groups-rings-fields

[next obsidian]

But often times you want to transfer statements about some sequence

Into one with objects you know better

And you can do this by having isos such that everything commutes like in this case

But I gave advice that when messing tensor products you should try and show isomorphisms via universal properties

[vestal snow]

is that what you meant by natural isomorphisms?

[next obsidian]

Sort of

A natural isomorphism is an isomorphism of functors

The natural iso gives you that sort of diagram for free

But you can get it by just showing it manually sometes

[vestal snow]

oh okay

[next obsidian]

I’d suggest proving things are iso using universal properties usually

But then EXPLICITLY FIGURE OUT THE MAP

Because like in this example, you neeeeeeed the definition of the map in order to show the diagrams commute

If you build up this stuff in your head about how things are isomorphic it’s really good because then you can reduce complicated problems like “injectivity of f (x) 1” into easier ones like “injectivity of f”

Eventually you’ll get muuuuch better at looking at some diagram of tensor products or whatever and just knowing it commutes

[vestal snow]

Alright thanks

[next obsidian]

Maybe what I said is confusing, but I’d just keep it in mind

[vestal snow]

I think I get the gist of what you're saying

[next obsidian]

Yeah my take away is simply

[vestal snow]

someone said something similar a while ago

That try to argue with arrows and diagrams instead of sets

[next obsidian]

Remember how things are isomorphic whenever possible

^^^^

This is basically me relating my experience

In AG I got a lot of maps like “the unique map making blah diagram commute”

[vestal snow]

The good thing with algebra is most of the isomorphisms are very straigtforward

[next obsidian]

Then I got a unique map in terms of that

And one in terms of that

So now the only way I can show this map is equal to something else

Is to show t makes some whack ass diagram commute

That has like 50 arrows everywhere

[vestal snow]

oof

[next obsidian]

But THE MAP HAD AN EASY DESCRIPTION

In terms of elements

I just got so wrapped up in “haha I get map for free” I never bothered to figure out what it actually was lmao

Don’t fall into that trap is all I wanna say haha

[vestal snow]

did you use atiyah macdonald too?

[next obsidian]

Usually the maps are quite simple, it’s the intuitive thing, the difficulty is showing that the intuitive thing actually gets you what you want

Sort of

I’ve done some@work out of it

And jumped around

But rn I’m going through Matsumura

[vestal snow]

I'm trying to get into algebraic number theory once I'm done with A-M

[next obsidian]

Gotcha, I’d talk to Zoph-Chan once you wanna do that

I don’t do NT at all

I just do AG

[vestal snow]

though universal algebra also seems very interesting

I did talk to him a while ago actually

[next obsidian]

Also Zoph started typing as soon as you said ANT lol

[vestal snow]

Really helpful guy

Hah

[next obsidian]

Yeah, anyway I’m off to bed, hope that cleared things up for ya.

[vestal snow]

Yup. Thanks

[next obsidian]

If you have other questions just post em in this chat and I’m sure me or others will swing by

And hopefully someone will answer

[vestal snow]

gn

[vestal snow]

Can someone show me how this diagram being commutative and f being injective implies that f (x) id is also injective?

Here A is the ring we're tensoring with

[knotty mason]

if f is injective and the diagram commutes, the other path is injective

so each part of it is injective

[vestal snow]

All I can see is that the restriction of f (x) id on the generators n' (x) a is injective

I can see the other path being injective from N' to N, but I don't see how commutativity implies f (x) id is also injective

This diagram is commutative and the bottom map is injective, but clearly the map at the top is not

[knotty mason]

oh my mistake, sorry!

[vestal snow]

np

I understand based on my discussion with magician why it must be injective, but I don't see how to prove it using the diagram

[chilly ocean]

Let (G, ∗) be a group. Prove that the map π : G −→ G defined by π(g) = g ∗ g is a
homomoprhism if and only if G is abelian.

Can someone help me with the first part

π(ab)=ab*ab right

[latent anvil]

@vestal snow the vertical arrows are isomorphisms

[chilly ocean]

But how do i continue to get π(a)*π(b)

[latent anvil]

@chilly ocean you're assuming π is a homomorphism, right?

[chilly ocean]

I am assuming G is abelian first@latent anvil

[latent anvil]

Oh okay

Well, write down what π(a)π(b) is using the definition of π

And try to show that equals abab

[chilly ocean]

First thing i did was
π(ab)=ab*ab

How to continue to get π(a)*π(b) is the problem

@latent anvil here

[latent anvil]

Yeah, that's why I'm suggesting you look at what π(a) π(b) is

It might give you an idea for how to show it's equal to π(ab)

[mild laurel]

What does it mean for a group to be abelian?

[chilly ocean]

π(a)=aa
π(b)=bb@latent anvil

[latent anvil]

right, so π(a)π(b) = aabb right?

So what you're trying to show is that aabb = abab

Does that make sense?

[chilly ocean]

There is the star operator
So π(a)= a star a
π(b)= b star b
So π(a)π(b) per what you're saying will be (a * a)(b * b)@latent anvil

[latent anvil]

Sure

To be clear, a*a = aa

They're two notations for the same thing

[chilly ocean]

Oh okay
Yeah makes sense

[vestal snow]

@latent anvil thanks. That makes sense

[latent anvil]

In general if you have a square like that with vertical maps bijections, then the top will be injective iff the bottom is

You can replace bijection with isomorphism and injective with monomorphism too

[vestal snow]

Yeah I just worked that out

It seems obvious now

[latent anvil]

this is why the thing magician was saying works

[vestal snow]

Maybe this is a signal I should head to bed

[latent anvil]

So like, what's your defintion of flat?

And do you know what an exact functor is?

[vestal snow]

Tensoring with it preserves exactness

[chilly ocean]

@latent anvil take a look at the picture please

[vestal snow]

But the book I'm using gave alternate definitions as well

[latent anvil]

Looks good to me squash

[vestal snow]

And I was using one of those

[chilly ocean]

@latent anvil thanks

[latent anvil]

Okay, so given a functor F which sends 0 to 0, we say F is exact if when you apply F to an exact sequence you get an exact sequence

Usually F will map modules to modules (more generally it can go between "abelian categories")

given a module M, tensoring with M gives a functor A-mod -> A-mod, and M is flat iff this functor is exact

What magician was saying works because if two functors are isomorphic, and one is exact, the other is as well

[vestal snow]

I think I get the general idea of what you're saying and I'll look at it more closely tomorrow morning

[latent anvil]

sure

[vestal snow]

Thanks for the help though

Category theory seems very interesting

[latent anvil]

I'm a fan

[errant drum]

Let $\phi$ be a continuous automorphism of $\mathbb{H} = \bR ^4$ which I've shown fixes the real axis so $\phi(r) = r$ for any real number. $\phi$ is linear so $\phi(a+b) = \phi(a) + \phi(b)$ and analogus identities for subtraction, multiplication and division have been proven
\
How would I show that $\phi$ preserves the dot product for purely imaginary quarternions
\
That is $\phi(q\cdot p ) = \phi(q) \cdot \phi(p)$

[cloud walrus]

Cosmicrayz:

[delicate mauve]

@errant drum have you shown that phi permutes {i,j,k}?

it's not hard

you can show phi(i)^2 = -1 so phi(i) is in {i,j,k}

similarly for the others

it follows trivially from this

[errant drum]

Are i,j,k the only square roots of -1

[delicate mauve]

yes

[errant drum]

Surely any rotation through any angle would satisfy this property

[delicate mauve]

well -i,-j,-k

also work

but that's it

Surely any rotation through any angle would satisfy this property

why

no it wouldn't

rotation through 20 degrees definitely doesn't satisfy it

[errant drum]

But rotation through 20 degrees is an isometry that is linear + satisfies the other rules?

