groups-rings-fields

[steel pulsar]

and I've already taken the time to think. I came back to see if anyone could explain this to me

again

[paper flint]

Then try to think more about what has been suggested above? Can you guess the number of permutations of S_n that fix 1?

Laying this out would simply be writing the solution

[steel pulsar]

I've already taken the time to think

btw, doesn't matter, I skip this question

[paper flint]

[slim kayak]

Any field element other than 0 is a unit tho

[stark helm]

so why f(x)=a0(constant) is reducible here?

[stark helm]

I mean f(x)=a0 reducible over field F?

[slim kayak]

It's most of the time a unit

[stark helm]

so you mean because a0 is not unit at a0=0, where 0 is irreducible over F?

[slim kayak]

Both reducible and irreducible elements are usually about non-units

Irreducible specifically requires the element to not be a unit. Could you post the actual statement?

[slim kayak]

Yeah, it's not irreducible since it is a unit

It's isn't reducible either

[stark helm]

so if there is no unit element, it is reducible right?

[slim kayak]

Unit means it is invertible

[stark helm]

so invertible means reducible right?

[slim kayak]

No, not necessarily. Reducible means it is the product of two non-unit

[celest furnace]

Try using lineararity of expectation

Define X_i = i is fixed by p

[stark helm]

So can I say because it is a unit, so it should be neither reducible nor irreducible?

[stark helm]

It seems somewhat weird that both reducible and irreducible require elements to be non-unit?

[slim kayak]

Oh, reducible doesn't require it

[stark helm]

but reducible require product of two non-unit right?

[slim kayak]

Yeah

If it didn't anything would be reducible

Like, given any element f you can express f as a*a^-1 f with a an unit

Which is stupid

And if ab is a unit, then so is a or just b. So units can't be reducible

[crystal vale]

In 8 question G/H is isomorphic to Z2?

[dull marsh]

Yeah

How would you prove that?

[crystal vale]

We map a Matrix to 0 which has positive determinant and a Matrix map to 1 which has negative determinant then kernel will be positive determinant matrix....

[dull marsh]

You got it yeah

[next obsidian]

I think it’s more clear or intuitive

When you see that Z/2Z is also {1,-1}

With multiplication

[crystal vale]

Yes

[next obsidian]

Then this is the map GL_n -> R\{0} -> Z/2Z

Sgn(det(M))

[crystal vale]

Okay

[still dew]

what is a nontrivial set of generators of S_A(set of permutations of A) where A is inifinite

Is it possible for {(k,k+1)} to generate S_N, N being natural numbers

i believe that any finite product of these will correspond to a permutation in some S_n for a large enough n but if we introduce a permutation that has like say infinite transpositions then it shouldnt be possible?

[true drum]

Those 2-cycles are enough to generate S_n. Another is <(1,2),(1,2,3,...i)> for all i>1

[celest furnace]

his question was more about an infinite set being permuted such as A = Z

What is S_N (N = natural numbers) generated by?

[true drum]

Oh I misread, sorry

[celest furnace]

But about this... permutations of an infinite set (such as N) would have size |N^N| >= |R| so any finite set of generators wouldnt work since you would get a countable free group i think

[celest furnace]

But yeah this is true {(k, k+1)} doesnt generate by this exact reasoning it cant be a finite product of those

And this fact shows you cant even have countably many generators

[still dew]

right

thank you

[celest furnace]

Happy to help

[still dew]

also

can we explicitly state a set of generators other than S_N itself

[celest furnace]

I think u can show this since it would be a countable union over k = length of the word and there would only be countably many words of length k

[celest furnace]

Idk how helpful this is since the set of generators would have same cardinality as |S_N| anyways (it turns out that |S_N| = |R|, not just >=

But you could include everything but (123) and you can get (123) = (12)(23) lol

[still dew]

right

[celest furnace]

So u could also drop all terms of the form (abc)

[still dew]

that

yes

thank you again

[celest furnace]

Of course !

[upbeat dirge]

sorry, what book are you using?

[dull tiger]

I have 4 groups $\mathbb{Z} \times \mathbb{Z} / <(6,12)>, \mathbb{Z} \times \mathbb{Z} / <(12,17)>, \mathbb{Z} \times \mathbb{Z} / <(1,6)>$ and $\mathbb{Z}$ and I'm trying to find the one, that is not isomorphic to the others by using the first isomorphism theorem (some also call it the fundamental theorem on homomorphisms). So I started by looking for homomorphisms between $\mathbb{Z} \times \mathbb{Z}$ and $\mathbb{Z}$ with respective kernels. I found $\phi(x,y) = 6x -y$ as homomorphism with kernel $<(1,6)>$, but then i got stuck. For a homomorphism with kernel $<(6,12)>$, $2x-y$ doesn't quite work, since it has a larger kernel than the one i want. But i can't seem to find a function that would work. So maybe that group is the odd one out, but for the kernel $<(12,17)>$ I already struggle to find a function, that sends 12 and 17 to 0. Is there an easier way to find those functions, than try and error? And is there a way to see faster, which one is not isomorphic to the others, than finding those homomorphisms?

[cloud walrus]

Marieeee

[uncut cloud]

In each case you have 2 generators as a Z-module, quotiented by a line(for the last one you can see Z as ZxZ /(0,1)).

If you were in the theory of R- vector spaces it would be clear that they are all isomorphics.

But here the question is when can you find another vector to complete the base as a Z module.
So you will find Z if and only if you have no common divisor

So the first one is the différent one

[dull tiger]

could you maybe explain the geometric approach here in more detail? I never thought of it like that. It seems like a good thing to enhance intuition behind the concepts

[uncut cloud]

Yes I am almost done eating

Are you familiar with Vector spaces ?

[dull tiger]

Don't rush please 🙂

[dull tiger]

Yes. I did a Uni class on linear algebra

[uncut cloud]

So here is the real vector case. You have the ambient space in red, and the space you want to quotient by in blue (not at scale but it won't be important for the real case). But here you have an important theorem which state the existence of a supplementary space (in yellow). So in each situation the quotient is clearly isomorphic to the yellow line, so you have in each situation a vector space isomorphic to $\mathbb{R}$.

[cloud walrus]

Critotomatic

[gleaming widget]

oh we have latex typesetting like this, ill redo my question

im starting to study abstract algebra this semester and decided to look at the curriculum and do some exercises (using John B. Fraleigh's A First Course in Abstract Algebra) and i got stuck on this:

Example 1.15 asserts that there is an isomorphism of $U_8$ with $\mathbb{Z}_8$ in which $\zeta = e^{i(\pi/4)} \leftrightarrow 5$ and $\zeta^2 \leftrightarrow 2$. Find the element of $\mathbb{Z}_8$ that corresponds to each of the remaining six elements $\zeta_m$ in $U_8$ for $m = 0, 3, 4, 5, 6, \text{and} 7$

in this example they show that $\zeta^2 = \zeta \cdot \zeta \leftrightarrow 5 +_8 5 = 2$ and so i was wondering if we just continue this to get the six other elements, i.e. $\zeta^3 = \zeta^2 \cdot \zeta \leftrightarrow 2 +_8 5 = 7,, \zeta^4 = \zeta^3 \cdot \zeta \leftrightarrow 7 +_8 5 = 12 - 8 = 4, ...$

[uncut cloud]

Now what happen in your case. Well the main difficulty is to find a supplement to the $\mathbb{Z}$ line generated by a vector $(a,b)$.
But does this supplement always exist ? if for instance $(a,b)=(1,0)$, it is clear that $(0,1)$ will do the job. But in general if you have a supplement generated by the vector $(c,d)$, then each point of $\mathbb{Z}^2$ will have a unique expression as $n(a,b)+m(c,d)$ for $n,m \in \mathbb{Z}^2$.
So you find a necessary condition !
The $2x2$ matrix generated by $(a,b)$ and $(c,d)$ must be invertible as an integer matrix. So it determinant must be invertible in $\mathbb{Z}$. So it must be $1$ or $-1$. So $ac-bd=1(-1)$. So you see that $gcd(a,b)$ must be $1$. And in fact Bezout theorem states the converse assertion.

