groups-rings-fields

[chilly ocean]

That's what I would do, throw in some elements of order 2 or 4 and hope they don't make any elements of "bad" orders

I'm not sure if there's a better way

Maybe a classification of groups of order 8 is useful

Off the top of my head my guess is they are Z8, Z4xZ2, Z2xZ2xZ2, and D8

[unique juniper]

i see

[chilly ocean]

Can't be Z8 obviously, not sure about the other ones.

I want to say I have done this problem before and that the subgroup is D8, but this is just off some shitty memory

Either way I think it shouldn't be too bad to actually work through it

[unique juniper]

yeah

but i cant work through larger groups

[chilly ocean]

Hm yeah, maybe it gets trickier/more annoying

[unique juniper]

thanks for helping

[unique juniper]

$G$ is a finite group, $N$ is a normals subgroup and $H \in Syl_p(G)$ show $\frac{HN}{N} \in Syl_p(G/N)$

[cloud walrus]

Yes

[unique juniper]

HN is a subgroup of G

|HN|/|N| is p^k

and er, HN must be a subgroup of G that contains N which it does

so HN/N is a subgroup of G/N

p^k divides (p^k r)/x

so must be a sylow p subgroup

this makes sense to me but i dont think its correct :/

anyone can tell me if correct or incorrect pls

[chilly ocean]

Something like that, but what if the order of N has a factor of (a power of) p?

But overall, order considerations are sufficient for this problem, yeah

[unique juniper]

that would be a contradiction?

because HN/N is a subgroup of G/N

so HN/N must divide G/N ?

and if x did divide p^k

then p^k cant divide |G/N| ... right?

[chilly ocean]

Not that x divides p^k, but it has some factors of p

[unique juniper]

but then if it had factors of p

it would divide p^k ?

i dont think i understood

what you mean

[chilly ocean]

I guess this is still unclear to me, what are r and x, actually?

[unique juniper]

$\frac{|G|}{|N|} = \frac{p_1^{k_1} p_2^{k_2}\dots}{x}$

[cloud walrus]

Yes

[unique juniper]

only focus on arbitrary p and combine everything else into r

[chilly ocean]

So x is the order of N and p^k r is the order of G? (And r has no factor of p?)

[unique juniper]

yes

yes

[chilly ocean]

But then why is the order of HN/N order p^k?

[unique juniper]

oof

yeah its not :/ mb

its actually $|H|/|H \cap N|$

which makes me stuck i guess

[cloud walrus]

Yes

[chilly ocean]

I think it's just a matter of writing the order of N as p^b * x or something

Including the number of factors of p into the order of N

[unique juniper]

wdym

nvm lol

[chilly ocean]

I haven't carefully done the calculations so I'm not 100% it works out,but eg HN/N should have "many" factors of p if you do the calculation

[unique juniper]

uh

i gtg

ill be back to read the help

[molten silo]

is every field a euclidean domain

[carmine fossil]

Yes

[molten silo]

nice

[chilly ocean]

every field is a field

[unique juniper]

:o

[quaint ivy]

even the https://en.wikipedia.org/wiki/Field_with_one_element?

[carmine fossil]

What about this one?

[unique juniper]

Looks like a British field

[cyan marten]

Suppose P, Q are projective modules, P is an extension of N by A, and Q of M by A, too. Then P + M = Q + N.

(Here A is the quotient)

Is there a nice intuition for why this is true?

I didn't really think it through, but I don't think it's obvious.

[mint seal]

projective extension of stuff is like.... adding.... somehow

😅

[eager schooner]

could someone help me out with this bad boy?

[mint seal]

jokes aside, what exactly is + here? direct sum of modules?

[solemn rain]

@eager schooner whats the problem

[eager schooner]

this is the answer, the problem I'm having is I don't understand why we don't consider (0,0,1) which seems to also give an infinite number of groups that aren't considered here

Ah yeah sorry the problem is basically to find a group that it is isomorphic to classify the factor group

[mild laurel]

They do consider (0,0,1)? i'm not sure what you're asking

[solemn rain]

i dont understad

[eager schooner]

Ah right

sorry

(1,0,0)

[cyan marten]

Yes.
If I understood you correctly, P is "close" to N + A and Q to M + A, in which case the result is close to being correct.

[eager schooner]

This makes it seem like it is at least z x z x z

[mild laurel]

because you can write that as (1,1,1) - (0,1,0) - (0,0,1) kinda

[eager schooner]

Well

That's confusing for this reason: if (1,1,1) has finite order but (1,0,0) has infinite order how are you characterizing them as basically the same

[mint seal]

I gotta 'fess up, I was just being facetious with that remark, I don't really know enough about projective modules to make heads or tails of your question

but if what I said was somehow helpful, awesome!

[mild laurel]

There's no classifying them as basically the same?

[cyan marten]

Well, I am not sure how often does this heuristic work, but thank you!

[latent anvil]

I don't have a good sense of it benedict

[eager schooner]

What I mean by that is we are saying that all the groups that could be generated by (1,0,0) * the normal subgroup are already generated by (1,1,1) * the normal subgroup - ah you know I think I see now, they are covered by the (1,1,1) * the normal subgroup - some (0,1,0) * the normal subgroup of which there are infinite possibilities

okay gotcha

[latent anvil]

I can give you a rephrasing of it but it's not really adding anything new

[thorn delta]

is benedict's question the same as this? Not familiar with their terminology

[latent anvil]

something like "the first Syzygy module is well defined up to stable isomorphism"

yes, it is kxrider

(unless I'm confused)

My first instinct is to think about vector bundles

But that's tricky because M isn't projective

[cyan marten]

Yes.

[cyan marten]

Oh, unfortunately I am unfamiliar with the terms. I don't know about vector bundles either.

[latent anvil]

well luckily I'm not saying anything new

I can try to illuminate why this lemma is relevant though

[cyan marten]

Do you want to use Swan's theorem?

[latent anvil]

Forget about vector bundles for rn

[cyan marten]

Great!

[latent anvil]

so suppose you have a module and you want to take projective resolutions

There are some modules which might be of interest

If you've fixed a projective resolution

actually, let me backtrack

[thorn delta]

(there's a name for this theorem and i remember it had a proof on the stacks project that was too slick for me to understand hmm)

[latent anvil]

Say we want to give a free resolution of M

this is sort of giving a presentation of M

right?

