groups-rings-fields

[formal ermine]

so f + (f) = 0 + (f)

[delicate orchid]

I disagree with this definition on a pedagogical level :pack:

[low void]

(f)={g*f : g in K[X]}}

[delicate orchid]

yeah now that's more like it!

:revupyourengines:

[low void]

f+(f)={f+g| g in (f)}

So what is happening here?

I=(f)

I still don't understand why this would be zero

It isn't even a number?

[low void]

Since g=h*f, we have f+g=f+hf=(h+1)f and since h+1 is also a polynomial if h is a polynomial, we have f+(f)=(f)?

[delicate orchid]

they're abusing notation a bit here, it should be (0+I)v = 0+I

[low void]

0+I is just I, right?

[delicate orchid]

yus

I wrote it as 0+I to emphasise the fact that it is indeed the 0 element of the ring K[x]/I

[low void]

AHHH

I somehow missed that

[feral umbra]

I am reading Gromov's 1981 paper "Polynomial growth and expanding maps" and want to prove the following lemma:

Let L be a Lie group with finitely many components and let G be an arbitrary finitely generated group. Suppose that for every number p =1, 2, . . ., there is a homomorphism G \to L such that its image is finite and has at least p elements. Then G contains a subgroup G' \subset G of finite index such that the commutator group [G', G'] \subset G' has infinite index.

This is the proof in the paper: Let q be as in Jordan's theorem. Take for G' \subset G the intersection of all subgroups in G of index at most q. It is clear that G' satisfies all the requirements.

I have been able to proof that G' has finite index in G. How do I proof that [G',G'] has infinite index in G'?

(Jordan's theorem states: For each Lie group L with finitely many components there is a number q such that every finite subgroup in L contains an Abelian subgroup of index at most q.)

[median pawn]

since o is noetherian, every ideal in o is finitely-generated

how do we see that a non-zero submodule J of K is a fractional ideal iff there exists non-zero c in o such that cJ \subset o is an ideal?

one direction seems fine. suppose J is a non-zero o-submodule of K and cJ \subset o is an ideal. As o is noetherian, cJ is finitely generated, say cJ = (a_1,...,a_m). then, J = (a_1/c, ..., a_m/c) is a finitely-generated o-submodule of K

what about the converse?

[agile burrow]

If your submodule of K is finitely generated, just clear the denominators of the generators

[median pawn]

ahh right! thanks

[dim widget]

This is a nice paper! The idea here is just simple group theory though: you get your G' in G of finite index. Now in all of your homomorphisms f_p: G \to L G has finite image with at least p elements, whence they each contain an abelian subgroup H_p \subset f_p(G) with index \leq q. Then f_p^{-1}(H_p) has index \leq q in G, so G' \subset f_p^{-1}(H_p), so f(G') \subset H_p which is abelian. Finally because [G:G'] is finite we see that f(G') is abelian of size \geq p/[G:G']. Since f(G') is abelian this shows (taking p large) that [G', G'] is of infinite index.

[feral umbra]

thank you so much!

[dim widget]

No worries!

[static needle]

Gaussian integers sorry i wasnt specific

[south patrol]

Hm I've been asked to show the polynomial x^{2^k} + 1 is always reducible over Fp, provided k > 1. I'm struggling a bit because clearly there needn't always exist roots (-1 needn't be a quadratic residue) but I've got the hint to consider the structure of (Z/2^n Z)^x

[lethal dune]

How to show integral elements of Q(t, √t³-t) over Q are Q itself?

It's not as obvious as it seems

[south patrol]

Yeah that seems interesting

Hm

[lethal dune]

There's also a second part to it that says show that this extension is not purely transcendental

which I have done

[dim widget]

Are you asking this as a question or as an exercise for other people?

[lethal dune]

no this one's a question

[lethal dune]

this might be an exercise

[dim widget]

So you can look at this field L as Frac(Q[t, s]/(s^3 - t^2)) and notice that this is finite over Q(s) which does not contain any finite extension of Q by Luroth. On the other hand suppose K \subset L was finite over Q, then K must be contained in a quadratic extension by degree counting. But K is also disjoint from Q(t) whence K must be contained in a cubic extension.

[south patrol]

Poggers

[lethal dune]

haven't seen Luroth, what's the statement

[dim widget]

Just that there are no subextensions of Q(t) except those isomorphic to Q(t).

That's also a nice exercise. Although showing the easier statement that there are no finite subextensions is much easier than the whole theorem.

[south patrol]

I wanna give that bad boy a go tbh

[lethal dune]

nice

[south patrol]

Also oop any ideas on dis

Perhaps I am overcomplciating it

[dim widget]

It has many proofs, including very elementary ones, so it's very pleasant to work out.

[south patrol]

This was 6 marks, and the previous part was to show x^4 +1 is reducible over F5 which is a one liner (for 5 marks) lol

[sonic coral]

is F_5 Z_5?

[slim kayak]

Is t some transcendental element over Q or is this about something else than usual field theory?

[sonic coral]

not the sully

[south patrol]

F_5 is common notation for it when thought of as a field yes

[dim widget]

So what are the finite extensions of F_p? How do you distinguish whether or not a polynomial is irreducible using the Galois group?

[south patrol]

Sure I will try wacking the gal group at it

[dim widget]

It is just a transcendental, but it's kind of harder than it looks. Not sure I see an elementary way to show that this is true for some totally general curve's function field.

I used a trick that only works for this particular equation.

[south patrol]

Oh wait I think I realise what do now

I shall give it a whirl thank

[long nebula]

Although be careful, F_n is not Z_n in general

Like for example, F_25 is not Z_25

[sonic coral]

yes Z_25 is not a field

[south patrol]

Indeed F_{p^n} can never be Z/p^n unless n = 1 since every (non-zero) element has additive order p in F_{p^n}

[slim kayak]

Given topos e-girls response... where did you even find that problem?

[novel parrot]

tips for this?

[south patrol]

Think about the proof that Z[X] is not a PID

The same idea works

[novel parrot]

oh

smart

(a,X) is principle

so forces a to be unit

[south patrol]

Nice

:)

[novel parrot]

thanks

[south patrol]

That was quick lol

Np

Yeah it's pretty cool

[novel parrot]

i did the problem showing Z[X] not principle recently lol

(2,X) not principle

[south patrol]

This is cool because it's like the start of some bigger theory in a sense lol

[novel parrot]

number theory?

number theory is considered algebra right?

[south patrol]

I mean dimension theory basically lol like studying how primes of R[x] relate to primes of R, I mean

[novel parrot]

oh

[south patrol]

So the point is like R[x] has more primes than R is some sense and a PID is essentially as close as you can be to a field without being one

that's the kind of intuition i mean lol

In fact I can illustrate what I mean lol

Are you fine with the fact that R is a field iff it has no non-zero proper ideals

(1) isn't prime

[novel parrot]

yeah

[south patrol]

Okay so the point is like, okay suppose R has a maximal ideal m which is non-zero

Well then the ideal m[x] of polynomials in R[x] with coefficients in m is also a prime ideal

[novel parrot]

sounds similiar to localization?

[south patrol]

But m[x] isn't maximal, since A[x]/m[x] is isomorphic to (A/m)[x] which is a polynomial ring over a field, hence not a field lol

[lethal dune]

assignment

[south patrol]

So m[x] is strictly contained in a maximal ideal M, say

[slim kayak]

What is the course about?

[south patrol]

But then we have strict inclusions 0 < m[x] < M of prime ideals in R[x]

Which is impossible for a PID

[lethal dune]

comm alg and galois

[south patrol]

noice

[novel parrot]

we could have just said that (X) is a maximal ideal

and then R[X]/(X) = R

and a field

interesting

if J is an ideal, then J^2 = {ab} where a,b in J?

[lethal dune]

context?

[novel parrot]

and in general we have J > J^2 > J^3 > ..

[lethal dune]

no

[chilly ocean]

closed under addition?

[novel parrot]

commutative algebra?

[novel parrot]

yes

J^2 is ideal

[lethal dune]

you wrote ab so I thought you are treating a,b as ideals

[novel parrot]

is this true?

[lethal dune]

strict? no otherwise yes

[novel parrot]

oh

ok

[lethal dune]

that's not what TT meant

[slim kayak]

For example if J is your entire ring they'd be all equal, so specifying the order relation is important

[novel parrot]

what did she mean?

[lethal dune]

you may not be able write all elements of J² as ab. what you do is take finite sum of the form a1b1+a2b2+...+anbn

[novel parrot]

ohhh

[slim kayak]

Krull dimension thingy, although not too hard to verify if you think about what it means for a principal ideal to be prime

[chilly ocean]

no that's not what i meant at all lmao

ryu nailed it

[slim kayak]

How is this not just a variation of the krull dimension of PIDs?

