groups-rings-fields

[rustic crown]

Remember that question we showing Q(alpha) is not Q(beta) for alpha = cbrt(2) and beta = cbrt(3)

[hidden haven]

Yes

[cloud walrus]

det

det

det

[hidden haven]

[cloud walrus]

det

[rustic crown]

same logic says c = 0

and once more gives b = 0

which can't be true.

[hidden haven]

Oh dang

Neat

[rustic crown]

[hidden haven]

Wait how did you get this?

[cloud walrus]

det

[hidden haven]

Nice

[rustic crown]

so just knowing the traces could help a lot, even in bad cases you would still get rational system of equations in a, b, c which we should be able to solve

[hidden haven]

Yeah it's pretty nice

Perhaps I'd study this stuff

[rustic crown]

yea pretty fun ❤️

[sinful mirage]

is anyone familiar with the induced representation?

I am having issues with understanding how 'fixing' the action on two specific kind of elements uniquely determines it

so the setup is that we are given a group G and a subgroup H. Knowing a representation of H, we want to determine a representation of G. To this end, we introduce an equivalence relation on G, such that $g \sim g' \iff g'=gh$ for some $h \in H$. This induces equivalence classes, which have representatives. let's denote the representatives of these equivalence classes with $r$, the set of representatives with $R$ and elements of $W_{r}$ with $w_{r}$. then ew define the representation space of G to be $V=\bigoplus_{r \in R} W_{r}$, where $W_{r}$ is isomorphic to $W$. So we are given a rep $(W,\rho_{H})$ and we want to get a rep $(V,\rho_{G})$. Then we fix how $\rho_{G}(h)$ acts(only for $ h\in G$)., that is given by $\rho_{G}(h):=\rho_{H}(h)$. we also fix how $\rho_{G}(r)$ acts, where r are representatives of the equivalence classes, that is by the isomorphism: $\rho_{G}(r): W_{e} \to W_{r}$

now the claim is that this uniquely determines the action for any $g \in G$

[cloud walrus]

ProphetX

[sinful mirage]

now the claim is that this uniquely determines the action for any $g \in G$

[cloud walrus]

ProphetX

[sinful mirage]

my trial is the following: $\rho_{G}(g) w_{r}=\rho_{G}(g) \rho_{G}(r) w_e=\rho_{G}(r'h') \rho_{G}(r) w_{e}$ and now i am stuck

[cloud walrus]

ProphetX

[sinful mirage]

if I would be able to say that $\rho_{G}(r'h')=\rho_{G}(r') \rho_{G}(h')$ I could proceed as: $\rho_{G}(r'h') \rho_{G}(r) w_{e}=\rho_{G}(r') \rho_{G}(h') \rho_{G}(r) w_{e}=\rho_{G}(r') \rho_{G}(h'r) w_{e}$ now $h'r$ is again an element $g \in G$, thus we can write $h'r=r''h''$. therefore $\rho_{G}(r') \rho_{G}(h'r) w_{e}=\rho_{G}(r') \rho_{G}(r'' h'') w_{e}=\rho_{G}(r') \rho_{g}(r'') \rho_{G}(h'') w_{e}=\rho_{G}(r'r'') \rho_{G}(h'')w_{e}$

[cloud walrus]

ProphetX

[sinful mirage]

this proof would be valid/would work if I could justify why $\rho_{G}(r'h')=\rho_{G}(r') \rho_{G}(h')$ and why $r' r'' \in R$

[cloud walrus]

ProphetX

[sinful mirage]

however,I can not see how I could justify those steps

am I going in the wrong direction with the proof?

[sturdy marsh]

The intuition is that the H part acts on the vector space and the "remaining" bit switches around the summands

any g in G can be written uniquely as g = rh for some r in R and h in H

(following your notation)

[sinful mirage]

yes,this i used

I am unsure though why \rho_G(rh)=\rho_G(r) \rho_G(h)

[sturdy marsh]

we're constructing rho_G

if rho_G existed

it should be a homomorphism

so that relation needs to hold

[sinful mirage]

yes,but this is circular right?

we want to construct a homomorphism

we can't assume it is a homomrphism a priori

[sturdy marsh]

we arent

we are defining it by that formula

[sinful mirage]

we assume that on equivalence classes it is a homomorphism and this should fix everything

[sturdy marsh]

hold on, lemme write something out

any g can be uniquely written as rh

and we DEFINE the action of g on W_e to be v maps to hv \in W_r

[tough raven]

More like showing there is a unique homomorphism satisfying all the special cases you have defined (namely h in H and r in R)

[sinful mirage]

yes,I used this too

you mean ew define the action of G on W_e to be the isomorphism W_e->W_r

right?

and we require this to be a homomorphism

i.e. \rho_{G}(r):W_e->W_r and we require this to be a homomorphism

and everything else should follow from this

[sturdy marsh]

right, and for any other summand W_r', we have gr' = r''h

and g acts on W_r' by v maps to hv in W_r''

[sinful mirage]

wait im a bit confused here

so you say that \rho_{G}(g) w_r=\rho_{G}(g) \rho_{G}(r) w_e=\rho_{G}(r'h') \rho_{G}(r) w_e

or?

that g can be rewritten as r'h'

$\rho_{G}(g) w_r=\rho_{G}(g) \rho_{G}(r) w_e=\rho_G{}(r'h') \rho_{G}(r) w_e$

[cloud walrus]

ProphetX

[sturdy marsh]

im not writing g as r'h'

gr' = r''h

for some r'' in R

and h in H

[sinful mirage]

yes,this can be rewritten,but where does gr' appear?

[sturdy marsh]

I want to define the g action on W_r'

[sinful mirage]

it is already defined for any $r \in R$

[cloud walrus]

ProphetX

[tough raven]

I don't really understand the construction TBH
Does H act on all of the copies simultaneously?
And does r in R send W_e to W_r only or maps every W_r' to W_(r'') where r r' in r''H?

[sinful mirage]

is $r'$ not in R in your notation?

[cloud walrus]

ProphetX

[sturdy marsh]

r' is in R

im just writing down the definition

[sinful mirage]

but the action of G on any $W_r$ is already defined by this

[cloud walrus]

ProphetX

[sinful mirage]

or?

so for any $r \in R$ this works

[cloud walrus]

ProphetX

[sturdy marsh]

arent we trying to define it?

[tough raven]

@ Brofibration if you understood the construction could you clarify this?

[sinful mirage]

I thought we defined how $\rho_{G}$ acts on H and how $\rho_{G}$ acts on $R$ completely,i.e. on all$ r \in R$

[cloud walrus]

ProphetX

[sinful mirage]

and we ask how this uniquely determines the action on $G$

[cloud walrus]

ProphetX

[sturdy marsh]

rho_G is a representation of G yes?

[sinful mirage]

yes

and we know how \rho_{G} acts on R,H completely

where R are representatives,H is the subgroup

[sturdy marsh]

rho_G(g) is a map of vector spaces

[sinful mirage]

so we want to see the action of rho_{G}(g) w_r for any g ,knowing how rho_{G} acts on W_r and h

yes

[sturdy marsh]

what is wr

[sinful mirage]

any vector in W_r

i.e. one of the copies isomorphic to W(the rep on H)

[sturdy marsh]

okay, we can write gr = r'h yes?

[sinful mirage]

yes

r'h'or anything

[sturdy marsh]

rho_G(g)(w_r) = (rho_H(h)(w_r)) \in W_r'

[sinful mirage]

how did you conclude this?

[sturdy marsh]

that is the definition

it might be easier for you to see this if you're doing it for sets instead of representations

say I have a set X with an H-action

can you build an induced G-set?

[sinful mirage]

to be honest and not to lie,I can't

[sturdy marsh]

it's the same idea

[sinful mirage]

I can't see why this would be the definition of rho_{G}(g)

I thought we want to define rho_{G}(g) using only what we have

[sinful mirage]

i.e. like this

[tough raven]

How does rho_G(r) act on W_r' for r' different from e?