You have a surjection of some free module onto M, that picks out. The generators

[cyan marten]

Well, I can see how a free resolution "of length two" is like a presentation. Do I have the right intuition?

[latent anvil]

yeah, that's the right idea

[chilly ocean]

guys i got a function, and i was asked to make its fourier serie

and also, to write down parseval identity

[latent anvil]

this is almost certainly not the right channel for that

[chilly ocean]

what do i have to right actually?

[latent anvil]

@cyan marten so maybe we have ... -> Fn -> ... -> F2 -> F1 -> F0 -> M -> 0

a free resolution

F0 -> M picks out some generators

and then the image of F1 -> F0 is sort of a module of "relations" between the generators

right?

and then maybe the higher things F_{n+1} -> F_n are like relations between relations

[cyan marten]

I see!

[latent anvil]

right!

so we call these modules syzygys

they're pretty cool

[cyan marten]

!

[latent anvil]

https://en.wikipedia.org/wiki/Hilbert's_syzygy_theorem

this is a big theorem to keep in mind

when thinking about them

[cyan marten]

Thanks!

[latent anvil]

so here's an issue

the projection F0 -> M is not uniquely determined

and its kernel isn't either

does that make sense?

[cyan marten]

Hmmm, if M is free but over a ring where the rank isn't uniquely determined, fot example?

[latent anvil]

well even in that scenario we could just add in redundant generators

[cyan marten]

Oh, you said the projection, not just F_0.

[latent anvil]

ah yeah

even then though

it's true that maybe we can take a minimal number

but I'm really interested in the map

(I'm really really interested in the kernel!)

(because the kernel is sort of the relations between generators)

like for a silly example, (x,y) = (x+y, y) in k[x,y]

right?

I guess these have the same kernel of the projection, hmm

[cyan marten]

Makes sense.

[latent anvil]

I would need to think harder to find an example where the kernel is nonunqiue

but I think it's pretty plausible

do you agree?

[cyan marten]

I feel so too

[latent anvil]

cool

so let's revisit your lemma

first thing, free is too much to ask for

this is a lesson of homological algebra

replace it with projective

so we want to look at the "first syzygy module" of a projective resolution of M

we're still sort of thinking of it as a module of relations between generators of M

still following?

[cyan marten]

The first syzygy module is F0?

[latent anvil]

No, it's the kernel of F0 -> M

ie the image of F1 -> F0

it's all the relations that the generators of M satisfy

[cyan marten]

Ohh, right.

[latent anvil]

so if you have two projective resolutions of M

[cyan marten]

I think so.

[latent anvil]

... -> P2 -> P1 -> P0 -> M
... -> Q2 -> Q1 -> Q0 -> M

we first maybe syzygy modules S of P. -> M and T of Q. -> M

these fit in a short exact sequence, pretty much by definition

0 -> S -> P0 -> M -> 0
0 -> T -> Q0 -> M -> 0

your lemma says that S (+) Q0 is isomorphic to T (+) P0

right?

[cyan marten]

Wow!

[latent anvil]

yeah!!!

this, i think, is the secret story of that lemma

[cyan marten]

Yes.

[latent anvil]

so like

if you think about a category

where X becomes isomorphic to X (+) P for any projective module P

[chilly ocean]

how can i use parseval identity over a fourier serie?

[latent anvil]

still not the right channel lol

try #advanced-analysis

@cyan marten so this is called the projectively stable category

iirc

(I saw this category in a talk once like 2 weeks ago 😆 )

and taking the first syzygy is a well defined endofunctor on this category

which is really really cool imo

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

[cyan marten]

Sorry, but I don't follow. Do you mean a fixed X, or that we have this as an identity?

[latent anvil]

we have this as an identity, sorry

have you seen the definition of the homotopy category?

this is sort of a similar idea

we're collapsing all the projective modules down to be trivial

[cyan marten]

I barely know the topological definition, but I read about it somewhere and I think I got the intuition behind it, and I can see the similarity

[latent anvil]

ah no i mean the homotopy category of chain complexes

[cyan marten]

Oh, I saw that just a few days ago

[cyan marten]

Isn't that a catastrophic cancelation?

[latent anvil]

yes!

which is why X becomes isomorphic to X (+) P for any P

and why this "first syzygy module" is now well defined

your lemma told us that it's well defined up to summing with a projective module, essentially

[cyan marten]

Ohh.. and since summing with a projective module isn't really summing in this weird category....

[latent anvil]

right!

furthermore, in certain special cases (like when your base ring is a group algebra, see the wikipedia article) this "first syzygy" functor is actually a self-equivalence!

so we get what's called a triangulated category

very cool stuff

(that I saw in a talk once)

(I am not an expert :P)

[cyan marten]

A self-equivalence is a "category automorphism," right?

[latent anvil]

pretty much, yeah

(but only an automorphism up to natural isomrophism)

(it may not have an on-the-nose inverse)

[cyan marten]

So just to make sure I understand, the first syzygy functor somehow extracts the relations between the generations in a unique or canonical way?

[latent anvil]

right!

and to make this well defined, we need to pass to this "stable" category

where summing with a projective module doesn't do anything

[cyan marten]

Ohh, I see.

[latent anvil]

but still, we're getting at something interesting here

like, I think this is the right context for the lemma

The vector bundle comment was mostly unrelated. A general slogan is that projective modules = vector bundles, by serre swan

But here we mix projectives and non projectives

So I think to phrase things geometrically you'd need to think harder about schemes than I'm comfortable with :P

anyways the stable category stuff isn't so important

The takeaway here is the idea of sygzygys and this application of your lemma

Oh also, if S is the first Syzygy of M wrt some resolution ... -> P2 -> P1 -> P0 -> M -> 0 then we have a resolution ... -> P2 -> P1 -> S -> 0

[cyan marten]

Me too lol

[latent anvil]

So it's sort of like moving the projective resolution forward by one

[cyan marten]

I see, but how is this relevant?

[latent anvil]

¯\_(ツ)_/¯

[latent anvil]

I just had this thought

[cyan marten]

Oh, sorry 😂

[chrome hinge]

Lmao im not into this stuff yet but Syzygy is a mathematician?