[novel parrot]

How should i start this?

[south patrol]

... that's exactly what it is lol

[slim kayak]

Getting mixed signals here lol

[novel parrot]

any (m mod IM) = r_1(x1 mod IM) + ... + r_n (xn mod IM)

[south patrol]

I am assuming someone asking about this hasn't covered dimension theory - that's why I did it in more elementary terms

[dim widget]

This is a really fun one.

[novel parrot]

looks hard lol

[south patrol]

Lol TTEG any uh further hints for that galois q i was doing

I'm not sure how to check if a poly is irreducible using the Galois group hm

[novel parrot]

im guessing the preimages of the x_i will generate M

[lethal dune]

lifting ideps

will be similar

[south patrol]

I mean my idea was basically to consider that the polynomial I gave, if irreducible, has splitting field F_{p^{2^{k}}

And play around w that being impossible but hm

[slim kayak]

I tried to be elementary, just name dropped krull dimension

[south patrol]

As did i lol, i mentioned dimension theory

[dim widget]

So if x is a root then the galois group acts on 1, x, x^2, x^3, ..., x^{2^k -1}

[south patrol]

Anyway les continue owo

Sure

[dim widget]

We are assuming that Frob_p acting on this set has order 2^k otherwise the polynomial is not irreducible.

[south patrol]

I mean okay so F_{p^{2^k}} is generated bby the roots of x^{2^k} + 1

[slim kayak]

I wasn't saying you weren't, it's all good. Moving on.

[south patrol]

And Gal group is yes just generated by Fröb and of order Z/2^{k}Z

yh

[dim widget]

So what is the order of multiplication by p in Z/2^k?

[south patrol]

Oh okay sure

Hm

Wait okay I was gonna say has order 2^k right lol

But that is wrong ig

[dim widget]

Addition by p has order 2^k for sure.

[south patrol]

Lol

[dim widget]

if p is not 2.

But first question: is multiplication by p invertible in Z/2^k?

Second question, what is the size of Aut(Z/2^k)?

Third question: are there any order 2^k automorphisms of Z/2^k?

[south patrol]

Yes, 2^{k-1} and uh no

i believe

[fleet pelican]

wait uh isn't p equal to 1?

[south patrol]

👀

Where did that come from

[fleet pelican]

sorry im crazy

[delicate bloom]

1 is my fave prime

[dim widget]

The rare "(0) is an integral domain" supporter.

[fleet pelican]

i literally alqways confuse Z/n with F_n

[dim widget]

Stephen Crowder at a desk meme: 0 is an integral domain, change my mind.

[delicate bloom]

love its field of fractions

[lethal dune]

Best algebraic closure in class

[dim widget]

Actually that is an interesting question, is the algebraic closure of the field with one element just $\mathbb{N}$ as a "vector space"?

[south patrol]

Lol

[lethal dune]

Literally everything becomes algebraic

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

Anyway potato, yes a group of order 2^{k-1} has no elements of order 2^k

[lethal dune]

[south patrol]

I mean yes

When I said 2^k I meant like according to the work we've done thus far, not answering in an absolute sense lol

Lol idk why my brain is exploding over this question in particular, maybe I need a break lol

[dim widget]

Oh well the point is this means that on the field $F_p(x)$ Frobenius has order $\leq 2^{k-1}$, so $F_p(x) \subset F_{p^{2^{k-1}}}$, so $x$ cannot have a minimal polynomial of degree $2^k$.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

If you know Nakayama's lemma from class it is one line.

[south patrol]

Hm

[south patrol]

Okay so why is this true, like why did you omit x^{2^k} = -1

I'm not really sure what you are doing tbh I am confusion 😿

I am go eat dinners I think but will be bach thankies

[dim widget]

Oh whoops I should've said it acts on the set -1, x, x^2, ..., maybe that's what was confusing you?

[chilly ocean]

hi there, I need some help multiplying these two permutations. the result should be a product of disjoint cycles.

(1 2 4)(3 6 5) * (1 6 5)(2 3)

I started off with 1 which maps to 6, which then maps to 5
What do I try to map next, 2? or 6?

[dim widget]

You should just follow an element until you get back to where you started, then try an element which you haven't tried before, it is deterministic.

So after you try 1 -> 5, you should try 5 (which goes to 2) then keep going until you get back to 1.

[chilly ocean]

ahhh

I see

thank you!

[lethal dune]

Nice, i was thinking some bs

[dim widget]

I am a bit confused if this is what they want because in this form it essentially "is" nakayama's lemma. They may want them to actually prove it themselves.

[proper moth]

Say you have a group G, where |G| = mp (p is prime). And we want to show there exists a subgroup of G with order p. I suppose Lagrange is not sufficient for this statement? (I suppose Lagrange can rule out certain subgroups being possible, but can it prove a subgroup with a certain order definitely exists?)

[formal ermine]

it's sylow time baby

[dim widget]

There is a partial converse to Lagrange called Cauchy's theorem, and a stronger one called Sylow's theorems.

For what you want it's just Cauchy's theorem.

[agile burrow]

I wonder if there are some analogues of Sylow for infinite groups

[rustic crown]

i've heard there is a profinite generalization of sylow, but never looked what it was

[dim widget]

Since profinite groups are determined by their finite quotients it's just the obvious bootstrapping.

[rustic crown]

so is it just like, given any profinite, there is a maximal pro-p subgroup?

and you obtain it by applying sylow to all finite quotietns and taking the inverse limit again

[dim widget]

"Given an infinite group G, for every prime p G contains a subgroup of order p^\infty"

[rustic crown]

what if G was countable tho

[agile burrow]

I was just spitballing lol, I've never really thought about it

[dim widget]

All countable profinite groups are finite.

[agile burrow]

I should work with profinite groups sometime

[rustic crown]

no i mean, if G was countable then you can't have a subgroup of order p^infty

[rustic crown]

oh yea i've seen a cute proof for this. you put a finite Haar measure on it and then poof, it can't be countable

[dim widget]

G is a subgroup of order p^\infty unless its finite 🙂

[rustic crown]

p^infty is uncountable

[dim widget]

Ehhh?

In what way?

[rustic crown]

p^infty >= 2^N

(det was joking >.<)

[dim widget]

I was too well-vaccinated as a child to understand jokes

[south patrol]

Aww tysm x

Actually seeing p^infty reminds me of a paper I read part of

I'll try to find it lol

[south patrol]

Okay dinners helped with the Galois problem I was doing lol so uh @ topos lol

[chilly radish]

You can define p-groups as groups where all elements are of order p^n, and then you can still talk about Sylow subgroups in infinite groups as maximal p-groups

iirc this theory works well for nilpotent groups

[agile burrow]

Very neat

[south patrol]

I guess the point is like okay for the moment assume $p \ne 2$ and say $f = x^{2^{\ell}} + 1$ is irreducible and has splitting field $K$ over $F_p$ and let $\alpha$ be any root of $f$ in $K$. Well on the one hand, we know that $\alpha$ has order $2^{\ell+1}$ in $K^\times$, but we also know - from the Fröbenius etc - that the roots of $x^{2^{\ell}} + 1$ must be given by $\alpha, \alpha^p,\dots,\alpha^{p^{2^{\ell-1}$. This implies in particular that $p^{2^{\ell-1}} \not \equiv 1 \mod 2^{\ell + 1}$. This is absurd for $\ell \ge 2$ because $(\mathbb Z/2^{\ell+1})^\times$ is not cyclic.

[cloud walrus]

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

[south patrol]

Lol my apologies

And the p = 2 case should be easier i think

For example because every finite extension of a finite field is separable

So there are no irreducible inseparable polynomials over F_p I suppose

[dim widget]

I think the idea is right but Z/2^{l+1} has cyclic multiplicative group. Also I don’t understand the occurrences of l+1 instead of l.

[south patrol]

Well $\alpha^{2^{\ell}} = -1$, so $\alpha$ has order $2^{\ell+1}$. And huh, pretty sure $(\mathbb Z/2^{\ell+1})^\times$ does not have a cyclic multiplicative group for $\ell \ge 2$

[cloud walrus]

potato

[dim widget]

Ah true. It’s different for 2

Okay I see what you’re doing, good job!

[meager jolt]

How do I show that $\mathbb{Q}(\mu_m) \cap \mathbb{Q}(\mu_n) = \mathbb{Q}$ for $m,n$ coprime?

