groups-rings-fields

[chilly ocean]

you should just assume commutative where necessary and keep reading for now

[sweet echo]

thank you:)

[chilly ocean]

what is the whole article?

looks like it could be up my alley

(symplectic, poisson geometry)

[sweet echo]

https://www.sciencedirect.com/science/article/pii/S0393044017301924

wait it might not be that

it is

its about putting "compatible" poisson structures on cluster algebras

based on work by Gekhtman et al

[chilly ocean]

all you can really get without commutativity is $$\mathrm{ad}_{a^{-1}} = -a^{-1}(\mathrm{ad}_a)a^{-1},$$ but that's still nice, and you still get the remark after "in particular"

[cloud walrus]

TTerra

[chilly ocean]

neat, thanks

[sweet echo]

thanks for the confirmation! you're right it's still nice

i got halfway there and then stopped because i couldnt see me getting to the commutative one

[chilly ocean]

i proved it by looking at {aa^{-1}, b} = {1, b} = 0

[sweet echo]

ah i wrote a^{-1} as aa^{-2}, i assume amounts to the same thing

just more tedious on my end

[chilly ocean]

same thing

[knotty frigate]

I'm just conjecturing but I'm guessing digraphs like these can only belong to Z_n or D_n

[chilly ocean]

$${1, b} = {1\cdot 1, b} = {1, b} + {1, b}$$ so ${1, b} = 0$, so using the fact that $aa^{-1} = 1$ gives $$0 = {aa^{-1}, b} = a{a^{-1}, b} + {a, b}a^{-1}.$$ multiply on the left by $a^{-1}$ to get $$0 = {a^{-1}, b} + a^{-1}{a^{-1}, b}a^{-1},$$ which is the same thing as $\mathrm{ad}_{a^{-1}} = -a^{-1}\mathrm{ad}_a a^{-1}$

[cloud walrus]

TTerra

[chilly ocean]

interpreted correctly, at least

the right hand side has a bit of notation abuse, i'm thinking of it as the thing taking b in P to -a^{-1} (ad_a (b)) a^{-1}

so you're really doing ad_a first, then multiplying on the left and right by a^{-1} and then switching signs

if you assume commutativity then indeed you can write this as ad_{a^{-1}} = -a^{-2} ad_a and no confusion would arise

i'm just being lazy and don't want to write the precise thing

[sweet echo]

yeah i get you

cleaner than my way

[solar glacier]

question

i need a walk through of this proof

Let $G$ be a group with $H,K$ subgroups. Suppose $H \subset N_G(K)$ then $HK$ is a subgroup. Ive already shown $HK$ is a subgroup iff $HK=KH$. So I want to claim that $HK=KH$

[cloud walrus]

MyMathYourMath

[solar glacier]

so i know one containment is trivial I cant seem to see which

it is KH \subset HK

the trivial direction

[cloud walrus]

MyMathYourMath

[solar glacier]

brb restroom break ILL BE BACK

im back

hang on another restroom break lol

5 mins

[chilly ocean]

[delicate orchid]

dude ate that taco bell

[solar glacier]

i did!!!!! lol howd u know

im almost back

[coral shale]

😠

[solar glacier]

ok im bCK

so how do I show HK \subset KH and KH \subset HK given H,K \leq G and H \subset N_G(K)

one direction is trivial but I cant see it :/

[delicate orchid]

what can you say about $hKh^{-1}$ given that you know $hkh^{-1} \in K : \forall k \in K$

[cloud walrus]

minecraftgrimreaper69

[solar glacier]

ive done this proof before i just cant think of the trick

[delicate orchid]

I wouldn't even call it a trick in this case

[solar glacier]

that h \in N_G(K)

[delicate orchid]

yeah

I'll let you think about it

[solar glacier]

ok

is KH \subset HK the trivial direction since H \subset N_G(K)

lemme think about this some more

[delicate orchid]

you can just show it directly without needing to worry about subsets

[solar glacier]

so say i have $x,y \in HK$ and i wanna show $xy^{-1} \in HK$ then set

[cloud walrus]

MyMathYourMath

[solar glacier]

$x=h_1k_1, y=h_2k_2$

[cloud walrus]

MyMathYourMath

[solar glacier]

with $h_i, k_i \in H,K$ resp.

[cloud walrus]

MyMathYourMath

[delicate orchid]

xy^{-1} is in HK because K is closed under inverses lol

it's a subgroup

[solar glacier]

is that because we write $xy^{-1}=h_1k_1k_2^{-1}h_2^{-1}$

[cloud walrus]

MyMathYourMath

[delicate orchid]

no it's because $y \in K \iff y^{-1} \in K$

[cloud walrus]

minecraftgrimreaper69

[solar glacier]

ah

its that simple?

[delicate orchid]

I mean, sure?

I don't know where you're going with this

[solar glacier]

where do we use that $H \subset N_G(K)$

[cloud walrus]

MyMathYourMath

[delicate orchid]

$hKh^{-1} = K$ for all $h$ was the thing I wanted you to realise

[cloud walrus]

minecraftgrimreaper69

[delicate orchid]

from this

also just realised my name is still minecraftgrimreaper69 one moment

[solar glacier]

but how to use this in proving HK=KH or HK \leq G

[delicate orchid]

$hKh^{-1} = K$ for all $h \in H \iff hK = Kh$ for all $h \in H$

[solar glacier]

i once did a proof where I showed HK \subset KH and KH \subset HK instread and i remember one direction being trivila

[cloud walrus]

Wew Lads

[solar glacier]

so take the union over all h \in H

to conclude that HK = KH

[delicate orchid]

yuR

[untold basin]

Hello, how can I prove formally that all the subgroups of S_3 are : himself, {id} and the subgroups generated by each element (separately ofc) ?

[solar glacier]

so since hK=Kh for all h \in H we conclude HK=KH

[delicate orchid]

there's only 6 elements lol just check

[untold basin]

I know, but elements =/= subgroups lol

[delicate orchid]

check the groups they generate is obviously what I meant

and ze answer will be revealled

alternatively, check that if you have a subgroup generated by two elements, where each one doesn't lie in the subgroup generated by the other, you have the whole group

that would be quickest I think

[untold basin]

That's what I wanted to avoid, a long proof

[untold basin]

🍔

[delicate orchid]

can't blame you lol

[formal ermine]

by lagrange the nontrivial subgroups must have order 2 and 3 respectively

and any group of prime order is cyclic

so it's just Z2 and Z3

[delicate orchid]

even nicer than my idea lol

[formal ermine]

do there exist field extension $k \subseteq m \subseteq l$ s.t. $l/k$ is normal but $m/k$ isn't?

[cloud walrus]

rectangle cube

[untold basin]

thanks !

[untold basin]

ZE answer

[delicate orchid]

ze answer...

