groups-rings-fields

[coral shale]

N^N -> R

N^Real -> R

or whatever ig

[delicate orchid]

R[[R[[X]]]]

[coral shale]

aaa

[delicate orchid]

we only need it in one variable though

for the funny infinite field of char 2 investigation

[coral shale]

what was this about again

ah yes

[delicate orchid]

ok since F_2 is a field we have so many nice facts about F_2[[X]]

number one, it's a UFD

[coral shale]

ED right

or not

wait

not

UFD but not PID

iirc

or wait am i trying to generalise facts about R[X] incorrectly hahaha

[delicate orchid]

wait uhh lets think

[coral shale]

This notion of factorisation for starters feels questionable to me...

[delicate orchid]

ah it's F_2 there's something nice here

I just need to think

[coral shale]

well I agree its an ID........

[delicate orchid]

yes I think it's a PID

[coral shale]

wat

hmmm

[delicate orchid]

consider (a,b) - if these are finite degree then it's identitical to R[X]

[coral shale]

yes.

[delicate orchid]

if these are infinite degree then a+b = a-b must be in the ideal

and a-b must be of finite degree

[coral shale]

says who

[delicate orchid]

because all higher order terms either have coefficients of 0 or 1

and both are infinite degree

[coral shale]

oh we're in F2

[delicate orchid]

yes exactly

[coral shale]

uhhhhhhh

[delicate orchid]

so an infinite number of terms must cancel

so they must be of finite degree

[coral shale]

I'm gonna have to think about this

[delicate orchid]

now, if one is infinite and one is finite, that might be a tad weird

[coral shale]

i dont find myself agreeing

101010101010101

1001001001001001

is not finite

[delicate orchid]

both finite and both infinite definitely reduce to a principle ideal

[delicate orchid]

it definitely is

[coral shale]

how is it wtf.

[delicate orchid]

oh wait you mean ...

put dots at the end man cmon sonnnn

[coral shale]

we are just subtracting sequences from each other

[delicate orchid]

hmm ok

[coral shale]

sry yes

[delicate orchid]

good point

would this ideal not then just be generated by the difference?

regardless of infinite degree?

[coral shale]

i havent got as far as you yet

thinking.

101010101010101...
1001001001001001...

So we are considering this yes

i uh ..

when u multiply these things..........

[delicate orchid]

I've googled it and apparently it's a very nice PID

that's no fun though so lets try and prove it

[coral shale]

Ill try this on paper later

thanks for the thought experiment tho

fun

Is it not an ED for sure...

[delicate orchid]

the suggestion seems to be to show that the ring is local

[delicate orchid]

I think it is

[coral shale]

is an ED?

how do we do euclidean....

101010101010101...
1001001001001001...

[delicate orchid]

I need to find that big chain of ring properties

[coral shale]

this or not?

[delicate orchid]

local rings aren't on there
FUCKkkk

oh wait

local rings don't even have to be integral domains

ok no clue on the euclidean domain front

[coral shale]

however the GCD function does exist

given its a pid

hmmmm

[delicate orchid]

I believe so yeah

since it's char 2 the GCD memes might be especially nice

cause if you're applying bezout's you only have to consider a-b and b-a

one of these has to be the gcd I believe

nevermind they're just negatives of eachother - the GCD is always a-b

[coral shale]

just mod it

wait

a-b = b-a

[delicate orchid]

yeah exactly

that's what I realised

[coral shale]

Btw does Zorns lemma not work

to show this thing is PID,. ED, whatever

[delicate orchid]

Zorn's lemma always applies I hope

I'm just thinking about why F[x] is a PID

maybe we can bootstrap a proof for the infinite case off of that one

[coral shale]

(f, g) = {xf + yg}

[proud bear]

the proof i know uses the fact it's a euclidean domain

[delicate orchid]

yeah that's the proof I know as well - I think

[coral shale]

I wrote the proof the other day, forgot it.

oh

does this work

waiit nvm

i was proving something else

xf + yg = gcd(f, g)
(xf + yg)(a) = gcd(f, g) (a)
0 = xf(a) + yg(a) = gcd(f, g) (a)

I cant remember what i was proving

[delicate orchid]

I dunno if we can use bezout's lemma btw

[coral shale]

but anyways, the argument should be similar to this

well yh

what does bezout apply to

ufd?

[delicate orchid]

PIDs

[coral shale]

not ufd?

oh only pids

[delicate orchid]

wait it's slightly weaker than a PID

[delicate orchid]

we just need the sum of two principle ideal domains to be a principle ideal domain which I think we can do using the fact that it's a ufd

I'm aiming for this first

[coral shale]

bezout is weaker yes

[delicate orchid]

if we can show this we can use your argument here

I think

[coral shale]

still sounds like a torture to get to

[sharp sonnet]

bezout + ufd will give you pid

[delicate orchid]

niceeeeee

I had a feeling

[sharp sonnet]

but the correct approach is to show formal power series are DVR

this isnt too hard

and then this gives you all the nice things

[delicate orchid]

a what?

[sharp sonnet]

discrete valuation ring

there is a unique maximal ideal

[delicate orchid]

I think I-

oh local domains

yeah I think I've shown (X) is the unique maximal ideal

[sharp sonnet]

yes it will be local and regular and all the nice things

[delicate orchid]

cause all degree 0s are units

[sharp sonnet]

those are the nicest rings that arent fields

[delicate orchid]

and all degree 1 or higher are in (aX), all of which are equvialent to (X) as a^-1 exists

so it's definitely a local domain

[coral shale]

So is this thing still a PID for a general field

F[[X]]

[delicate orchid]

I think so

[sharp sonnet]

yes

[delicate orchid]

we did it chat

[coral shale]

ok

[coral shale]

is it an ED?

