groups-rings-fields

[tall igloo]

id - id is not invertible

[karmic moat]

^

GLn isn't closed under addition

[fervent solstice]

idk how to give a hint that would not just give the answer up immediately

[dull ginkgo]

So if F is a field what kind of ring is F[X]

Congrats. Use it.

Explode

[fervent solstice]

but when considering quotient rings (esp. in polynomials) you usually try to express P € F2[X] as a sum of two things, one in the ideal you're taking the quotient with, one outside of it

[rocky cloak]

Do you know about polynomial division (long division)

[fervent solstice]

given that the thing that is in the ideal is going to be sent to 0 by the quotient map

[dull ginkgo]

The remainder is a valuation-minimal representative of a congruence class mod the ideal (namely the principal ideal) homie

[tall igloo]

How can you wiggle an n-deg polynomial in this ring

[rocky cloak]

But do you know what the result from doing it is?

Like what would you get if you divided a polynomial by x^2 + x + 1

[dull ginkgo]

In a pid P we can write out an element x in respect to y as
x = ay + r, where r is the remainder, essentially the smallest valuation element of the congruence class of x mod y

Yep, all the congruence classes

The impossible -> trivial pipeline

[tall igloo]

Average Grothendieck proof

[fervent solstice]

a problem is either unsolved or trivial

[karmic moat]

or both

[dull ginkgo]

Anyway hey chat so is the fact that $T \left( \bigoplus_I A_i \right) \cong \bigotimes_I T(A_i)$ due to the fact that the tensor product is like the coproduct in the category of $R$-algebras and the free/forgetful adjunction ?

[cloud walrus]

mizalign

[dull ginkgo]

Mobile Tex help

[tall igloo]

what is T? I assume the As are R-alegbras?

[dull ginkgo]

does the same work for $\bigwedge \left( \bigoplus_I A_i \right) \cong \bigotimes_I \bigwedge (A_i)$

[dull ginkgo]

Tensor algebra functor, free from R-mod to R-alg

[cloud walrus]

mizalign

[dull ginkgo]

Where it’s the graded comm tensor product

[rocky cloak]

The tensor product is only the coproduct in the category of commutative algebras.

So this shouldn't be true (unless you're just using the tensor symbol as the symbol for the coproduct)

[dull ginkgo]

Or the symmetric algebras with the abs-comm tensor product

[dull ginkgo]

Well the big o plus is going in, i.e coproduct in R-mod, and the tensor product is the coproduct coming out in R-alg

OH WAIT YEAH

You’re right

That would work for the symmetric algebra

[tall igloo]

Do you maybe want a tensor product over R

[south patrol]

Tensor product

[dull ginkgo]

I was trying to see if there is a functorial way of showing the Tensor algebra over a free module is free (or graded free)

[rocky cloak]

Yeah, it should work for the symmetric algebra, and also for the exterior assuming you're doing a tensor product of graded algebras (so you get the right signs)

[dull ginkgo]

Graded-comm right

[south patrol]

Are the R Algs c9mmurative

[rocky cloak]

Yeah

[tall igloo]

huh interesting. Would've thought the coproruct in R-algebras was the relative tensor product. Neat

[south patrol]

Oh i just said tensor product for no reason

Lemme read

[south patrol]

Yes it is the relative one

[tall igloo]

Ah good

[rocky cloak]

That should follow just from the adjunction

[south patrol]

Indeed we have little choice consideeing the initial object

[dull ginkgo]

Free as in admitting basis for the grading modules

[south patrol]

Compose the adjunctions bro

[rocky cloak]

Oh, you mean free as an R-module

[dull ginkgo]

Yes

I probably should’ve clarified

[south patrol]

Then just note free R algs have free underlying R module innit

[dull ginkgo]

real!

[south patrol]

At least in finite case

[dull ginkgo]

I should return to my algebra roots

[rocky cloak]

Yeah, I guess it's not really an abstract nonsense thing. You just have to look at what the free algebra is

[dull ginkgo]

Is there a category for which the universal enveloping algebra is free of

[rocky cloak]

free of what?

[dull ginkgo]

Like the adjoint to the forgetful from to R-alg

[south patrol]

Like u cam use the fact Sym^k takes free to free

[south patrol]

I mean which UEA do you mean

Whatever you mean this should be a tautology lol

[rocky cloak]

I'm not sure I understand the question

[south patrol]

Since free usually basixally means being im the image of a left adjoint of forgetful

[manic cairn]

oh hey there is a mod here rn so i can ask. a guy who has no special roles wants a thread here is it chill if i create one?

[dull ginkgo]

Like is the function from an R-algebra to it’s universal enveloping algebra a functor

[rocky cloak]

The other people with threads don't have any special roles at least...

[manic cairn]

dont you need at least VA to make threads here? something like that

[rocky cloak]

Idk, but you have very active

[rocky cloak]

Should be yeah

It's just A |-> A(x)A^op

[dull ginkgo]

I see

Thank you

[rocky cloak]

Don't think it has an adjoint though

[dull ginkgo]

I wanted to do that through the grading on the algebra lolz

[dull ginkgo]

Not exact I don’t think

[south patrol]

Wait wdym by UEA here

Usually i would take in a Lie algebra

[south patrol]

Oh this

Like for bimodules

[dull ginkgo]

Another very obvious question probably

Is a group representation literally just a group action on a vector space (I say “group action” generally to be a map to some automorphism group for some object )

[rocky cloak]

I'd say it's a linear group action on a vector space

[tulip otter]

a pirate learning algebra

[dull ginkgo]

How so?

Would a map G -> Aut(V) on a vector space allow a map from R[G] {group algebra} to End_R(V)?

[valid elbow]

Yeatte's Secret Thread

[karmic moat]

wait until the pirate start learning ag... then you could use the Sailing the Rising Sea thread in #algebraic-geometry ...

[rocky cloak]

Well if it didn't satisfy
g(v + w) = gv + gw
I wouldn't call it a representation, but I guess people disagree on the difference between representation and module

[rocky cloak]

Yes

[dull ginkgo]

Well that’s because G is mapped into Aut_R(G) I’d think, no?

Not just Aut(G)

[rocky cloak]

Right, but when you say "group action" I think
G -> Aut_Set(V)
not
G -> Aut_Rmod(V)

[dull ginkgo]

Ah sorry

[rocky cloak]

The latter being an R-linear group action

[dull ginkgo]

I am terrible with my terminology

If X is a C-object then I usually say a group action is a map G -> Aut_C(X)

[rocky cloak]

Anyway, not terribly important what you call things. Just as long as we agree what they are

[rocky cloak]

Which is where I would reach for the word representation instead

[dull ginkgo]

Smart, is that general usage for general categories

[rocky cloak]

Pretty much yeah

[lime warren]

someone out here ghost pinging my shit

[dull ginkgo]

another question, is this a proper isomorphism

Embed failure br

[rocky cloak]

That would be an isomorphism yeah. Not sure if "proper" is supposed to mean something specific here

[dull ginkgo]

Like of R-algebras

thank you!

That’s what I ended up with trying to understand mixed tensors

Before like, considering sections for the case of bundles :p

[dull ginkgo]

One more question. So lets say we have a monoid representation on a C-object A, i.e a monoid morphism f: M -> End_C(A).
If we have a subobject B of A such that f(M)B is a subset of B, then that B object is f-invariant

Would it be convention to call the representation irreducible iff A has no f-invariant subobjects?

[molten atlas]

Why is discussy leading me to this hellhole

Of all the places

[thorn jay]

I mean you could go to the other channels

Also calling this a hellhole but not discussy

[delicate orchid]

I wouldn't crash out if you said ts

[dull ginkgo]

from you that says a lot. thanks wew :3 :3 :3 bbg

[barren sierra]

Is there a way to prove the nilradical is the intersection of all primes without Zorn?