[woven obsidian]

Let $k=\mathbb{Q}, m = \mathbb{Q}(2^{1/4}), l = Q(i,2^{1/4})$

[cloud walrus]

AoiKunie

[woven obsidian]

More generally, let $l$ be any non-normal field over $k$ and $l$ its normal closure

[cloud walrus]

AoiKunie

[formal ermine]

normal closure?

[woven obsidian]

Essentially, the smallest normal field over k containing l

[formal ermine]

ah ok thanks

[solar glacier]

in showing if G/Z(G) is cyclic then G is abelian

the key is to first claim elements of G can be written of the form

$g=x^mz$ for some $z \in Z(G)$ and $m \in \Bbb{Z}^+$

[cloud walrus]

MyMathYourMath

[solar glacier]

since G/Z(G) is cylclic there exists an x \in G such that G/Z(G) = <x Z(G) >

then for g \in G

gZ(G) = (xZ(G))^m=x^mZ(G) which is true iff gx^{-m} \in Z(G) so it equals z for some z \in the center

thus g=x^mz

[formal ermine]

I'd then argue by (x^m z_1)(x^k z_2) = x^m x^k z_1 z_2 = x^k z_2 x^m z_1

[solar glacier]

yes that parts trivial

thanks

so i had the key claim right

[formal ermine]

yes

looks right to me

[solar glacier]

thanks

and for the backwards direction

if G is abelian then its equal to the whole center

thus the quotient is the trivial group which is trivially cyclic

right?

[formal ermine]

yes

[solar glacier]

sweet thanks

question regarding semi direct products

so the map $\phi: K \rightarrow \text{Aut}(H)$

[cloud walrus]

MyMathYourMath

[solar glacier]

is the kernel of \phi the elements that map to the identity map in Aut(H)

[coral spindle]

Of course

[solar glacier]

i.e., the maps from H to H defined by sending h to h

the isos

since the trivial element of Aut(H) is the identity map?

and kernel maps elements to trivial elements

is all this reasoning correct

[formal ermine]

yes

[cloud walrus]

MyMathYourMath

[solar glacier]

where G is the semit direct product of H and K

[cloud walrus]

MyMathYourMath

MyMathYourMath

[solar glacier]

how do I show for all g \in G that gxg^{-1} = x

[agile burrow]

What does the map phi actually do

[solar glacier]

sends elements of K to automorphisms of H

[agile burrow]

Which automorphism though

[solar glacier]

of H to itself

it just sends arbitrary elements of K to arbitrary elements of Aut(H)

[agile burrow]

It's just an arbitrary map from K to Aut(H)?

[solar glacier]

yes

its a homomorphism of course

oh

take phi of ghg^{-1}?

?

wait no

im trying to show x \in C_G(H)

[solar glacier]

its the map induces by the semi direct prod

induced*

[agile burrow]

Hang on, I'm putting away my toaster

[solar glacier]

lol ok

i need to show xhx^{-1} =h given that phi(x) is the identity on iso of H

but H is normal right

[agile burrow]

I'll need to write this out, but I want to say that this is specifically referring to inner automorphisms

Oh wait I think I see

Phi is induced by the semi direct product

[solar glacier]

yes

[agile burrow]

Does this imply phi is the map such that phi(x) maps h to xhx^-1

[solar glacier]

i thought if youre in the kernel phi(x) is the identity on H

[delicate orchid]

If I’m remembering my presentation theory right, yes it’ll map each element to an inner homomorphism

But I will check

[agile burrow]

Right, if x is in the kernel then it maps to the Identity

[delicate orchid]

*auto

[agile burrow]

But what does phi(x) actually do?

To an arbitrary element

I'm not explaining this well

[solar glacier]

its an auto of H?

[agile burrow]

Sorry this toaster is putting up a fight

[agile burrow]

My point here is that phi being induced by the semi direct product tells us what phi actually looks like

[solar glacier]

thats ok im gonna graba cup of coffee and think about this

[agile burrow]

As in we can specify what the map actually is

[solar glacier]

what is that

conjugation?

[coral shale]

(where do I start reading for what u guys are discussing)

[agile burrow]

Yes, conjugation in the semi direct product

So try proving that identity and then I think you will understand what I am trying to say

[solar glacier]

here

[agile burrow]

(to prove it, just explicitly conjugate elements in the semi direct product)

[solar glacier]

i think i got it

let x \in ker \phi then x \in K

and we need to show xhx^{-1} = h

but xhx^{-1}=(e_H,x)(h,e_k)(e_H,x)^{-1} and calculate using the definition of operation in semi direct product

[agile burrow]

Yes exactly

[solar glacier]

sweet thanks

[agile burrow]

Sorry I didn't explain very well earlier

But you got it

Semi direct products are cool

[solar glacier]

they have a twist in the operation lol

[cloud walrus]

MyMathYourMath

[solar glacier]

How do i use the fact that H is characteristic in showing ghg^{-1} \in H for all g \in G and h \in H

[agile burrow]

Is conjugation an automorphism?

[solar glacier]

yeah but tats a specific auto this is for any auto of G

[agile burrow]

Well if it holds for any automorphism then certainly it holds for conjugation

[solar glacier]

then

so conjugation by elements of G does the trick???

[coral shale]

???

[solar glacier]

plz more input 🙂

[coral shale]

you have a statement that holds for automorphisms
and u want to prove the same statement for conjugations

correct?

[solar glacier]

not the same statement but that ghg^{-1} \in H for every g\in G and h \in H

[coral shale]

but it is if we define f_g = ghg'

[solar glacier]

i.e. H is normal in G

[agile burrow]

Do you see that normality is equivalent to saying gHg^-1 = H

As sets

[coral shale]

oh nvm me ignore

[solar glacier]

yes

so phi of those is equal

[agile burrow]

So we have specific automorphisms phi_g where phi_g(x) = gxg^-1

What do we get when we apply phi_g to H?

[solar glacier]

you get back

um

gHg^{-1}

[agile burrow]

Right

But H is characteristic, so applying an automorphism of G to H yields...?

[solar glacier]

H itself

[agile burrow]

And so we can conclude that

[solar glacier]

that element is in H

[agile burrow]

No

We conclude that gHg^-1 = H

[solar glacier]

i.e. H is normal in G

[agile burrow]

Right, I guess more explicitly you say that gHg^-1 = H for all g in G

But do you understand why that equality holds?

[solar glacier]

because phi of H is gHg^-1 which is equal to H by assumtpion

oh H being characteristic

nice ok

[solar glacier]

thx again! I totally see it now! it wasnt too bad

i know I asked this but i just wanna make sure i got it right

show in G if x,y \in G the orders of xy and yx are same

does this work: Let $k$ be the order of $xy$ then $(xy)^k=x(yx)^{k-1}y=e$ $\Rightarrow$ $(yx)^{k-1}=x^{-1}y^{-1}=(yx)^{-1}$ thus the order of $yx$ divides the order of $xy$? i.e., $(yx)^k=e$.. and by symmetry were done?

