groups-rings-fields

[delicate orchid]

if I have to map out of C^op x C I'm just gonna leave...

[molten viper]

imma go back to rings and ideals yeehaw

[rocky cloak]

The homfunctor takes two modules A and B, and gives you a third module Hom(A, B)

[wraith cargo]

A functor is a map of categories

[molten viper]

oh interesting

[white oxide]

juicy

[rocky cloak]

And whenever you have maps f: A' -> A and g:B -> B' this induces a map Hom(A, B) -> Hom(A', B')

[wraith cargo]

And Hom maps two objects of a category to the set of morphisms between those objects

[topaz solar]

Hom(A, B) is just your morphisms A -> B, and in a lot of cases (modules, rings, and such) they can be considered as further instances of those structures

[molten viper]

wait wait wait, are those primes right?

[rocky cloak]

Yes

[molten viper]

wild

[rocky cloak]

Indeed

[crystal turtle]

bifunctor moment

[rocky cloak]

Contravariant in the first argument, covariant in the second

[crystal turtle]

two variable adjunction moment

[topaz solar]

A’ -> A can be thought of as pre-composing it with it

B->B’ is post composition

[rocky cloak]

It's like a mullet

[molten viper]

yeah idk what's going on here lol

[topaz solar]

f:A’ -> A means for g: A -> B we do indeed have gf: A’ -> B

It’s just function composition

[chilly ocean]

Could someone help me with this I’m a beginner at math

And possibly teach me how to solve them

[flat tree]

Please read #❓how-to-get-help

[crystal turtle]

not sort of, literally dual notion, in the sense of an adjunction

which is just this fact

[topaz solar]

Consider how a bilinear map looks like a linear function into linear maps

currying arguments and all

[crystal turtle]

🍛

[cloud walrus]

jagr2808

[rocky cloak]

Duality 🌈

[south patrol]

Lol asking about a (seemingly timed?) quiz on the wrong channel

[void cosmos]

imagine not being a master at tensor products lik me

ugh

...

amateurs

[south patrol]

Lol

[crystal turtle]

👀

[south patrol]

master of tensor products

[void cosmos]

yea literally grasped it first sight

its just so natural to me its as if i invented it

[crystal turtle]

i mean it's just like any other object defined by a universal property lmao

[void cosmos]

yea im joking

it was the hardest math topic for me in existence

[white oxide]

@void cosmos how far are u in hungy now

[void cosmos]

im in last chapter

structure of rings

u?

ur in injective/projective i think

[white oxide]

yea i need to read injective/projective for the third time and then do exercises

just started reading hom and duality and im lost

[white oxide]

damn

[void cosmos]

yea bro this section is looks hard

just looksh ard

but is actually the easiest of them no joke

[white oxide]

nah dude there's just so much composition and induced homomorphisms and verifications and i'm so overwhelmed

but yea ur prolly right on second or third reading it'll prolly get easier

[void cosmos]

yea it will get infinitely easier

[south patrol]

hungy

[void cosmos]

but ig it depends

on wdym by "hard" and "easy"

[white oxide]

and thinking about homomorphisms sending homomorphisms to homomorphisms is mind boggling

[void cosmos]

for me hard are proofs that involve a trick out of nowhere

i dont think this section has any tbh everything just follows from the defintion

but i think the notation gets confusing and u just get lost yeah

[white oxide]

im like going at two sentences per 5 minutes here

[void cosmos]

yea yea i see

[white oxide]

ye

it just gets tedious

[void cosmos]

ik what ur talking about

yea

but i mean u basically have phd level help for free here

so

u got this

[white oxide]

LMFAO that's 100% true

this server goated

[void cosmos]

yea

ppl here are super good

like

its like if im bronze they are diamond or smthing

btw the noetherian and commutative stuff is way easier

than this shit

the proofs are more slick but atleast u can have examples and test shit

lmfao

[white oxide]

lol yea i look forward to it

[void cosmos]

gl hf

[south patrol]

Lie algebra associated to the gneral linear group of the energy of a photon

[white oxide]

Lie algebra associated to the gneral linear group of the energy of a photon

[void cosmos]

?

crazy or what

ohh

gl

GL

yea

wtf is hf tho

[rocky cloak]

Plank constant times frequency I guess

I.e. energy of photon

[south patrol]

Yeah

[rocky cloak]

It might even make sense, aren't photons like irreducible representations of SU(2)

[south patrol]

though really energy is just omega

[rocky cloak]

Or is that just electrons

[south patrol]

hbar = 1 lol

[rocky cloak]

Pretty sure the standard model is based around representation theory of simple lie algebras somehow

[void cosmos]

why do you guys know physics

its literally like

made up

its not real

only real physics is actually chemistry

thats it

[rocky cloak]

Most of the stuff I know is made up

[south patrol]

I did physics in first year

[wraith cargo]

So did I
Except it was a QFT class and the prof was old mentally

[south patrol]

Rip

[wraith cargo]

So it was a challenge listening to him get lost every 10 minutes

[south patrol]

Lol

[wraith cargo]

For 90 minutes

[solar vessel]

90 minutes is crazy

[crystal turtle]

Imagine actually going to physics class lmao

[wraith cargo]

I actually liked physics a year ago
I wanted to be a physicist
But uh now I'm here!

Mathematical physicist

But still

Too applied for my taste

[rocky cloak]

I took a physics class once. It was mostly just normal math, but with horrible notation for everything

They even wrote $A_i^k B_k^j$ for matrix multiplication instead of just $AB$ like a sane person

[cloud walrus]

jagr2808

[solar vessel]

lmao

[void cosmos]

oppenheimer was so bad

godel tho

[south patrol]

Good

Yeah we had upper and lower indices in first lin alg course for 0 reason

[unkempt stream]

ADJOINT JUMPSCARE

[void cosmos]

yo

cdan someone tell me

the categorical terms

of the equality Hom(A tensor B,C) = Hom(A,Hom(B,C))?

[crystal turtle]

It's an adjunction

[void cosmos]

yea whts that

[crystal turtle]

Uhh

[void cosmos]

just for the knowledge

[crystal turtle]

How much category theory do you know lol

[void cosmos]

just definition of a cateogry

and a functor

and exact functor

if its alot then nvm

just was curious

[crystal turtle]

In short, I guess, it's a pair of functors F:C-->D and G:D-->C such that Hom(FX, Y) and Hom(X, GY) are in bijection with each other for every X in C, Y in D

[rocky cloak]

You just need to know what a natural isomorphism is first I guess

[crystal turtle]

Without categorical terms, the way I think of it sometimes is that it's a pair of functors F and G, going in opposute directions, so that F is "easy" to map out of, and G is "easy" to map into, in some compatible manner

[topaz solar]

Easy example: free & forgetful on groups

[wraith cargo]

Better example

Abelianization and forgetful

[topaz solar]

Every map from S -> UG is associated with F(S) -> G

So F is left adjoint to U

[crystal turtle]

In the hom-tensor case, this is with bilinear maps. A bilinear map induces a map out of the tensor product, and a map into the hom-space

[topaz solar]

[crystal turtle]

and this needs to be natural in both variables

[delicate orchid]

ofc you'd think this lol
unless you mean bad in the good way in which case I must agree

[rocky cloak]

The group ring is adjoint to the unit group
The grothendick group is adjoint to the forgetful functor from groups to monoids
The free whatever is adjoint to the forgetful functor from whatever to Set

[dim widget]

Keep the thread going everyone needs to say “better example” and give a universal algebraic object and the forgetful functor

[delicate orchid]

the group ring is adjoint to the unit group
since WHEN

[void cosmos]

wdym ofc i d think this

u think im not wise in the world of cinema?

im literally the goat

and it was boring af

[delicate orchid]

what year did citizen kane come out

[void cosmos]

2002?