[latent anvil]

Nah, just a word

i think it comes from some astronomical term

[viscid pewter]

it means 'to win at scrabble'

[latent anvil]

I was going to do a reading group on Syzygy stuff but something else came along

[cyan marten]

Thanks @latent anvil ! Although I now feel like doing algebra, but I have a statistics test tomorrow :(

[latent anvil]

aww

Good luck!

[viscid pewter]

tbh i thought you made the term up

temporary designation

[latent anvil]

lmao

[cyan marten]

I was trying to do it much more "mechanically."

[latent anvil]

well, I don't think any of this actually helps you prove it

[cyan marten]

We have a free module S such that P + S = Q + S, and this module is also free.

[latent anvil]

I remember having it as an exercise in my algebra class

[cyan marten]

It wasn't hard to prove.

[cyan marten]

(By the Mazur swindle)

This seems promising because it "links" P and Q.

[sturdy marsh]

linking P and Q is the right idea

[cyan marten]

But apparently it's not very useful.

[sturdy marsh]

but there are other ways of doing it

[cyan marten]

I did it by diagram chasing

[latent anvil]

Hi chm

I retained stuff from Matthew & Sarafina's talk

Apparently

(but not who the third speaker was)

(Kevin?)

[sturdy marsh]

yup

it was kevin

[latent anvil]

[next obsidian]

Yes it was

Also lol Sham

Did you know that this theorem they presented is proved in Weibel I think

[cyan marten]

Last question! A submodule of a finitely presented module is finitely generated, right?

[sturdy marsh]

fkn knew it

[latent anvil]

which theorem?

[next obsidian]

I assume this is the like

Projective things, surjection a

Then the direct sum with the kernel blah blah

Thing?

[latent anvil]

yeah it is

[sturdy marsh]

it's called schaunel's lemma

[latent anvil]

lmao owned

[sturdy marsh]

lol

[next obsidian]

Lol

“Actually Sándor it’s pretty easy just do...”

“And that’s how I invented Chmonkey’s lemma”

[latent anvil]

lol

[next obsidian]

Let's say you have a subring A < B

and an infinite subset S of B

then you can consider A[S], this subring of B, and the ideal I which is generated by S

if 1 is in I, then you can take finitely many elements x_i in I such that
1 = sum a_ix_i where a_i are in A[S]

[latent anvil]

What ring is I an ideal of?

A[S]?

[next obsidian]

A[S]

yeah

but I want to show I can choose a_i in A[x_1,...,x_n]

[latent anvil]

so we have another subring

A[x_1,...,x_n]

[next obsidian]

yup

[latent anvil]

And an ideal J of this generated by x1,...,xn

[next obsidian]

right

[latent anvil]

and we want to show 1 is in J

[next obsidian]

yup

[latent anvil]

What do we know about the ring extension A[S]/A[x1,...,xn]

is it integral?

like, we know that the extension of J wrt this extension is all of A[S]

[next obsidian]

Yeah

[old hollow]

Why wouldn’t the “natural bijection” be between H_i and A^n

Sorry this may be a dumb question lol

[next obsidian]

This is more apt for #point-set-topology

[latent anvil]

It's fine here lol

You are not one to talk

[next obsidian]

lol

[old hollow]

Lol

[latent anvil]

Here's a sanity check abs_0

Try it in P^1

or P^2

you'll realize that H_i is very small

(in general it's n-1 dimensional while A^n is n dimensional)

For this specific bijection, the map φ_i^-1 isn't defined in H_i

Since we divide by something which is zero on H_i

Sorry, maybe my thing about dimensions was more confusing than helpful

[old hollow]

Yeah I’m thinking in terms of n = 2

[next obsidian]

Think in terms of n = 1 first

Imo

[old hollow]

So A^2 is line a plane

[next obsidian]

H_i is just a point

[old hollow]

Oh alright

[next obsidian]

H_i is a copy of P^{n - 1}

[mild laurel]

The "obvious" bijection between A^n and H_i doesn't work because you'd want to send (0,...,0) in A^n to [0,...,0] in H_i but this isn't an element of P^n

[latent anvil]

it also isn't injective even if you remove the origin, since eg [a, a,..., a] = [1,1,...,1]

[old hollow]

Yeah my issue with it was that

I was thinking of n = 2, so A^2 would be a plane

And H_i would be a plane as well

[next obsidian]

no, it's a plane modded out by stuff

[old hollow]

So “obviously” there’s a bijection between them

Oh what

[latent anvil]

can you explain why you say this?

[next obsidian]

Let's say you've set the last coordinate as 0

[x,y,0] is still equal to [lambda x,lambda y,0]

so this is just P^1

[old hollow]

Yeah I was thinking “oh it’s 3d space but you set the z value to 0, so it’s a plane”

Wait

[old hollow]

Oh yeah that makes sense

[next obsidian]

that's why you want to enforce the z-value be non-zero, then you can just set z = 1

this means you can't scale any of the other values

since if you did you screw with z = 1

[old hollow]

Ohh you’ve still got the equiv relation on H_i so it’s not really a plane right

[next obsidian]

yeah

[old hollow]

Man the equiv relation seems so random to me

What’s the point of it

[next obsidian]

it fixes stuff

Think of P^1C

P^1C is the Riemann sphere

[old hollow]

I don’t even really understand projective space from the definition, like why call it P^n if it’s n+1 dimensional

[next obsidian]

having that point at infinity makes things nice

Well P^n is n + 1 copies of A^n

[old hollow]

Oh what

[next obsidian]

glued on top of each other so it's actually n dimensional

[old hollow]

That’s op

Wait what’s the equivalence relation for tho

I was intuiting it as making it lines thru origin not points

[next obsidian]

yeah, but that doesn't help me very much in higher dimensions. TBH idk, it just makes things work and that's good enough for me

xD

[old hollow]

Haha

Alright I’ll have to see this then

Ihhhhh ohhhohohohh

[latent anvil]

what's the thing that doesn't help chm?