[sinful mirage]

and work out the action using the given rules

[sinful mirage]

right,that I do not know yet

I only know that $\rho_{G}(r) : W_e \mapsto W_r$

[cloud walrus]

ProphetX

[sturdy marsh]

how would r act on W_r'

[sinful mirage]

my prof defined it as such

however,I can't see why r_0 r would be in R

[sturdy marsh]

r_0r is not in R

in general

[sinful mirage]

so why would this work?

[sturdy marsh]

it doesnt a priori

[sinful mirage]

so yes,the first step is constructing $\rho_{G}(r): W_{r'} \mapsto W_{r}$

[cloud walrus]

ProphetX

[sinful mirage]

at least this is what I see now

then we need to use thsi to construct $\rho_{G}(g) w_r$

[cloud walrus]

ProphetX

[sinful mirage]

since we don't know this yet

[sturdy marsh]

it doesnt necessarily map to W_r

that is the case only when r' = e

[sinful mirage]

okay,so $\rho_{G}(r) w_{r}$ doesn't map to $W_{r}$, where can it map then?

I could construct it as my prof defined above,but can't see where it maps

[sturdy marsh]

rr' = r''h

for some r'' in R

[cloud walrus]

ProphetX

[sturdy marsh]

and h in H

uniquely

yes?

[sinful mirage]

yes

[sturdy marsh]

it would map to W_r''

[sinful mirage]

wait,sec

you mean r'r=r''h,right?

this seems to be ok so far

and in our notation r_0=r'

r'r=r''h for some h in H

[sturdy marsh]

no r should be on the left

if we're trying to figure out what the r action on W_r' is

[sinful mirage]

$\rho_{G}(r) w_{r'}=\rho_{G}(r) \rho_{G}(r') w_e=\rho_{G}(r r') w_e$

[cloud walrus]

ProphetX

[sinful mirage]

right sorry

and rr' is an element in G,hence rr'=r''h for some h in H

yes

okay,so we have this so far

[sturdy marsh]

so then p_G(rr') = p_G(r''h)

[sinful mirage]

yes

[sturdy marsh]

=p_G(r'')p_G(h)

[sinful mirage]

this i can't see why

why this holds

[sturdy marsh]

it needs to be true

[sinful mirage]

since r''h is not in R

[sturdy marsh]

necessarily

if it exists

[sinful mirage]

but then this is extra structure than what we are provided

[sturdy marsh]

we know how h acts tho

[sinful mirage]

yes,we know how acts

but we don't know how (r''h) acts

since r''h is neither in H nor in R

[sturdy marsh]

if there is an action, it needs to be that way

heres an analogy

if I have a vector space V with a basis S

and another vector space W

and a map of SETS S ---> W

then this determines a unique linear map from V ----> W yes?

[sinful mirage]

yes

[sturdy marsh]

and how do we get this map?

the relation f(sum_i e_i) = sum_i f(e_i) NEEDS to be true IF a linear map V--->W which agrees with S ---> W exists

similarly in our case if a group exists, that relation NEEDS to hold

[sinful mirage]

yes,makes sense

[sinful mirage]

ok now I can see why this holds

how can I use this to construct further?

[sturdy marsh]

we just did it for r

write g = rh

and now if a G-action exists, we must have gv = rhv = r(h(v))

we know how r and h act

[sinful mirage]

so we know how $\rho_G(r)$ acts on any $W_r$ so far

[cloud walrus]

ProphetX

[sturdy marsh]

so this determines a G action

[sinful mirage]

why?

what is v here

[sturdy marsh]

vector

if an action exists, that relation needs to hold

by definition of action

[sinful mirage]

vector in which W?

any W_r'?

[sturdy marsh]

yes

[sinful mirage]

what is r(h(v))?

fisrt equal sign I see

[sturdy marsh]

okay my bad, I should have written it using your notation

[sinful mirage]

you mean $\rho_{G}(g) w_{r}=\rho_{G}(g) \rho_{G}(r) w_e$?

[cloud walrus]

ProphetX

[sturdy marsh]

p_G(rh) = p_G(r)p_G(h) NEEDS to hold if a homomorphism p_G: G ---> GL(whatever) exists

[sinful mirage]

yes,this I agree

but I can't see how to use this to construct the G action

[sturdy marsh]

so DEFINE p_G(g) := p_G(r)p_G(h)

[sinful mirage]

$\rho_{G}(g):= \rho_{G}(r) \rho_{G}(h)$, how does this tell me how I can act on $w_r$?

[cloud walrus]

ProphetX

[sinful mirage]

$\rho_{G}(g) w_r=\rho_{G}(r) \rho_{G}(h) w_r$

[sturdy marsh]

we know how r and h act on everything

[cloud walrus]

ProphetX

[sinful mirage]

$\rho_{G}(h)$ can't act on $w_r$

[cloud walrus]

ProphetX

[sturdy marsh]

h acts by the original rep

[sinful mirage]

oh

right

I thought only on $w_e$

[cloud walrus]

ProphetX

[tough raven]

We're only given how they act on W_e, but that fixes how everything acts on everything, yes

[sturdy marsh]

no all of the W_r are just copies of the same vector space

[tough raven]

Yes

But rho_G(h) isn't given on W_r

Even though they're isomorhpic

[sturdy marsh]

we're trying to define something

none of the other stuff exists a priori

[tough raven]

Specifically, if rho_G(h) and rho_G(r) are defined only on W_e, then
rho_G(g) (w_r) = rho_G(g) rho_G(r) (w_e) = rho_G(r' h) (w_e) = ( rho_H(h)(w) )_r'
where r'h = gr defines r', h.
So rho_G(any g) (any w) is fixed by rho_G(h), rho_G(r) only on W_e.

IK you probably said this, but I just got it.

[sinful mirage]

$\rho_{G}(r'h)(w_e)$ i see until this point

[cloud walrus]

ProphetX

[sinful mirage]

how to get last equal sign?

[tough raven]

Assume a representation exists, we're trying to show it's unique.

So we can assume the properties of representations

[cloud walrus]

Raghuram

[sturdy marsh]

it is a definition

which necessarily needs to hold if a representation exists

[tough raven]

Yes, from here it's by definition.

[sinful mirage]

ohhhhhhh,so $\rho_{G}(r'h)=\rho_{G}r' \circ \rho_{G}(h)$ and we know how$\rho_G}(h)$ acts on $w_e$

[cloud walrus]

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

[tough raven]

Yes, and how rho_G(r') will act on the result.

[sinful mirage]

yes makes sense

[tough raven]

So there's only one possible answer for $\rho_G(g) (w_r)$, which means only one possible representation.

[cloud walrus]

Raghuram

[tough raven]

I didn't figure out how to show that that is really a valid representation yet, though 😅.

[past temple]

given char k = p, how does one show that [k(t) : k(t^p)] = p

[sinful mirage]

okay,now I see why this holds accepting the definitions, thank you so much @sturdy marsh and @tough raven for the patience

[tough raven]

TBH I only understood it because of Brofibration's explanation.

[sinful mirage]

writing this in my latex notes after understanding it is such a relief

i've been struggling on this for so long

[hidden haven]

You dont need char k = p

k(t) = k(t, t^p) = (k(t^p))(t), try to find the degree of the newly adjoined element

[past temple]

oh

yeah t is a root of x^p - t^p

another question

for some reason my book doesn't explain this

in the context of perfect fields

what does the notation F^p mean?

where F is a field

[hidden haven]

set of pth powers in F

that turns out to be a field

if char F = p

because then its the image of the frobenius map

[past temple]

oh and the image must be another field

[hidden haven]

yep

you can also directly check it lol

That it is closed under operations etc

[past temple]

i see

[sinful mirage]

does cayley's theorem also work for lie groups?

i.e. any group is isomorphic to a subgroup of the symmetric group acting on G?