[cloud walrus]

MyMathYourMath

[solar glacier]

i have a big algebra qualifying exam coming up if anyone has any set of ring and group theory problems I could work on plz send them my way

[next obsidian]

that works

But it’s a little unclear

[solar glacier]

sweet thanks

[next obsidian]

I guess I would just finish it by like

Multiplying by (yx)

And then observe that (yx)^k = e

Might as well just finish it off yeah?

[solar glacier]

yes thats what i concluded

[next obsidian]

I think this also works:

[solar glacier]

i concluded this

[next obsidian]

(xy)^k = e

[solar glacier]

hitting w a y

[next obsidian]

So (xy)^-k = e

[solar glacier]

?

ah

[next obsidian]

Then (y^-1x^-1)^k = e

Hmmm

Idk

I feel like this would work

But maybe it does idk

[solar glacier]

i like my method of splitting xy^n out lol

and writing the inside as yx to power of k-1

which shows order of yx divides order of xy

[agile burrow]

That works - it's the same thing but I think it's easier to see that conjugate elements have the same order

[next obsidian]

Walter

[chilly ocean]

waltuh

[agile burrow]

chmonkey and tterra

[agile burrow]

I can DM you some of my algebra homework assignments if you'd like

[solar glacier]

that would be wonderful

[agile burrow]

specifically the ones on group theory and ring theory, but if you want to review modules and fields as well I can provide those

[solar glacier]

no no just groups and rings

thats all im being tested on

barely trying to wrap my head around groups and rings

[agile burrow]

Ok, one sec

[coral shale]

why not post here

[agile burrow]

I can do that too

[coral shale]

waw

[solar glacier]

whichevers easiest for you

[agile burrow]

incoming pdf bomb

[solar glacier]

LOL

[rotund aurora]

[coral shale]

someone pin this

[agile burrow]

i'm pretty sure most of these are from dummit and foote

[solar glacier]

@agile burrow thx so much!!

yeah i recognized some, could I ask you about solutions when stuck

[agile burrow]

Sure, I'll try to respond whenever I'm free

[solar glacier]

thanks, no rush whatsoeevr

[coral shale]

sure just handy whenever someone asks for some problems; does happen

maybe if i ping @uneven folio itll be done within this week

[next obsidian]

.pin

[next obsidian]

Chmowned

Shuri chmowned

[coral shale]

damn chmowned indeed

not me, but ryc

[formal ermine]

so I'm having kind of a problem right now

I think I maybe have done some highschool algebra wrong somewhere? idk

so I'm trying to find the galois group of L/Q where L is the splitting field of x^3 - 3

L = Q(cbrt(3), isqrt(3))

so I take an automorphism sigma

it fixes isqrt(3) and permutes the other one

[prisma ibex]

i\sqrt{3}?

[coral spindle]

You sure about that?

[solar glacier]

i have a proof solution for g to g^2 is a homomorphism iff G is abelian if you could check it

[formal ermine]

sigma(isqrt(3)) = isqrt(3)
sigma(cbrt(3)) = - cbrt(3)/2 + 1/2 i 3^(5/6)

now I want to find its order

[prisma ibex]

[formal ermine]

or is that already wrong

[prisma ibex]

yeah already wrong

[agile burrow]

just post here after rectangle cube gets help and someone (potentially me) can check it later

[formal ermine]

oh lmao

[solar glacier]

okie doke 🙂

[formal ermine]

so I like have no idea how to find the galois group of an extension lol

det explained it to me a lil yesterday already but it was using some 'heavy' machinery

my course has only introduced what galois group means

nothing more

[next obsidian]

What if you cube icbrt(3)

You definitely don’t get 3

[formal ermine]

why is the extension not Q(cbrt(3), isqrt(3))

[next obsidian]

.

[prisma ibex]

err I guess this is fine

[next obsidian]

Is this actually an equivalent way to write it down??

[formal ermine]

like the roots are cbrt(3) and -cbrt(3)/2 +- 1/2 i 3^(5/6), obviously we have cbrt(3). for the second one I pull out the first term, the +-, and a cbrt(3), so I'm left with isqrt(3)

[coral shale]

how does a sqrt get involved what

[coral spindle]

Idk why you wouldn't write it as Q(cbrt(3), zeta_3)

[formal ermine]

what's zeta_3

[coral shale]

3rd root of unity

[prisma ibex]

3rd root of unity

[coral spindle]

a primitive one, to be precise

[next obsidian]

Oh wait also sqrt(3)

[prisma ibex]

but what you have is equivalent to this

[coral shale]

anyways, we have a cubic, where does sqrt3 come from

[formal ermine]

uhh how I would I "realized" that I had zeta_3 there

[formal ermine]

3^(5/6) = sqrt(3)cbrt(3)

[prisma ibex]

the "obvious" way to write down the roots of x^3-3 is \zeta^a_3 * 3^{1/3}

so cube root of 3 times some 3rd root of unity

[formal ermine]

ahh because (zeta_3)^3 = 1?

[coral shale]

this is completely non-intuitive to me but sure ig

ngl im still half lost, but if ur sure

[prisma ibex]

yes, and then in this case the two primitive 3rd roots of unity are -1/2 \pm sqrt(3)i/2

so that's how sqrt(3) is showing up

[coral shale]

oh ok

[formal ermine]

right

[prisma ibex]

in general if you have x^n-b you can write the roots as like \zeta^a_n * b^{1/n}

[formal ermine]

yeh

so

what do I do next

[prisma ibex]

okay so you have your extension Q(3^{1/3}, \zeta_3)

and you have some subfields

Q(3^{1/3})

Q(\zeta_3 3^{1/3})

and Q(\zeta^2_3 3^{1/3})

these are all distinct

these are all cubic extensions of Q

on the other hand you have the subfield Q(\zeta_3), this is a quadratic extension of Q

[formal ermine]

I already showed that L has degree 6 over Q

and that it's galois

[prisma ibex]

okay then you know the Galois group is S_3

[formal ermine]

why isn't it C6

[prisma ibex]

well for one, you have a cubic

[coral shale]

it has to be a subgroup of s3

[prisma ibex]

so the Galois group is a subgroup of S_3

[formal ermine]

ah ok got it

[coral spindle]

It's kinda easier to see it as Dih(6) imo but each to their own

[prisma ibex]

also, if you look at the subfields we've written down, they should correspond to subgroups of Galois

[formal ermine]

what's Dih 6

[coral shale]

D6

[formal ermine]

wait

[coral shale]

dihedral on triangle

[formal ermine]

that's an aha moment for me right there

[prisma ibex]