[sharp sonnet]

all elements without constant term 0 are invertible

[delicate orchid]

god damn I wish I knew local domain implied all this stuff it would've made everything so much easier

[sharp sonnet]

so the unique maximal ideal is (X)

[coral shale]

Is there some way we can apply Zorn to the finite case to prove this stuff?

or is this nonsense

[sharp sonnet]

and then DVR gives you everything you can wish for

[chilly ocean]

wait what

i didnt know dvr is pid

[delicate orchid]

holy shit you aren't kidding

ignore the purple links OR ELSE!!!

[chilly ocean]

what falls apart if you have discrete valuation but not pid

oh

the contraction of ideals

ok

contraction of prime is prime

[delicate orchid]

ok so turns out F[[X]] is basically just F[X] minus maybe the Euclidean algo

[sharp sonnet]

its an euclidean domain via the discrete valuation

pretty sure

[chilly ocean]

its not a homomorphism tho bruh

oh okay

i went over that proof before

[coral shale]

How do you apply alg on 2 sequences?

🤔

[delicate orchid]

we posted the ring operations before

[chilly ocean]

yeah it def is

[delicate orchid]

and I wouldn't think about them as sequences, they're formal power series

[chilly ocean]

yeah

thats best way

[delicate orchid]

sequences you define multiplication term wise

that is not the case here

[chilly ocean]

is just polynomial multiplication

but on potentially infinite scales

[delicate orchid]

anyway! time to consider (1+x+x^2+x^3+...)

[coral shale]

I get that yes

[delicate orchid]

is this mofo irreducible(?!)

[next obsidian]

They are sequences tho

[coral shale]

but how do you gcd alg 2 infinite polynomials

[next obsidian]

Power series is the (x)-adic completion of the polynomial ring

[coral shale]

not in F2

[chilly ocean]

ye but ring structure isnt inf product

[coral shale]

i mean F in general

[next obsidian]

Which are Cauchy sequences

[delicate orchid]

yeah but it's far easier to think about actually multiplying the damn things if you think about them as infinite polynomials

[next obsidian]

It’s a unit

[chilly ocean]

ye

1+x

[next obsidian]

1-x

[chilly ocean]

minus probably

[next obsidian]

But yeah

[delicate orchid]

we're in F_2

so both are the same!

[coral shale]

what?

[next obsidian]

Doesn’t matter

[delicate orchid]

it is a unit though yeah

hmm

[next obsidian]

That equality is formal

[coral shale]

what???

[next obsidian]

It holds over Z

[coral shale]

whats its inverse?

[chilly ocean]

it has multiplication inverse

[next obsidian]

1-x

[chilly ocean]

1-x

[next obsidian]

Is its inverse

[coral shale]

ohhhhhhhhhh

aaaaaaa

nooooo what is going onnnn

[chilly ocean]

you can find inverse by doing divisom algorothm

[delicate orchid]

oh god I'm an idiot it's a local ring it only has one maximal ideal

that makes my job easier!

[next obsidian]

Any power series with invertible constant is invertible

This is a complete classification

And that’s why a power series ring over a local ring is local

[delicate orchid]

ok so finally wrapping back to the original question, wtf is F[[X]]/(x)

lets think about this one

[next obsidian]

It’s just F

[chilly ocean]

always will be

[coral shale]

i was about to say lmao

[delicate orchid]

god damn it I thought so

yeah we'll need a new approach shuri

maybe the ring of all sequences in F_2 works

[next obsidian]

What are you trying?

[delicate orchid]

construct an infinite field with char 2

[next obsidian]

Maximal ideals of this ring correspond to ultra filters

Oh yeah

That works lol o

Or just take the algebraic closure of F_2

[delicate orchid]

yeah that was my "back up" idea

[next obsidian]

Or like

[delicate orchid]

it would've been cool if F[[X]] worked

[next obsidian]

F_2(x)

Any field which contains F_2 works and there’s plenty

[coral shale]

ahhhh

the easy answer

[delicate orchid]

this one is cheating I'm afraid

[chilly ocean]

why

[next obsidian]

Okay then try Frac(F[[x]])

[delicate orchid]

ok NOW we're talking

[next obsidian]

I think this is formal Laurent series

[delicate orchid]

yeah it seems to be

crazy how that's a field... so wacky

[coral shale]

what is this frac thing

[next obsidian]

But in general Frac(A[[x]]) is not (Frac(A))((x))

[delicate orchid]

field of fractions

[latent anvil]

You take fractions

[coral shale]

F((X))

[latent anvil]

Like how you make the rationals out of the integers

[next obsidian]

Formal Laurent series

Is what F((x)) is

[coral shale]

ok ok

[delicate orchid]

ok maybe I do like ring theory

[next obsidian]

Note, you can only have finitely many non-zero negative degree terms

For formal Laurent series

[chilly ocean]

is it possible to have a polynomial ring with infinite indeterminants?

[latent anvil]

Yep

[next obsidian]

Otherwise multiplication doesn’t make sense

Yeah

[latent anvil]

Very useful counterexample

[chilly ocean]

any nice examples?

[next obsidian]

I mean

[latent anvil]

theres basically only one example

k[x1, x2,...]

[next obsidian]

Not quite…

But

[latent anvil]

You can have a different cardinality of variables

But that's it

[next obsidian]

Yeah

k[x1,…,] is useful for counterexamples

[chilly ocean]

when is this ever used though

[delicate orchid]

le standard non-noetherian ring counter example

[next obsidian]

For theoretical purposes, Z[A LOT] is a useful ring

[delicate orchid]

classic

[coral shale]

so we cannot formalize infinite laurent that goes both ways? 🤔

[delicate orchid]

what do you mean

[chilly ocean]

Z alot?

[next obsidian]

As is k[A LOT] for algebraic closure

[candid kiln]

[next obsidian]

You have to take arbitrarily many generators

[coral shale]

we cant have infinite number of negative powers

[next obsidian]

Hence “A LOT”

Yes

[coral shale]

for the formal laurent

[next obsidian]

The issue is when you multiply

[latent anvil]

Simply apply compactness

[next obsidian]

You’d have to do infinite sums

[latent anvil]

[next obsidian]

For Laurent series going in both directions

In say, R

You can make sense of this because infinite sums make sense

[coral shale]

yes that makes sense..