[cloud walrus]

Jeeves

[meager jolt]

where $\mu_n$ is the primitive nth root of unity ofc

[cloud walrus]

Jeeves

[lethal dune]

Some degree argument

[south patrol]

Oh cool, what was your idea?

[south patrol]

Oh lol I saw this on stack exchange

[formal ermine]

@south patrol what's your mse handle

[lethal dune]

planning to stalk

[south patrol]

Anyway, I'm p sure once you turn this into Galois theory this becomes CRT.

Unless I'm mistaken lol

[south patrol]

Uh should just be nuclear potato

[formal ermine]

how do you turn it into galois? fixed field?

[south patrol]

Considering them as subextensions of Q(ε_{mn}) and noting that an intersection of fields corresponds to the smallest subgroup containing both groups

||I believe basically Q(ε_m) and Q(ε_n) correspond to the subgroups (Z/m)^x x 1 and 1 x (Z/n)^x of (Z/m)^x x (Z/n)^x, from which the result follows quickly||

Perhaps I am making a dumb dumb though - i think it should be a bit more delicate but idk

[formal ermine]

nono I think that's right

[lethal dune]

you can also say [Q(α):Q] | m and |n so | 1 so for any alpha in the intersection, it must be of degree 1

[south patrol]

okay that is nice lol

Wait

Primitive root of order m has a degree φ(m) min poly tho, not m ryu, so i'm not sure how that'd work?

[lethal dune]

ah true

right cannot claim this

[formal ermine]

"without loss of generality, assume that phi(m) = m and phi(n) = n"
refuses to elaborate

[meager jolt]

Yeah if degree arguments like that worked I wouldn't be asking lol I hope

[dim widget]

What does work as @south patrol suggested is to look at the degree of Q(\mu_{nm})

[meager jolt]

Yeah I see 😦 I can't believe I missed that

[chilly radish]

||I think the groups should be backwards cuz you're looking at Gal(Q(e_nm)/Q(e_m)) which should be (Z/n)^x||

Not that it matters

[south patrol]

Yes true sorr6

[white oxide]

what was the purpose of the sentence before "Then"? (underlined in blue)

[agile burrow]

It's to show that the preimage of H contains N

[white oxide]

ah got it thx

[white oxide]

why is the join the smallest subgroup of G containing HN? if Ej is a subgroup containing HN, shouldn't it be smaller than the intersection of groups E1, E2, ... EN containing HN?

[agile burrow]

No, the intersection would be a subset of E_j

[white oxide]

hmm ok thank u

[white oxide]

what exactly do they mean by factoring the canonical map by a normal subgroup K of G?

(this is the third isomorphism theorem)

[south patrol]

That's what the diagram explains

Like you can split up the map into a composite

[lapis trail]

Anyone willing to help me out with tensors?

[novel parrot]

ask ur question

[lapis trail]

All I know about tensors is they are bilinear formssatisfy the universal property but the idea is too abstract for me to be able to answer a question like this

[formal ermine]

hint: C = R[x]/(x^2 + 1)

[lapis trail]

alrighty

[signal fractal]

How to show that if R is a commutative ring and $R[T]$ is the polynomial ring. $f=\sum_{i=0}^da_iT^i \in R[T]$ is a unit if and only if $a_0$ is a unit and $a_i$ is nilpotent for $i >0$. The tip I got on this exercise is to look at is to look at $g=\sum_{i=0}^eb_iT^i$ such that $fg=1$ and then prove $a_d^{k+1}b_{e-k}=0$ by induction. But I don’t really get how to get to the last equation and what it means. The only thing I know is that it probably relies on the fact that if a is a unite and x is nilpotent then a + x is also a unit.

[cloud walrus]

1000101

[south patrol]

Okay so like

One direction follows from the fact you mentioned at the bottom, pretty much immediately

For the other direction, well the hint is pretty inefficient

I think it's best to think about prime ideals - it gives a way cleaner solution

[signal fractal]

The easy direction is the one were I know that a_0 is a unit and the other coefficents are nilpotent and I just need to construct the g, right?

[south patrol]

That is the easy direction but yes it actually just follows from the fact

[signal fractal]

We haven’t talked about prime ideals yet 👍

[south patrol]

Oh okay lol

It's just that this is an exercise from Atiyah-Macdonald aha

[signal fractal]

It’s part of a linear algebra class and we just talked a little bit about ring theory

[signal fractal]

ah this is a famous book?

[signal fractal]

The problem I have is to show that it has to follow this property. Intuitively it’s very obvious, but it’s hard for me to get an idea how to formalize it.

[tribal moss]

Yeah, it's a classic graduate textbook in its subject.

[signal fractal]

Ah yes you’re right! Just looked at it it’s the second exercise. Now I’m interested to learn about prime ideals.

[dim widget]

I actually think it’s more instructive to write out the system of equations for an inverse personally! Fwiw

Because this same type of proof also shows up in other contexts

The “linear algebra” way is more flexible

[signal fractal]

I will try it

On a side note: Is it true that if $a,b,c \in R$ and $(a) \cap (b) = (c)$ that c is a lcm

[cloud walrus]

1000101

[dim widget]

The element c is the natural analogue of “lcm” even when such a thing may not exist.

In rings where an lcm does exist, it will be the element (c)

[lapis trail]

Is this why the bottom line is true?

[white oxide]

Thanks Gallian

[tribal moss]

That sounds a lot like Piet Hein. Is that poem properly attributed?

[white oxide]

nope

[tribal moss]

And yet the author probably expects his readers not to plagiarize stuff in their homework. Three cheers for double standards.

[barren sierra]

I'm lost about computing Ext and Tor

Ok so I want to compute $\text{Ext}_1^\mathbb{Z}(A, \mathbb{Z})$ where $A$ is a finite abelian group. So I think I start with a projective resolution of $\mathbb{Z}$ and then what?

[cloud walrus]

Spamakin🎷

[barren sierra]

I apply Hom(-, D)?

And then idk what to do from here

[next obsidian]

Ext is additive, so applying the classification you have to compute it for A = Z/nZ only

Now you have an easy resolution to use, now compute

Also, if you wanted to compute it via resolving Z you need to use injectives

You use projectives when resolving the first argument

[barren sierra]

wdym Ext is additive

[next obsidian]

It commutes with directs sums

Because Hom does

[barren sierra]

Oh

I see

And if I wanted to compute directly with Z, I use an injective resolution?

[next obsidian]

Yes, but use a free resolution of Z/nZ

There’s a stupid obvious one

[barren sierra]

0 -> Z - n -> Z -> Z/nZ -> 0 right?

I'm on phone (forgot laptop) so excuse that notation

[next obsidian]

Yeah

[barren sierra]

And then apply Hom and go from there? Lemme try that

[next obsidian]

I mean once you apply Hom you get a really funny thing

[barren sierra]

Uh

Is the answer 0?

Yes it took me this long but I double checked stuff and I'm hoping it's indeed 0

Hmmm maybe I fucked up

Ok yea I fucked up but I'm not sure how to compute the image of the pullback of multiplication by n on Hom(Z, Z)

is it just nHom(Z, Z)?

[prisma ibex]

yeah

[barren sierra]

Ok then I got it

[prisma ibex]

like the complex you end up getting is Z->Z with the map given by multiplication by n, so you should get Hom(Z/nZ,Z)=0 and Ext^1(Z/nZ,Z)=Z/nZ

[barren sierra]

I got 0 by accidentally assuming that applying Hom(-, Z) I'd maintain exactness and clearly no

[barren sierra]

Yea that's my homework, prove that's the case

But honestly all this homology stuff is brand brand new to me

So I'm still in "wtf even are these objects" mode

[prisma ibex]

yea

in general it can be pretty hard to do explicit computations with this complex or finding good resolutions

but the general recipe is the same

[barren sierra]

General recipe is still trippy

[prisma ibex]

for Ext it's kinda trippy because you can do a resolution on the right or on the left and it's annoying to remember which way things go

[barren sierra]

Wait you can do it in either side?

wut

[prisma ibex]

oh like if you're computing Ext^i(A,B)

you could take an injective resolution of B

and then apply Hom(A,-) to that

versus taking a projective resolution of A

and then apply Hom(-,B) to that

somehow the second of these is more natural or nicer in some ways, not least because projective resolutions are usually nicer to work with than injective resolutions

the fact that you can do both of these things and end up with the same answer is weird and not obvious

[barren sierra]

I may try to prove that

Probably a good exercise

And will let me wrap my head around things

[barren sierra]

For computing Tor^i(A, B)

Would it just be the dual?