S_3 has a unique subgroup of order 3, and 3 subgroups of order 2 that are all conjugate

[coral spindle]

Like if you just draw the roots what do you get right lmao

[prisma ibex]

the unique subgroup of order 3 corresponds to this quadratic extension

the 3 subgroups of order 2 correspond to these cubic extensions

[formal ermine]

yeh

thanks ng

[solar glacier]

question regarding sylow problems

say the break down is into primes with power 1 each say 2 times 3 times 5

if say n2=k can we say there are (2-1)k elements of order 2 and (3-1)m elements of order 3 if n3=m

but say we are given a group of order 24 which is 2^3 times 3

and i wanna assume G is simple this implies n2=3 and n3=4 (as being simple implies one of them is 1)

[worn wolf]

I forgot who is prepping for quals, but here was a hard one from my hw

Show a group of order 30 has a normal Sylow5 and a normal Sylow3 subgroup

[solar glacier]

that problem showed up on my midterm i got it right 🙂

[worn wolf]

Nice !

My professor told me the answer...

Actually when you went to his office hours he would just tell you the answer 😅

[solar glacier]

lol

[wooden ember]

Here you’ll then that at least 3*(4) + 3 elements of order a power of 2 and 4*(3-1) of order a power of 3 , which doesn’t really allow you to conclude a contradiction here

You could instead consider the conjugation action of G on the sylow 2 subgroups

Note that 24>3!

[untold basin]

Hello
How can I find the degree of Q(zeta_35) on Q ?
I found that it is either 12 or 24 but I am blocked here

[prisma ibex]

\phi(35)=24

[next obsidian]

Proof? Source? Source? Proof? Do you have a reference for that?

[untold basin]

Wait don't tell me that degree of Q(zeta_n) over Q is phi(n) xd

[prisma ibex]

it is yeah

[untold basin]

Is it because this extension is Galois and that the group of Galois is injected in (Z/nZ)* ?

[prisma ibex]

the Galois group is isomorphic to (Z/nZ)*

[untold basin]

Isn't it only for n prime ?

[prisma ibex]

no this is for any n

it's just that (Z/nZ)* isn't usually order n-1, it's usually smaller

[chilly radish]

Source: I made it up

[untold basin]

Mmmmh okay thanks

[chilly radish]

Honestly the way I know it is you first show the degree is phi(n) and then conclude that the galois group is isomorphic to (Z/nZ)*

Q(zeta_n) is the splitting field of the n-th cyclotomic polynomial. One shows that this polynomial is irreducible and integral and of degree phi(n), so the degree of the extension is phi(n)

this then shows that the aforementioned injection is in fact an isomorphism

[untold basin]

I don't know why but my prof only defined the cyclotomic polynomial for n prime and I searched on wiki and it is for any n

[untold basin]

I see

[chilly radish]

that's strange

it's not trivial to see that the cyclotomic polynomial is irreducible

but it's not too difficult

if you've shown this for prime n the proof for nonprime n should be similar

and follow by reduction

[untold basin]

Is it the same proof than for n prime ? Like we look at P(X+1)

[chilly radish]

oh

this is the proof that I learned

[untold basin]

Here we go xd

[chilly radish]

theorem 4.3 is this

[subtle flame]

in (Z,+) and (Q,*) inverse of 0 is 0?

[formal ermine]

(Q, *) can't exist

it must be (Q \ { 0 }, *)

because 0 doesn't have an inverse

[subtle flame]

a right, i was looking at ring examples

but it was (Q,+) not (Q,*)

but why (Q,*) can't exist?

[formal ermine]

because 0 doesn't have an inverse

[subtle flame]

okay but its still possible to have rational numbers equiped with operation *

[formal ermine]

no

(Q, *) isn't a group

(Q, +, *) is a ring, yes

[subtle flame]

yeah it is not a group

but i was thought when i write in brackets a set and an operation it just marks that (Q,*)

[formal ermine]

it usually means group

[subtle flame]

oh okay thanks

[chilly radish]

I mean, this is highly context-dependent

if you're asking about inverses sure

but in general just stating a pair of a set and an operation doesn't a priori have to specify a group

unless you say it does

[chilly radish]

yes, it's just not a group. It's a monoid

[ember field]

Not able to understand how this map works can anyone elaborate pls I understand that $d.d=0$ but when we apply $d^h+d^v$ on a module C_{p,q} we get C_{p-1,q}+C_{p,q-1} which is not in the complex. Can anyone please clarify this construction

[next obsidian]

A map from a direct sum is a map from each summand

Tot^+(C)_n is made up of a bunch of parts, each of which has d^h and d^v going to some parts of Tot^+(C)_n-1

So you use those maps on each summand

Similar for product, but now maps into products are maps into the individual ones

I guess this is the cochain version

So your example the arrows go backwards

But this is the idea

[ember field]

Is this due to the fact A+B+A=A+B in direct sums. What about products?

[next obsidian]

This isn’t true and I don’t see how it’s related

[ember field]

I meant A\directsum B\directsum C\directsum= A\directsum B\directsum

[next obsidian]

What

[chilly radish]

?

[next obsidian]

Look, it follows because the direct sum is a coproduct

If that doesn’t make sense, you define it on elements using basically the picture I drew

If that isn’t enough idk what else to say

[formal ermine]

not sure if this is the right place to ask this

but what is vague about "taking the reciprocal polynomial is an involution iff the constant term is not zero"

[south patrol]

Well it doesn't seem to make too much sense to me

A function f is an involution iff for all x, f(f(x)) = x

[languid hare]

Neither direction is true: 1/x an involution, whereas 1/(x^2+1) is not even invertible.

[formal ermine]

(f*)* = f iff the constant term of f is non zero

that's what I was trying to say

cc @languid hare

[languid hare]

Ahh, then it’s an odd statement altogether. I am at the airport rn though. I may be able to get back to this in a little bit. :3c

[white grotto]

can someone explain intuitively why a symmetric matrix has n real disctinct eigenvalues or vectors i am not sure?

[ruby sundial]

I dont see anything vague

This is perfectly valid statement

[south patrol]

to me it sounds like incorrect phrasing

it would be weird to say that e.g. differentiation of f is an involution if and only if f ...

[ruby sundial]

yeah its a condition under which the operation is considered involution

but then again thinking of the statement

isnt a problem the roots?

like if your polynomial has any roots wont they just be poles after reciprocating

and then the domain changes so you arent using same function

[tribal moss]

How would it be true at all? The reciprocal of a nonconstant polynomial is not a polynomial at all.

