groups-rings-fields

[summer path]

Do enough and you'll find both hard

[next obsidian]

I’m hard af. I’ve been raised on these streets

I eat nails for breakfast

[solemn dew]

prove it

[open sluice]

I think all math is good and we should appreciate the variety of fields in this subject

[solemn dew]

well thanks to abstract algebra we can actually go to details about fields

so every other mathematical fields should be thankful for this

not like @formal ermine who thinks analysis is harder

they just make a bunch of axioms of what a field is in chapter 1 of rudins book

pfft

[formal ermine]

I read "I'm hard af" and was like tf chmonkey?

[untold turret]

maximal subrings of a field wrt. strict inclusion are valuation rings?

[white oxide]

can somebody help me understand how the sentence starting with "if we" implies that the polynomial is solvable by radicals?

are they just saying we can construct the extension by radicals by adding, subtracting, multiplying and dividing to obtain the zeros of f(x) and appending it to the prime field F?

i'm a bit confused

like for example x^2 - 2 over Q

we take sqrt(2)

and obviously that raised to an integer multiple is in Q

and then we construct Q(sqrt(2))?

[untold turret]

localization (R-X)^{-1}R is local. Is X a prime ideal?

[white oxide]

any heroes?

[formal ermine]

X is an ideal?

[rocky cloak]

Yeah, you alpha_1 is some nth root, then you take some combination of addition multiplication, division etc of alpha_1 and take an nth root of that to get alpha_2 etc

[untold turret]

don't know!

[formal ermine]

wdym?

you asked the question

[untold turret]

it's not given

[white oxide]

ah ok that makes more sense thanks!

[untold turret]

It's a particular case of a larger proposition I'm trying to prove. Let U be set of units of R. (R-U)^{-1}R = Frac(R) implies R local?

[rocky cloak]

Is there any assumptions on X a priori? Like X being an ideal or R-X being multiplicatively closed?

[untold turret]

Suppose X is the set of units of R

[elder wave]

if it's mult closed it should be true right

[wooden ember]

This is false : take Z no?

[rocky cloak]

If X is the units, it's definitely not an ideal

[elder wave]

ah nvm

what i was gonna use depended on it being an ideal

then you can use the correspondence

[untold turret]

right on second thought it's absolutely not an ideal heheh

[elder wave]

of max ideals that have empty intersection with the localization subset

[warm wyvern]

chmonkey be chmonkin'

¯_(ツ)_/¯

[formal ermine]

my third favorite correspondence

[delicate orchid]

my favourite correspondence is the one I have with your mother

[untold turret]

uncalled for

why must everyone speak of and to his mother

[rocky cloak]

This is just true for any domain though right?

[delicate orchid]

dunno if anyone mentioned this but how are you localising here? if X isn't prime then R-X isn't multiplicatively closed

[untold turret]

yeah, just realized. if we localize using nonunits as multiplicatively closed subset, then we're just inverting all nonunits

[rocky cloak]

Just use universal property™

[delicate orchid]

b-b-b-b-b-b-b-b-but

[untold turret]

this is what i'm trying to make sense of

how the heck are they concluding R has only one nonzero prime ideal

[delicate orchid]

if all of the non-units form an ideal then you have a local ring

[white oxide]

maybe the wrong channel for this but i always get aired in #book-recommendations , but does anybody know a good text on galois theory for someone who already has had an encounter with the subject

[delicate orchid]

intersection of ideals is another ideal

now, how are they going from \subseteq to \eq is something I must ponder

[untold turret]

\subseteq to eq is automatic if nonunits are in a single ideal

[delicate orchid]

oh right, if that intersection is a proper ideal that contains R-U(R) they must be equal by maximality yeah

[untold turret]

but i don't get where the \subseteq comes from in the first place

[delicate orchid]

R-U(R) is the biggest a single ideal can possible be blah blah

[delicate orchid]

appending any inverse to any non-invertible element of R turns the entire thing into a field

this must have something to do with it

[formal ermine]

the stacks project page

https://stacks.math.columbia.edu/tag/09DU

https://stacks.math.columbia.edu/tag/0BMI

[white oxide]

oh coo lthanks

why sully

[elder wave]

i retract my sullies

i didn't read that you're already familiar

[formal ermine]

it's a consice summary of (most of) everything you need to know

the one on infinite galois theory is also nice

[white oxide]

epic i'll take a look at it after i finish my expository encounter with galois theory

[elder wave]

I'd still go for some textbook over stacks project

[formal ermine]

btw the infinite one requires you to know about limits and topology

[untold turret]

how can you bring yourself to believe this

[coral spindle]

it's a nice reference sometimes

[elder wave]

Yeah exactly

[white oxide]

i know a decent amount of real analysis does that count

[elder wave]

a reference

[white oxide]

granted i'm not good at it but i know the definitions and theorems and such

[elder wave]

not real analysis limits

[formal ermine]

like category theoretical colimits

[white oxide]

oh no clue what those are

so i should prolly review that beforehand

[formal ermine]

if you just wanna know "what they are" then I can recommend the definition in https://people.math.harvard.edu/~smarks/mod-forms-tutorial/mf-notes/galois-reps.pdf from personal experience

[white oxide]

how much category theory should i know beforehand

[formal ermine]

otherwise read a category theory book

[formal ermine]

only what inverse limits are lol

[formal ermine]

this also covers a bit of infinite galois theory btw

[white oxide]

all my category theory comes from a 10 page section in hungerford

[white oxide]

oh ok

[white oxide]

no exercises here huh

ig i can just scour for some online

[formal ermine]

ong ask chatgpt

(please don't actually do this)

[lyric furnace]

I hope this is the correct thread to post in.

TL;DR Given an ANN topology, represented as quiver , the resulting module of the path algebra could be reduced using homological algebra
i.e. clip off the bad parts of the resulting exact sequence of modules.

Then other optimization techniques could be used such as teleportation, on the least significant quivers by transporting the linear transformations associated with the vertices of the quiver representation.

IMO this sounds like an out-there and possibly not realistic in application especially taking account calculating homology groups

if anyone experienced could tell me im my rubber duck finally got the best of me and this is completely useless let me know. I've found papers using similar approaches not but quite tbis way

Note: the Torij functor would yield the r-modules of the path algebra re-representing the quiver

Someone tell my why I'm a moron

[white oxide]

aight bet

[mortal mortar]

what is an ANN topology, what is a Torij functor (I know the Tor functor), can you symbolically express your ideas because so far none of what you said sounds familiar at all to me and I dont know how many experts there are on quiver representations here. You might get a response in #category-theory theory as that is the natural setting for quivers

[lyric furnace]

linear transformations as morphisms and vector spaces (in the context of ANN(artificial neural networks)

for Tor I just mean Tor_ij subscript

i'd like to know if i'm completely dumb before typing something up, y'know? thanks for your reply 🙂 \

[white oxide]

how does the sentence that begins with "since a polynomial" imply that the smallest power of sigma_p^r leaving all the elements of K fixed is the nth power?

[rocky cloak]

If $\sigma_{p^r}^i$ fixes $\alpha$, then alpha is a root of $x^{p^{ri}} - x$, but this has at most $p^{ri}$ roots, so if it's supposed to fix all of K we need $i \geq n$.

[cloud walrus]

jagr2808

[formal ermine]

good morning @frigid lark

[frigid lark]

Morning

Go to sleep

[white oxide]

oh ok thanks that helps a lot

https://www.youtube.com/watch?v=Ct2fyigNgPY

yo this video is goated

[untold turret]

I'm misunderstanding something. Suppose I have a field k and a field k(x). Why must x be algebraic over k?

Does the following argument work?

