groups-rings-fields

[chilly ocean]

No, you do multiply each of the coefficients by 2

[lavish pike]

alright and then u get like a -8 whatever and that would turn into 3 right?

[chilly ocean]

Yeah, I'm not sure why they use -4 in the first place, I would use +1

3?

[lavish pike]

wait so it does turn int 2?

[chilly ocean]

-8+2(5)=2

[lavish pike]

lmfao.

idk im losin it

wait wuuuut

i thought that it was more like -8/5 leaves a remainder of 3

what is -8+2(5)

[chilly ocean]

Bro, do you know arithmetic?

[lavish pike]

apparently not

just make sure u dont become a teacher lmfao

[chilly ocean]

In Z5 we look at equivalence classes where we identify two numbers if they differ by a multiple of 5

So -8 = -3 = +2

[scarlet estuary]

-8/5 leaves a remainder of 3
a remainder of -3, which is 2!

note that -8 is 2 more than a multiple of 5

(the multiple of 5 in question being -10)

that's what remainder means

[rigid cave]

Anyone who knows what the Galois group of X^4 + 1 is? I am way too lazy to calculate it but I think that it is D4 or D8

over Q, that is

[chilly ocean]

What's D4? (Klein 4 group?)

[rigid cave]

No I mean the dihedral group of order 4

This is the dihedral group of order four, right?

So (12)(34) would be the "rotation" and the other one would be the reflection?

No, I mean the other way around. Right?

[hot lake]

so it's the Klein 4 group ?

(Z/2Z)² ?

[rigid cave]

Wait, the Klein 4 group is isomorphic to the dihedral group of order 4, right?

[hot lake]

I guess so

there aren't many groups of order 4

[rigid cave]

So the group in the picture above is D4, right?

[hot lake]

i guess ?

you mean the subgroup of S4 generated by (1,2)(3,4) and (1,3)(2,4) ?

in the picture

[rigid cave]

yeah exacly

V = { (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) }

taken from wikipedia so it must be the klein four group I think

[hot lake]

it's (Z/2Z)² yeah

[rigid cave]

Okay, thank you!

[hot lake]

you can also call it (Z/8Z)*

or (Z/12Z)*

[rigid cave]

what is the star thing?

[hot lake]

the group of invertible elements of a ring

the rings Z/8Z or Z/12Z here

[rigid cave]

oh, I didn't realise that!

[hot lake]

well there are lots of ways that a group might pop up

(Z/2Z)² should be especially ubiquitous given how small it is

[rigid cave]

Okay, I think I get it now. Thank you so much!

[pallid wing]

this is kinda obvious but im having trouble putting it formally

$if x\in {a_1,a_2,...,ak} \
f^2(x)=a_{i-2(mod;k)}$

[cloud walrus]

hanz

[carmine fossil]

let f be ($a_1 a_2 ... a_r$)
\newline Consider the disjoint cycle in $f^2$ containing $a_1$, elements in this cycle which contains elements of the form $a_{{1+2k} \mod r}$ for all k, which is to say that cycle is $f^2$ itself since all elements lie in that

[cloud walrus]

Drunknarwhal

[pallid wing]

thanks, this helps

[carmine fossil]

In general f is a r cycle implies f^k is a product of gcd(r,k) disjoint cycles of size r/gcd(r,k)

[pallid wing]

cool, danke!

[maiden ocean]

hrn

[cloud walrus]

Moth

[maiden ocean]

hints only pls

this is probably like quite obvious and im just being silly

[carmine fossil]

I mean you can't give a hint without giving away the answer

[maiden ocean]

if its that simple u can just like state one thing or whatever

its not a big deal just like. if there r details u dont have to go into them lol

[carmine fossil]

Take z=xy+x+y

[maiden ocean]

ah

yea

lmao

[cloud walrus]

slimvesus

[sturdy marsh]

is this not just the going up theorem?

[cloud walrus]

Brofibration

Brofibration

[sturdy marsh]

the going up theorem gives you a prime ideal in B with whatever properties you need

@chilly ocean

oh I just noticed that you're not assuming that the map is an inclusion

you can reduce it to the case where it is an inclusion

why is it an issue?

look at the map A --> A/ker

if I have I subset J in A/Ker

oh wait hold on

you just used lying over

use going up

you have q in Spec B

such that (b subset q) lies over (b^c subset p')

going up

okay if b is not prime

ignore that

quotient out by b

then you can just argue for prime ideals

B --> B/b induces a closed map

okay to clean the whole thing up

just pretend that b = 0 for now

and do the quotienting bit later

also quotient out by the kernel

so that the map is injective

and then fist prove the claim for an integral inclusion of rings R ---> S

and then use the fact that A ---> A/ker

and B ---> B/b

have the going up property

after you do it for the inclusion of rings, do not assume b = 0

okay lemme rewrite

1. Prove it for an integral extension of rings

2. Using the fact that quotient maps have going up, prove it for a general integral map A --> B

A iso to f(A)

isos have going up

[cloud walrus]

Brofibration

[sturdy marsh]

that's the only difference

umm, maybe I am confusing you

okay if you want to be super formal about the whole thing

say you have an injection f: A ---> B

then it factors as $A \xrightarrow{\simeq}f(A) \subset B$

[cloud walrus]

Brofibration

[sturdy marsh]

isomorphisms have going up

f(A) \subset B is integral

yes!

that's what I meant by A --> A/ker has going up

also, you should be able to reduce it to the case when b is prime

no, I was being stupid

but it shouldnt be a problem

yup

[lavish pike]

so finding all irreducible in a 2 degree polynomial in Z2 would involve an equation like $f(x)=x^{2}+ax+b$ and i would go through saying like b=0 then $f(x)=x^{2}+ax$ which then f(x) is reducible meaning b has to be 1 in order for it to be irreducible which then i just plug in 0,1 for x and solve correct

[cloud walrus]

NocuousNick

[lavish pike]

well not solve but like $f(x)=0^{2}+a0+1=1$ and $f(x)=1^{2}+a(1)+1=2+a=a$ which proves that $f(x)=x^{2}+ax+1$ is irreducible oh and that a=0 then $f(x)=x^{2}+1$ is also irreducible right?

[cloud walrus]

NocuousNick

[chilly ocean]

In degree 2 case, checking for roots is sufficient, yes

But why do you say a=0 case is irreducible, it has a root

[lavish pike]

well because $x^{2}+1=(x+1)^{2}=x^{2}+2x+1=x^{2}+1$ since we are in Z2 right?