[next obsidian]

Over arbitrary rings you’re screwed

[coral shale]

ah.

so we need to have some idea of convergence

analysis

[next obsidian]

Yes

[coral shale]

to make sense of them

ok

[next obsidian]

Which exists

But it won’t work in general

An infinite sum of 1s can’t make sense adically which is the usual way to make sense of convergence

But idk, someone’s probably invented something

[candid kiln]

adically as in p-adically?

[coral shale]

👌 learned a buncha new things 👀

[next obsidian]

That’s a special case yes

you define it the same way

[delicate orchid]

p-adics are a special case of a completion of a ring

[candid kiln]

[candid kiln]

Ooooo

[south patrol]

chap 10 atiyah macdonald moment

[delicate orchid]

a case I do NOT understand

[next obsidian]

The I-adic stuff is defined by two things being close if their difference is in a large power of I

[candid kiln]

Completion like completion of a metric space? (Sorry im too analysis pilled rn)

[delicate orchid]

completions are just that one thing I will never understand, it doesn't matter how many times I go over the defs they never make sense

[next obsidian]

And then you take legit completions

With regard to that

[south patrol]

it is related in that you're taking completions of topological groups but you don't have the metric space structure

[next obsidian]

The Hausdorff completion

[candid kiln]

ah, i see, idk about that

[south patrol]

well you define cauchy sequences in a topological way for the groups and then it's more analogous which is cute

[next obsidian]

It’s actually literally Cauchy sequences

And you do take a legitimate completion

[delicate orchid]

ohhhh yeahhhhh

[candid kiln]

if its cauchy sequences i can make sense of it

[next obsidian]

The algebraic definition for the completion is isomorphic to the Hausdorff completion

[delicate orchid]

cause they're close in high powers of the ideallll

ohhhHHH

[candid kiln]

[next obsidian]

I mean wew lads have you seen the algebraic construction of the completion?

[candid kiln]

someone spoon feed me algebra and topology

[delicate orchid]

yes

[next obsidian]

In terms of inverse limits

[delicate orchid]

I've tried to understand it many times

[next obsidian]

The idea is that those are sequences

[delicate orchid]

inverse limits are just so confusing

[next obsidian]

Because it’s an element of the product

And then you need the thing that like

the element in the n-th spot

Is the same as the (n-1)-th

After reducing mod I^n-1

This is legit just saying that

The difference lies in I^n-1

[delicate orchid]

so it's a series of quotients

[next obsidian]

So a single element

Is the Cauchy sequence

And the condition about thm agreeing

Is saying that the difference lies in higher and higher powers of I

[delicate orchid]

I get that elements are sequences from the construction of R from Q

that's the bit I actually get

[next obsidian]

Think about it for R[x]

And completing (x)-adically

If you look at the first like n-terms

[delicate orchid]

I think I need to think about it in F_2 or something really simple first

[next obsidian]

That’s a polynomial in n-degrees

The entire element represents the power series

[delicate orchid]

yeah (x)^n = (x^n) right?

[next obsidian]

Yup

[delicate orchid]

ok

[next obsidian]

Like the degree 0 stuff

Is just a constant

In degree 1

You have an a + bx

But you know after reducing mod (x)

It’s the same as the first

[delicate orchid]

hold on hold on, sorry boss you're going way WAY too fast

[next obsidian]

Degree 0 stuff is an element of R[x]/(x)

A constant

[delicate orchid]

yes ok

[next obsidian]

Ind degree 2 is

[delicate orchid]

I can see that

[next obsidian]

R[x]/(x^2)

It’s an a + bx

[delicate orchid]

so those are the first term in the sequences we end up with?

[next obsidian]

Right

[delicate orchid]

and these are the second?

[next obsidian]

As an element of the inverse limit

Yeah

[delicate orchid]

ok

[next obsidian]

But they like agree mod (x)

[delicate orchid]

yeah yeah

[next obsidian]

Then one above

[delicate orchid]

so like

[next obsidian]

a + bx + cx^2

[delicate orchid]

1+x => (1, 1, 0, 0....)

[next obsidian]

Right

But looking at this

This is defining

Sum_0^infininty a_ix^i

Each element in your sequence

Is a partial sum

Of the entire element which is the power series

[delicate orchid]

I see

I will now try and construct the 2-adics

[next obsidian]

But this is the same thing as

A convergent Cauchy sequence

The difference of consecutive partial sums

[delicate orchid]

yeah I can see that

[next obsidian]

Lives in higher and higher powers or (x)

[delicate orchid]

so eventually it'll become 0?

[next obsidian]

So in general the I-adic completion is just convergent power series

Not necessarily

[delicate orchid]

cause all polynomials in R[x] are finite degree? no?

[next obsidian]

No because it’s an infinite product

The things can continue to be non-zero forever

[delicate orchid]

so eventually they'll just be equivalent to 0 in R[x]/(x^n) for n > degree

hmm

[next obsidian]

No

Think about 1 + x + x^2 + …

If they eventually were 0

[delicate orchid]

yeah that's not in R[x] though

[next obsidian]

We’d never get an actual power series

Yeah that’s why we passed to this inverse limit

So we can make sense of one with infinitely many terms

[delicate orchid]

see this is the part that always just feels like magic

[next obsidian]

I mean it makes sense at each finite step

And an element of the inverse limit is an infinite tuple

Where everything can be non-zero

[delicate orchid]

there are no polynomials in R[x] that have infinite terms so by construction through quotienting by higher powers of (x) we must always get 0

so how can we construct sequences that do not satisfy this through this process

[next obsidian]

But that’s why we aren’t taking the same element

We’re saying they agree after cutting out all higher degree stuff

Like going from the n+1-th to the n-th

They agree above degree n

But we can always add stuff in higher degree

And keep fuckinf with the polynomial in higher and higher degree

[spice whale]

what are the infinite rings with char 2

[delicate orchid]

I put them in discussion

[spice whale]

thx

[delicate orchid]

oh hold on

so

under this completion

p -> (p mod x, p mod x^2, ...)
NOT p -> limit n-> inf p mod x^n

(for some notion of limit)

[next obsidian]

Yeah

[delicate orchid]

ok

I'll need to think about this more

[next obsidian]

But not everything looks like that

What you wrote down is why (in nice situations)

A embeds into A^

[south patrol]

how pathological a situation would you need for it not to embed?