Let k be a field and suppose k(x) is a field with x transcendental over k. Consider the embedding i:k -> F into the algebraic closure of k. Find some element in F that is not in the image of i, say y,, and extend i to k(x) by having x map to y. But then, k(x) is embedded in the algebraic closure of k, so it must be algebraic over k, so x can't be transcendental.

[vast quiver]

it is not true that $x$ must be algebraic over $k$. When you write $k(x)$, do you mean the field of rational functions over $x$ (i.e., $x$ is ``free'', rather than a situation like $\mathbb R(\sqrt 2 )$)? if that is the case, then $x$ is definitely not algebraic over $k$.

the place where your argument would fail is the step where you try to extend $i$ to $k(x)$. Unlike polynomial rings, you can't map your variable independently (i.e., the universal property of polynomial rings). Let me show why this extension business doesn't work, in contrast to polynomial rings.

if you have a ring hom $\phi: R \to S$, you can extend it to $\Phi: R[x] \to S$ by choosing an element $s\in S$ and defining
$$\Phi(\sum a_k x^k) = \sum \phi(a_k) s^k,$$
which maybe I'll abbreviate as $\Phi(p(x)) = \tilde{p}(y)$, where $\tilde{p}$ is $p$ but the coefficients are replaced by their images under $\phi$.

Motivated by this, in your situation you might try to define $i(p(x)/q(x)) = \tilde{p}(y)/\tilde{q}(y)$, right? This might not work because $\tilde{q}(y)$ might be zero for some $q$, which is equivalent to $y$ being algebraic. So, you cannot extend $i$ unless you are mapping $x$ to something transcendental (i.e., $y$ won't work)

[cloud walrus]

thejoesully

[untold turret]

I see! that makes a lot of sense, and yeah, I was thinking of defining it in a vaguely similar way, but working out the details turns out to not make a lot of sense

ty for the contrast with polynomial rings, that was particularly helpful

[vast quiver]

yeah no problem!

[untold turret]

in other words, when we're trying to extend the embedding $i:k -> S$ to $i:k(x) -> S$, with $S$ being the algebraic closure of $k$, and supposing we map $x$ to an element $s \in S$, since $s$ is algebraic and satisfies a polynomial with coefficients in $k$, $p(s) = 0$, we're in fact guaranteed an ill-defined function by virtue of $i(1/p(x)) = 1/p(s) = 1/0$, right?

[cloud walrus]

nrs

[vast quiver]

yup

[untold turret]

thanks, very clear explanation and super helpful!

[untold turret]

suppose at least one of x_1, ..., x_n is transcendental over the field k. then k[x_1, ..., x_n] is neither a field nor is it finitely generated as a k-module, right?

[topaz solar]

well, the finitely generated part shouldn't be hard to see since like 1, x, x^2, ...

if it were to be a field, you'd need an inverse for uhh x1 + ... + xn

[untold turret]

right you can make a polynomial of arbitrarily big degree

[untold turret]

hm, but why can't we find that inverse?

[topaz solar]

well in the case where it's just k[x] obviously it ain't happening

is k[x, 1/x] a field, and if so what could an inverse be for uhhh idk x+1/x

[untold turret]

right i see, thanks!

[topaz solar]

I mean obviously this isn't a proof but it's hard for any specific one to have an inverse unless you already add it in

then what about adding them, and that'll be troublesome

your transcendental term is gonna be pretty hard to have an inverse for without another transcendental I think

[untold turret]

yeah none of it is immediately obvious

probably is the contrapositive of zariski's lemma

but it's good enough for me that looking for inverses immediately seems like a hard task if possible

[topaz solar]

if you had an inverse to x_1 (which I'm assuming is your guaranteed transcendent) in terms of some expression on like k(x_i, ..., x_n) an algebraic extension then I think that's a problem to being transcendent?

[untold turret]

this is definitely the contrapositive to zariski's lemma

[topaz solar]

oh right

[untold turret]

do it for k[x] in fact and you're done

[topaz solar]

when random lemma escapes memory

[dim bane]

Hi I’m new here and not deeply well versed in field theory. To be clear, your proposition is not exactly the contrapositive of zariski’s lemma right? They are just closely related. I am reading the definition for the first time and trying to make sense of it.

I think the fact that k[x_1, …, x_n] is not a field if some x_i is transcendental is not part of the contrapositive of Zariski’s lemma

[untold turret]

i think the contrapositive is more something like, if k[x_1,...,x_n] is not algebraic over k, then either it is not finitely generated as a k-algebra or it is not a field

not exclusive or

[dim bane]

The Zariski’s lemma I’m reading on Wikipedia says if K is finitely generated as a k-algebra, then K/k is a finite field extension. The contrapositive to my thinking would be if K/k is an infinite degree extension then K is not finitely generated. This includes not only finitely generated transcendental extensions k[x_1, …, x_n] but also infinite degree extensions (algebraic or not). I am not sure if that is that useful of a distinction, though, your formulation seems more interesting.

[untold turret]

K needs to be a field otherwise the lemma doesn't do any work

i.e. the finitely generated k-algebra needs to be a field

[dim bane]

Ah. That was stated as one of the assumptions on the page I was reading but I didn’t write it down. But that makes sense, you can formulate the contrapositive without assuming K is a field, and you get something more like what you said

[vast quiver]

call R=k[x_1, ..., x_n]. You're right that "some x_i transcendental implies R is not a field" is essentially the contrapositive of Zariski's lemma because we definitely have R is a finitely generated k-algebra, it has generators x_1, ..., x_n.

now for "some x_i transcendental implies R is not finitely generated as a k-module" is also true. The property "being finitely generated as a k-module" is also referred to as R being a finite extension of k, which in particular implies R is an integral extension of k (this is proved by Cayley Hamilton, see Gathmann comm alg notes, prop 9.5). Then, in particular this implies each x_1,...,x_n is integral/algebraic over k, contradicting some x_i being transcendental

[frigid lark]

For this question, can we say anything more than G has an element of order either 0 or a multiple of n, where n is the degree of f?

[formal ermine]

that is a cumbersome way to say "let K be a number field that is galois and B its ring of integers"

[formal ermine]

what does the notation Irr(alpha, Q, X) mean

min poly of alpha over Q?

[frigid lark]

yeah

with elements in Mor(<X>, Q), where <X> is a monoid isomorphic to N generated by X

[formal ermine]

@frigid lark do you know dedekind's criterium?

idk how to actually solve this but I feel like it just screams dedekind

[frigid lark]

oh, do you think there is more to this than what I said?

That's kinda the reason I asked

[formal ermine]

yeah

though I'm not really sure if this gets us anywhere lol

[frigid lark]

also K is not necessarily a number field

As [K : Q] is not finite

[formal ermine]

oh

bruh

[formal ermine]

then you can't apply this

oop

[frigid lark]

like we could probably do a finite Galois subextension that contains alpha

so maybe not much is lost

[median pawn]

is a central subgroup a subgroup of the center of G?

[lethal dune]

yes

hence central

[delicate orchid]

not to be confused with centric

[median pawn]

what's that

[delicate orchid]

or fully centralised

[lethal dune]

[delicate orchid]

H is centric in G if C_G(H') = Z(H') for all H' conjugate to H

fully centralised is uhhh it has the biggest centraliser lol

C_G(H) >= C_G(H') for all H' conjugate to H

[median pawn]

ahh i see

[rocky cloak]

But K = Q(alpha) ...?

[frigid lark]

Also I don't see where the Dedekind criterium idea is going to

[frigid lark]

Ahh, yep

I did not see that, I'm going to revisit this question tomorrow

[rocky cloak]

So G is a group of order n, and the Galois group mod p is cyclic of order n. Then you just need a way to compare them.