[delicate mauve]

right

but why would that mean it squares to -1?

[errant drum]

Let me message you in a bit I'll read through my notes I might have confused myself

[delicate mauve]

nvm

i might be wrong

yeah x^2 = -1 uniquely charaterises unit quaternions

so you get that phi maps unit quaternions to unit quaternions

doesn't the fact that phi(qp)=phi(q)phi(p) follow from the assumption that phi is an automorphism?

what definition of automorphism are you using here?

@errant drum

[errant drum]

Yeah it does

I was saying division and subtraction also work

And yeah there are infinitely many square roots of -1

For instance (i+j+k)/sqrt (3) is another one

Also from the way the questions were structured if first asked me to prove phi preserves conjugation and phi is an isometry

If that helps

[delicate mauve]

yeah, I realized that all unit quaternions are square roots of -1

and vice versa

so is the problem solved?

[errant drum]

I'm still not sure how to prove it since obviously I could cite the fact isometries preserve inner products but I feel like that's not in the spirit of the question

Everything else was proven from quarternion arithmetic or using the fact phi was an automorphism

[delicate mauve]

what definition of automorphism are you using here?

[errant drum]

phi(ab) = phi(a)phi(b)

phi(a+b) = phi(a) + phi(b) etc

[delicate mauve]

right, so why doesn't it follow from this?

i don't understand what else you want to prove

we've shown that phi preserves purely imaginary unit quaternions and we have that phi(ab)=phi(a)(b)

what else do you want?

[errant drum]

Can you explain how phi(a dot b ) = phi(a) dot phi(b)

I'm completely lost here

Like what is the exact argument you're using

[delicate mauve]

phi sends purely imaginary quaternions to purely imaginary quaternions, do we agree on that?

x^2 < 0

[errant drum]

Yes

[potent condor]

i agree

well done

[delicate mauve]

So phi(Re(x)) = Re(phi(x))

and it follows trivially form this

[errant drum]

Ahhh ty for bearing with me

That makes sense

[delicate mauve]

this is by comparing both sides of phi(x) = Re(phi(x)) +Im(phi(x)) = phi( Im(x) + Re(x)) = phi(Im(x)) + phi(Re(x)) = Re(x) + phi(Im(x))

Re(x) + phi(Im(x)) = Re(phi(x)) +Im(phi(x))

[chilly ocean]

ok so everything makes sense... until the last line. how did they jump from b(a(x)) which i undertand, to "now start with 1 and trace out the cycle (12534).

the previous approach they described was the "pure cycle approach" and the answer was (12534)

is the "function list" approach dependent from the answer from "pure cycle approach"? what's going on here

[scenic sage]

so ok

look, b(x) is permutation which sends 1 to 2, 2 to 3, 3 to 4 and so on

a(x) is permutation sending 2 to 4 4 to 5 and 5 to 2

in b(a(x)) they produced a composition

[chilly ocean]

right

how do you get the final answer of (12534)

[scenic sage]

so look b(a(x)) sends 1 to 2, 2 to 5, 3 to 4, 4 to 1 and 5 to 3

[chilly ocean]

ok so you basically skip when there are the same two numbers in a row: so 12534

but then why not continute to 153 after 4?

bc then the numbers repeat and you just stop at the first 4?

ok fair enough, thank you @scenic sage

[scenic sage]

sorry

interned dead

@chilly ocean

1 to 2, 2 to 5, 3 to 4, 4 to 1 and 5 to 3
cyclically decomposed
Is (12 because 1 is sent to 2
(125 because 2 is sent to 5
(1253 because 5 is sent to 3
(12534) because 3 is sent to 4 and 4 is sent to 1

[chilly ocean]

ok, got it thank you 🙂

[opal vale]

Let $M,N$ be square matrices of order $r<4$. If both M and N have same characteristic polynomial and minimal polynomial then prove that M is similar to N.

[cloud walrus]

Manas:

[olive mirage]

Hint: Two matrices are similar if they have the same Jordan Canonical Form.

[chilly ocean]

where would algebraic structures go?

[next obsidian]

wdym?

If you mean what I think you do, I think this is the channel for it

[chilly ocean]

I think my professor made a mistake on this problem?
`Find the eigenvalues and eigenvectors of a linear transformation f of a vector space A = R^3 having in the standard basis in the following matrix:
|1 2 3|
|2 1 0|
|1 0 1|

Vectors:
|-2| |0| |2|
|2| |0| |2|
|1| |1| |1|

Values:
-1,1,3

Since they have provided both the vectors and values, that is what I'm supposed to solve for,defeating the purpose of the problem?

[next obsidian]

They provided all 3 of those?

Also this is linear algebra so it’s a better fit there

[chilly ocean]

they provided everything above... I'll move it

[next obsidian]

Don’t bother

Just next time I guess

[chilly ocean]

I hope there won't be a next time lol

[next obsidian]

I think the point is to associate the eigenvectors to their specific eigenvalue

So you can compute that by just multiplying them into the matrix lol

[chilly ocean]

right, so weird. Thank you

[vestal snow]

Can someone verify if my proof outline here is correct?

Prove that if p is a prime ideal of A, then p[x] is a prime ideal of A[x]

There's a few things I haven't justified:

1. The A-module homomorphism is also an A-algebra homomorphism
   2 Maybe the last past where I prove that A[x] (x) p is prime might be incorrect

I forgot to mention, we already know from a previous exercise that p[x] is isomorphic as an A-module to A[x] (x) p

[next obsidian]

I... think this right? The issue with going to elements is it becomes kinda grody

I feel the easier way to do this is to just note that (A/p)[x] is isomorphic to A[x]/p[x]

This is covered in like, D&F

And you can probably even get that statement in terms of tensor products

Just using commutativity and associativity of tensor products

[vestal snow]

dang

[next obsidian]

I guess the problem with trying to show that in terms of tensor products is you have to show everything actually works as algebras

Which isn’t really hard per se, but just extra work

[vestal snow]

yeah you're right

[next obsidian]

I think it suffices to just show it’s multiplicative on the product of any two simple tensors

And then you know how the map is defined + algebra structure is defined on both tensor products so it’s immediate

[vestal snow]

Got it

I was thinking about what you said yesterday and instead of working out the details as I go along, I'm leaving them for once I'm done with all the exercises of this chapter

[next obsidian]

That’s a bold move, but maybe a good one.