[next obsidian]

Non-finitely generated idea I think?

[fallow plume]

sorry to post this mid-discussion, but does anyone have a good text on applying group theory to solve problems?
I stumbled upon this and I realized that I never would've connected the two together and would like to see more/try my hand at some problems maybe
https://math.stackexchange.com/questions/3483093/a-groups-based-solution-to-an-imo-shortlist-problem-2005

[next obsidian]

Oh no

You need umm

[delicate orchid]

k[x1, x2, ...] my beloved

[next obsidian]

\cap_0^infty I^n

To be non-zero

Anything in there maps to 0

[south patrol]

Ah, yes

[next obsidian]

So you need it to be separated

[delicate orchid]

$\cap_0^\infty I^n$

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[south patrol]

Yeah, isn't there smth like the closure of {0} is given by the intersection of all the nbhds of 0 hence that

[delicate orchid]

ah I see

[next obsidian]

Uhh

No

Closure is the intersection of all closed sets

Containing it

But like you can see the topology is Hausdorff iff that’s 0

If it is, given any two things

x,x’ distinct

[south patrol]

I mean this

[next obsidian]

Wtf

[south patrol]

Not the general case of topology lol

[next obsidian]

I didn’t know this

[delicate orchid]

btw, ty for the explanation chmonkey that actually did help a lot

[next obsidian]

Swag

[south patrol]

Oh ok Atiyah Macdonald answered my question lol

[next obsidian]

Which one

[south patrol]

So yeah it's an embedding iff G is Hausdorff as you said, because of the kernel of the (candidate for the) 'embedding' being the intersection of all of those neighbourhoods of 0

[next obsidian]

Yeah

This gives you a way to like

Show if something is I-adically complete

Basically the existence of limits of Cauchy sequences

Translates to surjectiveness of the map A -> A^

And injectivity is if the topology is separated

[delicate orchid]

waittt a minute you ain't gonna try and get me on that alg topology shiet are ya

[next obsidian]

So you can show something is complete

By first showing it’s separated

And then showing that like, limits of Cauchy sequences already exist

in fact if you play around with it

It suffices to show that power series exist

Where power series means like

Sum a_n where a_n in I^n

This is the limit of the partial sums

Anyway, this is helpful because if you want to show something like

“If you’re I and J-adically complete, you’re I + J-adically complete”

Doing it this way is the easiest

[south patrol]

Oh cool

I was gonna ask when we actually would use I-adic topologies but considering how ubiquitous p-adics are that'd be silly xd

[next obsidian]

Unm

[south patrol]

kinda cool tho

[next obsidian]

So it’s kind of two ways

1: complete Noetherian local rings insanely good

Cohen Structure theorem OP

You can do a lot of hard commutative algebra leaning on this

Because you often reduce to power series over very nice rings

2: in algebraic geometry it makes sense for formal stuff

If you think of like

Okay this is hard to actually motivate but

The ring R[x]/(x^2)

Is formed of stuff a + bx

Where x^2 = 0

you can think of x as an infinitesimal

[south patrol]

I've done some basic classical alg geo if you need to use any terminology lol

[next obsidian]

And this is like “1st order differential information”

[south patrol]

ye

[next obsidian]

Okay so like

You can model tangent vectors as maps from this

Or I guess into it

On the scheme side it’s maps from

But that’s cuz it’s contravariant

Anyway

[south patrol]

ye

[next obsidian]

You can look at 2nd order info

By modding out by x^3

And 3rd…

And packaging all of it up you get power series

[south patrol]

Sure, that's cool

[next obsidian]

It’s kind of like… representing something via a umm

Maclaurin swries

Anyway but this is needed for like deformation theory

[south patrol]

Lol yes that was mentioned in the context in which R[x]/(x^2) was mentioned to me i think

[next obsidian]

Which is like, about looking at really really small ass perturbations of stuff

Right so those correspond to first order infinitesimal deformations

So like using this sort of stuff you can show stuff like

First order infinitesimal deformations of like

Oh fuck what was it

It was like if a closed sub scheme

Correspond to tangent vectors at the point of the Hilbert scheme corresponding to the closed subscheme?

I forget but like you can start to study geometry is the point

It’s hard to like

Describe

And I am only slightly acquainted with the stuff

[south patrol]

Yeah I imagine it's a bit beyond me rn with the schemes stuff but ye thankss

[next obsidian]

I guess the idea is just that it gives infinitesimal info

[south patrol]

I had it planned to learn chap 10 AMD today and now I have inadvertently learnt some whilst procrastinating on discord lol

[next obsidian]

Like okay

V(I) = V(I^n) for all n

Because they have the same radical

[south patrol]

Ye

[next obsidian]

The primes they contain are the same

So topologically

They define the same sub-thing

But with schemes they actually define different spaces

And V(I^n) is kind of like, larger

Since you know, I < J means V(J) < V(I)

[south patrol]

ye

[next obsidian]

So it’s larger but not in any way the like

Points of the space can see

And the completion sort of packages all that up at once

[south patrol]

Noice

interesting hm

[wooden ember]

I don’t see how gf=1 is easy? It would show that f is injective but you said that’s hard to show directly. Doesn’t f being surjective imply that fg=1? I don’t at all see how gf=1 follows