[next obsidian]

thinking of lines through the origin in like

[latent anvil]

Because it's probably the only way I can think of projective space

[next obsidian]

7-dimensional space

[latent anvil]

Based on past experience

[next obsidian]

Idk what that looks like anymore

[latent anvil]

ahh gotcha

Yeah I prefer the definition as a quotient

But sometimes you do want to talk about lines

@grassmannian moment

[next obsidian]

I'm still hardstuck on my A[S] problem sadge

[latent anvil]

I feel like thinking about that extension is the way to go

Not really sure though, sorry

[old hollow]

Also guys why did they use this strange bijection $$[x_0, \cdots, x_n] \mapsto (x_0/x_i, \cdots, \text{blah blah})$$

[cloud walrus]

abs_0

[latent anvil]

why do you find this strange?

[next obsidian]

this is saying you're fixing x_i = 1

[old hollow]

Idk why I find it strange

Oh

That’s it?

[next obsidian]

when you do that you get exactly n variables of freedom

aka A^n

[latent anvil]

Yeah I think it's easier to think of it as hitting a certain hyperplane in A^(n+1)

We're including [x0,...,1,...,xn+1] as (x0,...,1,...,xn+1)

okay so A[S]

We have this ring extension

What are nice properties of it

anything in A[S] can be written as an A[S]-linear combination of x1,...,xn

That's something

So it's finitely generated, but maybe not finite

I guess we're trying to prove it's finite?

[next obsidian]

Oh Sham I think I figured it out lmao. This is really stupid

[latent anvil]

oof

[next obsidian]

Just pick the y's later like

[latent anvil]

Start from the beginning

[next obsidian]

Okay, I think this is correct, a priori

you get to write 1 as
sum a_if_i with a_i in A, and f_i in A[S]

each f_i then decomposes as some sum
f_i = a_ijy_ij with a_ij in A, and y_ij in S (okay you might need to take powers of the y_ij but whatever)

so just take all the y_ij

[latent anvil]

Yeah, I think this makes sense

nice!

[next obsidian]

hurb

tfw stuck for 30 minutes

What threw me off was him writing

1 in sum y_iA[y_1,...,y_r]

I think if he said it was in A[y_1,...,y_r] then it would've been easier

[vestal snow]

If $yz$ $dx=y$ $dx$, then $z$ must be $1$ right?

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

I'm pretty sure that this is true, but this implies either the author of a paper I'm reading made a typo or I'm misunderstanding something

Also, y is not 0

[nova plank]

Context?

@vestal snow

Are all those elements invertible?

[vestal snow]

Yes

Those are elements of a field

[nova plank]

Then yes @vestal snow

[vestal snow]

So that means that this equality is only true when $m_i^{(\mu ')}-m_i^{(\mu)}=0$ for all $i$ (assume each $\alpha_i$ is distinct)?

[cloud walrus]

Have a Banana, Bitch

[leaden finch]

wheres the real analysis channel on here?

[chilly ocean]

#advanced-analysis

[leaden finch]

ty

[vestal snow]

Maybe I'm too tired to think, but this inequality does not follow from the statement right?

[sour plume]

Is \nu the same as \nu'

Yeah I think the important thing here is what nu and nu' have to do with one another

[vestal snow]

My bad

I'll post the entire screenshot

[cloud walrus]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[sour plume]

huh uh

,ti --set

[vestal snow]

I need to prove the claim

[cloud walrus]

Your timezone has been set to Europe/Amsterdam!
Your current time is 12:22 PM (CET) on Sat, 27/03/2021.

[sour plume]

what a world

sorry for that spam

[vestal snow]

I can assume theorem 1

I'm mostly just concerned about the inequality following from the last statement

[sour plume]

I thiiiink this is fine, lemme see

So the weird thing in the middle is definitely >= 0, I guess we agree on that

[vestal snow]

Yes

For the inequality to be true, we would need to have the $\sum$ equal to 0 as we can just choose $v' = t^{(\mu)} -2$

[cloud walrus]

Have a Banana, Bitch

[sour plume]

$ \nu' + t^{(\mu')} - t^{(\mu)} \leq \nu + t^{(\mu')} - t^{(\mu)} \leq t^{(\mu)} - 2 + t^{(\mu')} - t^{(\mu)} = t^{(\mu')} - 2$

why no compily uwu

[vestal snow]

$\nu' + t^{(\mu')} - t^{(\mu)} \leq \nu + t^{(\mu')} - t^{(\mu)} \leq t^{(\mu)} - 2 + t^{(\mu')} - t^{(\mu)} = t^{(\mu')} - 2$

[cloud walrus]

Have a Banana, Bitch

[sour plume]

So I'm just using that nu' is smaller than nu, and then the estimate that nu is smaller than t^mu - 2

(and i just wrote t^mu' and t^mu rather than the big fat sums as a shorthand)

[vestal snow]

wait I don't get it

[sour plume]

Which step? Or how this solves it?

[vestal snow]

One sec

[sour plume]

So as I understand it, the \nu' are a bunch of things smaller than \nu, and yeah, one of them may very well be equal to t^mu - 2; so in that case, the whole inequality turns into 0 <= 0

But all of the \nu' fulfil \nu' \leq \nu

[vestal snow]

Hold up

[vestal snow]

What were you trying to prove here?

[sour plume]

dis

I rewrote it with the t's so i wouldn't have to write the big sum

So the left-hand-side is equal to nu' + t^mu' - t^mu

[vestal snow]

Ah shoot

I just realized that t^mu was defined to be that sum

[sour plume]

Yah 😄 Maybe that was cheeky of me to do

But I guess that was the missing step for you then

[vestal snow]

Yeah it was

Thanks

If you don't mind

I had another question about the proof

[sour plume]

Sure I'll see how much I understand

[vestal snow]

"Using the theorem above, we just have to show..."

Why does showing this inequality suffice?

[sour plume]

looks messy, one sec

[vestal snow]

I'll send you the paper in case it makes it clearer

[sour plume]

So I guess what we're trying to do is using the new basis to represent every basis vector of the first basis

[vestal snow]

That's what I think too

[sour plume]

And that inequality is, I suppose, in some way necessary for all the elements in the sum to be contained in the new basis

[vestal snow]

what sum?