[cloud walrus]

NocuousNick

[chilly ocean]

yeah

[lavish pike]

so wait am i wrong? is x^2+1 reducible in Z2?

oh shit

i meant to say is not irreducible lol

gaht

[cursive temple]

x^2+1 is reducible in Z2

it is not irreducible in Z2

[delicate bloom]

cause it's characteristic 2, this is one of the few times you can say (x+1)^2 = x^2 + 1^2 lol

[native orbit]

it's easy enough to bash all the possible non-trivial quadratics that can exist in (ℤ/2ℤ)[x]

monic is the only choice, so there are 4 choices for the linear and constant terms

0x+0, x+0, 0x+1, x+1

the first three choices clearly reduce

the last one you just look at the evaluations at x=0,1

[uncut girder]

https://en.m.wikipedia.org/wiki/Field_of_definition

[uncut girder]

[latent anvil]

what's your ?

[leaden finch]

can someone help me with limits on how to find delta

[latent anvil]

also what does unconditional "variety" mean in this context?

[latent anvil]

is this the question you just asked in #advanced-analysis?

[leaden finch]

yes, i need help but no one has answered me 😦

[latent anvil]

someone did answer you

[leaden finch]

i'll check right now

[latent anvil]

also it's not an algebra question, so it wouldn't make sense to ask it in here, and you shouldn't post your questions in multiple channels

If you don't get any attention try pinging @ helpers after waiting at least 15 minutes

[leaden finch]

okay, but who do i ping though?

[latent anvil]

the helpers role

But anyway anticipation was trying to help with your q

[leaden finch]

okie got it

ty

[vestal snow]

Sure

What is your proof?

I think this should be pretty short

But it uses A LOT of machinery implicitly through Galois theory

Yup

This might be one of the "look at this challenging problem. Now look at this tool we developed. BAM" problems

Np

Okay I had a question about a paper I'm reading.

Assume this

and prove this:

Let $\alpha$ be an ordinary point. Then $\alpha$ is not a root of $g_{\mu}$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

I need to somehow show that $(x-k_1)^{\nu}(y-k_2)^{\mu}g_{\mu}(x)^{-1}$ has order $\nu+\mu$ at the point $\alpha$ for some $k_1,k_2\in k$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Which would follow if we could prove that $x-k_1$ and $y-k_2$ have orders 1 at $\alpha$ for some $k_1,k_2\in k$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

It seems like the obvious choice for $k_1$ is $\alpha$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

But there is no $k_2$ that makes $y-k_2$ divisible by $x-\alpha$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Oh and we also have a relation $y^p-y=\frac{f(x)}{\prod (x-\alpha_i)^{m_i}}$ where none of the $\alpha_i$ equal $\alpha$.

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Anyone know how I can fix this?

[wraith obsidian]

Have you tried turning it off and on again?

[chilly ocean]

ah, and by turning it off, you mean shredding the paper?

[vestal snow]

Got it. Commit seppuku until I'm reincarnated as someone smart enough

[long obsidian]

Any ideas would be appreciated but I understand if this post should not be here lol

[lethal cipher]

Quick question. Since this is a field (ring), what are the operations of addition and multiplication.

Is it simply just v(x+y) and v(xy) respectively?

I ask because it seems like v is one operation

[scarlet estuary]

theyre the operations from your field?

[vestal snow]

The same as R

Oh wait

My bad

[rustic crown]

if the group is non-commutative, then take conjugation, if its commutative then take phi(g) = g^k for some k coprime to size of group

[scarlet estuary]

assuming you maen v(x+y) and v(xy), the + and * operator just come from your underlying field

[long obsidian]

A field is like an ordered triple right (set, add, multi.)

[scarlet estuary]

v isnt + or * here

its a separate mapping

[lethal cipher]

Okay strange. So one thing I'll need to do is show that $A_v={x\in K| v(x)\geq 0}$ is a subring.

[cloud walrus]

dackid

[lethal cipher]

Is that possible if we don't know the actual operations for the field K?

[cursive temple]

For nonabelian groups the map $f_x : g \to xgx^{-1}$ is a nontrivial automorphism right

[cloud walrus]

Pappa

[cursive temple]

For some x

[long obsidian]

Oh your right about the conjugate actually. Thanks for the ideas ya'll

[cursive temple]

Constructing one for abelian groups isnt too bad either

[lethal cipher]

Because all we have to work with is the mapping v

[scarlet estuary]

why wouldnt it be

you still check the usual subring properties

[lethal cipher]

But we are checking them through this mapping 🤔

[scarlet estuary]

so?

have you ever proved, say, that the kernel of a homomorphism is a subring? (oops, ideal!)

similar idea here

[lethal cipher]

I'm confused why this does the job. Forgive me, I'm not familiar enough with these ideas. (Haven't taken abstract algebra yet)

[scarlet estuary]

are you familiar with the subfield test

[lethal cipher]

No, I am not.

What's the test?

[scarlet estuary]

if youre checking for subfields instead of subrings, additionally you must check that each nonzero x in S has a multiplicative inverse in S

(and that R is a field)

basically the idea is

a subring is a subset thats also a ring

but being a subset, it "inherits" some properties of its parent ring

addition and multiplication still commute, associate, and distribute

so it suffices to check that:

• addition and multiplication "make sense" (ie its closed under these)
• it contains 0
• it contains additive inverses

but we can simplify "it contains 0" further to "it contains some element"

since the other properties imply that a - a = 0 are in the subset

as long as a is

so to verify your set ${x \in L \mid v(x) \geq 0}$ is a subring

[cloud walrus]

Namington

[scarlet estuary]

we need to check the following:

[lethal cipher]

Okay, both 1 and 3 are pretty easy here.
I think I got number 2 as well.
Since v(-1)=0, v(x+(-y))=v(x)+v(-y)=v(x)+v(y)>=0

[scarlet estuary]

$v(x), v(y) \geq 0 \implies v(x - y) \geq 0$ \
$v(x), v(y) \geq 0 \implies v(xy) \geq 0$ \
there exists an $x$ with $v(x) \geq 0$

[cloud walrus]

Namington

[scarlet estuary]

v(x+(-y))=v(x)+v(-y)
why?

[lethal cipher]

v(1)=v(1*1)=v(1)+v(1) by property 2

[scarlet estuary]

[this is my nice way of saying "this isnt true"]

[lethal cipher]

Oh, I follow. Whoops

[scarlet estuary]

okay, so?

[lethal cipher]

$v(x+(-y))\geq \min{v(x),v(-y)}=\min{v(x),v(y)}$

[cloud walrus]

dackid

[lethal cipher]

My bad

[scarlet estuary]

right, since v(-1) = 0 follows from v(-1 * -1) = v(1) which must be 0

[lethal cipher]

Precisely

[scarlet estuary]

and of course the minimum of both of those is ≥0

by assumption

so v(x - y) ≥ 0 as well

[lethal cipher]

And since K is a field, it must have an additive identity, so there exists at least 1 x so that v(x)=0

[scarlet estuary]

the other two properties, as you observed, are easier

[lethal cipher]

Which takes care of property 1

[scarlet estuary]

yes

and being closed under * is immediate

[lethal cipher]

Yea, it's just due to Property 2

Now quick question: are we allowed to assume -1 is in K?

