groups-rings-fields

[solemn rain]

what are those

just F?

[latent anvil]

Invertible ones

yeah

[solemn rain]

ohh

invertible

yeaaa

does it break only when n =1

[latent anvil]

Well I don't know if it breaks when n = 1 or when n ≠ 1

but it seems unlikely in either case

[bleak abyss]

So yeah there are fields with non-isomorphic unit groups

So n=1 is wrong for sure

[latent anvil]

I was trying to think of one but failing

[solemn rain]

meeh i gotta go

sleep

this seems hivh level

as someone in the comments suggested words that i do not know yet

[bleak abyss]

Again it turns out to be wrong

[solemn rain]

wait

sloth center of GLn(F)

is F why

[bleak abyss]

F^{\times}

[solemn rain]

yea

y

why

[bleak abyss]

It's just scalar multiples of the identity

Yeah found a counterexample on MSE: https://math.stackexchange.com/questions/624289/examples-of-non-isomorphic-fields-with-isomorphic-group-of-units-and-additive-gr

Now GL_n(F) for n > 1 is gonna be tougher to break

[solemn rain]

Gl_n is isomorphic as group to Gl_m iff m=n (over any field)

[latent anvil]

Nice

That's very different lol

[solemn rain]

and GL(n) is F^multipliy

center*

[bleak abyss]

Wait yeah lol this just says like

GL_n(k) cannot be isomorphic to GL_m(K) if m\ne n

This says nothing about the fields

It says that like

GL_2(R) can't be iso to GL_3(C(t))

Or something

[solemn rain]

okaay

someone did acomment

suggested some big words that idk

should i post it?

okay i will

||It should be possible to identify the Borel subgroup as a maximal solvable subgroup and then compute the length of its derived series.||

idk what the fuck are those

what are ||borel subgroups||

ik the rest

[bleak abyss]

That's an AG thing

[solemn rain]

oh okay

[bleak abyss]

So a good bit above your head

[solemn rain]

yea

XD im supposed to learn AG soon hahaahaha

[latent anvil]

oh cool

I figured that the derived series or a central series or something would come up

[bleak abyss]

Not soon

[latent anvil]

but did not expect AG

[bleak abyss]

I feel like you should spend a good bit of time bulking up on easier background

[solemn rain]

yea yea i will not learn AG b4 reading analysis and topology texts atleast

AG is supposed to be grad right?

so yea this comes much later

ig

anyways

i will leave you to the problem

cuz i wanna sleep gl boys and you dont have to thank me for my amazing problems

[prime gale]

GL boys.😆

[tribal pasture]

Generally linear boys

[chilly ocean]

@solemn rain yo homeboy where did you learn group stuff from

[carmine fossil]

What is this secret code?

[woven obsidian]

I'm trying to understand this proof of cayley-hamilton for modules

Do they mean that Delta is the matrix with the element in row k, col i being $\varphi \delta_{ki}-a_{ki}$ ?

[cloud walrus]

AoiKunie:

[woven obsidian]

Because for a fix k,i $\varphi \delta_{ki} - a_{ki}$ is just a matrix with elements in A, right?

[cloud walrus]

AoiKunie:

[woven obsidian]

I figured it out

[leaden finch]

can someone help me with this one

this is my scratchwork but i got stuck

[woven delta]

@leaden finch suppose p is a product of 2 other integers

[leaden finch]

oh, why do we suppose that

[chilly ocean]

to find contradiction

[woven delta]

Also this belongs in #elementary-number-theory

[acoustic lodge]

hey im reading through harold pinters abstract algebra at the moment. It is very easy to read. What book should i read after this one? I want to focus on number theory to understand ramsey theory better. I think i have to learn galois theory as well, (the pic of my avatar). Any good recommendations? Lecture notes also fine but a lot of the ones ive found have mistakes in them

[prime gale]

Serge Lang's book "Algebra" is pretty standard for second Algebra course.

Also really cheap compared with most.

At least it was in second edition published by Addison Wesley. Looks like third edition is Yellow, thus expensive.

[acoustic lodge]

ty

[eager willow]

looking at the representation for the cyclic group order $p$ $C_p$ given by the generator $$A =\begin{pmatrix} 1 & 1 & 1\ 0 & 1 & 0 \ 0 & 0 & 1\end{pmatrix}$$ over $F =\mathbb{F}_p$.

I found the proper sub $FC_p-$modules that look like the spans of ${(1,0,0)^T}, {(1,0,0)^T, (0,1,0)^T}, {(1,0,0)^T, (0,0,1)^T}$. Are there any others? How can the module afforded by this representation be a direct sum of nonzero submodules?

[cloud walrus]

datorangeguy:

[eager willow]

p being an odd prime in case that wasn't clear

[golden pasture]

serre lang algebra

jacobson basic algebra is gud if pinter doesnt superset it

[smoky cypress]

What is model theory?

what are the prereq

and what's a good book on it

[prime gale]

Model theory is part of set theory.

[scarlet estuary]

ehhh

it's certainly part of formal logic

but you dont need sets to do model theory

[leaden finch]

can someone help me with this one. I proved it this way

[prime gale]

I guess so. But I've never heard of any models not based on sets. Though I never studied either set theory or model theory.

[scarlet estuary]

anyway, are you aware of the difference between semantic and syntactic entailment? @smoky cypress

[smoky cypress]

No

[scarlet estuary]

okay so

[smoky cypress]

I haven't really learned any formal logic

[scarlet estuary]

some theorems/statements can be derived directly from the logical rules of a theory itself

this is called "syntactic entailment/provability"

ie it comes from the syntax of the theory

when we start adding interpretations to these syntactic manipulations, such as interpretations of truthhood and falsehood

or when we start considering how we can, well, model these syntactical rules within a larger overall system

this is called the "semantic" level

and "semantic" entailment would mean something that is true in all models

obviously, if something is syntactically true, it has to be semantically true or your system doesnt make sense

(we call this "soundness" or "consistency")

[prime gale]

Isn't there a theorem saying they are the same thing?

[smoky cypress]

Well that's very vague, but I guess if I really want to learn about it I'll have to read some model theory book

[scarlet estuary]

the converse doesnt necessarily hold however

[prime gale]

Completeness?

[scarlet estuary]

completeness is that semantic entailment implies syntactic provability

it doesnt hold in all formal systems

[prime gale]

I see.

[scarlet estuary]

but it holds in, say, first order logic as a result of godel's completeness theorem

anyway

the study of syntactics is called "proof theory"

and the study of semantics is called "model theory"

roughly speaking at least

[smoky cypress]

oh ok

[scarlet estuary]

(naturally, in practice there tends to be a lot of crossover and stuff)

here's a more "concrete" example since so far i've been pretty abstract

you may be familiar with the peano axioms of arithmetic on N

[smoky cypress]

yeah

[scarlet estuary]

one important result is that these define a unique model of the naturals

if you have a set that follows the rules of the peano axioms, it must be the same thing as N

(up to notational differences)

[smoky cypress]

Yeah

[scarlet estuary]

what if we don't say that, however?

er

let me rephrase

what if we "weaken" the peano axioms?

[smoky cypress]

Like if the successor function is not injective

Then we'll get different things

Like Z/nZ

[scarlet estuary]

well, not quite

we could potentially get different things

Z/nZ would be a valid model of such an axiomatization

[smoky cypress]

Yeah

[scarlet estuary]

but N would also be a valid model

since both of these follow all the syntactic rules, once that axiom is removed

they just have different interpretations

in fact, i handwaved slightly when i said the peano axioms uniquely define N

depending on how you axiomatize PA, there could be multiple models

(speciifcally the first-order axiomatization permits multiple models while the second doesnt IIRC)

[smoky cypress]

Ok

[scarlet estuary]

however, any model of PA contains N

[smoky cypress]

Well I'm not exactly sure what first order logic means

[scarlet estuary]

(or something isomorphic to it)

[smoky cypress]

Ok

Yeah that makes sense

At least intuitively

[prime gale]

In fact, there is no way any set of first order axioms can characterize N.