[thorn jay]

As 1 is not nilpotent, the nilradical is never R when R is not the zero ring, so it implies the existence of a prime ideal and hence Zorn's lemma

[south patrol]

Existence of a prime ideal is strictly weaker than existence of a maximal ideal

[thorn jay]

Really? Darn

[south patrol]

Yeah

This is equivalently like the Boolean prime ideal theorem or the ultrafilter lemma (which is used for existence of non-principal ultrafilters)

[thorn jay]

I think you can get away with assuming that every ring has a prime ideal

[south patrol]

Yes you can i think

Let me think

Suppose f is not nilpotent and A non-zero. Then A[f^-1] is non-zero and by assumption has a prime ideal p. The preimage in A is a prime not containing f, which is all we need

Sure but the statement i gave is non-trivial logically

I mean it is not obvious that choice does not follow from ZF lol

[thorn jay]

Yus I was thinking of doing something like that

[south patrol]

Yee i guessed hehe

I love these kinds of arguments

[thorn jay]

Go go gadget localisation

Tbh

That'll get 0

[south patrol]

this works

[south patrol]

wdym

Oh lol

"do not assume ring has 1"

Is this D + F

[weary frost]

is this the same x for every a?

[thorn jay]

We don't want ax = a, we want ax = 1

[kind temple]

the ring here is not assumed to have a unit

[weary frost]

at this point its not established whether "1" even exists in the ring

*rng

[kind temple]

this should follow from ||injectivity of right multiplication by a||

[weary frost]

yea i see, if ax=a and assume bx =/= b then by injectivity abx =/= ab = axb = abx which is a contradiction

[kind temple]

yea, i think that works too. i was thinking that if we know a = xa, then ba = bxa => b = bx

[weary frost]

oh thats a cleaner way to phrase it : D

[kind temple]

only if c > 1

[weary frost]

since we have an identity we just use the corollary now?

ah true

[south patrol]

maybe i'm crazy bu like

isn't ||the product of all the non-zero elements the identity||

uhh

[thorn jay]

Prove it bbg

[south patrol]

i guess this must be correct but need to prove it lol

[thorn jay]

Nah what Abt elements with x^2 = 1

[south patrol]

oh yh

[thorn jay]

They won't vanish

[south patrol]

good point

lol

i was silly billy

[thorn jay]

Yo, my name is silly billy and I- AAARGRGGGHHH

[south patrol]

https://tenor.com/view/big-billy-big-wet-billy-gif-21086513

[barren sierra]

Oh nice

[dull ginkgo]

I have pondered this a bit tbh

To my knowledge the proof of Hilbert Basis theorem, that A[x] is noetherian iff A is, requires choice

But I think you can provde it using UFL with the lemma that every ideal is within a prime ideal

by considering the prime ideals of A[x] and the going up/down properties

It's dependent choice though :p

[south patrol]

Afaik it does not but some proofs given do need dependent choice

I should try and remember the proof I know lol

[dull ginkgo]

Weird how R is noetherian <=> R[X] is noetherian <=> R[[X]] is noetherian

is Noetherian-ness preserved under J-adic completion?

Seems to do with the leading/trailing coeff map

It is.

Thanks Atiya-Macdonald

A long while ago I wanted to see when the monoid ring functor preserves noetherian-ness but I basically ended up with the Hilbert basis theorem case

Because like for instance, if that property held for monoid M, then F_2[M] would have to be noetherian.

[south patrol]

Let $A$ be a noetherian ring, $I \subset A[x]$ an ideal, let $J = { \text{leading terms of polynomials in } I} $, which is an ideal of $A$. By assumption, $J = (c_1,\dots,c_n)$ for some ${c_i} \subset A$; pick $f_i = c_i x^{n_i} + \text{lower order terms} \in I$. Let $N = \mathrm{max}(n_i)$. Nowgiven nonzero $g \in I$, we can write it as $g = a_n x^n + \text{lower order terms}$, $a_1 \ne 0$. If $n \ge N$ then write $a_1 = \sum_i \lambda_i c_i$ for some $\lambda_i \in A$ and note $f - \sum_i x^{n-n_i} \lambda_i c_i \in I$ has strictly smaller degree. Moreover, $I \cap A[x]{< N}$ finitely generated as an $A$-module since $A[x]{< N} \simeq A^n$ as an $A$-module and is hence noetherian. If we pick generators $\omega_1,\dots, \omega_m$ for $I \cap A[x]_{< N}$ as an $A$-module, then we have $I = (f_1,\dots, f_n, \omega_1,\dots, \omega_m)$

[cloud walrus]

Prismatic Potato

[south patrol]

No choice

[dull ginkgo]

True

[thorn jay]

Very nicee

[south patrol]

Is this the standard proof? I was rusty with the last bit

[dull ginkgo]

I think choice is used for the equivalence of terminating ideal chains, finite generators, and every ideal set having a maximum

[south patrol]

in particular these hypotheses hold if R is f.g.

[south patrol]

Yeah. Here I am using "every ideal is finitely generated" which I suppose is the "strongest" in the absence of choice (?)

Well it definitely quickly implies any ascending chain stabilises

Don't think it implies every non-empty ideal set has a maximum in the absence of dependent choice (but with dependent choice, you can keep saying something isn't maximal and build a countable chain which must terminate by above)

hmm

Another funny thing I remember never being mentioned in some texts is how Noetherian wrt countable chains is equivalent to being Noetherian wrt arbitrary chains given choice

[south patrol]

i guess it's just cause "noetehrian wrt arbitrary chains" is of intermediate strength between countable chains and just arbitrary non-empty sets having maximal elements

[dull ginkgo]

This is real interesting.

Assume F_2[M] is noetherian. We can pull back the divisibility relation through the principal ideal map by saying x | y iff (x) includes (y), consider the family of ideals (x) for x in M\{e}, then this family by Noetherian-ness has a maximal element, i.e some maximal (x).

[south patrol]

oh sorry i misread

[dull ginkgo]

I’m trying to see what happens when R[M] is noetherian iff R is for each comm ring

I wonder if it somehow reduces to M being isomorphic (order isomorphic with the imposed divisibility monoid structure) to N^n

M must be commutative so F_2[M] also comm lmfao

I’ll ponder this in a bit

[tough raven]

I don't think so; the difference there is restriction to Boolean rings.

[tough raven]

If you really want to go down this rabbit hole, look into Noetherian-ness conditions in constructive mathematics.

[acoustic igloo]

I'm looking at the group on this page with id #52
https://en.wikipedia.org/wiki/List_of_small_groups#List_of_small_non-abelian_groups

it says Z_5 ⋊ Z_4, but we have to choose a homomorphism phi: Z_4 -> Aut(Z_5) right? I think if ker(phi) = {0} we get group #52, and if the kernel is {0,2} we get group #50 which is Q_20.

but i just assumed it would make a group when the kernel was {0,2} and i checked the orders of the elements with python, which led me to think it was Q_20, so idk if i'm right or not

[nimble folio]

\begin{exercise}
        Let $G$ be a finite group and $n > 1$ an integer such that $(ab)^n = a^n b^n$ for all $a,b \in G$. Let 
            \begin{equation*}
            \begin{split}
                G_n = \{c \in G \mid c^n = e\}, \h9 G^n = \{c^n \mid c \in G\}.
            \end{split}
            \end{equation*}
        You may take for granted that these are subgroups. Prove that $G_n$ and $G^n$ are normal in $G$, and $|G^n| = [G : G_n]$.
    \end{exercise}

[cloud walrus]

clubsoda14

[nimble folio]

I was able to show that G_n and G^n are normal subgroups, but I don't know how to show the second part

[covert cliff]

the map $\varphi: G \to G$ given by $\varphi(g) = g^n$ is a homomorphism

[cloud walrus]

Khush

[covert cliff]

then what does first isomorphism say?

[tidal schooner]

yes, that's correct!

[acoustic igloo]

thanks 🙏

[nimble folio]

i see

thank you

[rocky cloak]

I think this depends on how you define Noetherian without choice, as the usual equivalent definitions fail to be equivalent

[minor gazelle]

Guys is my stgi proven correctly? I got myself a bit confused since I was too addicted to those Lego proposition building so I tried to clear up my mind

[woeful sage]

Let R be a comm. ring with unity, I = (a, b) then can I write (a, b) = {ra + sb | r, s \in R}?

[velvet hull]

yes, you can quickly prove that they're the same

[woeful sage]

What do you use as the definition of (a, b)? intersection of all ideals containing {a, b}?

[velvet hull]

that definition works

[woeful sage]

What do you use as the definition?

RA?

