groups-rings-fields

[rotund aurora]

might be true

if p is a prime integer, then a and b are 1

isnt that fake tho

like take b=a

and consider a=(x,y) in k[x,y]

then x in a, but x not in (x,y)(x,y) I think

and x is prime

Just so you know, Isaacs uses the weird notation <, and to this day I still don't even know what it meant

and it's not proper inclusion like subset

I think in Isaacs $a<b$ meant $a\not\supset b$

[cloud walrus]

Croqueta

[rotund aurora]

crazy af

[glossy crag]

No, I'm asking why $\forall a,b(a,b>p\implies ab>p)$ implies $p$ is prime.

[rotund aurora]

ahh ok

[cloud walrus]

Ocean Man

[glossy crag]

I seriously doubt it, I read a lot of Isaacs and I'm pretty sure it's what I said.

Let me check

[rotund aurora]

tbh I dont remember, I just remember I had trouble with that notation xD

[rotund aurora]

ah yes yes

[rotund aurora]

still fake I think

take p=4

then the translation is $\forall a,b, (a\mid 4, ,b\mid 4\land a\neq 4,b\neq 4\implies ab\mid 4)\implies 4$ prime

[cloud walrus]

Croqueta

[rotund aurora]

the first condition is satisfied

because a and b is either 2 or 1

ah wait

nono

ok

it should be implies ab | 4 and ab!=4

so the condition is not satisfied

ah ok so yes that's in Isaacs

[void cosmos]

why are Z3 and Z2 not free modules?

free Z6-modules

[delicate orchid]

ew non-integral domain

[wraith cargo]

if they were they could be written as a direct sum of copies of Z6

[delicate orchid]

^

[void cosmos]

why cant be not?

ig its obvious if i think about it

but i cant write it down why

haha

[delicate orchid]

there's a theorem that states any principle ideal is a free module if and only if it's generated by a non-zero divisor

[wraith cargo]

since Z3 and Z2 are finite let's claim they can be written as a finite direct sum of Z6's
Then look at the cardinality of both sides

[delicate orchid]

but I cannot remember the proof for the life of me, it's something to do with there not being a unique linear combination of some element

completely unrelated to the question at this point because a simple cardinality argument does work yeah

but now this is annoying me

[void cosmos]

oh yeah obviously

mb

thank you so much

can someone give any motivation

for projective modules?

and their dual injective modules?

i understand the definition by now and can prove some basic stuff

like that every sequence of the form A-->B-->P-->0 is exact split

given P is projective

but the definition seems magic

[delicate orchid]

it's like free but only surjective/injective maps 👍

G is simple, if \rho(g) was scalar then g would be a non-trivial element in the centre of G

contradicting the fact that G is simple

I'm going off of what the proof says, it explicitly says that Z(p(G)) \cong Z(G)

although, no

let f be any homomorphism and z in the centre of it's domain, then f(gz) = f(g)f(z) = f(zg) = f(z)f(g) so f(z) is in the centre

now it's very possible for g to be in the kernel of rho, that is true

ah, we're taking \chi to be non-trivial

so there has to exist at least one element not in the kernel of chi

[agile burrow]

If you're interested in homological stuff, a module P is projective iff Hom(P, -) is an exact functor

Similarly, Q is injective iff Hom(-, Q) is exact

It's nice to characterize when functors are exact, so projective/injective already arise kind of naturally from this perspective

[void cosmos]

i am , it feels cool

what does - mean

• stands for any module?

[agile burrow]

Yes

[delicate orchid]

ah, got it - the rep has to be faithful, the kernel of a rep is a normal subgroup - G is simple

yeah it took me a lot of thinking

[agile burrow]

But yeah, projectives and injectives are fundamental to homological algebra

[delicate orchid]

who up resolving they projectives

[agile burrow]

That's a short resolution

[void cosmos]

homological algebra is the coolest math i did so far

after i finish this test , will love to pick up a text specifically for this field

or subfield

sadly its too topological motivated and idk much topology

😦

[thorn delta]

rip, was about to bring up vector bundles

[void cosmos]

u can bring up differential forms

i have an idea about those from lee

and tu

and de rham cohmology

hopefully homological alggebra has interesting problems

[thorn delta]

ive been really wanting to learn homological algebra too. there's something about the black magic we do in alg top i find very fascinating. hoping to get into weibel after i graduate

[void cosmos]

yea i think iam the other way around

idk much alg top

but would love to know homological algebra on its own

[agile burrow]

I also find homological algebra pretty interesting independently of topology/geometry

Idk, I managed to learn homological algebra before any topology just because I really liked algebra at first and I still find many of the constructions in homological algebra to be interesting in their own right

But it also nice to have motivation from other areas

[void cosmos]

does homological algebra do anything to analysis?

im learning functional analysis now

and would be very cool if there is any connection

[thorn delta]

if you're into some pretty high-brow operator theory, sort of maybe

[void cosmos]

this proof doesnt use the direct sum universal property right?

this is just saying P_i is projective so there is a map to A but then the map can be smalled down to the sum by using the projection?

like f:P_i --> A

then u would have fbar:sum(Pi) --> A ; fbar=f(pi(.)))

wouldnt this wokr?

work*

why is he using the inclusion

[agile burrow]

It does use the universal property of the direct sum

[void cosmos]

how please

[agile burrow]

You have a map from each summand, so the universal property yields a map from the direct sum

[void cosmos]

cuz i cant remember much of it tbh

ohh

yea i remembered now

[delicate orchid]

I managed to learn homological algebra before any topology
... wtf

[agile burrow]

I'm glad I helped you achieve this revelation

[void cosmos]

the direct sum is unique in the sense that if u have maps from each summand to the sum then they are "isomorphic compose-wise lmfao" to thee projections

right?

kinda like the product

[delicate orchid]

I suppose if you know what a "continuous map" is you don't need much else explicitly for simplicial homology

[agile burrow]

Dual to the product

[void cosmos]

so its the other way around

[delicate orchid]

it's the coproduct yeah

[void cosmos]

like if u have inclusions from the summands to the ddirect sum

?

if u can draw a diagram here i would really appreciate it

using tikz ig or whatever the namee

[agile burrow]

Yes, the direct sum comes equipped with maps from each summand into the direct sum

[delicate orchid]

it's actually both for finite sums isn't it?

[agile burrow]

The product and coproduct coincide for a finite number of modules, yeah

[void cosmos]

okay so this phi_i is like intuitively the "fake" inclusion

right?

[agile burrow]

I didn't even know what simplicial homology was lol

[delicate orchid]

did you literally just start with a general chain complex

[agile burrow]

All I knew was "chain complex go brrr"

[delicate orchid]

holy based

[agile burrow]

Yes lol

[delicate orchid]

galois

[agile burrow]

I have no idea what you mean by this

[delicate orchid]

\phi_i could literally be the trivial map

[void cosmos]

so does the proof go that the projectivity of one summand would induce a "fake map" to A so by universal property of the sum i must have the map to A?

[void cosmos]

by fake map i mean here the phi_i for example.