200x idk when

who cares

about old movies

lol

[rocky cloak]

Better example:
The free category is adjoint to the forgetful functor from categories to quivers

[delicate orchid]

... it was 1941...

[void cosmos]

wtf seriously

why would i know that

i can tell you avengers endgame tho

[delicate orchid]

because you claimed to like cnimea lol

[void cosmos]

( which is far better )

[delicate orchid]

oh right

so you don't actually like films, you like slop

as is to be expected

[void cosmos]

slop? bro endgame is number 1

[delicate orchid]

anyway back to talking to something interesting

[void cosmos]

go watch your black and white movies

lmfao film noir

[delicate orchid]

and you can go back to r/movies

[void cosmos]

im a mod there already

[delicate orchid]

guessing the action of the group of units on morphisms is just restricting them

[south patrol]

There is a lmfao film noir

[rocky cloak]

Ring homomorphism send units to units

[delicate orchid]

yeah exactly

ok

[void cosmos]

unexpected

[coral shale]

ring homomorphisms send subrings to subrings right

i just dk why anyone would care about subrings

[rocky cloak]

Indeed they do

[void cosmos]

no one does

[coral shale]

so sad.

well at least the image is a subring...

[void cosmos]

ideals are more importnat

[rocky cloak]

I think subrings are just too hard

[delicate orchid]

society if ring was abelian

[rocky cloak]

gnir

[delicate orchid]

ok I've proven that adjunction

that was a nice little exercise

[rocky cloak]

Similar thing works for monoids, so I guess that means the unit group of a monoid is the cofree group generated by it

Stretching the word "free" a little bit

[delicate orchid]

yeah it's basically the same argument

addition doesn't play too much of a role

[rocky cloak]

I just like the word cofree

[delicate orchid]

elements of the tensor product are cohomomorphisms

[rocky cloak]

Reminds me that epimorphisms should be renamed comonomorphisms

[unkempt stream]

By that def is abelianization technically a free functor?

[crystal turtle]

Basically

In some sense at least

But free -| forgetful is really the more proper sense of free

wait Ab --> Grp is kinda forgetful

[rocky cloak]

Also, the inclusion of Ab into Grp isn't really forgetful, it's just an inclusion

Like you're "forgetting" a property, but nothing about the structure

[crystal turtle]

I've seen it be considered a forgetful functor before, but yeah I agree

[rocky cloak]

If it goes in the other direction I would call it cofree

[unkempt stream]

????

OH

I THOUGHT ABOUT IT

[delicate orchid]

yeah there we go

[topaz solar]

Explains why you have bad taste

[void cosmos]

okay ur banned from the subreddit

[coral shale]

,,\bZ^n\oplus\bigoplus^k_{i=1}\bZ_{{p_i}^{\alpha_i}}

[formal ermine]

I used to be a mod for r/holup

[coral shale]

classification of finitely generated abelian groups right

but we have nothing for infinitely? D:

[void cosmos]

yea

[void cosmos]

ban him from there too please

[formal ermine]

so true

[topaz solar]

Reddit really lets anyone mod fr

[elder wave]

Marvel fans mad because Oppenheimer didn’t drop an unfunny one liner before nuking a country

[void cosmos]

loved the kurt godel cameo/teaser for oppenheimer 2 tho

after credit scene showing jon von neuman 😄

[dim widget]

@coral shale there is stuff for countably generated ones

But yeah it becomes very ugly in general and impossible for arbitrary abelian groups

[coral shale]

End(A) for any abelian group is a ring. But not all rings arise from this right

Like I can see for the same abelian group you can put multiple different multiplicative structures

[dim widget]

@coral shale you can’t even get F_{p^2} as the endomorphisms of an abelian group

[coral shale]

o found what i was after tho https://math.stackexchange.com/questions/181848/every-ring-is-isomorphic-to-a-subring-of-an-endomorphism-ring-of-an-abelian-grou

[dim widget]

Yes but that’s just the same as saying every group is a subgroup of the symmetric group

[chilly radish]

cayley's theorem

[dim widget]

But yes, you can let R act on itself, but it will not be all endomorphisms, though it will be all of those which preserve the R module structure

[coral shale]

yh i wasnt aware there was a cayley for rings ...

[warm ember]

how do you prove the map in the fundamental theorem of galois theory is actually a bijection im dum

[wicked patio]

Well, every subgroup of the automorphism group can definitely be associated with the field it fixes

[warm ember]

yes

but what if that field fixed more stuff

[wicked patio]

?

[warm ember]

like

suppose H \leq Aut(K/F) fixes E

o ye

[warm ember]

this is typo

what if H'>H also fixed E

[wicked patio]

so the thing you want to prove is

Iff E is the fixed field of H≤G

Then H is the automorphism group of E

[warm ember]

yes

how to prove

[wicked patio]

hmm

[formal ermine]

really annoying proof

[wicked patio]

It might be a strategy to just target the strongest result first

That is, try to show that the size of the automorphism group is always |G| divided by the degree of the field extension it fixes

Since it seems like less of a corollary and more like a fundamental part

[coral shale]

https://media.discordapp.net/attachments/1080590780416143490/1136102574819393676/2023-08-02_02.05.05.53.png

hmm just managed to prove this

so for modules its the same

And so if you have M over R, then you are seeing a copy of R inside End(M)

So really modules are over some subring of their endomorphism ring

I hope i got that right

subfield for vector spaces

[topaz solar]

Yee, Cayley but for rings/modules

[coral shale]

slowly getting it

so if we consider End(Ring) do we get anything interesting?

u get a 3rd operation right?

[coral spindle]

You get a group ig, I don’t know of any other cool structure on it

Aut(Z/nZ) is ofc (Z/nZ)* which is nice

But that’s really all we get there

[coral shale]

oh. was hoping for a different progression of hyper operations. Like +, x, ^ but something else

[coral spindle]

Hold on, I confused this for groups

End(Z/nZ) is in fact trivial

There’s only one

[coral shale]

ill have a go on paper when i get the chance 😵‍💫 but its nice to at least see if im working into a dead end or not

[coral shale]

well ig this is miserable for this one

ic, only 1 generator and f1 = 1

[coral spindle]

Yar

[chilly radish]

End as rings

Yea?

[coral spindle]

Of ring endomorphisms yeah

So the only structure I can discern this having is that of a monoid. Maybe it has some interesting properties idk

[topaz solar]

Well, can we get a ring with End(R) describable in a nice way that isn’t

Uhhh trivial

[coral spindle]

Le epic Galois moment

[topaz solar]

Right right just me being dumb

[coral spindle]

I wonder if H=D for these monoids

Apparently that’s something semigroup people care about

[crystal turtle]

H=D?