Or was that a bit of a stretch?

[golden pasture]

what is the additive inverse of 1

[scarlet estuary]

every element of a field has an additive inverse.

[lethal cipher]

https://c.tenor.com/ZFc20z8DItkAAAAM/facepalm-really.gif

[golden pasture]

tbf a lot of stuff in algebra once you relook at the definition it becomes immediate

[lethal cipher]

Okay, so -1 is just notation for the inverse of the identity

[golden pasture]

yesh

[scarlet estuary]

yes.

its not THE number -1

[golden pasture]

so if p=0 in K then -1=p-1

[scarlet estuary]

as in the real number

[lethal cipher]

Got it, so we didn't make any bold assumptions here.

[scarlet estuary]

it just means "negative of additive identity"

so -1 = 4 in Z/5Z for example

[lethal cipher]

That checks out yeah.

Alright, let me get this all down before I try to tackle the maximal ideal

Correct me if I am wrong, but I feel like it's important to show the additive identity is in there too

Or is that not necessary

[scarlet estuary]

oh, in this case you do need to show that

sorry, the formulation of the subring test i gave assumes rings do not require having unit

conventions vary by author

[lethal cipher]

Okay cool. Luckily it isn't hard, it's defined by the first property

So another thing my prof wanted me to do is try and prove that $U={x\in A_v|v(x)=0}$ is the units of $A_v$. What exactly would that entail?

[cloud walrus]

dackid

[lethal cipher]

I'm not sure what to prove, I honestly thought that'd just be the definition

[lethal cipher]

Nevermind, I got it. That wasn't bad

Question, how do you show a maximal ideal is unique?

[hot lake]

you show that it contains every ideal that is not R

(and so if there is another maximal ideal J, then I contains J, and by maximality of J, I = J)

[lethal cipher]

Perfect. That is very easy to show for this specific situation as $A_v\setminus U$ is the maximal ideal here.

[cloud walrus]

dackid

[chilly ocean]

I cannot understand why we must proof [x]_m = [y]_m. because afaik a congruence class is well-defined if it's invertible which means ([a]_m)^-1 has to be equal to [a]_n

you're not proving that [x]_m = [y]_m

you're proving that if [x]_m = [y]_m, then [x]_d = [y]_d

yes but [x]_d = [y]_d is in the same set. we must proof something like f([x]_m) = [y]_d

do you know what well-defined means?

yes, every single member of domain assigned to a single member of its range.

so we have exactly one answer for every input.

[chilly ocean]

how is this (second sentence) different from what they're doing in the picture you posted?

f([x]_m) = [y]_d means [x]_d = [y]_d.

ie

if [x]_m = [y]_m then [x]_d = [y]_d.

[chilly ocean]

why these two arguments are equal ? why f([x]_m) = [x]_d ?

what's f?

a function that maps every single element of domain to an element of its range

just any old function?

i thought you were using it to refer to the map in the picture you posted but okay

yes, old function

well, okay, then interpret what i wrote in terms of f being your map

so basically we must proof there exist [y]_m such that [x]_m[y]_m=1 then since d|m, such argument also exist in set d.

[chilly ocean]

^

i have just wanted to fit my invertible definition into that statement xD

thanks

[marble whale]

I've got a ring (R, +, ) and its ideal (I, +, ). I also have J := {x in R | xa = 0_R for every a in I} and I'm supposed to prove that J, + , is the ideal of the previously mentioned R.

I came to the conclusion that since x*0_R = 0_R, the ring J might be equal to R. But if the ring I contains anything else than 0_R, then J can only contain 0_R. Is my logic correct?

[mint seal]

hang on, just because x*0_R = 0_R that doesn't mean x is in J

for x to be in J it must have x*a = 0_R for all a in I

to prove J is an ideal, show it's closed under + and * and also j*r is in J for any r in R, j in J

[marble whale]

Hmm perhaps I made a mistaken in my explanation. So my idea was that either x has to be 0_R or a has to be 0_R. This happens for every a if R contains only 0_R or I contains only 0_R.
In the case of R ={0_R}, I is also {0_R} because I is an ideal of R and this case seems trivial.
In the case of I={0_R}, J would seem to be R. As x*a = 0_R when a = 0_R is true for every value of x

For those cases it should be easy to say that J is an ideal as R is said to be a ring and 0_R is the trivial ideal of a ring

[carmine fossil]

It's not clear why that should be true

[marble whale]

but hmm, a*x = 0_R only if either a = 0_R or x = 0_R and that happens only if I contains only 0_R or R contains only 0_R

I don't really know how much more specific I can be

[carmine fossil]

Is this an integral domain?

Otherwise only if part is not clear

[mint seal]

Your assumption about x and/or a is false

The ring can have zero divisors

[carmine fossil]

What is an example of ring R,such that J is not 0_R?

[mint seal]

consider R = Z/12Z = {0, 1, 2, ..., 10, 11} with + and * done modulo 12

it contains an ideal {0, 3, 6, 9}

[carmine fossil]

Ok,That works

[mint seal]

now, what is J for this ideal? All elements of Z/12Z such that if you multiply it by anything in {0, 3, 6, 9} you get 0

those are {0, 4, 8}

which is an ideal itself

ah

[marble whale]

I'm so confused

[mint seal]

btw, I can actually be any subset of R and you can form J exactly the same way and it'll be an ideal

[marble whale]

ah so searched for zero divisors and it seems like some matrices for example allow x *a = 0 even when x and a are not 0

[mint seal]

It's called the annihilator

[carmine fossil]

I think this is a far more interesting question: find a ring R such that the annihilator ideal is non trivial

Or it's probably obvious

[mint seal]

I think I gave one

The annihilator of {0, 3, 6, 9} in Z/12Z is {0, 4, 8}

[carmine fossil]

I didn't mean annihilator of some ideal in the ring

I meant the annihilator of ring

[mint seal]

Ohh

Well if R contains a multiplicative identity, its annihilator can only be {0}

[carmine fossil]

What about RNGs?