Also @terse crystal the reason I’m asking this is for an alg top question where I’m gluing a two cell along a loop of the form of this product

But I haven’t gotten to covering spaces yet so I’m not sure I understand how you’re approaching it

[spiral wolf]

I know I could do all of these problems by brute force expanding out alpha+alpha^3+…, that way, but I feel like there is some trick that I’m missing. I also don’t know what to Google as I’m not sure what this topic would be called. If anyone can point me in the right direction I’d really appreciate it

[chilly ocean]

I think the only times you'd really ever have to expand to cos and sin stuff are the ones where cos shows up explicity, and everywhere else you can stay in the alpha world

Especially by using that the sum of all nth roots of unity is 0

Although it's still a lot of computation lol

But that's just for the initial checks, I'm guessing the problems themselves can just rely on the identities from the checks

Which makes things go a lot smoother

@spiral wolf

And of course using that alpha^17 = 1

Also you probably noticed it too but just incase, small typo at the top, should be cos(2pi/17) + i sin(2pi/17)

[spiral wolf]

Yeah, that's what I meant. I could brute force expand a bunch of alphas, but knowing my professor, my gut tells me there should be some trick where I can stay in the realm of A,B,C...,L,M

Especially since he gives that chart of 10 identities

[chilly ocean]

Well there's one more thing that shows up and probably helps, alpha^k + alpha^{-k} = 2 Re(alpha^k) since they're conjugates

And the exponent is all mod 17 of course since alpha is of order 17 and this is a cyclic boi

Oh lol and 2 Re(alpha) happens to be exactly L, so maybe there's something there

Oh actually yeah, every single letter is just some sum of conjugates

[barren sierra]

I am being really dumb and struggling to find what this element is

so my methodology for this is as follows

D_20 is normal in D_40

I have this result I can use

[next obsidian]

r and s generate the group

So specificying where r and s goes determines the entire automorpjism

You just have to show it’s an automorpjism

[barren sierra]

hm

[next obsidian]

And then separately that it isn’t inner

[barren sierra]

yea so like showing it's not inner

is what I'm trying to do

[chilly ocean]

i forgot definition of inner

[next obsidian]

You can just enumerate the inner auto morphisms

And show mine of them send r and s to those same things

[delicate orchid]

an automorphism by conjugation

[next obsidian]

Just conjugate r by every element

And s by every element

[barren sierra]

is there a smarter way to do that?

[chilly ocean]

oh

[next obsidian]

Write a program to do it for you

[barren sierra]

I know i could do that

[delicate orchid]

most of the conjugations look very similar tbh

[chilly ocean]

yeah

[barren sierra]

I have already written said program lmao

[next obsidian]

You only have to check up to cosets of the center

Since things in the same closet generate the same inner automorphisk

So you can reduce it to a smaller number

[delicate orchid]

G/Z(G) iso to Inn(G) moment

[barren sierra]

yea

I guess

only 10 things to check

[dull roost]

Yo who can help me not even a hard question

Lol

[chilly ocean]

just ask

[delicate orchid]

[dull roost]

#help-14

[chilly ocean]

not abstract algebra

[dull roost]

True gg

[wooden ember]

Figured this out a bit similarly

[coral shale]

Back to the F[[X]] discussion we had a bit ago

for F[X] we can have an evaluation map

but this isnt really possible for F[[X]] is it?

Well that aside, are we able to construct a homomorphism from F[[X]] to F in general?

[thorn delta]

There is a homomorphism F[[X]] to F which sends a power series to the coefficient of X^0. I'm not sure if there is any sense in which we can have an evaluation map tho

[coral shale]

Thats a smart homomorphism 🤔

[chilly ocean]

Can someone let me know what topics should i study for this homework?

Like what topics in a textbook would i read

[tribal moss]

Looks like mostly group theory -- subgroups and quotients.

[rustic crown]

wait did i make an error?
so f sends all the letters to the corresponding cosets
and g sends the first n-1 letters to themselves and last one to the product of inverses of others.
g(f(w)) = g(wN)=w?
having a left inverse implies injectivity

[terse crystal]

And I solved it without using algebraic topology at all… and it seems it isn’t hard enough to require any algebraic topology …

Anyway I’m not good at algebraic topology so I don’t know any another solution,Just the pure proof within group theory. I only used topology to study subgroups following textbooks, but I haven’t used algebraic topology to study quotient groups

[tawdry crystal]

Does there exist a map from O(4) (the orthogonal group) to SO(4) (the special orthogonal group) so that whenever I am given a matrix in SO(4), I can turn it to SO(4) matrices?

[tribal moss]

There's a map alright: for example, you can divide the first column of the matrix by its determinant. But this doesn't produce a group homomorphism.

(In odd dimension, on the other hand, it works to divide the entire matrix by the determinant, since the determinant of an orthogonal matrix is ±1).

[hybrid island]

anyone have any ideas how to prove this

[chilly ocean]

diagonalize T - \lambda Id

[hybrid island]

not sure how to show that T - lambda * I is diagonalizable

[terse crystal]

CλIC^-1=λI
C(A+B)C^-1=CAC^-1+CBC^-1

[hybrid island]

what does ABC represent?

[hidden haven]

Another way to do it would be to use the fact that a matrix is diagonalisable iff there is an eigenbasis with respect to the corresponding linear transformation

Show that adding λI to a linear transformation maps an eigenvector to an eigenvector

So any eigenbasis for one is an eigenbasis for all others

[hybrid island]

okay that makes sense thank u

[barren sierra]

so I did 9

and the 4 solutions are 1, 2^n-1 - 1, 2^n-1 + 1, 2^n - 1

and these define some homomorphisms for semi-direct products

but how do I show that these are not isomorphic

so say x is one of those 4 solutions and we have the homomorphism from C_2 -> Aut(C_2^n) defined by mapping 1 to phi such that phi(1) = x

and this is valid by problem 9

Can I abuse the fact that since cyclic groups are abelian, every automorphism is outer and since automorphisms are unique defined by the mapping of the single generator that the groups are different?