[formal ermine]

maybe 8=D

[coral spindle]

Green’s relations, there are a few but H and D are two

I would bet that H=D for End(R) when R is Noetherian, but that is just a hunch

[red quartz]

oh

so should i just assume it for now

hmm i dont really like assumming non intuitive results

[formal ermine]

well you can just look up the proof

[limber sequoia]

say G is a soluble group. Then is $K(G) \cong G/K_{n-1}(G)$ ? ($K_n(G) = {e}$)

[coral shale]

I missed some steps. F a field,
"Galois monoid" from End(F) I presume
and
Galois group from Aut(F).
Cant see what breaks if we generalize to rings or like if we can no problem

[cloud walrus]

tomer_k

[rocky cloak]

What is K(G)?

[limber sequoia]

the commutant

and K_n(G) being K...K(G)

[rocky cloak]

I see, so K_n(G) is eventually the identity. But K(G) is usually not isomorphic to G/Kn-1

Just take S3 as an example

[coral shale]

what is commutant thonk

that isnt the centre or anything is it

oh commutator

K(G) = [G, G]
i presume

[limber sequoia]

thank you

[limber sequoia]

minimal subgroup containing all commutators

[coral shale]

i dont think it is a term

[limber sequoia]

thanks I'll keep it in mind, im working with some lecture notes so the terms might not be standard

[solemn dew]

how should one think here?

this is what i found:

but let's say that you can factor it into (a+bi)(j+bi), then would it be the product of those two or the one whose distance function is the biggest?

[coral shale]

what's J and what's i

[solemn dew]

i think it is called gauss integers

but of the form a+bi

where a and b are integers

[coral shale]

i didnt mean J[i].

Like does your book define these 2 symbols beforehand or something

Without context, that looks like an ordinary polynomial ring to me.

[wraith cargo]

Here's a hint
In a euclidean domain such as the gaussian integers, the gcd is the linear combination of two elements with the smallest non zero value of the norm

[wraith cargo]

It's Z adjoin the 4th root of unity

[coral shale]

why did they call Z J

yikes lol

[elder wave]

Jintegers

[dim widget]

Jahlentheorie über alles

[sharp sonnet]

Jawohl

[open sluice]

oh look, jermany is invading again

[lethal dune]

joj

[solemn dew]

x^4===1 (mod p) is it right?

was thinking about eulers theorem

[open sluice]

if x^2 = -1 then (x^2)^2 = (-1)^2

[solemn dew]

ahh alright thanks

are you allowed to do that ?

i thought only multiplication of Z and addition was allowed

[open sluice]

squaring is multiplication

a = b and c = d implies ac = bd right
so if a = b and a = b, then...?

[solemn dew]

a^2=b^2

😁

[static yew]

What is $\mathbb{Q}(\zeta_n)$ ? It's in a table with for example Q(sqrt2)

[cloud walrus]

Stevie-O

[glossy crag]

https://en.wikipedia.org/wiki/Cyclotomic_field

[solemn dew]

is there a smart way of doing this?

[wraith cargo]

Factor them

[static yew]

So Q plus an nth root of 1 apart from 1,-1?

Q(x) where x not in Q, x^n = 1?

[wraith cargo]

Ye
It's usually called the nth root of unity

zeta_n that is

The theory behind cyclotomic fields is very rich

Wiles and mazur worked in it a lot

[south patrol]

It's hot

[static yew]

It also mentions C_2
That is {-1,1} right? Group with operator being traditional multiplication?

[unkempt stream]

cannonball problem?

[static yew]

Wiles as in fermat's last theorem Wiles?

[wraith cargo]

Yes

[solemn dew]

what is that

im just thinking

guess a x

now you do polynominal division

[unkempt stream]

use euclid’s algorithm

gcd(x,y) = gcd(y,x mod y)

[solemn dew]

alright i'll remember this one

thanks

[glossy crag]

Yes, there is: the Euclidean algorithm

[frigid lark]

Wait over F the field of rational numbers, why not Q?

[smoky ivy]

If L/K is algebraic and separable of degree n, how do I prove that there are at most 2^{n!} fields N s.t. K \subseteq N \subseteq L

[south patrol]

Shouldn't the bound be better than that o

Do you mean "of separable degree n" or "separable and of degree n"

[smoky ivy]

and of degree n

[south patrol]

Well you can think of the corresponding group theoretic question

[rocky cloak]

Take x a primitive element of L, and f it's minimal polynomial. Then the Galois group of f is a subgroup of S_n so has at most n! elements. There is a bijection between subgroups and intermediate fields and any subgroup is a subset. There are 2^n! Subsets of S_n

[south patrol]

nice

[rocky cloak]

So the bound is definitely an upper bound

But you should be able to do much better

[smoky ivy]

i see

that's pretty nice

[south patrol]

hm i mean Gal(L/K) is a group of cardinality n so it has 2^n subsets and of course (much) fewer subgroups

or am i being dumb

[rocky cloak]

L is not a Galois extension

[south patrol]

Lol

Okay makes sense, sure

Because if you get a normal extension you get up to degree n! i suppose

[rocky cloak]

But it should be enough to consider subgroups that contain G(F/L), where F is the Galois closure

[south patrol]

and then it recovrs what you got i guess

Sad!

[rocky cloak]

Which seems like you should be able to reduce to 2^n somehow

[south patrol]

So I mean if we examine the proof that like there are finitely many intermediate fields, I'm pretty sure intermediate fields correspond to divisors of the minimal polynomial

right

[delicate orchid]

2^(n!) is like REALLY big so like it's probably not like bigger than that

[south patrol]

i mean 2^n! works lol we are making it smaller

[rocky cloak]

I'm just thinking group theoretically if H < G is a subgroup of index n, there can't be more than 2^n subgroups that contain H right

[delicate orchid]

sounds correspondense theorem-y

[glossy crag]

Many such cases!

[rocky cloak]

Like if you contain one element in a coset you have to contain the entire coset

[delicate orchid]

yeah, that's right

[south patrol]

yeah okay so what i said works in that yes if L/K an extension and x a primitive element then intermediate fields correspond to divisors of min poly of x

and so we can just count possible number of divisors

that gives us the 2^n bound right

[rocky cloak]

Think so

[solemn dew]

is this argument enough or does the equation not need to exist exist in K[x]?

[glossy crag]

The equation holds in K[x] by virtue of holding in F[x]

[south patrol]

yeah i think this is just a fairly trivial statement

[delicate orchid]

yes that's what he has to show

it really is just as simple as "thing in thing mean thing in other thing"

[rocky cloak]

Also any subgroup contains the identity, so that should bring the number down to 2^n-1

[delicate orchid]

sorry, write - not show

[south patrol]

Yup ofc

[solemn dew]

alrighty

[south patrol]

Can probable do more combinatorial stuff

But yeah

[delicate orchid]

ermmmm can we get a fact check on this please??