[scarlet estuary]

indeed, by completeness combined with incompleteness

we can make some model theoretic statements about nonstandard models of N

tennenbaum's theorem is quite an important one

models of PA cannot be recursive (nor can their addition/multiplication)

to give a concrete example of a nonstandard model of PA

consider the set of all sequences of natural numbers

N^N

call two sequences equal iff they agree except on a measure 0 subset

this is a nonstandard model of arithmetic with usual semiring operations

this is called an "ultraproduct" construction

[prime gale]

What is measure zero subset?

[scarlet estuary]

in this context, iff they agree except on finitely many entries

you might know the term "almost everywhere"

[woven delta]

That's not an ultraproduct then

[hot lake]

one important result is that these define a unique model of the naturals
@scarlet estuary whut

[scarlet estuary]

i clarified later

that that was handwaving

[woven delta]

Oh sorry

[scarlet estuary]

they dont define a unique model but every model contains N

up to iso

[smoky cypress]

Oh ok

[scarlet estuary]

@woven delta wdym

[smoky cypress]

What's the successor then?

[prime gale]

Add one to everything.

[hot lake]

all the operations are taken components wise

for the ultra product

[eager kindle]

What is a perfect finitely presented group?

[smoky cypress]

Ok

[hot lake]

remind us what is a perfect group

[woven delta]

Also could you clarify what you mean by measure on N? Do you mean like density or something

[hot lake]

zero measure here means finite subset i think

[scarlet estuary]

in this case its just a trivial measure

[woven delta]

Then how are you talking about ultraproducts?

[eager kindle]

I just recently learned what a finitely presented group, Zef. XD I have to summarize this paper that requires a lot more knowledge of group theory then I currently have

[hot lake]

looks like you need to go definition hunting

[woven delta]

Oh I thought you were handwaving measure, not the ultraproduct thing

[scarlet estuary]

i was handwaving zef's thing

about defining a unique model

[hot lake]

wait doesn't the ultraproduct need a nonprincipal filter or something wonky like that

[scarlet estuary]

which PA does not do

but yes im handwaving this too

and yes you need a nonprincipal ultrafilter

[woven delta]

Lol Zef that just means an ultrafilter which is an extension of the cofinite filter

It's not too wonky

[hot lake]

and you need axiom of choice to find one or something

[woven delta]

No

[hot lake]

really ?

[woven delta]

The ultrafilter axiom is weaker than choice

[hot lake]

that's kinda still not in ZF then ?

[woven delta]

Yeah

[scarlet estuary]

@woven delta anyway define a finite additive measure on your indices of the sequence in the "dumb" way, ie m(A) = 1 if A is in your ultrafilter, 0 otherwise. then definign equivalence by equal almost everywhere on the indices and quotienting by this gives you your ultraproduct

assuming your ultrafilter is nice enough

[woven delta]

Yeah

That's okay

[scarlet estuary]

thats what i was getting at

trying not to dive into too much explicit fuckery here

[woven delta]

Sorry for being pedantic lol

[scarlet estuary]

its fair

the notable thing here is that

this model of PA is (isomorphic to) the hypernaturals

ie the hyperreals >= 0 with fractional part 0

[woven delta]

Hmm

[scarlet estuary]

("fractional part" extended the natural way)

[woven delta]

I guess if you use the same ultrafilter to construct the hyperreals sure

I could see that making sense

[scarlet estuary]

so in practice this is like

N but it also contains "infinitely large naturals"

if you restrict your view to just N though

[woven delta]

You can actually explicitly write some infinite elements down

[scarlet estuary]

you have standard arithmetic

yep

[woven delta]

Using the fact that every cofinite set is in your ultrafilter

And Los' theorem

[smoky cypress]

Interesting

[woven delta]

Like an infinite element just mean a sequence that goes to infinity

I guess I should say something about Los' theorem actually

So that these statements make more sense

[prime gale]

Thanks. I learned about explicit construction of whacky naturals in first order PA. Is there any explicit construction of whacky naturals in ZF?

[smoky cypress]

I feel like I've asked this before, but what should I read to know basic logic

[slate forum]

https://imgflip.com/i/4eui8d

[prime gale]

Stalks?

[hot lake]

spectral sequences ?

[woven delta]

So Los' theorem says that in an ultraproduct, for a first order formula $\psi(x_1, x_2, ..., x_n)$ in n variables, the n-tuple of elements $a_1, a_2, ..., a_n$ of your ultraproduct satisfy $\psi(a_1, a_2, ..., a_n)$ iff the set of indices i such that $\psi(a_1(i), a_2(i), ..., a_n(i))$ is true in $M_i$ is in your ultrafilter

[smoky cypress]

sheaf

[cloud walrus]

Liquid:

[slate forum]

Sheaves, stalks, germs, kernel

[woven delta]

So in particular this applies to stuff talking about less than

So consider the fomula $x> n$ for each n

[cloud walrus]

Liquid:

[woven delta]

(where n just mean 1+...+1 n times)

Then a sequence which goes to infinity satisfies each of these formulas for all but finitely many indices

So therefore such a sequence a has $a>n$ for each n

[cloud walrus]

Liquid:

[woven delta]

And that's what we call an infinite element

This theorem also tells us ultraproducts where each structure we are indexing is the same has the same first order theory as that structure

So that's why this ultraproduct is a model of PA

@smoky cypress read enderton and marker

Enderton's book is called a mathematical introduction to logic and Markers book will come up if you Google Marker model theory

[smoky cypress]

Alright

Thanks

[smoky cypress]

Lmao

[scarlet estuary]

yeah sorry i shouldve clarified there that

that is a very misleading statement

lmao

you wouldve been justified

not just incompleteness but also

compactness

[thorn delta]

Let H < S_n such that [S_n : H] = 2. How would I show that H contains a 3 cycle?

[prime gale]

Doesn't. Take n=2.

[thorn delta]

Eh, okay. first suppose that S_n contains a 3 cycle.

[prime gale]

Are you assuming H is normal?

[thorn delta]

yea, kind of ig. Any subgroup of index 2 is normal.

[prime gale]

Yes--you are right. Dumb me. Okay. Let s be a 3 cycle. Then consider image of s under S_n->S_n/H=Z/2.

Got to be zero. Therefore s is in H.

[thorn delta]

interesting, I'm not sure i see why that is tho

[prime gale]

Did I make a mistake?

I never really studied group theory (as you can tell).

[thorn delta]

no, i couldn't tell. I just don't see any reason why the canonical map should take 3 cycles to the identity in S_n/H

[prime gale]

3 cycle has order 3, right?

Order of image must divide 3.

[thorn delta]

oh wow, true. Thanks

[cinder bone]

I completed a) and got r=12

but I can't figure out how to relate that to b)

I know if b and 21 are coprime, then 21k = 1 + b^(12) is true for some k

and i know 21p = (341)^99 + q is true for some p & q

using an online calculator, i know that the solution is q=20

but, not sure why

[cloud walrus]

IAmJon:

[upper inlet]

idrk number theory so you'll have to deal with the 341^3 on ur own ig

[thorn delta]

341 mod 21 is 5 and then use the relations Jon gave. Not sure if this counts as #groups-rings-fields tho

[upper inlet]

they were referred here cause they asked if they could ask about groups and modular arithmetic

It's fine tho because no one was using this channel

[cinder bone]

hmmm

i know 341 and 21 are relatively prime

okay so 341^3 mod 21 is 20

how can i relate that back to the relations Jon has given

use 341 as b and cancel somewhere?

yeah im big confused

[thorn delta]

u know that 341^99 = 5^99 (mod 21) since 341 = 5 (mod 21)

@cinder bone

[cinder bone]

im thinking it through

wouldnt that mean the answer is 5^99?