Oh right this is notation from D&F

I told you I'm on section 7.1

Yes I've learnt later material in my lectures

bro is talking about localization now

I gotta finish chapter 7 quick

which one?

[narrow temple]

H->HN/N being the map youre after that makes sense to me.
I personaly cant make sense of 1.4.9 as stated:
If N is a subgroup of H, then HN=H, so statement is H/N is a normal subgroup of H.
E.g. picking H=Z and N=3Z then youre saying Z/3Z is a subgroup of Z and to me this doesnt make sense as Z/3Z is not even a subset of Z.
Another thing is Im not sure how exactly you use proposition 1.4.6 to get the map H->HN/N from H->G, if you want you can clarify that a little.

[velvet hull]

almost correct, why can the division algorithm be applied for I?

[woeful sage]

I know what it is, I was just asking which is usually taken as the "definition"

I have that page open rn

[velvet hull]

the division algorithm only works when you fix an element that you divide by

but I consists of infinitely many elements

yep

that's what I was looking for

[next obsidian]

That doesn’t matter

You can just pick any f in I and then divide by f and get a representative for any g with norm less than the norm of f

[velvet hull]

I guess it wasn't clear what I was saying yep to lol

[languid trellis]

I am somewhat at a loss here. I'm currently trying to show the forward direction, that if a is a pm mod p, p = 1 mod 4, then so is -a.

I'm trying to show that the order of -a mod p is p-1. This would be enough to show that it is primitive. So suppose not, that (-a)^k = 1 mod p (k < p-1). Then k | p-1, so in particular k is even, so k = 2c, and (-a)^k = (-a)^2c = a^2c = 1 mod p. This would contradict a being a primitive root.

What's irking me though is that the very next question is to show that a is a primitive root mod p, where p = 3 mod 4, iff -a has order p-1/2.

I am not using the fact that p = 1 mod 4 in the 'proof' of the forward direction I gave. Can anyone advise on what could be going wrong?

[rocky cloak]

Im not sure how you're getting "k | p-1, so in particular k is even"

k=1 is not even for example

What you should notice is that
if -a has order k, then (-a)^2k = a^2k. So k must be a multiple of (p-1)/2

Then think about (-1)^((p-1)/2)

[languid trellis]

Yeah this was a mistep oops

[velvet hull]

yeah it's a pretty simple question

don't assume a norm exists, though

this works in UFDs

[languid trellis]

I really don't see how you're getting 'if -a has order k, then (-a)^2k = a^k'?

[velvet hull]

yeah, basically

[clear hatch]

This course is kicking me in the butt bro i have zero intuition for any of the material

Exam in one week I'm cooked

[rocky cloak]

Sorry it's supposed to say (-a)^2k = a^2k

So 2k must be a multiple of p-1 -> k is a multiple of (p-1)/2

[velvet hull]

by the correspondence theorem, ideals in the quotient biject to ideals in the PID containing the prime ideal

but all ideals containing the prime ideal are principal

it's a good theorem

[narrow temple]

Also just prime ideal = maximal ideal in a PID

[thorn jay]

genuinely the best isomorphism theorem

Y'all lattice haters

It's just a partially ordered set with meets and joins

Ragebaiting..

(a) / I = (a + I)

[velvet hull]

just use the prime implies maximal

that's even faster than the correspondence theorem

a field implies PID

[narrow temple]

what ideals does a field have?

[velvet hull]

because there are only two ideals

no need for ED

you really like that chain a lot

[karmic moat]

Dont overcomplicate it lol

[karmic moat]

https://tenor.com/view/truth-impact-third-impact-3rd-impact-truth-nuke-anime-meme-gif-9672669363915948084

[thorn jay]

Lol bro is coping hard

Flying cars would be a terrible idea tbh

[languid trellis]

If p is congruent to 3 mod 4, then p-1/2 is odd, so (-1)^((p-1)/2) = -1 (p). If p is congruent to 1 mod 4, then p-1/2 is even, so (-1)^((p-1)/2) = 1 (p)

So we need to extend this observation somehow, I'm thinking of using (-a)^k = -1^k a^k to calculate the order of -a.

So if we assume p = 1 mod 4, a is a pm, (-a)^k = 1 mod p, then

(-a)^k = -1^k a^k = a^k, (the last equality comes from (-1)^((p-1)/2) = 1 mod p, and that k is a multiple of p-1/2). This implies that k = p-1 by the fact that a is a primitive root.

For the case p = 3 mod 4, consider first the equation x^2 = 1 (p). The only inequivalent solutions are \pm 1. This implies that a^((p-1)/2) = -1 mod p, as it is a square root of 1 not equal to 1. So, (-a)^((p-1)/2) = -1^((p-1)/2) * a^((p-1)/2) = -1 * -1 = 1 (p), so -a has order p-1/2. (Again we use that k is a multiple of p-1/2)

[karmic moat]

Difficult enough controlling traffic on the ground

[thorn jay]

Istg

[karmic moat]

Would be hell doing so in 3D

[languid trellis]

if you disagree that correspondonce thm is the best then you need to do more algebra

[thorn jay]

Imagine a car crash at an altitude of >20 meters

[woeful sage]

They're called helicopters with wheels

damn sniped

[thorn jay]

Tech bro ahh idea

[karmic moat]

Have you seen what’s happening in california?

The self driving cars are getting stuck in gridlock and honking at each other

[south patrol]

Classic

[karmic moat]

https://youtu.be/Xvs0K1LG1ac?si=tTW2Si6aodxTlnkK

[tall igloo]

what if we had self driven cars and also they had pre-established routes and schedules and were also really big and had dedicated infrastructure

[thorn jay]

"almost" any car crash at that altitude is fatal

[karmic moat]

Wait that’s genius… and the self driven cars were really big and could hold a lot of passengers at once

[tall igloo]

i call it the trAIn

[thorn jay]

AND WHAT IF THEY RAN ALONG PREDETERMINED PATHS

[karmic moat]

Holy shit… go to silicon valley bro stop doing math

[thorn jay]

OMG I THINK WE MIGHT BE ONTO SOMETHING

[south patrol]

E = mc^2 + AI

E^2 = p^2 c^2 + m^2 c^2 + AI^2 real

[tall igloo]

anyways i bet the trAIns would never have any crashes and would be reliable and run on time

unless they are german for some reason

[thorn jay]

Ach nee, de trein is stukkie wukkie

[south patrol]

Uppsie wuppsi

[karmic moat]

Okay garmin video speichern

[thorn jay]

Oepsie woepsie, de trein is stukkie wukkie. We zijn heul hard aan t werk om dit te make mss kan je beter fwietsen owo

[south patrol]

Nice

[tall igloo]

yall thats dutch not deutsch 💀

[karmic moat]

Ze train haz gone kaput :(

[south patrol]

Dutch is cool

[thorn jay]

I am in fact average height

Blame my mother

[south patrol]

For a Dutch person

(7 foot)

[thorn jay]

She is a Belgian infiltrator

[south patrol]

250cm

[narrow temple]

This is from reids undergraduate commutative algebra chapter 1, which im mentioning because he tries to motivate the algebra by saying some things about how it ties to the geometric context.
For example talking about establishing K[X,Y]/(F) as "functions" on the set cut out by {F=0}.
Is that exercise some instance of something that can be rephrased in the geometric context?
He also had a short paragraph about the geometric picture of nilpotents e.g. Y in K[X,Y]/(Y^2), but I didnt quite understand the paragraph other than that the "functions" in K[X,Y]/(Y^2) carry more information than the ones in K[X,Y]/(Y).

[thorn jay]

She is relearning it but I cannot. Dropped it as fast as possible I hate that shitass language

[tall igloo]

based

[tardy hedge]

je parle francais aussi

[thorn jay]

Ye Canadian ass

[karmic moat]

For nilpotents, see eg Vakil’s “Foundations of Algebraic Geometry” section 4.2 “Visualizing schemes: Nilpotents”. Nice pictures there

[thorn jay]

(said with love)

[narrow temple]

Thanks, will check this out now

[karmic moat]

https://math.stanford.edu/~vakil/216blog/FOAGjul2724public.pdf

Here’s the pdf

Btw Spec A is the set of prime ideals of A

It has some topology on it called the Zariski topology which is why he gets to say like “closed points” and “irreducible closed subsets” and stuff but you can ignore that for now

[thorn jay]

To do b take the idea of a and generalize it

[karmic moat]

^ and maybe the name “Euclid” will ring a bell

[thorn jay]

Euclid's parallel postulate?? Will we have to prove the independence of some proposition from some set of constructed propositions satisfied by Q^x?????