[south patrol]

Very nice

[rocky cloak]

Assuming n>1

[south patrol]

i mean

[rocky cloak]

No wait, still true for n=1

[south patrol]

if n = 1 it still works

ye lol

[delicate orchid]

yeah the identity contains the identity

[south patrol]

what about degree 0 extensions

[rocky cloak]

I assume (C2)^d must be the groups with the most subgroups per number of elements

So that might give a good bound

[wraith cargo]

Mfw doing maf from hospital

[dim widget]

@wraith cargo oh noooo what happened???

@rocky cloak yes and you could, if you so desired, use the primes dividing the index and the sylow theorems to give even tighter bounds

[wraith cargo]

Psych ward moment

I read too much SGA and went nuts

Jk I'm getting admitted to their like daily program

[solemn dew]

is there any way to think of a homomorphism with a trivial kernel?

(c) here

having a hard time finding a homomorphism

[solar vessel]

a homomorphism with trivial kernel is an embedding as a subthing

in this case a subfield

[elder wave]

All field Homs have trivial kernel anyway

[delicate orchid]

these are finite sets of the same size, so a surjective map is automatically injective too

[solemn dew]

oh

[delicate orchid]

recall that the kernel of a ring homomorphism is an ideal, but we're mapping between fields - so the only ideals are (0) and the whole field

[solemn dew]

if all homomorphisms of fields have kernel 0 then all fields are isomorphic to each other?

[solar vessel]

no

homomorphisms usually don't exist/aren't surjective

[solemn dew]

okay then haha

but they are clearly isomorphic, so is there a way of finding a homomorphism in a smart way?

[crystal turtle]

well all finite fields of same cardinality are isomorphic, if that's what you mean. But otherwise not in general, no

and any two algebraically closed fields of same cardinality and characteristic, but that's cursed so we ignore that

[unkempt stream]

no. First isomorphism for rings says that

For f | A -> B
im(f) is iso to A/Ker(f)

but since field ideals (and thus kernels) are either (0) or the whole ring, there’s only two options for the kernel

(0) so it’s injective/mono, so an embedding, or the kernel is the whole ring, so it’s the zero morphism

same holds for simple rings btw, since kernels are bi-ideals

[unkempt stream]

so, there often isn’t one specific morphism. There can be many

this is especially apparent with embeddings

[static yew]

So my book has a table saying that for the field Q(sqrt2), an (the?) Automorphism group is C_2

Is that because C_2 is {1,-1} and and Q(sqrt2) is the same field as Q(-sqrt2)?

[unkempt stream]

Try coming up with the automorphisms

[formal ermine]

the

[unkempt stream]

me still vehemently stuck on artin’s lemma

[formal ermine]

me still stuck on sheaves

[static yew]

Does that mean the answer to my question is yes

[formal ermine]

no

[unkempt stream]

dear god

[static yew]

:/

[formal ermine]

also you probably mean the Galois group lol

[solar vessel]

I don't even remember what artin's lemma is

[formal ermine]

in this case it doesn't matter because Q has char 0 fwiw

but usually you talk about field automorphisms that fix the base field

i.e. Q

[unkempt stream]

there’s like two forms

the one I’m referencing is that

if G is a finite subgroup of the automorphism group of a field F, then F forms a finite extension over G’s fixed point field of degree |G|

[formal ermine]

@static yew to give a hint: either list all automorphisms or show that if sigma is an automorphism and x is a root of a polynomial f then sigma(x) is also a root

[red quartz]

if a polynomial f in F[x] has distinct roots then let K be the splitting field for f

suppose f has r roots

is Gal(K/F)=S_r?

also f is irreducible

over F[x]

[white oxide]

how does the exercise make sense/what am I missing here? for any r in R and n in Z, r1_s + n1_s = (0, r) + (0, n) = (0, r + n) so the first component will always be zero. but take any nonzero first component and you can't write it as such

since we're assuming that $S = R \oplus \mathbb{Z}$

[cloud walrus]

okeyokay

[topaz solar]

R doesn't touch the second coordinate ever

[white oxide]

I'm confused, sorry

[topaz solar]

@white oxide (r, 0) + (0, n) = (r, n)

[white oxide]

$r(0, 1) \neq (0, r)$?

[south patrol]

Lies

[cloud walrus]

okeyokay

[topaz solar]

because the underlying set is R x Z

[topaz solar]

the one on the right doesn't exist, and the one on the left can be taken as (r,0)(0,1) = (r*0 + 1*r + 0*0, 0*1)

= (r, 0)

[white oxide]

oh bruh right

oops

[topaz solar]

r is not an element of S

[white oxide]

right Z is not an R module but R is a Z module

[topaz solar]

you can make S into an R module

namely rs = (r,0)s

[topaz solar]

Also, multiplication doesn't just distribute, the Z module addition-y multiplication is iterated addition so it happens tp

[white oxide]

ye okay that makes sense

i think i got it now

thanks

[delicate orchid]

has anyone seen this notation before? I can't find anywhere that actually explains it

[topaz solar]

it better not be the image of [a, (-)] on the set C_s(T)

[red quartz]

if f=gh in base field F and let K and L be the splitting fields for g and h. let E be the splitting field of f. then is Gal(E/F)=Gal(K/F) x Gal(L/F)

?

[delicate orchid]

OK BUT WHAT IS [A, -] U MFS

I'm being gaslit by every mathematican at once

[topaz solar]

you know uh

[G, G]

[delicate orchid]

the commutator?

[topaz solar]

ye

[delicate orchid]

WHAT

[delicate orchid]

I repeat these sentiments

[topaz solar]

that's what I'm expecting, but idk

Fusion systems aren't my thing

[delicate orchid]

I'd agree with you if the other part was a morphism

but it isn't

[delicate orchid]

even in the original papers they use this notation without explaination

so I doubt it's a fusion theorem specific thing

[topaz solar]

well it sure seems to be a subgroup of Z(T)

[delicate orchid]

sure does!

[topaz solar]

so I'm thinking commutator

[delicate orchid]

yeah but they have notation for conjugating by an automorphism

unless you mean something else?

oh like

uhh

[topaz solar]

well in the commutator you multiply that by the conjugate of the element of C too

the uhh aba'b'

[delicate orchid]

yeah

hmm

so like t^{\alpha}t^{-1} or somesuch

[topaz solar]

so it "looks" like it'll be aca'c' for c in C_s(T)

[delicate orchid]

I'll see if that works, but if that's actually what it is that is so SO stupid

[topaz solar]

No idea what C_S(T) is here but surely it isn't something weird if they didn't define it :^)

[delicate orchid]

the centraliser of T in S

T is a normal subgroup of S

[topaz solar]

ah

yeah I'd expect some commutator action however you get alpha to work

[delicate orchid]

so what we're saying is that the doodad moves the centralisers to the actual centre

[topaz solar]

ye

not sure exactly how that would look, t^alpha t^-1 would be in TC_S(T) I think though so?

[delicate orchid]

this is part of the definition of being "strongly normal"

so is there an analgous statement for normal subgroups

[topaz solar]

why can't they just be normal and write it out

[delicate orchid]

why can't bobby actually motivate his definitions the fraud

[delicate orchid]

yeahhhhh ohhh ok

hmm

[topaz solar]

this looks like the thing being strong-ified

[delicate orchid]

yeah

so if g is in the centraliser of N in G then the dude is clearly in the centre of N

[topaz solar]

so we want t^a t^-1 central? or do we want something else like t^(bar a) t^-1?