[cloud walrus]

Critotomatic

Slimey

[rocky cloak]

Yes, that's correct. An isomorphism must preserve the operation, so multiplying by zeta on one side should give the same as adding 5 on the other.

[gleaming widget]

okay, i see, thank you

[uncut cloud]

So in your case you see that for $(12,17)$, $(1,6)$, you can find a supplement (generated by the coeffiecients of the Bezout equality), and so it is isomorphic to $mathbb{Z}$. But for $(6,12)$ you won't be able to find the supplement. So this case is suspect. In fact, as you can find a supplement for $(1,2)$, you will see that this quotient is isomorphic to $\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}$

[cloud walrus]

Critotomatic

[dull tiger]

Many many thanks, this helped me a lot!

[uncut cloud]

Your welcome. I struggle with this kind of questions in algébraic topology quit a lot so it is a pleasure to explain it

But the important point to notice is that Z^n quotiented by something isomorphic to Z is not always Z^n-1. You cas find some torsion

[dull tiger]

Are there some prerequisites on the quotient that still allow us to deduce that Z^n quotiented by something isomorphic to Z is Z^n-1?

i mean like some special cases, when the quotient has specific properties

[static glen]

Homomorphism theorem

Thinking of Z^n as a rank n Z-mldule might be helpful

[dull tiger]

okay, thanks

[uncut cloud]

It is exactly the same thing in higher dimension

If you take the line generated by (a_1,....a,_n), the quotient will be Z^n-1 if and only if you can complete this vector to form a nxn matrix invertibile as an element of M_n(Z)

So the condition of coprime is still necessary but not sufficient anymore

[gleaming widget]

returning to my example, how is it that we can know that $\zeta = e^{i(\pi/4)} \leftrightarrow 5$ lays the ground for an isomorphism between $U_8$ and $\mathbb{Z}_8$? ($U_n$ are the nth roots of unity, ${z \in \mathbb{C} | z^n = 1}$, and $\mathbb{Z}_n$ is a set (group?) under addition modulo n.)

do you just have to try? for example assume $\zeta \leftrightarrow 5$ then, $\zeta^2 \leftrightarrow 13-8 = 5, \ldots, \zeta^8 = \zeta \leftrightarrow 5$? because this all happens cyclically on a circle?

we are given that the elements of $U_n$ are given by
$$\cos\left(m\frac{2\pi}{n}\right) + i\sin\left(m\frac{2\pi}{n}\right),\quad m = 0, 1, 2, \ldots, n-1$$

in another exercise, i am asked to show why there is no isomorphism between $U_6$ and $\mathbb{Z}_6$ with $\zeta = e^{i(\pi/3)} \leftrightarrow 4$. here i just tried multiplying by zeta on one side and adding 4 to the other, but then i came to $\zeta^4 \leftrightarrow 4$, which means the pattern repeats before it should. is there another way of showing this? zeta is picked seemingly from the formula for the members of $U_n$, since it is what we get if you plug in $n=6$, but it doesnt grant an isomorphism

[cloud walrus]

Slimey

[dull tiger]

very good, thanks a lot!

[delicate orchid]

the order of those elements is different, isomorphisms preserve the order of elements

"the pattern repeats before it should" is saying that this map has a non-trivial kernel, and is thus not injective

[gleaming widget]

oh okay, so by trying, we find that the relation we are given isnt injective, and therefore not bijective, and therefore cant be an isomorphism?

[delicate orchid]

yeah

[gleaming widget]

is there another simple way of doing it that i am not seeing?

[delicate orchid]

I don't mean to be rude but you can't really get much simpler

you've found a non-identity element that maps to the identity, hence it's not an isomorphism

[delicate orchid]

or just do this, I guess this is easier

[gleaming widget]

the order of which elements do you mean?

[delicate orchid]

there are only two elements involved, 4 and zeta

[gleaming widget]

what do you mean by order?

[delicate orchid]

and this is where I check out

[gleaming widget]

ok well thanks

[sinful kite]

Don't worry about it since you are on the first chapter and they haven't defined a lot of stuff yet (haven't even defined groups yet!). What you noted is sufficient that the map is not bijective. When you'll do cyclic groups you'll look at order of an element and be able to use this language. For a sneak peek you can look ahead at the first para of the section on cyclic groups (section 6?)

[gleaming widget]

okay, i see, thank you

[frank cosmos]

i havent seen this idea of localization yet

[slim kayak]

The prime ideals of the localization are exactly those prime ideals that don't meet the multiplicative set. So if your prime ideals doesn't touch Z then you can know it's image is also prime (in Q[X]*)

[frank cosmos]

how come Z-modules <-> abelian groups but Z-algebras <-> groups? an algebra is supposed to be stronger than module right

[slim kayak]

are you sure it isnt saying Z-algebra <-> ring?

[tardy hedge]

PooP

[delicate orchid]

poop!

[south patrol]

Note a Z-algebra is just a ring with a map Z -> R into the centre, but there's always exactly one map Z -> R and it lands in the centre, so there is no difference in data between the two notions

[delicate orchid]

missing a "no"

[south patrol]

lol

Good point

thank

[delicate orchid]

missingno, if you would dare to be so bold

[south patrol]

LOL

Haven't heard that name in literally years

[tardy hedge]

wow same^

i remember watching creepypasta video from yuriofwind?

was that the guy?

[slim kayak]

You remember the names of creepypasta YouTubers?

[tardy hedge]

i remember yuriofwind cause i watched him a lot

he did gaming creepypasta vids

[south patrol]

This is funny. Reading a paper where they study certainn operations - the basic ones are "suspensions" so the author calls them "pensions" which i thought was funny

[tardy hedge]

BEN DROWNED

[south patrol]

But now post pandemic, does that mean suspensions are sus pensions?

[slim kayak]

Damn, I am disappointed Mr Erdapfel

[south patrol]

Herdöpfel

[delicate orchid]

lol?

[south patrol]

I am stuck on an algebra problem which I think just boils down to like a matrix bash

[delicate orchid]

that's all of them

[south patrol]

It's like uhhh

you know polynomial functors

[delicate orchid]

I do now

[cobalt heath]

Polynomil?

[delicate orchid]

wait yeah is that typo

[south patrol]

lol

[delicate orchid]

I can never be sure when u mfs are pulling shit like this

[delicate orchid]

anyway, never heard of them. Assuming it sends X to a coproduct of some F(X)^n

like a polynomial

damn nevermind the actual definition is so stupid I would never have guessed

[south patrol]

I think there are various different definitions, some of which are inequivalent

[delicate orchid]

no it isn't. You morons. Pick a better name

[south patrol]

yeah that isn't what i have

dw

This is basically uhhh

[delicate orchid]

hahahah I was close

ok nevermind the definition makes sense now

[south patrol]

say $F: \mathcal C \to \mathrm{Mod}R$ a functor, $\mathcal C$ symmetric-monoidal, and then we have all these various maps $F(X_1 \otimes \dots \otimes X_n) \to \bigoplus{k=1}^{n} F( X_1 \otimes \dots \otimes \widehat{X_k} \otimes \dots \otimes X_n)$, Call its kernel $\mathrm{cr}_n(X_1,\dots,X_n)$, which defines a functor $\mathrm{cr}_n:\mathcal C^{\times n} \to \mathrm{Mod}_R$

[cobalt heath]

Huh

[south patrol]

So basically the idea is uh

The cross effects vanishing corresponds to like

[delicate orchid]

I can see why they chose it now, it's to take fibres for some "exponential object" which replaces x^n in normal polynomials

hold on lemme absorb

[south patrol]

"oh we know about the functor given that we know what it looks like on fewer objects"

[cloud walrus]

potato

[cobalt heath]

Exponential object, that sounds quite screwy

[delicate orchid]

what's X_k hat here? is it like an evaluation?