Take projective resolution of B, apply A ⊗ -

Or equivalently

Take injective resolution of A, apply - ⊗ B

[prisma ibex]

well no, with Tor there's only one way

[barren sierra]

Aw

[prisma ibex]

tensor product commutes in a way that Hom doesn't

[barren sierra]

Ok I guess more homework, find a counterexample

Ok so I have to start with a projective resolution for Tor

[prisma ibex]

like if you're tensoring on the left versus on the right with a module it doesn't matter

versus if you Hom on the left versus right it changes the exactness properties among other things

and yeah for Tor you start with a projective resolution

of either A or B

so I guess there are two ways it's just they're both projective and now it's maybe a little clearer how things might be isomorphic

[barren sierra]

Tensor is additive right? So if I want to compute Tor^1(A, Z) for A finite abelian, I still only need to consider A = Z/nZ?

[prisma ibex]

yup

tfw every cohomology theory is just Ext in the right category

[barren sierra]

And then apply - ⊗ Z?

[prisma ibex]

yup, and then compute cohomology of the resulting complex

that's always the recipe, you resolve your object, discard the original object from the resolution, apply the functor, compute cohomology

omg so true

<@&268886789983436800>

lmao

[sand cradle]

agressive discussion of cohomology lmao

[frigid lark]

Forget to compute cohomology, compare answer with friend who also forgot to compute cohomology, panik

[fervent gyro]

Homological algebra is such an interesting topic. I'm currently soing some work trying to prove Serre's characterization of regular local rings using global dimension

[chilly ocean]

..............._
What's s(s with bar) in dihedral group

[lethal dune]

who knows

[chilly ocean]

grrj

[dim widget]

I think basically something incorrect! You really have to know that (Z/2^k)^* is not cyclic to solve this problem without really struggling.

[formal ermine]

fröbenius lol

[south patrol]

Ah ok

[south patrol]

What lol

[formal ermine]

mfw you say fröbenius but luroth

[south patrol]

Oh I didn't realise it was like

Lüroth

I was the cool kid who wrote Weierstraß and stuff

DER KRULLSCHE HAUPTIDEALSATZ

[dim widget]

German people parse umlauts like texit bot parses latex. If you put an umlaut in one part of a message and not another all of the text is indented with no spaces when they try to read it. It's actually quite hard on the eyes for them.

Pray for germans 🙏

[south patrol]

Ohvivmean I agree

Thoughts and prayers

[formal ermine]

and it's frobenius not fröbenius lol

[formal ermine]

do you know how to pronounce nullstellensatz?

[south patrol]

Oh lol didn't even realise

I think I mistakenly thought it was Fröbenius and then assumed people spelling it Frobenius were just lazy anglophones

[south patrol]

Nein. Die Wörter Null, Stellen und Satz kann ich zwar aussprechen, das Kompositum bringt mich aber zum Ausschreien, Weinen und Kotzen.

[novel parrot]

how can they assume that K is algebraically closed

[south patrol]

GCDs are unchanged under taking algebraic field extensions I'm p sure basically

[wooden ember]

Greatest common divisors don’t change across extensions

[novel parrot]

how does that help

[wooden ember]

Since you compute them by euclidian division

[wooden ember]

Well it’s a theorem about gcd’s

So we can compute that gcd in an algebraic closure without changing anything

[south patrol]

Or Euclidean division lol

[novel parrot]

How does that apply to this question though

[wooden ember]

I’m not sure this is the right way to show it I don’t see how you conclude in the end

Euclidian division is more transparent imo

[south patrol]

This is a general theorem about integral extensions

But yeah sure

[south patrol]

You can show the GCD in the alg closure is X-b

and then the GCD in the original field is also X-b

[wooden ember]

What’s the general theorem?

[novel parrot]

oh

[south patrol]

Wait now I'm paranoid I've made a mistake, I may have

[wooden ember]

I have a tendency to not see obvious things so dw

[south patrol]

Ah okay nah my theorem was mistated lol

[novel parrot]

how does this proof work though?

since we want to show that X -b is the GCD

X - c is another common factor, we must show that X -c | X - b

The way i understood

For finitely many lambda, X - b is not the GCD

since X - b and X - c is a factor

And that means for all but finitely many, X - b is the gcd

[warped viper]

Heyy! \

Here is my conclusion my paper on the ring of fractions. I'm not too confident with my words since it's a conclusion and not a proof, and especially in English. What do you think of it?\

(Let $A$ be a commutative ring and $\p$ be a prime ideal of $A$, let $A[(A\setminus\mathfrac{p})^{-1}]$ be the ring of fractions of $A$ over $A\setminus\mathfrac{p}$.)\

The fact that the ring of fractions of $A$ over $A\setminus\mathfrac{p}$ is local suggests that the information contained in this ring behaves particularly well around $\mathfrac{p}$. To be more specific, the concept of a local ring captures the idea that the elements of $A$ that are not invertible "behave" like zero. In the case of $A[(A\setminus\mathfrac{p})^{-1}]$, this means that the elements of $A$ that are not invertible in $A\setminus\mathfrac{p}$ are "localized" in $A[(A\setminus\mathfrac{p})^{-1}]$. We can see this like "smoothing" the behavior of $A$ near $\mathfrac{p}$, where the non-invertible elements of $A\setminus\mathfrac{p}$ are "approximated" to zero.\

This is the reason some sources call the ring of fractions the "localization".

[cloud walrus]

Overfull hbox, badness 1000
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[barren sierra]

Can someone help me see where I am going wrong with this. I want to show that for a finite abelian group $A$ we have that $\text{Tor}{1}^\mathbb{Z}(A, \mathbb{Q} / \mathbb{Z}) \simeq A$. Without loss of generality by classification we consider $A = \mathbb{Z} / n\mathbb{Z}$. So I have the following projective resolution
$$
0 \to \mathbb{Z} \xrightarrow{n} \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to 0
$$
and now I apply $- \otimes{\mathbb{Z}} \mathbb{Q} / \mathbb{Z}$ and recover
$$
0 \to \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Q} / \mathbb{Z} \xrightarrow{1 \otimes n} \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Q} / \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Q} / \mathbb{Z} \to 0.
$$
So now to compute $\text{Tor}_{1}^\mathbb{Z}(A, \mathbb{Q} / \mathbb{Z})$ I want to compute
$$
\text{ker}( 1 \otimes n) / \text{im}(0) \simeq \text{ker}( 1 \otimes n).
$$
And here I'm stuck. In theory my end goal suggests $\text{ker}( 1 \otimes n) = \mathbb{Z} / n\mathbb{Z}$. However the kernel of the identity map $1$ is $0$. So $\text{ker}( 1 \otimes n) = 0 \otimes$ something $= 0$ and so I don't see how I recover $\mathbb{Z}/n\mathbb{Z}$. I know I'm doing something wrong but I can't see what.

[cloud walrus]

Spamakin🎷

[rustic crown]

the map should be n⊗1, but that doesn't matter... 1⊗(1/n) would be in the kernel eitherway

[barren sierra]

oh oops yea that's backwards

[barren sierra]

wait wdym

[rustic crown]

1⊗(1/n) would be mapped to n⊗(1/n) = 1⊗1 = 1⊗0 = 0

[barren sierra]

right but I'm talking about the other direction

wait

[formal ermine]

ah rip

maybe this will help: https://youtu.be/t1CRjNriCR8

[barren sierra]

Oh I need to use the fact that for abelian groups $G$ we hvae $\mathbb{Z} \otimes G \simeq G$ right?

[cloud walrus]

Spamakin🎷

[rustic crown]

yea, that would simplify the complex

so you'll have the map Q/Z --> Q/Z given by multiplication by n

and by definition Tor_0 is coker and Tor_1 is kernel of it

[barren sierra]

right

so then I need to show ker of that is Z/nZ which is kinda obvious but idk how Z/nZ appears as a subsett of Q/Z formally

like the intuition is there for sure but I can't state it precisely which is bad

[formal ermine]

how would you even tensor non abelian groups

[barren sierra]

well I'm viewing them as Z modules

implicitly

you need a ring of some sort for tensors but that's neither here nor there right now

[rustic crown]

can you find an element in Q/Z that has (additive) order n?

[barren sierra]

1/n

Oh and that's a subgroup duh

[rustic crown]

yep, so <1/n> is the copy of Z/nZ that you're looking for

[barren sierra]

Tysm

[rustic crown]

[lapis trail]

Isn't the answer 2??

[rustic crown]

assuming by 1^(1/n) you mean zeta_n = exp(2 pi i/n), those two fields are equal. zeta_34 = - (zeta_17)^9

[south patrol]

Indeed the nth cyclotomic extension has degree phi(n) and phi(2n) = phi(2) phi(n) = phi(n) for n odd for example

[lapis trail]

Ah ok

I do recall that now that you mention it

[coral shale]

thats horrifying notation...