If B = Z[alpha], then p remains prime in B. So G acts on B/p, thus the two Galois groups are the same. But if B is bigger... I don't know. Maybe over can show that Z[alpha] is invariant under the action of G...

[frigid lark]

I think we might have an onto him from G to the Galois group of f mod P, given by $\sigma \mapsto \bar{\sigma}$, $\bar{\sigma}(\bar{\alpha}) = \overline{\sigma(\alpha)}$ where $\bar{\alpha}$ is just $\alpha$ reduced mod p

[cloud walrus]

parrottea

[frigid lark]

By the Galois group of f mod p, I mean $Gal(\mathbb{F}_p(\bar\alpha)/\mathbb{F}_p)$

[cloud walrus]

parrottea

[glossy crag]

Can't you say G is cyclic of order n? f irreducible mod p => by Dedekind's criterion pB is prime => there exists a unique element \varphi\in Gal(K/Q) such that \varphi(x)=x^p mod pB for x\in B. This element necessarily has order n, therefore it's a generator of Gal(K/Q). But you do need the machinery of Frobenius elements for this (assuming I'm correct), idk how one would come up with it himself.

Nvm, you need to know p does not divide |B:pZ[a]| in order to apply the criterion. Idk if irreducibility in general implies pB is prime.

[glossy crag]

If L/K is Galois and E/K is any extension, how to show the restriction homomorphism Gal(LE/E)->Gal(L/K) is open (in the Krull topology)?

[opaque tartan]

We have found that for every $u \in \ker \psi$ there exists a $u'' \in \ker \psi$
with $u * a = a * u''$. \
(1) Why, for a fixed chosen $a \in G$, is the mapping
$\psi_a : G \rightarrow G, u \mapsto u = a^{-1} * u'' * a$
an isomorphism from $(G, *)$ to itself, i.e.
an automorphism? $\psi_a$ is called an inner automorphism of $G$. \ \
(2) Why does $\psi_a$ map every normal subgroup $U$ of $G$ onto itself?

[cloud walrus]

Sciencenjoyer

[opaque tartan]

So,hi I need a little help for (2). I'm new to this topic, so I'm struggling a bit

[topaz solar]

i'd like to ask your answer to one

but for 2

[opaque tartan]

huh

[topaz solar]

what's the definition of a normal subgroup

I feel like since it mentions ker \psi theres some context but the statement of (1) looks odd

[opaque tartan]

okay let (G,**) be a set and a element of G. Let U be a normal subgroup of G then: a *** U = U * a

why is it italic lol

[topaz solar]

that what asterisks do

[opaque tartan]

now its thicc

okay let (G,+) be a set(edit: group) and a element of G. Let U be a normal subgroup of G then: a + U = U + a

[topaz solar]

do you have that g U g^-1 = U

[opaque tartan]

yes I have read online, that these are equivalent

[formal ermine]

do you mean a group?

[opaque tartan]

oh yeah sorry

[topaz solar]

yes you know it's true but is this proven already

[opaque tartan]

nope, I haven't even seen that definition in my book yet, I think

[topaz solar]

obviously if you had it, it trivializes this proof, and you should see why this is what it's asking

[opaque tartan]

Oh that is very cool

[topaz solar]

it's exactly asking \psi U = U

right?

[opaque tartan]

What is?

[topaz solar]

this question?

[opaque tartan]

You mean (2)?

[topaz solar]

yes, the question at hand?

how would you propose you prove it

given that Ua=aU

[opaque tartan]

Is that something obvious?

[topaz solar]

yes

because that's what it's asking, that \psi_a(U) is U?

[opaque tartan]

Oh, right

my bad

[topaz solar]

what have you tried my guy

or how do you think you should do it

[opaque tartan]

Would it help to show that \psi_a is surjective (even though that was already done in (1))?

[topaz solar]

Well, you’ve already shown that right?

[opaque tartan]

ye

[topaz solar]

So just use it instead, where would you propose using it

[opaque tartan]

What if I let v be an element of U and then calculated \psi_a(v) = v, why is the only solution because psi_a is bijective?

[topaz solar]

but psi(v) will probably not be v

[untold turret]

thanks a lot for the reference!

[cloud walrus]

The great Sharp

[opaque tartan]

It would mean that psi_a maps every normal subgroup of G onto itself

[topaz solar]

Well, yes that’s a literal statement

[void cosmos]

are u guys done

[topaz solar]

But, it will not be the identity usually

[void cosmos]

or wait

[topaz solar]

So it won’t have psi(v) = v

[void cosmos]

questions on noetherian stuff is in #algebraic-geometry ?

[topaz solar]

It should fit there too yes

[opaque tartan]

oh that is a good thought!

[topaz solar]

So rethink what it’s asking

[opaque tartan]

For every element v of U there is an element v' of U such that psi_a(v') = v.
That should be better

[untold turret]

@topaz solar k field, x transcendental. then k[x]/(x) -> closure of k has x map to 0, so x is nonunit

[topaz solar]

And that v’ in U implies v in U

[opaque tartan]

Thanks, I keep missing something

[topaz solar]

Consider using how Ug = gU to try and show it

[opaque tartan]

I'm already on it

[topaz solar]

As an aside, I do hope you know what Ug = gU says is equal

[opaque tartan]

It means that the operation is commutative of course

just kidding

[topaz solar]

[opaque tartan]

lool

I know

@topaz solar It's easier than I thought. I showed that for every v' there is a v such that psi_a(v') = v. I do not need to show the opposite as well, or do I?

v' and v being in a normal subgroup again

[charred crescent]

can someone help me understand the subgroup generated by two subsets, i.e. <A,B> ?

i think it consists of all words from A union B

[lusty marlin]

This is the same as the subgroup generated by the set A ∪ B

That is, <A,B> = <A ∪ B>

[topaz solar]

You want to show psi(U) = U, and ofc there’s such a v because it’s a function G->G, you need to show v is in U

[charred crescent]

okay cool, so the elements look like words from A ∪ B

[opaque tartan]

So that will do amirite?

[topaz solar]

no

So, take v in U, psi(v) = ava' right?

or uh

a'va whatever direction

[opaque tartan]

yes

e is btw the neutral element

[topaz solar]

then there exists, dependent on v, a term v' in U so that va=av'

[opaque tartan]

yes

[topaz solar]

so can't just take arbitrary v' in U at the start like you do

[opaque tartan]

I chose the v and v' to be the ones with that property (without mentioning because lazy)

[topaz solar]

too bad order matters

so you let v in U, then exists v', (rest of it)

[opaque tartan]

yes, I am sorry

[topaz solar]

so you've shown psi(U) is a subset of U

you do know psi_a has an inverse, namely, psi_a'

[opaque tartan]

I did know that

[opaque tartan]

Couldn't I just write down the same steps of equality, which I used in the screenshot, for the other direction?

[topaz solar]

[topaz solar]

pretty similar, or just reuse it

[opaque tartan]

No it was a joke, I was brain-afk

[topaz solar]

and then note \psi_a (\psi_a'(x)) = x

[opaque tartan]

Why would this imply that x \in U?

[topaz solar]

it doesn't, but maybe you can use this fact yourself

and add it to what you've shown

[opaque tartan]

Yes, got it know, thanks

[white oxide]

how does this example follow from the theorem? the theorem states nothing about the ring R

or rather how do we know the existence of the Z6-module isomorphism

[topaz solar]

Well, part (iii) of the theorem

is the projective part used here

As for the isomorphism, you can like

[delicate orchid]

What’s da iso doe????

[topaz solar]

compute it directly even???

Z_2 x Z_3 ~ Z_6 as groups right?