[delicate orchid]

no it like, directly correlates to this fusion theoretic definition

[topaz solar]

ah aight

[delicate orchid]

morphisms in a fusion system are like conjugation in a supergroup

[topaz solar]

makes sense

[delicate orchid]

well I'm glad it makes sense to at least one of us

[topaz solar]

well, making anything out of it is way different than "yeah thats a sensible morphism"

[delicate orchid]

I could use so much groupoid theory if there weren't inclusion maps bro it's unreal

[topaz solar]

uhhh just transfer to semigroupoids gg

[formal ermine]

@topaz solar hiii

[coral shale]

baa

[topaz solar]

hello my child

Have you come to confess your sins (not knowing determinants)?

[south patrol]

Knowing about determinants is the true sin

[crystal turtle]

who needs determinants

[solar vessel]

me

[chilly ocean]

everyone

[ornate vessel]

Are there any natural numbers in Z[w]/(2-5w) where w=cis(2pi/3)

[coral shale]

i bet 0 or 1 might

[ornate vessel]

I get that there arent

Only 0

Idk

Wait

I meant

In (2-5w)

[coral shale]

why isnt 2...

[ornate vessel]

The generated ideal

[coral shale]

ok...

[rocky cloak]

What do you mean by natural number in this context?

[formal ermine]

<@&268886789983436800>

[ornate vessel]

I mean natural numbers in (2-5w) ideal of Z[w]

[rocky cloak]

Ah, okay

[formal ermine]

hey jagr did you get a tag from the mod ping again?

[rocky cloak]

No, but I had the app open, so... I dunno if it matters

[formal ermine]

the most moderatorest moderator

[sharp sonnet]

jawohl

[rocky cloak]

Well (2-5w)(2-5w^2) = 39 at least

So any multiple of 39 is there

I would guess that's the best you can do

[ornate vessel]

Is 2-5w^2 in the ideal?

Why multiply by it

Xd

[rocky cloak]

The ideal consists of all multiples of 2-5w

So you can multiply by whatever you like

[ornate vessel]

Is 2-5w^2 multiple of 2-5w

[rocky cloak]

Don't think so

[ornate vessel]

Ohh

Ok but

We can multiply with all elements of Z[w] right

[rocky cloak]

Yes

[ornate vessel]

But 2-5w^2 isn't

In zw

[rocky cloak]

Yes it is

Z[w] = {a + bw + cw^2 } in this case

In general it consists of all polynomials evaluated at w

[ornate vessel]

I thought it was a+bw

Hmm

[rocky cloak]

That would not form a ring

[coral shale]

(a) = aR

note.

[coral shale]

min poly is cubic

[ornate vessel]

Ohhh

Then

Is Z[w]=a+bw+cw^2+dw^3+ew4 if w generates C5

[coral shale]

Z[w] === Z[x] / (min poly of w)
u seen or not seen

[ornate vessel]

I kind of see

[coral shale]

this min poly degree n

u r taking all polynomials
mod this minimal polynomial

And so what you end up with are polynomials degree n-1

they can reduce to it

[ornate vessel]

Ok I see

I was confused with Z[]

But I got it now

We need all possle

Multiplications

So it can form a ring

And all sums

So it's a polynomial polynomial of the elemnt

But it's given as intersection of all subrings that contain Z and the element

Xd

It makes sense

Min subring that contains Z and the element

Then what is the notation with () instead of []

[coral shale]

smallest field

[ornate vessel]

So it has all inverses of polynomials

[coral shale]

mm.

[ornate vessel]

but

It's the same form right?

Polynomials of the elemnt

[coral shale]

[ornate vessel]

Xd

[rocky cloak]

Yeah it will look like f(x)/g(x) where f and g are polynomials and g(x) is nonzero

If x is algebraic then F[x] = F(x)

(F is a field)

[ornate vessel]

😮

Ok

Idk these are confusing

Buts it's fine I got it 🙂

[rocky cloak]

Anyway, for your problem. If

x(2 - 5w) is an integer then applying the automorphism which swaps w and w^2 doesn't change it, so

x(2 - 5w) = x'(2 - 5w^2)

Multiplying these together gives 39 xx', and xx' is an integer. Since 39 is square free the square root is at least 39, so that is indeed the best you can do

[rocky cloak]

Actually I'm stupid w^2 = -1 - w

So it does actually equal a + bw in this case.

Doesn't change my calculations, but yeah

But 2 - 5w^2 = 7 + 5w if that's easier to deal with

[mint girder]

what

[ornate vessel]

🙂

Oh right

[vestal snow]

Classify all groups of order 33^2. Here's what I think should work. All such groups are abelian. To see this, we first need to show that n_3 and n_11 are both 1. n_11 being 1 follows easily from Sylow's theorem. Using Sylow, we get n_3 is either 1 or 121. Let's say we somehow show n_3 = 1.

Note that the the 3-Sylow and 11-Sylow subgroups will both be abelian as they are have order p^2. Since the Sylow subgroups are normal, G is isomorphic to their product. Thus, G is abelian.

Any ideas on how to show n_3 = 1?

[delicate orchid]

You mean 122 right

Perhaps use the other part of this sylow theorem, n_p divides the index of the sylow p-subgroup in G

I’m trying to remember how it goes for p^2q^2

[glossy crag]

I feel a little unsure doing induction that involves pairs of indices, e.g. proving Jordan-Holder (with compseries of lengths n and m). Is it valid in such situations to proceed by induction on min(n,m) (to sort of it give it a "uniformity")?

[white oxide]

i feel like such a fucking idiot right now, can i get another hint besides the ones they provided to show that ker zeta \subset Im theta?

okay i'll probably return tomorrow, i've spent over an hour on this problem lol

it's prolly cuz i'm tired and all the symbols and shit

thanks tho

[south patrol]

I mean ye fair it is not the most conceptual / obvious thing in the world

hope break help

[static yew]

I seek your guidance on doing some light trolling

6÷2(1+2)

if you were in, say, F_5, what would you consider the 6 to be? an indeterminate/transcendental?

[frigid lark]

1

[static yew]

but 6 (from ℤ) doesn't exist in F_5

[upper pivot]

is this the light trolling?

[open sluice]

5 := S(4) = 0
6 := S(5) = S(0) = 1

[static yew]

5 doesn't exist in F_5 either 😛

just 0,1,2,3,4

I know that when I just blindly reduce things modulo whatever y'all always call me out on it

[frigid lark]

Alr then, a homomorphism from the multiplicative monoid generated by 6, to F_5, sending 6 to 1

[static yew]

right, that's an option but it's not required to be that way, is it

[upper pivot]

this is clearly a troll, lets stop engaging. And also <@&268886789983436800> troll alert

[static yew]

meh

[ivory trail]

you know how there is a unique map from Z to any unital ring

[static yew]

right but I'm going for ridiculously abstract here
like I've been learning about Q[sqrt2] which is (a, b in Q -> a + b sqrt 2)

I think I said that right

[ivory trail]

sorta

so what's the issue

[static yew]

so if you start in a universe where the only digits are 0,1,2,3,4
6∙2^(-1)∙(1+2) is basically the same as x∙2^(-1)∙(1+2) because there is no 6
which means the expression doesn't actually exist in F_5, just like sqrt(2) doesn't exist in Q.
As 2/3 + sqrt(2) needs Q[sqrt2], you would need F_5[6] for the expression to have meaning.

now, you can (and did) define 6 as such:
6x = x6 = x
6 + x = x + 6 = 1 + x
which would make to be congruent to 1, which... if I remember correctly... would make it isomorphic to F_5[x]/(x-5)? Which would just be F_5 then
but I could **also say that 6 is transcendental and F_5[6] is of the form a + 6b (in which case the original expression is 6∙4)
and I could say 6 is congruent to something other than 1. that'd be 4 different possible values - {0,2,3,4} and produce

a bunch of rings that, while pretty much useless, are mathematically sound, no?