[lapis trail]

Yeah he's really something this professor 😂

[rotund aurora]

btw

I think Gauss actually uses the square root notation in modular arithmetic

like sqrt(-1) to denote an x st x^2=-1 mod something

and like roots of 1

[south patrol]

Based

[barren sierra]

That's awful

[south patrol]

Cyclotomic polynomials.

[rotund aurora]

Dont mind my weird pencil annotations

[south patrol]

Oh lol went down the rabbit hole looking at proofs of irreducibulity of cyclotomic polus

Weird

[south patrol]

Oof

[formal ermine]

we did that in alg 1

it was horrifying

[south patrol]

Which proof(s) did you do

I saw a weird proof which I really enjoyed but it's long if you write it all out properly and rigorously lol

[lethal dune]

I find the proof nice tho

reducing mod p and claiming simple root or something

[south patrol]

Ye

That seems the standard wa

[rotund aurora]

I remember it being pretty natural and nice

like its no joke

[white oxide]

my fault it is

lol

Piet Hein, "T. T. T.," Grooks (1966)

[formal ermine]

@south patrol

the reply didn't work

[solar shore]

wanted to let everyone know who helps me with questions here that i got a 97% on my last abstract algebra midterm

feelin good abt the final now

[south patrol]

Death by subscripts

[solar shore]

y'all goated fr

[formal ermine]

this is the proposition btw

@south patrol

the reply didn't work once again

F_0[x] = Z[x] => F_0 = Z confirmed

[south patrol]

Inch resting

Fr

Nah interesting lol

The way it's written reminds me of my German lecturer lol

The mod p stuff

Idk I haven't seen others write it like that

[low void]

http://math0.bnu.edu.cn/~liuym/Book-algebras-and-representations.pdf

Can someone help me understand p. 33, especially Definition 2.7 and Proposition 2.9?

I don't understand proposition 2.9 because in order to define a(v)=X·v we need to define ·, but · seems to be defined in terms of a, this seems circular.

I don't understand this :/

[white oxide]

Are the possible orders of the elements of A6 1, 2, 3, 4, and 5?

wait nvm

[chilly radish]

You are starting a priori with a K[x] module V, and you are defining a 'new' structure on it via definition 2.7, and you wanna show that this new structure is actually equivalent

xv is a prior defined in V

Alpha is just the multiplication by x map

[low void]

So we're basically not explicitly defining · here?

[chilly radish]

Yes

It's already defined

[low void]

Ah, I see. So Definition 2.7 is, at first, just an "example" of a K[X] module that later appears to characterize all K[X] modules?

Definition 2.7 says "That's a way to define a K[X] module"

Proposition 2.9 says "Actually, you can characterize all K[X] modules that way using alpha=Xv"

[low void]

Still a bit weird to me ... if we say that it is already defined, are we saying that it is defined in the manner of definition 2.7?

Like is there a ß: V->V somewhere in the background that we're not explicitly mentioning?

[chilly radish]

Well, technically every K[x] module structure can be obtained this way, but that's not really important

Well yea that's pretty much that

[low void]

"can be obtained in this way" means "there are others, but we can reduce them to this way"?

[chilly radish]

It means this is alwahs an equivalent definition

That's in fact what the proposition says

Like, the relevant homomorphism IS alpha in this case

That defines the module structure

The module structure is given to you from on high, but it can always be reduced to 'what is K doing to V' and 'what is X doing to V'

By linearity

[low void]

You mean what "K[X] is doing to V"?

[chilly radish]

And the question 'what is X doing to V' is answered by 'it acts as a K-module homomorphism'

No I mean K separately and X separately

Like, the action of K[X] on V is defined by what is x . v and what elements of K do to an element v

[low void]

Hmm

[chilly radish]

I feel like my answers are just confusing you further, i'm not sure how to phrase this in the best way for you to understand, sorry

[low void]

If we know how g acts on v, it is no suprise we know how X acts on v. The converse is the interesting part.

[chilly radish]

Well, every polynomial is just a sum of elements of a_ix^i right

[low void]

That we know from the way X acts on v, we know how any g acts on v

[chilly radish]

If we know how x acts once then we know how ot acts i times as well

And we know that the module action is distributive

[chilly radish]

So if we knoe how K acts and how X acts, you knoe how X^i acts, and so you know how a sum of a_iX^i acts

Is what I mean

[low void]

If we're saying that V is a K[X]-module, we know there is a ·: K[X]xV-->V which fulfills the necessary axioms. And, in proposition 2.9, I should not confuse this · with the Definition 2.7 where we say that g · v:=g(alpha)(v). Is that correct?

Because otherweise, if we set alpha(v)=X·v, we'd have alpha(v)=X·v=g(alpha)(v) where g=X.

[chilly radish]

Yes

You don't know how \cdot acts in this case

But you can think of using alpha to define a NEW action

g*v= g(alpha)(v)

[low void]

Okay, thanks, that's what was confusing me

[chilly radish]

And then showing that in fact g*v=g.v

Where . Is your original action

[low void]

I think I understand

[chilly radish]

Great!

[low void]

I feel like this presentation is kind of misleading however

[chilly radish]

In what way

[low void]

Because of this

But maybe I am the problem 🤣

@chilly radish To sum up, we have to multiplications:
(i) g·v=g(alpha)(v)) as in Definition 2.7
(ii) * which is not further defined and associated with the K[X] module V in Prop. 2.9

Next, we're trying to show that V=V_alpha by using alpha(v)=X*v (* is the multiplication in (ii)).
And then, we're relating (i) to (ii). g*v=g·v?

g·v:=g(alpha)(v)

[chilly radish]

Yes

Pretty much

[low void]

HOW is this obvious 🫠

But I believe I understand now ...

"It reams to check that for any polynomial g we have g*v=g·v"

[chilly radish]

Yes

[low void]

Thank you! @chilly radish

[chilly radish]

You're very welcome

[formal ermine]

a module is projective iff every ses splits iff left morphism is an exact functor

is there any nice condition like the middle one for an injective module too?

like a module is injective iff right morphism is an exact functor

but I'm looking for one similar to that every ses splits

[rustic crown]

what does it mean that left morphism is an exact functor?

[formal ermine]

the R-mod P is projective iff Hom_R(P, \cdot) is an exact functor

[rustic crown]

P is projective iff 0 --> M --> N --> P --> 0 splits
similarly
Q is injective iff 0 --> Q --> M --> N --> 0 splits

[formal ermine]

thanks

[agile burrow]

You should prove this because it's very similar to the proof of the definitions of projective being equivalent

[rustic crown]

didn't know this was called left-morphism

[formal ermine]

I made that term up

lol

[rustic crown]

i just call it covariant hom :p

[formal ermine]

ah right makes sense

[formal ermine]

will do

[agile burrow]

det did you learn spectral sequences

[rustic crown]

only read the intro in vakil

which had no proofs

[agile burrow]

[rustic crown]

.<

[agile burrow]

I may work through the last section of aluffi and supplement with vakil

But really I think I just need to do lots of computations

[rustic crown]

det tried reading some AG today

me ded

its so hard to read a book from somewhere in the middle >.<

[formal ermine]

what book did you read

[rustic crown]

hartshorne

[rotund aurora]

does aluffi make any sense if you dont know algebraic topology

?

[rustic crown]

wanted to read about surfaces for a talk i need to give

[rustic crown]

yea fully.

[agile burrow]

Yeah, aluffi is entirely self-contained

[slate tide]

It's honestly so worth to learn as many different spectral sequences as possible to play around with, cause some contexts are easier than others

[rotund aurora]

yes, but for the sequences stuff and categorical stuff, would it make sense?

idk just curious

[agile burrow]

Yes, I learned all of the homological algebra I know from the last chapter of aluffi

[rotund aurora]

but whats the point of homological and cohomological stuff

[agile burrow]

Makes sense, the only context I've seen them applied in is alg top for some homotopy group computations but I should diversify

[rotund aurora]

doesnt this come from like differential forms for example, and serre duality?

[agile burrow]

I mean the main build up of the last chapter is proving existence of derived functors and all of the machinery behind that

But once you have the tools it opens up a lot of other areas

And you can talk about those in a purely algebraic context, though motivating them might be more natural from topological settings and stuff as you suggest

[rotund aurora]

so the applications things he talks about here are like topological and geometric things?

(this is the first paragraph of the homo chapter)

[agile burrow]

Yeah, lots of applications to topology, geometry, and number theory to name a few off the top of my head

[rotund aurora]

where does homo and cohomo things show up in NT?