[white oxide]

oh right

oops

[topaz solar]

[delicate orchid]

Le Chinese remainder whatever

[white oxide]

anyways doesn't part (iii) of the theorem require that Z6 is a free module

[topaz solar]

R is a free R module my guy

[stone depot]

Regarding cauchy theorem,
suppose o(G) is prime then we know its cyclic and element has order of prime number p
if not prime and if its even then,
every finite even order group has a element of order 2 which is a prime itself

if its odd then we know say for element x then x^o(G) =e
we can prime factorize it and
(x^p1) ^(p2*p3... pn) = e
is this logic correct
sorry for not using latex

[empty rose]

no it isn't

[dim bane]

if you have an element $x$ for which $x^{|G|}=e$, doesn't that imply the group is cyclic

[cloud walrus]

suremark

[stone depot]

i was searching for u everywhere

sorry I was not able to give reply on time
but there is a theorem every finite even order group has a element of order 2

like u can pair element to its inverse but there still be one element left in even order group so 'e' can't be its inverse as its identity element neither any other element so it should have order 2

[empty rose]

...huh yeah i guess that works

[tribal moss]

That is a good argument, yes.

[empty rose]

although that's an element of order 2, not an element of order p

[stone depot]

but its a prime cauchy theoram states if p | o(G) then there exits x belong to g whose order is p

[tribal moss]

This is true, but it looked like you were asking whether you had a convincing proof of that theorem in general.

[empty rose]

yep, an element of order p

an element of order 2 is not an element of order p unless p happens to be 2, which it might not be

[dim bane]

your logic doesn't handle prime power divisors, but even if it did you can't assume there exists x such that x^|G|=e. that's not even true for finite abelian groups

[empty rose]

for instance if you have a group of order 12, 3 is a prime number and 3 | 12, so cauchy's theorem in this case would say that the group has an element of order 3

[stone depot]

I am not saying that a finite group which has order equal to odd is cyclic
I am just saying we can prime factorize its order into
p1 * p2 * p3 say
then
x^p1 if its not prime then it should be in the group
we continue like this
(x^p1) ^p2
if its not e then it should also be in the group
and finally
(x^(p1*p2)) ^p3
it should be e

according to this

[empty rose]

your argument would prove that it has an element of order 2, which is true but doesn't prove cauchy's theorem

[tribal moss]

That is actually true for all x in a finite group G -- it's Lagrange's theorem.

[dim bane]

er, you're right. I was confusing ord(x) = |G| and the property x^|G|=e

[empty rose]

...oh right yes it is
but what they- yeah

for the rest of the proof they gave to work requires that what they actually assumed is the order of x is |G|

[stone depot]

see this please sorry for wrong rotation

[empty rose]

,rccw

[tribal moss]

The trouble with this argument is that okay (x^(p1p2))^p3 = e, but it might be that it's because x^(p2p3)=e itself, and then you haven't found an element of order p3 after all.

Effectively you're arguing that the finite group contains an element whose order is some prime that divides |G|.

[stone depot]

but x^p2 must belong to G

[tribal moss]

Cauchy's theorem promises more: it says that you get to choose in advance which prime factor the order must be.

[stone depot]

say x^p2 = z
z must belong to G
then z^p3 =e
please correct me if i am getting u wrong

[tribal moss]

This is true, but that just means that z has order either 1 or p3.

If z=e (which is quite possible), then it doesn't have order p3.

[stone depot]

I think i got my mistake I am not proving for that specific p which divides o(G)

thank u so much

[white oxide]

If we consider any chain ${h_i} \in \mathscr{S}$, would $\mathscr{H:} \bigcup_{i \in I} \text{Dom }h_i \to J$ be a legitimate upper bound for the chain?

[cloud walrus]

okeyokay

[void cosmos]

hello

[void cosmos]

yes

are u done? @white oxide

[white oxide]

yea I checked it and it worked

[void cosmos]

okay

can someone walk me through the proof of a module having a composition series implies having the ACC?

a composition series ( for an R-module A) is a series A=A_0 > A_1 >A_2...>A_n such that Ai/Ai+1 is simple

this would have length n+1

and A_is are submodules

[delicate orchid]

The decomposition series is a maximal length chain of submodules isnt it?

[void cosmos]

idk whats a decomposition series

not mentioned in the text

[delicate orchid]

So if the composition series is finite length so are all of the other chains

Sorry I meant composition

[void cosmos]

okay np

[void cosmos]

can a module have a composition series of infinite length?

or does that not make sense?

[delicate orchid]

Maximality is guaranteed by the quotient being simple if that’s not clear

[delicate orchid]

Good point

[void cosmos]

yea cuz the proof in the text( which i could not follow ) just assumes its finite of length n

[delicate orchid]

I’d presume it has to be finite then

[void cosmos]

yea its weird

but i cant think of any examples or anything

i just got the definitoin

definitoin*

lmfao

definition*

[delicate orchid]

An example of a module with an infinite decomp chain?

[void cosmos]

maybe like Z and just ideals of multiples

[delicate orchid]

Just take any vector space of countable infinite dimension

[void cosmos]

like 2 4 6

[delicate orchid]

Mixed signals boss

[void cosmos]

okay

forget this question

let it for later

[next obsidian]

(6) is not contained in (4)

[void cosmos]

@delicate orchid assume its finite length

whatn ow

[delicate orchid]

:jawdrop:

[south patrol]

Anyway yes I can

[void cosmos]

yea anyone plz

[delicate orchid]

Then what I said holds I believe

[south patrol]

1 sec

[empty rose]

(what's the ACC?)

[formal ermine]

ascending chain condition

[void cosmos]

no i meant

[next obsidian]

Jordan Hölder says the length of a module is invariant under what composition series you take. So if finite, they’re all the same, if infinite they’re all infinite

[void cosmos]

2Z and 4Z

[next obsidian]

That’s literally the same thing

[void cosmos]

6Z and 4Z

[next obsidian]

6Z = (6)

And again, (6) is not contained in (4)

[void cosmos]

yea it should be like 16Z 8Z 4Z 2Z Z

i think?

[next obsidian]

Yes

[void cosmos]

yea cool

mb

but this is not composition tho

its just a normal series

[next obsidian]

How is this not a composition seties

[chilly radish]

Yes this is called a finite length module

It's a special property

[next obsidian]

The quotients are Z/2Z

[delicate orchid]

those quotients are clearly simple

[chilly radish]

That's mean

[void cosmos]

wait do u not include the last Z part?

[next obsidian]

What

[void cosmos]

like Z/4Z

[next obsidian]

What

2^n/2^n+1 = 1/2

[delicate orchid]

re-read the definition of a composition series

[void cosmos]

yea i got it

[south patrol]

Hello

[void cosmos]

this isn ot a compsoitoin series tho : 16Z 8Z 4z z

4Z Z*

right?

[next obsidian]

I mean if you skip 2Z then yes

[void cosmos]

yea

yea yea mb sorry brain dmg

[void cosmos]

he is alwaays mean to me cuz im not that smart

he should be punished 😦

[south patrol]

Anyway so like you can just show by induction that the length of the composition series is the maximal length of any strictly ascending chain of submodules

[chilly radish]

I meant it's mean to the modules

[void cosmos]

what

lol

[chilly radish]

Calling them simple

[south patrol]

it should be like immediate from definitions that this is the case

idk if anyone said

[void cosmos]

what composition series

[chilly radish]

You can immediately know it's not one because it's not maximal

[void cosmos]

oh you mean A has one now

[south patrol]

Any compostiio nseries

[next obsidian]

Moldilocks?

[south patrol]

I mean they all have the same length as has been previously stated

[void cosmos]

and then this compositoin series must have length greater than any ascending chaing of submodules

[next obsidian]

[south patrol]

cf. [Chm23]

[void cosmos]

right?