[prisma ibex]

No

[sinful mirage]

so only for finite groups right?

[prisma ibex]

So even if we take Cayley’s theorem to be about finite groups

There is an analogue of this for finite dimensional Lie groups but it’s more complicated

[sinful mirage]

can you give an example of an infinite dimensional lie group please?

the dimension of the lie group is the dimension of the manifold right?

how can one even define an infinite dimensional lie group

[prisma ibex]

You can use things like Banach manifolds and so on

Anyways

Cayley’s theorem is in some sense a theorem about the regular representation of a (finite) group G

So you can look at the analogous regular representations of Lie groups

[sinful mirage]

yes I heard about this but only in compact case

for instance if the group is not compact then there a lot of issues arise(at least in phys)

[prisma ibex]

Right the compact case is Peter Weyl, the non compact case is hard

[tough raven]

Oh, BTW, if you proved that it was a valid representation, could you share that proof as well?

[prisma ibex]

In the compact case the regular representation of G on L^2(G) decomposes into unitary representations

In the non compact case you have to consider possibly nonunitary representations

[sinful mirage]

do you have a 'path'/references to understand the proof of the Peter WEyl theorem? I know it's a long journey,given my knowledge,but right now i'm following a cousre on rep theory of finite groups and lie algebras(fulton harris)

[sinful mirage]

sure,will do if I wrote it all up in LaTeX

[prisma ibex]

There’s more stuff you can say like the tempered representations are exactly those in the support of the Plancharel measure in this spectral decomposition and the discrete series representations are those with positive measure

So this might be bad advice but imo

Finite group representation theory is very different from representation theory of Lie groups and Lie algebras

[sinful mirage]

(I know nothing neither about functional analysis,nor harmonic analysis)

[prisma ibex]

So much so that the former is not necessarily a prerequisite

[sinful mirage]

I thought finite group rep theory would be a good point to start with

[prisma ibex]

In some ways yes

In other ways not really the situations are genuinely just very very different

The later situation is a lot heavier on analysis obviously

Functional analysis is definitely a prerequisite but again like

You can learn this stuff by just diving in and forcing yourself through it

[sinful mirage]

which is a 'self contained' reference for mathematicians in rep theory of lie groups?

Brian Hall does mostly lie algebras

[toxic scaffold]

you know they're closely related right

[prisma ibex]

It depends, do you care just about finite dimensional representation theory or the infinite dimensional setting as well

[sinful mirage]

in finite dimensions yes,in infinite i'm not sure if lie's correspondence theorems work

the infinite dimensional one is actually the most important one cause that appears in qft/qm

[prisma ibex]

Right yea

[sinful mirage]

do actually lie's correspondence theorems work in infinite dimensions?

i.e. that every rep of a lie algebra can be lifted to a rep of a simply connected lie group?

[prisma ibex]

For infinite dimensional Lie groups not really stuff starts to fall apart

I’m talking more about like

[sinful mirage]

reps of finite dimensional lie groups on infinite dimensional spaces

in physics I doubt there appear infinite dimensional lie groups themselves

[prisma ibex]

You’re looking at representations of a finite dimensional Lie group acting on an infinite dimensional vector space

Yea

I mean Lie’s theorems are more about the actual Lie group/Lie algebra correspondence

There is some extra tricks you have to do in the infinite dimensional case

For instance (g,K) modules in the sense of Harish Chandra

[sinful mirage]

but then do they hold for infinite dimensional reps of finite dimensional lie groups?(just the result-getting there will take time)

[prisma ibex]

Essentially yes there’s just some funny business with (g,K) modules and all that

Or like having to put some more adjectives on things sometimes

But things basically work as long as your representations aren’t anything atrociously behaved

(The adjective is usually to only look at admissible representations)

[toxic scaffold]

in physics the reps. of classical groups are for the most part the only Lie group reps. that get used afaik

[prisma ibex]

Mhm

[sinful mirage]

which book would you recommend @prisma ibex ?

[prisma ibex]

I’ll be honest I’ve never found a really good book that does everything here

Or rather I’ve never really looked for one and instead just learned a lot of this stuff from bits and pieces online

Harish Chandra’s original papers are good just kinda outdated

Vogan’s Unitary representations of reductive Lie groups is good

That’s probably the best source I know of I suppose

[toxic scaffold]

Anyone looked at Weyl's books?

[untold sapphire]

Surely for the infinite dimensional case you're better off just studying the Universal Enveloping Algebra?

Since it contains all representations of the Lie Algebra

[delicate bloom]

I read his book 'Space, Time, Matter' several years ago, pretty fun, but past that nothing lol

[untold sapphire]

Also you don't have to deal with non associativity

And you've got a PBW basis, etc

I think this is a fairly common approach

Also you get to learn about Hopf Algebras, which are fun

[sinful mirage]

there are some nice mathphys lecutres notes on that but very advanced

[untold sapphire]

Dascalescu et al have a nice book with solutions

Really the hard part is getting used to coproducts

[toxic scaffold]

I guess I still haven't properly finished group theory 101 😦 what I know about Lie groups and representations is little informal bits I've picked up from studying physics

I got the point in the abstract algebra course where they pivot to fields and rings

[prisma ibex]

Imo the fastest way to get into this is just worry about the totally general theory later and start with examples

[delicate bloom]

yeah dive in, you're not gonna break your neck cause it's too shallow or anything lol

[prisma ibex]

Understand the representation theory of R and U(1) really well (this is Fourier theory) and then understand SL_2(R) really well

[toxic scaffold]

in physics we do reps of SO(1,3) a lot

[sinful mirage]

in physics we need reps of poincare

[prisma ibex]

Honestly if you can get SL_2(R) nailed down really well a lot of the other examples will get a lot easier

[sinful mirage]

but doing reps of poincare is... nightmare I guess

[prisma ibex]

The SO(3,1) situation is very similar to the SO(4) situation

[toxic scaffold]

we did Poincare rep. on Fock space

[prisma ibex]

Usually the next handful of examples would be unitary groups and orthogonal groups in low dimension

And some spin groups in low dimension

For physics this is most of what you need anyways and you can just learn by example rather than going too far into the general theory

Perhaps someday I will know the general theory I just know a few small groups really well lmfao

SL_2 is essentially the single most important nontrivial example, and I know Sp_4 really well and that’s it

Oh SL_3 I guess

That’s it lol

[toxic scaffold]

I've wondered about stuff like the weight of a representation. The wikipedia page seems a bit dense for me.

[prisma ibex]

Ah yea this stuff is good

The SL_2 case is the simplest

The finite dimensional representations of SL_2 are classified by a single integer, the highest weight

The generators e and f of the Lie algebra act as weight raising/lowering operators

the remaining generator h doesn't change the weight but it acts with eigenvalues the weights of the representation

[toxic scaffold]

I don't understand 😦

What are e and f? Any two generators of a set of 3 generators? How to they raise and lower? How does this relate to eigenvalues? Maybe it's too much right now

[prisma ibex]

if you're in the Lie algebra sl_2 of 2x2 matrixes with trace 0

a standard choice of generators of this 3-dimensional Lie algebra is e, f, h given by

these satisfy [h,e]=2e, [h,f]=-2f, [e,f]=h

actually any elements {e,f,h} of a Lie algebra g satisfying [h,e]=2e, [h,f]=-2f, [e,f]=h is called an sl_2 triple

[toxic scaffold]

I can see e and f have different rank than h, is that figuring in here

[prisma ibex]

I don't know if this is totally related?

anyways

now for each n≥0 there's an irreducible representation V_n of sl_2 of dimension n+1 and highest weight n

I guess on the Lie algebra side it would be like

V_1 is the standard representation of SL_2 (the action of SL_2 on a 2-dimensional vector space)

and V_n=Sym^nV_1

in terms of the weights what does this look like

well the element h is semisimple and it acts on V_n with eigenvalues -n,-n+2,...,n-2,n

(so in particular this decomposes V_n into n+1 1-dimensional eigenspaces according to the eigenvalue by which h acts)

the elements e and f move things between different eigenspaces

e increases the eigenvalue by 2 (weight raising) and f decreases the eigenvalue by 2 (weight lowering)

Maybe one really nice way to realize this is like

V_n is isomorphic to the vector space of homogeneous polynomials in 2 variables x and y of degree n

e is say yd/dx and f is say xd/dy

[sinful mirage]

@tough raven actually checking that it really is a representation is very tedious

i'm stuck

we are given $\rho_{G}(g) w_{r}:= \left(\rho_{H}(h') w_{e} \right)_{r'}$ and we should check that this is a rep

[cloud walrus]

ProphetX

[sinful mirage]

that is $\rho_{G}(g g') w_{r}=\rho_{G}(g) \rho_{G}(g') w_{r}$

right?