[mint seal]

that I don't know, I wonder

[marble whale]

Well, in this case if a*x = 0_R and a and x don't have to be zeros, then J can be said to belong to R. But I'm not really sure how I would used that to prove that J is an ideal. I just see a set which might or might not be closed

[carmine fossil]

A group is always closed

[marble whale]

well, a set which might or might not be closed

[carmine fossil]

Show wrt +,J is closed

[mint seal]

Take two elements of J and show that their sum and product still have the right property to be in J

[carmine fossil]

and if a is in J xa and ax will also be in J for all x in R

[mint seal]

yeah, and that

[carmine fossil]

Closure under product is a specific case of this

So,You don't have to do that specifically

[mint seal]

you do have to check that

or oh I see what you mean

yeah

[marble whale]

c, d in J
c+d in J since (c+d)a = ca + da= 0_R + 0_R = 0_R
cd in J since (cd)a = c(da) = c*(0_R)= 0_R

[mint seal]

yup

actually there's a slight flaw in the second argument

it's not that da = 0_R

but da is in I

so c(da) = 0 as required

In the first part you assume c,d both in J

but in the second part you take c in J but d is anything in R

[marble whale]

I guess I could get around that by saying cd in R so cd = g, g in R. Thus if g*a=0_R can be found, then multiplication function is also closed in J

[mint seal]

you basically just have to erase the step where you said c(da) = c*(0_R)

instead, you can say da = something in I

using the fact I is an ideal

and then since c is in ann(I), c(da) = 0_R as desired

where I called J by the notation ann(I) to emphasize its property

[marble whale]

hmm that works if d*a is every value in a, in other words, I is only 0_R

[mint seal]

it's ok, a is totally freely chosen from I

the argument begins by considering (cd)a

where a is arbitrary but in I

and we conclude this equals 0

so cd is in J

[marble whale]

Almost got this, I'm so close

ah got it. Thanks for the help. Algebra has proven to be my weakness or perhaps I just don't get my teacher

[mint seal]

np, I had a tough time with it too back in the day

[mild laurel]

which implication in particular?

uh

I think Lang here is trying to prove the example using the proposition, not proving the proposition

Oh, I mean, N_H contains H, so since H has index 2, N_H must have index <= 2

yeah

I mean (A : B) >= 1 is always true for any subgroup B of A

sure

[mint seal]

yup

[mild laurel]

Yeah

[mint seal]

that's like the thing that comes up in Cayley's theorem I think

there's also a conjugation action (?)

[mild laurel]

I mean

You can think of an action of a group G on a set X as a group homomorphism from G to sym(X)

He's using what he said earlier

And only looking at the action on H and it's one other conjugated subgroup

Yes

[mint seal]

I agree that mentioning the kernel is normal is not really using anything from before

I'm also not sure why it has to have index 2

I think he emphasizes that it's normal because the whole idea is to get that H is normal as a contradiction

[mild laurel]

Well, the kernel can't be everything right

You can use the first isomorphism theorem here

This homomorphism is onto, so G quotient the kernel must be Z/2Z

So the kernel must have index 2

Yeah sure

Yeah sure that works too

[mint seal]

hm that's no good

H is always a normal subgroup of N_H

but N_H need not be normal in G

I don't see why N_H can be viewed as a kernel

you said ker of conjugation of H

not sure that works

I liked how you had it

G/H

as a set, it's the set of cosets of H

since H is index 2, there are 2 of those

hmm

[mild laurel]

The problem is that

[mint seal]

maybe it happens to be true when H is index 2

[mild laurel]

An element is in the kernel of the group action homomorphism if and only if it acts trivially on all the elements

[viscid pewter]

bigger than index 2?

[mild laurel]

So let all the conjugate subgroups of H be denoted by X or something

An element in the kernel of the map G -> sym(X) acts trivially on all the elements of X

Whereas an element of G could act trivially on H (and so be in it's normalizer) but not act trivially on some of the other conjugate subgroups

[viscid pewter]

i think it's only index 2

special case

i mean clearly not every subgroup is normal

ok so assuming the index of the normaliser in G is 2

we have a group idk X where X is just {H, gHg-1} where g isn't in H

from the group action of G on H

by conjugation

which corresponds to a homomorphism from G to uhhh something? i'm a bit unclear on this tbh

but anyway the kernel of this homomorphism is H

how is it not a group

[uncut girder]

Does this work for any group?

[viscid pewter]

for any group, a subgroup with index 2 is automatically normal

[uncut girder]

Or only finite groups

[viscid pewter]

oh uhhhh

[uncut girder]

Does it work for lie groups?

[mild laurel]

not if you want closed subgroups I don't think so

[viscid pewter]

ok so any element ghg-1 in gHg-1 can be thought of as a representative of gHg-1, and clearly gHg-1 is closed right

mood

oh, lang, that fits

... i concur

but anyway any element of either H or gHg-1 can be done in the form of ghg-1, and combining two you'll surely end up in one of the two again

[uncut girder]

If you start with a closed subgroup, wouldn't the conjugates be all closed subgroups?

[viscid pewter]

i think it's like, H and gHg-1 together are the whole group

so necessarily any combination of elements from either or both is in one of the two of them? hmmmm

yeah idk what i'm talking about

[mild laurel]

this only rlly works for index 2

2 is special here bc if you fix one of the elements, you have to fix the other too basically

whereas this might not be true for S_3 and up

this goes back to what I said earlier about how you can fix one of the elements but maybe not all of them in general

If you have an element of S_2

that fixes one of the elements, it must also fix the other element

[viscid pewter]

completely different tack, or well related but different:
i think the way i learned this result is like. you have two cosets of H in G, 1H and gH left, and H1 and Hg right; 1H = H1, so gH = Hg, so H is normal

that seems simpler

not quite sure how to relate it to that proof up there, i'm not the sharpest at normal subgroups and things

[next obsidian]

For the index 2 thing I think showing gH = Hg is the easiest way to show its normal

[cloud walrus]

mirzathecutiepie

mirzathecutiepie

[viscid pewter]

i think the homomorphism is just from G, not G dot H

[mint seal]

I think so

[mild laurel]

yes

ehhhhhhhhhhhhhh

f(x) is a permutation on the set {H, gHg^-1}

so what does it mean to say that f(x) = H

[mint seal]

x being in ker f should mean f(x) is the identity permutation

[viscid pewter]

it's a restriction of the original permutation g sending H to gHg-1 right

[mint seal]

which means xHx^-1 = H

and x g H g^-1 x^-1 = g H g^-1

[viscid pewter]

tbh i really really hate this proof

[mint seal]

I think we're getting our permutation via a conjugation action

[viscid pewter]

would it have killed lang to explain a little more

does he sleep hugging the trees he saved

[mild laurel]

just read a diff book?

im p sure you read this book at my recommendation

[mint seal]

when hitting H with conjugation by x, there are only two possible outcomes, H itself or "the other one" which we can notate g H g^-1

for some g

this is because H is index 2

so that's how the permuting comes about

some x leave H alone, that's the identity permutation

and others switch H with gHg^-1

[viscid pewter]

ok so it's like

given that H is index 2, we've sorta kinda more or less shown that N_H is index 2 in G? i think?

and since H is in N_H, H = N_H

wait

nope that's not it

wait no

we wanted to show that N_H = G because H normal in N_H?? i'm so confused, ignore me, sorry

[mint seal]

cracks knuckles

[viscid pewter]

the action of G on the set of all subgroups is a homomorphism from G to the symmetric group of all subgroups of G

restrict the action to the orbit of H, just {H, xHx-1}; the action on this orbit is just basically flipping one to the other or keeping it the same, so S2, or Z2; what's important is it's a group

so there's a homomorphism from G to Z2; the stabiliser of H in the conjugating action is all h in H, the stabiliser of xHx-1 of the conjugating action is all h in H again? (edit: this is because as all h in H sends H to H, h must send xHx-1 to xHx-1 to be a permutation)

so the kernel is just H; kernels are normal, H is normal in G

i think that's it

f me

[mint seal]

why is this part true? "the stabiliser of xHx-1 of the conjugating action is all h in H again?"