[south patrol]

omit it

[delicate orchid]

Hom(X, Y)

[delicate orchid]

oh ok. Kummer theorem

[south patrol]

So basically it's like

[cobalt heath]

Yes.. ah wait. Sometimes that is in the category, huh

[delicate orchid]

sometimes!

[south patrol]

if cr_n vanishes then we know what happens when we put in an n-fold tensor product, given that we know what it looks like on (n-1)-fold tensor product

which is sort of meant to be analogous to being polynomial of degree < n

[delicate orchid]

oh

[south patrol]

like polynomials being determined by a few points

But also like, the nth tensor power is polynomial of degree n

[delicate orchid]

potato can you see how this is related to our discussion of coskeletal simplical sets from yesterday lol

[south patrol]

Not explicitly but in terms of vibes yes

[delicate orchid]

reminds me of the bar resolution kinda

[south patrol]

like you can reconstruct smth given partial info

[delicate orchid]

more specifically

you're omitting a term - that's a face map

[south patrol]

True yes actually very good observation hm

[cobalt heath]

Yea that looks like derivation map (that's all I know)

[south patrol]

I wonder if there is a way in which yes this is just a simplicial thing

[delicate orchid]

which makes sense considering I bet you're doing cohomology with this

[south patrol]

But yes the context I'm in uses simplicial sets things

lol

[delicate orchid]

you're doing a coproduct of a bunch of (n-1)-simplices

[south patrol]

Yeah derived functors

[delicate orchid]

let me cook one sec

it's like the bar resolution but you're not just A^\otimes n-ing it

if all of the terms are equal and you ignore the F then it just is

anyway we're getting side tracked I think

[cobalt heath]

Maybe one can draw Simplex-shaped diagram for this

[delicate orchid]

still don't know what these are btw. Don't really care either

[south patrol]

Wait I messed up LOL

i think

maybe

[delicate orchid]

anyway I get this definition now

kind of. I still think the nlab one is more intuitive to me

like I'm still not 100% seeing how you'd call this "polynomial", but I definitely can with the nlab one

[south patrol]

No it should be

(+) rather than (x), I was a dumbass

[delicate orchid]

OH

[south patrol]

It's that like

[delicate orchid]

right yeah

[south patrol]

The source I was reading views Mod_R as a symmetric monoidal category and doesn't use the normal direct sum notation

But it's symmetric monoidal under direct sums lol

But it makes way more sense now

Lol

[mighty kiln]

[south patrol]

And so then the point is like

if F is an additive functor then it is polynomial of degree 1

Lol

Okay this makes way more sense lol

[delicate orchid]

the point is that you're doing a bunch of sums in lower variables and then combining them together using the monodial product in C? nvm whoops

or is the direct sum on the outside an honest to god coproduct

[south patrol]

Oh wait no I'm okay so

Okay lol

So the direct sum is a direct sum yes a coproduct

And the tensor is fine

But the example to have in mind is C = Mod_R with (x) given by direct sum

lol

Oopsies

[delicate orchid]

ok I should also clarify, F is the polynomial functor here - not it's kernel?

[south patrol]

Sorry yes I've written it badly - F is the functor, cr_n is the "nth cross effect", and F is said to be poly of degree <= n if the n+1st cross effect vanishes

(in which case all the higher cross effects vanish too)

[cobalt heath]

Hmmmm

[delicate orchid]

I'm not on enough crack for this shit

[cobalt heath]

Too much abstract nonsense?

[delicate orchid]

where are the triangles

I just cannot see how this is a polynomial, sorry

[south patrol]

So for $F = (-)^{\otimes N}:\mathrm{Mod}_R \to \mathrm{Mod}R$, $F$ being poly of degree $\le n$ i.e. F(X_1 \oplus \dots \oplus X_n) \to \bigoplus{i=1}^{n} F(X_1 \oplus \dots \widehat{X_i} \oplus \dots \oplus X_n)$ being injective for $n > N$ corresponds to the fact that the sum $(x_1 + \dots + x_n)^N$ is determined by the lower powers of sums of $x_i$

[delicate orchid]

right, that makes sense - and what I think I was aluding to up here

I just... wouldn't call that "Polynomial"?

[south patrol]

I guess

uhh

[delicate orchid]

it's a skill issue on my part

[cobalt heath]

Ah wait, so coproduct of X_i's, not tensor product?

[south patrol]

Nah on my part lol

hm

yh

[delicate orchid]

the tensor product was just a symmetric monoidal product in C

[south patrol]

uhhh

[cloud walrus]

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

[south patrol]

yh

[cobalt heath]

Yea

[delicate orchid]

and for R-mod you get TWO!!! woahhhhh

[cobalt heath]

Huh???

[delicate orchid]

a ringad is a ring in the category of endofunctors

ok, semiringad in this case

[slim kayak]

what is this even for 😭

[delicate orchid]

wait hmm

[cobalt heath]

Polynomials where

[delicate orchid]

can we just do n = 1 lol?

right yes that's additive ok I see that at least

I think what's confusing me is this looks like an element of R[x_1, ..., x_n] but it also has to be degree n?

I'm sticking with this one omegalul. Concrete categories only for me

[cobalt heath]

How do you determine (x1 + x2)^3 by the lower powers of sums

[south patrol]

Basically if you're interested in derived functors of non-additive functors then like

Polynomial functors are the easiest next case lol

[south patrol]

It's the other way round, it's like

(x_1 + x_2 + x_3)^2 from (x_1 + x_2)^2 etc

[cobalt heath]

Ah other way round

[slim kayak]

I think ill stick to homological algebra then

[delicate orchid]

this is homological algebra

[cobalt heath]

This was not clear for me, hm

[south patrol]

tbf this is homological algebra lol

[delicate orchid]

(see ⁠get-advanced-access to use this channel) topics that you'd learn in a first introduction to abstract algebra, pertaining to groups, rings, modules, and fields.

[slim kayak]

i thought non-additive functors have to do with additive cats

[cobalt heath]

When you want derived functors for non-additive functors

[slim kayak]

Okay, doesnt preserve biproducts

[delicate orchid]

wait. @south patrol is geometric realisation a polynomial functor

[south patrol]

e.g. symmetric powers

[unkempt stream]

scary

[delicate orchid]

no. nevermind

forgot there was a quotient

[cobalt heath]

Idk how

[frank cosmos]

is this summary right? modules are abelian groups which are like vector spaces over a ring and algebras are rings such that the r-operation is compatible with the ring multiplication

[delicate orchid]

symmetric polynomial decomp is magic

[south patrol]

wdym by r-operation

[delicate orchid]

the map R -> End_R(A) ig

[frank cosmos]

yeah

[south patrol]

Sure

[frank cosmos]

it gives a map R x A -> A

[cobalt heath]

Symmetric poly where

[unkempt stream]

Mmm delicious curry

[delicate orchid]

x_1+x_2 dear

[cobalt heath]

But that is missing x3

[delicate orchid]

I hate that guy

:3 -> x3 may he rest in peace

[unkempt stream]

*foghorn BWAAAAAAUUUUUUH*

[frank cosmos]

is there some word for "in between" objects. like a ring with R-module structure, but where the 2 multiplications arent compatible so it isnt an algebra

[unkempt stream]

make one lol

[delicate orchid]

nobody cares about such an object ngl

[unkempt stream]

true!