341 modulo 21 = 5 is equivalent to 341 ≡ 5 (mod 21) so wouldn't 341 ^99 modulo 21 = 5^99 be equivalent to 341^99 ≡ 5^99 (mod 21)

[thorn delta]

yes, so now you have (5^12)^8 * 5^3 = 5^99 so there is more ez simplification you can do

[cinder bone]

that's definitely not 20 though

im really confused lol

[thorn delta]

um yea it should be

[cinder bone]

5^99?

[thorn delta]

what do you have so far? How are you not getting 20

[cinder bone]

im confused where i implement the fact that b^12 ≡ 1 (mod 21)

and how 5^99 wouldn't be the straight up answer

since that's 341^99 mod 21

[thorn delta]

well, 5^99 > 21 so the answer will reduce further.
The point of writing (5^12)^8 * 5^3 = 5^99
is that we can use part a) to compute 5^12 mod 21

[cinder bone]

okay so i have (341^12)^8 * 341^3 ≡ (5^12)^8 * 5^3 (mod 21), (5)^12 ≡ 1 (mod 21), and (341)^12 ≡ 1 (mod 21)

im thinking i have to somehow get to 341^2 ≡ 4 (mod 21)

[upper inlet]

Right, so what does that tell you about the value of (341^12)^8*341^3

mod 21

because mod is multiplicatively distributive iirc, you can simply replace 341^12 with 1 in that expression

[cinder bone]

so then I have 341^3 ≡ 5^3 mod 21

[upper inlet]

did you see what I wrote?

[cinder bone]

yeah.

[upper inlet]

do you see how that simplifies the expression?

[cinder bone]

(341^12)^8*341^3 = 341^3

[upper inlet]

Ok just making sure we're clear on that

[cinder bone]

mod 21

[upper inlet]

so then I have 341^3 ≡ 5^3 mod 21
I believe you can say this

[cinder bone]

and similarly (5^12)^8*5^3 = 5^3

[upper inlet]

yep

and 125 is small enough so that you could easily do the remaining calculation by hand

,w 125 mod 21

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by IAmJon#5792

[upper inlet]

badda bing badda boom

[cinder bone]

okay sweet

thanks

[upper inlet]

np

[thorn delta]

i.e. 5^3 = 25 *5 = 4 * 5 = 20 mod 21 since 25 = 4 (mod 21) just to get more use out of those rules

[upper inlet]

oh yeah

I keep forgetting how mod. arithmetic works lol

[thorn delta]

what is the homomorphism from S4 to S3 induced by the action of S4 on D_4^k?

[woven delta]

By congugation

An action on a set with 3 elements is the same as a map to S_3

[thorn delta]

oh so when they say an action from S4 to D_4^k, they mean an action from S4 to {D_4^1, D_4^2, D_4^3}?

[woven delta]

Okay do you know any sylow theory?

[thorn delta]

nope

[woven delta]

Oh ok, this sort of thing is exactly what the sylow theorems are about

But okay anyway there are 3 subgroups of order 8 which get congugated to each other by any element of S^4

This gives you a map to S^3, where you send an element to the permutation that acts like that element does on those 3 subgroups

[thorn delta]

right, okay i see

[cloud walrus]

inhyak:

[thorn delta]

@gleaming egret n^{1/b} must be rational, and since the nth root of a prime is irrational for n >= 2, you have to have that p^b divides n for each prime p dividing n.

you just want n^{a/b} to be rational though?

hm, but if b is negative then n^{1/b} won't be an integer in general. Lets move to #help-8

[thorn delta]

okay, back to this question i was on earlier. I've been super distracted lol

I'm not quite clear on the very last part.
"Neither will serve as each leads to quotient group too small to act transitively on three elements."

I know what it means for a group to act transitively, but I only know the very very basics of group actions, so I don't quite understand why this is important.

[leaden finch]

anyone good with abstract algebra?

[thorn delta]

@leaden finch what's your question?

[leaden finch]

im struggling on a problem and i dont get the concept

[thorn delta]

alright, what's the question exactly?

[leaden finch]

im having trouble understanding

[smoky cypress]

That's more of a number theory problem

[leaden finch]

is it?

[smoky cypress]

Yeah but we can do it here

So, every integer is congruent to 0, 1, 2, or 3 modulo 4

Right?

[leaden finch]

thats bc of the congruence mod?

[scarlet estuary]

im not sure what that means

it's because of the definition, yes

anyway:

• what's 0^2 mod 4?
• what's 1^2 mod 4?
• what's 2^2 mod 4?
• what's 3^2 mod 4?

[leaden finch]

0 , 1 , 0, 1 ?

[scarlet estuary]

right.

what does that tell you about congruences of squares?

every integer is congruent to either 0, 1, 2, or 3 mod 4

and when we square one of those, we get either 0 mod 4 or 1 mod 4

can you "fill in the gaps" to complete the proof?

[leaden finch]

wait.. which gaps

[scarlet estuary]

mainly just justifying why we can make conclusions based on the square of the congruence

rather than having to square the number before taking it mod 4

[leaden finch]

hmmm we can say that were are subtracting?

[chilly ocean]

@solemn rain yo homeboy where did you learn group stuff from

[solemn rain]

bed time stories

[chilly ocean]

lmao stop trolling man

[thorn delta]

Bumping my question since the activity here kinda died last night. I'm still not quite clear on the last part of this proof. I'm stuck on the part where it says "Neither will serve as each leads to quotient group too small to act transitively on three elements."
I know the definition of a transitive action, but idk how that is supposed to apply here. It talks about S4/ker(f) acting transitively on 3 elements? What action are they talking about exactly?

[solemn rain]

do you mean the part that S_4 acts on D_4 by conjugation?

if so

then remember the orbit-stablizier theorem

where the action is conjugation

[thorn delta]

okay, idk orbit-stabilizer theorem, but I am referring to the very last few sentences, where it explains how the "quotient group would be too small to act transitively" if the normalizers had order >8

[solemn rain]

[G : G_x] = |O_x|

this is the orbit stablizier theorem

the proof is simple you just define a map from G to O_x

f such that f(g) = g*x where * denotes the action

f is surjective

now G_x is a subgroup here

f(g) = f(h) implies that g is h in G/G_x

so the function that sends gG_x to g*x is a bijection

hence same number of elements

@thorn delta i think this is what the proof meant by too small

it doesn't satisify this: [G : G_x] = |O_x|

got it?

[thorn delta]

not really. why wouldn't [G : G_x] = |O_x| hold?

[solemn rain]

it would?

it always holds

[thorn delta]

right, but then what do you mean by this?

it doesn't satisify this: [G : G_x] = |O_x|

[solemn rain]

the orders

of G

wouldnt satisify this

i think thats what they mean that G is too small

[thorn delta]

I must not know enough about group actions for this to make sense. I don't really even understand how the action of S4/ker(f) (assuming that's the quotient its talking about?) on 3 elements is defined, much less why it needs to act transitively.

[solemn rain]

say we have a set A

and G acts on the set A

say N is a normal subgroup pf G

sorry power went off

anyways just define the action just how you would define it in any group

u dont need an explicit one to see

wait oh i just read

yea just compute orders

24/12 is e

2*

2 cannt act on 3

transitively

@thorn delta

[thorn delta]

but idek why we care about it acting transitively?

[solemn rain]

do you know what transitive meana

means*

[thorn delta]

the action has only one orbit

[solemn rain]

okay

cuz the action must be transitive 2 elemente cant act

on 3

orbit stab?

got it?