I guess because this action is faithful?

or.. it just wouldn't be a permutation if h sent xHx^-1 AND H itself to H

so h must send xHx^-1 to itself

[viscid pewter]

i think that's it

[mint seal]

seems legit

I'm reading the proof from D&F now

[viscid pewter]

i think their proof is different and better

[mint seal]

looks like they use the usual left action instead of conjugation action

[viscid pewter]

so much shorter

[mint seal]

it uses that "every left coset is a right coset" characterization of being normal

[viscid pewter]

yeah, that's standard for them

it's a group

and there's only one group of order 2

[mint seal]

D&F aren't doing a contradiction though

ah, yeah earlier they assume N_H is not all of G, i.e. H is not normal

[viscid pewter]

yeah, so that all checks out right

[mint seal]

what's the cool thing about conjugation and rubik's cubes?

like they let you swap certain squares around without messing up everything else, or something

I never quite got the intuition

[wraith obsidian]

Because with that you e.g. only need to learn an „algorithm“ for an edge-three-cycle on the top layer and suddenly you're able execute all permutations of three edges by doing

• setup moves moving the three edges to the top layer
• the three-cycle
• the setup moves in reverse order

On the one hand, this is practically used in blindfolded speedsolving due to its simplicity in permutation.
On the theoretical side, it's a really good tool to „invent“ algorithms or prove certain constructions are possible I guess, since you only need to prove that the edge three cycle exists, and by conjugating that way you suddenly get arbitrary permutations of the „same kind“, but just acting on different pieces, for free

The rubik's cube is a weird example because it's been so heavily studied, but just imagine someone would give you a different puzzle; a strategy like that would be quite helpful in figuring out what you can do and in finding out structural decompositions of the underlying group

Or you can just ignore all that and mumble something about „permutation group theory“ and „something something Schreier-Sims“ which I know nothing about

[vital fossil]

mind if i interupt with a question? or is there a question going on right now

[wraith obsidian]

Not that I'm aware of, I was just reading backlog

[vital fossil]

ah ok

i think this question is quite simple

but im not good at algebra

i was just wondering if what i wrote is correct

just for

this highlighedt part

i believe this is how modules work right? in this course we're not expected to know all the details and im abit too lazy to go through the details myself since ibut i recall them being completely linear algebra methods

[mild laurel]

yeah I don't think that's quite right

[vital fossil]

how come?

[wraith obsidian]

I find it a bit confusing here to use square brackets for equivalence classes as well as the ℤ-submodule generated by certain elements

or am I misreading what ℤ[a-b,a+b] means

[vital fossil]

Z[a-b,a+b] means all the possible combinations x(a-b)+y(a+b) with z,y in Z

[mild laurel]

Like a-b is in Z[a-b, a+b] but not in Z[2b]

[vital fossil]

but im not saying Z[a-b, a+b]=Z[2b]

or does what im saying imply that?

[mild laurel]

oh, you changed it to Z[a,a] on top? Or is that a typo?

[vital fossil]

not a typo

[a]=[b] in this quotient space

since its quotiented by Z[a-b,a+b]

[mild laurel]

I don't think it makes sense to quotient Z[a] by Z[2b] then

Z[2b] isn't a submodule of Z[a]

[vital fossil]

here [a]=[b] so it is

[mild laurel]

I mean

That's the equivalence relation

But it's still true that Z[2b] is not a submodule of Z[a]

[vital fossil]

how about this?

[mild laurel]

yeah I think this works

[mint seal]

I started an FTL run and named my crew Lang, Dummit, and Foote

they all died in a fire before the end of sector 1

RIP the Conjugation

[mild laurel]

is that game fun? I've considered getting it

[next obsidian]

Bruh

FTL is super good

[mild laurel]

hurb

[next obsidian]

It’s also cheap

[vital fossil]

thanks

😄

whats FTL?

[mint seal]

it's amazing and the Multiverse mod is practically a full blown sequel

it's a very challenging sci-fi themed roguelike, pausable combat strategy game

[loud root]

Hi can I have some help explaining something?

I'm trying to understand this proof but one line seems too much for me

there's two statements in this sentence and both confuse me

[mild laurel]

The ideal I is maximal, so since I + (x) is an ideal that is bigger than I, I + (x) must be all of R

[loud root]

Ah okay, that makes sense

[mild laurel]

Since I + (x) is all of R, 1 is in R so 1 is in I + (x)

[loud root]

?I+(x) is {a+b : a in I and b in (x)} right

[mild laurel]

yes

they use the fact that (x) = {xy : y in R} in the last part

[mint seal]

isn't that a definition in some sources?

[loud root]

So I+(x) is also equal to Union_{y in R} I+xy

so 1 is in at least one of these elements of the union

i don't know why my brain needs me to break it down into so many small pieces

thanks @mild laurel

[mild laurel]

no problem

that's how it is when you're learning things for the first time

nice

[mint seal]

it's feeling pretty good I think

I wouldn't say it's a waste even if you still don't completely understand it

it's good to go through these things

ahh

nice, yeah that's a good observation

I feel about 90% sure there's no mistake in your writeup

I've been going back through algebra myself recently, it was always a tough spot for me

definitely didn't understand all this on the first pass through

[wraith obsidian]

My visualization also relies on the „invariant“ part

I don't even think of subgroups as „complicated objects“ or made up of different things
I just think of them as „large elements spanning some area in the group“
So in my mental animation the conjugation action throws an element at it and sorta shifts that area around (together with all of the group)

And if it's normal, it's like just staying where it is

Yeah

Just remember that the fruit has an identity that it doesn't lose after the conjugation action

so it doesn't get pushed too far around lol

And when we're talking about left multiplication then it's sorta like a pancake elevator

since the cosets partition our group

Well, if you start with food metaphors I will certainly not stop

[mint seal]

if monads are burritos, what mathematical objects are flautas?