[cloud walrus]

ProphetX

[sinful mirage]

this is what I have so far

[tough raven]

That's why I asked

[tough raven]

Yep, and inverses too.

[sinful mirage]

what do you mean inverses?

if rho is a homomorphim

OH NOOOOOO

[tough raven]

This is all the uniqueness part?

[sinful mirage]

yes

[tough raven]

Right

[sinful mirage]

now I should show that it exists

[tough raven]

Well, just expand definitions IG.

[sinful mirage]

but i don't see in my final result where g appears

[tough raven]

r', h depend on g

and r

[sinful mirage]

wait sec

this is ym final result

in the first line should be w_e right?

[cloud walrus]

Raghuram

[tough raven]

Ye
Assume a w without a subscript is _e

[sinful mirage]

sorry for being a burden my brain is slow i'm processing and trying to write it in latex myself to see

i don't see where you got thes e conditions from

[tough raven]

OK for inverses we know $\rho_G(g^{-1}) \circ \rho_G(g) = \rho_G(g) \circ \rho_G(g^{-1}) = \rho_G(e)$. So it should be enough to show that $\rho_G(e)(w_r) = w_r$. And indeed $er = r e$, with $r \in R$ and $e \in H$, so $\rho_G(e)(w_r) = (\rho_H(e) (w))_r = w_r$, so we're done.

[cloud walrus]

Raghuram

[tough raven]

That's from the formula we derived in uniqueness right?
$\rho_G(g) (w_r) = (\rho_H(h')(w))_{r'}$, where $r', h'$ are such that $r' h' = gr$.

[cloud walrus]

Raghuram

[tough raven]

The pair r', h' exists and is unique because of how R was constructed as choosing one representative of each coset gH.

[sinful mirage]

ohh just relabeled

for g_2

since now we have rho_{G}(g_2)

right

[tough raven]

Yes

Since r'', h'' was on the way

[sinful mirage]

yes right

[sinful mirage]

why do we know first line?

ah since we proved it is a homomorphism

from 1)

right?

so I understand the proof for multiplication

[tough raven]

We proved that it's multiplicative.

[sinful mirage]

yes

[tough raven]

And used that, to reduce the inverse requirement to identity requirement.

[sinful mirage]

therefore we can say rho_{G}(g) rho_{G}(g^-1)=rho_{G}(e)

[tough raven]

Then prove the identity part.

[tough raven]

Exactly.

[sinful mirage]

ohh you meant you want to prove $\rho_{G}(e)=id_{w_r} w_r=w_r$ right?

[cloud walrus]

ProphetX

[sinful mirage]

i.e. it maps identity to identity

the property of homomorphism that it maps identity to identity

[tough raven]

$\rho_G(e) = \operatorname{id}_V$

[cloud walrus]

Raghuram

[tough raven]

But it's sufficient to show that $\rho_G(e) (w_r) = w_r$ for any $r \in R, w_r \in W_r$, since $W_r$ spans $V$.

[cloud walrus]

Raghuram

[sinful mirage]

yes

[sinful mirage]

r'h'=er

where did you get re from?

so how does $\rho_{G}(e) w_r=w_r$ imply $er=re$?

[cloud walrus]

ProphetX

[sinful mirage]

$\rho_{G}(e) w_{r}=\left(\rho_{H}(h') w_{e}\right)_{r'}$, where $er=r'h'$

[cloud walrus]

ProphetX

[sinful mirage]

can you elaborate please why we need the inverse condition?

for a homomorphism isn't the inverse condition always satisfied?

[tough raven]

That's what a homomorphism is?

I mean

[sinful mirage]

a homomorphism is by def $\rho(g_1 g_2)=rho(g_1) rho(g_2)$

[tough raven]

Maybe there's a theorem that multiplicative implies the inverse condition

[cloud walrus]

ProphetX

[tough raven]

But I wasn't sure so I proved it.

[sinful mirage]

ahh ok I see

can you elaborate how you got er=re?

i see er,but not re

[tough raven]

er = re?

What?

e is the identity of G

[sinful mirage]

er=re here

[tough raven]

By definition, e(anything) = (anything)e = anything

[sinful mirage]

yes thats right

[tough raven]

Oh

[sinful mirage]

but why you need both directions?

[tough raven]

So er = r
re = r
hence er = re

That was to calculate \rho_G(e)

So $\rho_G(g)(w_r)$ requires us to find $r \in R$ and $h \in H$ such that $rh = gr$ (such r, h guaranteed to exist uniquely).

[cloud walrus]

Raghuram

[sinful mirage]

oh

[tough raven]

er = re (because both are r) and $r \in R, e \in H$, the latter because $H$ is a subgroup

[cloud walrus]

Raghuram

[sinful mirage]

right

[tough raven]

So r, e must be the elements of $G$ we were looking for. That's it.

[cloud walrus]

Raghuram

[sinful mirage]

right makes sense

[sinful mirage]

and a very last question. here when you said we also have,did you use that g_1 g_2 r=r'' h'' h'?

i.e. you said that there is $\rho_{G}(g_1 g_2)w_r=\rho_{H}(h''')w_e)_{r'''}$

and then you said $g_1 g_2 r= r'' h'' h'=h''' r''' \implies r'''=r''$ and $h'''=h'' h'$?

[cloud walrus]

ProphetX

ProphetX

[sinful mirage]

or how did you conclude how $\rho_{G}(g_1 g_2) w_r$ looks like?

[cloud walrus]

ProphetX

[sinful mirage]

this is how I would do it but i'm not 100% sure this is correct

[tough raven]

It should be r''' h''' not h''' r'''
But yes

[sinful mirage]

right

sorry

ok makes sense,FINALLY

thanks so much for the patience

[tough raven]

g1g2r = r'' (h'' h') = (something in R) (something in H)
And that representation must be uniqeu

[sinful mirage]

after a whole day finally the proof works

[fiery berry]

so i have to show that if t is transcendental over K, then there exists an unique isomorphism $\gamma : K(t) \rightarrow K(t)$ such that $\gamma (k) = k \forall k \in K$ and $\gamma(t) = \frac{1}{t}$ My progress so far: Since t is transcendental the evaluation homomorphism $ev_t : K[x] \rightarrow K(t) $ is injective, thus we can extend it to a unique isomorphism phi from Q(K[x]) to K(t) with phi(p/q) = p(t)/q(t) via the universal property of the quotient field . I am a little stuck at this point, i dont even really know whether im heading in the direction. Can someone pls give me a hint ?

[cloud walrus]

chrisply

[fiery berry]

or well i know now that phi would already fulfill the first requirement namely that phi(k) = k for all k in K

[chilly ocean]

Tbh I didn't fully read what you wrote but I think it should basically follow from writing an element of K(t) as p(t)/q(t) (at least, uniqueness follows, forgetting about existence)

[chilly ocean]

Maybe this is not formal enough, idk

What is Q(K[x])?

