I'm even getting that much right, it's not clear to me what's compelling about that property?

[next obsidian]

Imagine if you walked 10 steps to the right

Then 10 steps up

And this was different than walking 10 steps up

Then 10 steps to the right

That’s how I justify it to myself

[verbal harness]

That's commutivity though

[next obsidian]

Oh yeah

Maybe I’m a

[verbal harness]

Also holonomy lol 🙂

[next obsidian]

I mean you could say

10 steps right

15 steps right

Versus 15 steps right then 10 steps right

As looking at 10 + 5 + 10

Also have you ever tried to write down non-associative stuff?

I just can’t do that to myself, I love myself too much to write all those parenthesis

[verbal harness]

The best/simplest examples I know are {0,1,2} over |a-b| (which doesn't form a latin square, so I guess isn't a loop) and the small order 4 loop I found which does.

[verbal harness]

Non-commutative stuff with left- and right-whatevers sounds just as messy to me though haha

[next obsidian]

Well I also only live in a commutative world

[sturdy marsh]

[next obsidian]

Not with groups, but for groups it’s alright

For rings tho pftttt

[verbal harness]

Rotations of a sphere is a fine non-abelian citizen in my 3d graphics programming world heh 🙂

Marc ten Bosch says "stop using quaternions for that and use rotors" and I go ksdjkflashdgh

[lethal cipher]

I am a bit stuck on this. How do I find limitations on a homomorphism here?

[next obsidian]

Write these as a quotient of Z[x]

Then use that Z[x] is a free algebra

And then that maps from a quotient are those from the thing quotienting by, with kernel containing the thing you quotiented by

[thorn delta]

1 has to map to 1, and you can conclude a lot from this

[lethal cipher]

We have never discussed what a free algebra is at any point

[lethal cipher]

How so?

[next obsidian]

The map is determined by where you send x

And you can send x to whatever you want

Additivity determined what all of Z has to go to

Then by multiplicativitu determines where x^n goes from where x does

So then by more multiplicativitu where mx^n goes

Then that determined by additivity where every polynomial goes, just from where x does

You have to do some work I guess to show that any choice of where x goes is valid

[lethal cipher]

Ngl. I do not understand this approach at all

[lethal cipher]

On another note, I think one thing this tells us is that Z maps to Z

[thorn delta]

ye

so that restricts things quite a bit

[lethal cipher]

Well, one thing we do know is that a either maps to itself or -a, since those are the only automorphisms of (Z,+)

And I do believe that the restriction of phi by Z is a bijection

So then how does b need to be restricted then?

[thorn delta]

apply the definition of a homomorphism to an arbitrary a + bsqrt(3)

[lethal cipher]

Ummm, phi(a+b sqrt(3))= phi(a)+phi(b sqrt(3)). Not sure how this helps.

Well... phi(b sqrt(3))=phi(b) phi(1*sqrt(3)), so it looks like b is resfricted in the same way

[serene radish]

Is g(1, 2, 3)(4, 5)g^{-1} = (g(1), g(2), g(3))(g(4), g(5)) ?

g is from S_5

I just need a quick yes or no

it is

I just wanted to save some time

but nvm

[thorn delta]

right, i.e. you only have to choose where sqrt(3) maps now

[barren sierra]

My only thoughts are

the only subgroups are ones of size p or q

or trivial and whole group also I guess but those don't matter

but idk how to leverage that

[serene radish]

Do you know Cauchy's theorem?

@barren sierra

[barren sierra]

no

[serene radish]

Cauchy's theorem states if we have a group of finite order and a prime p such that p | |G|, then there exists an element in G with order p

[barren sierra]

hm ok makes sense

[lethal cipher]

Well it only has one option: sqrt(-3)

[serene radish]

Can you proceed from there?

[thorn delta]

no, not necessarily

[lethal cipher]

Oh?

[next obsidian]

You don’t need that

@barren sierra

[viscid pewter]

it helps a lot tho

[next obsidian]

Oh wait

You totally need that

[viscid pewter]

surely

[next obsidian]

To grab the elements

[lethal cipher]

Well, it's got to at least be some multiple of sqrt(-3)

[viscid pewter]

yeah

[next obsidian]

To then smash

Chmonkey moment

[thorn delta]

think about relations that phi(sqrt(3)) is going to have to satisfy

big hint: || (phi(sqrt(3))^2 = phi((sqrt(3))^2) = phi(3) = 3 ||

[lethal cipher]

Wait, but the only two solutions for that are sqrt(3) and -sqrt(3)

But those are not elements in our second ring

[thorn delta]

ye

[lethal cipher]

So does it need to map to an integer?