[thorn delta]

I'm not sure I am understanding what you are getting at. It sounds like you are saying the same thing in just slightly different ways.
IF S4/ker(f) acts transitively on 3 elements i.e. |O_x| = 3, while |ker(f)| > 8, then yea, I do see that you get the issue of [G : G_x] = 2 != 3 = |O_x|.

But I'm just not seeing why we need a transitive action to conclude that S4/ker(f) is iso to S3.

[solemn rain]

i dont think u need to

lmao nvm idk wow

is it because S_n acts transitively

i mean thats S_n/something]

not sure what difference does this make

yea S_n acts transitively always

i think

so thats why

@thorn delta

i really hope someone fixes the mess im saying

[thorn delta]

hmm? You don't mean that EVERY action of Sn on a set is transitive, do you? Not sure I am understanding oof

[solemn rain]

no

the natural action

of Sn

like the one you would think of

is transitive

[thorn delta]

right, okay, i know that. Still not really connecting the dots though lol. We're not talking about Sn acting on {1, 2, ..., n} transitively by evaluation, we are talking about S4/ker(f) acting on 3 elements. Its clear that if S4/ker(f) doesn't act transitively on 3 elements, then it can't be iso to S3, but I don't think that is the point you are trying to make

[solemn rain]

N must be ker(f) the normalizer must be of order 12 or 24 hence cant satisfy acting transtiively

hence iso to S3

the question taht i cant answer is why do we have to have a transitive action

my probable answer is that S naturally acts transitively

and the action by the quotient group is just same action as the group

but i am not sure and idk and im sleepy

[thorn delta]

i think i follow so far, but yea idk either. ill come back to it another time maybe. thanks for explaining some of it tho.

[royal yarrow]

I would recommend moving that to either proofs and logic or discrete math

maybe precalc

[obsidian path]

I understand that since it's induction it's not as intense but.. I really believe this is trickier than your average induction question

Ok, I'll just move it and delete my messages from here then. Apologies

[royal yarrow]

My objection wasn't that it wasn't a hard problem, but that it's not abstract algebra

no worries

[obsidian path]

no I understand gotcha, ty!

[old hollow]

If you union the elements of a quotient group, does it give you the original group again

[woven delta]

If you take the union of all the equivalence classes yes

That's part of the definition of an equivalence class

[old hollow]

Oh lol

Is an equivalence class a set?

I’m just confused why it’s called a “class”, is that a mathematical object or term

[carmine fossil]

A class is identical to a set,but if it were a set it would violate a set axiom

[old hollow]

Which set axiom

[carmine fossil]

Some set axiom

[scarlet estuary]

erm

[carmine fossil]

Set until not

[scarlet estuary]

i'd assume you mean restricted comprehension

but in all common cases equivalence classes are sets

i dont think theres any particular reason why theyre named as they are

[old hollow]

Oh

Thanks

[scarlet estuary]

you can certainly construct an equivalence class that would be "larger than" a set, e.g. groups being equivalent iff the element with maximum order has the same order

or whatever

but practically speaking i havent really seen this

[old hollow]

Ah

[dusty oracle]

A more trivial example is for all sets, x ~ y iff a tautology is true

Then the equivalence class of any set would be the class of all sets

[jagged salmon]

Hi everyone, is anybody here involved in researching complex valued neural networks? Specifically multicomplex? I'd love to see if you could point me to some interesting literature that isn't so easy to find.

If you have anything over bicomplex or higher orders, I'd be especially interested 🙂

[hot lake]

multicomplex ??

[jagged salmon]

It's a number system in Hypercomplex.

You might have heard of one of the more well known hypercomplex numbers, such as Sedenions

They're super interesting, i'd recommend reading about them

[hot lake]

not having associativity is kinda eeeeh

well idk any literature about complex valued neural networks

maybe you could make quaternions useful for movements through space

[jagged salmon]

Yea there's a lot of literature for quaternion neural networks

And split-complex

And not having associativity is a bit unsettling initially, but the group I work with has libraries built in many programming languages to replicate any functionality that multidual and multicomplex numbers have, so I'm not worried about the number systems themselves, rather their behavior in neural networks

[slate forum]

let X be an irreducible affine variety, then we define A(X)=k[x]/I(X), and O_X is the structure sheaf on X that assigns to each open set U the rational functions on U whose denominators don't vanish anywhere on U, why is it that O_X(X)=A(X)?

[prime gale]

Are you asking why (upsidedown L)(Spec_A)=A?@slate forum

[slate forum]

I have no idea

https://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/alggeom-2002.pdf

This is the book

it's page 18

[prime gale]

That is, for any ring (commutative with unit) (upsidedown L)(Spec_A) = A.

[slate forum]

18-19

I don't know what upside down L is

[prime gale]

Same as H^0.

[slate forum]

?

zeroth cohomology?

[prime gale]

Yes.

[slate forum]

of what?

[prime gale]

Also same as O_(Spec_A)(Spec_A).

That is, if X=Spec A, then O_X(X) = H^0(X,O_X) = (upsidedown L)(X).

[slate forum]

What homology are we using?

[prime gale]

Well, those three things are all equal whether or not X=Spec A. But if X=Spec_A, then they are all equal to A.

[slate forum]

We haven't introduced homology at all yet

[prime gale]

Well, it is kind of tautological that H^0(X,O_X) = the others, because that's how we define cohomology.

[slate forum]

Like, all I have is this:

and

[prime gale]

I think what you are asking is why, if X=Spec_A, that O_X(X)=A. I don't remember how to prove, but it isn't hard. Certainly Harshorne contains a proof. It isn't just a tautology, but it also isn't hard.

[slate forum]

We haven't even talked about Spec_A

[prime gale]

Your X is Spec_A.

[slate forum]

Like this is just about an irreducible affine variety

[stoic rose]

Your X is Spec_A.
@prime gale This is not schemes, just varieties

[prime gale]

Everything I've said is still true for affine varieties.

[slate forum]

I have no idea what a scheme is

At least, not yet

[stoic rose]

If I understand well you are asking why O(X)=A[x1,..,xn]/I where X is the variety V(I)?

[prime gale]

Don't have to know what a scheme is. But you more or less already know what one is, if you know what structure sheaf is.

[slate forum]

uhh

I'm asking why O_X(X)=A(X)

[stoic rose]

What is A(X)?

[slate forum]

coordinate ring

[prime gale]

Okay. Let me look at your definitions.

[slate forum]

k[x_1...x_n]/I(X)

[stoic rose]

Ok yes I see

[prime gale]

So you are asking, if A is an integral domain, why the intersection of A_p=A, for p prime ideal of A, right?

[slate forum]

$\mathcal{O}_X(U):={\frac{f}{g} \mid g(u)\neq 0 \forall u\in U}$

[cloud walrus]

HelixKirby:

[slate forum]

I tihnk

and $A(X)=k[x_1,\dots x_n]/I(X)$

[cloud walrus]

HelixKirby:

[prime gale]

Whoops. Again I am being too general.

[slate forum]

The fact that things get more abstract than what I'm already saying blows my mind

[prime gale]

I think you are looking for Hilbert's Nullstullensatz. Or whatever. My spelling isn't that great.

[slate forum]

I don't think so

[stoic rose]

So roughly O(X) is the set of polynomial functions X->k. Such a function is given by a polynomial on all the affine space that we can restrict to X only. However, we want to count polynomials that agree on X, so we have to quotient out the ring of all polynomials of the affine space k[x1,...,xn] by those who vanish on X, i.e., I(X)

Does this makes sense?

[slate forum]

these are essentially polynomials on A^n, but we restrict them to X?

[prime gale]

You are trying to prove that A=intersection A_p, where p is a maximal ideal containing I. A_p = {f/g, such that g(p)!=0}.

[slate forum]

Perhaps? I don't think so though

[prime gale]

Isn't this literally your statement?