[karmic moat]

True…

[thorn jay]

Or probably any permutation of the set of primes

Don't see any reason why that should fail

Then I'm guessing it won't be an isomorphism

Won't be a homomorphism

f(8) * f(2) = 14 * 2 ≠ 49 = f(16)

[velvet hull]

oooooo

[south patrol]

Also like you can check that 2 = 0 here and i = 1 so there are only two elements

[velvet hull]

what's the basis?

[potent condor]

You’re correct

Yes

Lmao how do u know that

[elfin wraith]

Yur

[narrow temple]

Thanks that was a helpful read

[karmic moat]

np

[tardy hedge]

I guess in a PID so that irreducible elements are prime, and prime ideals are maximal ..

[velvet hull]

it is true that F[x] is a euclidean domain, but I don't see how that's useful here

so?

PIDs can have finitely many primes

just pick a field

sure, but then you need to prove that there are infinitely many maximal ideals

and that does not sound easy

go back to primes

hint - ||proof by contradiction||

[thorn jay]

||Can this be done with field extensions stuff?||

[velvet hull]

ahh, that's a pretty cool proof, I think it works

||if there are finitely many primes, adjoin all the roots of those primes to get a finite extension K of F that is algebraically complete. But then K is finite which is not possible||

[thorn jay]

||yes that's the proof I had in mind I wanted to ask "can this be done without field extensions lolololol||

[velvet hull]

wait, I don't think this proof is watertight

in particular it's true that ||every polynomial in F[x] splits in K, but I'm not sure that every polynomial in K[x] also has to split||

[south patrol]

Or ||use Euclid's proof lol. I believe it goes through verbatim||

[thorn jay]

||alternatively, there must be an irreducible and hence polynomial for every degree as there are finite fields for all orders p^n, thus there are infinite primes in F[x]||

[south patrol]

or is this what hchan meant

You can even explicitly count how many irreducible polynomials of degree n there are I believe lol

and then just show that sum is oo

which is funny

[velvet hull]

can someone show this? I think that's a gap in the proof

[south patrol]

it'll be that algebraic over K means algebraic over F, hence in K

[velvet hull]

oh, right

lmao basic galois skill issue 💀

[thorn jay]

Istg

[south patrol]

The proof of this is much harder in the infinite field case

[thorn jay]

Get better at transitivity

[thorn jay]

Proof of what

[south patrol]

that there are infinitely many irreducible polys in F[x]

[velvet hull]

what do you mean

look at the hint

[south patrol]

oh i didn't realise that was already mentioned. but ye i was making a joke lol

[thorn jay]

Infinitely many primes :D
With degree > 1 D:

[velvet hull]

I wasn't talking to you lol

[south patrol]

lol addition is easier than multiplication so let's only show addition works

[thorn jay]

So me

[south patrol]

easier just to note it is self-inverse tbh

ye

though like ig there is not much to this

what does "prove directly" even mean here lol

It seems hard to find a way to prove it indirectly

[thorn jay]

There really isn't, it's like basically reproduction of definitions

[dull ginkgo]

I think it was an exercise in Jacobson to show that the number of irreducible polys of a given degree can be shown to be the Dirichlet-convolution-inverse of some nice sequence

Yeah n*Irr(n,p) is the inverse of p^n lmfao

[tall igloo]

yes this is equivalent to there being such an injective group homomorphism

well dont do that, you should trust your own argument!

why do you dismiss a group of order 21 being in A_10? its been a while since ive had to think about alternating groups haha

i see, so youre just checking that you can reduce the question to checking this

[minor gazelle]

That’s a wrong proposition I deleted it already. I didn’t check subjection (🫣🫣) i have been quite addicted to games but i am fixing them now 🥰

[wicked patio]

There is a group of order 21 in A_10 though...

[tall igloo]

[deleted to not give away the answer]

sounds right

[wicked patio]

yes

[minor gazelle]

I bought a book though it looks elementary enough for foundation, coming up random proposition to prove usually lead to deep confusion afterwards. I thought it had been a good idea saving money 😭

[rocky cloak]

There are ways to get books for free though. Like the library, using free resources or methods of more disputed legality.

[minor gazelle]

I have too much discipline issues with e book 🫣🫣

[rocky cloak]

How so?

[minor gazelle]

I got my book with a discount so I guess it’s worth kinda.. and algebra is fun 🥰

I don’t know if it’s e book I just feel like why would I even read it

[rocky cloak]

Yeah, if you like books by all means. I meant if you need to save money in the future.

[south patrol]

Print it out using a public library or smth ;chad:

textbooks too expensive

[minor gazelle]

Indeed, i usually like instant-food style study

Like I watch a bit of YouTube video and I felt so confident and then I messed up, and ebook made it even worse for my poor discipline 🫠🫠

[south patrol]

Lol wdym

[minor gazelle]

Like 10 min YouTube vid and inspired feel so great and then do puzzles and reveal later that there’s a lot of gaps in knowledge

[rocky cloak]

I'm still a little confused. Does the simple fact that it's an ebook dismotivate you or is it something about the format?

[minor gazelle]

It’s just that… I am not disciplined enough 😭😭

[south patrol]

Ye

Or is it just so you arent tempted to use the computer for other things

I understand that lol

[rocky cloak]

A useful trick could be that a homomorphism G -> S5 is the same as an action of G on a 5-object set.

So one can think a little about what such possible actions could be

[south patrol]

Q8 is a country not a group

[rocky cloak]

If you know what the 2-sylow subgroup is then that would solve it yeah

[karmic moat]

Yeah this one was awesome

[tall igloo]

unironically want to do this for HA.pdf

[karmic moat]

W

[rocky cloak]

Kuwait

[south patrol]

Lmao

[rocky cloak]

Pronounced the same

I'm not sure where 3 enters into the picture.

The idea is Q8 has order 8, so if there is such an injection then Q8 would be the 2-sylow subgroup.

But this path presumes you know (or can determine) the 2-sylow subgroup

[south patrol]

Wdym

[rocky cloak]

Yes the 2-sylow subgroup is the same

[south patrol]

Oh lol

[rocky cloak]

Yes the 2-sylow subgroup has order 8

[south patrol]

I mean sure

I was confused cause u also said 8 = 2^3

And assumed that 3 had smth

[rocky cloak]

S4 has a subgroup of order 8

[south patrol]

Lol im so confused

[rocky cloak]

I think I'm confused as well

[south patrol]

Wwhat do you mean by this other than that that the 2-sylow group has order 8

[delicate orchid]

because it's a 2-group

[south patrol]

You always say p-sylow

Not p^n - sylow

[coral spindle]

8 is not a prime

[rocky cloak]

They are indexed by primes

[coral spindle]

Q_8 also has a Sylow 7-subgroup. It is of order 7^0.

[south patrol]

But now yhis makes sense

[rocky cloak]

It would be kinda inconvenient if you need to change the name of the p-sylow subgroup depending on it's size

[delicate orchid]

lets stop bickering about semantics and show that the sylow is D_8 already

[rocky cloak]

I much prefer the thinking about group actions of Q8 argument, but if this is the path we're going...

[delicate orchid]

my favourite group is the Sylow 71-subgroup of the monster

[delicate orchid]

we bringing out the Burnside ring for ts one?

actually no wait I like this better too

[rocky cloak]

I won't, you may if you like

[thorn jay]

Found the problem interesting and an approaching it like this dw 💖

[delicate orchid]

I have examined the table of marks in my third eye

dunno what this means cause there definitely isn't an isomorphism between a group of order 8 and of order 120. You need to show that Q_8 isn't the 2-Sylow of S_4

[coral spindle]

A while ago I found the smallest n such that Q_8 embeds in S_n. I cannot remember what it was or how the proof went. I think it may have been n=8... but I don't recall.