[sour plume]

uuuh wait no yeah

it's a product and not a sum

So I guess we're representing every term of the old basis as a single element of the new basis? How odd

[vestal snow]

I don't think that's possible

unless a,b=0

[sour plume]

hmmm

But "base" really means "basis of a vector space" here right

[vestal snow]

yes

[sour plume]

oki i need some thonkin time then

[vestal snow]

Thanks for the help so far

If you manage to figure it out, feel free to @ me or dm me

[sour plume]

Oooh okay I think I see

So we look at this equation

[vestal snow]

yeah

[sour plume]

aw shit maybe i don't

sorry i got excited

i'll ping you

[vestal snow]

it's okay

Thanks

[sour plume]

So, the way the proof is written, the only thing that's happening is that we're expressing elements of the new basis in terms of the old basis -- but it's absolutely not surprising that we can do that, since, yeah, we already know that we can express all holomorphic differentials in terms of the old basis. Because it's a basis. So as stated the proof is kinda useless

But! We can just modify it a bit: Rather than expressing (x - a)(y-b) g_mu in terms of x y g_mu, we can perform the same proof by replacing x and y selectively with x + a and y + b

[vestal snow]

I don't think I follow

Is the proof not valid?

[sour plume]

I think it's written down in a silly way

But the core idea can probably be saved

[vestal snow]

What is the core idea?

[vestal snow]

We need to do the opposite right?

[sour plume]

So here, the first line tells us "we can express (x-a)^nu (y-b)^nu g_mu as linear combinations of terms x^nu' y^mu' g_mu"

This is almost an expression of the new basis elements in terms of the old basis elements, but we still need to make it so that the index of g_mu is equal to the exponent of the x

And that's what the second line does: So with that down, we understand that (if the exponents follow all the inequalities that we need), we have expressed the new basis in terms of the old basis

[sour plume]

yeah exactly

[vestal snow]

I think I get it

[sour plume]

But doing the opposite looks exactly the same: We just need to instead do the same thing to $x^\mu y^\nu g_\mu(x) = (x + a - a)^\mu (y + b - b)^\nu g_\mu(x)$

[cloud walrus]

Lartomato

[vestal snow]

Yeah

[sour plume]

This would've been so easy to write down properly but somehow people who wrote papers in the 80s were all on the craziest fucking drugs

It's all unreadable garbage

(hyperbolic but holy fuck i never go into these kinds of papers with high expectations)

[vestal snow]

Yeah my advisor said that this paper would not be accepted at any journal today lol

[sour plume]

I wish you the best of luck fighting through the rest of this

[vestal snow]

Thanks

[sour plume]

But hey, you learn how to become a mathematical mind-reader when working with papers like this

[vestal snow]

I'm planning on rewriting it

[sour plume]

If it's important enough, might be worth it

[vestal snow]

So the inequality wasn't even that relevant lol

Except for the set builder thing

Bookkeeping indexes

[sour plume]

Oh the inequality was important too, to make sure that the exponents all match up

[vestal snow]

Yeah

But the heart of the argument was right there

[sour plume]

Yah

[vestal snow]

I kinda feel silly after this lol

[sour plume]

dat's okay

[vestal snow]

This is like one of the standard tricks with polynomials

[sour plume]

Yeah that's also the thing

Somehow a lot of old papers manage to write down simple ideas as the craziest, most cryptic hieroglyphs

Maybe they just thought differently about the stuff back then

But yeah, don't worry, you're gonna learn a lot from that I imagine

(also if you are indeed tired, maybe go to bed and continue on a better day -- people overestimate their productivity when tired)

[vestal snow]

Garcia to a Kindergartener: The rationals are just the fractional field of the initial object in the category of rings

[chilly ocean]

Yes it is luckily weekend, so as much sleep as possible is attainable without worrying about waking up for class/work/etc (unless you have work on weekend 😞 )

[vestal snow]

Probably so

Thanks for the help my man

[sour plume]

No problem fam

Stay healthy and hydrated

[carmine fossil]

Why does anyone care about group iso theorems other than the first one?(I am referring to
G/Ker(T)=H if T is a surjective hom from G to H)

[viscid pewter]

because they're just really neat

[carmine fossil]

Like I remember using second iso exactly once while doing D and F

never used 3rd iso. the lattice one was kinda neat,tho

[rigid cave]

So I have problem that I am kinda stuck on. Let $\alpha$ and $\beta$ be algebraic elements in the field $F$. Assume that the minimal polynomial to $\alpha$ over $F$ has degree $m$ and let the minimal polynomial to $\beta$ over $F$ be of degree $n$. Assume that $m$ and $n$ are relatively prime. Show that $[F(\alpha, \beta):F] = mn$.

[cloud walrus]

Older Sister

[rigid cave]

I know that I can use the tower law, $[F(\alpha, \beta):F] = [F(\alpha, \beta):F(\alpha)] [F(\alpha):F]$ but I do not know how to show that $[F(\alpha, \beta): F(\alpha)] = n$

[cloud walrus]

Older Sister

[oblique river]

Try writing out the tower law in both possible ways

And youre right, you cant immediately conclude that [F(a,b) : F(a)] is n, but it is some integer, so you can conclude that [F(a,b) : F] is divisible by m

@rigid cave

[rigid cave]

Ok, so like applying the tower law on both F < F(b) < F(a, b) and also F < F(a) < F(a, b)?

[rigid cave]

So I get something like this: $[F(\alpha, \beta):F(\beta)][F(\beta):F] = [F(\alpha, \beta):F(\alpha)][F(\alpha):F]$ which gives me that $[F(\alpha, \beta):F(\beta)] = [F(\alpha):F]$ and vice verse since they are relatively prime?

[cloud walrus]

Older Sister

[oblique river]

Yep

Also yep to your other statement

[rigid cave]

Okay, thank you so much!

[oblique river]

Np and gl

[chilly ocean]

Hi

I’ve a question

$a \prec b$

[cloud walrus]

Bany, um futuro phideas

[chilly ocean]

What mean “\prec”?

Id remember

The “subgroup of”?

[viscid pewter]

i think it's just a set theory thing

ordinals

a precedes b

[magic owl]

It's a generic symbol for an ordering normally

it depends on the context you see it in

often it shows up when < has a different meaning in the same context

You can use it to denote any ordering you want but we can't tell you what it might mean in your context w/o more info

[chilly ocean]

Okay... no problem

[magic owl]

Yes

[oblique river]

The only group for which those are equal is the trivial group

[next obsidian]

Wtf is Perm(G)?