[delicate orchid]

why are you calling it fake

[agile burrow]

Very strange choice of terminology

[delicate orchid]

^

[void cosmos]

its my intuition for it

[agile burrow]

Then I have some doubts about your intuition

[void cosmos]

😦

[agile burrow]

Because it's just a map lol

Like any map

[void cosmos]

yeea ik but its fake as in

okay i mean like

fake inclusion map

now better?

[agile burrow]

No

[void cosmos]

its like trying to be inclusion but its not

[agile burrow]

But it isn't trying to be inclusion, it's just a map

[void cosmos]

my intuition is

[delicate orchid]

[\begin{tikzcd}
& {A_1\oplus A_2} \
{A_1} && {A_2} \
& M
\arrow["{i_1}", from=2-1, to=1-2]
\arrow["{i_2}"', from=2-3, to=1-2]
\arrow["{\phi_1}"', from=2-1, to=3-2]
\arrow["{\phi_2}", from=2-3, to=3-2]
\end{tikzcd}] given this set up

[void cosmos]

that the direct sum with it's inclusions/projections are unique up to composition

[cloud walrus]

Wew

[delicate orchid]

[\begin{tikzcd}
& {A_1\oplus A_2} \
{A_1} && {A_2} \
& M
\arrow["{i_1}", from=2-1, to=1-2]
\arrow["{i_2}"', from=2-3, to=1-2]
\arrow["{\phi_1}"', from=2-1, to=3-2]
\arrow["{\phi_2}", from=2-3, to=3-2]
\arrow["{\exists! \phi}"{description}, dashed, from=1-2, to=3-2]
\end{tikzcd}] there is a unique map induced

[cloud walrus]

Wew

[void cosmos]

that is if u can find other inclusions from each summand then u can find a function that maps the real one to the fake one

[delicate orchid]

nice one texit you fraud

[void cosmos]

ie both are homeomoprhic

homomorphic*

[agile burrow]

Regarding this, you need each summand to be projective to induce a section from the direct sum. The universal property of the direct sum begins with the data of a map from each summand A_i to M and then induces a map from the direct sum to M such that the triangle commutes for each i

[delicate orchid]

it's just telling you that you can draw an extra arrow on ur funny diagram, and that arrow is the only one that makes the diagram commute

[void cosmos]

yea so the projectivity gives us the phi_i

[thorn delta]

the phi_i you could say are like "pieces" of phi in the sense that phi restricts to phi_i on each summand

[void cosmos]

from here

and by the direct sum property weee get our phi

[thorn delta]

you have the same formal setup for disjoint unions

[craggy lichen]

why cant there be a transitive subgroup of S_4 of order 6?

[delicate orchid]

there are no projective objects in that image , but I think I know what you mean

[delicate orchid]

orb-stab

I think

[craggy lichen]

how does it follow from that though

[delicate orchid]

if the subgroup is transitive then the action on 4 elements will have a single orbit of size 4
4 does not divide 6

I'm assuming that's the action you're talking about, tbf

I'm too tired atm sorry boss

[craggy lichen]

tyty

[dreamy chasm]

can anyone help me find the order of this element?

[white oxide]

i don't really understand the line about choosing representatives and how that shows that p(alpha) = 0, can somebody help me?

like what happened to the alpha on the second page

they didn't plug it in

also how does that show <p(x)> = 0 lol

did they just take p(x) as the representative?

or what representative did they choose rather

[delicate orchid]

keep multiplying x+(x^2+1) by itself until you get 1+(x^2+1)

[white oxide]

oh nvm

got it

[dreamy chasm]

thanks I’ll try this

[wind locust]

the underlined bit doesn't really make sense, right? or am i being stupid?

[cloud walrus]

I Abhor Hatcher

[wind locust]

i feel very eepy can someone check this for me

Fr obviously stands for the coproduct

[dreamy chasm]

anyone have any ideas?

or a general approach to solving a problem like this

[pastel cliff]

old question - should this not have been e1 \otimes e2, e2 \otimes e1 in the 2nd and third col's instead

[thorn delta]

umm, maybe

im not sure if there is a more canonical choice here

[pastel cliff]

should it not matter?

actually i guess not

basis vector ordering doesnt really matter

[thorn delta]

generally i think people will pick some kind of lexographical ordering of the indices for their basis, but idt it matters that much as long as you make your arbitrary choices clear

[pastel cliff]

if i understood correctly this should be the second col

question repost

[thorn delta]

yeah, if your basis is ordered like e1e1, e1e2, e2e1, e2e2

also, i just realized i was being inconsistent with the ordering of my basis here my bad

[pastel cliff]

i didnt pick up on that, that did seem to be the first col

[thorn delta]

its the first column for your choice of basis, not mine

[pastel cliff]

ahhh okie

i think im getting this at least

tensors weird

[thorn delta]

nice

[vagrant zinc]

$\alpha^{-2}=\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \
2 & 5 & 3 & 1 & 4
\end{pmatrix}=\alpha$

[cloud walrus]

Awuita Fria

[vagrant zinc]

chicos ¿es esto correcto?
guys is this correct?

[pastel cliff]

not enough context

[vagrant zinc]

look

exercise i

[pastel cliff]

how do you take inverses of elements of the symmetric group

my own question but: is $\C \otimes_\Z \C$ isomorphic to $\C$?

[cloud walrus]

not sebbb not stμ₂dying

[pastel cliff]

i showed that Q \otimes Q = Q already

but im not sure about this one

[vagrant zinc]

I was guided by this so I came to the conclusion of the first image I sent.

[prisma ibex]

nope

how about showing $\mathbb{C}\otimes_\mathbb{R}\mathbb{C}\simeq\mathbb{C}\times\mathbb{C}$

[cloud walrus]

nGroupoid

[white oxide]

by 0 they mean the additive identity element in E right? namely <x^2 + 1>

since x^2 + 1 + <x^2 + 1> = <x^2 + 1>

[wicked patio]

yeah

= 0 + <x^2 + 1> (0 in F)

[white oxide]

okie thank u

[white oxide]

how did they get from the part underlined in blue to beta = b0 + b1alpha + ...

i included the theorem and the proof leading up to it for context

[pastel cliff]

does the \otimes_R vs \otimes_Z matter that much

[prisma ibex]

yeah loads

[pastel cliff]

oh then maybe that changes things

bc my questions is about Z-modules specifically

did that first part already

[quartz quiver]

I'm trying to prove exercise 54 here. So far I got that: proceed by induction on number of prime factors. assuming G is not abelian, by exercise 53 it is not simple, then there exists a maximal normal subgroup $N$. Let $|N| = p_{\alpha_1} \cdots p {\alpha_n} = k$. By proposition 16 in the text, the automorphism group is isomorphic to $(\mathbb{Z} / k\mathbb{Z})^{\times}$ and $|(\mathbb{Z} / k\mathbb{Z})^{\times}| = k - 1$. By induction hypothesis every subgroup is cyclic and hence abelian. So $C_G(N) \geq N$, so by maximality (and since $C_G(N)$ is normal) it is either $G$ or $N$. If $C_G(N) = G$ then $C_G(N) \leq Z(G)$ and so $|G / Z(G)| \leq |G / N|$ so by the inductive hypothesis $G / Z(G)$ is cyclic so $G$ is abelian, a contradiction. I'm trying to rule out the case where $C_G(N) = N$. If so then $G/N \sim N_G(N)/C_G(N)$ so is a subgroup of the automorphism group of N. Then $p{\beta_1} \cdots p_{\beta_m} \mid (p_{\alpha_1} \cdots p_{\alpha_n} - 1)$. So does this divisibility contradict the stated properties of the primes (that $p_i$ does not divide $p_j - 1$ for all i and j) or am I on the wrong path?