[south patrol]

That's what i just said yes

gotem

and weakly greater than sure

[void cosmos]

why is that tho

[delicate orchid]

because the quotients are simple

[south patrol]

Well it should be quick from definitions

[delicate orchid]

meaning each submodule is maximal in the previous one

there's nothing "Inbetween" you could pick to make things longer

[void cosmos]

yea right

yea right cuz if u assume there is one larger

then it would contradict that the simplicity of like

the left and right submodules

[hidden haven]

Do you know that if A and C are Noetherian and
0 → A → B → C → 0
is exact then B is Noetherian?

[void cosmos]

of the original compositon series

[void cosmos]

yes

can prove it too

[south patrol]

I think it's important to show the numerical element to this though

[hidden haven]

Nice, you can use this to prove that having a composition series → Noetherian

[hidden haven]

True

[next obsidian]

Finite length = Artinian + Noetherian

🤡

[void cosmos]

yea this is true

[delicate orchid]

"if the things are finite then they are finite" w0ah

[void cosmos]

or not finite length but having a composition series gives both ACC and DCC

[next obsidian]

Wew, kill yoursef

[delicate orchid]

MODS

[next obsidian]

[void cosmos]

??

wtf is worng with this guy

[hidden haven]

He's too based for this server

[delicate orchid]

too based for tv

[next obsidian]

I need to move to truth social

[void cosmos]

u need to move to league of legends

lol

anyways

potato okay

now i think the contradiction is to find some chain of submodujles having larger length?>

[south patrol]

Anyway so say $0 = M_0 \subsetneq \dots \subsetneq M_n = M$ is a composition series of an $A$-module $M$ and let $0 = M_0' \subsetneq \dots \subsetneq M_k' = M$ be any other chain. We can - for example - just refine this to a composition series by adding extra stuff if necessary, then by Jordan Hoelder this must be of length n, and so $k \le n$.

[cloud walrus]

potato

[south patrol]

Tbh like Jordan Hoelder is easy enough for modules so we may as well appeal to it and I think proving it by hand would just be near identical

[void cosmos]

idk jordan holder

[hidden haven]

Schreier refinement will be less nukey

[south patrol]

Jordan Hoelder says that any two composition series of M are of the same length

[hidden haven]

As it is a lemma to Jordan holder

[south patrol]

and moreover, the composition factors are the same up to permutation

Isn't Schreier refinement more for groups?

[delicate orchid]

oh my golly gosh it's the free groups guy

[south patrol]

For modules this is pretty easy to prove

[hidden haven]

I've only seen it for modules

[south patrol]

oh lol

[void cosmos]

oh

i proved it but for groups

but that was long time ago and i dont remember any shit detail

[south patrol]

It's nicer in this case cause everything is a normal subgroup lol

[hidden haven]

Normalize groups

[void cosmos]

yo

Mat_n(D) is going to have always a composition series right?

[formal ermine]

yo

[void cosmos]

given D is a division ring?

cuz ideals of this are of the form Mat_n(I) where I is an ideal

but D has no ideals?

[delicate orchid]

are we viewing Mat_n(D) as a D-module

or as an algebra

[void cosmos]

no

a ring

[rocky cloak]

What do you mean composition series as a ring?

[void cosmos]

just thinking of the ideals as submodules

like a composition series of modules of over that ring

[delicate orchid]

there are other submodules that aren't ideals though

Mat_{n-1}(D) comes to mind

[rocky cloak]

So submodules correspond to left ideals (or right). But you seem to be talking about two sided ideals

It has more left ideals than just M(I)

[void cosmos]

yea

im talking about two sided ones

[rocky cloak]

Right, but then I don't think the notion of composition series really makes sense

But it is true that M(D) is a simple ring

[void cosmos]

wait so i would need two composition series?

one for left R-modules and one for right R-modules?

to show its both noetherian and artinina?

[delicate orchid]

the most general I've seen composition series get is over abelian categories and Ring isn't abelian

[rocky cloak]

There's a distinction between being left noetherian and right noetherian, and same for artinian yes

[void cosmos]

okay

[chilly radish]

I think the thing you're missing is that in a noncommutative setting the left abd right module structures are not isomorphic. So there's not two composition series, rather a composition series for each module. Most R modules don't have a bimodule structure tho where the left and right multiplication are different, so they don't have two composition series in a natural sense

Generally when working with modules over noncomm ring left modules and right modules are distinguished

Results still hold for both classes but not necessarily 'both at once' iygwim

[void cosmos]

yea i get u

okay look at this example

the ring of all 2x2 matrices of (a,b) , (0,c) where a is an integer and b and c are rationals

this is right noetherian but not left

i suppose to show this is to find a compoistion series?

[chilly radish]

A finite one yes

[void cosmos]

is it the same composition series for Mat_n(D)?

as in like

M_i = R(e_1+e_2+..+e_i) where e_i is the matrix that has 1 in (i,i) and 0 elsewhere?

R is the ring?

[paper aurora]

if L = Zx+Zy is a lattice (x,y in C, forming a rank 2 basis over R), a proof I am reading claims that the number of sublattices of L with index p is p+1. why is this?

[frigid lark]

I think Lang mentioned frobenius elements in an example, but gave all the theorems in more generality.

Which lead to my confusion

I ended up just satisfying this proposition

And then the Galois group of fhat over Z/pZ will be cyclic of order n

and then a counting argument got me my answer

[weary owl]

https://www.youtube.com/watch?v=ZUhjaNl0Twg

[elfin prairie]

can someone walk me through this proof?

I am a bit confused about the structure of the proof

perhaps also some words on what this proposition is trying to convey (I assume it's trying to say that a product of n elements of set S is uniquely defined)

<@&286206848099549185>

[hot lake]

It says that all the ways to compose n elements all give the same result

[cunning lake]

I'm reading a thing and it explained conjugation as basically just 'renaming' the elements of a group, using the symmetric group as an example. this made me wonder, is it true that all isomorphisms from a group to itself are conjugation? if this is false could someone please show me a counterexample, if this is true could someone please give me a hint on how to prove this because I want to try and prove it

[hot lake]

Isomorphisms from a group to itself are called automorphisms

The ones coming from conjugations are called inner automorphisms

[dim bane]

automorphisms via conjugation are called inner automorphisms. but there are also outer automorphisms sometimes

[hot lake]

They form a subgroup of the group of all the automorphisms

[cunning lake]

makes sense that they form a subgroup

[hot lake]

I believe S6 is an example where there are other automorphisms

[cunning lake]

ty for the googleable term

[hot lake]

It's a normal subgroup

And the quotient group is called the outer automorphisms group

So you are asking whether this quotient group can be non trivial

And it can

[cunning lake]

neat

[delicate bloom]

is there anything we can say about the automorphism group of a group's automorphism group?

and so on, looking at automorphism group of that automorphism group

in particular, when do these terminate

[warm ember]

how do i do this

i tried induction on s

base case trivial

then if ker pi_H is nontrivial its easy

but what if ker pi_H is trivial

[rocky cloak]

This is the same as just asking about subgroups of Z^2 of index p. They will necessarily apear as kernels of surjective group homomorphisms to Z/p.

There are p^2 homomorphisms from Z^2 to Z/p (just by picking two places to map x and y). One of those is 0, so is not surjective. There are p-1 isomorphisms from Z/p to itself, so we have overcounted by that amount.

So there are p^2 -1 / p-1 = p+1 subgroups.

[rocky cloak]

In an abelian group conjugation is trivial, so for example Z/3 would be an example.

[rocky cloak]

So presumably p^s+1 still divides the order of G.

If H is normal, then G/H contains an element of order p, which you can lift to g and adjoin to H.