[delicate orchid]

here's my hint: all subgroups of Q_8 are cyclic

[delicate orchid]

it definitely isn't 6

and therefore isn't 7, I believe 8 tbh. Once you have C_2 \wr C_2 \wr C_2 you should be able to embed loads of stuff

[rocky cloak]

It is indeed 8, I guess that raises an interesting question:

Which groups of order n don't embed into Sm for m<n?

[delicate orchid]

nvm it's obviously 8 hahahahahaha

regular representation

[coral spindle]

I think this is very hard to answer

I remember looking this up, there is a name for this invariant

and it's quite hard to compute at least

[delicate orchid]

cyclics of prime order take it or leave it

[coral spindle]

I think I deleted my notes on this :(

[delicate orchid]

ok yes that's an injective map, why is it a group homomorphism

[delicate orchid]

for p-groups we might be able to think about Q-representations and lift back to the burnside ring in a controlled manner

well they aren't because the 2-sylow is D_8 and not Q_8

hence why I said show that the 2-sylow isn't Q_8

or try to show that it is ig if you want no external info

[delicate orchid]

I gave you a hint

[delicate orchid]

I'd have to remember what the degrees of the idempotents to say anything meaningful from here but the gist is that the linearisation map B(P) -> R_Q(P) sending a P-set X to Q[X] is split surjective

and there's a section that preserves the obvious grading on both corresponding to the idempotent associated to the trivial subgroup. So this question is probably related to groups where none of the rational representations are faithful

@coral spindle am I close lol

ur staring at an element of order 21

it absolutely does

interesting to think about all of the possible cyclic groups you can get in S_n

nice little combinatorial problem for you

and me too tbh

[coral spindle]

I haven't been following sorry

[delicate orchid]

that's ok 💔 I know you hate me anyway 💔 it's fine 💔 I won't cry that much 💔

[sonic coral]

the order of an element in S_n is the lcm of the cycle types

assuming it’s disjoint

[coral spindle]

It's easier to calculate the smallest S_n containing some C_m rather than the largest C_m contained in some S_n

[delicate orchid]

there's an element of order k in S_n if and only if k is the lcm of the parts of some partition of n. So for 10 we have "the boring ones", then ones like 7+3, 7+2+1, 5+4+1 - giving elements of order 21, 14, 20 to give an example

yur

[sonic coral]

and then you just need to make sure it’s even since you care about A_n

[delicate orchid]

are you sure lol

[coral spindle]

yes 👁️

[sonic coral]

the first definitely sounds easier than the second

[delicate orchid]

they both seem like they'd be in the same complexity class unless I'm stupid

[coral spindle]

for the latter, you just factorise and add

for the FORMER

rather lmao

I suppose given that factorisation is hard...

but the other way around is kinda tricky for human beans

[delicate orchid]

you just factorise
Jarvis google "NP-complete"

wait is it complete or just NP

now I'm actually going to have to google

I have trolled myself

[rocky cloak]

Figuring that out might be good progress on that million dollars

[white oxide]

Is the second answer no? For instance, $7 \mathbb{Z}$ is a maximal ideal of $\mathbb{Z}$, but the ideal generated by $(4)$ in $(\mathbb{Z}/7 \mathbb{Z})[x] \cong \mathbb{Z}[x]/7\mathbb{Z}[x]$ is proper, hence not a field and so $7 \mathbb{Z}[x]$ is not maximal

[cloud walrus]

okeyokay

[white oxide]

Wait nvm (4) might not be right lol

[coral spindle]

(Z/7Z)[x] is isomorphic to Z[x]/7Z[x]... can you justify that?

[white oxide]

Yeah I proved it earlier for the first part of the problem

[coral spindle]

Oh wait it's 7(Z[x]) you mean, not (7Z)[x]

[white oxide]

O I meant (7Z)[x] lol

I proved that A[x]/p[x] \cong (A/p)[x]

[coral spindle]

Am I losing it folks

[white oxide]

this is what sweet potato claimed

[coral spindle]

I am full of it

[thorn jay]

Yes

[coral spindle]

So 4x9 = 5x7 + 1

so I infer that 4x9 = 1 in (Z/7Z)[x]

So I can see that (4) = (1)

Not so sure about that being proper

[tardy hedge]

great pfp

[white oxide]

Lol I forgot elementary facts about the units of Z/nZ

anyways I'm sure there's a proper ideal...

[rocky cloak]

So it's true that the polynomial ring isn't a field.

But (4) isn't a proper ideal, not sure where you got that.

[thorn jay]

My profile banner is red because of how red this mf guy is 😭

[rocky cloak]

(x) would be a proper ideal though

[coral spindle]

I think (x) would be a good thing to look at

[tardy hedge]

bro is always crying

[coral spindle]

Sniped wtf

[rocky cloak]

4*2 = 8 = 1

[coral spindle]

🗿

[white oxide]

Wait is this true for any field lol

[coral spindle]

Listen it's very late

[white oxide]

wait nvm

no proper ideals in a field

Oh wait every nonzero element of Z/pZ for p prime is a unit huh....

[rocky cloak]

That is correct

[thorn jay]

That tends to happen in fields

[white oxide]

lol

[tardy hedge]

what happened to JP!?!?

i miss him already

[thorn jay]

Idk I think this pfp is just better in general 😔

[tardy hedge]

who dat

dat aint JP

[rocky cloak]

Might be what makes it better

[tardy hedge]

jp is the best canadian

[thorn jay]

Artist called The Caracal Project, I saw him live B2B with IMANU and Buunshin which was insane, and took that picture

[tardy hedge]

#2 is drake

[rocky cloak]

Nathan fielder is my top contender

[thorn jay]

Kian is the best Canadian

[tardy hedge]

:0

[elfin wraith]

Jordan Peterson?

[tardy hedge]

Yas

[elfin wraith]

I just saw the #2 is drake and so I’m assuming/praying you’re taking the piss

[delicate orchid]

#1 justin bieber

[elfin wraith]

Holy 2012

[thorn jay]

My #2 favorite Canadian is Jonald

[wispy light]

John Petrucci is American, though.

[south patrol]

Goat

[nimble folio]

\begin{exercise}
        Suppose $G$ is a nontrivial finite group and $H,K \unlhd G$ are normal subgroups with $\gcd \left( |H|,|K| \right) = 1$.
            \begin{enumerate}[label = (\arabic*),itemsep=1pt,topsep=3pt]
                \item Define a nontrivial group homomorphism $\phi:G \rightarrow G/H \times G/K$.
                \item Prove $G$ is isomorphic to a subgroup of $G/H \times G/K$
            \end{enumerate}
    \end{exercise}

[cloud walrus]

clubsoda14

[nimble folio]

What does it mean to prove that G is isomorphic to a subgroup of G/H x G/K?

[misty holly]

Exactly what it states

That there's a subgroup of G/H x G/K which is isomorphic to G

[nimble folio]

yea i figured it out

its just first isomorphism theorem

[misty holly]

Indeed

I did this exact question a few days ago

A fun problem is figuring out exactly when G is isomorphic to G/H x G/K

[tall igloo]

H = 1, K = G 👍

[rocky cloak]

The canonical map should be an isomorphism iff G = HxK, but they can be isomorphic in different ways I guess

[cinder onyx]

Quick question, I know that the kernel of any surjective homomorphism from a commutative ring to a field is always a maximal ideal, but is the converse also true? For any maximal ideal I of a commutative ring R, does there exist some surjective homomorphism φ:R->F s.t. kerφ=I?

[tribal moss]

If it's just an arbitrary isomorphism we can even have G=H×H and let H and K both be H×1. Seems to be tricky to imagine a nice general condition that example satisfies.

[tribal moss]

If I is maximal, then R/I is automatically a field, yes.

[cinder onyx]

oh god I'm stupid nevermind 😭

thank you

[tribal moss]

Good occasion to internalize that "surjective homomorphism" is the same thing as "quotient by the kernel".