[ruby sundial]

yeah thats what im saying

[south patrol]

Well also not sure what reciprocal means

I assumed they might mean uh

[ruby sundial]

but i think they mean reciprocal for power series

[south patrol]

reversing order of coefficients

[ruby sundial]

ig now i see some ambiguities with the question

[tribal moss]

Okay, that are two different interpretations that would both be true with a restriction to nonzero constant terms. :-)

[ruby sundial]

if they did mean power series then the statement is true regardless of constant term

[tribal moss]

Well, the reciprocal of 0+1x doesn't have a power series around 0.

Unless you count Laurent series, but then it becomes ambiguous which annulus to take the reciprocal of the original function on.

[summer notch]

if we have some field K and some extension L/K, given any automorphism f of K, there will be some automorphism g of L (not a member of Aut(L/K), it doesn't fix K pointwise) such that f(k)=g(k) for all k in K?

feels intuitively correct but idrk why

[oblique river]

Zorn’s lemma can be used

[formal ermine]

reciprocal as in f* = x^(deg f) f(1/x) (I stated this in the sentence before, forgor to do here on discord tho, mb)

[south patrol]

ok phew

lol

Well I think the fact that this question has spawned lots of confusion is evidence enough that it is a bit vague XD

[formal ermine]

my corrector marked the "iff a_0 is nonzero" part as vague

that's what confuses me

[chilly ocean]

grader moment

[formal ermine]

he also took a point away on something but didn't explain why

[chilly ocean]

maybe they wanted you to prove it instead of just state it

[formal ermine]

I proved it

[chilly ocean]

i dunno

sounds like a grader problem and not a math problem

maybe you shoulda said "constant term" instead of "a_0"?

[sweet echo]

what breaks it being an involution when the constant term is zero? feel i am missing something

[chilly ocean]

it's not even clear to me what exactly you wrote

[chilly ocean]

do it twice and see what you get

[formal ermine]

this probably

[chilly ocean]

if f is of degree n then you get x^n f(1/x). if this if of degree m, then you get... what?

and what is the degree of x^n f(1/x)?

think about it

[sweet echo]

im with you, i messed up my example of x^2 + x

[oblique river]

You dont say what a_0 is

[chilly ocean]

i feel like this is a little silly but it also makes sense

like, what else could it mean? lol

reasonable from the grader though

[oblique river]

Would anyone write something like

a_0x^2 + a_1x + a_2?

[chilly ocean]

i just remembered that's a thing sometimes

yeah okay fine

grader justified

[elder wave]

if the course constantly ( ) uses a_0 as constant term (mine does for example) then i find it silly

[tribal moss]

The grader is probably fighting the idea some students have that certain letters automatically have a meaning that connects them to each other, which doesn't need to be stated explicitly.

[chilly ocean]

too late buddy

[elder wave]

is this revenge

[chilly ocean]

in a lighthearted way

[tribal moss]

I would say just because there's a polynomial called f somewhere, we can't expect that the coefficients of that particular polynomial are automatically what is called a0, a1, a2, ...

[elder wave]

either way the grader should have specified what he's deducting points for

[formal ermine]

yeah on that note

[oblique river]

I mean, isnt this why?

[formal ermine]

I was gonna ask about this case too

the solution is correct

I verified it using wolframalpha

[elder wave]

oh god it's that pirate problem

[chilly ocean]

nooooo you have to do it by hand nooooo

[formal ermine]

yes lmao

[oblique river]

Its not just about being correct, you have to actually communicate it well too

[formal ermine]

I did the chinese remainder theorem by hand if that counts

[oblique river]

And idk how well you did that because i cant read german

[formal ermine]

I mean

I did it the same way as in a)

and that got full points

but b) didn't

[elder wave]

maybe ask the grader

[formal ermine]

scary

[elder wave]

we can only speculate especially the people who don't speak german

graders are just students as well most of the time

[formal ermine]

yeah

I will write him an email later then I guess

[elder wave]

[summer notch]

how 😮 by like "successively" adding elements or something?

[tribal moss]

Successively adding elements is (intuitively) what Zorn's lemma does for you, under the hood.

As a user you just get delivered a maximal extension of some f to an automorphism on some subfield of L. All you now have to do is show that that subfield needs to be all of L, or the solution you got wouldn't have been maximal.

[lavish sigil]

Let $k$ be a field. Then obviously, $k[t^2,t^3+t]$ is a subring of the polynomial ring $k[t]$. It is correct that it is not equal to the whole ring, right? How would I show this, formally? I am pretty rusty on algebra in general...

[cloud walrus]

gustavn64

[summer notch]

ok so let S={L' subfield L such that f extends to L'} is what you're saying?

Zorn's implies S has some maximal element which must = L?

[lavish sigil]

Intuitively, to get $t$ you would have to take the second generator, but then you have to cancel the $t^3$, which should be impossible, I think

[cloud walrus]

gustavn64

[tribal moss]

The elements of the partial order would be the actual extensions of f to automorphisms of fields between K and L, ordered by set inclusion.

[summer notch]

how do you order functions by set inclusion

what you said doesn't make sense to me

[tribal moss]

View a function A -> B as a subset of A×B.

[summer notch]

oh 🤔

hmm

[tribal moss]

That gives the same ordering as saying "f R g iff g is an extension of f", but saying "ordered by set inclusion" is customary when using Zorn's lemma -- and it suggests how to find an upper bound for a chain, namely to take the union of all the sets and showing that produces a function with the specified properties.

[summer notch]

yea i see now

that makes sense

[lavish sigil]

Also, if it were isomorphic to $k[t]$, then the singular affine plane curve $y^2=x^2-x^3$ would be isomorphic to the smooth curve that is the affine line itself.

[cloud walrus]

gustavn64

[tribal moss]

Actually, I don't think this is true. For a counterexample consider K=Q[sqrt2] and L=Q[sqrt(sqrt(2))]. If we let f be the nontrivial automorphism of Q[sqrt2] which interchanges sqrt2 and -sqrt2, then there's nowhere in L for g to map sqrt(sqrt(2)) to such that its square is f(sqrt2) = -sqrt2.

[summer notch]

the proof is correct to show that zorn's lemma applies, right? the problem is that M (the maximal element) might not equal L

it would normally go something like "if M neq L then there is some a in L not in M, so M(a) <= L, then blah blah shows that M(a) also works, contradicting maximality", but in fact M(a) might not work

[chilly radish]

Yea you might.not always be able to extend

This always works if L and K are both galois over some base field F

I'm not sure if you can weaken the assumption further

In fact I think if L is normal over some subfield of K, and the automorphism fixes that subfield that's enough

I have to think about this tho

I'm assuming algebraic in either case

The question is if you need galois or if just normal is enough

[formal ermine]

I don't understand "This field is changed only for the steps involving the computation of an nth root."

and what the sequence has to do with them being normal

[summer notch]

fields are closed under addition, subtraction, multiplication, and division, so you won't obtain new fields by doing those operations

[solar glacier]

is it true that if H is normal and K is any subgroup that HK is a subgroup

i know if K is normal and H is any subgroup its true

[formal ermine]

I'd think so

[tawny pine]

doesnt matter which one is normal

[formal ermine]

yeah if either H or K lie in the normalizer of G then HK is a subgroup

[solar glacier]

ok thats what i was thinking thanks

[solar glacier]

thx

[formal ermine]

right

and what about the normal thing?