[serene radish]

Hey if I'm asked to find blocks of a subgroup of S_n what should be my action?

Just conjugation or?

[thorn delta]

well is has to map to something in Z[sqrt(-3)] whose square is 3

hint: || the set of all your options here is empty ||

[lethal cipher]

So wait, does that mean no such homomorphism can exist?

[thorn delta]

ye

[lethal cipher]

And then I think the same issue happens for homomorphisms in the other direction

[thorn delta]

yea i believe so. A similar line of argument shows that sqrt(-3) has to map to something whose square is -3

[lethal cipher]

Actually now that I think about it, not necessarily

phi(3) can be 3 or -3

Because the two automorphisms in Z are n ->n and n-> -n

So I dont think it breaks like we think.

On the other hand, it does tell us that the identity function on Z doesn't cut it, but... n-> -n will

[thorn delta]

phi(n) = -n is a group automorphism of Z

phi(n) = -n is not a ring automorphism of Z, if that's what you were trying to claim.

[lethal cipher]

Ah good point. So we are only restricted to the identity map of Z onto Z, which cannot work

[barren sierra]

yea no I'm stuck still. I'm considering the cosets of G : <x> where the order of x is p

so there are q of those

since Lagrange's theorem is a thing

and then

¯_(ツ)_/¯

[proud bear]

if |x|=p and |y|=q, whats |xy|?

[lethal cipher]

@next obsidian I was struggling to understand your approach, but were you drawing the same conclusion?

[next obsidian]

This is exactly the point

A map from Z[x] is about where you send x

[barren sierra]

pq

but like

[next obsidian]

In the quotient x is sqrt(3)

And by property of quotients you need the image of x, call this y

[barren sierra]

hm

[next obsidian]

To satisfy y^2 = 3

Which is impossible

[barren sierra]

how so I get that it's cyclic @proud bear

[next obsidian]

Group of order pq

Has element of order pq

[lethal cipher]

That makes a lot of sense. Out of curiosity, how do we know x is sqrt(3)? I am not sure I understand that

[next obsidian]

The isomorphism

[barren sierra]

oh

oh

[next obsidian]

Z[x]/(x^2 - 3) to Z + sqrt(3)Z

Sends x to sqrt(3)

I mean this is what quotienting by x^2 - 3 does

It says x^2 = 3

So x is sqrt(3)

[barren sierra]

wait no that's what I'm trying to prove

[next obsidian]

Lmao

[barren sierra]

so I can't just say that

god I wish

[next obsidian]

That isn’t what you asked

But

[barren sierra]

I know G is order pq

[next obsidian]

You need to use they commute and have coprime order

Then just do it manually

[lethal cipher]

So how did you realize the group you were quotienting by was x^2-3?

[barren sierra]

and there are subgroups of order p and q

[next obsidian]

Cuz this is how you adjoin any sort of algebraic thing like this

[next obsidian]

if orders of g and h are coprime and they commute order of gh is the product of their orders

Prove this as a lemma

You’ve already introduced an element of order p and one of order q so their product generates the group by the lemma

[next obsidian]

My point here was sqrt(3) is algebraic so just like for field extensions you can make the extension by that via this quotient of polynomial ring construction

Using the minimal polynomial

Once you see this a few times you just know this is what to do ¯_(ツ)_/¯

[lethal cipher]

So I've seen this used to show C is isomorphic to R[x]/(x^2+1), and I understand why that is true. But I will admit I am having a harder time recognizing how to apply this for a more general range of problems (like you are doing here)

[next obsidian]

You mean R

Not Z

[lethal cipher]

Yes, sorry

[next obsidian]

This is just formalizing the idea that maps are given here by specifying where, in this case, sqrt(3) has to go

Subject to saying it must go to something squaring to 3

Really I guess you don’t need the full power of this for this problem

You showed directly it would have to go to something squaring to 3 which can’t exist

But writing like I did would show that a map Z[sqrt(3)] -> R for any ring R is exactly

A choice of an element in R of square = 3

Anything satisfying this defines a map from that ring by just sending sqrt(3) to it and then just extending linearly and multiplicatively

The point here is just that maps Z[x] -> R are given by sending x to anything you want, call this element a

x is a free variable, just like a basis vector for a vector space is, you can just send it to whatever

Then maps Z[x]/(f(x)) are just those maps Z[x] -> R whose kernel contain f(x), which is to say you need a to satisfy f(a) = 0

[lethal cipher]

Okay, I think I am understanding more. Thank you.