I think this could be a rewarding vein of research

[languid meteor]

hey, could anyone help me figure out with how exactly alpha and beta are given under the isomorphism in (a)? Im not sure how to interpret that

like if I restrict the inverse of the isomorphism to (m,0) then I get m?

[chilly ocean]

Ah, I hate this problem

[languid meteor]

yeah it sucks

[chilly ocean]

I think it is a named theorem, like such and so's lemma

[languid meteor]

think its the splitting lemma

[chilly ocean]

Ah yeah

[languid meteor]

any decent info I can find on it is in category language

[wraith obsidian]

What do you mean by „how α and β are given“. Isn't this what's written in (a)?

Oh, okay, there is one layer in between, I see.
They mean that α is the composition of m↦(m,0) followed by the isomorphism → L(+)N

[mint seal]

so which implication are you wanting to do at the moment?

[languid meteor]

like it says they're given "under the isomorphism"

im trying to do (a) => (b)

ah I see

[latent anvil]

wtf I this is one of the best problems why would you hate it

angery sham

[chilly ocean]

Because it's weird

And arrow stuff is bad

[languid meteor]

so that m element lies in L right?

[wraith obsidian]

should be so, yeah

[chilly ocean]

Admittedly I am not that algebraic, so Im sure there are much worse arrow problems

[languid meteor]

so I have gamma(alpha(m)) = (m,o)

[wraith obsidian]

In general, for me it helps to think in terms of the words „embedding“ and „projection“.
Trying to find a section of the projection M→N is trying to find an embedding of N into M that is undone by the projection.

[mint seal]

yeah, that's good

stick N into M+N in the most natural way imaginable

[wraith obsidian]

as god intended

[languid meteor]

like send n -> (0,n)?

[wraith obsidian]

Almost, now you're in the direct product L(+)N
You need to land in M

[mint seal]

I think I get (a) -> (b) and (a) -> (c)

but I don't see how to get any other one to complete the chain

[wraith obsidian]

it's the same trick as they did in (a), you just need to ||throw the isomorphism at it||
Now you have ||a map N→M, which is of the right type, as you want a section to the projection M→N which is defined in (a)||

[languid meteor]

,rotate

[languid meteor]

im pretty sure this should do it?

then yeah (a) -> (c) is very similar

I dont think I can prove (b) => (c) so I might need to show (b) or (c) => (a) instead

[mint seal]

yeah

[wraith obsidian]

I think the core concept would be that ||Having an imbedding by the first arrow and a section in the second essentially gives you an internal sum in M where exactness gives you that it's direct|| or something like that

[mint seal]

I guess to construct the isomorphism M -> L+N ...

assuming you have a map M -> L

you put that map in the first 'slot'

[languid meteor]

ahh that does make sense actually

[mint seal]

and then use \beta for the second slot

and the fact the sequence is short-exact will imply this really is an isomorphism

yeah I think lux was saying essentially this

I was thinking out loud on (c) -> (a)

[languid meteor]

yeah im pretty sure its given that alpha is injective and beta is surjective, so there must be something there to make clear the bijectivity

[wraith obsidian]

to get from section and retraction to the sum notice that wherever we have an „arrow pair“, whether it's left or right, we can compose to get ||a projection p of M „into itself“ since it's idempotent|| and then ||1-p is also an idempotent giving us a direct sum decomposition into Ker p (+) Ker 1-p||
If I didn't totally forget how ||idempotents|| work

that's how I remember this whole shidazzle

[mint seal]

1-p?

[wraith obsidian]

*identity

[languid meteor]

so I can use the section/retraction to split any m in M into a sum of an element of L and an element of N

then just show direct sum that way

or is there a more sophisticated method

[mint seal]

that sounds good to me

I don't quite see the details

like say we have the retraction from (c)

$\rho: M \rightarrow L$ such that $\rho \circ \alpha = \mbox{id}_L$

[cloud walrus]

ManifoldCuriosity

[mint seal]

and we also know alpha is injective, beta is surjective, and ker beta = im alpha

[languid meteor]

im trying to write m in M as m = m + (something) - (same something) and then show that (something) is in L and (m-(something)) is in N

but to no avail so far

[mint seal]

I wanted to consider the map given by

$m \mapsto (\rho(m),\beta(m)) \in L \oplus N$

[cloud walrus]

ManifoldCuriosity

[mint seal]

I can show it's onto

given $(l,n) \in L \oplus N$

[cloud walrus]

ManifoldCuriosity

[mint seal]

mmm wait, maybe I don't see this

yikes, it's worse than I thought

yea nvm my attempt at a mapping, it's harder than that

[languid meteor]

im trying it the other way around where you map L + N -> M by (l,n) -> alpha(l) + sigma(n)

this is for (b) => (a)

[maiden ocean]

im trying to show that if Spec(A) is disconnected A is isomorphic to the product of two non trivial rings

so its pretty easy to take this and find ideals a, b with V(a cup b) = varnothing and V(a cap b) = Spec(A) which the latter is equivalent to a cap b subseteq nilradical

but im not sure what to do from here?

like i was thinking u could do maybe a/(a cap b) x b/(a cap b) or something

but idk how to get that a cup b cover A

if they even do

[latent anvil]

What does the former equality say?

V(a cup b) being empty I mean

[maiden ocean]

im not sure tbh

it means a cup b isnt in any prime ideals of A

but it doesnt mean that it cant be in, like, the union of prime ideals of A

so idk how we can confirm that it contains a unit or smth

[sturdy marsh]

if an ideal is in a union of finitely many primes, it is in one of them

[maiden ocean]

oh right!! ok so that means it must contain a unit and thus 1 right

so fkA cup fkB = (1)

[sturdy marsh]

A cup B isnt really an ideal

[maiden ocean]

er

right

[sturdy marsh]

but the ideal generated by it is A+B

[maiden ocean]

mhm

[sturdy marsh]

so yeah you get A+B = R

what do you know about A \cap B

[maiden ocean]

its contained in the nilradical

[sturdy marsh]

yeah sorry I had to leave for a second

[maiden ocean]

so er if we do f: (x, y) |-> x -y the kernel is x = y so its A cap B

ah hmmm

can we like. quotient out A by A cap B

and then it still sums to R

[sturdy marsh]

ignore A cap B

look at the ring Z/6Z

what does Spec Z/6Z look like?