[late matrix]

R -> R/I by quotient map

oh someone answered mb

[cinder onyx]

Yeah I was being an idiot lol

wait hangon

hello

[late matrix]

lol hi mati

[misty holly]

I meant the canonical map

[molten jewel]

how fundamental is group theory in modern math?:

[tall igloo]

yes

[molten jewel]

any good intro resource? i'm in the middle of bachelors degree in math

[thorn jay]

howve you not gotten group theory then

with the postgrad role lmao

[molten jewel]

you need the postgrad role to access some areas of the server

[thorn jay]

its all good lol i dont care it was just kinda funny to see someone with the postgrad role asking how important group theory is

[tardy hedge]

Dummit and foote ig

[molten jewel]

thnks

[tardy hedge]

Yea there are a lot of intro algebra books

[tall igloo]

huh interesting I think of the roles as your position academically lol

[thorn jay]

well ill say that i am nowhere near the level of my role lmao

[tardy hedge]

I dont have postgrad role

[thorn jay]

i am the sussus amogus

[tardy hedge]

Im pre sure i have access to channels

[molten jewel]

idk how to remove

[tardy hedge]

Its ok Kidd

You’re just a Kidd, after all

[molten jewel]

a very poor kiddo

[thorn jay]

damn congrats on the child, after all a joke that horrible can only be made by a dad

[tardy hedge]

Thank you 🥹

[tall igloo]

also postgrad always confuses me since I see it and think I'm talking to postdocs 💀

I say this having the role and not being a postdoc

[south patrol]

||Physicists could do this||

[velvet hull]

I have no idea how I got the pending postgrad role

[south patrol]

Me neither

Jk.

[tall igloo]

||Also some applied or CS-y math people lol||

[south patrol]

But i think basically when you join and if u say ur interests then it signs you up automatically

Maybe though dont most maths courses have group theory in curriculum lol

Hm

[tall igloo]

idk an embarrassing amount of math majors from my undergrad never learned group theory

[south patrol]

Oh lol this is a mandatory course for us

[thorn jay]

i think knowing basic abstract algebra is important

[south patrol]

Actually our physics LA course ostensibly makes you learn what a group is

Or did anyway

[thorn jay]

just as is knowing analysis

[tall igloo]

if the things are like applied or cs stuff then id partially agree but like,,, idk the point of an undergrad math degree is to give you exposure and basics in multiple areas of math

[kind temple]

i disagree with this

[tall igloo]

What alternative do you propose

[tall igloo]

based

[south patrol]

I assumed it would be everywhere ngl

[thorn jay]

how do they know without trying group theory

[south patrol]

How do you really know what you are interested in at the start of a degree

Like you probably know basically nothing about maths if you begin

[thorn jay]

everyone should get a taste of all sorts of math so they can make an actual informed decision

[south patrol]

At least i did lol

[tall igloo]

Sets are "modules" over the """field of one element""" so if you care about anything that's a set you should care about modules and therefore about groups

[south patrol]

And now i am a phd student i know nothing again

[tall igloo]

mant such casss

[tardy hedge]

My undergrad math major no algebra required either

[tall igloo]

Me when I need to do literally any computation

[tardy hedge]

Honours math program had all tht stuff

[thorn jay]

chat should i wipe my memory so i know as much as everyone else when i start uni in a couple months

[south patrol]

Real

[tardy hedge]

Yes

Bro wtf classes r u taking in uni

Dont tell me intro to abstract algebra

[south patrol]

You should learn some SERIOUS computations

[tall igloo]

"someone probably has a spectral sequence somewhere that computes this" - me, 5 times a day

[south patrol]

There is only one computation relevant to me and you can prpbably guess what it is @ hatfuel

[thorn jay]

havent decided yet, wanna talk to the student advisor and see what higher year classes i can take

[kind temple]

wait ur in hs?

[south patrol]

that's why enpeace knows about UA

Jk sorry easy poke

[kind temple]

crazy

[tall igloo]

Is it the fibanocci stuff lol

[south patrol]

But i am impressed

[south patrol]

No i mean serious math

Lol

[tall igloo]

lmaoo

[south patrol]

You can probably guess tho

[tall igloo]

Is it the morita stuff you were saying earlier

Idk how computational that is

[thorn jay]

i already talked to her and she said its okay i just have to sit down and talk to her abt it which is awesome

[south patrol]

Nah i mean like

THH(Fp) lol

[tall igloo]

ah true

We love THH(Fp)

[south patrol]

I am confusion about smth now

[thorn jay]

truly the easiest branch of math

[south patrol]

But every time i wonder about smth which is possibly not hard i am scared it is well-known and Ill embarrass myself

Roughly derived algebraic geometry

[limpid plover]

Is contemporary abstract algebra a good book?

[tall igloo]

REAL

[south patrol]

Uhh integrated masters, tho i did learn it in ug

[south patrol]

Phew

Lol

Not just me!

[tall igloo]

i think it is like,, almost everyone lol

[south patrol]

True lol

I have smth rn where it is like seemingly claimed without proof in a paper if you squint and unravel difficult notation

So idk if it counts as proved or not

Im not american

[thorn jay]

man assumes everyone to be american

hoW

i am just chronically online i think

[tall igloo]

To be fair Britain shouldve been annexed by America

[thorn jay]

everyone should have been annexed by the Netherlands

maybe that'll fix y'alls infrastructure

[tall igloo]

it's so cruel there's a whole country of people who have to use that accent every day

[tall igloo]

Lmaoo yeah

[thorn jay]

yeah but thats fucking Amsterdam 💀

one of the most expensive cities i think

and a shithole anyways

never in my life would I want to live there

is it the best or just the most elite because its in amsterdam

Nijmegen is great

[thorn jay]

great culture and underground scene

[south patrol]

Yh

Though usually ig you say BSc for sciences

Tho i do not have one

[south patrol]

Real

[south patrol]

It gets really tiring

Bro is an expert

Ah ok

Rip

Still surprised so many voted for it bruh

[tall igloo]

i spent one week in Cambridge UK earlier this summer and decided I was glad they rejected me for grad school LOLOL

[south patrol]

I am v surprised they rejected you but then there are also no homotopy theorists there

Presumably ORW

[tall igloo]

I'm not surprised, Oscar has very high standards. Also I think they want a masters degree

[south patrol]

Ah ok

I also applied to him lol but again not rly homotopy theory

[south patrol]

But yeah i imagine itd be the masters

Idk

I enjoy the slander

[tall igloo]

Well I also had to submit some like research question and idt it's one Oscar cared about 💀

[south patrol]

Oh bruh

I do not remember doing that lol

Oh no i thjnk i did

[tall igloo]

i applied to Cambridge UK just in case some other dude (you know who) stayed there but mans wasn't even involved in the admissions process

[south patrol]

It is funny tho cause in the UK like often they would ask u for likw a thesis topic and plan or smth

But then math dept specifically will say dw

Nah

[south patrol]

E.g. Warwick had this iirc

And for my uni i had to give a thesis topic but i just said "Homotopy theory / AG" and apparently that is now my working thesis title

[tall igloo]

the two genders

[south patrol]

😭

And Durham lol

And Warwick

Yeah but they wanna sneak in

Lol

[tall igloo]

is imperial a real school or just a business school

[south patrol]

Real school

Very good

[tall igloo]

huh interesting

not really 💀 somehow that's all the PR of theirs I've seen

[south patrol]

Lol

But yeah is v strong for like sciences and math

[tall igloo]

hUH

[south patrol]

Idk about finance sorry lol. But for economics stuff sure i think it is v much up there

[south patrol]

Yeah it is v good lol

Arguably best in uk for engineering for example

Ah chem eng

💀

Oh LOL

bruh

Nah what i was gonma day is chem eng imperial iirc was like the highest requirements of any degree im the UK

[tall igloo]

this is why math good. Hagoromo chalk is shelf stable

[south patrol]

Because they would basically keep upping the requirements if you did more than 3 a levels

I mean A level requirements

[tall igloo]

cries in Python code calculating the universal formal group law

[south patrol]

Oh yeah i guess Oxbridge medicine has high GCSE (or equivalent) requirements

But not the highest A level

Yeah probably

I think the vibe is like at Cambridge you are much more likely to get an offer but then STEP more than cancels that out

I should add that i applied to (and did) Physics for a year lol

Before swapping to maths

Nah

Same as ug

Though it is common to move begweem the two unis

And just stayed

I meant cause u went bruh

But i mean a lot of phd students here did ug at cam

And vice versa

[tall igloo]

wow imagine getting accepted to grad school at your undergrad

[south patrol]

I was the year that did not do exams

[south patrol]

Tbf i think in the US there is more of a bias against staying on than here

That is my impression anyway?