[agile burrow]

Galois cohomology seems to be a popular phrase

[rotund aurora]

ahhh right

I think Ill wait to study these things

first I wanna see them in nature 🏞️

[agile burrow]

Fair enough

[slate tide]

I would totally look at the Atiyah-Hirzebruch SS, the K-Theory calculations are really good for getting comfortable. It's also not too bad to show cool results like how rational K-theory and even degree cohomology with rational coefficients are isomorphic via the chern character or something. But other than that, the Serre SS, the Lyndon-Hochschild-Serre SS, and the Cartan-Leray SS for a cover are so nice too.

[agile burrow]

Ah yeah, I've done a bit of lyndon-hochschild-serre so I'll review that

But thanks, I'll try to get around to these

[slate tide]

Yeah Caran-Leray is a slight generalization where you mix group cohomology with a space that the group acts on to compute quotient spaces

[south patrol]

AHSS for K theory hehe

Compoot K theory of RP^n/RP^m I dare you

[slate tide]

Okay LOL

I know you need to abuse d_3 being stable to deal with that, but still lol

[formal ermine]

stupid question, but what do they mean with "replacing M by 0"?

[agile burrow]

Quite literally just taking the projective resolution of M and placing a 0 where the M is

[lethal dune]

[formal ermine]

how is H_0(P) = M if we're removing M from the sequence

[agile burrow]

In the projective resolution, exactness implies that M is the cokernel of P_1 -> P_0

When we replace M with 0, H_0 of the resulting sequence is P_0 / im(P_1 -> P_0), where P_0 is the kernel of P_0 -> 0

and the quotient is precisely the cokernel of P_1 -> P_0

[formal ermine]

I'm confoosed, do the morphisms stay the same?

[agile burrow]

Yes, all the other morphisms are the same

[formal ermine]

except the one from P_0 to M?

ah

yeah ok I see where I went wrong

I thought the sequence should still be exact

thanks walter

[south patrol]

Lol yeah was a joke, the only way I know to compute it is comparing with CP^n/CP^m and stuff

[agile burrow]

np

[formal ermine]

fuck I ran out of paper

[south patrol]

Idk any non-trivial AHSS computations besides that though eek, isn't it rly rough in the real case for example

[lethal dune]

AHSS abbreviation for?

[slate tide]

Yeah it is rough for the real case but you can still get information in certain degrees for relatively simple spaces (speaking from what I've done which isn't very much). And yeah even in the complex case it's still hard for spaces that aren't like spheres or CP^n's (but low dimensional manifolds are plausible to me), but I think these still give a feel for it at least.

[slate tide]

Atiyah-Hirzebruch spectral sequence

[lethal dune]

[slate tide]

I definitely saw an MO post about computing some stable homotopy groups of RP^2 using AHSS which was cool

[lethal dune]

[slate tide]

And I guess just pulling up a list of cohomology theories gives you a lot of examples to play around with

You can show how oriented cobordism agrees with singular cohomology in low dimensions for example, which is pretty cool IMO. It just gives you a context to work with spectral sequences versus just running something like a base-change Ext SS or really any Grothendieck SS (except if you do algebraic geometry or something like that)

[hollow mica]

Tensors are useful objects in their own right (differential forms), but I've often heard in their algebraic study something along the lines of "tensors [or more specifically the universal property] allow[s] us to convert problems about multilinear maps into problems about linear maps, i.e. we can use linear algebra." What is an example of a problem about multilinear maps that can be solved in this fashion?

[prisma ibex]

maybe it's worth mentioning that this passage is almost certainly referring to the universal property that tensor products satisfy: it literally turns multilinear maps into linear maps in this sense

[dim widget]

I’m not sure about “when it’s useful.” Almost any proof using linear algebra could be translated into multi linear algebra in one line. But it puts everything on the same footing which is pleasant

[prisma ibex]

it's just meant to say that any time you have a bilinear map out of a product, you can turn it into a linear map out of a tensor product. Linear algebra is about vector spaces and linear maps between them, so by doing this you've brought multilinear maps back into the natural setting for linear algebra

you can come up with particular examples for how this would be handy, so long as you can come up with particular examples of bilinear maps that you care about

[delicate orchid]

tensor products are essential to the existence of blackholes :pack:

she doesn't know

[formal ermine]

unaware

[chilly ocean]

clueless

[south patrol]

Hm so I was wondering smth. The way the lecture notes I was following constructs induced reps as follows: let $H \sub G$ be a subgroup and $W$ a $k$-linear rep of $H$. Then we form the tensor product $k[G] \otimes_k W$ of $k$-modules, turn it into a $(kG,kH)$-bimodule by $g \cdot (x \otimes w) = gx \otimes w$ and $h \cdot (x \otimes w) = xh^{-1} \otimes h \cdot w$ and then take the $H$-coinvariants

[formal ermine]

topos, I can't believe you don't already know this at (insert age here). aren't you ashamed of yourself?

[cloud walrus]

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

[south patrol]

How does this fit in with more general theory? i can't seem to find much about this construction at all...

It seems a little perverse to tensor over k and then add structure for example

[formal ermine]

hi mrean

[south patrol]

But idk if this is a special case of some more general construction that I've not seen

[dim widget]

Wouldn’t it make more sense to mention entanglement if you were going to talk about something physical which needs tensor products?

[delicate orchid]

she doesn't know

[formal ermine]

entanglement is an immediate consequence of what wew said

[prisma ibex]

maybe this is just saying that this tensor product naturally carries this structure, which is fine

[delicate orchid]

stop taking anything I say seriously I'm the court jester

[south patrol]

hm ok

[prisma ibex]

I mean somehow one part of what is happening that is quite natural is like

[south patrol]

I realise now that basically what he's doing is like

[hollow mica]

can I see one / a sketch perhaps

[south patrol]

In a sense he's just splitting up the "normal" construction of a tensor product over a non-comm ring into first quotientien out stuff to make everything k-linear and then quotienting out even more stuff

[prisma ibex]

like the tensor product is the natural way to go from k[H]-modules to k[G]-modules, but then if H is already a subgroup of G it's maybe not so surprising that you should be able to extract some other k[H]-module structure from this new k[G]-module

[south patrol]

Yeah sure

Weird I can't find much on this online though

[dim widget]

I think it’s just the usual definition of induction via tensor products as in eg Fulton and Harris but phrased in a slightly less common way.

[prisma ibex]

there's a tiny bit about this written in Remark 1.14 of Bellovin's notes: https://rmbellovin.github.io/teaching/induced.pdf

[south patrol]

hm how so

[prisma ibex]

appears right after the more standard construction

[south patrol]

I mean all the constructions are equivalent ultimately ofc so like

[dim widget]

They just break it up into tensor product, then push out

[south patrol]

But yeah I guess sure the H-coinvariants should be thought of as corresponding to orbits under the action of H on G

[rustic crown]

[dim widget]

I mean fundamentally quotient if by that left action of C[H] is what it means to take \otimes_C[H] so it’s just the usual definition

[south patrol]

Indeed

[dim widget]

Wait so what was the question?

[south patrol]

Hm that's not quite this though lol

Wait hm

Idk it was more like has anyone seen this kinda construction / is it a special case of smth more general

And the answer sort of seems to be know lol, like it's close to but not the same lol

[formal ermine]

sorry for the stupid question but I don't quite get where we get the cochain complex from

[south patrol]

This is confusing me though like why opt for smth different to what books seem to do typically xd but fair

[delicate orchid]

it says directly below the complex

[prisma ibex]

yeah I mean the definitions are equivalent, I don't know that I've really seen this particular phrasing of it that much

but they're only a few steps removed from each other

[dim widget]

It is, if you have a right space M for a group and a left space N you can take a space M x^G N which is the push out of the two spaces along this action

That’s a very general construction

[south patrol]

Yup sure

[formal ermine]

but P is ... -> P_2 -> P_1 -> P_0 -> 0, no?

[delicate orchid]

that functor is contravariant is it not

[south patrol]

Anyway thank i am ok now

lol

[formal ermine]

oh I'm stoopid

the N is on the right

sorry

[south patrol]

I need to find some good examples of induced reps though because I have literally seen a single example where they are useful for computations aha

[delicate orchid]

apology not accepted.... u will be fined

[delicate orchid]

wait, really?

[south patrol]

I know ultimately they are useful theoretically e.g. Brauer induction theorem and I've used Frobenius reciprocity

Yeah Wew lol

[delicate orchid]

have you never computed one?