[delicate orchid]

at that point just make the category of such objects and use general notions lol

[cobalt heath]

Idk I gotta consider it as dimension computation on vectorspace

[south patrol]

I'm pretty sure Absta like

so

(x_1 + x_2 + x_3)^2 can be written as like

(x_1 + x_2)^2 + ... + (x_1 + x_3)^2

minus like

[unkempt stream]

SLBY category (silly billy algebras)

[south patrol]

x_1^2 + x_2^2 + x_3^2 or whatever

[cobalt heath]

Wait

[south patrol]

So yeah i think this is just theory of symmetric polys as you say

[cobalt heath]

I thought you were only allowed to use (x1 + x2)^2

Anyway it makes sense now

[delicate orchid]

so why aren't they called SYMMETRI- ohhhhh it's a symmetric monodial product

ohhhhhhhhh

[south patrol]

wait maybe i'm silly

idk

[unkempt stream]

Symmetric polynomials in N variables of degree n is generated by nCr(n + N - 1, n) i think

[south patrol]

Sleepy

[delicate orchid]

burrrppppp

[south patrol]

lol

Need to think

[unkempt stream]

as a vector space

[delicate orchid]

WHAT

[south patrol]

aren't those the integer valued ones

[unkempt stream]

wait no

homogeneous

not symmetric (big dumb)

[delicate orchid]

I don't even know what that notation means

[mighty kiln]

Why is it be called polynomial functor

[south patrol]

ig you can plug in 0

[unkempt stream]

I would say binomial coefficient but many see the combinatoric form N choose n

[delicate orchid]

oh you mean the dimension

[mighty kiln]

Is it cardinality be polynomial

[unkempt stream]

i don’t want to latex, mobile lol

[delicate orchid]

sum of exponential objects

[mighty kiln]

Oh

[delicate orchid]

rename this channel to #the-hot-box with all this :pack:

[cobalt heath]

How

[mighty kiln]

That be funni way to polynomial

[south patrol]

as in like (x_1 + 0)^2 = x_1^2

wiat no i got myself confused i think lol

hm

[unkempt stream]

I have never heard of a polynomic functor and I am scared

[cobalt heath]

I mean, how is this legal in this setup

[south patrol]

I don't think it is idk

Maybe I shouldn't've pursued this analogy

[unkempt stream]

goofy

[south patrol]

i don't think they're like too common

but i could be wrong

I only know about it cause smth i was doing very explicitly used a polynomial functor

[unkempt stream]

flashback to the time I tried to formally prove R[X] for com ring R is a graded ring

unfun

working with the universal property

[south patrol]

Oh wow this paper is a bit old lol

[slim kayak]

category theory was 2 weeks old when this was published

[delicate orchid]

why the fuck are they taking a semidirect product of a group H and a F[G]-module. How are they even doing this

[unkempt stream]

functor algebra AAAAAAA

[south patrol]

lol

[delicate orchid]

https://www-users.cse.umn.edu/~webb/Publications/CategoryAlgebras.pdf jumpscare

[south patrol]

who's they

[delicate orchid]

bobby

[south patrol]

lol

ofc

[delicate orchid]

unrelated quip

[south patrol]

QUip

maybe we should move to #advanced-algebra or smth

lol

[slim kayak]

what even is the definition of graded youd use in such a scenario

Idk, polynomials fit more into #prealg-and-algebra

[south patrol]

LOL

[delicate orchid]

ok I'm just going to assume they're passing the map given by H -> Aut(G) (presuming it's that way around) to a permutation of the basis of F[G]? this is cringe

[south patrol]

Well just give X grading 1 innit

[delicate orchid]

he should have explained this

[cobalt heath]

I had an intuition why when the kernel vanishes, it would be called degree n poly, but then I forgot

[delicate orchid]

polynomials are only 0 at n points? uhhh

[slim kayak]

I am not sure how to picture that if your description of R[X] is given by an universal property

[unkempt stream]

that’s the “fun” part

Mainly came down to proving R[X]/(X) is iso to R, and doing some module maps and using the fact that (X) and powers of it are principal

[delicate orchid]

ok no now it's just bullshit. The generalised fitting subgroup of G is a module over the group algebra? This is stupid
literally doesn't even type check anymore

[south patrol]

Hm

[unkempt stream]

Mainly just extending the identity morphism of R and sending X to 0 to get a morphism from R[X] to R

[cobalt heath]

What are you guys reading

[unkempt stream]

asserting uniqueness, showing (X) is in the kernel of that morphism

and then doing some other uniqueness arguments to show that the kernel must be in (X)

i.e they’re equal

there’s probably a way easier way but that was the way I thought of doing it in the moment

That uses the fact that Ker(a) lies in Ker(b o a) for any ring morphisms

[unkempt stream]

this just came to my head first because of it being used in some salamander lemma proof i tried

[south patrol]

Lol consider canonical map $R[x] \to R$ using universal property as you said, let $I$ be the kernel and say that $f$ is of degree $n$ if $n$ is maximal with $f \in I^n$

[cloud walrus]

potato

[south patrol]

then you need to check that that gives you a nice decomposition i guess

Like the sum should easily be direct but then you need to check it actually sums up to R[x]

i think

[unkempt stream]

i just used powers of the ideal (X) being (X^n) and just considering (X^n)/(X^n+1) as a module

And using the quotients to decompose

[unkempt stream]

canonical maps are unique maps, right?

i always get that term mixed up

[coral spindle]

No, there is significantly more than one map R[x] → R; canonical essentially means 'without making arbitrary choices' — potat probably meant the (this time unique) map sending x to 0.

In fact this is most certainly what they meant

[cobalt heath]

Wish I could do
R[x] -> R x Z
r |-> (r, 0)
x |-> (0, 1)

[unkempt stream]

that’s what I meant by it being the unique extension (valuation)

that sends X to a given value in the codomain

[cobalt heath]

Nvm, this does not even make sense

[unkempt stream]

which is unique by the universal property

[slim kayak]

I am wondering what a nice definition of the universal property of poly ring would be. A ring A together with a set-inclusion {X_i} -> A, and a ring inclusion \iota sending R into A so that for any map f: R -> S one has an unique map g : A -> S so that the following two conditions hold?

1. g \circ \iota = f
2. the image of the X_i in A get send to select elements s_i in S

[unkempt stream]

Let R be a commutative ring (with unity)

For any ring morphism R into P, and any element x of P, there is a unique morphism extending it from R[X] into P that sends X to x

is the def I am using

though I think you need to restrict P to be commutative with unity, or at least have R in it’s center (an Algebra) if you’re considering it for rings

that uniqueness of the morphism is what I used the most

[crystal turtle]

R[x] is a representing object of the functor A --> Cring(R,A) x U(A)

Where U: CRing --> Set is the forgetful functor

[cobalt heath]

What is A

[crystal turtle]

A is a (commutative) ring

[unkempt stream]

uniquely the identity and setting X to 0, we have an epimorphism from R[X] to R, and an endomorphism, t, on R[X] by using the embedding.

Now consider the projection, p, from R[X] onto R[X]/(X). This sends X to 0, and uniquely extends a map from R to R[X]/(X). It is the unique morphism to do so.

However t fixes the embedding of R into R[X], so composing the projection of R[X] onto R[X]/(X) over the endomorphism t ALSO extends the above-mentioned map and sends X to 0. Thus: p o t = p

Ker(t) must lie in Ker(p o t) = Ker(p) = (X). But because X is in Ker(t), then (X) must lie in Ker(t). This means (X) = Ker(t), and by First Isomorphism Theorem, R[X]/(X) is iso to R

what remains is that the ideals (X)^n/(X)^n+1 in R[X]/(X)^n+1 are isomorphic as modules to R. This map is actually similar to the definition of the Zariski cotangent space

you have to show that X is not a zero divisor

this is an absolutely rediculous way of doing things lmao

[frank cosmos]

this seems to simple to be correct: the image of phi is a submodule and hence either 0 or N

and then phi is injective since the kernel is either 0 or M

this proof is much longer though https://en.wikipedia.org/wiki/Schur's_lemma

[agile burrow]

You're correct

The module theoretic proof on Wikipedia just looks longer because they prove that the image and kernel are submodules

[frank cosmos]

any hints

[delicate orchid]

Schur's lemma

[frank cosmos]

but here only 1 module is simple

[delicate orchid]

recall the definition of an endomorphism

[frank cosmos]

so then End(M)=Aut(M) \cup 0

[frank cosmos]

it is trippy that integral domains are isomorphic to principal ideals of themsleves (as modules wrt the ID)

[delicate orchid]

chat what does this mean

[frank cosmos]

(if the ideal is (x), x!=0, you have the obvious map r->rx which is surjective by principal ideal defn and ker=0 since integral domain)

[frank cosmos]

R integral domain. I is a princiapl ideal. R = I as R-modules

[delicate orchid]

so every ring is the 0 module over itself?