[cloud walrus]

Darylgolden

[quartz quiver]

<@&286206848099549185>

[wraith cargo]

They showed that every power of alpha higher than n-1 can be written as a linear combination of lower powers

So any element of F(alpha) is of this general form

Since otherwise there could be higher power terms but we ruled them out

[quartz quiver]

(figured it out, $|(\mathbb{Z} / k\mathbb{Z})^{\times}| \neq k - 1$ but $\varphi(n) = (p_{\alpha_1} - 1) \ldots (p_{\alpha_n} - 1)$. Then $p_{\beta_1} \cdots p_{\beta_m} \mid (p_{\alpha_1} - 1) \ldots (p_{\alpha_n} - 1)$ means $p_{\beta_1} | (p_{\alpha_1} - 1) \ldots (p_{\alpha_n} - 1)$ and by Euclid's lemma $p_{\beta_1}$ divides one of them, a contradiction.)

[cloud walrus]

Darylgolden

[plucky flicker]

If I’m given a representation of G1 and a representation of G2, then the tensor product of the representations will be a representation of G1 xG2, right?

Also, is it true that every representation of the direct product is the tensor product of the corresponding representations?

[wind locust]

Hey, can anyone help me out? I just don't see why the highlighted bit is true

[lethal dune]

I don’t even see the highlighted part

[formal ermine]

it's trivial

[wind locust]

lol, hang on it's uploading

[woeful sage]

Would you guys consider dummit and foote an entirely undergrad book or a undergrad book with some graduate level topics?

[lethal dune]

Latter

But treatement is undergrad level

[wind locust]

I'm still lost

Any help would be appreciated

[broken stirrup]

i know weird question, but can you let me know the firt letter of your name?

im also reading the same book, same topic

and i know someone who doesn't like Hatcher... made me think its not coincidence at all

maybe we are taking the same class :d

[wind locust]

Lmao

[wind locust]

It's Y

[broken stirrup]

gotcha

i know who you are

I'll expose your identity now

[wind locust]

But I reckon any half decent mathematician can recognize the problems with that abhorrent book

Oh

Please do

[broken stirrup]

next one is i

[summer path]

there's this wonderful thing call direct messaging, where you can sort this out.. but this is entertaining so please continue

[tribal moss]

No, please keep such chatter out of the topic channels to the extent possible.

[feral agate]

For the application of the first isomorphism theorem I noticed that in solutions of problem it is sufficient to show that the homomorphism given is surjective to a set to conclude that that set is the image of the homomorphism. Why is this the case? Can't you have that the image is a set that is bigger than the set which surjective the homomorphism is surjective to?

[delicate orchid]

The image is the largest set that a function is surjective onto - you’re right in that you can take subsets of the image and the function would still be surjective onto that set

I think that’s what you were saying anyway

[feral agate]

Am I missing anything in (2c) where it's concluded that the image of f is the same as the set that f is surjective to?

If I want to apply first isomorphism theorem correctly, how do I actually show that a set is the largest set that a function is surjective to.

[empty rose]

well the codomain of f is Z x Z/2Z, so its image isn't going to be larger than that
if it's also surjective, then its image is therefore exactly that

[delicate orchid]

I assumed they were asking how you proved it was surjective lol

[feral agate]

Ah I see

[feral agate]

Yeah, that's what I meant. I understand it now, thanks.

[oblique matrix]

Is there any use in considering the infemum of the set of upper bounds instead of supremums or the supremum of the set of lower bounds of the set of upper bounds etc. in an ordered set?

[delicate orchid]

if n is an integer then write 2a-b = n for some integers a and b (which is always possible). Then it’s possible to pick a to be even or odd as 2a will be even regardless, which fixes the parity of b (in order to ensure that 2a-b = n still) thus we can find elements that map to (n, 0) and (n, 1), so it’s surjective

[oblique matrix]

In case the supremum doesn't exist, it may provide a useful generalization

[oblique river]

You should try to prove that these dont really give you anything interesting. Let S be a subset of a totally ordered set X and let U be the set of upper bounds of S. Then sup U is the maximal element of X, if one exists, or is otherwise undefined. The supremum of the set of lower bounds of U is equal to sup S, which is equal to inf U

So if sup S doesnt exist, then neither do those other things

[woeful sage]

yea you can prove sup A is the inf of set of all upper bounds of A

and vice versa (also wrong channel lol)

[tulip hawk]

how to solve it?

[rustic crown]

whut have you tried so far uwu?

[tulip hawk]

hint it use isomorphism theorem

but I can,t do

[delicate orchid]

have you managed to prove it's reflexive and symmetric?

those are much easier than transitivity

[tulip hawk]

reflexive and symmetric is trivial :p

[delicate orchid]

I agree

anyway when I see a quotient by an intersection my mind jumps to the 2nd isomorphism theorem

[rustic crown]

make a dwawing

[oblique matrix]

What channel should this be in? I thought posets were an algebraic structure

[rustic crown]

the upper 4 thingies are finite, can you show bottom 2 are also finite?

[woeful sage]

oh posets

nvm I thought you were talking about sup and inf in an analysis context

[tulip hawk]

we may use problem 5

[rustic crown]

oh finitely generated is a little different from finite

[tulip hawk]

[flint crater]

Bit confused so decided to ask here. Is a representation ring a Grothendieck group?

[tulip hawk]

you are right

[agile burrow]

Yes, it's the Grothendieck ring of the category of finitely generated modules over the group algebra

Or actually

Hmm maybe I'm not entirely sure about what I'm saying, but if char F doesn't divide |G| then the category is semisimple so every module is projective

In which case I think the representation ring agrees with the Grothendieck ring?

[flint crater]

Ahhh alright. My book states The Grothendieck group of the finite generated k[G]-modules is called representation ring and thought it mean that a representation ring is a group which means it has a weird name lol

[agile burrow]

Oh

I mean it inherits a ring structure via tensor product

[flint crater]

Ahhh got it thanks

Also small side question. Does Z^5 mean ZxZxZxZxZ or Z/5Z?

[delicate orchid]

former

[flint crater]

Former means first right?