[south patrol]

Well I guess as an exercise like once

[delicate orchid]

this is the problem with u lalaland mathematicians u don't ever actually DO anything u jsut talk shite

[south patrol]

But for actually computing character tables I usually just use orthogonality and things lol

and that's what everyone seems to do idk

[agile burrow]

Lalaland mathematician

[south patrol]

Or like inflate characters

[delicate orchid]

yeah lifting is more normal but sometimes the group is simple!!

[south patrol]

more normal xd

[slate tide]

Lalalangland mathematician

[delicate orchid]

xdxddxxdfxdxddxxdxddx

[south patrol]

Hm are there any good examples to do stuff with

like where induction helps you find new reps

[delicate orchid]

induce from a normal subgroup see what the affect of tensoring with C[G] IS

[dim widget]

Maybe try inducing a character from an index two subgroup?

[south patrol]

Yeah I imagine that is good but isn't that pretty special but yeah

[delicate orchid]

wow it's what I said but cooler... wow.... ok.... yeah wow....

[south patrol]

Stuff like D_n works well I guess

Maybe I just don't have enough groups in my mind lol

[dim widget]

The galois representations of Cm elliptic curves defined over Q are of that form

It’s useful to recognize

[south patrol]

actually imma read more fulton-harris

saur good

[south patrol]

Oh interesting

[prisma ibex]

automorphic induction of Hecke characters

[delicate orchid]

hmmmmmmm index p subgroup.... how delectable my dear...

[formal ermine]

is this typo? should this say nth cohomology group?

[dim widget]

Yes, depending on their definitions. The most natural thing to call it would be cohomology.

[formal ermine]

okay thanks

thought I was going crazy

[dim widget]

I mean it’s just an indexing thing, it would be just as logical to call it the H_{-n} of some complex in negative degrees 🙂

[next obsidian]

Why is f(1) = 1 lol?

[hollow mica]

Is this a valid proof that there exists no isomorphism between (Q, +) and (Q, *):
Let f: (Q, +) -> (Q, *) be an isomorphism. Then necessarily f(1) = 1so f(2) = f(1 + 1) = f(1) * f(1) = 1, contradiction.

[next obsidian]

The identity of Q under addition is 0

Anyway, consider what would need to map to -1 inside of (Q,*). Also, I think you want (Q\{0},*) if you want both to be groups

[hollow mica]

Oh oops

no yea I got the -1 proof I was just wondering if this was a simpler proof

[next obsidian]

Ah

Well, no

Haha

[hollow mica]

yeah rip

[next obsidian]

I think this is really the simplest proof

You don’t even need to write anything down

Just notice that (Q,+) has no finite order elements other than 0

But (Q\{0},*) has an order 2 element

[hollow mica]

Right

[static needle]

How about H^n (X)? Isnt that one the most widely adopted notation?

[lapis trail]

Will someone explain this to me?

[delicate bloom]

What have you tried?

[lapis trail]

I'm studying up on the frobenius automorphism now

[lapis trail]

Gal(F_{2^10}/F_2) is generated by f(a)=a^2 I think. But I don't know what to do with that here

[white oxide]

what is the use of the first isomorphism theorem? i haven't seen it used anywhere else in my book (at least the factoring part; G/H iso to image of G is everywhere)

by factoring part i mean in that a homomorphism from G to G' can be factored into mu gamma where gamma is the canonical homomorphism from G to G/K

[white oxide]

uh

yes

that's what it is

oh shit

my fault you're gonna say something else

my bad

[summer path]

it's a very commonly used tool in showing things are isomorphic, i think it's probably best to see how useful it is by doing some problems

also what textbook are you using?

[white oxide]

hm ok that makes sense

[white oxide]

fraleigh

[summer path]

hmm i remembering using that textbook before, but i forgot for what and im not too familiar with its exercises :(

[white oxide]

aww yeah that's fair, it's very gentle and the isomorphism theorems are in the advanced group theory section

but thank u tho

[summer path]

i guess if you wanted to, you could prove the second, third, fourth iso theorems as exercises; idk which ones are which for you, since every book seems to have a different permutation of them lol

[next obsidian]

This is like, the most useful result in all of algebra

You quickly completely avoid writing out “by the first isomorphism theorem” because you use it so often and it becomes so engrained in your mind

I forget this is a theorem sometimes

[summer path]

although if you really want to write something, just write FIT, otherwise it's just too many characters lol

[next obsidian]

I would have no idea what that means, it is clearer to say nothing

[summer path]

i meant that usually when people are learning it for the first time, on homework people feel the need to say they used theorem xyz, but I have often seen "First Isomorphism Theorem" abbreviated at F.I.T. in some classes

[rustic crown]

strictly speaking, the factoring part isn't the first iso theorem. it's the universal property of quotients more specifically, but they're pretty much the same so doesn't matter. In particular, the universal property characterizes quotients, and so you can think of it like an equivalent definition, therefore it's super hard to not use it lmao. In the start you don't think about it much because all the maps you define is done explicitly via elements, but after a while, you would just find it easier to define the map in a few steps by using universal properties alone.

[dawn latch]

the distinct left cosets of 9ℤ in 3ℤ are 3+9ℤ and 6+9ℤ. do i include 9ℤ itself?

[sonic coral]

a set is always a coset of itself

[frigid lark]

Or just not given a name in Lang

But we do have a Tate theorem (and conjecture)

And several proofs credited to Tits

[smoky ivy]

this is actually quite cool, i never thought of the first isomorphism theorem categorically

[rotund aurora]

Andrew Tate ???

[dawn latch]

how do i find the cosets when the group is finite? (in a cayley table)

[frigid lark]

No, just A Tate

[dawn latch]

can someone give me a starting point?

[frigid lark]

We know that if a,b are in the same left coset of H, say aH, then a and b differ by an element in H, say a=bh for some h in H. Also we know that the size of any two cosets are the same

Also it should be obvious since H is a group that H is so a coset of H

[dawn latch]

ok, i get this part.

[frigid lark]

The first statement is also an equivalence class

[dawn latch]

hmm ok, for instance is wH a coset of H?

[frigid lark]

Sure

[blissful crystal]

but also notice that w is in H

[frigid lark]

And it is equivalent to H

[dawn latch]

ohhhh

that makes sense

so it's basically, elements that are not in H * set of H? (* is the binary operation)

sorry if obv question

[frigid lark]

But then you can see that xH and yH are the same

[dawn latch]

that's true... but it didn't say that i have to provide the distinct cosets right? (or am i just wrong)

[frigid lark]

Explicitly xH={xh | h in H}

[frigid lark]

At your level for understanding, I would write out the entire sets

[dawn latch]

ok, noted

[static temple]

(btw side note: this group is of card 6, there are only two groups with this cards: Z/6Z and S3. You see that the tab isn't symetrical thus the operator is not commutative: it's thus S3. You can explicitely find the x, y... that make the correspondance with S3, and the correspondance is not unique :DDD.||you can also see that the number of correspondance is the number of ways to fix the transpositions, which is 3!=6. well, Aut(S_3) = S3, which is quite beautiful :>>>|| )

[blissful crystal]

unsure if i should ask this here or in #algebraic-geometry but now that school is done for the year, im planning to attempt going through atiyah and macdonald. I dont expect to work through all chapters, or finish all problems in a given chapter. For people who have done a chunk of that book though; is there anything i should know before i get into this? any important stuff the book doesn't cover? honestly anything about this endeavor is appreciated.

[ivory trail]

this is an incomplete answer but i liked eisenbud's exposition better, but you can probably use both (what i did) or use one to supplement the other

[quartz quiver]

https://math.stackexchange.com/a/626429/834966 I'm trying to follow this hint but am not sure how to prove that ${1} \trianglelefteq H_1 \trianglelefteq \ldots \trianglelefteq H_k$ is a central series. I get that $H_{i+1} / H_i \cong (H_{i+1}/Z(G)) / (H_i / Z(G))$ but this isn't enough, right?

[cloud walrus]

Darylgolden

[blissful crystal]

thanks

i think i looked through all the preface/intro stuff in eisenbud a few months ago, but i thought that it was too beyond me

i guess that might have changed recently

[rustic crown]

[dim widget]

Indeed there's a whole category theoretic notion called a "regular category" which is essentially a category in which a generalization of the 1st iso theorem holds.

[formal ermine]

english language and their numbers

1st iso in german is called homomorphism theorem

and 2nd/3rd are called noetherian isomorphism theorems

[south patrol]

Based

I call them "fundamental theorem of algebra", "weird lattice boy" and "cancel fractions"

[formal ermine]

yeah but like fr the last one is just common sense

ong why would you make it look like a fraction if it doesn't behave like one

[dim widget]

I agree, oh ny god what is wrong with anglophones?