[summer notch]

if you want the sequence of fields to be normal, you can't just add in some nth root of c, as x^n-c (or x^n-c divided by the nth roots of c that are already in the field) might not split

you have to add in the roots of unity to make F_i/F_{i-1} a normal extension

i.e. you make F_i the splitting field of x^n-c

[celest cairn]

$\sigma: \sqrt[4]{2} \mapsto -\sqrt[4]{2}.$
$\sigma: $i$\sqrt[4]{2} \mapsto $i$\sqrt[4]{2}.$
$\tau: \sqrt[4]{2} \mapsto \sqrt[4]{2}.$
$\tau: $i$\sqrt[4]{2} \mapsto -$i$\sqrt[4]{2}.$
What is $\sigma\tau($i$\sqrt[4]{2})$?

[cloud walrus]

Sapphire Gaming

[formal ermine]

it's the composition

sigma(tau(isqrt4))

btw I'd write them down like this instead:
sigma : sqrt4 -> -sqrt4
sigma : i -> i
tau : sqrt4 -> sqrt4
tau : i -> -i

[solar glacier]

question, how do you know which method to use Sylows theorem? i.e., when do you use the argument that the group ends up with more elements than there were originally in the group and when do you say "the remaining constituttes a sylow p-subgroup which is unique"

and say the order has a prime in its decomposition with expoennt equal to just 1 and say n_p=k

only in this case can you say that theres (p-1)k elements of order p

[celest cairn]

So, would $\sigma\tau($i$\sqrt[4]{2}) = $i$\sqrt[4]{2}$?

[cloud walrus]

Sapphire Gaming

[untold basin]

Hello, what is the minimal polynomial of $\mathbb{Q}(\zeta_9 + \zeta_{9}^4 + \zeta_{9}^7)$

[cloud walrus]

Christophe*

[coral shale]

primitive element theorem might be relevant

nvm dont think so

[untold basin]

I just realised I don't need it

I'm still curious tho

[next obsidian]

This is fixed by z -> z^4 so I think it generates the sub field corresponding to that auto morphism or something like that

Like, it satisfies x^4 - x

Then go from there

[untold basin]

I know the degree by the way

It is 2

I did all the Galois stuff

[next obsidian]

Okay so it’s like

[untold basin]

I determined that [Q(zeta_9) : Q] = 6, its Galois group is isomorphic to Z/6Z
So all the strict subfields are the fix ones

[coral shale]

is this a standard trick

[next obsidian]

x(x^3 - 1) = x(x - 1)(x^2 - x + 1)

So you see which one of these it satisfies

[untold basin]

Wait I don't follow you

[south patrol]

Ig the generator is the orbit sum of zeta_9 under the map sending zeta -> zeta^4 and orbit sums r common

[untold basin]

How did you see that so easily ?

I know that the minimal polynomial is of the form X² + aX + b

[coral shale]

you lost me with a few words

orbit sum ig

[untold basin]

a := zeta_9 + (zeta_9)^4 + (zeta_9)^7

[coral shale]

that sum under different mappings of zeta?

[untold basin]

I did a² = 3((zeta_9)^2 + (zeta_9)^5 + (zeta_9)^7)

I know that (zeta_9)^6 + (zeta_9)^3 + 1 = 0

[south patrol]

If a group $G$ acts linearly on a vector space $V$, then for each $x \in V$, $\sum_{g \in G} g \cdot x$ is fixed by $G$

[cloud walrus]

potato

[untold basin]

But I can't combine that so that my X² + aX + b becomes 0

[coral shale]

hmm

[untold basin]

I THINK I FOUND IT

[next obsidian]

Actually idk if this is true lol I forgot that raising powers isn’t linear lol

But I did find the auto morphism which fixes it

[south patrol]

So in particular what we can do here is take the Galois group of $\mathbb Q(\zeta_9)$ and then consider the subgroup generated by $\sigma$ mapping $\zeta_9 \mapsto \zeta_9^4$ and then the orbit sum (from above) is the generator

[next obsidian]

All auto morphism of Q(zeta_9) are given by raising zeta_9 to some power

[untold basin]

I did it

[next obsidian]

So I guessed 4

And it worked

[untold basin]

We also have the one that send to zeta_9 to the 8th power

[cloud walrus]

potato

[south patrol]

Yes, complex conjugation I guess

[untold basin]

Oh right

Wait wtf I found that X² + aX is the minimal polynomial for all a 🧍

[south patrol]

uh

[untold basin]

z = ((zeta_9) + (zeta_9)^4 + (zeta_9)^7)

I am searching a and b in Q such that
z² + a.z + b = 0 i.e. 3((zeta_9)² + (zeta_9)^5 + (zeta_9)^8) + a ((zeta_9) + (zeta_9)^4 + (zeta_9)^7)) + b = 0

But I know that 1 + (zeta_9)^3 + (zeta_9)^6 = 0 because X^6 + X^3 + 1 is the minimal polynomial of Q(zeta_9) over Q

[next obsidian]

#bots message

Am I dumb?

This says this element is 0

Lol

[untold basin]

So I have that it is equivalent to
3(zeta_9)² ( 1 + (zeta_9)^3 + (zeta_9)^6 ) + a * zeta_9 * (1 + (zeta_9)^3 + (zeta_9)^6) + b = 0 <=> b = 0

[south patrol]

wait yeah

[untold basin]

am i dumb

[south patrol]

isn't it zero

cause like

[next obsidian]

I mean I tried thinking about t geometrically

And didn’t see why it was 0 obviously

But

[untold basin]

it is this : 1 + (zeta_9)^3 + (zeta_9)^6 times zeta_9

[next obsidian]

It looks like it is

[south patrol]

I mean multiply it by zeta_9^3

and you get the same thing

[next obsidian]

Oh my god

Bro that’s obviously 0

[south patrol]

oop

[untold basin]

wait I don't understand anything

[next obsidian]

The sum of two conjugate roots of unity

[south patrol]

so uh

[next obsidian]

Is -1 right?