[delicate orchid]

yeah this is weird phrasing, the set of f.g. k[G]-modules form a semiring under direct sum+tensor product and then I'd say we take the grothendieck completion of that semiring

and yes former means first

if they meant Z/5Z they're write Z/5Z

the Z_n notation for cyclic groups and it's consequences have been a disaster for the human race

[flint crater]

Hahahah that is why I asked. I thought about Z_n and the possibility of there being a typo

[formal ermine]

what is the difference between a generic and a separable polynomial

[elder wave]

one is generic, the other is seperable

👍

[formal ermine]

https://cdn.discordapp.com/attachments/929807268994744340/1093978547565105182/unknown.gif

[delicate orchid]

https://cdn.discordapp.com/attachments/335352580777443338/1097501383441264690/Ftx6534XoAEqh9d.png

is a generic polynomial not just like

[formal ermine]

non zero discriminant

my prof said they're different

but like

they feel like they're the same

[delicate orchid]

yeah they do feel like the same lol

[pliant raptor]

How can I show that $(x-1)$ is a maximal ideal of $\mathbb{R}[x,y]/(xy)$? I've tried to show that their quotient is a field by trying to find a map from $\mathbb{R}[x,y]/(xy)$ to $\mathbb{R}$ with kernel $(x-1)$, but I couldn't do it

[cloud walrus]

ImHackingXD

[delicate orchid]

I'd just prove directly that if you append any other element of R[x,y]/(xy) to (x-1) you'll end up with the entire ring

[rotund aurora]

use correspondence/third iso whatever the name is

[delicate orchid]

ah that's another good idea

[rotund aurora]

you just want to show that R[x,y]/(xy,x-1) is a field

[pliant raptor]

I've thought of that, but it doesn't seem obvious to me what the inverse image of $(x-1)$ through the projection would be

[cloud walrus]

ImHackingXD

[rotund aurora]

you are thinking about it the wrong way

x and y are algebraically independent over R, and of each other

and now you introduce two relations

x-1=0

xy=0

so x=1

but then xy=y=0

so you are just left with R

[pliant raptor]

ahhhh

ok

Thanks

[pliant raptor]

So, $(\mathbb{R}[x,y]/(xy))/(x-1) \cong \mathbb{R}[x,y]/(xy,x-1)$ because of the third isomorphism theorem right?

[cloud walrus]

ImHackingXD

[rotund aurora]

yes

[pliant raptor]

ok

[rotund aurora]

it doesn't matter the order in which you quotient

[pliant raptor]

ok, thank you very much

[sly storm]

Hello. Can anyone here explain in a simple and intuitive way and in detail why in the hyperbolic plane it is generally impossible, starting with an arbitrary circle of area less than 2\pi, to construct (with straightedge and compass) a quadrilateral with all sides and angles congruent having the same area, and vice versa, that is, starting with an arbitrary quadrilateral with all sides and angles congruent, it is impossible in general to construct a circle having the same area?

[delicate orchid]

nope

[charred bison]

if [F(a) : F] is an odd number, how do you show that F[a] = F[a²]?

[south patrol]

Hint: try le tower law

[charred bison]

hmm

for [F(a²) : F(a)]?

[junior atlas]

Looking for help on part of a proof. Prove that for any prime $p>3$ then $ab^p - ba^p$ is divisible by $6p$.

[cloud walrus]

ElHefe

[junior atlas]

I need help on proving divisibility of 3. I can get 2 and p, but I need 3 so we can say 2x3xp is also divisible and therefore 6p.

[quiet pelican]

Proof by exhaustion works
Consider the cases where a = 0, 1, -1 mod 3 and where b = 0, 1, -1 mod 3 with every pair of combinations

[junior atlas]

@quiet pelican Is there another way that's more elegant? 2 and p are neatly done by even odd definitions, and some algebra. I can do it by exhaustion still though

[quiet pelican]

Actually, note that it’s trivial if 3 divides one of a, b, otherwise a^2 = b^2 = 1 mod 3, and p > 3 implies p is odd, so a^p = a, b^p = b mod 3

[junior atlas]

Do you mind explaining why a^2 = b^2 = 1mod3? Thank you for the help!

[quiet pelican]

Fermat’s little theorem for p = 3

[delicate orchid]

or just 1^2 = 1, 2^2 = 4 = 1

[charred crescent]

what exactly is a groupoid? I have read the definition as a group with a partial function replacing the binary operation, and also as a category in which every morphism is invertible.

i can repeat the definition, but i still don't really know what groupid means on an intuitive level

im also not sure if this belongs in the category thr channel

[void cosmos]

yo

proving pi J_i is injective --> each is

[formal ermine]

I'm by no way good at category theory, but I like to think of it in this way: a group is a one-object category in which all morphisms are isomorphisms, a groupoid is any category in which all morphisms are isomorphisms. so it's kind of a "generalization"

[delicate orchid]

I feel like the category theoretic definition follows more naturally from a group. You can encode the data of a group into a category of one object, with one isomorphism on that object for each element of the group. A groupoid is just this but with more objects

u know what illuminator I'm getting u banned for cyberbullying me AGAIN

[formal ermine]

sniped

xoxo

[void cosmos]

suppose B-->Pi_J_i is the map that we want to find but for J_i

call it h

cant we define hbar:B-->J_i by just pi_i(h(b))?

why wouldnt this work

pi_i would be the projection map

[charred crescent]

@formal ermine @delicate orchid the thing is, im not sure what is meant by "a group is a category with one object". does this mean that the group G is the object? if so what are the morphisms? if not, does it mean that there is some object on which G acts? if this is the case, is it similar to the notion of group actions and what exactly is the object?

something has yet to click with this concept

[south patrol]

Well

The group is the category

Maybe I can give some motivation lol

So like for example you can think of GL(n) as the category with one object R^n and all the maps R^n -> R^n being incertible linear maps

[fleet pelican]

The morphisms are the elements.

[south patrol]

But in general we can just abstract away the object

[fleet pelican]

The object is the things the group acts on, e.g. the underlying group as a set.

[charred crescent]

ohhh so each object in a group in a sense "permutes" every other object in the underlying set?

[fleet pelican]

Element of the group. There is only one object. Basically yeah.

[charred crescent]

element** yes

[fleet pelican]

The automorphisms needn't be trivial. In a group they just need to be invertible

Permutations are a decent picture to have in mind.

[charred crescent]

my attempt to define: let (G,.) be a group. then fix some a in G. define p_a : G -> G as the morphism given by a. then for any g in G, p_a(g) = a.g

?

and you can do that for any a in G

[fleet pelican]

You should be more abstract, less explicit.

The object isn't G, the category is.

[charred crescent]

the category is the object?

[fleet pelican]

One potential picture is G as a set, concretizing the homomorphism G --> S_|G|, in general it can be any G action, possibly an entirely abstract one.

[fleet pelican]

The category is the group, the object is abstract.

[charred crescent]

i guess what i am getting hung up on is how can the object not matter? is it not necessary to define what the elements of G (morphisms) are acting on?

or is it defined this way so that the object can be anything

[fleet pelican]

Because there can be multiple examples of the object, and the category acts the same.

What is important is the morphisms and they behave.

[charred crescent]

okay fair enough

[charred crescent]

this does make a bit more sense now though. thank you all 🙂

[void cosmos]

yo anyone ?

[high fable]

πr²

[formal ermine]

<@&268886789983436800>

[wraith cargo]

I would maybe say to use the universal property of the direct product

That's a first idea at least

[void cosmos]

okay can u help me

with why this is wrong

[high fable]

Is that illegal?