right you said not zero

I still don't buy this, lol. Z and Z/2Z are equal as Z-modules?

right but they're the things ok

[frank cosmos]

mb nonzero yeah

[delicate orchid]

no it's fine, you made sure to include it lol

[unkempt stream]

what

[delicate orchid]

but I just can't bring myself to believe it

[frank cosmos]

do you mean Z and 2Z

[delicate orchid]

I did, yeah

[unkempt stream]

trying to understand what’s going on lol

[delicate orchid]

every integral domain is isomorphic to it's non-zero ideals as R-modules

[frank cosmos]

principal ideals

PIDs are isomorphic to all non-zero ideals

[formal ermine]

Huh

[frank cosmos]

my intuition for this kind of stuff is pretty bad

i keep imagining vector spaces which is 50/50

[formal ermine]

can't you take like

[delicate orchid]

no no no this can't be true. pick a principle deal I, then R/I = R/R = 0

Z/2Z isn't 0

[formal ermine]

Z^2 and then Z x {0}

[delicate orchid]

integral domains only

[languid trellis]

true its {0, 1} smh mr phd man

[delicate orchid]

but Z \cong 2Z

I don't work with infinite objects for reasons like this. They're fucking stupid

oh yeah you can quotient out by something isomorphic and not get 0? moronic

[cobalt heath]

Not PID

[formal ermine]

this is true for the ring of integers of a number field

[unkempt stream]

Funny

[frank cosmos]

wtf

[cobalt heath]

There is non-bijective injection 2Z -> Z 👀

[frank cosmos]

the embedding, yeah

[formal ermine]

@delicate orchid so you can't take Z as an example

[formal ermine]

wait no this isnt true

theyre just finitely generated

but in the case of K = Q

theyre isomorphic to Z^order

[delicate orchid]

Z[x_1, x_2] and (x_1) \cong Z can't be the same because the former is free of rank 2 and the latter is free of rank 1

[frank cosmos]

Z is an integral domian

[delicate orchid]

no, that's as algebras you idiot. Think

[formal ermine]

so if K = Q and O_K = Z then all nonzero ideals in O_K = Z are isomorphic to Z

[next obsidian]

What’s going on

[unkempt stream]

isomorphic in the Z-module sense?

[delicate orchid]

they'd be Z-modules then yeah

[frank cosmos]

.

[delicate orchid]

do I need something awful? Non-noetherian?

[next obsidian]

For what Wew?

[unkempt stream]

yes :)

[formal ermine]

https://commalg.subwiki.org/wiki/Principal_ideal_is_isomorphic_to_integral_domain_as_a_module

[next obsidian]

Ike

[formal ermine]

this says as a module over the integral domain

[next obsidian]

Like

[delicate orchid]

I don't buy it

[next obsidian]

If x is any regular element (has no zero divisors)

The map R -> (x) given by multiplying by x

Is an isomorphism

[delicate orchid]

yes this is the proof given before

[frank cosmos]

i said this

[next obsidian]

And R-linear

[delicate orchid]

I don't buy it. It doesn't sit right with me

[next obsidian]

Bruh

[frank cosmos]

^ hence trippy

[rose prism]

how is this surprising

[cobalt heath]

Lmao

[delicate orchid]

imagine someone walked up to you and said every group was isomorphic to each of it's normal subgroups in some sense. I'd look at them like they were a martian

[frank cosmos]

because of this

^

[next obsidian]

This is just how they are bruh

[rose prism]

and if someone walked up to me and said Grp behaves the same as RMod

[next obsidian]

I guess this is a quick proof that a finite integral domain is a field

[rose prism]

id look at them like they were a martian

[cobalt heath]

Well I mean, R-module isomorphism is quite lenient

[unkempt stream]

integral domains are cancellative lol. So if y is in (x), there must be a unique c such that cx = y. The isomorphism is the map that sends y to that c or vice versa I guess

[frank cosmos]

well, its still an isomorphism of abelian groups

but yah

[delicate orchid]

ok I buy this for Z. So I buy it for particular principle ideals via Z being inital lol

[unkempt stream]

i don’t think for general integral domains it’d hold though

only if PID?

[frank cosmos]

the theorem is for principal ideals of integral domains

so some ideals of IDs and all ideals of PID

[cobalt heath]

R-Mod is a category, maybe R is initial in this category?

[unkempt stream]

oh then it makes complete sense due to cancellativity

[frank cosmos]

yea thats the proof i gave

[delicate orchid]

the trivial module is the zero object, so no

[frank cosmos]

it is also final i believe

[delicate orchid]

I said zero, that implies both inital and terminal

[frank cosmos]

oops

yeah

[unkempt stream]

Also wew since when were you mod

[delicate orchid]

abelian category moment

[cobalt heath]

Ah ye

[frank cosmos]

shits nicer since u dont need 1->1

@delicate orchid how do you think of algebras over a ring

whats ur intuition/visual picture

[delicate orchid]

quotient of a polynomial algebra

specifically, a set of symbols with relations on them

[frank cosmos]

whats a polynomial algebra

[unkempt stream]

Tensor algebra? Yah queen 💅

[delicate orchid]

people can come at u with that "durr ring with a map R -> End(A)" but like, who? ykwim

[delicate orchid]

R[x_1, ..., x_n]

[frank cosmos]

oh i just say polynomial ring

[delicate orchid]

well yeah but we're viewing it as an algebra

[frank cosmos]

sure

what do quotients of this look like then

[unkempt stream]

by setting certain relation-defined ideals to 0 by setting minimal polynomials of these ideals to 0 if your ring is neotherian

If it’s not noetherian then dive in a woodchipper or something

[frank cosmos]

are all submodules of it ideals?

[delicate orchid]

ok also, that's commutative algebras

but I do view algebras usually as quotients of the appropriate free algebra

[unkempt stream]

I know modules have a tensor algebra

that’s the adjoint of the forgetful from R-alg to R-mod

[delicate orchid]

monad algebra

[unkempt stream]

the fuck is a monad

[delicate orchid]

monoid in the category of endofunctors

[unkempt stream]

WHAT

[cobalt heath]

Huh

[delicate orchid]

construct the tensor algebra but with a general symmetric monoidal product instead of the tensor product

[cobalt heath]

Ahh

[unkempt stream]

what

[delicate orchid]

a monad algebra is actually different, sorry

too many words that don't mean anything

[cobalt heath]

Yea idk where the endofunctor would be appearing here

[unkempt stream]

yeah

scary

[plush tree]

How can U(1) be an Isometry group for a Riemannian manifold?

[delicate orchid]

circle :3

[plush tree]

but an isometry group has to be generated by some Killing vector right?

[delicate orchid]

circle! :333

there's a pretty obvious action of U(1) on R^n

[plush tree]

U(1) seems like it has to be generated by a scalar infinitesimal transformation

[delicate orchid]

is your manifold rotationally symmetric in any sense

[plush tree]

U(1) is a rotation in the complex scalar plane, and riemannian geometry is real, so good point about that as well

[south patrol]

hm is there

[delicate orchid]

spin it round

[south patrol]

okay sure lol

[delicate orchid]

oh yeah

[south patrol]

I wasn't sure if there was any particular one you had in mind

[delicate orchid]

for n > 1

sorry

[south patrol]

No dw i didn't think about that lol

I was more just unsure if yeah there was one canonical one

[delicate orchid]

nah, just some plane

[south patrol]

Like on C^n there is one obvious action of U(1)

[unkempt stream]

like a record baby

[delicate orchid]

yurrrr

[south patrol]

But sure yes on R^n there are tons of natural ones

ye

Choose an embedding of C in R^n and then let U(1) act on that

Overcomplicated

[cobalt heath]

Natural ones?