PAT lol

[rocky cloak]

Is that not common?

[south patrol]

This feels like pushing it but yh ig we do compared to Europe

[tall igloo]

yeah depends on the school and major but yeah

[south patrol]

This convo is just reminding me of high school stress and people obsessing over uni admissions tho

😭

I did not like the competitiveneess

Can i share something embarrassing

I just made some stock using vegetable scraps etc

And wanted to drain it

[tall igloo]

Me when I go to a grad application panel and they're like our undergraduates are the best students out there!!! And then the next sentence is and that's why it's a dual edged sword applying here for grad school since we'll have higher expectations since we already know u 💀

[south patrol]

But instead of draining into a bowl i just drained it down the sink without thonking

[rocky cloak]

Derp

XD

[south patrol]

Yikes

[south patrol]

Yeah

Main annoying thing is i had simmered it for like 20 mins

Lol

[tall igloo]

oof free dinner for the sewer rats

[rocky cloak]

Me when I pull the wrapper of my ice cream then throw the ice cream in the trash

[south patrol]

It's chewsday innit bruv gonna eat some stuff in the sewers

[south patrol]

Have you done this

[rocky cloak]

I have yes

[south patrol]

Condolences

[rocky cloak]

(but I fished it out)

[tall igloo]

It was chewsday yestaday guvna ah ya nuts

[rocky cloak]

Nobody noticed

[south patrol]

Hatfuel do u know about the E_oo cotangent complex

I am curious about smth

Actually it may be ok

[tall igloo]

oh dear I was reading this section in HA today while half asleep

so I know Of it but I don't know it

[south patrol]

Ah it is just how like

If you have a map of ordinary commutative rings u get the topological / E_oo cotangent complex as well as the algebraic / normal cotangent complex

And there are results on comparing them

But if u are not familiar dw lol

[tall igloo]

huh yeah I am not familiar

[south patrol]

Maybe i should talk to a SMA about this

[south patrol]

Ah well the cool / weird thing is that there is a comparison map which is like 3-connective or smth lol

But not an equivalence usually

It is very related to the difference between S and Z

[tall igloo]

I think I saw something in HA near the cotangent section that claimed E_oo rings are morally like boring rings but wiggled a bit

[south patrol]

In the sense that also if you are over Q then it is an equivalence

[south patrol]

Yeah lol

[tall igloo]

What category do these complexes live in

[south patrol]

A-modules

[tall igloo]

yeah this is wild to me

[south patrol]

I think the idea is like

Higher htpy groups should be nilpotents right

[tall igloo]

In what sense?

[south patrol]

Uhh tbf this is of course not literally true in general lol but with examples like the sphere it is true

[tall igloo]

oh well sure

kid named MU

[south patrol]

But like i think this can be a good heuristic

And then you have things like a map of E_oo rings is an equivalence iff cotangent complex vanishes and iso on pi_0

[south patrol]

Actually a better thing is like

The postnikov filtration is made up of square-zero extensions

At least for connective stuff to make myself safe lol

[tall igloo]

ya sure

[south patrol]

And ig that is the idea behind this

[tall igloo]

cries in chromatic

[south patrol]

Tbf connective should not matter here but ye

Here is a fun question

Uhh wait can i remember this

[tall igloo]

So the cotangent complex of an Einf ring R is an R-module you said?

[south patrol]

Yee

[south patrol]

Oh it was like

What ring is it where Tyler Lawson showed it is not an E_13 ring or smth lol

BP?

[tall igloo]

Yes

[south patrol]

But it is E4?

I wonder what the precise k is lol

[tall igloo]

well, we should be careful, Tyler showed itnfor p=2

[south patrol]

Ah sure

Did Senger did it at other p

[tall igloo]

Yeah and Andy also did it for almost all BP<n>'s

There's a few that are still wide opens

[south patrol]

Nice

[tall igloo]

And some that were already known to be E inf

[south patrol]

Lol nice

Quite a snappy thesis with a very nice result

[tall igloo]

I actually forget how Tyler did it, I know Andy's thing was some power operation nonsense

As most of these things go

[south patrol]

i think tyler also did power ops

idk how else one can do this

😭

[tall igloo]

yeah the E infty operad is a horrible horrible object

[south patrol]

compute the space of structures and show it is empty lol

Ngl operads in the formalism of HA still scare me lol

Well just p technical and it is written so generally

[tall igloo]

this is why I have never read that part of HA.pdf and just learn how to form sentences with the vocab words surrounding this lolol

[south patrol]

Lol

[thorn jay]

whats HA?

[tall igloo]

Its food enough for me since I usually only need "admits a frobenius"

Higher algebra, book by Jacob lurie

[thorn jay]

1500 pages D:

[tall igloo]

I kinda wanna run an indexing program on it first, Ctrl F'ing it is a pain

[thorn jay]

abt 3 years ago i think

nah

i know a bit of integration, basics of analysis

i do know linear algebra cuz i just picked it up along the way, and now I know basic point set

but besides that mostly focused on algebra

its cool

[south patrol]

Chad

[south patrol]

Ah ok lol

Frobenius... beautiful

Hmm

[tardy hedge]

they have more classes in other topics?

[thorn jay]

no module theory 😔

[wraith cargo]

Chad absolute frobenius vs. chud relative frobenius

[tardy hedge]

ok

[thorn jay]

universal algebra

teach one

[tardy hedge]

rip

[thorn jay]

define "real music"

[tardy hedge]

ragebait

[south patrol]

lol

Chungus $\delta$-ring structure

[cloud walrus]

Prismatic Potato

[elfin wraith]

She frobenious on my structure till she make quantum computing interesting

[delicate orchid]

U need to put enough distortion and bass EQ on the master so the entire thing becomes a 40hz square wave

[thorn jay]

actual phonk producers

honestly have done that a couple times its funny as hell

esp when you put a limiter after

[fervent solstice]

ok uh

i have little experience with commutative algebra

I'm trying to formalize the same thing as "Y divides F means that there exists H such that F = YH" but with ideals
so something like "(F) is a subset of (Y) means that there exists I such that (F) = (Y) inter I"

but how could I express such an ideal I? using the generators it's simple to take I=(H) but this implies using noetherianness

[coral spindle]

I = (F)

[fervent solstice]

yeah also that

[coral spindle]

You should frankly not be thinking of it this way.

[fervent solstice]

idk how to not lose information

[coral spindle]

To contain is to divide in ideal land -- this isn't because there is some parallel in divisibility, but because if a principal ideal contains another principal ideal, this is precisely to say that the generator of one divides the other

So you should take ideal inclusion as a "multi-element" divisibility

(rather than some kind of direct translation)

[thorn jay]

well I mean, ideals were at first discovered to study divisibility right

[coral spindle]

They were defined to encapsulate precisely this "multi-element" divisibility idea.

I could elaborate on this but it's not terribly interesting.

Anyway

[coral spindle]

Besides, when you see actual prime decomposition of ideals down the road, the thing you need to look at is the product of ideals rather than intersection really

[thorn jay]

primary decomposition is in terms of intersection :P

[coral spindle]

thank goodness I'm thinking of prime, rather than primary, decomposition

[fervent solstice]

The reason I was thinking about that is that Fulton's Algebraic Curves proves the unicity of intersection numbers (in a very simplified setting, i.e. over k[X,Y]) having defined them as something that depends on a pair of polynomials (those who generate the curves that are intersecting). Since curves really only depend on polynomial ideals and not on single polynomials, I was trying to figure out a proof for a definition of intersection numbers that'd make them depend on ideals and not single polynomials

However there's an induction step down the road where the polynomial is split into two (the F=YH business). I was trying to figure out the analogous for ideals

[thorn jay]

but is it unique

🤨

[coral spindle]

I'd really prefer fewer peanut gallery comments

[coral spindle]

Well this seems very serendipitous, because you seem to be looking precisely for a factorisation of ideals

In algebraic geometry this is really saying that every variety decomposes into a finite union of irreducible varieties, which I think is what you are needing for your proof

Indeed this is the Lasker–Noether theorem about the primary decomposition of ideals

[fervent solstice]

So this would work only thanks to k[X,Y] being Noetherian?