[obsidian sleet]

Lets not troll pls

[void cosmos]

now we know that the product is injective

[high fable]

K sorry

[void cosmos]

so given 0-->A-->B exact there exists an R-module homo B-->product call it h such that g(h) = f

g is the map A-->B

and f is the map A-->product

cool?

[wraith cargo]

Hmmmmm
I was thinking of maybe using the fact that 0 -> I -> A -> B -> 0 splits when I is injective

then you get some homomorphism of SES

[void cosmos]

i still dont have that

this is right after the definition

[wraith cargo]

I mean isn't that an equivalent definition lol
But OK let's use your definition

[void cosmos]

cool?

[wraith cargo]

Actually that might work lol

[void cosmos]

now why cant i just define the map from B to each J_i

by just projecting the image at product

to each J_i

why wouldnt that work

why do i need universal property of produc

t

[wraith cargo]

So OK by your definition

there is a unique map B -> prod J_i

such that the diagram commutes

[void cosmos]

no

no uniqueness

[wraith cargo]

???

oh sorry

not unique

but there is a homomorphism

[void cosmos]

yes

its just not unique

[wraith cargo]

yea sorry

I think abt universal properties too much

[void cosmos]

yea no worries

i just wanna know why my shit doesnt work

[wraith cargo]

I mean I think the proj_i \circ h should work no?

[void cosmos]

yes

thats what i did

but i got lost in what the fuck commutativity means

like which ocmposte that

i always switch them

[wraith cargo]

Ahh

OK so

[void cosmos]

its so annoying

[wraith cargo]

Your diagram would look like

[void cosmos]

sorry for bad handwriting:

[wraith cargo]

0-> A -> B
| /
ΠJ
|
J_i

[void cosmos]

yeeaa

now

[wraith cargo]

OK so the vertical map is proj_i \circ g

[void cosmos]

to show that h bar : B-->J_i ; hbar(b) = pi_i(h(b)) works

i need to show what

[wraith cargo]

ye ye

[void cosmos]

forget about homomorphism

which this compose that

would prove the diagram

is commutative

..

[wraith cargo]

OK so

here's the idea

let's call your new homomorphisms
f_i = proj_i f
h_i = proj_i h
And g stays the same because that's the map A->B

We need to show that

f_i = h_i \circ g

[void cosmos]

why is f_i proj_i f

ohh its cuz its a new diagram now

[wraith cargo]

yea

[void cosmos]

wait let me draw it and tel lme if its right

now i want to show that this hwole diagram commutes with J_i being the new module cuz thats what i want to prove

so this long left vertical line would accumlate to pi_i o f right

?

[wraith cargo]

yes

[void cosmos]

and i would then want to show that my new map call it h bar satisfies , g o h bar = pi_i o f

?

that would say the diagram is commutative?

[wraith cargo]

The LHS should be flipped!

[wraith cargo]

yess

[void cosmos]

wdym

[wraith cargo]

g is first

then hbar

so hbar \circ g

[void cosmos]

u sure?

[wraith cargo]

Look at your diagram

we're starting at A

you first take g to B

[void cosmos]

yea fuck

yes yews

[wraith cargo]

okeww

[void cosmos]

how do u not get confused

?

do u jus follow the money

follow the elements*

and choose what makes sense?

[wraith cargo]

Idk I usually imagine diagrams in my head lol

I did a lot of category theory so I kinda got used to diagrams
But that's a skill you def learn over time

[void cosmos]

okayy

painful skill

but okay

[wraith cargo]

lol but once you learn it it's great

[void cosmos]

okay so it follows directly right?

h bar o g = pi_i o h o g = pi_i o f ( by commutativityy of first diagram )

and thats it?

road haog

[wraith cargo]

yep yep

[void cosmos]

okaay cool

got it

tysm

[wraith cargo]

np

[void cosmos]

ig now the other implication

would need the universal propeprty

ie showing each J_i is injective then the product is

[wraith cargo]

uhhh I would say so yes

[void cosmos]

yea cuz injecitivty gives us the fake maps

yea yea cvol

ty

[wraith cargo]

npnp

[molten viper]

So, I'm working on this problem, and I've gotten to the final part. Is it enough to essentially say "if F were a splitting field, x^3-2 would be factorable into linear factors, but it isn't as the cube roots of unity are not in F"?

the "any polynomial" part is what's a little confusing

My justification for that is that F / Q is an algebraic extension

[lethal dune]

doesn't sound right.. at least your argument is insufficient. F being a splitting field doesn't mean it should split x³-2

try to give more justification

[molten viper]

hmmm

It seems circular to say that if it is normal then the orders would be equal

It also seems that any splitting field which contains F would actually be larger than F

I'll discuss the problem with my professor

[lethal dune]

oh nvm I got confused with splitting and separable, your argument seems fine

It's almost correct but show there's no cube root of unity in F

[molten viper]

Gotcha

Thanks much

[void cosmos]

can someone give an example of a module such that the direct product is not projective

[lethal dune]

Z×Z×Z×..

not projective over Z

[void cosmos]

why

what happens if we choose our B to be Z

do we get a contradiction?

[lethal dune]

let me think if that's actually correct lol

[void cosmos]

i would really appreciate if we could walktrhough the proof from the definitions

this kind of stuff really helps me understand more

examples.

[lethal dune]

yeah it's true, it relies on the fact that submodule of a free module over PID is free

also the module ∏Z is not free over Z (check MSE for explanation). But projectives are submodule of a free module so ∏Z can't be free

[void cosmos]

sadly still didnt cover modules over PIDS

[lethal dune]

keep this example in mind then

[void cosmos]

cant we prove it from the definition?

or is it hard

[lethal dune]

finite prod of projective is projective btw

because it's just direct sum

[lethal dune]

both the statements aren't trivial to prove but you can try

[void cosmos]

okayy

tysm

[delicate orchid]

I imagine it's a direct generalisation of the nielsen-schreier theorem?

oh no wait it'll be way easier than that as you just ||need n-s theorem for abelian groups||

[lethal dune]

what's NS theorem?

[delicate orchid]

time to use some lateral thinking to solve this mystery

or wait is that logical thinking

hmmmm

mysteries induce mysteries

believe it's from here - unless you're confused about something else?

it's just decomposing v into the basis B_{S^nV} I believe

yeah cause it's orthonormal

if you have any orthonormal basis you can do this

you can write any vector $v \in V$ as $v = \sum_{u \in \mathbb{B}} \langle v, u \rangle u$ if $\mathbb{B}$ is an orthonormal basis of $V$

[cloud walrus]

Wew

[delicate orchid]

yus

same goes for the alternating space

np

[valid night]

Translation: A [representantensysteem] for the left coset of $H$ in $G$ is a subset $S\subset G$ that contains exactly one element from every left coset of $H$ in $G$. If $S$ is such a [reresentantensysteem], then $|S|$ = $[G:H]$, and $G$ is the disjunct union of the subset $sH$

[cloud walrus]

Kroros

[valid night]

Does anyone know the English term for representantensysteem?