[wooden ember]

Yeah it does I was just confused at first cause you said it was ezdy right after saying showing f is injective is hard, but doesn’t more time showing fg=1 which just follows from surjectivity. I did it this way too

[wooden ember]

Ah okay sorry i just assumed that’s what you were doing: I couldn’t quite understand the picture

[chilly ocean]

I don't think this argument works. It seems to me that the numbers of elements x with x^2 =0 are all distinct (if I didn't make mistake)

[barren sierra]

Oh so like for one automorphism, there are some number of solutions such that x^2 = 0 but for another automorphism there's another number of solutions?

Interesting

[chilly ocean]

Exactly yes

[wooden ember]

I’m not entirely sure of what I’m about to say but I think semidirect products are isomorphic iff the map to automorphisms are the same up to conjugation

I know the if is true but not sure about the only if

[chilly ocean]

I don't think it's true...

[barren sierra]

Yea that doesn't sound right

[chilly ocean]

As I know it's a much complicated problem

[barren sierra]

Something about outer or inner automorphisms is relevant but I forget the terms

[wooden ember]

Yeah no it’s not true

It’s only an if statement

Would have been too nice

Okay there is a special case where it’s true

Namely where the twisting group is Z and the images of 1 are conjugate in outer automorphisms

Oh and you also need a hypothesis on the other group not surjecting on S

But I suppose if your semidirect product is nice enough that you have a presentation, you should be able to find some different characteristics of each group

[upbeat juniper]

does the first isomorphism theorem hold for rngs as well?

I feel like it's yes, because the proof for rings didn't make that much use of 1 so it should carry over

[hidden haven]

The induced map in the first isomorphism theorem always exists

It may not be an isomorphism in general

In the situation of the sticker, it will always be injective

Because you have quotiented by the kernel

It will always surject onto the image

But it is not an isomorphism in cases like topological spaces

But it is always a bijective homomorphism onto the image in the situation of the sticker, so in structures with only operators and constants, (+, *, 0) in this case, it will be an isomorphism onto the image

[hidden haven]

Quotients are defined so that this happens

[upbeat juniper]

yeah I was thinking you can just do the coset construction and then show it satisfies the required properties

[hidden haven]

Yes

In general the coset construction is replaced by general equivalence relations

Like you quotient sets by equivalence relations rather than cosets

The problem with cosets would be, when you have different numbers of operations then how do you know which operation's cosets to take? What if none of the operations is invertible?

Works for rngs since + is still invertible

Probably works I mean

[terse crystal]

Which part?

I send it again with the error fixed:

The core part is showing TG’/G’=KG’/G’ since they are both kernel of G/G’—>H/H’

Or is it symbol? K is the kernel of π:G—>H, x bar represents the equivalence class xG’ in G/G’ for x from G

T is the minimal normal subgroup containing x_1…x_n

[wooden ember]

How does the final chain of isomorphism’s allow you to conclude G’ n T = G’ n K?

[terse crystal]

The former is contained in the latter

Say you have two normal subgroups H_1 and H_2 of G, H_1 is contained in H_2. Then if the canonical homomorphism G/H_1—>G/H_2 is isomorphic then H_1=H_2

[wooden ember]

Oh right

Thanks I get it now

[terse crystal]

👌

[wooden ember]

I was mainly confused by notation the first time around

[terse crystal]

Yeah I should mention in word that K is the kernel, I just gave the exact sequence.

[wooden ember]

Yeah I got it for K but I thought T was just a subgroup (not normal closure) and that the next sequence including T was also meant to be exact

[terse crystal]

I see

[wooden ember]

Thanks again

[terse crystal]

NP

[terse crystal]

sorry maybe you have noticed that it still need to be modified the third line from bottom, it doesn’t really make any sense right…😂. So It actually should be you have an isomorphism φ: T/G’nT—>TG’/G’=KG’/G’—>K/G’nK For any x bar=x G’nT from T/G’nT, φ(x G’nT)=x G’nK in K/G’nK. So x is contained in G’nT iff x bar=1 iff φ(x bar)=x G’nK=G’nK=1 bar iff x is contained in G’nK … so G’nT=G’nK

So just delete the third line from bottom,The rest remains

[wooden ember]

Oh lol yeah, I didn’t question the 2nd isomorphism theorem form

[terse crystal]

I didn’t notice it was wrong at all😂

[wooden ember]

Me neither haha

[terse crystal]

Gladly it can be fixed😂

[untold cloud]

Hi, for a ring R, and ideal I, what does it mean to reduce R module I?

[hidden haven]

Quotient ring R/I

[untold cloud]

Does that mean it is just a surjection between R and R mod I

[hidden haven]

That is the quotient map yes

To reduce R modulo I would mean taking the quotient ring R/I

But maybe it would depend on context

[untold cloud]

I see i see

Thank you!

[lethal dune]

this is actually a sticker now

[hidden haven]

All of this latest sticker news gets posted in #cats

Subscribe to keep up

[delicate orchid]

No, no I don’t think I will

[lethal dune]

why no #dogs

[hidden haven]

We have asked mods to at least give us a thread

and enable t!dog

even tagged mniip and metal and manan separately

[lethal dune]

🙈

[hidden haven]

They just and do nothing

[lethal dune]

lol

[hidden haven]

I hate the mods

[lethal dune]

[delicate orchid]

Mod purge mod purge

[lethal dune]

we need new mods

[hidden haven]

nitezba mod

[lethal dune]

[wooden ember]

Apply for mod moldi

[coral shale]

modilocks?