[ivory trail]

if you were to, for some reason, treat 6 as a formal symbol devoid of any context, then i suppose a bunch of stuff like this works out

but when people write 6 they almost never do that

[coarse kestrel]

iirc, if you assume 6 is transendental (it's not) then the elements of F_5[6] would be a_0 + a_1(6) + a_2(6)^2 +a_3(6)^3 + ... because it cannot be a finite field extension.

[static yew]

oh that's for sure

[static yew]

oh good catch! thanks, I screwed that up.

Since the "gotcha" expression from the original meme "6÷2(1+2)" only has one "6" we never get beyond that, but yeah. The difference between my F_5[6] and Q[sqrt(2)] is that sqrt(2) is the root of a polynomial in Q[x] and 6 isn't the root of a polynomial in F_5[x].

one day I hope to have internalized this properly

[ivory trail]

wait

if you're approaching this from a completely formal point of view

[static yew]

I was trying to get as formal as I could

[ivory trail]

then writing "Q[sqrt(2)]" is technically an abuse of notation

because if "sqrt(2)" is a formal symbol then this should just be Q[x]

but it's not

[static yew]

Q[sqrt(2)] shorthand for Q[x]/(x^2-2) right?

[ivory trail]

we write Q[sqrt(2)] as shorthand for Q[x]/(x^2-2)

[static yew]

woooo I remembered correctly

[ivory trail]

and sometimes if we're treating it as a subfield of a bigger field like Q[sqrt(2), sqrt(3)] we also have a chosen inclusion map

so when you say

sqrt(2) is the root of a polynomial in Q[x] and 6 isn't the root of a polynomial in F_5[x]
the former is true because it follows via this very mild abuse of notation, and the second is actually false because 6=1 in F_5

unless you write beforehand that "6" should be treated as a purely formal symbol, like "x"

which no one does for good reason

but then this is all technically informal reasoning

[coarse kestrel]

it's the root of x-1 or x+4, whichever way you want to think about it

[ivory trail]

for why the two rings are "different"

[static yew]

that's what I was getting at: inside of F_5 there is no 6
formally, if you're in F_5, there's no way a 6 could come about. Yes, a perfectly reasonable (and typical) interpretation is to assign 6=1

my point was formally I am not obligated to do that

[ivory trail]

actually you were just comparing the difference in adjoining an element to a base ring

formally symbols can mean whatever

[coarse kestrel]

3+3? 3 * 2? 4+2? 2 * 2 + 2? (2+1) * 2?

[ivory trail]

but everyone will be confused if you say that 6 doesn't make sense in F_5

because they understand it to be notation for 1+1+1+1+1+1

or perhaps the image of 6 in the canonical map Z -> F_5

[open sluice]

we normally define 6 as 5 + 1, where 5 is defined as 4 + 1

[ivory trail]

technically the "3" in Z and F_5 are different things

[open sluice]

so 6 does make a good amount of sense

[ivory trail]

so when you say "in a universe where the only digits are 0,1,2,3,4" you haven't actually been formal here

i was able to understand that you mean the elements of F_5, but then you have to just say that

but those aren't the same elements ("digits") from Z

because they belong to different rings

[static yew]

I'm not saying it doesn't make sense, I'm saying it doesn't exist
it's like in F_2 there is no 2.

F_2 is generally canonically defined as Z/2Z, which is a set {0,1} with associated + and * operators
there is no 2 in that set
in order to have an expression in F_2 that includes a '2', that '2' came from outside the set, and therefore the expression isn't really in F_2

just like "x + 1" isn't in F_2. it's a polynomial with coefficients in F_2. That polynomial does exist in F_2[x], though

(it also exists in one of the common representations of F_4)

[ivory trail]

F_2 is generally canonically defined as Z/2Z, which is a set {0,1} with associated + and * operators
this is incorrect. Z/2Z = {0 + 2Z, 1 + 2Z}

[static yew]

oh, good point

[coarse kestrel]

If you're going with that, then you have to figure out what 3+3 is in your field since closure requires 3+3 and 3 * 2 be elements of your F_5 field.

and, i'm pretty sure, if you assign it any value other than 1 you'll run into problems.

[long nebula]

This is untrue

6 is defined to be 1+1+1+1+1+1 which is 1 in F5

[long nebula]

this ^

[vestal snow]

Why 122?

[chilly radish]

Yea why 122

[vestal snow]

n_3 needs to divide 121, so it can be 1, 11, and 121. We eliminate 11 because it is 2 mod 3

[coral steeple]

In question 7a, does HK=KH mean that for all hk in HK there exists k'h' st hk=k'h' or does it mean hk=kh? It should be the former right?

[coarse kestrel]

hk=k'h'

[coral steeple]

Yeah that makes more sense

[chilly radish]

Yea it's equality as sets

[coral steeple]

I'm stuck trying to show HK is in KH

If HK leq G

The other direction is easy: fix h_0, k_0 and choose any h,k. Then h_0khk_0 is in HK by closure, so it is equal to some h'k'. Multiply to see that kh is in HK

[rocky cloak]

I don't think this is true. The automorphism group of C11xC11 has order 120*110 which is a multiple of 3. Thus there is a semidirect product between C11xC11 and C3.

Edit:
As for clarifying all of them, shouldn't be too hard to see that all such groups are semidirect products, and that up to conjugation there is only one nontrivial group homomorphism from C3xC3 / C9 to Aut(C11xC11)

[chilly radish]

Why is your first inclusion true

Wait sorry my bad I see it now

Just replace every h with k and vice versa

To show the other duration

Direction

[coral steeple]

Really? But we don't have that KH is a group, I'll try again though

[chilly radish]

Oh I'm sorry

You're right

You can't use closure

A priori

[coral steeple]

Right

[chilly radish]

Remember the formula for the inverse of a product

[coral steeple]

Hmm I did try something along those lines but again couldn't get it to work in this direction

[cloud walrus]

person2709505

[rocky cloak]

For hk in HK you know that it is the inverse of some other element in HK

[rocky cloak]

And your argument shows that inverses are contained in KH

[coral steeple]

Thanks, that's very intuitive and helpful

[delicate orchid]

chat....

chatters it's happened

[open sluice]

groumps rimgs and fielmds

wait i never learned modules tho

[delicate orchid]

how did you never learn modules you silly silly individual

[long nebula]

modules are rings acting on groups

so we're good

[worthy solar]

Can I get a book recommendation for this subject.(intro level).