Maybe that’s S_G

Yeah, you have to send e to e

Which is why what Buncho said is true

A fun thing tho is to see when Aut(G) is Perm(G\{e})

?!?!?!

[paper flint]

Lol

[next obsidian]

This is impossible if G is finite

By size reasons

[magic owl]

im confused

yes but ur the one confusing me

lol

what is G/{e} + 1

oh its not 'or sumn'

you know thats true for a finite group hahaha

[next obsidian]

It’s true for infinite too...

[magic owl]

i dont believe in cardinal arithmetic

[next obsidian]

This mf

Bruh

Not a quotient

\ is set difference

xD

[paper flint]

Lmao

Your brain on algebra

[hot lake]

did someone write G/{e} when they meant G\{e}

[next obsidian]

This is what I said

Notice I did \

I did \ I tell you!

Oh lol

Those are the same for every infinite group

By size considerations

I am talking about Aut(G)

Aut(G) acts on the non-identity elements

When is it all of S_{|G| - 1}

Is the question

So when do you have every possible permutation of non identity elements

The answer IIRC is not very often

[hot lake]

I don't think it's possible

[next obsidian]

It is possible

I think |G| <= 4 tho haha

[hot lake]

yeah something like Z/3Z or (Z/2Z)²

I was thinking about infinite groups

[next obsidian]

I don’t know if when I proved it I ruled out infinite groups

Or if I only looked at finite

Mizra don’t look at this

|| you get every non identity element is the same order, so in the finite case prime. This tells you it’s a p-group so it has non-trivial center so it’s abelian. Then apply classification, and do whatever you need to finish||

[next obsidian]

I meant Mirza

No haha since then G\{e} is empty

And you have exactly one automorphism

[hot lake]

you also have exactly 1 permutation of the empty set

[next obsidian]

...

[hot lake]

xD

[next obsidian]

Fair

I guess that’s true

Like the empty permutation would say

“This is what you do to non identity elements”

And there are none

So you just do nothing

And we know you fix e

[rigid cave]

Is there any "easy" way to show that $\frac{x^p - 1}{x - 1}$ is irreducible over $\mathbb{Q}$?

[cloud walrus]

Older Sister

[rigid cave]

Like a way to use Eisenstein criterium?

[delicate bloom]

yeah plug in x+1 for x

[rigid cave]

But how do I use Eisenstein on $\frac{(x+1)^p-1}{x}$

[cloud walrus]

Older Sister

[cursive temple]

just simplify

[delicate bloom]

think about the binomial expansion

[rigid cave]

cool pfp btw

[delicate bloom]

ty ty

also worth mentioning the constant term is when x=1 in the original polynomial which is just 1+1^2+1^3+...+1^{p-1}=p

[chilly ocean]

@rigid cave did you get it?

[rigid cave]

well I am honestly not that comfortable with the binomial theorem. I am still in highschool...

[chilly ocean]

lmao but youre doing AA

[rigid cave]

But I think that this becomes $x^{p-1} + px^{p-1} + \cdots$

[cloud walrus]

Older Sister

[chilly ocean]

maybe this helps

[delicate bloom]

well if you don't know the binomial theorem by heart you should do that first

$$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$$

[cloud walrus]

Merosity

[magic owl]

I still google it sometimes

[chilly ocean]

I used to always google it

[delicate bloom]

since in our case n=p means when 0<k<p that you have p! in the numerator divided by smaller factorials less than p, so they never can divide out the p

[chilly ocean]

but now taking discrete so I kidna had to remember it

[delicate bloom]

which meets the requirement for Eisenstein

[chilly ocean]

full solution

[delicate bloom]

you can derive it by thinking in terms of the binomial expansion literally counting the combinations

[rigid cave]

ooohhh

[delicate bloom]

(x+y)^3 = xxx + xxy+xyx+yxx + xyy+yxy+yyx + yyy

[golden pasture]

ohgod

oh

i thought we were in noncommutative

my heart

almost died

HAHAHA

[delicate bloom]

see how if you start to think about order mattering you're literally counting the number of ways?

[rigid cave]

yup

[delicate bloom]

haha I am thinking of them as being noncommutative temporarily

[rigid cave]

okay, I got it now. Thank you so much!

[delicate bloom]

if you have more questions about it just ask no pressure

[rigid cave]

d(-_^)

[cloud walrus]

slimvesus

[oblique river]

yes; S^G = S \cap A^G. 1 is in both of those on the right so it's in S^G

np and gl

[hot lake]

did they forget to define G_s ?

G_s = {g in G | gs = s} ?

are you missing that f is surjective ?

ah wait it says

[mild laurel]

it's just surjective by definition?

[hot lake]

"a bijection onto the orbit Gs"

[mild laurel]

An element of Gs is gs for some g, so f(gH) = gs?

[hot lake]

well not really, but you have to show that the image is Gs

which is uuh obvious

?

f itself isn't surjective, but they are saying its image is Gs

and since it's injective

you get a bijection from G/H to Gs

[chilly ocean]

Question: Prove that if m|n (m divides n), then the function π_m,n :Zn -> Zm is well-defined.

the function?

how we should proof if (x,y1) (x,y2) exist then y1=y2 in this case ?

[chilly ocean]

π_m,n

[latent anvil]

nate, could you define that function for us?

so that we're all on the same page

the notation π_m,n isn't universal

[chilly ocean]

just wait a moment

$\mathrm{\Pi}{m,n} : \mathrm{\mathbb{Z}}{n} \longmapsto \mathrm{\mathbb{Z}}_{m}$

[cloud walrus]

Nate

[chilly ocean]

is our function

what does the function do to elements of Z_n

we know its domain and codomain. what does the function actually do?

i can take guesses but it's better if you tell us

our hypothesis is if m|n. it just assigns integers from set n to m

is the function $\pi_{m, n}(x + n\bZ) = x + m \bZ$???