Just replace O_{X,p} with A_p. Just another name for same thing.

[stoic rose]

Yes, for example suppose you want to give a regular function on the parabola defined by y-x^2=0. This function is a polynomial in x and y, say x^4. Note however that the function x^4 is the same as, for example, the function y^2 if we restrict our view on the parabola (because, there, y=x^2). This is why we quotient by I(X)

[slate forum]

Could I think of I(X) as the kernel of restriction of polynomials on A^n to X?

[prime gale]

Yes.

[stoic rose]

You are trying to prove that A=intersection A_p, where p is a maximal ideal containing I. A_p = {f/g, such that g(p)!=0}.
@prime gale I think the commutative algebra just obscures it. I know that book and, at least up to this point, the emphasis is on geometry, I don't even think that localizations are defined in generality

[slate forum]

So, the coordinate ring is to be thought of as polynomials restricted to X

I guess?

But these have no denominators

[prime gale]

Anyway, you are trying to prove that A = intersection O_{X,p}, right?

[slate forum]

I don't know what A is

A=A(X)?

[prime gale]

Yes. A=A(X).

[slate forum]

O_{X,p} are functions that locally look like rational functions where the denominator does not vanish at p

Is that right?

[prime gale]

So it suffices to prove that every g(x) of degree >=1 vanishes at some point on X.

Well, mod I, anyway.

[stoic rose]

So, the coordinate ring is to be thought of as polynomials restricted to X
@slate forum Yes, though you ideally want to think of the variety X by itself and not too much as embedded in the affine plane. This is why you want to know some intrinsic properties of it, such as the regular functions defined on it (or its open sets). This is a bit counterintuitive when you discover algebraic geometry for the first time but it is a quite important abstraction.

[slate forum]

So, don't think of it as living in an affine space?

How is its topology defined then?

[stoic rose]

Well at some point you have to consider it as a subset of the affine space but once you know its intrinsic structure (topology, sheaf) then you can think roughly only about this structure and not about the actual points in the affine space

So it suffices to prove that every g(x) of degree >=1 vanishes at some point on X.
@prime gale I'm not sure but I think he is more looking for an intuitive explanation than a rigorous proof, which I think follows kind of directly from his definitions (correct me if I'm wrong)

[slate forum]

Ok, so here's my big question, why is it that regular functions on X, which are rational functions whose denominators don't vanish on X, are the same as A(X)=k[x_1,...x_n]/I(X)

[prime gale]

I think there is something to prove. @stoic rose I think he is more or less proving that if A is a ring, then H^0(O_{Spec A})=A.

[stoic rose]

I think there is something to prove. @stoic rose I think he is more or less proving that if A is a ring, then H^0(O_{Spec A})=A.
@prime gale The book doesn't use schemes, that will jsut be more confusing

[prime gale]

Ok, so here's my big question, why is it that regular functions on X, which are rational functions whose denominators don't vanish on X, are the same as A(X)=k[x_1,...x_n]/I(X)
@slate forum You mean "don't vanish anywhere on X", right?

[slate forum]

The book has not defined schemes yet

yeah

That's what I mean

the denominators are never zero anywhere on X

How is it that I can write a fraction f/g = h/1, where h is in A(X)?

Since I think O_X(X) is by definition a subset of Frac(A(X))

[prime gale]

That is the statement that if g is not contained in any maximal ideal, then g is a unit.

[slate forum]

it is?

[prime gale]

Well, if you have f/g with g a unit, then you have your h, right?

The statement that g vanishes at p is just the statement that g is in the maximal ideal p.

[stoic rose]

Ok, so here's my big question, why is it that regular functions on X, which are rational functions whose denominators don't vanish on X, are the same as A(X)=k[x_1,...x_n]/I(X)
@slate forum Ok I kind of cheated before because I said that regular functions where polynomials on X. Actually we can have rational functions in k(x1,...,xn) but we can show that for every such rational function that doesn't vanish on X, we can find a polynomial that agrees with it on X, so we can basically suppose everything is a polynomial

[slate forum]

That they're rational is unnecessary?

[prime gale]

Isn't that what I just proved?

[stoic rose]

Every rational function that don't vanish on X is equivalent to a polynomial (in the sense that the rwo agree on X) In some sense, we can kind of kill all singularities outside of X by multplying by a suitable polynomial that is 1 on all X and 0 on the singularities

[slate forum]

I don't know why this is so confusing to me

oh?

How so?

This is what I have to work with

So $if f/g\in \mathcal{O}_X(X)$, then we have $g(p)\neq 0$ for all $p\in X$, there's apparently some way for me to make $f/g=h/1$.

[cloud walrus]

HelixKirby:

[stoic rose]

I would need to think a bit for the proof of this but intuitively you can see that we can demand that a polynomial is 1 on X and 0 on all the (finitely many) singularities and then kind of interpolate

[slate forum]

And apparently, that $g$ does not vanish on $X$ anywhere means that $g$ is a unit in $A(X)$?

[cloud walrus]

HelixKirby:

[prime gale]

Yes.

[stoic rose]

Yes, its inverse is 1/g, which is well defined because g doesn't vanish

[slate forum]

but that's not a polynomial

[stoic rose]

That's a regular function

[prime gale]

It might be one.

[slate forum]

where is 1/g taking place?

in the fractions?

or in A(X)?

[prime gale]

For example, in the ring R[x]/(x-1), 1/x is a polynomial.

[slate forum]

wat

If g were not a unit, then there's a maximal ideal in A(X) that contains it, this corresponds to an ideal containing I(X) in k[x_1,...x_n]

[prime gale]

Well, in the ring R[x]/(x-1), x=1, so 1/x is a polynomial.

[leaden finch]

can someone help me with this one

[prime gale]

@leaden finch Sorry--too hard for me.

[slate forum]

Isn't this number theory?

[leaden finch]

modern algebra

idk how to prove it

[slate forum]

uhhh, just do what it says?

[stoic rose]

Since you restrict your attention to the variety X, there is not a single quotient of polynomials in k[x1,...,xn] that can represent a given regular function. In the exemple I gave above y^2 and x^4 represent the same function on the parabola. In am's example, 1/x correspond to the polynomial 1 on the variety x=1 (which is here just a point)

[slate forum]

for all cases here it's either 0 mod 4 or 1 mod 4

[prime gale]

@stoic rose Well, it is just a point if R is a field.

[slate forum]

Ok...

[leaden finch]

yes, but am i on the right track?

[slate forum]

Seems fine?

[stoic rose]

@leaden finch this would be better suited in #elementary-number-theory , plus we are already having a discussion

[leaden finch]

how do i know which ones one no congruent?

[slate forum]

uhh, if it has a remainder of 1, then it's 1 mod 4, if it has no remainder it's 0 mod 4

But uhh, yeah, on Z(y^4-x^2), then sure, x^2=y^4

[stoic rose]

@stoic rose Well, it is just a point if R is a field.
@prime gale yeah we are talking about varieties over fields here, let's not go into more complicated things

[slate forum]

so.... why is it that g is a unit?

How do I show that?

[prime gale]

How are you defining I(X)?

[slate forum]

The polynomials that vanish on X

[prime gale]

How are you defining X?

[slate forum]

X is an irreducible affine variety

[prime gale]

How are you defining affine variety?

[slate forum]

an irreducible algebraic set

I just say irreducible in the definition of affine variety because apparently people don't always require them to be irreducible

[prime gale]

This is really nullstullensatz.

[slate forum]

nullstellensatz has so many different forms

I wouldn't be surprised

[prime gale]

You are wondering why, if you have a variety defined by {f_i}, then I(X) = ideal generated by {f_i}, right?

[slate forum]

The only ones I'm familiar with are maximal ideals are of the form (x_1-a_1,...,x_n-a_n) and that I(Z(J))=r(J)

I don't think so

[prime gale]

Oh. So you already accept that I(Z(J)) = <J>?