[vestal snow]

A lot of the details repeat. For example, tensor of direct sum is the direct sum of the tensors

[next obsidian]

I could see it going well, or going horribly tbh

[vestal snow]

even when the direct sum is not finite

[next obsidian]

Oh yeah for sure

A lot of stuff with tensor products for me

was like doing exercises and in the exercises you come across some sort of thing where you want something to be iso

Like the tensor of direct sum thing

And so you prove it once, and show that it makes a diagram commute the way you want it to

And then you sort of have added a tool to your toolbox

These sorts of tensor product rules are super useful

And just building up as many as possible is what you wanna do

Knowing how to write localization in terms of a tensor product

How to adjoin a variable via a tensor product

Etc

[vestal snow]

I guess the difference between your approach and mine is that I'm crossing my fingers that by the time I'm done, the tools I've used are actually true

[next obsidian]

That’s why I said it could go badly, or well

Honestly most of the time if it seems true, it is

In my experience

[vestal snow]

The beauty of algebra

[next obsidian]

I think it has something to do with tensor product being universal

[vestal snow]

Plus, I can always ask someone on here to verify it for me

[next obsidian]

The fact that it’s impossible to work with with elements most of the time means the only way you can work with it is via these sorts of “obvious” stuff

Where obvious is the math version where obvious doesn’t mean you saw it immediately

But that after you see it you go “okay yeah, that’s the only way this could’ve gone”

[vestal snow]

"it looks too good not to be true, therefore it is"

[next obsidian]

Practically haha

[vestal snow]

unfortunately, I learned the hard way this doesn't apply to nigerian prince scammers

Well, thanks for the help. I'll get back to math

[next obsidian]

👍

[chilly ocean]

um could someone guide me on what to do for all three parts?
like for a, list permutations that correspond to rotations thru angles 120 or 240 degrees
um, how do you know what's 120 or 240 degrees?

[next obsidian]

if $0\to M\to N\to L\to 0$ is exact (modules) and $M$ and $L$ are free is $N$ also free?

[cloud walrus]

Mathemagician:

[latent anvil]

yes, L is projective so the sequence splits

@next obsidian

[next obsidian]

ohhhhhhj

hmmmmmst

that makes sense lmao

So I guess

[latent anvil]

you can probably phrase it really concretely

In terms of bases

[next obsidian]

I did this proof and part (a) was to show something, and part (b) something else

I had the above situation

and it would've finished part (a)

instead I did some whacky other thing and proved a lemma to get (a)

and my lemma is part (b)

lmao

[latent anvil]

Lmao

Does b depend on a?

[next obsidian]

Oh the lemma is, assume that M is finite, N is finite presentation, then L is finite presentation

nope

[latent anvil]

Nice

oh yeah that makes perfect sense

[next obsidian]

given the same exact sequence

[latent anvil]

pick a presentation for N

[next obsidian]

yeh

[latent anvil]

Add in the generators of M

[next obsidian]

yeah

[latent anvil]

As relations

That's a finite presentation of L

[next obsidian]

tfw you see it immediately...

I wrote out a diagram

and its cooler because the presentation

[latent anvil]

I mean yeah I would prove it with a diagram

[next obsidian]

becomes diagonal maps 😎

[latent anvil]

But I think the proof would be similar

Ugh I want to do algebra

[next obsidian]

You are

[latent anvil]

research is too hard

[next obsidian]

:^)

[latent anvil]

this is true lol

[next obsidian]

fucked up algebra, but algebra nonetheless

[latent anvil]

Oh you meant my research

Maybe

It doesn't feel like algebra

It feels like category theory

[next obsidian]

oh

algebra

jk

[latent anvil]

lies

I emailed the person who wrote the book

That was confusing

it feels bad to do this

[next obsidian]

like with a question on a proof?

[latent anvil]

yeah

I mean the proof was "by inspection"

He used those words

[next obsidian]

Oh yeah, dude I've been there

The difference is

you can't just text Mannigfaltigkeitsbegeisteter and ask him why it's true

since you're Mannigfaltigkeitsbegeisteter

[latent anvil]

Lmaoooo

This is true

[opal vale]

Let M,N be finitely generated modules over a principal ideal domain A. Which of the following conditions imply that M and N are isomorphic?

(i) $M^n$ is isomorphic to $ N^n$ for some positive integer n.

(ii) $M \oplus P$ is isomorphic to $N \oplus P$ for some finitely generated A-module P.

[cloud walrus]

Manas:

[latent anvil]

both imply it

Do you have any ideas for how you night prove i? @opal vale

It's worth noting that both can fail for A not a PID or M, N not finitely generated

[opal vale]

I think i have to use fundamental theorem

[latent anvil]

yeah, that's probably a good idea

[scenic sage]

Semetey?

[latent anvil]

?

[scenic sage]

nvm

[next obsidian]

Semetey

[next obsidian]

Commutative algebra is kinda cool. If you have a faithful module (Ann(M) = 0) over a commutative ring A which is Noetherian, then A is Noetherian since you can embed it into M^n

More generally if M is noetherian then A/Ann(M) is Noetherian

[latent anvil]

that makes sense

[clever violet]

This is a little dumb

but I can't seem to be able to proof why the order of g^x is p-1/gcd(p-1,x)

I know its true

but I don't know how to reason through it

[solemn rain]

wdym

[knotty mason]

what does it mean by invertible? multiplicative or additive

[solemn rain]

idk if its multiplicative

if it is then i dont see how i would do it

tbh

[knotty mason]

what you know is that 7 = 0

[solemn rain]

if its additive then

[knotty mason]

so you also have stuff like 4 = -3

[solemn rain]

1+1+1+1+1+1+1 = 0

so keep removing 1's

and u will get ur 5 invertible elements

right?

[knotty mason]

i think so yeah

[solemn rain]

yea

but for multiplicative

i tried to htink

every element is just a sum of 1s

but then 7 is prime so

u cant like have two things multiplied together (of 1s)

[knotty mason]

it's not true that every element of a ring will be 1+1+..

[solemn rain]

and go to 7 ones aka zero

[knotty mason]

for examples the Z[i] ring

[solemn rain]

uh oh

yea i have no idea for multiplicatrive

but for additive i think its what i said

[knotty mason]

but maybe sticking to the values generated by 1 would be good

3*5 = 15 = 7+7+1 = 1

[solemn rain]

whats 3 and 5

[knotty mason]

just 1+1+1 and 1+1+1+1+1

[solemn rain]

how do u know those are in the ring

oh

yea i was thinking about this

but i missed it

i was thinking (1+1+..)(1+1...) some times

yea yea cool

[hot bridge]

oww this is my question

😄

can you explain me ? @knotty mason

[upbeat juniper]

R contains an isomorphic copy of $(\mathbb{Z}/7\mathbb{Z})^\times$, and all of its elements have a multiplicative inverse

[cloud walrus]

some username:

[hot bridge]

this means all of this elementes invertible right ?