[untold basin]

it is 00:00 here i am becoming crazy

[next obsidian]

No this obviously isn’t always true

Uhhhh

What’s rucking cos(120)

Isn’t that -1/2

Yeah

[south patrol]

and so this is just 0

[coral shale]

chmonk chmonk lmao

[cloud walrus]

potato

[next obsidian]

Yeah potato good good good

[untold basin]

yes i know and how it helps

[south patrol]

Uh

the field is just Q

[next obsidian]

the minimal polynomial is

Like…

[south patrol]

what even is the min poly in that case

[untold basin]

:OOOOO

[south patrol]

lol

x -1

[coral shale]

x

[next obsidian]

No

It’s ducking x

[untold basin]

wait what

[south patrol]

lol

well

[next obsidian]

My guy

[south patrol]

no generators

lol

[next obsidian]

The element you’re adjoining

Is 0

[untold basin]

WHAT

[next obsidian]

That’s what I linked

[untold basin]

LOOOOOOOOOOOOOOOL

[next obsidian]

And what we’ve been saying for the last like 5 mins

[south patrol]

lol

[untold basin]

THATS WHY IT FIXES MY THING

[coral shale]

F[x]/(x) = F gg

[untold basin]

I WAS LIKE OK I FOUND A THING

[south patrol]

lol

[untold basin]

it is 0

[south patrol]

rip

[untold basin]

xdddddddddddddddddddddd

[next obsidian]

Lol

[untold basin]

pls help

[south patrol]

oh okay yeah

it satisfies x^2 + ax for all a

lol

nice

[next obsidian]

So actually

[south patrol]

so it is 0

[next obsidian]

I was right

[south patrol]

yes

[untold basin]

XDDDDDDDDDDDDDDDDDDDDD

[next obsidian]

This does satisfy x^4 - x

[south patrol]

lol

[next obsidian]

I did factor that wrong tho

It’s x(x-1)(x^2 + x + 1)

[south patrol]

Okay actually so like this is also worth pointing otu like lol

[coral shale]

wait what the answer is 0?

[south patrol]

orbit sums give you 0 half of the time

[coral shale]

fuck

[south patrol]

lol

[next obsidian]

It’s just that the factor is x

[coral shale]

i thought it was that quadratic

[next obsidian]

Not the quadratic

[coral shale]

bruh wtf

[next obsidian]

[south patrol]

gg

[untold basin]

omg 💀

[next obsidian]

Chmowned

[formal ermine]

why?

[cloud walrus]

potato

[south patrol]

for all n

[untold basin]

I need to find the elements of Q(zeta_9) that are fixed by <zeta_9 |-> zeta_9 ^4>

[south patrol]

Oh lol

[coral shale]

Qs in it

heheh

[untold basin]

don't spoil me

but how should I think

[next obsidian]

Isn’t this kinda like

All the zeta_9^k are independent over Q

For k = 0 through 8

[coral shale]

you should think whats definitely in it

Q

[south patrol]

lol

[coral shale]

then think what else might be there

[south patrol]

i mean isn't this an extension of degree 2

[next obsidian]

So write a generic element as a_0 + a_1zeta_9^1 + … a_nzeta_9^8

[south patrol]

so you just need to find a single non-trivial element

[untold basin]

yes

[next obsidian]

Apply the raise by 4 map

Equate coefficients

Or are the zeta_9^k not independent

Oh obviously not

[untold basin]

?????????

[next obsidian]

Lmfaooo

[south patrol]

like

[next obsidian]

We exhibited a relation

When we found that sum that became 0

Wait

[untold basin]

am i being trolled

yes

[south patrol]

hm

[next obsidian]

Add me as friend

Now

[south patrol]

i wonder if this can just be guessed though lol like

[untold basin]

That's what I tried and I ended up putting fucking 0 xddddddddddddddddddddddddddd

[south patrol]

well

wait no

it isn't of degree 2 lol i'm pretty sure

[untold basin]

Wait

[south patrol]

i mean think about it more

[untold basin]

Are we talking about

[next obsidian]

Do you know Galois theory?

[untold basin]

the elements of Q(zeta_9) that are fixed by <zeta_9 |-> zeta_9 ^4>

[untold basin]

):

[celest cairn]

$\sigma\tau($i$\sqrt[4]{2}) = \sigma($-i$\sqrt[4]{2}) = $-i$(-\sqrt[4]{2}) = $i$\sqrt[4]{2}.$

[cloud walrus]

Sapphire Gaming

[south patrol]

interesting you did i in text

[formal ermine]

you can't pull out the -i like that

[south patrol]

i mean fair

[south patrol]

Try to find what this subgroup is first

[formal ermine]

sigma(-i ftrt(2)) = sigma(-i) sigma(ftrt(2)) not -i sigma(ftrt(2))

[south patrol]

okay so like

you said the size was 3 i guess or smth and this is not correct

it isn't multiplication by zeta_3, it is taking it to the power of 4

[untold basin]

Size of what

[south patrol]

the size of the subgroup that map generates

idk like

check that calculation

[untold basin]

the group <zeta_9 |-> zeta_9 ^4> is of order 3

[south patrol]

what is your working for that?

so let's call that map like σ or smth

[untold basin]

i just compose zeta_9 |-> zeta_9 ^4

[south patrol]

so we don't go insane

[untold basin]

yep

[south patrol]

σ^2 sends zeta_9 to what

[untold basin]

σ(zeta_9^4) = σ(zeta_9)^4

[south patrol]

= zeta_9^16 ye

[untold basin]

yep

then again

64

modulo 9

1

[south patrol]

uhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

wait lol

[untold basin]

potato?

[south patrol]

@next obsidian are we dumb

i mean u and me chomkney

like

[untold basin]

zeta^64 = zeta^63 * zeta

(zeta_9^9)^7

=1

[untold basin]

aren't you confused about order and index

[celest cairn]

ok I still got i ftrt(2).

[south patrol]

no i'm not

[untold basin]

Idk how you call it but like the big group divided by the actual group (the orders)

[south patrol]

I just

[untold basin]

do you want me to explain all my Galois stuff quickly?

[south patrol]

wait yeah chmonkey okay so

wait nvm

[formal ermine]

yes, it's correct

[celest cairn]

Cool, thanks

[formal ermine]

np

[south patrol]

Okay nah so basically Chmonkey and I got into some confusion but it's all fine now lol

Your working is fine.

the group is of order 3, so this is an extension of Q of degree 2

So we just need to find a single element not in Q

and the "obvious" thing (the orbit sum) doesn't work

[untold basin]

I was making a paint ):

[south patrol]

okay lol

So yes, there is an "easy" element fixed by σ

[untold basin]

wait i'm thinking of this

are you using primitive element theorem

[south patrol]

No

I'm saying like

[untold basin]

there is an algebraic element on q that is fixed by sigma

yes

[south patrol]

if K/F is an extension of degree 2 and x in K \ F, then F(x) is a subfield of K strictly larger than F which forces F(x)=K

so in fact to find the field you just need to find any element of the fixed field of <σ> which isn't in Q

And you can just guess one

[maiden heath]

Hi peeps, i'm trying to show that the restriction of a representation is still a representation. This is what I have. However my lecturer told me that the restriction of rho is not necessarily a linear map. I thought that restriction of a function preserves linearity. Why is that not true?