[slate forum]

is that the radical?

if so, then yes

[prime gale]

Sorry--I mean I(Z(J)) = radical(<J>).

[slate forum]

that's a version of Nullstellensatz

If a polynomial vanishes on the common zero locus of a set of polynomials, then some power of it is in the ideal generated by those polynomials

[prime gale]

If you accept that, then any function that doesn't vanish anywhere on Z isn't contained in any maximal ideal containing I. Therefore is a unit in k[x]/I.

[slate forum]

wuh

baby steps

[prime gale]

Do you accept that if f is contained in a maximal ideal containing I, then f vanishes somewhere on Z?

[slate forum]

uhhh

$f\in m \supset I$, then $Z(m)\subset Z(I)$

[cloud walrus]

HelixKirby:

[prime gale]

I take that as a yes?

[slate forum]

I don't know, I'm trying to see why

so $r(m)\supset r(I)$

[cloud walrus]

HelixKirby:

[prime gale]

What's r?

[slate forum]

but $r(m)=m$

[cloud walrus]

HelixKirby:

[slate forum]

radical

so $r(I)\subset m$

[cloud walrus]

HelixKirby:

[prime gale]

You don't really have to worry about radicals when comparing stuff with prime ideals.

[slate forum]

Are we saying $I$ is prime/

[cloud walrus]

HelixKirby:

[prime gale]

No. But m is prime.

[slate forum]

ok, so yeah, $r(I)\subset m$

[cloud walrus]

HelixKirby:

[prime gale]

r(I) is contained in p iff I is contained in p, for prime p.

Anyway, that is a digression.

Back to the question, Do you accept that if f is contained in a maximal ideal containing I, then f vanishes somewhere on Z?

[slate forum]

not yet

is Z(I)=Z?

[prime gale]

Well, Z(m) is contained in Z(I), right? And f vanishes on Z(m).

[slate forum]

I'm sorry, this is probably getting frustrating

yes

[prime gale]

So if f vanishes on Z(m), and Z(m) is contained in Z(I), then can you conclude that f vanishes somewhere in Z(I)?

[slate forum]

sure, Z(I) has Z(m) in it

[prime gale]

So you agree with statement?

[slate forum]

I think s

o

[prime gale]

Back to this statement: any function that doesn't vanish anywhere on Z isn't contained in any maximal ideal containing I. Therefore is a unit in k[x]/I.

[slate forum]

that's the contrapositive

so yes?

[prime gale]

Do you accept that if f is not contained in any maximal ideal, that f is a unit?

[slate forum]

if it were not a unit it would be contained in a maximal ideal, which is a contradiction

so yes

Why did I have to go back and forth so much between affine space and the ring?

[prime gale]

Still not sure that is what you were trying to prove, but you have proven that if g doesn't vanish on Z(I), then g is a unit in k[x]/I, right?

[slate forum]

yeah, I think that's it then?

I wanted to show that O_X(X)=A(X)

So, I guess that's it?

[prime gale]

Sounds like that is unwinding definitions + Nullstullensatz.

[slate forum]

I guess, but if I couldn't do that

I don't know how I'm gonna be able to do anything

jesus

[prime gale]

I guess it didn't help that I started with all the wrong definitions.

[slate forum]

So, instead of directly showing g was a unit, we instead went around and showed that it wasn't contained in any maximal ideal of A(X)

[prime gale]

@slate forum By the way, you mentioned about things getting more abstract than this. It gets scary.

[slate forum]

I like abstraction

[stoic rose]

Ok, so here's my big question, why is it that regular functions on X, which are rational functions whose denominators don't vanish on X, are the same as A(X)=k[x_1,...x_n]/I(X)
@slate forum Let me try an explanation. Suppose you have a rational function f/g that does not have poles on X. We want to show that 1/g is equivalent to a a polynomial in the quotient k[x1,...,xn]/I, i.e., that g becomes a unit in this quotient (this is the commutative algebra way of saying there is a polynomial that agrees with 1/g on X). The fact that g does not vanish on X can be translated into saying that there is no point on which both I and g vanish, i.e. V(I+<g>) is empty, i.e. I+<g>=<1> by nullstellensatz. Thus I and <g> are coprime. By chinese remainder theorem, we can thus find a p in k[x1,...,xn] that is congruent to 0 mod g and 1 mod I. This is a multiple of g that is a unit in the quotient k[x1,...,xn]/I, so g itself must be a unit in that quotient (I is I(X))

[slate forum]

I just didn't know it could get worse

[stoic rose]

I typed too slowly but that's an alternative explanation 😋

[slate forum]

oh my

g is non-vanishing, f can though right?

[stoic rose]

Yes, actually we don't really care about f

[slate forum]

ok, just making sure

[prime gale]

f can be zero. I would say something more clear like "g is nowhere vanishing". non vanishing can be ambiguous.

But I know what you mean.

[stoic rose]

Oh yes that was a typo, corrected

[slate forum]

Ok, I tihnk I see it from that too

Jeez, was completely going about this wrong

[stoic rose]

Now you understand

[slate forum]

I wanted to take f/g = h/1, then show f=gh mod I(X), that is f-gh vanishes on all of X

though, I didn't actually have a candidate for h

but I guess this means that I need to solve for h, so I need to know that I can invert g

[stoic rose]

Yep exactly

[prime gale]

radiateur gives you a way to construct h. I guess my way doesn't.

[slate forum]

since g is nowhere-vanishing

on X=Z(I)

we have $Z(g)\cap Z(I) = \emptyset$

[cloud walrus]

HelixKirby:

[slate forum]

is that right?

[stoic rose]

Z is the set of zeros?

[slate forum]

and this translates to $Z((g)+I)=\emptyset$

yes

[cloud walrus]

HelixKirby:

[stoic rose]

I'm used to V but yes 👍

[slate forum]

r((g)+I)=(1)

How did we get rid of the radical?

[stoic rose]

If r(J)=1 => 1^n is in J for some n => J=(1)

[slate forum]

oh duh

ok, then we get (g)+I = (1)

[stoic rose]

Yep

[slate forum]

so there are polynomials $a,b\in k[x_1,\dots x_n]$ and $h\in I(X)$ such that $ag+bh=1$, so $ag-1 = bh$, so $ag=1 mod I(X)$

[cloud walrus]

HelixKirby:

[slate forum]

at long last

However, this was all for something basic, jeez, I feel so dumb

[prime gale]

I think you phrased it very well.

[stoic rose]

I don't think you're dumb, juggling between geometry and algebra can be very confusing, I myself struggled a bit to find the proof

[stark salmon]

hey there, been working on this example sheet, last question has been really thought, not sure where to start tbh, anyone have any tips on how one could go about solving this ?

Let G be a group that acts on the set X and let H be a subgroup of G. We define the set : \\ $X^H = {x \in X | hx = x$ for all $h \in H}$ \\ ie. the points of X that are fixed by H. Show that if H is normal, then the group action of G on X leads to there being an action of G/H on $X^H$.

[cloud walrus]

Frogu:

[carmine fossil]

What does it mean to say there is a group action of A on B?

[prime gale]

Just define the obvious group action, then show that your definition is well defined and is a group action.

[eager bobcat]

Where to start with 36 ... I have proven to myself that the remainder of division of 2 multiples of 3 is a multiple of 3

Trying this right now but not sure what to do

I feel like I have to get rid of the b coefficient but not sure how 2

[chilly ocean]

@solemn rain yo homeboy im learning group stuff from dummit foote

[chilly ocean]

yo so additive Q and Z aren't isomoprhic, but if we changed ordering,on let's say Z, could we find the isomorphism? It feels like there should be one (but it wouldn't be integer group then I guess)

[sharp sonnet]

change the ordering?

find the isomorphism? but there is none

[hot lake]

isomorphic as groups or as ordered sets ?