[fiery berry]

uniqueness follows but not that gamma(t) = 1/t right ?

the field of fractions of the polynomial ring K[x]

[chilly ocean]

Erm, I'm not sure I follow. I was just meaning that if gamma is determined on k and on t, then it must be determined on all of K(t), simply gamma(p(t)/q(t)) = p(gamma(t))/q(gamma(t))

[fiery berry]

hm yeah that makes sense

[chilly ocean]

Existence should follow too, we can write gamma(p(t)/q(t)) = t^n p(1/t)/t^n q(1/t) for n large enough to make the numerator and denominators polynomials. And presumably isomorphism properties are not hard to check

[fiery berry]

i think i'll have to think about this for a bit

[chilly ocean]

Maybe you would want your evaluation to be at 1/t instead of t? And something like Q(K[x]) iso to K(t) in the obvious way, so that chaining things together you get the iso from K(t) to K(t) via t maps to 1/t

[fiery berry]

oh yeah that sounds really good

ok im pretty sure this will work, ill work out the details now. Thanks for your help 🙂

[sturdy mirage]

I'm trying to understand something pretty basic... if $E$ is the splitting field of some polynomial over $\mathbb{F}_p$ with $p$ a prime, is the characteristic of $E$ the same as for $\mathbb{F}_p$? Specifically, is that characteristic $p$?

[cloud walrus]

reking

[golden pasture]

yup, recall how we define the characteristic

1+1+... p times ... + 1 = 0

[sturdy mirage]

Yeah, that's what I figured. The lecturer must have made some small mistake, which always makes me start doubting myself

[chilly ocean]

there arent field homomorphisms between fields of different characteristic anyway

[sturdy mirage]

what about the zero homomorphism

[delicate bloom]

there's no zero field

annoyingly/conveniently we define field homomorphisms to send the multiplicative and additive identities from one field to the other

[sturdy mirage]

ohhh okay

[chilly ocean]

why annoyingly?

i think makes much more sense

[delicate bloom]

well annoying in the sense that I have to be the guy saying "ackshually..."

conveniently because it makes much more sense lol

[sturdy mirage]

so you could only have a zero morphism as a ring homomorphism, from a field to a ring

[chilly ocean]

zero morphism from a field to another field is not a field morphism

but its a rng morphism

[sturdy mirage]

i am the "but what about the trivial case!" person

[chilly ocean]

i am every case i can think of and everyone hates me

[sturdy mirage]

i don't know you enough to hate you, yet!

[tough raven]

This doesn't prevent the zero field from existing?

Just means there are no morphisms out of it. You can have a morphism into it.

[sturdy mirage]

isn't that just by definition of field needing to have at least a 0 and a 1..?

[chilly ocean]

in a field 0 is not equal to 1

so theres no zero field

[rustic crown]

i want my ring maps to send 1 to 1

[golden pasture]

in the 0 ring, 0=1

but i think ring maps need 1 to 1 as well typically right?

else your 0 ring has a map to all rings

[rustic crown]

yea, that's what i feel is nicer.

[chilly ocean]

right, replace ring morphism by rng morphism in my message

[rustic crown]

[chilly ocean]

wtf is rng, are you mad?

[golden pasture]

ring without i

ring without identity

ring without 1

[chilly ocean]

You can now shut.

[chilly ocean]

rings have 1

rngs don't

and yes you can dm me

[rustic crown]

rigs don't have additive inverses (or if you want it doesn't have negatives)

[chilly ocean]

" " don't have any structure

[proud bear]

you sure? they seem to be some sort of space

[oblique river]

Pls dont call them rigs

They are semirings

Which sounds much better than rig

Rng is enough of a meme already

Monoid is still used though

[snow flint]

rng is randum number generator 🙂

[oblique river]

Monoid is a semigroup with identity

[chilly ocean]

there are groups and groupoids, rings and ringoids, ??? and monoids

[sturdy mirage]

how do you pronounce rng

[golden pasture]

r n g

[chilly ocean]

is $\sum_{1\leq j \leq n} x_j^2$ reducible over $\mathbb{C}$ for any $n\geq 3$?

[cloud walrus]

Carla_

[latent anvil]

I think x^2 + y^2 + z^2 is irreducible?

so let $A = \C[x,y]$ and consider $f(z) \in A[z]$ defined by $f(z) = z^2 + (x^2 + y^2)$. If $f$ factored then $x^2 + y^2$ would be a square in $A$. If $f$ were a square it would be the square of a linear polynomial, $f = (ax + by + c)^2 = ax^2 + abxy + by^2 + cax + cby + c^2$. comparing coefficients we see $a = b = 1$ but $ab = 0$, a contradiction

[cloud walrus]

shamrock

[latent anvil]

oh sorry you were asking if this is the case for all n hmm

so this comes down to whether $x_1^2 + \ldots + x_n^2$ is a square for any $n \geq 2$

[cloud walrus]

shamrock

[latent anvil]

And I think the same logic as for the n = 3 case I posted above shows this is impossible

oh whoops sorry i screwed up, it should be $a^2, ab, b^2, ca, cb, c^2$

[cloud walrus]

shamrock

[latent anvil]

so $a^2 = b^2 = 1$, still contradictory with $ab = 0$

[cloud walrus]

shamrock

[latent anvil]

anyways I think the same logic works in general. if $x_1^2 + \ldots + x_n^2 = (a_1 x_1 + \ldots + a_n x_n + b)^2$ then comparing the $x_i^2$ coefficients and the constant term gives $a_i^2 = 1$ for all $i$ and $b = 0$. but we have a term like $a_1^{n-1} a_2 x_1^{n-1} x_2$ in the final square on the right which is nonzero, while the $x_1^{n-1} x_2$ term is zero on the left

[cloud walrus]

shamrock

[latent anvil]

@chilly ocean is this making any sense?

[fringe dawn]

I am stuck showing something that should be relatively simple. I'm working with k=Z_p and vector spaces over k. I've showed that there are exactly p k-endomorphisms for k (basically showed that an endomorphism is given as a multiplication). I'm trying to show that up to isomorphism (in the category of k-vector spaces) there is only 1 object that has p endomorphisms. Any ideas?

[oblique river]

Maybe you could prove that the endomorphism ring of a vector space acts transitively

Well maybe the zero vector is odd but

If v is any nonzero vector in V, and w is any vector in V, possibly zero, then there is an endomorphism of V which takes v to w

(This might require the axiom of choice for infinite-dimensional vector spaces? But maybe not)

[fringe dawn]

hmm

[oblique river]

For finite-dimensional vector spaces it’s easy since endomorphisms are just matrices

[fringe dawn]

yeah

[latent anvil]

(I think this probably requires choice in the infinite dim case)

[oblique river]

That is my instinct as well

[fringe dawn]

Wouldn't it be possible to just pick a finite dimensional subspace?

[oblique river]

You need to define the endomorphism@on all of V

And so you probably need to do something like

Pick a basis for your fd subspace and then expand that to a basis of V

[fringe dawn]

Yeah but I can literally project everything that isn't in the subspace to 0?

[latent anvil]

here's something mindblowing: it's consistent with ZF that there's a vector space V over Z_2 with no nonzero linear maps V -> Z_2 (https://mathoverflow.net/a/79437/136523)

[oblique river]

No you cant

You can project a subspace to 0

But not the complement of a subspace

[fringe dawn]

Oh right

Yeah got ya

[sturdy marsh]

you can if you have an inner product tho

right?

[oblique river]

You need to define a “complementary subspace”

[fringe dawn]

Yeah, exactly

And that's not an issue in the finite dimensional case

[oblique river]

Isnt picking an inner product the same thing as picking a basis

Or at least picking an inner product also requires AoC

[sturdy marsh]

yeah that sounds right

[oblique river]

This is nuts btw lol

[latent anvil]

Yeah haha

[next obsidian]

Just pick a basis element omegalol

[latent anvil]

also what are you saying about an inner product brofibration?