[lethal dune]

modiji

[hidden haven]

Imagine being mod

[lethal dune]

imagine being pure mathematician

[hidden haven]

now what

[frail zealot]

[hidden haven]

[delicate orchid]

Zased

[hidden haven]

Is he still muted

[lethal dune]

he was muted? lol

[hidden haven]

Muted by bot because he posted a gif and gif link had bad word

[lethal dune]

lmfao

probably he's watching us rn

[hidden haven]

Hi nit

He was muted 24 hours ago

[delicate orchid]

Another 24 hours

[lethal dune]

lol what 48 hr mute by bot

[hidden haven]

bruh

[delicate orchid]

Who are the rules for?! Us?! Or the people in power?!?! The proletariat must RISEEEE

[lethal dune]

[hidden haven]

Zased

[rose shard]

Can someone help me w this one

[delicate orchid]

I’d start by computing the powers of each element (multiple each matrix by itself until you get the identity) - then try and combine all those powers

Just find all the different results you can get by combining those two matricies basically

[rustic crown]

you can think geometrically if you like that more...
the first matrix sends x to -y and y to x, which is a clockwise rotation by pi/2
and the second matrix fixes x, but sends y to -y, so it's reflection about an axis
once you see this, showing that the group is D4 isn't hard

[delicate orchid]

I’ve got a question as well actually because I appear to be skill issuing massively

I’m getting that the galois group of Q(sqrt(2), sqrt(3)) is S_3 but shouldn’t it be C2 x C2?

[hidden haven]

Yes

[delicate orchid]

I can’t seem to get the Klien 4 though

Cause i seem to be able to get every different permutation of the set {sqrt(2), sqrt(3), sqrt(6)} as an automorphism

Which is why I’m getting S_3

[hidden haven]

sqrt 2 and sqrt 3 decide sqrt 6

[delicate orchid]

Yeah I know but I can still just swap them around right?

[hidden haven]

Also sqrt 2 can't go to sqrt 3

because those don't satisfy the same polynomials

Polynomial equations should remain true under homomorphisms

[delicate orchid]

I’ve gotten no where near considering how these affect polynomials

I barely even understand the formal definition of a galois group tbh

[hidden haven]

F

It is the group of automorphisms of the extension

[delicate orchid]

F = Q in this case

Yes I know

[hidden haven]

Damn

[delicate orchid]

That is a collection of words that, when separated, I understand

[coral shale]

the roots can switch with their own conjugates only

xx = 2
xx = 3
are your 2 equations

[delicate orchid]

What on earth do you mean by conjugate?

Cause this book has been talking about group representations into C so do you mean complex conjugates?

[hidden haven]

So how do you compute Galois groups right now

[hidden haven]

no

[delicate orchid]

I don’t have a clue how, if I did I wouldn’t get S_3

[hidden haven]

Conjugates of a over F are the roots of the minimal polynomial of a over F

[hidden haven]

fair enough

[delicate orchid]

Oh ok so they’re kinda similar in that they give isomorphic field extensions?

[hidden haven]

yes

[delicate orchid]

Ok got that part

[coral shale]

i mean sqrt conjugate

what do we call these things

[hidden haven]

negative

[coral shale]

nahhhh

[hidden haven]

Lol

[coral shale]

that cant be right

😂

[hidden haven]

additive inverse

[coral shale]

ok jokes aside what about 1+rt2

and 1-rt2

[delicate orchid]

Ok so we can’t map sqrt 2 to sqrt 3 because reasons

[hidden haven]

because field homomorphism

[delicate orchid]

Lets see if I get K4 now

[waxen hedge]

conjugates are … orbits through the action of the (absolute) Galois group
Two algebraic numbers x and y are conjugates if and only if they have the same minimal polynomial (over the base field)

[coral shale]

i shall call them R conjugates henceforth

as opposed to C conjugates

[delicate orchid]

This actually helps

Oh of course the galois group acts naturally on a field extension

[hidden haven]

You have map Q → Q(sqrt2, sqrt3). Every automorphism of this larger field will be an extension of this homomorphism. You have to count the number of extensions. This is how one usually finds Galois groups

[delicate orchid]

So we can get to that extension via first extending by sqrt(2) and then sqrt(3) and then also doing it the other way around, right?

[waxen hedge]

If you look at C as an algebraic extension of R, you can see that the definition are consistent, since the Galois group of C/R is just the identity and the conjugation

[hidden haven]

yes

So now you have reduced to the problem of counting extensions when adjoining 1 element 😌

[delicate orchid]

And trivially the galois groups of each of the intermediate extensions are just C2? I hope?

[hidden haven]

Not trivially seeing what you have and haven't proved so far

[delicate orchid]

Intuitively then

[hidden haven]

For me yes, but can you manage that?

[delicate orchid]

There’s kind of only one option for the automorphism

[hidden haven]

But yes those are C_2 and then the group for the extension turns out to be the product of the 2

[delicate orchid]

I should find a book that works through more examples tbh

Cause this is just lists of theorems

[coral shale]

Can we get most abelian groups by considering Gal Q(...)

I feel like if you have a cardinality of generators above Reals you fail

else you can

[hidden haven]

Inverse Galois problem

[hidden haven]

What book

[coral shale]

lol nice

im thinking about this the wrong way cus i felt they had to be abelian 🤔

[hidden haven]

Confused with higher homotopy groups?

or this

[coral shale]

just elementary error probably

hmmmmmm

Groups formed from direct product are abelian

right?

[delicate orchid]

Groups rings and Galois theory by victor snaith

It’s a literally who book I borrowed from my uni library

[hidden haven]

Direct products of abelian groups are abelian

[hidden haven]

You can try Milne

It good and short and examples 😌

[delicate orchid]

Structure theorem by beloved

[coral shale]

ofc smhmh

[delicate orchid]

Sure why nottttt

[coral shale]

Why an I thinking the galois group is always a product of C_k

[delicate orchid]

In slightly better news I have managed to prove that the identity map is the only field automorphism of R

[waxen hedge]

that's not totally related to your question, but there is a hard theorem ("Knonecker-Weber theorem") which says that every abelian extension of Q is contained in the cyclotomic extension (take the subfield generated over Q by all of the roots of unity)

so every abelian Galois group Gal(K/Q) is a subgroup of the Galois group of the cyclotomic extension, which is the projective limit of the Z/nZ with n >= 2 integer

[tawny pine]

@delicate orchid

[coral shale]

Is galois of Q(rt2, rt3, rt5)

C2xC2xC2 ?