For part a), it is clear that phi(I) is an ideal of phi(R), but I am not so sure if we can ensure it is an ideal of S.

[next obsidian]

Think of a really extreme example

Let S have very few ideals

Then you’ll see it can’t be true

[lethal cipher]

I mean, the most obvious ideal is just {0}

[next obsidian]

No, think of what S have few ideals

There’s a very large class of rings with an extremely low number of ideals

[lethal cipher]

Something is clearly jumping out at you. I can't think of such a ring

[next obsidian]

Any ring always has 2 ideals

(Unless it’s the 0 ring in which case they agree)

What if those are the only ideals

[lethal cipher]

Oh, 0 and S

[next obsidian]

And if those are all you have?

Can you think of a ring with this property?

What if I wrote K instead of S

Or F

[lethal cipher]

F makes me think of fields

[next obsidian]

And?

That’s good because

[lethal cipher]

and I do believe it is restricted to just those

[next obsidian]

Yes

Because unit in I means I = everything

Okay

What’s your favorite field

[lethal cipher]

R sticks out at me

[next obsidian]

Sure

I prefer C but R is good too

So now we want a ring, we’ll call it S now because we used the letter R

A map S -> R whose image isn’t an ideal

Cuz S is an ideal of S so just whatever

We’ll show the image of the entire ring isn’t an ideal

All you need is the image of S to not be 0, or not be all of R

Anything stick out?

[lethal cipher]

Well, why not let S be Z let the map be the inclusion map

[next obsidian]

Yup

This is what I had in mind

[lethal cipher]

Sweet! Now what about for b)

[next obsidian]

Inverse image is better than images

I’ll just say that and let you figure out if it’s true or not

[lethal cipher]

I've come to learn that from topology

[next obsidian]

It’s true in algebra too

[lethal cipher]

Hmmm, so I see that if we take r and j (j in the preimage of the ideal J and r in the ring) then we have phi(r)phi(j)=phi(rj)...which is in J

So the answer is a clear yes, because rj is in the preimage of J

Okay, last thing. I've been thinking about this problem a lot. Personally, showing multiplication is closed in N(R) is easy, but I am really struggling with showing addition is closed, let alone a subgroup of (R,+).

Trust me, I definitely see that it is going to be used. I am just not sure how it is helpful

[ (a+b)^n=\sum \binom{n}{k} a^{n-k}b^k]

[cloud walrus]

dackid

[lethal cipher]

well my thought is the product of the powers for a and b (say r and s)

but why would be able to conclude that a^(rs-1)b=0 with that?

well that is concerning

The problem is that a^-1 isn't guaranteed to exist, so we can't just say a^(rs-1)=a^(rs)a^(-1)

[limpid edge]

am I dumb or a^n = 0 -> a^m = 0 for m >= n ?

[lethal cipher]

Oh shoot! You are right!

[limpid edge]

and n = 2*max(r, s) is good enough i think

although the choice doesn't really matter

[lethal cipher]

I do like that one more. Because then it is very obvious that either n-k or k is greater than max(r,s)

Ty for the help. I don't know why I was not piecing that together

[limpid edge]

r + s also fits nicely

[lethal cipher]

I am not too sure about that. It seems like 2max(r,s) is the magic number

[limpid edge]

(x + y)^(r + s) = x^(r + s) + ... + x^r y^s + ... + y^(r + s)

in fact i think r + s - 1 is best

your "minimums" (middle terms of expansion) have exponents of (r, s - 1) and (s - 1, r)

[rotund shoal]

God idek why but I find cyclotomic extensions insanely hard

anyone able to help me out with this one?