[delicate orchid]

I'd call it a transversal

a left transversal specifically

also coproduct for disjoint union... I approve

[valid night]

Yes I finally found something about it

Thank you very much

It's almost impossible to google maths stuff in Dutch

[delicate orchid]

no worries

[sonic coral]

can i get a hint to show that 1 is the only element with a multiplicative inverse in a ring where r^2 = r

[south patrol]

well if r is a unit and r^2 = r...

[summer path]

Use contradiction

[south patrol]

lol no need for contradiction

[summer path]

Or that

Lol

[sonic coral]

unit means it has a multiplicative inverse right

[lethal dune]

@south patrol maybe you can try this one. My solution wasn't elementary, see if you can find one easy argument for this.
Show that the only integral element ℚ (t, √(t³-t)) over ℚ are ℚ itself. also this is not a purely transcendental extension

[delicate orchid]

indeed

so we have some r^-1 such that r^-1r = rr^-1 = 1, and also the fact that r^2 = r

and now with their powers combined how can you conclude r = 1

[lethal dune]

is it true that only ring satisfying that property are ∏Z/2?

[sonic coral]

maybe r^-1*r^2 = r^(-1+2) = r = 1

idk if that’s how the powers work in rings

[wraith cargo]

Hom_k(V,W) is the set of G-equivariant maps?

[delicate orchid]

yeah that's it

[sonic coral]

and how did we know r was invertible

[delicate orchid]

what would be the point if r wasn't invertible

we're showing 1 is the only invertible element with r = r^2

[sonic coral]

oh duh

[dusty verge]

You can bypass that with associativity either way

Or rather, it does work like that, as a consequence of associativity

[sonic coral]

how could you show that using associativity

[dusty verge]

r^(-1)r^2 = (r^(-1)r)r

[sonic coral]

oh right

[dusty verge]

Man that was brutal to type out on mobile lol

[south patrol]

Oo

[dusty verge]

What's an integral element?

Also my field theory is super rusty. Q(x) is just Q[x]/x right?

[void cosmos]

no

Q(x) is Q adjoin x

{a+bx | a,b in Q}

[south patrol]

That is just Q

[south patrol]

No

[void cosmos]

😦

[south patrol]

Q[x] is polynomials over Q

[dusty verge]

Oh right, it's Q(f(x)) is Q with the roots of f stitched in

[south patrol]

Q(x) is rational functions over Q, the field of fractions of Q[x]

[dusty verge]

That it?

[south patrol]

Idk what you mean

[dusty verge]

Okay. I think I'm following

[lethal dune]

[void cosmos]

damn i got mega confused now

[south patrol]

Bleak

[dusty verge]

So F(whatever) is rational functions of whatever?

Gotcha

[south patrol]

Hi Joe

[lethal dune]

~~joe who? ~~

[dusty verge]

What's an integral element then?

[lethal dune]

element that is a root of a monic polynomial in Z[x]

[void cosmos]

wouldnt this suppose that x is trasncendtal

[south patrol]

That is what the notation Q(x) typically means

[void cosmos]

i meant the usual for example Q(sqrt(2))

wouldnt this the more typical

be*

[south patrol]

then yours only holds in a special case

Where the extension is of degree 2

[void cosmos]

i thought yours was the more special

[dusty verge]

Consider x a placeholder for a polynomial

[void cosmos]

cuz it would be true when x is some transcendtal

say pi

[dusty verge]

For the sake of convenience

[lethal dune]

if x is algebraic then Q(x)=Q[x] anyway

[void cosmos]

yea true

[south patrol]

I mean like

If someone asks what Q(x) means im gonna say the rational functions over Q lol

Without further context

[void cosmos]

haha ig i would ask whats x

[lethal dune]

Q evaluated at x

[wraith cargo]

that would be x(Q) 🤓

iirc there's a book that uses this function notation lmao

absolutely horrible

[formal ermine]

some rep theory book uses that too

[delicate orchid]

reps and chars of finite groups by liebeck and whatshisface uses xf for f(x)

unless it's a character then they use f(x)

or a rep

[molten silo]

pls help

I think the radical <x^3,y^2,z> is <x,y,z> which is maximial. What about the other ideals?

[lament dawn]

!help

[flat tree]

Please read #❓how-to-get-help

[lament dawn]

Or #groups-rings-fields

[delicate orchid]

mig...

boss...

[molten silo]

ehh?

[delicate orchid]

I dunno how to tell you this

[lament dawn]

Oh...

Uh.

[delicate orchid]

@molten silo ur good boss

[lament dawn]

Demonstrate the correct absorption property

Haha sorry

[dusty verge]

K any field?

[molten silo]

yes

i still dont get this

[delicate orchid]

do you mean prime ideal by primary ideal?

[dusty verge]

Nah, I had to look it up but it's close

[delicate orchid]

because that first ideal is definitely not prime I don't think

[dusty verge]

Primary is xy in p implies x^n or y^n in p

[delicate orchid]

right ok now I believe it's primary

[molten silo]

they are different

prime implies primary

[delicate orchid]

yur yur

[dusty verge]

Isn't that second polynomial irreducible?

[delicate orchid]

certainly looks like it

[molten silo]

i think so

my teacher ddnt give answer to these questions unfortunately

[dusty verge]

Wait what if K is Z/2Z

Hmm, nvm

[delicate orchid]

I mean the first one is just intuitavely primary and it's annoying me I can't formalise it

[molten silo]

for the first one

[dusty verge]

Let xy in ideal. Then three cases, one of them divides

[delicate orchid]

the only problematic things are x*x^2 and y*y which cause it to not be prime but still allow it to be primary

[dusty verge]

In each case we good

[molten silo]

i think the radical is maximal

[delicate orchid]

actually yes, of course it is

the radical is <x,y,z> is it not

[dusty verge]

It is

[molten silo]

i think it is

[delicate orchid]

boom

one down

[molten silo]

two to go

[dusty verge]

Second one,.the polynomial is probably irreducible

[molten silo]

i think so

[delicate orchid]

the 2nd one is maximal ri- yeah

[dusty verge]

Did you recently cover the generalized Eisenstein criterion?

[molten silo]

last year

[dusty verge]

That's what Google said to use for multivariates

I don't wanna deal with it, but I bet it works

[molten silo]

i cant remember what is says, but i do think its irreducible

[dusty verge]

Could also argue by contradiction that it can't have a linear factor

[delicate orchid]

could argue by "grrr it's obviously irreducible"

no cross terms so how can factor

[dusty verge]

Wait what's the analogue for the ftoa for multivariables?

[delicate orchid]

uhhh nullstellenstatz?..... ???

[fleet pelican]

nullstelensatz?

[dusty verge]

Is it max degree linear factors, or sum degrees or is there not one?

[molten silo]

its noetherian

the field i mean

[dusty verge]

Ooh I just sinned

[fleet pelican]

plz repent

[dusty verge]

And mistook irreducible with prime but it's okay cause it was right

[molten silo]

i dont understand what you mean?