[sturdy marsh]

defining complements

if you have an inner product then you get orthogonal complements

[oblique river]

If you have an inner product you can define “orthogonal complement” snd then do exactly what you want, namely, project the complement of a subspace to 0

[latent anvil]

I think you'll need to start with a closed subspace to make this work

[oblique river]

It’s finite dimensional

[latent anvil]

oh sorry I missed that we were in finite dimensions

[oblique river]

No, it’s a finite dimensional subspace

[latent anvil]

oh I see what you mean

yes

that makes sense

[oblique river]

Also we’re working over F_p

[latent anvil]

wait then what even is an inner product lol

just a nondegenerate bilinear form or smth?

[oblique river]

Yeah, just “whatever you get if you choose a basis”

[latent anvil]

oh sure

[oblique river]

Although thats not nondegenerate

[latent anvil]

wait how so?

[oblique river]

If you have a p-dimensional spce

Then (1,1,...,1) paired with itself

Is 0

[latent anvil]

non degenerate just means <v, w> = 0 for all w implies v = 0

[oblique river]

Right?

[latent anvil]

which is still true

[fringe dawn]

Actually I don't think I even need to go that far.

[latent anvil]

because you can take w to be the basis vectors

[fringe dawn]

I think in my case it's enough to argue that any vector space is a free module

[oblique river]

Oh nvm i tjought it meant <v,v> neq 0

[latent anvil]

I agree that things can be self-orthogonal though

[oblique river]

That requires the axiom of choice

But yes i think the conclusion of this discussion is

Just assume all your vector spaces have bases

And then it’s easy

[latent anvil]

ah I found what I was looking for

it's not quite a vector space over Fp with automorphism group order p

but you can have a vector space over F2 with automorphism group cyclic of order 3

(in the absence of choice)

same post but the next answer: https://mathoverflow.net/a/49404/136523

[fringe dawn]

yeah I never said I don't need the axiom of choice 😄

[latent anvil]

okay algebra time

wait sorry Pseudo are you done with your question?

[fringe dawn]

Sure

[latent anvil]

it seemed like it but I wanted to check

cool

[fringe dawn]

okay algebra time
My question was an algebra question 😄

[latent anvil]

sorry haha

I meant "my algebra time"

[fringe dawn]

😄

[latent anvil]

I was answering a functional analysis question in another channel

so im switching gears

[fringe dawn]

Sure, I'll let you do your thing 😄

[latent anvil]

okay so let $(R, \mathfrak{m})$ be a noetherian local ring of depth $\geq 2$. Suppose there's a ring $R'$ and an element $f \in \mathfrak{m}$ with $R \subseteq R' \subseteq R_f$ (feel free to assume $f \notin \mathfrak{m}^2$ if you want). Further suppose $R'$ is finite over $R$

[cloud walrus]

shamrock

[latent anvil]

This should be impossible and I'm trying to figure out why

chmonkey has a proof using local cohomology but I don't want to learn local cohomology

my current line of thought is whether $R'$ free, i.e.\ $\mathrm{depth}_R R' = \mathrm{depth} R$, i.e.\ $R'$ is flat

[cloud walrus]

shamrock

[latent anvil]

note that $\mathfrak{m}^r (R'/R) = 0$ for some $r$

[cloud walrus]

shamrock

[latent anvil]

also using the short exact sequence $0 \to R \to R' \to R'/R \to 0$ and computing some ext groups we get $\mathrm{depth} R' \geq \mathrm{depth} (R'/R)$ and $\mathrm{depth} (R'/R) \geq \mathrm{depth} R - 1$

[cloud walrus]

shamrock

[latent anvil]

wait...

shouldn't $\mathfrak{m}^r (R'/R)$ force $\mathrm{depth} (R'/R) = 0$?

[cloud walrus]

shamrock

[latent anvil]

i think so :o

AHHHH

exciting

but I know $\mathrm{depth} R' \neq 0$, so $\mathrm{depth} R' = 1$

[cloud walrus]

shamrock

[latent anvil]

nvm, the thing I was thinking of to bound any depth by the depth of R only applies in the finite projective dim case

okay it still works

okay cool I have a proof

I applied the depth inequality to the SES wrong

what you really get is $\mathrm{depth}(R'/R) = 0$ and $\mathrm{depth}(R'/R) \geq \min(\mathrm{depth} R - 1, \mathrm{depth} R')$. but I know $\mathrm{depth} R' > 0$, so $\mathrm{depth}(R'/R) = 0 \geq \mathrm{depth} R - 1$

[cloud walrus]

shamrock

[latent anvil]

thus $\mathrm{depth} R = 1$, contradiction

[cloud walrus]

shamrock

[tough raven]

For a n-dimensional space V, the endomorphisms of V form an n^2-dimensional space.
For an infinite-dimensional space V, the endomorphisms form an infinite-dimensional space.

[tough raven]

Is this the same as choosing a basis?

Axiom of Choice

[rustic crown]

Say you mapped t to some u. Then you get an injective map F(t) --> F(t) whose image is F(u). This is an automorphism if and only if the degree of F(t)/F(u) is 1.

Yea

[hidden haven]

First try to prove finiteness by finding a polynomial that t satisfies over F(u)

Then use gauss lemma to prove its irreducibility

Yeah

You have to prove its irreducibility in F(u)[x]

[rustic crown]

its probably nicer to be a bit more explicit and say g(x) * u - f(x) in F(u)[x]

[hidden haven]

For a UFD (I think) R, a monic polynomial being irreducible over R implies its irreducible over fraction field of R

Not sure if I'm stating this correctly

[rustic crown]

(content 1 is sufficient... don't need monic)

[hidden haven]

Content 1?

[rustic crown]

gcd of all coefficients is 1

[hidden haven]

Right

Makes sense

Lol there are others?

[rustic crown]

there are like a dozen of them lol

[hidden haven]

Yeah that's the same one

Just specific to Z

Instead of UFD R

[rustic crown]

another one that comes to mind is the lattice point criterion for checking if something is a quadratic residue mod p or not

[hidden haven]

pro stuff

[rustic crown]

#elementary-number-theory

[hidden haven]

Never done that

[rustic crown]

I started liking algebra cause of number theory

[hidden haven]

I started liking algebra cause of upendra

[rustic crown]

uppu is

why do you think that?

[hidden haven]

Why does that follow from u being transcendental?

[rustic crown]

since you realized that u is transcendental... you can do that!

what exactly did you use here?

[hidden haven]

Why does that follow from u being transcendental?

You have a polynomial in F[x,u]

Even if it is linear in u, it may reduce

[rustic crown]

yep.. that's what i was trying to say

[hidden haven]

For example ux is linear in u and in x but reduces as u times x

Ye

[rustic crown]

yep

same thing

[hidden haven]

Yeah but better to say F[u][x]

Because you have a polynomial over F[u]

[upper pivot]

i mean eh its all isomorphic

doesnt really matter

[hidden haven]

But once you have the polynomial and are trying to show irreducibility then you can treat as whatever

[latent anvil]

good pfp johm

[hidden haven]

Intention is different is all

[upper pivot]

thanks shamrock

[hidden haven]

We are writing the minimal polynomial of an element over F[u]

[upper pivot]

sure i guess, small change but i see your point

[latent anvil]

squirtle was your question answered? I'm having trouble following the chat history

[tough raven]

At first I thought this was some new form of trash talk

[latent anvil]

(I'm asking because I have a question but don't want to be rude)

[rustic crown]

well almost...

[latent anvil]

I'll wait then

[rustic crown]

so do you agree that f(x) - u*g(x) is irreducible over F[u][x]?

a simple way to see this would be to use Gauss lemma again. if you look it as a polynomial in u then the gcd of its coefficients is 1. But its irreducible over F(x)[u] as its linear. so we're done!