Other than gallian

[long nebula]

Dummit Foote

[delicate orchid]

bobby fusions Artin

[worthy solar]

🫡 I will take a look

[wraith cargo]

NOOO WHERE IS THE SHIT POSTING CHANNEL

[summer path]

Still there

[worthy solar]

Oh is that what this channel is? But at the advanced level (turns out it's a thread)

[long nebula]

😭😭

[prime sundial]

how long have you all been thinking about splitting the algebra channel

i never thought it needed to be separated

[south patrol]

A while lol

Abs alg has a lot of (on a broader scale) easier stuff which means there isn't a good place for more involved qs ig

[long nebula]

Yeah it's been a while

Quietly behind the scenes, lots of arguing

[delicate orchid]

it was not quiet it was loud and very violent

[summer path]

All this means is that we can shitpost even harder here now

[south patrol]

Lol

Now we need to split topology/alg top into point set, homology and pi1, advanced alg top, homotopy theory

[worthy solar]

You don't happen to have the isbn off the top of the dome

[delicate orchid]

no of course i don't

the pdf is the first google result when you search "artin algebra"

[worthy solar]

😫 I think I figured it

[wet zodiac]

wha happen

why was algebra separated into 2

[worthy solar]

New update

[open sluice]

algebra 2

[wet zodiac]

algebra 3

ok so now we got the advanced channels (#real-complex-analysis #groups-rings-fields)

and the advanced advanced channels (#advanced-algebra #advanced-analysis)

[delicate orchid]

honestly this mofo could go in early university

[worthy solar]

Don't forgot stats

And number theory

[summer path]

That's probably true

[open sluice]

this is one of the last 3 math courses majors take at my school

[worthy solar]

What are the others

[wet zodiac]

your math majors are weak

[open sluice]

yes

[open sluice]

1 semester real analysis and then a capstone

[worthy solar]

sounds like my uni :/

I'm taking analysis this fall

[delicate orchid]

algebra is mistreated so much bro it's unreal

I was fortunate enough to be introduced to groups in my first semester

[worthy solar]

Are you in the US?

[open sluice]

it was my second semester for me

ya

and i'm pretty happy i learned groumps early

carried some bits of my linear algebra

[worthy solar]

I looked at it for a bit

[crystal turtle]

Hmm what's up with these new chanmels

[worthy solar]

Stopped on cycles

Will return soon

[worthy solar]

So what do you have to take alot of non math courses

[open sluice]

yes but I mean I don’t think that’s exclusive to my situation

[open sluice]

tbf most of the students here are actually math ed

[wet zodiac]

LAME...

[delicate orchid]

1984

[open sluice]

good book

[wet zodiac]

if my university says "no" to the courses i want to take

[delicate orchid]

"ur gonna pay us $9k to become a [thing]?? too bad you have to spend at least half of that doing random bullshit"

[wet zodiac]

im just gonna say "screw your random bullshit classes" and transfer

[ashen heron]

monoid homomorhisms need not map identity to identity?

[delicate orchid]

depends on your definition

to me they do

[wet zodiac]

i am here to learn eldritch horrors beyond my comprehension

not take random classes

[delicate orchid]

such a dumpster fire nation, anywho

[delicate orchid]

google "cosimplicial space"

[wet zodiac]

category theory

[delicate orchid]

no they're actually very topological objects

but then again what isn't :smoked:

[wet zodiac]

but yeah transferring seems very

appealing

[worthy solar]

What if your course work doesn't transfer. I guess long run it's better

[wet zodiac]

if you wont let me take the classes i want so you guys make more money by throwing random classes

you're not gonna get any money from me in the first place

[open sluice]

i got bad news for you

[worthy solar]

What is math ed? Like becoming teacher?

[wet zodiac]

yea

[open sluice]

yes K-12 teacher

I mean algebra is probably helpful for them in theory but most of them don't know how to learn it properly so it doesn't even matter

[ashen heron]

monoid is just a semigroup with identity to me, and with a monoid homomorphism i mean a map phi: X to Y, phi(x.y) = phi(x).phi(y)

[delicate orchid]

the perfect graph on 12 nodes.... that's like 100 million edges dawg...

how u gonna teach all them edges...

[delicate orchid]

all definitions I've ever seen require identity being maped to identity

[open sluice]

we've got 4 years, gotta make it count

[delicate orchid]

this is not guarenteed by phi(xy) = phi(x)phi(y)

like it is with groups

[crystal turtle]

That's always the definition of monoid, just some people include the requirement that f(1)=1, some might not

[delicate orchid]

you have to specify it seperately

[ashen heron]

oh

[wet zodiac]

wait

[crystal turtle]

(I haven't read much about monoids, so I cannot speak to actual conventions)

[wet zodiac]

ring homomorphisms dont gaurantee that phi(1)=1 if it isnt specified?

damn

[delicate orchid]

yur

[ashen heron]

my book doesn't and i am yet to see any examples of homomorphisms to come up with a suitable counterexample

ring homomorphs don't guarantee

0 mapping

[wet zodiac]

WHY....

WHY MUST YOU BE MEAN TO ME MATH...

[delicate orchid]

because not all elements are invertible

prove it for groups and find the exact moment you need inverses lol

[crystal turtle]

Yeah for rings I've seen some books not require preserving 1

[delicate orchid]

do those same books also deal with non-unital rings perchance

[crystal turtle]

Yeah lmao

[ashen heron]

the function R to R', a mapsto 0_R' is a ring homomorphism

[delicate orchid]

non-unital monoids

[ashen heron]

although my book also defines rings to not have unity by default

but that doesn't matter anyway

[wet zodiac]

ah i see

phi(1) may not necessairly be invertible

[ashen heron]

you need the property cc = c implies c = e

[wet zodiac]

yeah

[delicate orchid]

yeah ring homomorphisms always map idempotents to idempotents

but we want 1 -> 1!!!!!!!!!!!!!!!

[ashen heron]

is learning some sage a good idea for someone early in algebra?

[delicate orchid]

none of those stinky 1/|G| \sum_{g \in G} g\chi(g) bastrards

[delicate orchid]

it can be good for generating examples and checking answers to your questions I suppose

[crystal turtle]

Semigroup time

[wet zodiac]

oh no

they're taking away more properties i have been taking for granted

[crystal turtle]

Magma time

[delicate orchid]

it's ok nobody works with semigroups except the people in computability theory (weirdos)

I work with normal objects like groupoids but instead of isomorphisms everything is a monomorphism instead

[glossy crag]

Oh wow, the channel finally got split

[crystal turtle]

Left cancellable semi-groupoids?

[delicate orchid]

sure why not

[lethal dune]

grf

[solemn dew]

hint?

[crystal turtle]

Work through the definitions

Like just check the axioms, not hard but might be a bit tedious

[ashen heron]

associativity is the only hard one isn't it? the hint is to change the indices in the double sum, notice sum with columns first then rows is the same as rows first then columns

[crystal turtle]

Associativity is literally always the hard one

[south patrol]

Or just say "R[x] is by definition a ring"

[crystal turtle]

"it's in the name"

[rustic crown]

(or for any monoid M define R[M] and use it for M = free monoid on x :p)

the work is the same, might save you some work in the future

[crystal turtle]

Wonder if there's a way to apply Eckmann Hilton for this to get associativity for free

[elder wave]

In this economy?