[south patrol]

I silly

[delicate orchid]

"pick a plane"

[south patrol]

Ye lol

[unkempt stream]

plane in the ass

[delicate orchid]

WHAST

[south patrol]

This is funny

turns out i like misundrstood a paper

and looked at it again like 6 months later and understand it fine now

[unkempt stream]

true

[south patrol]

wasted time first reading it

[delicate orchid]

me with first year ug group theory

[south patrol]

fire

It was one of those cass where like

Idk if this is common

but you can't tell where in the paper they actually state the main result lol

[unkempt stream]

imagine taking group theory smh my head

[south patrol]

[cobalt heath]

Huh, isn't that bad writing?

[delicate orchid]

my problem with papers is that my adhd is so debilitating I literally cannot read them all at once

or even an entire result

or even a proof

[unkempt stream]

may I suggest cocaine?

[slim kayak]

wdym?

all at once?

[delicate orchid]

like without either needing to swtich between two different papers or just getting distracted

[plush tree]

I guess it could be some kind of gravitational wave that is invariant under rotations in the complex plane. e^{\i \theta} g_{ \mu \nu} is kinda like a local version of conformal invariance...

[delicate orchid]

oh you're a physicist

[slim kayak]

gravitational wave? why not a perfectly fine generic wave unless lorentzian geometry is important here

[plush tree]

oh, U(1) was an isometry group

[slim kayak]

I guess medication isnt an option here?

[plush tree]

so the lie derivative of the metric along tangent vectors of the orbits needs to be 0

the orbits being circles in the complex plane... which only intersects with riemannian manifolds

[delicate orchid]

I was medicated. Gave me psychosis

[cloud walrus]

Plazzi

[delicate orchid]

anyway

[south patrol]

Ngl you should probably move to #diff-geo-diff-top if you want more help x17

Probably better people there to help

[delicate orchid]

the inner automorphism group can be abelian and G not be

Inn(D_8) = C_2 x C_2

[slim kayak]

Sorry to hear that

[delicate orchid]

it's ok, it probably wasn't the cause but it's known to trigger it early in people predisposed to it

[south patrol]

I think

Wait

Okay so

[delicate orchid]

use the fact that you can write any element g in G as hx^t for some fixed x in G, t in Z, h in Z(G)

[south patrol]

How did you conclude "G = < axa^-1 > "?

What are a and x

[chilly ocean]

Thanks

[chilly ocean]

t in Z?

[delicate orchid]

an integer

sorry, x isn't in Z(G). I wrote that wrong

just some fixed x in G

with h in Z(G)

let me amend

[cloud walrus]

Plazzi

[south patrol]

But what are a and x

[delicate orchid]

oh yeah that notation you wrote is complete nonsense to be blunt

[south patrol]

It seems your notation has confused you

[chilly ocean]

x is fixed and a is any element of G

[south patrol]

Like there are you are writing that G is cyclic

[delicate orchid]

write $c_a$ for the map $x \mapsto axa^{-1}$

[cloud walrus]

W;3w Lads Tbh

[delicate orchid]

then Inn(G) is the group of all c_a

[south patrol]

But I don't see how that follows from anything you've done thus far

Like you say Inn(G) is cyclic and then say G = <axa^-1>

How did you go from one to the other

[south patrol]

I'm not sure what you mean by this

Like did you just pick any two random x and a in G

[chilly ocean]

x is an element in Z(G)

So ax=xa for all a in G

[south patrol]

Okay, how does that tell you G = <axa^-1> i'm confused

[chilly ocean]

It doesn't im just dumb

Inn G is the Set of all inner automorphism

And i didn't read set

[south patrol]

it is also a group

[delicate orchid]

no it's a group

that is like

[unkempt stream]

same!

[delicate orchid]

so far from the problem here

[delicate orchid]

stay strong king

[unkempt stream]

S_6 is the weird symmetry group with a nontrivial outer automorphism group right

[delicate orchid]

step one: show that for any g in G, we can write g as hx^t for some h in Z(G) and a fixed x in G

[delicate orchid]

indeed

what about it

[unkempt stream]

i never understood that

[delicate orchid]

it's a really weird coincidence that just happens due to some obscure group action

[unkempt stream]

i’ve read before it relates with like, SL(2,Z) having abelianization Z/12Z

which is also uber specific

[chilly ocean]

I already read the solution of the book and it's quite similiar to your hint. I Just thought maybe im right too

[delicate orchid]

and the numerical relationships required for it to happen again just don't occour for larger values

[south patrol]

Oh yeah wew did you see this fnny thing like

[delicate orchid]

this one is strange but I'm more comfortable with it now

[south patrol]

If G is non-abelian, is Aut(G) non-abelian

[unkempt stream]

how so?

[warped sonnet]

so Im currently checking out fields, and im confused as to what exactly polynomials over fields (or rings in general) means in theory... for example polynomials of degree 2 over a field GF(3) (i.e finite field with 3 elements); I dont understand how one would visually imagine that, or just the idea behind it

[unkempt stream]

it’s hard to visually imagine it

[delicate orchid]

I cannot think of a single example where it isn't

[south patrol]

Yeah so uh

[delicate orchid]

at least not finite groups

I don't care for infinite

[south patrol]

It is possible for it to be abelian for finite groups

[delicate orchid]

oh ok

shh

[south patrol]

But no examples known until like 1975

[delicate orchid]

lemme think more

[south patrol]

so bizarre

[celest furnace]

Does Aut(G) contain an isomorphic copy of G?

[delicate orchid]

no

[south patrol]

no e.g. bcause of what I just said

[delicate orchid]

Aut(C_2 x C_2) = S_3

[warped sonnet]

hmm i see, also as variables for these polynomials we can only use elements of the field right?

[chilly ocean]

But i was wrong 😦

[celest furnace]

Wow so G can be non-abelian and Aut(G) abelian?? Crazy

[unkempt stream]

What are you tryung to visually imagine?

[delicate orchid]

can you link this source/paper potato

[warped sonnet]

idk im just trying to wrapp my head around the idea of having polynomials over rings

[south patrol]

Bear in mind too that polynomials aren't functions, or at least not in the way you might want

But that is freeing in the sense

[unkempt stream]

polynomials themselves aren’t functions yeah, but they do have valutions which are

[delicate orchid]

they're just normal polynomials but instead of numbers infront of the x you have an element of the ring

[unkempt stream]

and those valuations are morphisms

[south patrol]

that you can just view them as formal expressions like a_0 + a_1 x + ...

[delicate orchid]

you two are going too high level here

[south patrol]

Oh lol

[delicate orchid]

but yes, they're just formal expressions like that

[south patrol]

I thought mine might help in that it is quite fundamental to what they are

[delicate orchid]

up to a_n x_n, notably

can't go on forever

[unkempt stream]

The silly?

[warped sonnet]

this already helped guys, thanks (:

[tender wharf]

some textbook authors define them as sequences

[delicate orchid]

sequences doesn't help build off of the already established intuition for polynomial multiplication though

[tender wharf]

treating them as polynomials leverages the intution behind polynomial multiplication you have though

yeah

[delicate orchid]

yrurrrrr

@south patrol Kummer theorem

[south patrol]

What

[delicate orchid]

polynomial multiplication mentioned

[rose prism]

it is just the convolution on \ell^2(Z)

in fact

c_00

[south patrol]

polynomial multiplication is just Day convolution of the multiplication on the field

[delicate orchid]

the worst part is... it is...

[rose prism]

no its the best part

[delicate orchid]

nuh uh

[south patrol]

well ig power series

[unkempt stream]

that’s how fast fourier transform is used to make low complexity multiplication algorithms

Lol

[rose prism]

yes

[delicate orchid]

snoorrrreeeeeee

[unkempt stream]

”low complexity” with a rediculous constant

cannot spell

[rose prism]

https://hal.science/hal-02070778v2/file/nlogn.pdf

[delicate orchid]

SNOOOORREEEEE

notice how they don't put the constant

suspicious that

it is interesting that this is at all possible though

[celest furnace]

WHAT

[daring nova]

you know 3b1b made a video on that

right

pretty sure it was 3b1b

[delicate orchid]

he made a video on convolutions

[daring nova]

many

from the probability series

[rose prism]

do you think he used the O(n log n) integer multiplication alg to animate it

[glossy crag]

Is there a set's worth of groups of a fixed (infinite) cardinality?