[carmine fossil]

Are Q and Z isomorphic as sets?

[chilly ocean]

yes

as groups I mean

Z should still be a group if we make it dense right

like if we preserve the ordering from Q

yeah I know

[hot lake]

well ordering has nothing to do with the group structure

so idk what you mean with reordering

it sounds like you are asking "if I paint this triangle blue, can it become a square ?"

[chilly ocean]

yeah nvm for some reason I thought the operation might change

and as you said comparing elements doesn't mean anything

[prime gale]

Actually, you can change the color of letters by adding and removing parts. https://en.wikipedia.org/wiki/Synesthesia

[hot lake]

xD

[chilly ocean]

ok I think I know why I understand group action: it's a map from G x A to A, where G group A some set - but for me it feels like g*a might not be element of A

maybe I haven't seen enough examples, but just seems weird to take a product of completely different structures and map them to just some set

[prime gale]

Your map has to satisfy associativity condition.

And also, identity has to act as identity on A.

[chilly ocean]

does associativite imply the product ga is in A?

or is it kinda assumed or to check when defining a map?

[prime gale]

I mean that you need these two conditions in addition to what you mention.

[chilly ocean]

I just don't see how those two conditions would imply the image is in A, feels like first you define a map, then you check if the image is in A, then you check those 2 conditions

[prime gale]

Don't worry about it. If you continue to study math, you will see billions of examples of group actions on sets.

[chilly ocean]

I have to worry about it

[prime gale]

It isn't something obscure.

I mean, don't worry about not seeing enough examples. You will see plenty.

It is super important topic in math.

[chilly ocean]

Are you saying if you have some group G and set A, any mapping from GxA satisfying those two conditions maps to A?

[prime gale]

You are assuming that g*a is in A. That is part of your definition. You don't prove it from your definition. Though if you have an example and want to prove your example is indeed a group action, then you need to prove it.

[chilly ocean]

ok.

[prime gale]

No. I am saying you left those two conditions out of your definition of group action.

[chilly ocean]

yes I did because that's not what I was confused about, but now its clearer

[prime gale]

But don't worry about not having enough intuition about what a group action is. You will see plenty of them and get the intuition.

[chilly ocean]

btw would you write $(f \circ g )(x)$ as $f(g(x))$ or $g(f(x))$?

[cloud walrus]

Godel:

[chilly ocean]

One of the books I'm using uses the 2nd one and it triggers me so much, slows me down every time it appears

[carmine fossil]

Which one?

[prime gale]

Huh? Never heard of it being second one. Weird book.

I have heard of (fg)(x) = g(f(x)) though. And also (fg)(x) = f(g(x)). But with the o, it is never ambiguous.

[prime gale]

That all looks fine to me. (gof)(p) = g(f(p)) = (fg)(p). All pretty normal.

[carmine fossil]

(fg)(p) is not same (fog)(p)?

[prime gale]

When people write (fg)(p), sometimes they mean f(g(p)). Sometimes they mean g(f(p)). But when people write fog(p), they meaen f(g(p)). Never g(f(p)).

When people write (fg)(p), sometimes they mean fog(p). Sometimes they mean gof(p). I always use the o, so it is not ambiguous.

If you are thinking about the right action of a group G on a set A, then (fg)(a) = g(f(a)). If you are thinking about a left action of a group G on a set A, then (fg)(a) = f(g(a)).

[carmine fossil]

Ok,In that sense

Also,Why would you ever use right actions? Can't you just use left actions in every single case?

[prime gale]

Yes. But sometimes one is more convenient to talk about. You can convert a right action into a left action with the group isomorphism g->g^{-1}.

[hot lake]

that's not really a group isomorphism

[prime gale]

Sorry. You are right. Shouldn't have said group isomorphism. But it is a way to convert right action to left action.

[woven delta]

It is a group isomorphism

Between G and G^op

[vestal snow]

Is the subgroup (x^2,y^3) normal in the free group F({x,y})?

[prime gale]

Doesn't seem like it. Doesn't contain x^-1y^3x.

[vestal snow]

Hmm okay.

Can someone help me out with formalizing my proof for this?

C_n means Z/nZ in this notation

I have everything I need basically

Suppose H is a group with maps f: Z_2 --> H and g: Z_3 --> H

Define a function from G as x goes to f(1) and y goes to g(1)

How do I prove that this function from G to H is well defined?

[prime gale]

What's H?

[vestal snow]

Any arbitrary group with morphisms from Z_2 and Z_3

We're trying to prove that G is the coprod of Z_2 and Z_3

[prime gale]

I see. Using definition of coproduct.

[vestal snow]

I was trying to write G as a quotient group of the free group

But that didn't work

[prime gale]

Since construction of coproducts is easy for finite groups, I would just use construction.

[vestal snow]

Yeah, but how would you show well defined?

Like we want to get a function from the coprod into H

And you always run into the problem of proving well defindedness

[prime gale]

To show it is well defined you only need to show that x^2=0 and y^3=0, right?

That is, (image of x^2)=1_H and (image of y^3)=1_H.

[vestal snow]

I get what you mean intuitively, but when formalizing it, don't we need to show that representation of an element does not change where it goes through our function?

[prime gale]

Huh?

[vestal snow]

Like if we have some product of x and y = some other product of x and y

We need to show they go to the same element

[prime gale]

You have map from F_{x,y}. You have defined a map from F_{x,y} to H. To show that this descends to a map from G, you only need to show that image of x^2=1 and image of y^3=1.

[vestal snow]

Oh is this a theorem?

[prime gale]

I wouldn't call it a theorem...

[vestal snow]

intuitively it makes sense

But I feel like I need to prove it

[prime gale]

More the definition of "G is subject only to relations x^2=1 and y^3=1."

[vestal snow]

Hmm okay

Thanks

[slate forum]

Can anyone parse this notation?

9 (i)

[prime gale]

Square means dot. That is, the thing you plug into your function.

[slate forum]

I understand that

But what goes in?

Like earlier it says (f,v):=f(v), but now it's supposedly defining a map from V to k by taking a function as input?!

[prime gale]

v_i^* is the function that takes a vector to its v_i coefficent.

[slate forum]

Yeah, that's usually what it does

But it looks like it's defined funny here

[prime gale]

Could be. If (a,v) means put standard inner product on the vector space with basis and evaluate inner product of a and v, then it looks correct.

[slate forum]

But it didn't say anything about an inner product

[prime gale]

I dunno. I just have those two pages. That is just an interpretation that can make it correct. But in any case, you understand the concept, right?

[slate forum]

Yeah

Just gotta make sure

[prime gale]

Or it could be a sucky book. But as my friend pointed out, you need to read sucky books, because papers are usually really sucky compared with books.

[chilly ocean]

i can't find an explanation on why cycle decomposition loops back

and when it does and doesn't loop back

[prime gale]

Unless you provide more context, nobody's going to know what the hell you are talking about.

Are you talking about symmetric groups?

Maybe some weird homology thing?

[chilly ocean]

sorry brand new to this

but i mean function decomposition of permutations

[prime gale]

WTF does loop back mean?

[chilly ocean]

loop back is when f^m(x) =x

[prime gale]

Are you talking about permutations on a finite set?

[chilly ocean]

ya

[prime gale]

Then it'll always "loop back".

[chilly ocean]

yeah but like why

i can't find the explanation

on the internet anywhere

[prime gale]

Well, eventually f^a(x)=f^b(x), for some a and b with a!=b because set is finite. Only finite number of permutations. Infinite number of integers. So they can't all be different.

Then, since f is invertible, f^(a-b)(x) = x.

[chilly ocean]

and since we're talking about permutations, it'll be bijective right?

[prime gale]

Yes. f is bijective because f is a permutation.