[dusty verge]

Okay, the guaranteed way to show it's irreducible is that it's linear in x, reducing in y would give a sqrt(x), and reducing in z you can't get three linear factors

[molten silo]

i see

[dusty verge]

I was thinking that because the poly is irreducible the ideal must be prime, but that just means the ideal is irreducible

[molten silo]

okay that makes sense

[dusty verge]

Third ideal isn't primary, because it's not maximal but it's its own radical, right?

[molten silo]

what is its radical?

[delicate orchid]

k[x,y,z]/(x-y^3-z^2), we have x = y^3-z^2, y^3 = x-z^2 and z^2 = x-y^3 => y^3 = y^3-2z^2 => z = 0 => x = y^3, so this is iso to K[y] but I can't get it to be a field (skill issue)

[dusty verge]

Itself I think

[delicate orchid]

not sure why I replied to that post

yeah third ideal is it's own radical, all of the generators are linear

[dusty verge]

Then it's not primary, since it's not maximal

[molten silo]

so if they are all linear then they are equal to its own radical

but how do you show its not maximal?

[dusty verge]

Contained in (x, y, z)

Wait no

[delicate orchid]

the noetherian implication is only one way

[dusty verge]

Forget what I just said I'm stupid

Nah it's both I think

I might be misremembering wikipedia let me double check

[delicate orchid]

[dusty verge]

I misremembered lmao

[molten silo]

i think its one way

[dusty verge]

Yeah the if if was a different property

[delicate orchid]

also, is it? how is x+1 in there lol

[dusty verge]

Oh speaking of my misremembering, k[y] has no zero divsiors

[molten silo]

wait is the radical prime?

[dusty verge]

I thought it worked like x + x^2 okay?

[delicate orchid]

true, so it's prime at least

which implies irreducible because I said so

[dusty verge]

Nah that proves that the ideal is primary

[delicate orchid]

yurrrr

[molten silo]

yes

[delicate orchid]

god there are a lot of things for me to keep track of here

[dusty verge]

Well also that

[delicate orchid]

we want an implication going the other way don't we

[molten silo]

damn

you sure its prime?

[dusty verge]

https://en.m.wikipedia.org/wiki/Primary_ideal

[molten silo]

if not the contradiction works

[dusty verge]

Second thing in properties

[delicate orchid]

wait I'm skill issuing hard on number 3

[delicate orchid]

K[x,y,z]/(x+1, y-1, z+1) is clearly iso to K so the ideal is maximal

[molten silo]

Use second definition maybe

oh

[dusty verge]

Oh wait yeah lmao

[molten silo]

i think that works

[dusty verge]

Now go write it up real pretty

[delicate orchid]

yeah, an ideal generated by only linear polynomials is obviously maximal apologies fellas

[molten silo]

thank you boys.

[next obsidian]

Wew is noob

I was on cooldown so I couldn’t fix the slop you were laying down

[dusty verge]

I wonder why

[molten silo]

yeah yeah

sure thing

[next obsidian]

I was on the 10 min cooldown for when you join the server

[dusty verge]

I wonder why

[delicate orchid]

nice "honorable" role stinky

[next obsidian]

Not my fault ur a noob

:dvacat:

[delicate orchid]

problem got solved didn't it

:blazed2the9s:

[dusty verge]

Can I just say, lang is an asshole?

His book is hard :(

[wraith cargo]

lang is a notorious aids denier it is evident in his algebra book

[dusty verge]

True he never once mentions aids in it

As far as I know I'm only like 100 pages in

Help

[coral spindle]

You took a break too?

[coral spindle]

Yes, Lang just wrote books for himself. Good luck with it.

[next obsidian]

🚬

[dusty verge]

Thanks I took like a year break so I should get back to it

My two days here have revealed to me that I need to freshen up on ring shit

[frigid lark]

(Entire rings are also called integral domains. However, linguistically, I feel the need for an adjective. "Integral" would do, except that in English, "integral" has been used for "integral over a ring" as in Chapter VII, §1. In French, as in English, two words exist with similar roots: "integral" and "entire". The French have used both words. Why not do the same in English? There is a slight psychological impediment, in that it would have been better if the use of "integral" and "entire" were reversed to fit the long-standing French use. I don't know what to do about this.

Although the aids stuff is probably horrible

But I love his algebra book

Well as much of it as I have read

Also if you got through groups it might be a bit easier for a bit

[warped viper]

Let $A$ be a ring with unit and $S$ subset of $A$ such that $1_A\in S$ and for all $s,t\in S$, we have that $st\in S$. I'm going to define two equivalence relations. Let $(a,s),(b,t)\in A\times S$, i claim that if $(a,s)\sim_1(b,t)$ then $(a,s)\sim_2(b,t)$ and conversely. \bigskip

First relation :
\begin{center}
$(a,s)\sim_1(b,t)$ if there $\exists u,v\in A$ such that : $au=bv\in A$ and $su=tv\in S$.
\end{center}
Second relation :
\begin{center}
$(a,s)\sim_2(b,t)$ if there $\exists u,v\in A$ such that : $au=bv$ and $su=tv\in S$.
\end{center}
Since $A$ is a ring and $a,b,u,v\in A$ it is always true that $au\in A$ and $bv\in A$ so writing $\in A$ or not does not change anything. But my teacher insists that my equivalence relation isn't right (the first one). I can accept that he wants to use the second since he decides what he wants my project to look like, but to me they are equivalent so i don't understand why he insists, am-i wrong that they are equivalent ?

[cloud walrus]

Overfull hbox, badness 1000

[next obsidian]

Aren’t these literally the same thing?

[reef trench]

Hey guys, I am rather unsure of this question, because I know the answer intuitively, but I am struggling to make the proof rigorous

The question is asking that if we have a set X, and the set of all functions that map X to X, then this set of functions form a semi-group under functional composition.

I know that functional composition is associative, but I honestly have no clue how to show that other than "This is basic knowledge, lel"

[delicate orchid]

show the image of each x \in X under (ab)c is equal to the image of the same x under a(bc)

is how I'd do it

[reef trench]

Of course, lol

XD

[wraith cargo]

||(ab)c(x) = ab(c(x)) = a(b(c(x)) = a(bc(x)) = a(bc)(x)|| solution to check after you've tried it

[reef trench]

Yep, that's actually what I did!

thanks guys

[dawn latch]

in group isomorphism

how were these mappings derived?

[delicate orchid]

you're literally just relabelling the table

or are we

[dawn latch]

I'm sorry, but how

[delicate orchid]

right ok this group is S_3

we're not, my mistake

which automorphism is it

ignore me for now

a = Id b = (123) c = (132) d = (12) e = (13) f = (23)
u = Id v = (12), x = (13), y = (23), z = (132), w = (123)

[dawn latch]

i don't understand, unfortunately

[delicate orchid]

yeah ok, so what this is is a relabelling of the table followed by some reordering of the rows/columns that's what's going on here

[dawn latch]

what are those (123) (132) ...

[delicate orchid]

yeah I said ignore me

but now unignore me

I was just translating that awful cayley table into an actual group

[dawn latch]

ohh

how did you get b = (123)

[delicate orchid]

the choice of whatever d,e (or v, x) are determines what everything else is (this is because those two elements are generators of the group)

I knew they had to be order 2 by looking at the table (d*d = a, e*e = a)

just as a question, is this an example of a group isomorphism you've been given?