[hidden haven]

Yes

[coral shale]

then where does my thinking go wrong D:

how do we get non abelian

[tawny pine]

the identity is the only order preserving one

[hidden haven]

splitting field of x^5 - 2

[coral shale]

i will look ty

ohhhhhh

ohhhhhhhhhajsjsjs

thats S5 right

waitt

[hidden haven]

No

[coral shale]

ok no ima think carefully about this one 👋

[delicate orchid]

Are you talking about mapping 1 -> -1?

[coral shale]

D10?

[hidden haven]

It is ||a semidirect product of C4 and C5|| I think

[coral shale]

more guesses

ok i shouldnt be guessing around with no paper lul

[hidden haven]

It is also the only ring homomorphism

[coral shale]

at least ik my intuition is off somewhat

[tawny pine]

its also the only ||insert blank||

[hidden haven]

I thought you were correcting him

[delicate orchid]

The degree of a field extension is just the dimension of the field extension taken as a vector space over the original field right?

[hidden haven]

Yes

[delicate orchid]

God I wish this book had answers to these exercises

[hidden haven]

Guess what Milne has

[delicate orchid]

So we’re extending Q by some third root of unity and it’s telling me it’s a degree 2 extension

[coral shale]

3 surely

[tawny pine]

i invoke poes law

[hidden haven]

It's correct

[coral shale]

wut D:

[delicate orchid]

But elements are clearly of the form of a+br+cr^2 so it should be 3?

How

Is it because it’s all generated by just r?

[hidden haven]

Think again

[hidden haven]

Yes but not uniquely

[coral shale]

oh wait is this just C

[hidden haven]

The 3 powers are a spanning set

Not a basis

[delicate orchid]

Right cause they’re not linearly independent

[hidden haven]

Extensions of Q not R

[delicate orchid]

In the field extension anyway

[hidden haven]

And why are they not

[delicate orchid]

But they’re absolutely independent as vectors over Q surely

[coral shale]

Argh man im too new to this stuff

[hidden haven]

Are they

[delicate orchid]

There is no rational number that maps one root of unity to another yes

Other than -1 -> 1

[hidden haven]

Linear independence is not about a single root

It's about linear combinations

[delicate orchid]

Yes but there’s only 2 complex roots in this case

[hidden haven]

Add them

yes

no

[south patrol]

modulo some sign errors

[hidden haven]

Lol

You had it right in the second attempt

yes

[south patrol]

Yes

[coral shale]

what is going wrong here 🤔

[hidden haven]

This is fine

You have not identified the minimal polynomial correctly

Remember to check irreducibility

[coral shale]

omg 😂

Back to z^2-5

So I believe there are 5 elements in the galois group

[hidden haven]

wait is that what I said

x^5 - 2

[coral shale]

mistype

[hidden haven]

Is what I meant

ok cool

[coral shale]

yes 5 elements right?

or is that wrong for starters

[hidden haven]

Nope

[coral shale]

uhhhh

ok so call the first root w

w, w^2, ..., w^4, w^5(=2)

Now I am looking for the Q-homomorphisms

[hidden haven]

I would recommend first trying to figure out the degree of the extension

[coral shale]

Its 5

x^5 - 2 is minimal

[hidden haven]

[coral shale]

god D:

ok I will check properly

[hidden haven]

We are taking the splitting field which means we need to put in all the roots of this polynomial

[coral shale]

I have just started what splitting fields are... but I feel like I'm not lacking the knowledge to do this right?

Just done almost no exercises

I was under the impression the splitting field is
Q(5th root of 2)

[hidden haven]

Nope

[tribal moss]

That is a subfield of R

[hidden haven]

You won't get the other roots in it

[coral shale]

2^(2/5)^5 = 4

thank you

this makes so much sense now lmao

[coral shale]

I see......

ok i will have a go at this when im more awake

why did i think all 5 roots were real....

[delicate orchid]

sorry chat my phone died

right before moldi gave me the answer as well

I did work out the geometry of the situation just so happens to give you a real value that's rational in the meantime

but cmon mfs really expecting me to notice that cos(2/3*pi) is rational like WHO?

[hidden haven]

pranked

[delicate orchid]

at least it makes sense to me now 😌

[hidden haven]

You could also notice that x³ - 1 is not irreducible

[delicate orchid]

connecting these ideas back to polynomials is way beyond me unless I think about it for like an hour straight

like, yeah that's the minimal polynomial of this field extension - I think

but the fact that's irreducible correlating to the degree being one lower is a connection I haven't yet made

although knowing that does make it nice and easy to generalise this to other roots of unity extenstions

so, say

the degree Q(r^5=1) is 4?

[hidden haven]

The point is that that's not the minimal polynomial

[hidden haven]

Yes

[delicate orchid]

ok so, perhance, perhaps, mayhaps, is the minimal polynomial x^2+1?

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

[hidden haven]

Nope

[hidden haven]

[delicate orchid]

this entire field of study is a scam

[hidden haven]

8*4 = 10 moment

[delicate orchid]

more like

I feel like a high schooler walking in on a module theory lecture

[hidden haven]

I'm trying to say that your polynomial multiplication is wrong

[delicate orchid]

oh yeah lol

idk it was just a guess

[hidden haven]

F

[delicate orchid]

oh wait it's x^2+x+1

[hidden haven]

+2?

Ye

[delicate orchid]

ok and since that's degree 2 we get a degree 2 extenstion?

[hidden haven]

Yes

[delicate orchid]

because... they're both called degrees Q.E.D

[hidden haven]

[chilly ocean]