@upbeat juniper

[vestal snow]

Suppose $N_0$ is a submodule of $N$ and $\sum m \otimes n'$ is $0$ in $M \otimes N_0$, then $\sum m \otimes n'$ is $0$ in $M\otimes N$

Is there a way to prove this without using the construction of the tensor product

Using just the universal property

[woven delta]

Where does N_0 come into this?

[vestal snow]

Whoops my bad

[woven delta]

Just edit the original message

[vestal snow]

Suppose $N_0$ is a submodule of $N$ and $\sum m \otimes n'$ is $0$ in $M \otimes N_0$, then $\sum m \otimes n'$ is $0$ in $M\otimes N$
Is there a way to prove this without using the construction of the tensor product
Using just the universal property

[cloud walrus]

Have a Banana Bitch:

[vestal snow]

Given that the $n'$ are contained in $N_0$

[cloud walrus]

Have a Banana Bitch:

[woven delta]

I guess my intuition would be to draw the diagram for the tensor product of $M\otimes N_0$ with a balanced product $M\times N_0 \to M \otimes N$

[cloud walrus]

Liquid:

[woven delta]

This induces a map $M\otimes N_0 \to M \otimes N$

[cloud walrus]

Liquid:

[vestal snow]

Maybe we can do it using the fact that every bilinear map from M x N can be restricted to a map from M x N_0?

And then we can think of the natural inclusion of M x N_0 into M (x) N as a bilinear map that must factor through M (x) N_0

Oh my bad I didn't see your messages earlier

[woven delta]

Lol

[vestal snow]

Was that what you were saying?

[woven delta]

Yeah basically

You have to show that it factors though

[vestal snow]

Isn't that the universal property of the tensor product?

that any bilinear map factors?

[woven delta]

Any bilinear map from MxN

I would have to draw the diagram to verify this

One sec

[vestal snow]

if we restrict any bilinear map from M x N to M x N_0, we'll get what we need, won't we?

[woven delta]

Okay actually the map from $M\otimes N_0 \to M \otimes N$ does it without any special consideration

[cloud walrus]

Liquid:

[woven delta]

Because the bilinear map $M\times N_0\to M \otimes N$ sends $(m, n)$ to $m \otimes n$

[cloud walrus]

Liquid:

[vestal snow]

Is this what you're saying

[woven delta]

We are just defining the map as the restriction of the map M\times N \to M\otimes N

[vestal snow]

the different i represent the canonical mapping into the tensor products

[woven delta]

Yeah

[vestal snow]

alright

[woven delta]

From the fact that the diagram commutes you get that the simple tensors in the above one get mapped to the simple tensors in the below one

So that gives you that the sum of simple tensors in the M \otimes N_0 get mapped to the corresponding sum in M \oplus N

So if the sum is 0 in M \oplus N_0

Then it is 0 in M \oplus N

[vestal snow]

The last part is just using the properties of homomorphisms right?

[woven delta]

Yeah

[vestal snow]

Cool

Thanks for the help

[woven delta]

I'm not convinced the map is an injection though

(cause you drew the injection arrow)

[vestal snow]

Yeah I was thinking about that. Good thing we don't need injectivity

[woven delta]

Yeah!

[vestal snow]

actually

you're right

in general it is not an injection

[woven delta]

Do you have an example?

[vestal snow]

I can't think of an example, but here's how I can justify it. If you remember during the construction of the tensor product, we mod the free module of M x N by a submodule D which is generated by a certain kind of elements

let D_0 denote the submodule corresponding with tensoring with N_0 and D denote the submodule tensoring with N.

[woven delta]

Okay I see the problem that may come up

[vestal snow]

yeah

In general, D_0 does not equal D

so you get things not getting killed mod D_0 that get killed mod D

seems like after a week of banging my head against the wall, tensors, exactness, and flatness are starting to click

Thanks for your help though

[latent anvil]

have a banana, you're basically proving that M (×) — is functorial

And then applying that to the inclusion

[next obsidian]

That’s have a banana bitch to you

[next obsidian]

Spectral sequences r easy right, it’s just like reading pages of a book

I can read books

[old nimbus]

🤔

[olive mirage]

they're a not too complicated idea utterly ruined by the fact that you have to write them down 😛

[next obsidian]

Lmao

You don’t like 3 indices?

[olive mirage]

not on functions!

[next obsidian]

You know, I have worked through a bit of A-M, and a bit of Matsumura, and I must say I prefer Matsumura much more

I think Matsumura assumes more, or at least some stuff A-M develops in the start of the general theory of modules is relegated to the appendices of Matsumura, but I like the amount of stuff Matsumura covers. I find it much more interesting

I'm basically gonna throw every cool result from Matsumura in here primarily so Shamrock can see it lmao

If $I_1,\dots,I_n$ are such that $A/I_i$ is Noetherian and $I_1\cap\dots\cap I_n = (0)$ then $A$ is Noetherian

[cloud walrus]

Mathemagician:

[next obsidian]

It's sort of like saying Noetherian is a local property, but in a different way then the "every localization at a prime is..."

[mild laurel]

I think it might be a bit tough to explain what division algebras are. You need to talk about rings and fields, and then talk about algebras over fields

and then division algebras are special examples of algebras over fields

Yeah

It's the set minus operation, it means Q without the 0 element

It seems like they do it because you can't divide by 0

[upbeat juniper]

@hot bridge yes the nonzero elements in the isomorphic copy of $\mathbb{Z}/7\mathbb{Z}$ (there are 6 of them) are invertible

[cloud walrus]

some username:

[next obsidian]

Could someone check to see if this proof makes sense to them? $\$

Let $A$ be a local ring such that the maximal ideal $\mathfrak{m}$ is principal and $\cap_{n > 0}\mathfrak{m}^n = (0)$. Then $A$ is Noetherian and every non-zero ideal of $A$ is a power of $\mathfrak{m}.\$

Let $\mathfrak{m} = (s)$, then $\mathfrak{m}^n = (s^n)$ so it suffices to show every non-zero ideal of $A$ is a power of $\mathfrak{m}$ to show $A$ is Noetherian. Given any non-zero $a \in A$, there is a maximal $n$ such that $a \in \mathfrak{m}^n$, if not then $a \in \cap_{n > 0}\mathfrak{m}$ and would be $0$. This implies that we can write $a = us^n$ where $u$ is a unit i.e. not in $\mathfrak{m}$ else $a \in \mathfrak{m}^{n + 1}$. Let $I$ be a non-zero ideal, and take any non-zero $a \in I$, recall we can write $a = us^n$, now let $n$ be a minimal such $n$ so that there is a non-zero $a \in I$ with $a = us^n$. Every element of $I$ can be written as a multiple of $s^n$, so that $I \subseteq \mathfrak{m}^n$, but we see that $u^{-1}a = s^n \in I$ so that $\mathfrak{m}^n \subseteq I$ so that $I = \mathfrak{m}^n$, and thus we are done.