[dawn latch]

no, my apologies

i found it online

[delicate orchid]

ah then this is a good exercise! See if you can show that this is indeed a group isomorphism

[delicate orchid]

you've got the fact that it's bijective for free

[dawn latch]

what was given to us is this:

[delicate orchid]

right I see

[dawn latch]

ok, I'll try

[delicate orchid]

one mo

[delicate orchid]

if we do a little re-arranging it becomes much clearer what your map should be

every isomorphism will be some rearrangement like this followed by a "relabelling"

[molten silo]

Guys any ideas?

[delicate orchid]

what ring is x/y in here, K[x,y]/(x^2-y^3) also?

[delicate orchid]

ah ok it's from the localisation of K[x,y]

that now makes sense

[molten silo]

yes

i think

[delicate orchid]

I'm pretty sure K[x,y]/blah is iso to a subring of the localisation so it actually makes sense to ask if an element is integral over it in that case

is it not a root to f(a) = xa-y^2

fair warning it's 2am here

[pastel cliff]

how do you find the commutator subgroup of a thing

[molten silo]

my brain

[dim widget]

xa-y^2 isn't monic

[pastel cliff]

prime math hours

[delicate orchid]

fuck you're right

uhhhh

[molten silo]

how is x/y in k[x,y}/blah

[dim widget]

But this is a good exercise to think about

x/y isn't in it

That's the point

[delicate orchid]

what's the group

[dim widget]

r \in R then showing that r is integral over R would just be showing that it satisfies x - r which is trivial

[pastel cliff]

but also wanna know in general

i know what abelianization is

but is there a general way of doing it

[dim widget]

x/y naturally lives in the fraction field of k[x, y]/x^2 - y^3

[delicate orchid]

fuck yes you have a presentation I don't have to actually THINK

[pastel cliff]

god forbid

[delicate orchid]

do you just want the abelianisation

or do you need the derived subgroup explicitly

because one of these answers will be great and the other will make me go to bed

[dim widget]

If you want the answer it's just that since x^2 = y^3 then x/y satisfies Z^2 - y trivially. But you should think about what is going on geometrically

[molten silo]

im not sure i understand.

[pastel cliff]

yeah this thingy

but i assumed i needed to do the commutator subgroup man thing first

[dim widget]

So like if you have an integral domain R (like say the integers Z), you can form its total field of fractions (for Z it is Q, the rational numbers)

[delicate orchid]

shove [a, b] as an extra relator, this is the presentation of the abelainisation

[dim widget]

This is the set of elements a/b where a, b \in R and b is not zero, modulo the obvious relations

[delicate orchid]

you can then use this to simplify the other relators down

[pastel cliff]

you mean like

just say that ab = ba

and see what happens to the relations

[delicate orchid]

Ab(pi_1(X)) = <a, b : a^2b^3, aba^-1b^-5, [a,b]> = <a, b : a^2b^3, b^-4, [a,b]> = <a, b : ba^{-1} = a, b^4, [a,b]> yes - this smells like Z[i]

[dim widget]

im not sure i understand

[molten silo]

i think i understand now. for k[x,y]/<x^2-y^3> are all the polynomials that satisfy x^2-y^2 =0

[dim widget]

No, not exactly...

k[x, y] is all polynomials

<x^2 - y^3> is imposing the relation x^2 = y^3

[pastel cliff]

this is the full question lol

[molten silo]

ok how does this imply x/y is in the set?

[dim widget]

This belongs probably more in the algebraic topology sub. You could do it with just algebra but it's more convenient to do it with graph theory.

[delicate orchid]

or just bash out that presentation into a smith-normal form which I'm not doing at 2:15am

[molten silo]

ok

[dim widget]

x/y is the ratio of two such functions.

[dim widget]

There isn't a Smith normal form for noncommutative groups, a, b don't commute.

[delicate orchid]

[a, b] is a realtor I took the abelianisation

[pastel cliff]

realtor

[delicate orchid]

As that’s what sebbb was asking

[pastel cliff]

it is for alg top lol

[delicate orchid]

If ur gonna pull me up on each typo ur gonna type a novel by the time this convo is done

[pastel cliff]

funni

but it looked more like an algebra question so i pooped in here

[dim widget]

It's fine, I guess the thing to observe is that now you are quotienting out by aba^{-1}b^{-1}, aba^{-1}b^{-5}, a^2b^3

[dusty verge]

oh thank god

[dim widget]

so combining the first two relations we see that b^4 is trivial

then a^2b^{-1} is trivial because a^2b^3 and b^4 are

So that implies that a^2 = b.

[delicate orchid]

This is where I got stuck lol

Oh yeah DUH

[dim widget]

So the group is just generated by a which satisfies a^8 = 1

But probably it would be more entertaining to draw some pretty pictures.

(so the answer is Z/8 for the record).

[pastel cliff]

maybe a silly question

[dim widget]

^Shoot

[pastel cliff]

dont we get that b^-4 = id? not b^4

[delicate orchid]

Same thing

[dusty verge]

I hate math problems that have big annoying equations

[delicate orchid]

Take inverses of both sides

[pastel cliff]

mmm yes

[dusty verge]

Why can't all math just be nice simple to look at things that make your brain scream?

[dim widget]

All math is simple to look at

as long as you don't have to understand it.

[pastel cliff]

just look at it lmao

[dusty verge]

Yeah but long bs equatoins are annoying

Honestly numbers are the worst part of math

Wait that's not true cause analysis exists

[delicate orchid]

Based based based

[wraith cargo]

Number theory moment bruh

[delicate orchid]

Non-rigorous 1570-2023

[dusty verge]

At least number theory is cool in theory

[delicate orchid]

Number theory bad until it becomes ring theory

Then it’s DOG WATER

[pastel cliff]

it becomes ring theory???????

[dusty verge]

ideals are just a generalization of divisibility

[delicate orchid]

What are rings a generalisation of

[wraith cargo]

Tru!
Congruence can suck my dick i do not care when a congruence can be solved modulo 2^2^k

[delicate orchid]

Oh, yeah. Z

[dusty verge]

My last semester of my masters I had to do a class on number theory, and the analytic shit was brutal

[pastel cliff]

stop it

i forgor

[wraith cargo]

Me when pi(x) > log(log(x))

[dusty verge]

I don't think we got to that one, but we did prove the prime number theorem with the sieve of some guy

Not erastosthenes, it was way worse

Oh right the selbert sieve

[delicate orchid]

Initial object in Ring for a reason buddy :blazing8s: :blazed2the9s:

[pastel cliff]

blazed2the9s

[dusty verge]

Is there a channel for throwing around random problems on here?

I'm bored

[delicate orchid]

Just hop between advanced channels ig

[pastel cliff]

just post in the relevant channel

[dusty verge]

Oh I wanted problems to try and solve

Or rather, ask people details of and then get no where on

[pastel cliff]

look up qualifying exams

[dusty verge]

Nah I'll just torture myself with lang I guess