I kinda need to just talk it through with someone cause idk where to start ngl

It's dummit and foot 14.5.8

oh wait nvmind I think I got it

[barren sierra]

is solving this just using first iso theorem?

yea it is

neat

[rotund shoal]

Hey so can someone check my logic real quick? Trying to find the galois group of x^3 - x + 1:

so first of all it's clearly reducible over Q (standard trick since f(r/s) = 0 implies r and s divides 1 implies the possible roots are simply 1 and -1 which aren't roots)

thus since f has degree 3, we have that the galois group is isomorphic to A3 or S3

and since the discriminant is -23 and isn't a square, we get that the Galois group is S3

[hidden haven]

Looks fine assuming discriminant was calculated correctly

[rotund shoal]

Alright thanks!

[obtuse kernel]

are upper level linear algebra questions appropriate here?

[chilly ocean]

yes

[obtuse kernel]

great - I just have a quick one right now

[chilly ocean]

as long as it's not something like row reduce this matrix then it also should be fine here

[obtuse kernel]

is the independence of two subspaces the same thing as being disjoint?

range of T

well except for 0

i think that's how we've been using disjoint

cool thanks

[chilly ocean]

just admit you can't solve the problem slimvesus

[obtuse kernel]

wait am I crazy or does the complement have to be N?

actually, nvm

[lethal cipher]

So question: part b isn't quite finished. How do I show that the preimage of J is a subgroup under addition. I am not so sure I see that

[woven delta]

Suppose a, b have f(a)\in J, f(b) \in J

[lethal cipher]

Oh nevermind

I see it

If we take j and j' in the preimage, then phi(j-j')=phi(j)-phi(j'). Since both elements are in the ideal J, phi(j)-phi(j') is in J

So I was able to show N(R) is an ideal, but I am having a bit of trouble figuring out N(Z_72)

One thing I do know is all of the elements relatively prime to 72 cannot be in it

[woven delta]

So if we are looking for numbers a so that a^n is divisible by 72 it may help to look at the prime factorization of 72

And the prime factorization of a

@lethal cipher

[lethal cipher]

Yep, I've been thinking about that. 72=2^3 *3^2

I believe because of this, our contenders are restricted to multiples of 6

Actually, that's exactly what it is

[woven delta]

Yes

This method works in general for the ring Z/nZ

[lethal cipher]

Okay, so let me know if this reasoning makes sense.
Let 6*n be an arbitrary multiple of 6. Then 3x72xn^3 is (6n)^3

And so (6n)^3=0 since it is a multiple of 72

[woven delta]

Sure, that's fine

[lethal cipher]

Okay cool. Thank you

[dense mulch]

If I need to find all nonisomorphic abelian groups of order 3^6, how would i do so

I figured that it would make more sense to look for all the isomorphic abelian groups, all the possible iterations of products of powers of 3

But I'm not sure how there would be any nonisomorphic abelian groups

[viscid pewter]

Z_9 and Z_3 * Z_3 are both abelian of order 9, but they aren't isomorphic (one has an element of order 9)

[dense mulch]

AH yes, thank you!

I was not keeping elements of order x in mind, thank you so much!

Hold up actually

So are all abelian groups of order 3^6 non-isomorphic then?

[viscid pewter]

??

i mean

ok so consider (Z_3)^6

[dense mulch]

The thing that's tripping me up is that, by the Fundamental Theorem of Finite Abelian Groups, any products of powers of 3 should be isomorphic to G (the finite abelian group of order 3^6)

ok im listening now lol

[viscid pewter]

that's not how the fundamental theorem goes

so for 3^6

you could have (Z_3)^6

you could have (Z_9)^2 * (Z_3)^2

you could have (Z_729)

there is no unique abelian group of order n unless n is prime, or 1, i think

[dense mulch]

wait did you mean (Z_9)^4 * (Z_3)^2?

[viscid pewter]

no

that would have order 59049

[dense mulch]

ohh yeah I see

Ah, the confusion was why (Z_3)^6 was different from

Z_3^6

Thank you once again for clarifying

[wind locust]

hey there

can we have a free R-module of rank n, with a linearly independent subset of cardinality greater than n? I know it's true when R is an integral domain, but I'm not sure what happens when R is just a commutative ring with 1 . I found this https://math.stackexchange.com/questions/1205340/what-would-be-an-example-of-free-module-such-that-cardinality-of-linearly-indepe but didnt really understand the explanation

[chilly ocean]

what about the explanation doesn't make sense

[wind locust]

I don't really see how such a linearly independent A would induce an R-monomorphism

[chilly ocean]

if A is a linearly independent set with m elements then it generates a submodule isomorphic to R^m