[cloud walrus]

Mathemagician:

[latent anvil]

they're a not too complicated idea utterly ruined by the fact that you have to write them down 😛
@olive mirage

that's such a mood with my research stuff right now

You should Google the operad axioms if you have a moment (especially if you state them in a general category...)

Turns out the thing that's been blocking me for a week is that the book I'm reading had a typo in its indices

Fun times

Also magician the Matsumura posting is appreciated lmao

@next obsidian proof looks good to me

[next obsidian]

Swag

[next obsidian]

Shamrock, another cool thing. If A is a noetherian ring then any surjective ring endomorphism is an isomorphism

[next obsidian]

Also here's something, A is noetherian iff all finite A-modules are of finite presentation

[latent anvil]

Why do we need noetherian?

Oh a ring endo might not be an A module endo

that's pretty sick

Both of those

Can you give a proof sketch or are they messy?

So like one direction of the noetherian iff result is obvious

The other is spookier

because you need finite generation to apply it in the first place

[next obsidian]

No it's easy easy

[latent anvil]

Oh okay

[next obsidian]

So for the first one, simply look at successive ker phi^n

if it doesn't start 0 it keeps growing

[latent anvil]

Ahh yeah I've done this in linear algebra

[next obsidian]

(To make it easiest I actually did phi^{2^n})

[latent anvil]

like no modules at all

Not this exact problem

but the technique is familiar

[next obsidian]

For the latter, for one direction you use this lemma

if 0 -> K -> M -> N -> 0 is exact, K is finite, M is finite presentation then N is

[latent anvil]

Right, we talked about this earlier

[next obsidian]

If something is finite it is surjected onto by A^n

Which is Noetherian

so K is finite

And A^n is finite presentation

[latent anvil]

what?

Oh are you proving the easy direction?

[next obsidian]

Assume Noetherian

[latent anvil]

yeah I got that

[next obsidian]

Actually the other direction is easier tbh

It's just sneakier

[latent anvil]

oh really?

Huh

[next obsidian]

Take I not f.f.

f.g.

[latent anvil]

I saw the finite => finite presentation immediately

[next obsidian]

Assume A not Noetherian

[latent anvil]

right

[next obsidian]

I not f.g.

Take M = A/I

Lemma

0 -> K -> M -> N -> 0 is exact

N being finite presentation and M finite

[latent anvil]

wait I'm confused

One sec

[next obsidian]

implies K is finite

[latent anvil]

so assume M finitely generated implies M finite presentation for all M

Yes?

[next obsidian]

Yes

[latent anvil]

Oh I missed the I in the first line

[next obsidian]

And that A is not Noetherian

[latent anvil]

I just read "take not finitely generated"

[next obsidian]

oh lol

Yeah so I is an ideal of A

not f.g.

[latent anvil]

Yup

Okay now I'm on board

[next obsidian]

Take the sequence 0 -> I -> A -> A/I -> 0

[latent anvil]

Right

[next obsidian]

If A/I is finite presentation since A is finite

I would be finite

[latent anvil]

Oh I didn't realize that held

Finite presentation => any presentation is finite?

Err, not exactly

[next obsidian]

Yup

sorta

[latent anvil]

because you can add extra generators

[next obsidian]

Actually that is true

[latent anvil]

But if it's like, exact

No?

[next obsidian]

If M is finitie presentation

for any surjection A^n -> M

the kernel is finite

not free

but finite

[latent anvil]

Yes, this is what I meant by "but if it's like, exact"

Which wasn't very clear

[next obsidian]

Ah

[latent anvil]

I was saying F -> F' -> M -> 0 might have F not finite rank even if F' is

because you can add extra generators

But if it's exact at the left

Then the left thing is f.g.

[next obsidian]

Ahhh yeah yeah

[latent anvil]

so the extra generators have to be kind of redundant

[next obsidian]

That would be ridiculous yeah

[latent anvil]

Anyways I still didn't realize that

That's pretty cool

[next obsidian]

Yeah so like

Look at this

https://mathoverflow.net/questions/1788/does-finitely-presented-mean-always-finitely-presented-answered-yes

I was literally just reading this before hand

But look at Brian Conrad's answer

It is mega not simple

But he describes some like general process I'm gonna take a more careful look at later

It seems like something that will eventually be a useful technique

[latent anvil]

Oh I think we discussed this on Twitter

This result

[next obsidian]

who is we

[latent anvil]

We not being me and you

just me and people on Twitter

[next obsidian]

and which result?

[latent anvil]

there was some very general discussion of it

The finite presentation thing

[next obsidian]

ohhh

gotcha

I mean it's sorta natural to wonder

and it's cool to see it works

[latent anvil]

https://twitter.com/schala163/status/1259957142674604032?s=19

-chloe (@schala163)

@sarah_zrf @mathemensch @BiCapitalize off topic but here's a fun (read: not fun) problem: If G is finitely presented, that means it has a presentation with finitely many generators and finitely many relations.

Prove that ANY presentation of G with finitely many generators al...

Twitter

Ah unfortunately one person has gone private

so it's a little hard to follow

[next obsidian]

Oh I saw bits of that

I remember Sarah saying compact

[latent anvil]

Yeah I understand why I didn't get the connection

Because this was for groups

Compact didn't mean compactness theorem though!!

[next obsidian]

Yeah, I kind of

read a bit

and was like "okay wtf"

and just left out of the thread

[latent anvil]

I do not understand everything I read on Twitter

I'm emailing the author of a textbook two days in a row

Because I think his correction has flaws

tfw

Gonna be embarrassing when he does a very simple computation and resolves it all for me

[next obsidian]

I mean

If this stuff is so new and niche

I think he might be happy to have someone care about it

And maybe even more to care enough to be that careful about it

¯_(ツ)_/¯

[latent anvil]

that's fair

I also thought the same thing yesterday

[next obsidian]

Did you see I shared my like favorite problem

[latent anvil]

And it turned out to be a legitimate error

[next obsidian]

and it was S^G

[latent anvil]

No I didn't

[next obsidian]

And zoph freaked out haha

[latent anvil]

Where was this?

[next obsidian]

i forget maybe chill?

[latent anvil]

Had he seen it before?