[cloud walrus]

(T*Terra, dq^i \wedge dp_i)

[chilly ocean]

no

then what is it

can you like

send a picture

or share what you're working out of

actually I have faced this question without more explanations. I thought maybe there is something obvious that I don't know.

so you don't know what pi_{m, n} is

no

are you working out of a textbook? is there a list of symbols in the back?

maybe it's defined in an earlier problem?

i have just found it in exercises and doesn't make sense for me

are you working out of a textbook?

can you share what you're working out of?

found it. it relates to Chinese remainder theorem.

let me focus on it. I'll share my answer

[viscid pewter]

pls post it

[chilly ocean]

$\mathrm{\Pi}{m,n}([a]{n}) = [a]_{m}$

[cloud walrus]

Nate

[chilly ocean]

lmao

[chilly ocean]

that's this

yes, the equivalence class of a with respect to the equivalence relation of congruence modulo n

[chilly ocean]

to prove that, we just have to say a+nk = a+mk, m=n for all pi_m,n([a]_n)=([a]_m) therefore, for all (x,y1)(x,y2), y1=y2 for all y1,y2 member of Z.

[native orbit]

suppose I wish to show that ∛2 is not in ℚ[∛5]

why is it sufficient to show that ∛2 is not in ℤ[∛5]

I saw this on a post and I am not immediately able to accept this sufficiency

[chilly ocean]

@chilly ocean was it correct ? xD it seems you are a master in this field

[native orbit]

this kind of feels like irreducible over ℚ ⇔ irreducible over ℤ

except in the extension

but I do not see why it is “obviously” true

[mild laurel]

@native orbit it should be since the third root of 2 is an algebraic integer and since Z[third root of 5] is the ring of algebraic integers of Q(third root of 5)

[native orbit]

right but what result is that

relying on

[mild laurel]

I'm not sure what you mean

The algebraic integers of Q(∛5) are exactly Z[∛5]

So if ∛2 is in Q(∛5), it must be in Z[∛5]

[native orbit]

oh

i see

thank you for your service

[next obsidian]

This is false

Consider 1

[mild laurel]

?

[next obsidian]

Exactly

[mild laurel]

r u trolling or

[next obsidian]

Yes

[mild laurel]

ok

[next obsidian]

Why aren’t you piss color btw

[mild laurel]

i can see why u didnt get into columbia

[next obsidian]

Actually also more generally

How do you know the algebraic integers is Z[cube root 5]

[mild laurel]

sagemath told me

[next obsidian]

Like the algebraic integers aren’t always just like, adjoin whatever you adjoined to Q

Oh

Lol

[mild laurel]

yea

[next obsidian]

I see

Very well, have a nice day

Does anyone know anything about the ramification theory of valuations? If so, where does it appear, and why is it important?

[latent anvil]

I was going to reply with the algebraic integer thing but then I decided I was too lazy to check what the integers of Q(cbrt(5)) were

sage good

[mild laurel]

@next obsidian it's kinda like the ramification of primes

[next obsidian]

wtf does that mean

[mild laurel]

do you know what ramification of primes are

[next obsidian]

no

Ramification to me means

[mild laurel]

so like

2

[next obsidian]

fails to be a covering map

[mild laurel]

factors as (1-i)^2

up to units

[next obsidian]

okay

[mild laurel]

ramification is when you have anything to a power higher than 1

since ur a geometry bro

[next obsidian]

in Z[i]?

[mild laurel]

yeah

in Z[i]

[next obsidian]

does 5 ramify

[mild laurel]

no

[next obsidian]

🤔

[mild laurel]

5 is what

(2 - i)(2+i)?

[next obsidian]

so you factor into irreducibles in Z[i]

?

[mild laurel]

Right

[next obsidian]

So can't you only ramify up to degree 2

like isn't there something about irreducibles in Z[i[

like if you're 1 mod 4

you break up into two things

and that's all that can happen

[mild laurel]

well for Z[i] yes

[next obsidian]

Oh so you consider other extensions

[mild laurel]

only up to degree 2 but in general

[next obsidian]

I bet this is secretly

[mild laurel]

you can get higher

[next obsidian]

geometric ramification too

[mild laurel]

yes

[next obsidian]

So how is valuation related

[mild laurel]

the inclusion Z -> Z[i] gives you a map from spec(Z[i]) -> spec(Z)

[next obsidian]

Yeh

[latent anvil]

oh so if we do the galaxy brain thing and think of Q as the field of functions on Z or Q(i) as the field of functions on Z[i] we're sort of asking when a function on Z has a double root after lifting it to a function on Z[i]

[mild laurel]

the ramification of primes is exactly geometric ramification

so for number fields at least

[next obsidian]

how is Q functions on Z

expalin

[mild laurel]

all valuations comes from either p-adic valuations from some prime p

[latent anvil]

it's the local ring at the generic point

it's the function field of Spec Z

[next obsidian]

yeah but

how are these functions

[latent anvil]

Oh I meant functions on Spec Z, not Z

[mild laurel]

or is an infinite valuation coming from the usual absolute value

[latent anvil]

They're meromorphic/rational functions

[next obsidian]

Like

okay

I don't see how you can tell order of vanishing

but I guess this is like

via valuations

lol

haha

[latent anvil]

For holomorphic ones or polynomials you can define it by divisibility

Right?

[next obsidian]

Sure but like

[latent anvil]

how big of a power of (x-a) are you divisible by

[next obsidian]

p/q evaluated at (n)

should be like

looking at

p-bar/q-bar yeah?

and like maybe q is in (n) so it's all fuck but

[latent anvil]

right

How is this different from the case of the order of a zero of something in k(x)

you still mod out and lose the information

[next obsidian]

actually I think this is why valuations measure order of 0

For C

[latent anvil]

and just get 0

[next obsidian]

you get a valuation for every point

[latent anvil]

Sorry for interrupting your explanation zoph

[next obsidian]

by factoring out (z - a)^n

and then n is your valuation for that function

[mild laurel]

no its good

idk if chmonkey wants any more explanation

[next obsidian]

i do

So how does

a valuation and it ramifying

have to do with this prime ramifying business

[mild laurel]

so for every prime, you can form the usual p-adic valuation

[next obsidian]

idk what this is...

oof

Oh wait I think I know a thing

[mild laurel]

uh

[latent anvil]

chm didn't you do like I-adic completion stuff. Don't you have an I-adic valuation there?

or an I-adic absolute value

[next obsidian]