[crystal turtle]

Now that I think about it, that would be much worse than just checking associativity LMAO

[wet zodiac]

did they actually kill the stuff from the rings...

[delicate orchid]

the natural killomorphism

[coral shale]

who dared to do this

no no no

whattttttttttt

[delicate orchid]

what now

[coral shale]

#groups-rings-fields
#advanced-algebra

we can't just do this, all the lore...

[hot abyss]

dang how dare you call me not advanced smh (i’m not)

[long nebula]

I wanna KILL, KILLLL

[teal vessel]

proving that upper triangular matrices of GLn(F) constitute a subgroup is making me reach all the way back to the index notation form of matrix multiplication and it makes brain hurty

[solar vessel]

there's a nice way of thinking about it

[rocky cloak]

Maybe characterize upper triangular matrices as those with certain invariant subspaces instead

[teal vessel]

there is, but I'm trying to be formal. I know what I'm trying to do but you can't exactly draw an "arbitrarily sized triangular matrix" on a whiteboard the way I want

[solar vessel]

ye

think like this

[teal vessel]

ngl it's been too long for that statement to have too much meaning for me. I'm probably dimly aware of it but that string of words is not behaving the way I want.

[coral shale]

formal:

jk, but i dont feel like the index notation should be that bad

[teal vessel]

it's not, I just had to remember whether it's row-column or column-row for the indices lol

[coral shale]

i remember it by

[teal vessel]

technically either way I'm basically doing a mirror image of each other, so at worst I'm just doing a lower triangular matrix instead

[coral shale]

11 12 13 14...
21
31
41
.

like 'count normally'

[teal vessel]

I thought so, but my brain said "mathematics is intuitive unless an arbitrary decision is involved"

[coral shale]

well i find worthwhile to remind myself which way every so often

[rocky cloak]

So let V_i be the space spanned by the first i basis vectors. Then a map f is upper triangular if f(V_i) < V_i

It's not very hard to show that the set of invertible maps for which f(V)=V forms a subgroup

So invertible upper triangular matrices is just an intersection of these subgroups

[teal vessel]

I'm actually having fun with this index notation tbh. I've managed to show that for $A=XY, a_{ij}=\sum_{\alpha = 1}^n x_{i\alpha}y_{\alpha j}$ and that a member of that sum is significant if, and only if, $i \leq \alpha \leq j$

[cloud walrus]

GoldenPhoenix

[teal vessel]

this immediately implies that the lower half of the matrix must all be 0, because i>j

in other words, the product of (upper) triangular matrices is always triangular.

[teal vessel]

I frankly don't remember the proof, but I know that the inverse of a triangular matrix is also triangular (of the same type), therefore it has both closure and inverses, and the identity matrix only has nonzero elements in the case of i=j, therefore i ≮ j and so it has identity

thus the upper triangular matrix set in GLn(F) (as well as the lower triangular set) constitutes a subgroup.

me: proves subgroup
textbook: now do it again but for triangular matrices with a unit diagonal.

oh, not too bad, just tweak the argument a bit. We know that for any entry on the diagonal, it has a value of 1, and for the product of two matrices, the row of the left matrix will have N-i possible non-zero elements to the right of the initial 1, and the column to be multiplied with will have N-i necessary zero elements below the 1 at the diagonal. By doing that swively thing that you do for matrix multiplication, it happens that all non-zero elements after the 1 in the left matrix will line up with zeros in the right matrix. Because of the definition of triangular matrices, the roles will be reversed on the other side, thus the only non-zero summand would be 1x1=1, therefore any diagonal entry of the product of two of these matrices will be 1. Thus we have closure.

now I just need inverses, but something tells me that shouldn't be too hard.

(at least if I remember the proof for triangular inverses)

[delicate orchid]

inverses are very easy if you remember what the eigenvalues of an upper triangluar matrix are (and already know that the upper triangular matrices form a group, which you do)

[night onyx]

I don't know if this picture helps but it's a way of visualizing how to get the (i,j) entry of AB for matrices A and B, and it makes seeing why UT * UT = UT for upper triangular matrices a bit clearer

[teal vessel]

I mean, for any arbitrary triangular matrix (with nonzero determinant, else it has no inverse), the inverse can be found by using GJ- elim with an appended identity matrix, using back-substitution, which then will only change the upper triangle of zeros (leaving the diagonal untouched except by scaling by the diagonal entry). Thus the inverse of the triangular matrix must populate only the upper triangle, and the diagonal entries should be entry-wise multiplicative inverses. Therefore if the original has 1 in every diagonal entry already, 1 inverse is 1, so the resulting matrix will also have 1s.

[delicate orchid]

the eigenvalues of a upper triangle matrix are the diagonal entries, so our matrix has eigenvalues all equal to 1, and therefore so does its inverse which we know is an upper triangular matrix, and therefore has 1s along it's diagonal

[teal vessel]

like I have often said: I'm stupid, just fast.

[delicate orchid]

me too

[coral shale]

checkout einstein summation notation

or whatever its called

(omit \sum, 2 indices imply a summation and some extra details)

[static yew]

Insanity coming in from the old abstract algebra channel

Now we have a philosophical consideration

So F_5[sqrt6] is the ring F_5[x]/(x^2-6) -> F_5[x]/(x^2-1)

Which is an F_5 adjoined with a primitive root of unity

[rocky cloak]

So usually F[a] = F[x]/f(x), where f is the minimal polynomial of a

[delicate orchid]

the primitive root of unity you're adjoining is just 1

[static yew]

Is it?

[rocky cloak]

x^2 - 6 is not the minimal polynomial of sqrt(6), because that's just x - 1

[delicate orchid]

6 = 1

[coral spindle]

The isomorphism you claim does not hold, in part because x^2-1 isn't irreducible

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

F_5[x]/(x^2 - 1), therefore, is not a field.

[static yew]

I know it's not a field

[rocky cloak]

If you want to adjoin another square root of 1, that's fine. But the resulting ring won't be a field

It will be F5xF5

[coral spindle]

You claim it's isomorphic to F_5[sqrt 6] yet you say you know it's not a field, curious <insert image here>

[rocky cloak]

(by Chinese remainder theorem)

[static yew]

I never claimed F_5[sqrt6] was a field

Q[sqrt2] isn't a field

[coral spindle]

Q[sqrt 2] is a field

Q[sqrt 2] = Q(sqrt 2)

[static yew]

What's the multiplicative inverse of sqrt2?

[rocky cloak]

Sqrt(2)/2

[coral spindle]

^

[static yew]

Hrmmm then I clearly didn't grasp the subtleties between [] and ()

[coral spindle]

There is no rule that says Q[ alpha ] is never a field

[rustic crown]

Boytjie

[coral spindle]

Hey det

[rocky cloak]

F[a] is the smallest ring containing F and a. Now this could or could not be a field

If a is algebraic it will be a field

[coral spindle]

Iff, in fact :)