If H is not normal you can do the same trick to the normalizer N(H)/H. You just need to show that p divides this.

Consider the action of H on G/H by left multiplication. What does is mean for an element to be fixed by this action? What do you know about group actions of p-groups?

[rocky cloak]

It seems that if you allow transfinite iteration, it always terminates

https://arxiv.org/abs/math/9808014

And if it reaches the trivial group in finite time, then it's one of these cyclic groups

https://oeis.org/A117729

[coral spindle]

Cool!

[glossy crag]

Why does the degree of the composite LL'/K divide the product of degrees of L/K and L'/K? I can see that it's less than or equal to the product, but how to show divisibility? It may be assumed the degrees are coprime, if that helps.

[void cosmos]

yo

any PID is noetherian right?

cuz consider the union of the chain?

and then it stablizies with the union?

and this argument can be used to show that

i dont need it to have only one generators

as in lilke

if the ideals are finitely generated

then its the same argument rihgt?

[elder wave]

All ideals fin gen is equivalent to noetherian

[chilly radish]

under dependent choice

[agile burrow]

In fact, you need to assume the degrees are coprime since the divisibility doesn't hold in general. But yeah, both deg(L/K) and deg(L'/K) divide deg(LL'/K), so if the degrees are coprime then their product divides deg(LL'/K). On the other hand, you've shown deg(LL'/K) is less than or equal to the product, so they must in fact be equal.

[glossy crag]

Lol i forgot that the degrees of L and L' must divide LL'. Thanks

No, wait a minute, I don't know if the degrees are actually coprime.

Here p does not divide n

[rocky cloak]

Not true in general if a and b are two different cube roots of 2, then Q(a) and Q(b) both have degree 3, but Q(a)Q(b) has degree 6.

If the extensions are normal you should be able to do it by comparing the Galois groups.

[glossy crag]

Ah, you're right, the Galois group of the composite embeds into the product of the galois groups. Thanks.

[void cosmos]

any hints with number 4?

[solemn dew]

does anyone understand:

[frigid lark]

Why would sticking on a generator each J_i not work?

[void cosmos]

the chain will not be infinite?

wdym sticking a generator

on each J_i

[topaz solar]

There is no finite set generating it, so why not take a (minimal) infinite set of generators

[limber sequoia]

what is an anti-isomorphism? In the sense of "T -> T* is an anti isomorphism" (T being a linear transformation and * the adjoint)

[frigid lark]

Like let (a_i) be a minimal family of generators for I. Then let J_i = (a_1,...,a_i)

[limber sequoia]

it seems to follow from (UT)* -> T*U*

[frigid lark]

Basically what sharp said

[topaz solar]

I mean you don’t want duplicates yada yada

But yeah each is an ideal containing the next

[frigid lark]

An argument I can't be stuffed making rn, but should exist

[void cosmos]

wait this should work

why doesnt it work

[topaz solar]

It does?

[void cosmos]

oh

i thought he was asking me why it doesnt work

[topaz solar]

As long as youre careful about it anyway

[void cosmos]

yea yea it makes perfect sense

and its easy lol

im so stupid haha

tysm

[topaz solar]

Corollary, if I is finitely generated

[void cosmos]

then this eventually stablizies i think

by the union

[topaz solar]

You have some issues if it doesn’t

[void cosmos]

wdym

[topaz solar]

I mean, clearly a similar argument doesn’t work

[frigid lark]

If I is an ideal of R, a Noetherian ring, and if a minimal generating set of I is {a_i}. If you have a chain J_1 subset J_2 subset ... whose union is I, then each a_i will sit in some J_s_i, then as {a_i} is finite, there is some J_n which will contain all {a_i}, and hence I

[formal ermine]

f(ab) = f(b)f(a)

an anti morphism from A to B is a morphism A to B^op

[limber sequoia]

thanks

[void cosmos]

yea

got it

thanks guys

ur op

[solemn dew]

how do you make a 2-sylow group of S_{8}?

i tried to but it is wrong

[rocky cloak]

[n/p] is just how many multiples of p less than n there are, [n/p^2] is how many multiples of p^2 etc.

[rocky cloak]

Any (non-sylow) 2-subgroup is normal in a bigger 2-subgroup. So you could sort of iteratively construct bigger and bigger 2-subgroups by taking normaliseres.

So maybe you start with an 8-cycle, s = (1 2 3 4 5 6 7 8). And consider another 8-cycle in the same subgroup like s^3 = (1 4 7 2 5 8 3 6).

Then find an element t such that tst^-1 = s^3 like (2 4)(37)(6 8). Then you got yourself a group of order 16. And then continue like that. A bit tedious I suppose.

The easier method is of course to just know a head of time that an answer is ||<(12345678), (15)>||

[delicate orchid]

this group is too small to be Syl_2(S_8)

$S \in Syl_2(S_8)$ has order $|C_2 \wr C_2 \wr C_2| = 2(2^3)^2 = 2^7$, that group is order $2^6$

[cloud walrus]

ALPERIN'S FUSION THEOREM

[delicate orchid]

I believe you're missing the action of something like (13)(24)

[karmic moat]

what does \wr mean

never seen that before

[coral spindle]

Wreath product

[karmic moat]

waow

ty

[delicate orchid]

you can get a really cool presentation for these groups by considering their action on an p-ary tree

[formal ermine]

🤓

[delicate orchid]

why are you here

[formal ermine]

waiting for my flight

I don't even know the gate yet

ffs

[rocky cloak]

Right, counting is hard. Throw in (24)(37)(68) or whatever

[molten viper]

So, I'm trying to wrap my head around the definition of Artinian rings. Could someone give me an example and a non-example or provide a place I could read more about the subject?

[coral spindle]

Any field is an artinian ring. The integers are not an artinian ring.

[molten viper]

hm

So, the integers aren't because I can always take subsets of ideals that are themselves ideals

[delicate orchid]

consider the chain of ideals (2^n)

[molten viper]

right? e.g. (1) subset (2) subset (4) subset etc

[delicate orchid]

yeah exactly

[molten viper]

or superset

pretend I said superset

whereas with a field, you only have 2 ideas anyways (1) and (0)

ideals*

[coral spindle]

An exercise following up from that: prove that an Artinian integral domain is a field

[molten viper]

I did see that as a relevant theorem, I'll give it a try

[coral spindle]

You may also want to try showing that given a field k, the ring M_n(k) of nxn matrices over k is an Artinian ring

[molten viper]

matrices are my bane, so maybe

[coral spindle]

You may need to get up to speed with them if you want to study rings

[molten viper]

well I understand them

I just don't like them

[molten viper]

Re: this,
I'm going the route of assuming A (my ring) has some element which is not a unit
so (x) \neq A

I also know x isn't a zero divisor (because A is a domain)

so perhaps the contradiction comes from (x) superset (x^2) superset (x^3) ... being an infinite descending chain of ideals?

[coral spindle]

So if you can show that is indeed descending then you're done

[molten viper]

hmmmmm

well, let's do indunction on n in (x^n)

induction*

A \superset (x) is our assumption, so given (x^n) can we show (x^n+1) is a superset relationship

[coral spindle]

You're phrasing this as a question, but it's your job to answer it

[molten viper]

I'm talking to myself :P

Well, since this is a monomial ideal, we're done on that front because a \in (x^n+1) is clearly in (x^n), as a = bx^n+1, so a = (bx)x^n

at each successive step you're removing the elements of the form bx^n

But it's really counter-intuitive that you're never going to "reach zero"

but, the integers are precisely a non-Artinian domain

[topaz solar]

It’s artinian so what about your sequence

[molten viper]

well that's the contradiction

our sequence must end somewhere

but I believe that implies x is nilpotent

which is a contradiction

[coral spindle]

It does not imply that x is nilpotent.