I mean what are the prime ideals

[maiden ocean]

uh its like. the ideals generated by a where a divides 6 right

(2) and (3)

[sturdy marsh]

yes

[latent anvil]

ah sorry moth I am very tired and forgot about this the moment I replied

[sturdy marsh]

can you write Z/6Z as a product

[latent anvil]

I'll let brofib explain

[maiden ocean]

uh (2) cap (3), x, y |-> x + y no?

wait

sorry lol

(2) x (3)

this is an isomorphism of (2) x (3) onto Z/6Z

but it only works cause the intersection is trivial doesn it

[sturdy marsh]

not really

(2) isnt a ring

think chinese remainder theorem

[maiden ocean]

er right Z/2Z and Z/3Z

oh! is it A/a x A/b

ahaha i got tripped up because i was thinking idempotents

[sturdy marsh]

you arent quite done

[maiden ocean]

ah hm. so we do A/a x A/b given by mapping... x + fkA and y + fkB to x + y + (fkA + fkB) = x + y?

uh is this well defined

wait lmao

no hold up

[sturdy marsh]

the map from A --> A/a x A/b has a kernel

[maiden ocean]

right, a cap b, no?

[latent anvil]

yup

But what do you know about a cap b?

[maiden ocean]

um its contained in the nilpotent

[latent anvil]

so not enough to say it's trivial

[maiden ocean]

right

[sturdy marsh]

so you have a product decomp of A/Nil

[maiden ocean]

ya

[sturdy marsh]

so this gives you a nontrivial idempotent in A/Nil

[maiden ocean]

uh. do we have to do the e x (1 - e)

[sturdy marsh]

well we first need to find an e in A

you have an e in A/Nil

[maiden ocean]

um ok so we have e + fkN = e^2 + fkN, that is (e - e^2)^n = 0

[sturdy marsh]

right, let d = 1-e, which is also idempotent

[maiden ocean]

er. i guess expanding it out to e^n(1 - e)^n = 0 is easier, then

yea

[sturdy marsh]

try playing around with e and d

did you figure it out?

[maiden ocean]

idts : |

[sturdy marsh]

what does this tell you

you have e^nd^n = 0

try showing that (e^n) + (d^n) = A

because then this would give you a product decomposition of A

[maiden ocean]

yea

i mean i guess im supposed to show e^n is idempotent

agh

[sturdy marsh]

no show this

what is (1-e)^n?

[maiden ocean]

uh 1 - en + (n choose 2) e^2 - ...

[sturdy marsh]

yeah okay that was maybe a bad approach

[maiden ocean]

hshshs

[sturdy marsh]

so in the quotient we have e + d =1

[maiden ocean]

mhm

[sturdy marsh]

uh wait lemme try writing this down

[maiden ocean]

np

[latent anvil]

oof

so where are we at

e^n and d^n are idempotents of A which sum to 1 in the quotient?

[sturdy marsh]

e is an idempotent in the quotient

d = 1-e

we have e^nd^n = 0 for some n

[latent anvil]

gotcha

[sturdy marsh]

oh wait there cannot be a common maximal ideal containing them

[latent anvil]

right, agreed

I thought you just looked at like e^(2n) lol. Been a while since I did this problem

or wait...

(e+f)^(2n)?

who knows

[sturdy marsh]

this should imply that they are coprime right?

[latent anvil]

(e) and (f)?

what about (x) and (y) in k[x, y]

[sturdy marsh]

(e^n) and (f^n)

[latent anvil]

Err wait sorry

nof the example I was thinking of

hmm

Yes I agree

[sturdy marsh]

if e^n + f^n was not the whole thing

[latent anvil]

The sum can't be contained in any maximal ideal

yup

[sturdy marsh]

then we get a maximal ideal containing them

[next obsidian]

based problem tbh

[latent anvil]

yup

[sturdy marsh]

so they should sum to 1

[latent anvil]

sorry, I am really ragged tonight

yeah I agree

[next obsidian]

I saw it from the middle and was like "yooooo... "

[sturdy marsh]

and chinese remainder then gives A/e^n x A/d^n

[maiden ocean]

:suffering:

[latent anvil]

so to backtrack

[sturdy marsh]

oh wait there's a gap

[latent anvil]

We're essentially saying that r((e^n, f^n)) = r(r(e^n) + r(f^n)) = r(r(e) + r(f)) = r(e, f) = 1?

[sturdy marsh]

in the argument

[latent anvil]

oh, hurb

[sturdy marsh]

(e) and (d) are coprime

[latent anvil]

oh yeah I switched from d to f

I feel like this works though

[next obsidian]

I'm not sure what step you're at, but you definitely looked at (e + f)^2n

[latent anvil]

so start with some ideals, choose e, f in them with e + f = 1, then since the product of the ideals is nilpotent we have e^n f^n = 0 for some n. By looking at primes containing them we see r((e^n, f^n)) = r(r(e^n) + r(f^n)) = r(r(e) + r(f)) = r((e, f)) = (1)

so (e^n, f^n) = (1)

so we have coprime ideals (e^n), (f^n) whose product is 0

[sturdy marsh]

r(e^n) = r(e), yes

[latent anvil]

yes

[sturdy marsh]

I was being stupid

[latent anvil]

np

I was being more stupid earlier

with me "wait (x) and (y) aren't contained in a common maximal ideal of k[x, y]"

Anyways this is a really really nice proof

Better than whatever I've done in the past

[next obsidian]

Wait are you looking at radicals and doing radical arithmetic to show that if e,f generate (1) then e^n and f^n do? It's as easy as taking a prime containing both

[latent anvil]

what

[maiden ocean]

hurb

[next obsidian]

like if p contained (e^n,f^n) it contains e,f so contains (1)

so not prime

[latent anvil]

oh yeah sure chm

That's how you prove the radical arithmetic

[next obsidian]

oh okay

[latent anvil]

like all those identities

[next obsidian]

I was thinking you were like trying to do weird crap with r(r(shit))

[latent anvil]

Are best proved by looking at primes containing

I was lol

[next obsidian]

but why

haha

[latent anvil]

radical arithmetic is nice

Idk why you avoid it

[next obsidian]

fair

it solved that earlier problem

[latent anvil]

@maiden ocean this is my proof sketch

well, brofib came up with the proof

but lmk if this is confusing

It starts from choosing ideals a, b with a + b = 1 and ab = 0

I think moth is still digesting this problem

[maiden ocean]

i believe i understand it

and the existence of such ideals follows from Spec(A) being disconnected

[latent anvil]

right exactly

I did this problem originally with ugly binomial theorem nonsense

but radical arithmetic is honestly so nice

@chilly ocean I think you're good now

[maiden ocean]

the rest of the problem is rly easy

im good from here

[latent anvil]

Yee

[maiden ocean]

ty @sturdy marsh and @latent anvil

🫂

[latent anvil]

I am so tired

and apologize for any inconvenience I may have caused

please do not tell my supervisor

yes to the second question

for the first, a lot of the applications of the class formula are about centralizers and commutativity and so on

Which are encoded by the conjugation action

[next obsidian]

The conjugation action is the first action you're really going to study a lot of, and the normalizer, centralizer, etc. are all intimately connected to that

[latent anvil]

Lang is discussing this before to explain how to interpret G_x in the specialized formula, probably

[next obsidian]

Which one is G_x

that's the like

stabilizer of x right?

g such that gx = x

then yes

this is also common language for it

It strikes me as being motivated by groups acting on topological spaces, it just kinda sounds like a word that belongs there

@latent anvil opinion?