[dim widget]

No

But up to isomorphism yes

[coral spindle]

There's even a set's worth of magmas of a given cardinality up to isomorphism wowee

[glossy crag]

Why then can we speak of Ext(G,A) for a G-module A, the set of equivalence classes of extensions of G by A that induce the given G-action on A, and its isomorphism to H2(G,A)? Just use instead of an extension its isomorphism class?

[dim widget]

because there are a sets worth of equivalence classes

[glossy crag]

Why.

[coral spindle]

Dude you can just get the set of all operations on the structure

[dim widget]

The group is on the underlying set which is the product of G with A

So there are no set theoretic issues if you view it that way

[glossy crag]

Yes, all extensions have cardinality GxA.

[dim widget]

Yes and up to equivalence all of the sets are equivalent to G x A with some exotic multiplication

[coral spindle]

Like choose your set G. Then the set of all functions G x G → G is a set. Then just require these to have the group laws. Boom, you've got a set of all possible groups with set G, which must in particular have all equivalence classes represented.

[glossy crag]

What you're saying conradicts what nG+ said.

[coral spindle]

No it doesn't come on

[coral spindle]

As she said right here, up to isomorphism there is!

[glossy crag]

All group extensions ahve cardinality GxA, fix it, then boom.

[coral spindle]

That's because I'm not getting all group extensions. I'm getting all CLASSES!

[dim widget]

The point is that up to equivalence it doesn’t matter which set of size G x A you pick, so you might as well just take that set itself. Then you get a description in terms of cocycles which you are referring to with the cohomology stuff

[glossy crag]

Ok, but is an equivalence class itself the size of a set?

[dim widget]

It depends on how you define it

If you define it as a bunch of maps m: (G \times A)^2 \to G \times A with some properties then yes obviously

[coral spindle]

It's just as good to choose a representative

[glossy crag]

See, I don't get this: you're basically partitioning the "set" of all extensions (which you're saying is not a set) into equivalence classes and somehow it's fine to then consider the set of equivalence classes (the partitions), but not the original "set" itself that they partition.

[coral spindle]

The barrier is precisely the fact that we can't fit all sets into a set. For example it is impossible to put every group of cardinality 1 into a set, because that would require us to have a set of all singleton sets. Does that make the problem clear? This has nothing to do with groups.

[dim widget]

That’s totally fine to do!

For instance

There is a proper class of finite sets

But I am fine to say that there are countably many equivalence classes with bijection as an equivalence relation

[glossy crag]

That's a good example, I suppose. I know it's about sets, not groups, that's precisely what I'm struggling with

[glossy crag]

Why ?!

Why does this reduction process yield a set's worth of stuff.

[dim widget]

Because by definition a finite set is in bijection with {1,…,n} for some n which may be 0

And there are only countably many of those

[glossy crag]

That's fine in the finite case, but how would you do the infinite case?

[glossy crag]

(that's what I was really thinking about here)

[coral spindle]

Are you asking for some like, non-set-theoretic way of seeing that there is a set's worth of equivalence classes? I wouldn't hold your breath lmao

Like you just have to come up with a construction

I described one way to do this above

[dim widget]

I mean it’s a tautology that for any cardinal there is a unique set with that cardinality up to bijection

[glossy crag]

Man, I hate set theory.

Is there even a point to it? It feels like it's a "turtles all the way down" subject that is not grounded on anything whatsoever.

[dim widget]

I don’t think it’s very important until it suddenly is

[coral spindle]

are you gonna ask for a foundation that doesn't rely on anything?

[dim widget]

it’s definitely not a thing that an undergraduate has to worry about

unless they find it interesting

In the end good mathematics should really be foundations agnostic, and dealing with set theory is just some kind of chore that you occasionally have to do in very niche circumstances

[cobalt heath]

There is always type theory, or category theory for you

[glossy crag]

le undergraduate category theorist

Man, I wish I had that meme chart saved, can't find it now.

[coral spindle]

Why do people keep calling category theory a foundation for mathematics in the same sense of set or type theory? Makes no sense to me.

A complaint to the void — if you respond seriously I will kill an innocent person on the street

You will be held responsible

Legally

Or I'll just be real mad who knows

[cobalt heath]

But do you also hate category theory

[glossy crag]

No, because categorical problems don't come up (in anything I've read, at least).

[dim widget]

there are not categorical problems, only categorical solutions

[glossy crag]

Oh never mind, I do have it, thankfully.

[barren sierra]

because they hear that type theory and set theory are foundations and pattern match the word "theory" + the idea that it's a higher abstraction of some sort

[glossy crag]

@dim widget on the subject of set theory, one thing I've always idly wondered about is what a class is (e.g. the class of objects of a category). I assume it can't just be semantic wordplay (not calling a set a set), so how does it work?

[unkempt stream]

categorical considerations

[dim widget]

I dunno

I probably read the definition at some point but I forgot it

[glossy crag]

We are doomed.

[formal ermine]

i know a class as like

[dim widget]

it’s something you put in papers when you talk about certain objects so people know you went to school

[formal ermine]

you have some objects right

and a property that all of those objects share

[unkempt stream]

is there any other classical derived functors besides like hom’s Ext and tensor’s Tor

[formal ermine]

dont ask me about a rigorous definition tho

[cobalt heath]

I did hear that in set theory.. you have issues like

Some sets that you want to define is too large

[formal ermine]

theres like sheaf cohomology i think

[cobalt heath]

So that it is no longer a 'set'.

[unkempt stream]

i mean just over simple abelian categories

[formal ermine]

ah dunno

[cobalt heath]

So you give it another name and call it 'class' 💀

[glossy crag]

Yeah I've heard of stuff like this, but how does it work? Just say "uh ackshually this is a class and it has all the properties a set can't have, suck it haters"?

[cobalt heath]

I dunno tbh, maybe blame Cantor?

[glossy crag]

I won't, poor guy had it bad enough (institutionalised). I'll blame Hilbert instead, the hack.

[dim widget]

over arbitrary abelian categories this is all that is guaranteed to exist. Even Tor doesn’t necessarily make sense a priori, unless you just mean the derived functor of the coproduct

[glossy crag]

"No one shall eject us from the paradise Cantor has built for us"

[cobalt heath]

IIRC Russell developed the concept of type theory, but we still do not use it regularly

[dim widget]

I looked it up and in ZF there actually is no definition Lmao

[dim widget]

choice doesnt help

[coral spindle]

Type theory is actively researched in many ways, just like set theory is, but it is not as popular a foundation of mathematics as set theory is amongst mathematicians in general. In an interestingly contrarian twist, it is very popularly used in proof synthesis and verification software.

[cobalt heath]

Yea. Hmm, contrarian twist..

[cobalt heath]

Do you have an opinion about why it is not so popular?

[coral spindle]

Yes, I think it's because (1) set theory was around and used for decades before type theory, and (2) type theories can provide some barriers to doing certain things which can feel relatively constraining for people used to set theory.

[delicate orchid]

that's how I feel about set theory

it's obvious what I mean. Stop crying

[cobalt heath]

I see, I agree. Maybe type theory is formalized too late

[rocky cloak]

At least in first order logic, a class is any collection defined by an unbounded comprehension.

So all sets with such and such property, is a class. From that perspective, there's no difference between a class and a property.

[delicate orchid]

my comprehension is disgustingly bounded

[rotund aurora]

I think it's basically like {x such that P(x)} where P is some property. The point is that this needn't be a set, therefore it is not subject to the axioms of sets (specially, the epsilon relation). So given a class C it does not make sense to ask if C belongs to some other class, so you avoid contradictions. But I think it is fine for a given set x to ask if x in C or not

and in particular, when you quantify in ZFC you are quantifying over sets, not proper classes

[cobalt heath]

Imo while it has limitations, type theory is great in that it makes you to be more explicit. Weird identifications have been quite hard for me.