[topaz solar]

Well if each ideal is just the whole thing that works too

[molten viper]

hm

so

there has to be some N such that (x^n) = (x^n+1) for n > N

if it is Artinian

[wraith cargo]

That's Noetherian

Oh eait

No lol

[molten viper]

But in the context of this argument

Because if we stop at 0, that's a contradiction

("reaching zero" as I called it earlier)

So the only other option is that you have some point where it stabalizes

But since it's a principle ideal, this implies that x^n+1 | x^n

right?

[coral spindle]

principal*

[molten viper]

I was about to ask

[coral spindle]

x^n+1 does divide x^n for some n. Now what does that imply.

[molten viper]

(I said monomial earlier, but I meant principal)

[molten viper]

doesn't that imply x is a unit?

or that x^n = 0

[coral spindle]

It does

[delicate orchid]

what's the ring here, k[x]?

[molten viper]

just some artinian ring

[coral spindle]

No, we know that x^n is not 0, because we are in a domain.

No!

[topaz solar]

Artinian integral domain (field)

[molten viper]

sorry, artinian domain

[coral spindle]

commutative!

[delicate orchid]

all rings are ocmmutaitaivioeivoe

[topaz solar]

Ah right I missed the commutative adjective

[coral spindle]

Grumble grumble artin-wedderburn

[molten viper]

when did we say it was commutative?

[coral spindle]

An integral domain is a commutative domain

N.b. a noncommutative ring can't be a field!

Unless you're French

[molten viper]

I didn't know this

I might be using some terms interchangably that I shouldn't

anyways

We assumed x wasn't a unit

so this is a contradiction

[coral spindle]

Indeed

So in fact x is a unit

[molten viper]

(thus A is a field)

All this to try and figure out what the hell a hilbert series is

smh

[topaz solar]

How horrible of an artinian domain (non-commutative) can you get

[delicate orchid]

uhhhhhhhhh

I can't even think of a non-com artinian domain (because Mat_n doesn't work )

@coral spindle gimme one

[coral spindle]

uuhhhhhhhhhhhhhhh

Quaternions?

[delicate orchid]

immensely boring

[topaz solar]

L unfortunate

[delicate orchid]

H[x]/(some fuckin 239084759385 degree irreducible)

[coral spindle]

[molten viper]

Surely being anti-commutative is a nice-ish property

[coral spindle]

But it's not anticommutative

[delicate orchid]

H is

[coral spindle]

1*2 = 2*1 in the quarternions

So clearly not anticommutative

[molten viper]

shit

[delicate orchid]

https://cdn.discordapp.com/emojis/1120823648975073350.webp?size=80&quality=lossless

[coral spindle]

I love that pic of him

[delicate orchid]

:uponthewitnessing:

[molten viper]

Is that fucking Gotham

the chess guy

[delicate orchid]

yes

[coral spindle]

Course it is

[molten viper]

amazing

[coral spindle]

It has to be

[molten viper]

um anyways

so like

[delicate orchid]

:jubulantarcsofheaven:
https://cdn.discordapp.com/emojis/1124379186245668894.webp?size=80&quality=lossless

[molten viper]

is being anticommutative a nice property?

[delicate orchid]

yeah it's pretty swagggg

[coral spindle]

I guess so

[molten viper]

"nice" in the sense that it makes things a little well behaved

[delicate orchid]

reminds me of - \rtimes C_2 if you know you know

[coral spindle]

But I might point out that the only anticommutative rings are of characterstic 2 (i.e., they have 1 = -1)

[topaz solar]

skew commutative

[delicate orchid]

this sucks. I hate this

[molten viper]

I feel like I'm getting into the deep levels of the Algebra Icerberg

[coral spindle]

But yes, we do indeed like Lie algebras

[delicate orchid]

there ain't no triangles yet

[molten viper]

what

[delicate orchid]

you gotta get to the triangles

[molten viper]

what do you mean by triangles

[delicate orchid]

cohomology

and like

[molten viper]

OH

[coral spindle]

Triangle man triangle man
triangle man hates person man
they have a fight
triangle wins
triangle man

[delicate orchid]

no I mean literal triangles. Subsets of R^n

not commutative triangles

[topaz solar]

Oh oop

[molten viper]

I feel like

[topaz solar]

Simplicies

[delicate orchid]

although yeah universal properties are swag

[molten viper]

I will eventually end up there

cause this is kinda fun

[delicate orchid]

I know about them but the lord knows I cannot compute anything to do with them

[celest furnace]

Legendary poem

[molten viper]

I'm very in over my head ngl

[formal ermine]

lmaooo

[formal ermine]

.pin

[hidden haven]

You would usually want graded commutativity instead of anticommutativity, in which you have a graded ring such that multiplying homogeneous things commutes up to a sign that depends on the degree of the things you are multiplying

[molten viper]

oh wild

[coral spindle]

It's from a song just so yall know https://www.youtube.com/watch?v=vOLivyykLqk

[delicate orchid]

aka Z[x]

[topaz solar]

Z2 graded

[delicate orchid]

wait.... let him cook actually...

[hidden haven]

No?

[molten viper]

I'm kinda trying to learn about graded rings so

[hidden haven]

xx = xx not -xx

It is commutative graded not graded commutative

[delicate orchid]

wait, I was talking about graded nonsense not anticommutativity

OH

lol

[topaz solar]

When do we have graded commutative

[molten viper]

Reminds me of uh

matrix determinants

[hidden haven]

Any kind of cohomology theory should give you graded commutative products (if it gives you products)

[topaz solar]

Z as Z2 graded

[delicate orchid]

any metacyclic group with a ring structure :wickedwizardweed:

[topaz solar]

Yeah that’s unintelligible

[hidden haven]

So cohomology of spaces, group cohomology etc

[delicate orchid]

good

[hidden haven]

Because the tensor product of chain complexes has to have certain negative signs

[delicate orchid]

yeah cohomology has that nice thing where you take two n-cycles or whatever they're called and then you smack em together and boom 2n

[hidden haven]

So true

[molten viper]

I love having no clue what's going on lmfao

[delicate orchid]

you and me both

[topaz solar]

I don’t know cohomology but I believe it

[hidden haven]

[delicate orchid]

I can see this clearly if we take the bar resolution

[hidden haven]

Lol basically graded commutative is the natural notion

Whenever you have differentials

So chain complexes, spectral sequences etc

[topaz solar]

Well it’s unintelligible to me but it’s less aberrant than opetopic types

[hidden haven]

Which are common things

[delicate orchid]

wait a minute

spectral sequences are like 2d right

[hidden haven]

Ye

[delicate orchid]

hmmmm...

[molten viper]

I just got an email and thought my thesis director got our meeting time wrong

[delicate orchid]

so are cosimplical spaces....

[molten viper]

It was in fact some spam

[delicate orchid]

I'm doing some cookin rn chat lemme think

[hidden haven]

By space do you mean topological or simplicial set

[delicate orchid]

simplical set

I don't care for topologies

[hidden haven]

Ye sure that's kinda 2d

[delicate orchid]

how do I make one into the other there MUST be a way

[topaz solar]

My simplical fu is lacking but I don’t think they’re quite so easily transferred

[delicate orchid]

I know next to nothing about both of these constructions

[hidden haven]

There is a functor SSet → Ch(Ab), not sure how that generalizes to 2d

[delicate orchid]

yeah that's what I was thinking

just put another c at the start of both of them

csSet -> cCh(ab)

[hidden haven]

Wait do you wanna turn a bisimplicial set into a double complex or a spectral sequence

[delicate orchid]

I'm not quite sure what I want to do just yet

oh wait yeah

it would just be a double complex

FAIL

[hidden haven]

Ye

SS too much structure

[delicate orchid]

very sad

[molten viper]

Yo, followup question

[delicate orchid]

actually it's not sad cause I can write "^{\bullet}" which makes me feel important

[molten viper]

A ring like k[x] isn't Artinian in general yeah?