[south patrol]

Also Christophe, if you want a more general approach @untold basin uh

Every $g \in G= \mathrm{Gal}(\mathbb Q(\zeta_n)/\mathbb Q)$ is given by sending $\zeta_n \mapsto \zeta_n^k$ for some $k \in (\mathbb Z/n\mathbb Z)^\times$ and in fact the corresponding map $G \to (\mathbb Z/n\mathbb Z)^\times$ is a group isomorphism which I think can help you like think about this

[cloud walrus]

potato

[untold basin]

We haven't proved in my course that it is an isomorphism

Only injection

[south patrol]

Ah okay

[untold basin]

I asked today

[south patrol]

Well injection is easier and then the key thing is just showing that |G| = φ(n)

Which is equivalent to showing that the nth cyclotomic polynomial is irreducible

(do you know what those are)

[untold basin]

Yep, except that we only defined the cyclotomic polynomial for n primes

[south patrol]

I can explain, it's not gonna be helpful for this problem so it's not cheating lol

Ah okay

[untold basin]

I got curious so I know it but it's not in my course

[south patrol]

Ah okay

[untold basin]

My prof said it's a lot of arithmetic

[south patrol]

Oh like to prove it is irreducible?

Yeah it is a bit of a pain actually

Sad times

One thing that is relatively easy though is showing that the cyclotomic polynomials for p^n are irreducible

basically you just do the same argument as for p

But yeah the general case is a slight pain

[untold basin]

(zeta_9)^7 ?

[south patrol]

and a proof I need to review xd

Uhh

[untold basin]

euh

[south patrol]

7*4 = 28 = 1 i guess

so no

[untold basin]

forget

[south patrol]

But try uh (zeta_9)^a

lol

[untold basin]

I am ded xd

[south patrol]

what do you want to be the case

with this a

[untold basin]

a - 4a mod 9 = 0

(zeta_9)^3

o

m

g

i determined that Q(zeta_3) is a subfield

And i'm searching for it ?????????????????????

kill me xddddddddddddd

@south patrol thanks

[south patrol]

wdym you're searching for it

Oh

lol

so you'd already found a proper subfield of Q(zeta_9)^sigma but lol xd

okay nice

Yeah zeta_3 was what I had in mind too

[agile burrow]

I'm having a bit of trouble with this exercise. I'll refer to \Gamma as G and \Gamma' as H.

In the hint, the first isomorphism follows from Shapiro's lemma. 6.9 states that the n-th cohomology of G with coefficients in the coinduced module M is isomorphic to the coinvariants of D (x) M where D = H^n(G, ZG). (Here, D (x) M has the diagonal G-action where G acts on D via g * u = u g^-1.) Finally, Exercise 4b shows that if G is finitely generated and [G : H] = infinity, then the coinvariants of any coinduced H-module is 0. Ultimately, I should be able to show that H^n(H, M') = 0.

I think what I need to do here is show that D (x) M is also coinduced, but I'm not too sure how to see that. I haven't used the fact that H^n(G, ZG) is finitely generated as an abelian group yet.

[cloud walrus]

walter

[agile burrow]

But I don't think this uses the fact that H^n(G, ZG) is finitely generated so I'm not entirely sure

I'm pretty sure the above doesn't work because the bimodule structure doesn't work out the way it needs to

[ruby sundial]

is there a property for algebras where all subalgebras its are finitely generated

because finitely generated algebras can have not finitely generated subalgebras

[chilly ocean]

smells noetherian-y

🧐

except "algebra" could mean something non-associative, of course, depending on conventions and stuff

so...

[ruby sundial]

yeag

i am meaning associative though lol

we in algebra discussion

[chilly ocean]

lie algebras btfo

[ruby sundial]

i dont know how to make this distinction efficiently but here it goes

if the ring the algebra is over Noetherian but the algebra is not or vice versa and are either possible?

my guess is that yes the first is possible

it is possible to have an infinitely generated algebra over a noetherian ring but not a “noetherian algebra” over a infinite dimensional ting

where noetherian algebra means all subalgebras are finitely generated

and i guess yes the answer is that noetherian is this condition we want

also ig a side question is this

for rings S in R, krull dim S < R iff R noetherian?

[ruby sundial]

these arent intended to be answered, sorry for littering thoughts

[maiden ocean]

https://mathoverflow.net/questions/253283/algebras-whose-subalgebras-are-finitely-generated MO post on them

[ruby sundial]

ty

[tender wharf]

Is this even possible? Isn't
\begin{align*}
<(2,0)> \oplus <(0, 2)> &= {(2k, 0) \mid , k \in \bZ } \oplus {(0, 2k) \mid , k \in \bZ} \
&= { ((2k_1, 0), (0, 2k_2)) \mid , k_1, k_2 \in \bZ }
\end{align*}

[coral shale]

is what possible

[cloud walrus]

DerpZ

[tender wharf]

<(2,0)> oplus <(0,2)> does not look like a subgroup of Z oplus Z

[coral shale]

huh

isnt

like

u have 2 subgroups

their direct product

or sum

[tender wharf]

yeah direct product

[coral shale]

will be a subgroup

so...?

thats like a thm

[tender wharf]

oh its asking about the

ah

[coral shale]

direct product must result in a group

and that lives inside original group so subgroup

[coral shale]

Anyways, this is wrong, think carefuller

[tender wharf]

its an

external direct product

that symbol is used in this book as the external direct product

[coral shale]

huh

[tender wharf]

so I was confused

if it's the internal direct product then yeah

[coral shale]

what difference does that make

[tender wharf]

isn't

[coral shale]

the external and internal are the same up to isomorphism

[tender wharf]

oh they're asking about the isomorphism

[coral shale]

if its external then it just means that product is isomorphic to some internal

[tender wharf]

I mean yeah <(2,0)> is isomorphic to <2>

[coral shale]

anyways anyways

to be clear do u see where the error is

[coral shale]

idk wym

[tender wharf]

nvm nvm notation

not immediately

i actually don't see the error

[coral shale]

also, the last product has to be internal

otherwise it makes no sense

the object on the last line is not a group

not closed

[tender wharf]

oh

[coral shale]

ie. some error

[tender wharf]

right right

yaeh I should've taken

x from <(2, 0)> and y from <(0, 2)> does that fix it?

[coral shale]

huh

[tender wharf]

as in

[coral shale]

in plainer words

[tender wharf]

G oplus H is just {(g, h) | g in G h in H}

[coral shale]

ok yes nvm

i was about to say its the even coords

thats an internal direct sum

[tender wharf]

book notation confusing sadge

[coral shale]

in this context right