[rustic crown]

(just wanted to be explicit about where we used f and g are relatively prime)

[hidden haven]

Yes because when you view ug(x) - f(x) as a polynomial in x it has degree max{deg f, deg g}

From what det did above

Using Gauss lemma twice

You want to show a polynomial is irreducible in F(u)[x], so you use Gauss lemma to say that irreducibility in F[x,u] is enough, then Gauss lemma again to say irreducibility over F(x)[u] is enough (using f and g coprime here)

F(x)[u]

Because it's then a linear polynomial in one variable over a field

[rustic crown]

no no. that's a field nothing is irreducible in a field.

over and in are very confusing

[hidden haven]

Are units not considered irreducible?

Also is 0?

[rustic crown]

._.

[rustic crown]

yea, you can think of irreducibles corresponding to maximal principal ideals.

[hidden haven]

[latent anvil]

units aren't irreducible because if they were unique factorization in a UFD would fail, in that a unit can be factored as an empty product or as itself

thanks, i will take all the credit

okay so

[hidden haven]

Right

[latent anvil]

let $f : X \to Y$ be a finite type scheme map and suppose $Y = \mathrm{Spec} A$ is affine. let $B = \Gamma(X, \mathcal{O}_X)$

[cloud walrus]

shamrock

[rustic crown]

hard stuff

[latent anvil]

i have been hartshorning recently

is $B$ a finite type $A$ algebra?

[cloud walrus]

shamrock

[latent anvil]

by the sheaf condition, B embeds into a product of finitely many finite type A algebras (take any finite affine open cover of X)

but subrings of finitely type rings may fail to be finitely generated :(

(even over a noetherian ring!)

@AlgebraicGeometry

@ helpers

[bleak abyss]

Hmm let's see

[latent anvil]

the map being finite type means there's a cover of X by affines {Ui} where the map Ui -> Y (of affine schemes) is dual to a finite type ring map

[bleak abyss]

Gotcha

Hmm so to do special cases

What can we say if X is affine?

[latent anvil]

then yes

the map being finite type is equivalent to the domain being quasicompact (finite cover by affines) and for any affine open in the domain, the functions on that are a finite type A-algebra

like, "P happens on a finite affine cover" = "there is a finite affine cover and any affine in the domain satisfies P"

where P(U) = the map dual to U -> Y is a finite type ring map

[bleak abyss]

So if Y = Spec(Z) then we're just asking that B is finitely generated right?

[latent anvil]

Y = Spec A

[bleak abyss]

I'm saying A = Z

To make life easier

[latent anvil]

Wait what's Z

Oh

[bleak abyss]

Integers

[latent anvil]

Like the ring

lol

Yeah, B is a finitely generated ring

(not necessarily a finitely generated abelian group, though)

[bleak abyss]

Tru

Okay so I think this should hold if Y=Spec(field) right?

[latent anvil]

why is that?

[bleak abyss]

You said B embeds into a product of finitely many finite-type A-algebras

[latent anvil]

yup

[bleak abyss]

And I feel like here the linear algebra is probably nice? Something something shit splits something something

Or hmm maybe it's not as obvious as I thought

[latent anvil]

relevant example: k[xy, xy^2, xy^3, ...] is a non-finite type subalgebra of k[x,y]

even though k[x,y] is finite type

[bleak abyss]

Rip

[sturdy marsh]

it's probably not true

[latent anvil]

https://mathoverflow.net/a/48831/136523

I agree brofib

I just can't think of a counterexample

[bleak abyss]

Hmm so can we turn this into an example? Thing is idk examples of schemes aside from Spec and maybe I can guess what Proj looks like

[latent anvil]

yeah so like, P^n does not work

so im out of examples

(jk but my other goto non-affine examples are like, infinite disjoint unions, and those won't be quasicompact)

[bleak abyss]

Sadly quotients are finitely generated lol

So trying some weird subspace shenanigans won't work

[latent anvil]

ping me if you think of an example (you too @sturdy marsh)

we also do have way more structure here than like the wild subring of k[x,y]

we know geometric info

[sturdy marsh]

what if you mess up the grading on a polynomial ring

[latent anvil]

blocked

[sturdy marsh]

essentially something like this

you want something like that to be the intersection of all the degree zero parts of localizations

[slender sable]

question 7 anyone?

[rustic crown]

sage: A = matrix(ZZ, [[1,0,0],[1,0,1],[0,1,0]])                                 
sage: A.characteristic_polynomial()                                             
x^3 - x^2 - x + 1

This tells you A^3 - A = A^2 - I

do you see how you can show that now?

[slender sable]

ya and then you just generalise for n or some other process involved?

[rustic crown]

just induct on n

[slender sable]

cool cool

[cloud walrus]

det

[frank fiber]

Is every group of order n with an element of ordern n isomorphic to $\mathbb{Z}/n\mathbb{Z}$?

[cloud walrus]

Gustavo Ortega

[chilly ocean]

i don't know, is every such group so?

what are your thoughts?

[frank fiber]

if $G$ is a group of order $n$, with an element $g$ of order $n$, a isomorfism $f$ between $G$ and $\mathbb{Z}/n\mathbb{Z}$ should have $f(g)=[1]$

[cloud walrus]

Gustavo Ortega

[chilly ocean]

why?

[frank fiber]

beacuse [1] is the element of order n in $\mathbb{Z}/n\mathbb{Z}$

[cloud walrus]

Gustavo Ortega

[frank fiber]

and so $f(g^n)=f(e)=[1]^n=[0]$

[cloud walrus]

Gustavo Ortega

[chilly ocean]

can you actually define such an f though?

it might help to write the group G as {e, g, g^2, ..., g^(n-1)}

[frank fiber]

all the groups $G$ of order $n$ with an element of order $n$ are cyclic?

[cloud walrus]

Gustavo Ortega

[chilly ocean]

ye

of course

[frank fiber]

by hipotesis $G$ has an element of order $n$, but why all the other elementsof $G$ have finite order?

[cloud walrus]

Gustavo Ortega

[chilly ocean]

the subgroup of G generated by g has n elements. G has n elements. so G is the subgroup generated by g.

[chilly ocean]

it's a finite group...

[rustic crown]

maybe think like this? you only have one element at your hand g. what do you do with it? only thing that is possible is to multiply it with itself a bunch of times. so you start getting g^0, g^1, ..., g^(n-1). Since order of g is n, these are all distinct elements! so you got all the elements of the group you started with.

[frank fiber]

so, i think the isomorfism betwen $G$ and $Z/nZ$ is $f(g^a)=[a]$

[cloud walrus]

Gustavo Ortega

[rustic crown]

yep that will turn out to be an isomorphism! but its not the only one...

[frank fiber]

thanks

[rustic crown]

[chilly ocean]

is the galois group of splitting field of an irreducible polynomial of degree n over a fixed perfect field K always the same?

[oblique river]

fields of characteristic 0 are perfect

[chilly ocean]

ik?

[rustic crown]

what are you comparing it with?

[oblique river]

so then the answer is no?

there are irreducible degree 3 polys over Q with different galois groups

unless you are taking the same irreducible polynomial and calculating its galois group at multiple different times

in which case the answer is yes: the galois group doesn't depend on the time of day you calculate it

[rustic crown]

how would you rigorously prove this though 🤔

[chilly ocean]

example?

[oblique river]

min poly of z + z^6 where z is a primitive 7th root of unity

has galois group Z/3Z

any cubic with two complex roots and one real root has galois group S_3

[chilly ocean]

do those have same degree?

[oblique river]

yes

lmao

that's the poitn of the example

[chilly ocean]

im dumb forgive me

can a poly of degree 4 have galois group isomorpic to S4 then?