Isotropy gropu

[latent anvil]

yeah that's more commonly used there

idk the etymology

But its what's used in eg riemannian stuff

iirc

oh mirza group actions really shine when they're acting on spaces

It's beautiful

idk of you've seen any topology

[next obsidian]

quotients

by group actions

[latent anvil]

If not I will withhold

[next obsidian]

sometimes you get a stack as a result

[latent anvil]

sure, so chmonkey is mentioning a cool application

Have you seen projective space?

I look at it in the mirror daily

ah that's okay

Something to look forward to :)

anyways yeah group actions in topology/geometry are great

[next obsidian]

Is there a different super common group action

I always go to like

R\{0}

into S^1

err

R^2

but that's also just another scaley boy

[latent anvil]

yeah, matrix groups

@next obsidian have you seen a construction of the grassmannian

[next obsidian]

lol

[latent anvil]

Everyone should follow @grassmannian on Twitter btw

[mint seal]

orbifolds seem cool

[next obsidian]

By a cover of open subfunctors

yes

[latent anvil]

For fun facts about k-planes in R^n

I also want to learn about orbifolds MC

[next obsidian]

Orbifold

[latent anvil]

they seem neat

anyways chm

[next obsidian]

I should just say fuck smooth manifolds

[latent anvil]

Bad construction

[next obsidian]

and learn orbifold theory

[latent anvil]

you should not

[next obsidian]

Yeah there's a construction in early chapter of ISM

but it's not good

[prisma ibex]

Nah that’s based

[latent anvil]

That one fucking sucks

[next obsidian]

you want the cool one later

[latent anvil]

Let me explain the good one

[prisma ibex]

Orbifolds good

[next obsidian]

nG moment

[latent anvil]

So chm

What is a plane

[prisma ibex]

[latent anvil]

Or k-plane

[next obsidian]

it just A^n

[latent anvil]

no clearly it's an equivalence class of orthonormal bases

idoit

[prisma ibex]

Good answer!

[next obsidian]

I knew that

I was testing you

[latent anvil]

When do two bases have the same span?

[next obsidian]

When they are in the same orbit of the matrixy action

just write one in terms of the other

duh

[latent anvil]

when you can transform one into another by like the orthogonal group or some shit acting on the right

that's pretty much exactly it, yeah

[next obsidian]

[latent anvil]

So O(k) acts on the space of orthonormal k tuples in R^n

[next obsidian]

I see where this is going

[latent anvil]

or something

I'm not your boss

and hey this is a nice group action

[next obsidian]

~

[latent anvil]

Nice enough that the qutoeitn is a manifold

tada

[next obsidian]

~

[prisma ibex]

I don’t think the orthogonal group is what you want here

[latent anvil]

Fuck off

[prisma ibex]

[next obsidian]

Yeah you want SO3(k^pi)

[latent anvil]

What's wrong with a good old fashioned orthogonal group

I may being saying everything wrong here I am not operating at maximal efficiency

[prisma ibex]

You want a certain parabolic subgroup to get a classical Grassmannian of k planes

[latent anvil]

hmm

[prisma ibex]

If you take the Borel subgroup of upper triangular matrices you get a complete flag variety

[next obsidian]

your mom is a certain parabolic subgroup which enables you to get a classical Grassmanian of k planes

OWNED

[prisma ibex]

The upper triangular matrices are the ones preserving complete flags

[next obsidian]

I have heard these words from Julia once

Matrices bad am I right?

[latent anvil]

yeah ng I don't see the problem

we have a principal O(n) bundle of the stiefel manifold over the grassmannian right

doesn't this present the base space as a quotient of the principal bundle?

Which is the construction I was giving chm

Am I going wrong somewhere @prisma ibex ?

[prisma ibex]

Well okay I guess the thing you’re quotienting by is a product of orthogonal groups

Maybe you’re right and are just thinking of it in a different way

[latent anvil]

No, I'm pretty sure I'm quotienting by O(k)

[prisma ibex]

I’m most familiar with Gr(k,n)=O(n)/(O(k)xO(n-k))

[latent anvil]

right, that's not the construction I'm giving

[prisma ibex]

Ahh okay carry on then

I think these are equivalent though

[latent anvil]

you know how the grassmannian is a classifying space for O(k)?

the infinite grassmannian

[prisma ibex]

Ahh right okay you’re thinking about it that way

[latent anvil]

and the universal bundle is the space of orthonormal frames

[prisma ibex]

Okay yea that’s fine

[latent anvil]

yup exactly

But in the finite dim case

[mint gulch]

Hey, I have been reading Introduction to Lie Algebras of Karim Erdmann and Mark J. Wildon and the motivation of the chapters is classifie all finite dimensional Lie Algebras, I think it's interesting but, why is important to classify them?

[mild laurel]

classifying things is like a lot of math

classification of finite dimensional vector spaces for example, tells you that there's one vector space of each dimension up to isomorphism and that's super useful

Maybe another important example is classifying finite fields

Since we know that all finite fields are isomorphic to F_q for some prime power q, we can directly work with that instead of working with just the definition of finite field

[mint gulch]

Ok, I see, thanks

👍

[chilly ocean]

I'm not sure what is a period, but I suspect this is based on the definition of exponent. Exponent means lcm of orders of elements right?

So the reasoning of the second statement is: if no element had order (period) divisible by p, then the minimal exponent of G has no factor of p. But this is a contradiction because the order of G divides a power of its minimal exponent

[carmine fossil]

Since G is abelian,period is just LCM of orders of each element

And if there is no element whose order is divisible by p,LCM won't contain p

So,G can't divide a power of exp

The period has to divide order of each element

Because a^n=1 implies ord(a) divides n

[nova plank]

What is the period? Order of subgroup generated/smallest exponent that gives identity?

[chilly ocean]

Overall there is nothing tricky here, just some fundamental theorem of arithmetic type stuff

[nova plank]

Okay

[carmine fossil]

p may divide |G|, but there may be no element of order p

In the general case

Yea,mb

That is true in general

For abelian, period is also LCM of order of generators

Abelianness is needed to ensure G/H is a group

[plucky flicker]

ZxZ subrings are nZxmZ and D={(a,a):a \in Z}. My question is that is there any other kind of subrings?