[coral spindle]

Yes... but you should think of Noetherianity as an extremely reasonable requirement. Unless you are trying to say that Noetherianity is the key here, in which case yes

[fervent solstice]

Yes, I guess I'm also still trying to make up a motivation for (thinking about) ideals in non-Noetherian rings

but that solves it, thanks

[coral spindle]

Well while this particular geometric interpretation no longer applies, like I said you still have a perfectly reasonable interpretation of ideals as things that describe generalised 'multi-element' divisibility

nothing changes on that end

I don't know if people do algebraic geometry with non-noetherian rings

[fervent solstice]

what do people even do with those?

[coral spindle]

Idk, cry?

We do care about some non-noetherian rings, I'm exaggerating. They're just trickier to deal with.

[south patrol]

Non-noetherian rings appear a lot in mixed/positive characteristic geometry and surrounding math

Because of things like https://en.wikipedia.org/wiki/Fontaine's_period_rings?wprov=sfla1

More generally like these perfectoid rings are important

[south patrol]

Also in this vein like fortunately this statement only requires the space X being Noetherian, and Spec A being a Noetherian space is much weaker than A being Noetherian. So there are more examples (Silly ones include k[X1, X2,...]/(X1^p, X2^p, ...))

(Remark: annoying thing here is that a scheme being Noetherian means a different thing to its space being Noetherian lol)

[coral spindle]

That's crazy I didn't know that Spec A being Noetherian didn't mean A was

That's really cool

[south patrol]

(Note again should be Spec(A)'s space being Noetherian aha)

[south patrol]

I guess I now wonder though how this relates to Lasker--Noether

like because the link between Lasker--Noether and decomposition into irreducible subvarieties works for (classical) algebraic varieties and kinda uses the Nullstellensatz etc

ig the general thing is that irreducible components of Spec(A) correspond to minimal primes, and then Lasker--Noether tells you about those in quite a strong way

[nimble folio]

Let $G$ be a group, $a \in G$, and $|a| = n$. I'm trying to show that $|a^k| = \frac{n}{\gcd(k,n)}$. I was able to show that $\left( a^k \right)^{\frac{n}{\gcd(k,n)}} = e$. I'm having trouble on the rest of the proof: \nl
            
            \noindent Let $0 < m < \frac{n}{\gcd(k,n)}$. Then $m\cdot \gcd(k,n) < n$. Since $|a| = n$, it must be that $\left( a^{\gcd(k,n)} \right)^m \neq 0$. \nl
            
            \noindent This is where I'm stuck. We showed in a previous problem that $\langle a^k \rangle = \langle a^{\gcd (k,n)} \rangle$, so it follows that:
                \begin{equation*}
                \begin{split}
                    |a^k| = |\langle a^k \rangle| = |\langle a^{\gcd(k,n)} \rangle| = |a^{\gcd(k,n)}|.
                \end{split}
                \end{equation*}
            But is this enough to deduce that $\left( a^{\gcd(k,n)} \right)^m \neq 0$ implies $\left( a^{k} \right)^m \neq 0$?

[cloud walrus]

clubsoda14

[nimble folio]

Nvm I'm dumb

[thorn jay]

well, see it this way: Noetherianness asks for finitely generated ideals, but the space Spec(A) being Noetherian means that every radical ideal is finitely generated as a radical ideal, which is to say that it is the radical of a finitely generated ideal. So if you have a non-Noetherian ring with a sparse amount of prime ideals then this should not be too hard to imagine

though i guess you could hope on the niceness of commutative rings to get the equivalence of the two, alas..

[tardy hedge]

Peanut gallery? Lol

[thorn jay]

i was kinda being an annoying ass needlessly interrupting conversation

lol

[gusty thistle]

does anyone have a hint for => ? I dont really see what I could use. Of course g is conjugate to g^{a'/gcd(a',#G)} but is there some sort of result about conjugacy that I should be aware of that allows you to go further?

[tribal moss]

|g| divides |G| so an exponent that is relatively prime to |G| is also relatively prime to |g|.

Wait, sorrry, I'm stupiding. That's the wrong direction.

Instead, how about using Bezout's identity to add a multiple of |g| to a' until you reach something coprime to |G|?

[gusty thistle]

ill try it

[tribal moss]

Correction, not Bezout.

But just CRT should let you find a multiple of |g| which added to a' will eliminate every prime factor of |G|that was not in |g|.

[gusty thistle]

so the goal is to find an x such that gcd(a'+ x |g|, |G|) = 1 ?

I see how that would solve the problem hmm

[tribal moss]

Yes.

[gusty thistle]

Do you have a hint on how to rewrite that into a system of congruences ?

[tribal moss]

You could look for a number that

• is 0 modulo |g|, so so you can write your number as x|g|
• is 1 modulo every prime that divides both |G| and a', so this prime will not divide a'+x|g|
• is 0 modulo every prime that divides |G| but divides neither a' nor |g|, so this prime still won't divide a'+x|g|

[prime sigil]

How does one prove that
Z + Z*i / (2-i) is isomorphic with Z_5 using the first morphism theorem for rings? I've tried to find a ring morphism that projects 2-i onto but I'm stuck. ( Z is the set of whole numbers)

[dull ginkgo]

Is this a problem from Jacobson

[tribal moss]

Is that Z[i]/(2-i) ? Usually one would read the slash as binding tighter than the +, but that won't make sense here.

[prime sigil]

It's to denote the quotient group

[tribal moss]

Also note that "the first morphism theorem" means different things for different authors. See the helpful table at https://en.wikipedia.org/wiki/Isomorphism_theorems#Note_on_numbers_and_names

[tribal moss]

Yes, but usually one would parse your phrasing as Z + (Z*i/(2-i)), which doesn't make sense.

[prime sigil]

The one I'm referring to is theorem A

[prime sigil]

Ah yes, I didn't use brackets because they aren't used in my course book but it would be
(Z + Z*i) / (2-i)

[rotund aurora]

try finding a map Z[i]-->Z/5Z

[tribal moss]

And Z[i] is Z[x]/(x²+1)

[gusty thistle]

ok that works that is pretty clever

i need to learn more elementary number theory haha

what made you think of that ?

[gusty thistle]

no dummit & foote (ch. 18. section 3)

[tribal moss]

Fuzzy intuition there must be a way to write g^a' as g^a for some qualifying a, plus (as shows in my posts) a few rounds of trial-and-error to find an actual rigorous reason that would be true.
[this should have been a reply to "what made you think of that"]

[gusty thistle]

cool

thank you

[unreal sedge]

Hello guys, I'm studying introductory abstract algebra on my own and encountered this example of a binary operation. I was wondering what a good example could be to supplement this? The example I'm thinking of is the following:
Let $X$ be the set containing all real polynomials. Then, we define $* : X \times X \rightarrow X$ as a binary operation that composes two elements of X into another polynomial $p(x)$ in X.

[cloud walrus]

whyhello

[velvet hull]

sure, that's a working example

[unreal sedge]

Alright thanks

[velvet hull]

although it's pretty specific

[unreal sedge]

Wdym by specific?

What would be a more "General" example?

[velvet hull]

because Map(X) includes any possible map X -> X

and you're only considering polyomials

[unreal sedge]

Oh I see

[velvet hull]

which is a very specific subset of the functions from R to R

[unreal sedge]

So the general case would just be $f : X \times X \rightarrow X$

[cloud walrus]

whyhello

[tribal moss]

But it's definitely useful to be aware that polynomials are closed under composition!

[unreal sedge]

I see.

[velvet hull]

one way to interpret what you are doing, when you quotient out by an ideal, is that you are imposing the relations given by your ideal on the base ring

so, for instance, quotienting out by (2-i) in Z[i] is exactly the same as declaring that i = 2 in Z[i]

and think about that for a moment

[unreal sedge]

i love algebra guys it might just be better than analysis

[broken pollen]

have some guts and say this in an analysis channel

[tribal moss]

It only makes it somewhat confusing that i is a particular complex number which we know is different from 2 ... which I why I suggested writing Z[i] as Z[x]/(x²+1) instead, so there are fewer preconceived notions to ignore.

[unreal sedge]

heh...

[prime sigil]

Would the function given by
Z[i] → Z/5Z: a+b*i ↦(a^2 + b^2) mod 5 be the one to use?