[next obsidian]

Or one of the discussions

he saw it in the FB group

and couldn't ever solve it

[latent anvil]

Lolllll

[next obsidian]

and brought it to his math club I think

Then I said James Baxter posted it originally after you posted it to reddit

[latent anvil]

Lmao 10/10

[next obsidian]

and then it devolved into James Baxter said he likes casual sexism for like 3 min

[latent anvil]

lmaoooo

sounds right

[next obsidian]

or something to that effect idk

[latent anvil]

I thought he was banned like last week

[next obsidian]

They said they think so too

And maybe he came back in on a different account or something? idk

But like

Liquid made a hella funny joke that he got banned for asking very specific analysis questions

[latent anvil]

loooool

[next obsidian]

and I was like "I could see that given what I've seen on FB"

I'm gonna keep Matsumura posting

so I guess you're gonna see highlights of Comm Alg

[latent anvil]

h*ck yeah

[next obsidian]

Fractional ideals make my brain bleed

[latent anvil]

oh wait one thing first

[next obsidian]

I don't like them, they're weird

[latent anvil]

people were talking about my research topic in AGS!!!

[next obsidian]

I saw that

[latent anvil]

also, don't those show up in ANT?

[next obsidian]

I didn't know that's what you did but i saw your reply

[latent anvil]

I think Thomas told me that's what the class group was or smth

[next obsidian]

yeah I posted in the NT channel

asking

and yeah Zeta said that

[latent anvil]

Yeah I'm trying to generalize quantum groups ig

oh lol

[next obsidian]

that's the most basic way to describe them

But there's better ways to think about them

[latent anvil]

I just caught up

Maybe bug Thomas?

[next obsidian]

Maybe haha

[latent anvil]

For help understanding them

[next obsidian]

Idk if they'll continue showing up

Like, for what I'm doing rn

If I do ANT and they crop up there

I'm sure I'll just get used to them there

And I don't think AG uses them really? At least for what I'm currently doing so I'm not sweating over it

But like apparently invertible fractional ideals are projective

and like

🆗

I should start doing AG again

I just haven't wanted to figure out that one problem

We discussed like a week ago lmfao

[latent anvil]

yeah I think the vakil study group is staring late June

2021

[next obsidian]

Probably

Lol

I'm having fun doing Matsumura

[latent anvil]

That's awesome!

[next obsidian]

I like it a lot more than A-M

It makes me actually excited to be learning comm alg

It's pretty fun

I just like seeing all the like upgraded versions of stuff

that like normally would be like "one can prove more generally... but this is beyond the scope of this text"

[latent anvil]

Lmaoooo

[next obsidian]

And all this like "if this, then A is Noetherian"

I now know many more ways to show something is Noetherian

Which is like, cool

[latent anvil]

Yeah I am probably putting algebra in general on the backburner next year

[next obsidian]

Damn

[latent anvil]

Which is sad

[next obsidian]

So like, just 508 and 509 right?

oh unless

you graduate early

in which case ???

[latent anvil]

Well yeah my current plan is no 508 which is such a meme

Because I wanted to take that so badly

[next obsidian]

RIP

[latent anvil]

here's my current plan

Psych in fall will be moved, I'm not taking it twice

[next obsidian]

Is 548 the riemannian thing

[latent anvil]

this would fulfill all my reqs

Yeah

No

It's bundles

547 is riemannian

[next obsidian]

oh

[latent anvil]

I think I will take 509, but I could take complex instead...

[next obsidian]

I mean 1 quarter?

Oh wait

it'll be the first quarter?

[latent anvil]

Which one?

[next obsidian]

umm

like it's the complex that's a part of the series right?

Since they flipped when they happen

[latent anvil]

Yeah 524

Will happen in spring

[next obsidian]

Since for a second I thought it would be the riemann surface

And I was like wut

[latent anvil]

Oh lol no I'm not enough of a chad

I would be taking complex geometry

[next obsidian]

So like

[latent anvil]

But Lee won't be teaching it

[next obsidian]

you do cover more than 336 did

But the first half or so of it is stuff you saw before

but from a different development path

[latent anvil]

The thing is, it'll be two years since I took 336

[next obsidian]

True

[latent anvil]

I've already forgotten most of it

[next obsidian]

Like I don't think I even once

Used the CRE

[latent anvil]

whereas you took it right after

Wow

[next obsidian]

Tbh

[latent anvil]

anyways this schedule looks pretty reasonable on paper and I kind of want to shoot for it

I wouldn't be TAing

or doing research or anything

[next obsidian]

So like

what's the alternative

Say you go with the original plan

you had

[latent anvil]

Don't take CS next year

Take all of algebra next year

Maybe do research instead idk

[next obsidian]

I'm just thinking like

what would you do your senior year

[latent anvil]

ANT, AT, senior thesis

[next obsidian]

Like, what, do combo?

[latent anvil]

Maybe AG take 2

[next obsidian]

Hmmm

[latent anvil]

and also CS

Yeah gabe I'm taking bundles next winter

Not AG bundles

diff geo bundles

[next obsidian]

Also we should switch rooms

Maybe like math-discussion

[latent anvil]

Oh yeah lol

[cloud walrus]

Manas:

If M and N is finitely generated module over the local ring (A,m), can we say $$ (M\otimes N)\otimes A/m =  (M\otimes A/m)\otimes  (N\otimes A/m)$$

[hot lake]

so aren't you basically asking if $A/m \otimes A/m = A/m$ ?

[cloud walrus]

Zef Klop 🍃 🌿 🌻:

[latent anvil]

Are you tensoring over A or A/m on the right hand side?

[opal vale]

Ok got it. Thanks

[opal vale]

Suppose M be a module over a Noetherian ring A and $ Ass M = Ass A$. Then have M to be a free module??

[cloud walrus]

Manas:

Suppose M be a module over a Noetherian ring A and $ Ass M = Ass A$. Then have M to be a free module??

[opal vale]

<@&286206848099549185>

[chilly ocean]

what is Ass?

[opal vale]

Set of all associated primes

[wooden ember]

nice

along with the sexy primes, math is full of surprises

[elder valley]

doesn't every elliptic curve defined over a finite field have complex multiplication because of the frobenius endomorphism?

[oblique river]

yes

[elder valley]

and if i understand correctly by the classification of the endomorphism algebra, the endomorphism ring has at most 1 other independent endomorphism. is there some way to determine what that morphism is?

[oblique river]

other than frobenius?

I'm not sure but I'll think about it

[hot lake]

the dual of frobenius ?

[elder valley]

yeah. the classification is either Q, an order in an imaginary quadratic extension of Q, or an order in a quaternion algebra over Q. the second case would be just frobenius, but i'm wondering what the other endomorphism is in the third case

[oblique river]

can't the dual of frobenius be equal to frobenius?