[unkempt stream]

i mean for like the general common abelian categories, like arbitrary-R-modules, are there other classical derived functors besides the Hom-Ext and Tensor-Tor examples

those are the two I usually hear about

[dim widget]

there are other ones that may or may not exist depending on what R is

but if you just know you have an abelian category those are all you get essentially

[unkempt stream]

Is there any examples

[dim widget]

completion at an ideal of R can have an associated derived functor

Similarly there is the related completed tensor product operation

[next obsidian]

Don’t even talk about derived completion

I don’t dare venture into that part of the stacks project

[dim widget]

@unkempt stream actually forget what I said non exact completions don’t exist

[next obsidian]

Isn’t it not always even the same as deriving the completion functor???

[dim widget]

this is also true

[unkempt stream]

torture

[delicate orchid]

moronic

what's the point HUH

we doing cohomology with completions HUH??? craaazzyyy talk

[next obsidian]

You will satisfy the Mittag Leffler condition and LIKE IT

[unkempt stream]

categorical violence

[dim widget]

But if the ideal is finitely generated then it does all work out nicely

[next obsidian]

https://tenor.com/view/chipi-chipi-chipi-chipi-chipi-chapa-chapa-cat-cat-chipi-gif-3982040863797256894

[delicate orchid]

I am haunted by visions of a manor in a park

there's someone in the window. They look sad

[topaz solar]

So, jagr described one as a first order definable “collection” of all sets which satisfy so and so property, but here’s a better one

Suppose you have a model of set theory

M, we call it

A class X is a subset of M

The particular case jagr refers to is a definable class, I.e. x in X iff phi(x)

[topaz solar]

Note: this “subset” is not necessarily an element of M (or identifiable with one), and in fact probably isn’t

Our M exists in some metatheory because there’s 0 harm really assuming we have one

So it’s like taking a subset of a group/its underlying set

Anyhow, obviously you can’t quantify over classes because they’re not even points in M

But you can still manipulate em if they’re definable (or similar) because you have properties “visible” to the inside

[topaz solar]

I think this is kinda impossible, at least if you ask for arbitrarily large things wrt set theory, since involving very large cardinals dooms your ability to get by without any foundations relevance

cf Whitehead’s problem

Since
Combinatorics of large cardinals -> the foundations controlling them is involved

Especially since ZFC kinda doesn’t control them much as is

[south patrol]

oh i mean i agree lol

[topaz solar]

It’s not that distinct from type theoretic things since ya know, syntax <-> semantics^op-ish stuff

[cobalt heath]

Oh no, now you are legally responsible

[dim widget]

I don’t know in what way it’s impossible… I agree that logically speaking we need some theory of foundations I just don’t think that the mathematics that interests me depends on which system we pick

[topaz solar]

The ones that interest you maybe, but things like Whitehead’s problem really shouldn’t be surprising that they pop up

But idk what stuff you do so

*for those in the audience, whitehead’s was something like Ext^1(X, Z) = 0 implies X free abelian iirc?

[dim widget]

yeah I don’t think one should be surprised because there are formal arguments that will show you that there absolutely must be such statements in any consistent axiomatic system and some of them may asked by normal mathematicians

[topaz solar]

Yep

The particularly egregious ones in ZFC being combinatorics of cardinals, because those are like always independent

Namely because they all imply large cardinals or one of the forcing axioms or smth

[grizzled crane]

How do I prove an equivalence relation partion the set into disjoint classes?

[celest furnace]

Well you first need to show the whole thing is a union of equivalence classes

Then show two equivalence classes are the same iff they their intersection is nonempty (equivanelty, if they have an element in common, they are the same!)

There we go

[grizzled crane]

suppose R is a equivalnce relation, then equivalence class of x is [x]={y: y related to x}

I suppose this is an equivalence class

[celest furnace]

Yes thats the definition

[grizzled crane]

Should I start with a set?

[celest furnace]

What

[grizzled crane]

like finite or something ? how do I go by showing union is the set?

[celest furnace]

You need to show that each element x in the set is in some equivalence class

[grizzled crane]

any element x is clearly in [x]

[celest furnace]

Exactly

[grizzled crane]

sort of the same way we proved coset properties ?

[celest furnace]

Yeah exactly

In fact you can derive a lot of the coset properties from properties about equivlance relations

[grizzled crane]

right

so now I have to show the union of these classes is the set X

[celest furnace]

Like aH = bH iff they have an element in common i.e. ah_1 = bh_2 equivalently b^{-1}a is in H

[celest furnace]

Youve already done that

Each class is contained in X and X is contained in the union of the classes

So now you just need to show they are "disjoint" by showing that if they have nontrivial intersection they are equal

[grizzled crane]

oh right

by some contradiction

?

[celest furnace]

U can do it directly

[grizzled crane]

hint please?

[celest furnace]

Suppose that [x] cap [y] is nonempty, and let z be in [x] cap [y]. Then z is related to x and z is related to y so...

Think about what condition would show that [x] = [y]

[grizzled crane]

both should be subsets to each other

[celest furnace]

What would show that [x] is a subset of [y]?

[grizzled crane]

pick any x, show x is realted to y aswell

[celest furnace]

Exactly!!!

So can you use z now somehow?

[grizzled crane]

I’m guessing, one way we can show [x] is subset of [y] and by symmetric property the other way

[celest furnace]

Yeah

[celest furnace]

But how do you do this?

[grizzled crane]

since there is an AND

any z, is in both the sets?

or say intersection is always the subset

[celest furnace]

Ill give you a hint

[grizzled crane]

sure

[celest furnace]

Since z is in the intersection y is related to z and z is related to x

[grizzled crane]

can I use transitivity?

[celest furnace]

Why couldnt you?

Exactly right!

[grizzled crane]

yeah its an equiv relation

Makes sense now

Thanks a lot

[celest furnace]

Ur welcome

[grizzled crane]

I really appreciate your effort and patience

Have a great day

[celest furnace]

Haha thank you, you too !

[steel pulsar]

If I have a group for example: Z/3Z, Z/4Z, Z/5Z etc...
Is there a fairly simple and understandable method to determine the subgroups of each?

[daring nova]

for Z/nZ yes

for any group, like Sn, no

[daring nova]

these are nice because cyclic groups are the simplest

[steel pulsar]

which ?

[daring nova]

the subgroups of <g> are the <g^k>

[steel pulsar]

For example for Z/4Z what would g and k be?

[daring nova]

g = 1 in Z/nZ for example

writing this additively, it's the <k>

i.e. the multiples of k

when k is coprime to n, it still generates all of Z/nZ by Bezout's identity

otherwise it doesn't

[steel pulsar]

wait, I don't really understand examples like that, but there is something simpler that I understand

If d is is a divisor of n, then Z/dZ is a subgroup of Z/ nZ

for me that it's simple

[daring nova]

well, rather, it's isomorphic to a subgroup

n/d (Z/nZ) is a subgroup of Z/nZ

[steel pulsar]

so this sentence is not correct?

[daring nova]

how do you inject Z/dZ into Z/nZ
To claim it's a subgroup

[steel pulsar]

I don't know

[daring nova]

then it's a bold claim to say something is a subgroup when you can't even argue it's a subset

[daring nova]

that's why I'm saying in this case, you send them to multiples of n/d
they're all distinct until d (n/d) = n = 0, so it's a cyclic group of order d
Hence it's isomorphic to Z/dZ

hence you use the fact d | n to inject it

[runic galleon]

are these cyclic groups ?

[coral spindle]

Z/nZ is a cyclic group of order n, yes.

[runic galleon]

the diff between these two are that one is an order of the whole group and the other is the order of just an element from a group

but what does an order of a group means ?

[coral spindle]

Your first screenshot just defined that

[runic galleon]

yea my bad

i mean what does an order of an element in a group means

[runic galleon]

so the number of elements in the set G

[coral spindle]

Your second screenshot defined that

I am confused by this question. You show us the definition, and then ask what the definition is. If you have some more specific confusions about the definitions, you should try to make them clear when asking.

[runic galleon]

i'm sorry, but i just want to know if it has a relation with the total , partial order relations

[coral spindle]

No.

[runic galleon]

oh

okay