[oblique river]

yes

you can get S_n for any n

maybe a better example is, for any prime p, x^(p-1) + x^(p-2) + ... + x + 1 ahs galois group Z/(p-1)Z

but x^(p-1) - x - 1 (or maybe I want +x ,I don't remember) has galois group S_(p-1)

The galois group of x^4 + 1 over Q is (Z/2Z) x (Z/2Z) but the galois group of x^4 + x^3 + x^2 + x + 1 is Z/4Z

[cloud walrus]

det

[rustic crown]

pretty memorable example even though i don't know the proof

[oblique river]

oh that's a nice example, my go to is $x^n \pm x - 1$ but I can't ever remember if it should be + or -

[cloud walrus]

Buncho Bananas

[chilly ocean]

its weird to me cuz when you adjoin first root, its degree n, and you can permute between those. but why can you permute between the others

[oblique river]

I don't know what you mean

also I'm not sure why that would imply that all galois groups would be the same

[chilly ocean]

Can we say that the union of two equivalence relations is not necessarily a equivalence relation, because for example Z_2 U Z_3 has 0 in two different equivalence classes, which is impossible?

[oblique river]

please don't interrupt

[chilly ocean]

Sorry

eh actually nvm i was doing stupid confusion i think

[oblique river]

just because the roots have degree n doesnt mean that there is a cycle of order n

for example, the roots of $x^4 + 1$ are $\frac{\pm 1 \pm i}{\sqrt{2}}$ where the two signs need not be the same

[cloud walrus]

Buncho Bananas

[oblique river]

the galois automorphisms will permute $\pm i$ and $\pm \sqrt{2}$

[cloud walrus]

Buncho Bananas

[oblique river]

i.e. they all have degree 2

what is true is that the galois group will act transitively on the roots

but that doesn't mean that there's a single element whose powers act transitively

a nice fact is that every abelian group shows up as the galois group of some irreducible polynomial over Q

(and the degree of that poly is the size of the abelian group)

[chilly ocean]

not true for nonabelian?

[oblique river]

of course there are also lots of nonabelian groups which show up

it's unknown

whether or not all finite groups are galois groups over Q

there's a much weaker (and not that interesting when you look at the proof) statement that "for any finite group G there is some extension of fields E/F whose galois group is G"

but it's an open problem whether or not you can take the base field to always be Q

(a very interesting open problem, in my opinion)

it's called the "inverse galois problem"

some things are known -- for example, every solvable group I believe is known to be a galois group over Q

for an example in the other direction, the matrix group SL_2(F_13) is not known to be a galois group over Q, in the sense that we don't have any known examples with that galois group and can't prove that such an example exists

(at least I think that was true as of like 5 years ago)

[chilly ocean]

very interestimg, thank you

Damn thats interesting

[rustic crown]

The proof for finite abelian group was pretty cute

[oblique river]

hmm nvm maybe SL_2(F_13) is known, but it's not known to occur infinitely often? I'm having trouble parsing these references, but it does seem like the group SL_2(F_27) is definitely still unknown

[rustic crown]

that's pretty weird... I thought people would have brute-forced all groups which are small enough

[oblique river]

SL_2(F_27) has something like 27^3 elements in it

(give or take)

which is a lot

[rustic crown]

btw can we say something about G if we know N and G/N are galois group over Q?

[oblique river]

I don't necessarily think so. Think about it in the other direction -- if we know that G is a galois group over Q and N is a normal subgroup, then we get that G/N is a galois group over Q\

but we don't get that N is a galois group over Q

I don't know for sure, but it doesn't seem right

[rustic crown]

i was just trying to see if you could reduce the inverse galois problem to only finite simple groups

[oblique river]

yes I understand your motivation

I think there is maybe a hope that something like that could work

[rustic crown]

what about some special cases like if A and B are galois groups over Q, can we say A x B or the semidirect product of A and B are also galois groups over Q?

case of direct product seems true.

[oblique river]

let's say that there are "enough" extensions of Q with galois group G/N

if we take the compositum of our N-extension with one of our G/N extensions, we migh thope to get a G-extension

but this is why we need "enough" G/N-extensions

for example, in your direct product case, it is true if your A-extension and B-extension have trivial intersection

meaning their intersection is Q

[rustic crown]

doesn't composite give you only subgroup of the direct product?

[oblique river]

oh yeah you're right you would only get the direct product, so this definitely wouldn't work

yeah this seems even less likely to me now

[rustic crown]

yea so if there are "enough" such extension then direct product is done

[oblique river]

I don't think you could do much better than that in general

[rustic crown]

[oblique river]

I'm not an expert in the IGP though

so everything i'm saying is just sort of from general experience doing algebraic number theory

[rustic crown]

i just started reading some alg nt

[oblique river]

welcome to the club

:)

[rustic crown]

btw do you recommend some book? i started with Marcus Number Fields

[nova plank]

mirza is reading this too

[oblique river]

I learned first out of chapters 1 and 2 of neukirch

I dont have a copy of marcus

looking at the toc of marcus

it seems reasonable

though maybe a little fast in the beginning, depending on your background

[rustic crown]

chapter 1 was pretty nice

[torn sinew]

if GF(2) = {0,1}, then what are the elements of GF(2^3)? (or how do you construct them)

i found this picture, do you just keep adding the original elem until set is complete?

[rustic crown]

ah no, that would be pretty tedious. but yea that is one away. list down all 8 elements. then list down all ways to add and multiply and look at the ones which form a field.

a prototypical way to get a bigger field from a smaller field F is to first look at the polynomials over F, this is usually denoted by F[x] if the indeterminate is 'x'. and then look at some irreducible polynomial f(x). If you consider the set of remainders which you get on dividing things in F[x] by f(x), then that set forms a field structure. This field is denoted by F[x]/(f)

also you can check that in this case size of this new field is |F|^deg(f)

yea so if you want to find a field of size 8, you start with GF(2) and then look for an irreducible cubic. x^3 + x + 1 and x^3 + x^2 + 1 are the two irreducible cubics.

[torn sinew]

so you just look for polys that have a remainder when you divide them by x^3 + x + 1?

(in gf(2))

[rustic crown]

yea so when you divide by a cubic the remainder has smaller degree its of the form ax^2 + bx + c where each of a, b, c are either 0 or 1

the operations on this set are just normal addition and multiplication, but then at the end find the remainder with that polynomial again

[golden pasture]

alternatively F_{p^f} is the splitting field of x^{p^f}-x over F_p, so you can factor that and pick a factor of degree f to form your extension field

So for p=2, f=3, we have x^8-x=x(x+1)(x^3+x+1)(x^3+x^2+1)

[golden pasture]

which precisely correspond to this

[torn sinew]

i see, thanks both

[ancient night]

Do I understand correctly that I can define a polynomial ring over a polynomial ring over a polynomial ring etc? If so, how can I call this polynomial?

[oblique river]

it's a multivariable polynomial

[chilly ocean]

(R[x])[y] = R[x, y]

[ancient night]

No, I mean that the ring over which the polynomial is defined is itself polynomial in that very indeterminate

[oblique river]

R[x][x]? that's... not really a thing

that's just R[x] again

[ancient night]

The article on wikipedia about polynomial decomposition mentions that (x^2+1)^3 + (x^2 + 1) is also a valid polynomial https://en.wikipedia.org/wiki/Polynomial_decomposition

[oblique river]

like, the expression $p_0(x) + p_1(x)x + p_2(x)x^2 + \cdots p_n(x)x^n$, where the $p_i(x)$ are polynomials in $x$, it itself just a polynomial in $x$

[cloud walrus]

Buncho Bananas

[oblique river]

of course it is, you can expand it out if you'd like

oh, so you're just composing polynomials

[ancient night]

So $(x^2+1)^3 + (x^2+1)$ and $x^6+3x^4+3x^2+1+x^2+1$ are equal as polynomials?

[cloud walrus]

JohnDark

[oblique river]

yes

for the same reason that (2^2 + 1) and 5 are the same number

[ancient night]

Are we talking about polynomials or about polynomial functions?