[hot lake]

under that classification iirc the dual of frob would be the complex conjugate of frob in whatever imaginary quadratic extension

I don't remember if I have seen cases with a quaternion algebra

[oblique river]

it's a quaternion algebra if and only if the curve is supersingular

which can certainly happen

[olive mirage]

which was Serre's proof you couldn't get a cohomology theorem for curves over finite fields with coefficients in Z

[elder valley]

so how do you find an endomorphism $\psi$ that's $\mathbb{Q}$-linearly independent over the $\mathbb{Q}$-algebra generated by frobenius and 1? for a supersingular curve over positive characteristic

[cloud walrus]

Auvera:

[vestal snow]

How do I prove that if I is generated by pure powers, then it is irreducible?

[hot lake]

you mean if I is a monomial ideal and is generated by pure powers then it is irreducible ?

[vestal snow]

yes

[kindred mist]

Silly question, can someone explain why in F[x], (x) intersect (x + 1) is nonzero (where F is a field)? Does this depend on the field or is it nonzero for any field?

[hot lake]

it contains x(x+1)

[kindred mist]

yeah

idk why I didnt think of that, thanks

[next obsidian]

Who wants to help with some (maybe hard) commutative algebra?

Here's the setup, we have integral domains A,B and A is a subring of B

We know that there are finitely many primes of B which lie over (0), and that B is a f.g. A-algebra

Show that Frac(B) is a finite extension of Frac(A)

I think I can show that Frac(B) is algebraic over Frac(A)

[uncut girder]

Lie over A means contains A?

[next obsidian]

Whoops

I meant lie over (0)

As in p \cap A = (0)

So showing Frac(B) is algebraic over Frac(A) is easy since any element is a/b with a,b in B, so they are integral over A => algebraic over Frac(A) => quotient is algebraic over Frac(A)

But now I need to somehow show that the extension is finite :/

Maybe I can do something like, show that any set of n linearly independent elements of Frac(B) descends to n unique primes of B which lie over (0)?

That would give it to me

[uncut girder]

I doubt that's true

[next obsidian]

I don't really see any other way to go about this haha'

And apparently this statement is true since someone told me this is how I solve a Hartshorne problem 🙃

Grub help me

[uncut girder]

Do you have examples of integral domains A contained in B such that B has more than just the zero ideal lying over (0) in A

[next obsidian]

Z and Z[i]

I think

the ideal (i)

[uncut girder]

(i) = Z[i]

[next obsidian]

wut no

[uncut girder]

Because i is a unit

[next obsidian]

hmmmmmmmmmmmmmssssssssssstttttttttttttttttt

Oh yeah

true true

what about (2i)

Isn't prime

wait no it is

[uncut girder]

Intersect Z is (2) which isnt (0)

[next obsidian]

yeah

Fudge

These things must exist, otherwise the whole concept of generically finite seems really stupid

At least between integral affine schemes

Okay this is really contrived but what about (0) inside of any ring?

Or like a field contained in a ring?

You can't intersect the field since then you'd contain a unit

Or you can consider like Z[x], there are prime ideals corresponding to irreducible polynomials and they intersect Z trivially

By dimension stuff

In fact that is an example with infinitely many primes lying over (0) I think

Just take (x - a) over all a in Z

[uncut girder]

I see

[next obsidian]

;w;

I think Frac(Z[x]) is Q(x)?

Either that or Z(x), but I'm pretty sure it's Q(x)

If it is Q(x) then this shows the hypothesis about the finitely many primes lying over (0) is needed

Since Q(x) is transcendental over Q even though Z[x] is a finitely generated algebra over Z

[uncut girder]

What about (Z/pZ) and (Z/pZ)[x]

[next obsidian]

I think it's the same thing

there are infinitely many primes in (Z/pZ)[x] right?

I think?

Actually yeah it does, F[x] always does

you can just emulate Euclid's proof of infinitely many primes in Z

[olive mirage]

if F is finite then F[x] has an irreducible polynomial of eveery degree

(this is equivalent to the existence of finite fields of all prime power orders)

(I'm not arguing with @next obsidian 's correct proof, btw. Just giving a stronger result)

[next obsidian]

is it like x^n - 1

Since the splitting field has order p^n

[olive mirage]

x^n-1 is never irreducible

[next obsidian]

Wait, then how do you form the finite fields

Oh is it cyclotomic stuff?

I am really bad with finite fields

[olive mirage]

well, use an irreducible polynomial of degree n.

[next obsidian]

[olive mirage]

but a trick is if you want to make the field with p^n elements, then x^(p^n)-x splits in that field by Lagrange's theorem

and, not so hard to prove, it is the splitting field of that polynomial over F_p

[next obsidian]

Oh sure

take a derivative

[olive mirage]

yeah, it depends what you're already assuming you know

but that is one way to prove existence and uniqueness

that set has order p^n and is a field, inside kbar

but you can use that plus mobius inversion to get a formula for how many irreducibles there are of each degree

[uncut girder]

Hm nvm

[olive mirage]

When I was an undergrad I was reading Berlekamp and they give an incredibly tedious proof of the existence of irreducibles of each order, and I found the mobius inversion proof and wrote it up and sent it to one of my profs, and was very heartbroken when he replied with "Yes, this is correct, this is the standard proof!"

[next obsidian]

😦

I remember on the first day of calculus I went home and read the textbook and noticed how they used limits, so I just tried to emulate it and found the derivative of x^n and when I told my prof he told me that that's a derivative and I'm not allowed to use it on tests until we showed that rule

lol

[woven delta]

Lol

I can barely do derivatives tbh

[dawn kiln]

tbf it's annoying, but it kinda makes sense

[olive mirage]

I always isolate one test where people can't use derivative rules, because it is annoying for students, and I want to minimize the torture.

[dawn kiln]

for our discrete maths course, we weren't allowed to use the fundamental theorem of arithmetic for a lot of our proofs which while annoying, forces you do do things without assuming a more powerful theorem

(even if you already know the proof)

[olive mirage]

it is really hard to appreciate that the fundamental theorem of arithmetic is highly non-trivial and has a multipart proof.

[dawn kiln]

we covered the proof in our class

[olive mirage]

but understanding the parts of the proof leads to a lot of neat stuff

yeah I was just talking about it pedagogically haha

[dawn kiln]

tbf the more egregious tool that we weren't supposed to use to begin with was induction lol

[scarlet estuary]

urm

what axiomatization of the integers were you using

that doesnt have induction lmao

[dawn kiln]

it was a bit of a strange course, and we did use it in later proofs

[scarlet estuary]

or was it just a matter of "dont do it for now, we'll cover the methods later"

[dawn kiln]

ye

[next obsidian]

??? lol

[dawn kiln]

but i personally think it's too fundamental to leave out for that long lol

[scarlet estuary]

i mean its literally an axiom of N

[next obsidian]

Unless you construct the integers explicitly and show induction is valid

[scarlet estuary]

(or a bunch of axioms)

[dawn kiln]

the course is run by the CS department, it's a bit of a mess as it currently is

[scarlet estuary]

hard to get more fundamental than that

[next obsidian]

I don't see how that benefits anyone lmao

[dawn kiln]

so it's less about starting from the minimum assumptions and working up, and more about teaching compscis to write proofs 🤷‍♀️

[next obsidian]

Yeah lol, induction isn't an assumption

It's just built in lol

[scarlet estuary]

yeah but surely induction should be one of the first proof techniques covered

[next obsidian]

^^^

Hell compsci people do induction

on like binary trees or something I don't really remember

[dawn kiln]

yeah tons of things

[next obsidian]

I just remember my comp sci friends telling me about it

[uncut girder]

Guys

[dawn kiln]

yeh sry

[uncut girder]

Wrong channel

[dawn kiln]

will move to #discussion

[uncut girder]

I think this implies the only ideal lying over (0) is the zero ideal in B

Say P is an ideal lying over (0)

And x is an element of P

x is algebraic so satisfies a minimal polynomial f(t) with coefficients in A.