[wind locust]

hm, I guess that's true

but if the answer is that simple why do every textbook complicate the matter by requiring R to be an integral domain

anyways, thanks a lot, this was rly obvious ig

actually, wait

[wind locust]

i dont think that is necessarily a contradiction, no? A free submodule N of a free R-module M need not have rank less than or equal to that of M

ah i'm being stupid, ofc thats not a contradiction, but that is how we induce the monomorphism

and the monomorphism would just be the inclusion map from N to M where N is the submodule right?

thanks again

[chilly ocean]

epimorphism and monomorphism are words too big for me

[wind locust]

haha apparently they were also too big for me because what I meant was a monomorphism not an epimorphism, i.e. an injective linear map

sorry its 5am here

[chilly ocean]

yes it's just inclusion

[wind locust]

yeah alright

thanks a lot!

[prisma thunder]

Let $F$ be a field of characteristic $p$. Let $K = F(\alpha, \beta)$ where $\alpha$ is transcendental over F and $\beta$
is transcendental over $F(\beta)$. Let $L$ be a splitting field of $(x^p - \beta)(x^p - \alpha)$.
\begin{enumerate}
\item Prove that $[L : K] = p^2$.
\item Prove that $\gamma^p \in K$ for all $\gamma \in L$.
\item Prove that $L/K$ is not simple
\end{enumerate}

[cloud walrus]

eM

[prisma thunder]

We haven't had the chance to cover transcendental extensions extensively essentially (heh)

typo: beta should be transcendental F(alpha), not F(beta), as that doesn't make sense

I would have guessed L is F adjoined pth root of beta, pth root of alpha, and a pth primitive root of unity; but, since F is characteristic p, no such primitive exists (really, 0 = x^p-1 = (x-1)^p, so any pth root of unity in char p is just 1), so L is F adjoined pth root of beta and pth root of alpha.

Consider the degree [F(pth root of alpha, pth root of beta):F(alpha, beta)]. Hmm.

Let me go ahead and split this, by multiplicativity of towers, as
[F(pth root of alpha)(pth root of beta):F(pth root of alpha)(beta)]
times
[F(beta)(pth root of alpha,): F(beta)(alpha)]

Both of these should p but I'm trying to see why

[ancient rivet]

my abstract algebra course was alright... it was kind of abstract... but i was really hoping we were going to learn how to multiply sounds and divide by colors

now THATS abstract

7 times 🟩 = -3 with remainder: 🧩

[next obsidian]

I think basically you’re using transcendentals to artificially make purely inseparable extensions.

That’s what the gamma^p thing is about

It then gives you that the extension is a degree of prime power

And you basically have to just show that like, you can’t get there with just a single p because something like beta doesn’t pop up just from modding out by alpha - x^p

I think that proves it’s not simple too basically, like my intuition is that you can write this like artificially doing a degree p purely inseparable extension via alpha, and then again via beta

If you find a source for some stuff on purely inseparable stuff I think you’ll be able to find things to prove this

There’s stuff in Isaacs’ book on grad algebra

And the other sources I know are like the stacks project and Bourbaki Vol II on algebra, maybe like chapter 5 or 6?

[waxen hedge]

Might be in Lang but I've not read that part

[hidden haven]

Multiplication with remainder

[lilac trench]

for subgroups H and K, where H is a subgroup of N(K) and K is a subgroup of N(H) and the intersection of H and K is trivial, does everything in H commute with everything in K

[proud bear]

hkh^-1k^-1=(hkh^-1)k^-1=k'k^-1\in K
hkh^-1k^-1=h(kh^-1k^-1)=hh'\in H
so hkh^-1k^-1\in H\cap K so it equals 1

[lilac trench]

wow

[chilly ocean]

has anyone here ever actually completely finished "abstract algebra" by dummit and foote?

how long did it take?

[lilac trench]

sounds like spain ||without the s||

[unique berry]

I'd consider it but it's not necessary even at the grad level

Just as long as you can do all the problems

My guess though would be 1 section per day would make for like 6 months I think

[chilly ocean]

I thought it'd take a year

[unique berry]

Well realistically yes

I'm assuming you're completely focused on D&F

And doing every problem

But realistically something like that would take at least a year

[chilly ocean]

wait completely focusing on D&F and doing it would take a year?

[next obsidian]

To do every problem?