I think this is related to that

[latent anvil]

just take I = (p)

[next obsidian]

but I never described what the number was

like

we described convergence of cauchy sequences

but I never saw an actual like metric

[latent anvil]

it's just the highest power of p you're divisible by

[next obsidian]

or whatever

lol

okay

that's easy

[latent anvil]

(the absolute value takes the reciprocal of this)

[next obsidian]

Oh

you're just like looking at the highest value of I^n you're in

err

no like

the highest one you're still in

wait that's what I said

okay I see

this is just the valuation for a DVR lol

I'm not certain that this gives a valuation tho

since then you should be a DVR I think since this thing is gonna be a subgroup of Z

So we should be able to turn any ring into a DVR by doing this I-adic thing

So I looked it up and

somehow apparently looking at the highest power of p you're divisible by does give a valuation

but this makes me think that Z should be a DVR...

oh no I see

actually the valuation ring associated to this has to have (p) be the maximal ideal

so that it's actually Z_(p)

which is a DVR

[mild laurel]

yeah when you localize you get a DVR

[next obsidian]

makes sense okay

I am on board

[mild laurel]

so you get a valuation for all the prime ideals of your ring

[next obsidian]

yup

[mild laurel]

and for number fields at least, you also get valuations coming from the usual absolute value

[next obsidian]

Value group a subgroup of R^+

[mild laurel]

So for any embedding K -> C, you get a valuation coming just from the usual absolute value on C

[next obsidian]

yee

[mild laurel]

ostrowski's theorem tells you this is all the valuations

[next obsidian]

why are these valuations important tho

[prisma ibex]

They encode all the possible completions

[next obsidian]

of Z?

Or of a number field

[prisma ibex]

Of Q yea

[next obsidian]

of Q?

[prisma ibex]

Or more generally of any number field

[next obsidian]

Q shouldn't be able to be completed right?

[prisma ibex]

What

[next obsidian]

Okay let me back up

lmfao

I-adically

it cannot be completed in anyway

[prisma ibex]

What?

[next obsidian]

I mean

the only ideal of Q is 0

but I remember that

You know... you can complete it... as a topological space

in different ways

[prisma ibex]

Yes

The completions corresponding to the finite primes are indexed by the nontrivial prime ideals of the ring of integers

[next obsidian]

right

[prisma ibex]

Q has nontrivial completions Q_p for p prime and R

The former correspond to the closed points of Spec(Z)

The latter corresponds to a point at infinity in some sense

[next obsidian]

Yes

[prisma ibex]

The trivial completion corresponds to the generic point

[next obsidian]

right

So what you're saying is... Spec Z is the Riemann sphere

or the projective line

[latent anvil]

what's the trivial completion, ng?

[prisma ibex]

It’s analogous to A^1

[latent anvil]

I'm reading this conversation in the sense of metric completions

With different metrics on Q

Not really sure if that's what you mean

[prisma ibex]

You have to compactify Spec(Z) in some mysterious sense to get something analogous to P^1

And yes this is what I mean

[latent anvil]

Ah, so just use the discrete metric to get a "trivial completion"

I see

[prisma ibex]

On Q you either have various ultrametrics for each prime or the usual metric

[next obsidian]

ultrametric...

[delicate bloom]

metrics that satisfy the strong triangle inequality

[thorn delta]

Anyone have an idea of what i've done wrong here? I assume i'm incorrect since I didn't use that the degree of K(u)/K is odd

[mild laurel]

You have your inclusion of fields backwards

K(u^2) is a subset of K(u)

since everything in the first is in the second

[thorn delta]

oops, thanks

[native orbit]

i mean you can explicitly write u as a rational expression in u²

Let M(x) be the ℚ-minimal polynomial of u, and M(x) = E(x) + O(x) (even and odd parts)

odd polynomials always have 0 as a root, so E(x) = xU(x)

E(u) + u U(u) = 0

u = -E(u)/U(u)

polynomials of strictly even degrees have the identity A(x) = B(x²) where B has the same coefficients of half the degree

Hence, u = R(u²)/S(u²), which is in ℚ(u²)

Therefore, we have ℚ(u)⊆ℚ(u²) and the trivial inclusion ℚ(u²)⊆ℚ(u), so the extensions are identical

[next obsidian]

Pretty sure you can only take Q-coefficients if its characteristic 0

[native orbit]

oh i misread

[lavish pike]

so i gotta find the monic $f(x)=3x{^5}-4x^{2}+1$ in $Z_5$

[cloud walrus]

NocuousNick

[lavish pike]

whihc like u just factor out the three or whatever and get the c=3 and g(x) is just f(x) with a 3 takin out and 4/3 and 1/3 but then im supposed to find the inverse of 3 in $Z_5$

[cloud walrus]

NocuousNick

[lavish pike]

so like how do i find the inverse of 3 in Z5

[chilly ocean]

Guess and check

[lavish pike]

but like how tho lmfao it is just going right over my head rn

[mild laurel]

What does something the inverse of 3 mean

[lavish pike]

1/3

but isnt 1/3 in Z5 just 1/3?

[mild laurel]

how is 1/3 an element of Z5?

[lavish pike]

so elements fo Z5 have to be like 0,1,2,3,4

[chilly ocean]

1/3 is totally an element of Z5

[lavish pike]

so why does the remainder being 1 make it the inverse in Z5

[lavish pike]

yea my bad forget shit yuh know simple mistake.

[next obsidian]

I mean if you meant

the inverse of 3

[chilly ocean]

"a has inverse b" means ab=ba=1

[next obsidian]

then it exists as ||number here||

but like 1/3 looks like you're talking about the image of 1/3 mod 5Z

[chilly ocean]

Wow, you just gave him the answer

[next obsidian]

which doesn't exist

[lavish pike]

lol i figured it out since the remainder is supposed to be one $3*2=1$ in Z5

[cloud walrus]

NocuousNick

[lavish pike]

so the inverse of 3 in Z5 is 2

alright this make 0 sense lmao

Find a monic associate of $3x^{5}-4x^{2}+1$ in $Z_5$

[cloud walrus]

NocuousNick

[chilly ocean]

You basically had it

[lavish pike]

yea so u multiply it by 2 by why do u only multiplly the -4 and 1 and not the x^5