If the constant term of f(t) is 0, then B contains zero divisors, as you can write f(t) = t•g(t), and 0 = x•g(x).

So the constant term is some nonzero element of A, call it a.
But then f(t) -a=t•g(t) for some g(t) nonzero, and x•g(x) is in the ideal generated by x, and is equal to -a which is nonzero.
So P actually lies over more than (0), contradiction.

@next obsidian

I dont know why finitely many primes living over (0) would be an assumption in the exercise if uniqueness of ideal lying over (0) can be proved from algebraic

Does that make sense?

[next obsidian]

Oh shit, I realize my proof is mega wrong for the fact that Frac(B) is algebraic over Frac(A) lmao

I erroneously used finite type extension => integral but that's just straight up wrong

That's only true for a finite extension, i.e. finite as a module

Take like Z[x] over Z, Z[x] is generated by 1 and x over Z as an algebra, but clearly x - 1 is not integral over Z, it doesn't satisfy any non-zero polynomial in Z[t]

let alone monic

So your proof holds if B is a finite extension of A because then you know that x does actually satisfy a polynomial with coefficients in A

@uncut girder

Also sorry I was taking a jog

[next obsidian]

This is the snake lemma

[vestal snow]

How do I prove that Ker$(d) \subseteq$ Im$(\tilde{v})$

[stone fulcrum]

List of things I really don't understand yet: Exact sequences

[next obsidian]

Okay so banana

You know how the map is made right?

oh wait that part isn't too hard I think

Showing exactness at the first map from the cokernel is harder I think

Wut

[cloud walrus]

Have a Banana Bitch:

[next obsidian]

Do you know how the map is defined?

[vestal snow]

Yup

I'll share what I have tried

One sec

[bleak abyss]

I think this is epsilon less general actually

[next obsidian]

Yeah

you don't need two of the 0s

but then you don't get the 0s at the end of the les

[bleak abyss]

Yeah this is a thing where I'd prefer to have it as two separate statements

[next obsidian]

Same, but A-M do it that way

[bleak abyss]

Yeah I'm criticizing A-M here

[next obsidian]

It sort of can matter tho but I'll just tell them. So for the first row, you don't need that first 0, but if you don't have it the LES doesn't start with a 0

Likewise you don't need the last 0 on the second row, but then the LES doesn't end with a 0

[vestal snow]

Suppose $m''\in$ Ker$(d)$. Then $f''(m)=0$ for some $m\in M$.

$f(m) = u'(n')$ for some $n'\in$ Im$(f')$. Say $n'=f'(m)$ for some $m'\in M'$

[cloud walrus]

Have a Banana Bitch:

[vestal snow]

We need to somehow show that $m$ is in Ker$(f)$

[cloud walrus]

Have a Banana Bitch:

[woven delta]

Oh so for this sort of thing there's a nice diagram you can draw

[next obsidian]

So take something in ker gamma which is mapped to 0, call this a

[bleak abyss]

Oh tbh I didn't even see that extended exact sequence with ker(f) and coker(g') lmao

[next obsidian]

so a is in C, then you pull it back to something in B up to something in ker g, so call this element b

[bleak abyss]

I just saw them say "Oh if f is injective then the map ker(alpha)->ker(beta) is

I prefer this tbh

Packages everything in one go

[next obsidian]

You then map it to $\beta(b)$, and you know that $g'(\beta(b)) = 0$ by commutativity so that you can pull it back to something in $A'$, call this element $c$

[cloud walrus]

Mathemagician:

[next obsidian]

Now we know $c$ maps to $0$ in the cokernel

[cloud walrus]

Mathemagician:

[next obsidian]

By definition of the snake map

Right?

[vestal snow]

@next obsidian Are you defining the map $d$?

[cloud walrus]

Have a Banana Bitch:

[next obsidian]

Yeah I'm working through how it's defined

since the next step involves going back up

[vestal snow]

I already know the definition

[next obsidian]

Sure, but I had to set it up

[vestal snow]

Oh okay

[next obsidian]

$c$ maps to $0$, so that it is in the kernel of that map to the cokernel, but by exactness there exists a $d \in A$ such that $\alpha(d) = c$

[cloud walrus]

Mathemagician:

[next obsidian]

Now remember we had a $b \in B$, note that $g(b) = g(f(d))$ by commutativity

Make sense?

We pulled back our original $a \in \ker\gamma$ to this $b$, and $b$ is defined UP TO SOMETHING IN KER $g$

[cloud walrus]

Mathemagician:

[next obsidian]

This last point is mega important

[vestal snow]

One second I'm having some trouble parsing this stuff

[next obsidian]

Sure, try and work it out until you agree with what I've written so far

[cloud walrus]

Mathemagician:

[next obsidian]

I edited something since I actually fucked it up a little

[vestal snow]

When you pull back something from C to B, how do you know its in Ker g?

[next obsidian]

You don't

So $g$ is surjective

[cloud walrus]

Mathemagician:

[next obsidian]

so there exists something which maps to $a$, but I mean it's only well-defined up to something in Ker $g$

[cloud walrus]

Mathemagician:

[next obsidian]

So I got a $b \in B$ such that $g(b) = a$, but actually if $k \in \ker g$ then $g(b + k) = a$ as well

[cloud walrus]

Mathemagician:

[next obsidian]

This means I can sort of wiggle $b$ up to something in the kernel

[vestal snow]

I don't get what you mean by "well-defined up to something in Ker G"

[cloud walrus]

Mathemagician:

[next obsidian]

and everything works fine

[vestal snow]

Ah so you can add by something in the kernel without changing where it maps to?

[next obsidian]

Yup

So I show we can actually WLOG that $b$ is mapped to 0 by $\beta$ by adding something onto $b$ such that it now maps to $0$

[cloud walrus]

Mathemagician:

[next obsidian]

so i can replace b with b + that something

and everything still works fine since b + something still maps to a

But you'll see what I mean in a sec, this is why I took the time to grab some $d \in A$

[cloud walrus]

Mathemagician:

[next obsidian]

So I'll repost the image, and then do everything that we've done up till now