[chilly ocean]

ok thank you

[prime gale]

No problem. Sorry if I sound a bit rude.

[chilly ocean]

nahh

you're good

[prime gale]

It's just that if you start inventing your own language, nobody is going to know what you are talking about unless you explain your new language.

[chilly ocean]

LOL

yeah

i just heard it on youtube

[slate forum]

@prime gale it's Rotman's homological algebra

[golden pasture]

f is a root of unity

[thorn delta]

there is a typo here, right? sigma(x_1, x_2, ..., x_n) should be (x_2, x_3, ..., x_p, x_1) and not x_2x_3...x_px_1

[timid hull]

yeah i think so

[steady axle]

Suppose we have a projective resolution P -> Z -> 0 of G modules with finite Z rank and we apply Hom( _ , Z) then we get 0 -> Z -> Hom(P, Z) - (1) . Why is (1) exact again?

[prime gale]

Isn't Hom always left exact? That is, isn't all that other information you have irrelevant?

You can prove your statement with nonsense directly from fact P->Z is epi. Let f: P->Z. Let g in Z = Hom(Z,Z). Assume image of g in Hom(P,Z) is zero. That is, assume g o f: P->Z is zero. Then since 0 o f:P->Z is also zero and f is epi, then g=0. Thus Z->Hom(P,Z) is injective.

@steady axle

[steady axle]

Isn't Hom always left exact? That is, isn't all that other information you have irrelevant?
@prime gale projective resolution is ... -> P_2 -> P_1 -> P_0 -> Z -> 0 , how does left exactness do the job?

[prime gale]

I see. You mean P is not just one module.

[steady axle]

no it is not. additionally in ... - > P_3 -> P_2 -> P_1 -> P_0 -> Z -> 0 P_is are projective

[prime gale]

I haven't thought this through, but cohomology of (1) is derived functors of Hom, right?

[steady axle]

it is Hom(_, A) where A is a G module

[prime gale]

So you are asking why R^iHom(_,Z)(Z) = 0, right? I am just rephrasing, not adding anything.

[steady axle]

I am not very comfortable with the language of derived functors. What I want is when we consider Z dual of ... - > P_3 -> P_2 -> P_1 -> P_0 -> Z -> 0 why is it still exact?

If it helps the context is definition of Tate cohomology via complete resolutions

[prime gale]

I don't think it is just some easy nonsense. I have actually never looked at a left derived functor before--let alone this one.

Is this how you define Tate cohomology? That is, Tate cohomology is derived functor of Hom(_,Z)?

[steady axle]

No.

There is a way (called splicing maybe) to connect P ->Z->0 and (1) (its Z dual) to get a very long sequence then apply Hom(_,A) and define homology groups in usual way

[prime gale]

Looking at wikipedia, the cohomology groups are Ext(Z,Z).

[steady axle]

i.e. ker/im

this is one way another way is to use norm map between oth homology group and 0th cohomology group to stitch together long exact sequences of homology and cohomology to get vey long exact sequence

the 1st one is the one i am trying to learn

2nd one is easier

[prime gale]

And wikipedia says Ext^i(A,B)=0 if A is projective. But I guess that is all cheating.

That is, I am probably using tons of results that you are learning this stuff in order to prove.

Wait a minute, isnt your statement trivial?

Z is projective, so the trivial resolution is a projective resolution. The cohomology doesn't depend on which resolution you pick. Cohomology is zero for trivial resolution. So it is zero for your resolution. Therefore exact.

Trivial resolution is 0 -> Z -> Z -> 0.

[steady axle]

but Z is not free over Z[G]

[prime gale]

What is G?

[steady axle]

group, finite in this case : )

[prime gale]

I see. I thought you were talking about Z modules--aka abelian groups.

[steady axle]

oh sorry no.

[prime gale]

So Z[G] is non abelian group ring?

[steady axle]

yes

[prime gale]

Wow--I never even thought about non commutative rings.

Looks like I am totally wasting your time. But you taught me something.

[steady axle]

no i thought about some things which i had not earlier so it was helpful

[prime gale]

This stuff is normally defined for "abelian categories" which all goes out the window.

[steady axle]

which stuff?

[prime gale]

Everything I've learned about abelian categories.

Sorry--I mean derived functor stuff and injective or projective resolutions is usually defined for abelian categories. So everything I've learned goes out the window.

Maybe your stuff is an abelian category. I have no idea.

[steady axle]

well we are interested in cohomology of modules which is an abelian category so this stuff is in fact in abelian category setting afaik

[prime gale]

I guess you are looking at Ext groups. I have no idea why they are zero.

[steady axle]

why are u convinced that they are 0

[prime gale]

It is just rephrasing your statement in terms of derived functors. Not adding anything.

That is, definition of Ext groups is cohomology of (1). It turns out you can also define Ext groups as right derived functors of Hom(Z,_). So if you can show that for any injective resolution, that if 0->Z->I is exact, then 0->Z->Hom(Z,I) is exact, then you have shown Ext is zero, and thus Hom(P,Z)->Z->0 is exact.

I think this stuff is true for all abelian categories, so it should work in your case. But I have no clue about group rings.

How does Z[G] act on Z? Every element of g acts as identity on Z?

[steady axle]

How does Z[G] act on Z? Every element of g acts as identity on Z?
@prime gale yes

[woven obsidian]

How do I view k as a k[x]-module?

By letting the "multiplication by scalar f(x)" be f(x) * a=f(0) * a? For some a in k

[steady axle]

By letting the "multiplication by scalar f(x)" be f(x) * a=f(0) * a? For some a in k
@woven obsidian yes

[woven obsidian]

Thanks

[oblique river]

@woven obsidian there are lots of ways -- for any b in k you can have f(x) act on a by multiplication by f(b). The only one that makes the sequence in the problem compatible with the k[x] action is if you choose b=0

since the kernel is (X)

[loud kettle]

Man rings were never introduced to me as necessarily commutative. When I took my undergrad class in algebra rings were always non commutative, surprising to find that so many were introduced to rings as commutative

[solemn rain]

pussy mathematicians

[chilly ocean]

@prime gale hey can i ask you a Q about yesterdays question that i asked you

[prime gale]

@chilly ocean You can ask. If it is too hard for me, I won't be able to answer.

[chilly ocean]

actually nvm i got it

thx

[smoky cypress]

If f is a surjective homomorphism and m is a maximal ideal of the codomain then is preimage of m a maximal ideal of the domain

[prime gale]

If your ring is commutative with unit, then I think so.

Easy to prove with just definitions.

[restive shard]

Hi. Could anyone recommend me an interesting book on the applications of abstract algebra to analysis?

[ocean magnet]

is that what algebraic topology is? 👀

[solemn rain]

@cinder bone what do you think

[cinder bone]

yes? im inclined to think

something about ad-bc=1

prevents me from throwing a matrix full of zeros in there

but idk how to prove

[solemn rain]

so let x be an element in S

how would you define x^-1

normally

and once you do

[scarlet estuary]

are you familiar with the determinant

[solemn rain]

does this fit in the set

what does an inverrse mean ?

for an element

[cinder bone]

i should be able to multiply by a matrix to get the identity matrix

[solemn rain]

yea

an inverse for x, call it x^-1

should satisify xx^-1 = x^-1x = identity

now try to see how this element looks like

i know its a 2x2 matrix with coefficints from R

whats missing is that is the determinant 1

[cinder bone]

yeah i get that

[solemn rain]

if x has det 1 does x^-1 have det 1

given x*x^-1 = identity matrix?

[cinder bone]

i dunno thats what im wondering

hmm

[solemn rain]

okay first as some1 askd

are u familiar with the determinant?

determinat properties etc?

[cinder bone]

no

i know what it is but nah

im just trying to prove this thing is a group

or not i suppose