[frigid lark]

By FTA you meant trivial corollary of Liouville's Theorem right?

[summer path]

legendre symbol

[south patrol]

Nah I'm joking like first isomorphism theorem

FTA's Galois theory proof is cutest anyway imo

[frigid lark]

It is

Just saw it yesterday

And then said I ain't doing Canal ever again

[frigid lark]

Wait, then this means that the SIT and the TIT are trivial corollaries of a trivial corollary of Liouville's Theorem

[warm crow]

Here, R_p = { a/b | a is in R, b is not in P }, right?

[rotund aurora]

Let G be an abelian group of order n, and define a function f by f(m)= # of elements in G of order m

you can recover G from f obviously

but I wanted to write an algorithm to do that, and all Im thinking of is very "brute force"

so idk if there would be a nice way to do that

cuz first, Id obtain the prime factorization of n

then you do something

and it suffices to implement the algorithm for the case when n is a prime power

like you could try every single possibility lmao

nvm I think this might be simpler

you look at the highest order, say its p^k. Then you know G=(Z/p^kZ)^r*H. Say the highest order in H is p^q, then to find r you want to look at how many elements of order p^q are there

and proceed

[frigid lark]

F(p)>1, F(2)>1

Oh and F(n)=0 for n > p

[formal ermine]

I thought this was only true if G is simple

[rotund aurora]

G is abelian

[frigid lark]

Oh, is it finite?

[rotund aurora]

ah yeye

[dim widget]

n < \infty

[rotund aurora]

I actually wonder like

if you look at this function f for an arbitrary group (not necessarily abelian) of fixed order, how many isomorphism classes are there with the same f

like I dont think G can be recovered from f in general

[frigid lark]

There are some non abelian cases

[dim widget]

I don't know if there is going to be a satisfying answer for this, even with just two composition factors it's very complex.

[frigid lark]

Or are there?

[frigid lark]

Does this give you S_p?

Cause F(2) isn't necessarily a transposition

Oh, and I'm already assuming it embeds into an appropriate symmetric group

[rotund aurora]

it would be funny to compute f for S_n

[dim widget]

What do you mean by F(p) > 1 and F(2) > 1? Those are very loose conditions and it's not the kind of detail that @rotund aurora was asking for

[south patrol]

I think I am going insane with some Galois theory lol

I shall ask in a second lol

[rotund aurora]

wait I think you could do that, or something close

[dim widget]

It's a pretty simple exercise to do so.

[rotund aurora]

yeh think I did something similar some time ago

[frigid lark]

I'm trying to guarantee a transposition and a p cycle.

But yeah, I think this is way too loose to do anything with

[south patrol]

Okay basically what I was confusion about was like uh

So let's take $\alpha = \sqrt{1 + \sqrt{7}}$ and consider the normal closure of $K:=\mathbb Q(\alpha)$. P sure this is just $L:= \mathbb Q(\alpha,\beta)$ where $\beta = \sqrt{1 - \sqrt{7}}$, and that we have extensions $L/K/\mathbb Q(\sqrt{7})/\mathbb Q$ all of degree 2

[cloud walrus]

potato

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

[south patrol]

Now I want to say that $\mathrm{Gal}(L/\mathbb Q) \simeq \mathbb Z/2 \times \mathbb Z/2 \times \mathbb Z/2$, because we can basically send $\sqrt{7} \mapsto \pm \sqrt{7}$, then map $\alpha \mapsto \pm \alpha$ and then $\beta \mapsto \pm \beta$, breaking it up as I did into extensions of degree 2

[cloud walrus]

potato

[south patrol]

But this seems wrong, since then there are 4 elements fixing beta, even though Q(beta) should be a degree 4 extension of Q I guess

Perhaps I am just messing up part of the correspondence lol

[frigid lark]

Don't you have some autos that send alpha to beta?

[south patrol]

Yeah agreed since they are roots of the same irreducible poly

But now I am confused what was wrong with the reasoning above with extending isomorphisms progressively

[dim widget]

Wait what are you trying to do? Compute the Galois group of Split(Q(sqrt(1 + sqrt(7)))?

I missed the question.

[south patrol]

Uh assuming split means like the normal closure yes

:)

[dim widget]

Okay so why would it be Z/2 \times Z/2 \times Z/2?

[south patrol]

Okay so my reasoning was you can use this lemma on extending isomorphisms where like

Given some (separable) extension F(a)/F, you can extend any automorphism of F to one of F(a) by sending a to any root of the min poly of a over F

And so I was like well I can take automorphisms of Q(sqrt(7)) (send sqrt(7) to itself or its negative), then do the same with Q( sqrt(1 + sqrt(7)) and then with that whole splitting field thing

And that would give me 8 = 2^3 different auotmorphisms, all of which are all of order 2 and commute so that we would wind up with Z/2 x Z/2 x Z/2

[frigid lark]

Are you sure this is how it goes, cause you have an automorphism of f, which is not necessarily the identity, then your automorphism doesn't preserve Irr(F, alpha)

[dim widget]

So you have these extensions Q(sqrt(1 + sqrt(7)))/Q(sqrt(7)), what effect does conjugation on the bottom extension have on the minimal poly of the top extension?

[south patrol]

What exactly do you mean by conjugation here?

[dim widget]

Like the unique nontrivial element of Galois

[south patrol]

Which seems to be what I said

Hm

[frigid lark]

Ahh, isomorphism, not auto

[south patrol]

Yeah that's more general but here they were automorphisms anyway

[dim widget]

Yes but the key condition is about the effect of \phi on the minimal polynomial

[south patrol]

Uh oh

Okay yes that is the thing I was being dumb about, thank you for nipping this in the bud lol

Lol

Okay!

Thank you Topos_Theory_E-Girl and Parrot Tea

[dim widget]

This looks like a good exercise! A lot of people kind of neglect these explicit galois theory exercises.

[south patrol]

Any tips on being a Parrot and/or an E-girl

[frigid lark]

I need to stop saying stuff here

I don't help

[south patrol]

I'm rusty with Galois lol haven't done enough compuations in a while

[formal ermine]

can someone pls clarify what this is asking

[south patrol]

Nah it was foine Parrot dw, I mean you made an excellent correction

[dim widget]

Set up your patreon early!

[south patrol]

Aha what

Nice.

[frigid lark]

Don't

[south patrol]

Actually yeah so it'd be easiest to compute the Galois group, presumably, by just sending alpha to any other root and then sending beta to like lol where it has to go

[delicate orchid]

The two sequences aren’t isomorphic

[south patrol]

Oh, I see where my issue is coming from lol

[agile burrow]

Two extensions are said to be equivalent if there exists a morphism between the middle groups such that if you put an identity morphism on the left and right, the whole thing commutes

[south patrol]

Almost all of the Galois theory computations I've done have had minimal polynomials unchanged by the other elements of the Galois group

[formal ermine]

thanks

[south patrol]

Stuff like Q(sqrt(2), sqrt(3))

[dim widget]

They're definitely isomorphic! Just not equivalent.

[delicate orchid]

Same thing

I’ve never heard the term “equivalent” tbh

[dim widget]

It's important! It means that the maps on either side are the identity map.

[delicate orchid]

Why is a backwards implication compared to categories that’s bad

[dim widget]

Rather than arbitrary isomorphisms.

[delicate orchid]

I believe this should be changed immediately

[delicate orchid]

Nvm I have but only for group extensions- now it makes sense

[south patrol]

Oh nice so like

The Galois group is actually non commutative

So in fact centre size 2 lol

nice

[formal ermine]

I am trying to show that Ext^1_Z(Q, Z) = 0. for this I need to show that Q is not a projective Z mod, so I tried finding a ses that doesn't split. e.g. 0 -> Z -> Q(mu_1, mu_2, ...) -> Q -> 0. now I'm stuck trying to show that Z oplus Q isn't isomorphic to Q(...)

[agile burrow]

I'm confused

I'm pretty sure the Ext group is nonzero

[formal ermine]

but Q is not a projective Z mod

[agile burrow]

Correct

If it were projective, then the Ext group would be zero

[next obsidian]

I’m confused

On multiple levels

[formal ermine]

oh lmao

I got the iff wrong

[agile burrow]

There are easier ways to see Q is not a projective Z module than finding exact sequences which don't split, but do whichever you feel most comfortable with

[formal ermine]

Q is not projective because Q oplus Q is isomorphic Q which isn't free?

[agile burrow]

I mean sure, but also you could just say Q isn't free lol

Projectives over a PID are free

[formal ermine]

ah

didn't know that

[agile burrow]

Good exercise