[rocky cloak]

It is not

[delicate orchid]

yeah it isn't

[topaz solar]

When would your cosimplical simplical set be a spectral sequence

[rocky cloak]

Don't even need the "in general" part

[molten viper]

but it seems something like k[x]/J is

well

no but it is sometimes

[delicate orchid]

that mofo is noetherian tho

[topaz solar]

Depends on the J

[molten viper]

Yeah depends on the J

which is something that's relevant to my thesis

but it's always Noetherian if k is a field or even a Noetherian ring

[rocky cloak]

If k is a field, and J is not 0, then it will be artinian

[molten viper]

Oh interestin

that's actually more general than I thought

[rocky cloak]

yes Hilberts basis theorem

[delicate orchid]

this is interesting, it's obviously true if J is prime but I didn't know it would hold so generally

[molten viper]

but that's not the case in terms of like, k[x_1,...,x_n]/J, J has more restrictions for this to be Artinian

[rocky cloak]

If J = (f) and f had degree n, then k[x]/J is n dimensional

Finite dimensional implies artinian

[delicate orchid]

true, I suppose

I can't see it very clearly though

dw I'll think about this on my own

[rocky cloak]

Yeah this usually won't be

[molten viper]

start by considering like, J = (x^n)

[delicate orchid]

yeah that one's obviously artinian

OH right yeah

[molten viper]

I would have a harder time proving it for a general f

[delicate orchid]

eventually when you hit something with an x^n in it it will "collapse down"

so it's n dimensional

sick

[molten viper]

The (probably a bit too un-rigorous) way I think about it is like

if you mod out by f, and f = x^n + stuff

then x^n = -stuff

[delicate orchid]

yeah that's what I'm saying

[molten viper]

and you can just substitute

[delicate orchid]

that's what I meant by "collapse down"

[molten viper]

Ahhh

[delicate orchid]

the x^n turns into something of degree n-1 or less

[molten viper]

Yeah this Does Not Work in general for k[x_1,...,x_n]

[delicate orchid]

yeah

still noetherian albeit

[molten viper]

but if J contains enough nice fs, it will work

[coral spindle]

I hope you're not looking for an Artinian ring that isn't Noetherian!

[delicate orchid]

oh no no no no

[molten viper]

eg. J = (x_1^2,...,x_n^2), it is Artinian

[coral spindle]

Anyway, finite-dimensional algebras are always Artinian because finite-dimensional vector spaces are Artinian as modules

[white oxide]

how does the displayed math show that the group is isomorphic to the group G_n under multiplication modulo n?

[coral spindle]

But simply: dimension go down

[delicate orchid]

yeah if you turn all of the non-units into nilpotents it's definitely gonna be artinian

[molten viper]

Well yeah, but is there a simple condition for when it's finite dimensional

"simple"

[delicate orchid]

I'm sure Dr. 2808 has something cooking as we speak

[molten viper]

who lol

Just for example, J = (x_1) doesn't give an Artinian ring for n > 1

I know we're overloading n but sh

[rocky cloak]

Not impossible: [Z, Q; 0, Q] is (right) Artinian, but not (left) Noetherian

[delicate orchid]

what on earth does that notation mean

[white oxide]

oh nvm rm \equiv 1 mod n oops

[rocky cloak]

Matrices with entries from Z, Q and 0

[delicate orchid]

LMFAOOOOOOO

[coral spindle]

Matrices with integer entries top left, rational entries top right and... damn so fast

[rocky cloak]

(in relevant locations)

[delicate orchid]

bro really on that $\bZ \rtimes \bQ$

[cloud walrus]

wew

[delicate orchid]

I got the semidirect backwards

AGAIN

[molten viper]

what does \rtimes mean 😱

[delicate orchid]

semidirect plopduct

[coral spindle]

Oi wew

[rocky cloak]

If J contains x_i^N for all i and some N

[delicate orchid]

stfu I need to read about triangles made out of backwards triangles

[molten viper]

yeah, but I don't think that's exhaustive

[coral spindle]

Normal subgroups are like $N \unlhd G$, so since $N$ is a normal subgroup in the semidirect product, the triangle points towards it in the same way: $N \rtimes G$

[cloud walrus]

Borty

[delicate orchid]

oh I'm well aware

[rocky cloak]

It's not necessarily so simple to check I guess

[molten viper]

If it's a monomial ideal, it's simple

[coral spindle]

Idk if you're aware, I only speak because I like the sound of my own voice ||I'M JOKING SORRY||

[molten viper]

once you leave that realm, it gets harder

[coral spindle]

you mean a principal ideal right

[delicate orchid]

same

[topaz solar]

Crocodile eat da bigger one

[molten viper]

We're in polynomial rings, I mean ideals generated by monomials

[coral spindle]

OK

[coral spindle]

I just remembered you said monomial earlier ;)

[molten viper]

Could be a cool thing to try and figure out a condition on J which implies P/J is finite dimensional

[topaz solar]

What if we had a polynomial f_i in k[x_i] and looked at J=(f_1, .., f_n)

Could have a similar collapse like earlier?

[molten viper]

I believe that would work yes

[rocky cloak]

I doubt you can do much better than J containing x_i^N

[molten viper]

I mean, sure

but like, maybe some condition on the generators

[rocky cloak]

Maybe something using Krull dimension

[molten viper]

this is funnily relevant to my thesis work

[rocky cloak]

Like if (f_n+1) contains a prime in k[x]/(f1, ..., fn)

[molten viper]

I'm not familiar with Krull dimension

[rocky cloak]

Krull dimension is the longest chain of prime ideals

And a noetherian commutative ring is artinian iff the Krull dimension is 0

[molten viper]

interesting

[rocky cloak]

I believe there is some theory that relates the Krull dimension of quotient of polynomial rings to the polynomials

[molten viper]

I.... should read up on this!

[rocky cloak]

But I don't remember exactly that the conditions should be

[molten viper]

Yeah for the problem I'm working on for thesis stuff

We've been in an artinian ring this whole time

A quick search tells me "the dimension of the affine variety defined by an ideal I in a polynomial ring R is the Krull dimension of R/I." @rocky cloak

[rocky cloak]

I think it's just that the Krull dimension of K[x1, ..., xn]/(f1, .., fm) is larger than n-m

So you need at least n generators

[molten viper]

Give me a reference, I'm begging you 🙏

[rocky cloak]

Yeah, prime ideals correspond to irreducible closed sets, so dimension of a variety corresponds to Krull dimension

[molten viper]

I really need to review my definitions of variety and stuff

I have a book lying around

that I think I left in my apartment which I am not at

[rocky cloak]

https://math.stackexchange.com/questions/27145/why-are-artinian-rings-of-krull-dimension-0

Atiya Macdonald theorem 8.5

[molten viper]

Thank you

Wait a minute...

how are we defining length

are we saying like, A subset B has length 1?

[delicate orchid]

wait didn't you prove this earlier lol?

you showed commutative + artinian = field right

[molten viper]

I showed artinian integral domains are fields

[delicate orchid]

right, slightly different then

[rocky cloak]

This actually proves it yes

[molten viper]

Oh!

[rocky cloak]

Assume A is artinian and has a prime p. Then A/p is an artinian integral domain

Hence field, hence p is maximal

[molten viper]

It's a chain of prime ideals

not a chain which "starts with" a prime ideal

got it

[rocky cloak]

The converse is probably trickier