groups-rings-fields

[barren sierra]

vs (G, *)

[delicate orchid]

not what I'm getting at but very true

[barren sierra]

oh

[coral shale]

I'm not sure the 2 are isomorphic

(are they?)

[delicate orchid]

it's additive because you literally just add the rationals in the group lol

[chilly ocean]

you talked about multiplicative groups

[barren sierra]

for arbitrary groups, the notation is whatever

[delicate orchid]

(no)

[barren sierra]

but like for common groups like the rationals

[chilly ocean]

not about the multiplicative group of the rationals

[barren sierra]

we all know + and * mean different things

[coral shale]

(is this trivial (lul))

[chilly ocean]

implying multiplicative and additive is a group property

[barren sierra]

and also in a ring ofc they mean different things

[delicate orchid]

forgive me for dropping a word or two I'm using shift+tab as a space bar

[barren sierra]

oof

[coral shale]

Purely in the context of groups, it makes not much sense to talk about additive/multiplicative groups

I think.

[chilly ocean]

i forgive you

[coral shale]

It is only when you relate to rings that it means something

[delicate orchid]

anyway chat what were we talking about

[coral shale]

.

convince me the additive and multiplicative groups aren't isomorphic lul

[chilly ocean]

of rationals?

[coral shale]

i guess

[delicate orchid]

(-1) has order 2 multiplicatively

[chilly ocean]

actually

[coral shale]

or hopefully something more general 👀

[delicate orchid]

it is infinite order additively

[chilly ocean]

i think its true in general

[coral shale]

niceeeeeee

easy trivial

[delicate orchid]

easy money

[chilly ocean]

in a ring R additive and multiplicative groups are not iso

iirc

[barren sierra]

in a ring, multiplication may not be a group even

[chilly ocean]

spamakin

[barren sierra]

see: the integers

[coral shale]

field, field.

[barren sierra]

oh

[chilly ocean]

ring of units

[barren sierra]

Ah

[chilly ocean]

group*

[coral shale]

or that

[delicate orchid]

yeah it's even nicer for finite fields you can just use a cardinality argument

$|F^{\times}| = |F|-1$

[cloud walrus]

Wew Lads Tbh (200 🍇) ✓

[delicate orchid]

so there cannot be a bijection

[coral shale]

amazing.

[chilly ocean]

do for infinite rings now

[coral shale]

Oh crap i thought u were meming lmao

u said finite, nvm

[barren sierra]

infinity - 1 = infinity QED

jk jk

[chilly ocean]

indeed infinity-1=infinity so that proves nothing

[coral shale]

Can we use the -1 argument 🤔

[delicate orchid]

not for infinite fields

[coral shale]

what about infinite field of characteristic 2

[delicate orchid]

you do however always have that (-1)(-1) = 1

[coral shale]

COULD they be isomorphic?

[delicate orchid]

ohhh NOW we're talking

I'm gonna need to think

[coral shale]

we need all the elements to be self inverse multiplicatively

[delicate orchid]

uhh

[coral shale]

and then we're surely done.

if this works.

or maybe not.

[delicate orchid]

is the set of all sequences in F_2 a field

nah nvm that approach won't work

[coral shale]

https://proofwiki.org/wiki/Additive_Group_and_Multiplicative_Group_of_Field_are_not_Isomorphic

D:

[delicate orchid]

yoooo

[barren sierra]

proof by proofwiki

my favorite

[coral shale]

chara 2 was relevant

knew it totally

[delicate orchid]

I see the word monomorphism
🚪 🚶‍♂️

[delicate orchid]

yeah it's the only case it's not completely trivial

very good idea to think about char 2

[coral shale]

yh now i formally see in my head why thats obvious lul

a + a = 0 needed, duh

for f(-1)

[barren sierra]

similarly epimorphism

[delicate orchid]

they basically just reduced the case of any char 2 field to F_2 and then used my cardinality argument

[chilly ocean]

whenever you see the monomorphism, the epimorphism sees you

[delicate orchid]

epimorphisms https://cdn.discordapp.com/attachments/359052581022203914/940656085893799936/unknown.png

[pastel cliff]

gets funnier every time

[chilly ocean]

What are some cool facts about epimorphisms of epimorphisms?

[coral shale]

they aren't even closed in any way as a structure are they

not in a way I can think of for now

if theres no structure where are the morphisms 👀

[barren sierra]

trippy fact of the day, nothing in the definition of ideals says that ideals are subrings. The fact that ideals are subrings is just a consequence of the definitions >_>

[coral shale]

im confused, what.

there are 2 instances of ideals that are a subring, no?

non-proper non-trivial ideal isn't a subring

[barren sierra]

An ideal I of a ring R is a subgroup of R such that for all x in I and a in R, ax is in I

that;s the definition of an ideal

but you can prove that it's a subring easily from that

[coral shale]

Oh wait you're talking about rngs

🤦

imagine.

[delicate orchid]

rings don't have 1 moment

[barren sierra]

imagine thinking rings have to have 1

[coral shale]

We only study r1ngs

[barren sierra]

and excluding the ring of even integers and such

[coral shale]

the fcks a rng

[coral shale]

u got me, ig u win

[delicate orchid]

the ring of even integers

it's an ideal

grrrr

[coral shale]

rng

[barren sierra]

it's a ring

[chilly radish]

It's also a ring

[coral shale]

grrrrrrrrng

[delicate orchid]

different definitions to me grrr

[barren sierra]

1 is overrated

who needs it

[coral shale]

r1ng

[chilly radish]

It's only an ideal if you consider it as a subset of Z

But it lives on its own

[barren sierra]

name one thing that 1 is useful for

you can't

[coral shale]

If I ever write a set of notes, ima use r1ng everywhere in it.

[barren sierra]

lol

[chilly radish]

I'm sure that won't get annoying

[barren sierra]

[coral shale]

u write it the same

lel

handwritingly

[barren sierra]

are you dotting your 1's

[coral shale]

If I did this everywhere no one will probably notice lul

more better

[delicate orchid]

no 0

[coral shale]

All ideals are rongs

So Frac(R) is the ring of fractions...

What is Frac(Z/4Z)?

I'm missing how Frac( ) is defined 🤔

[sturdy marsh]

frac is only defined for domains

Z/4Z is not a domain

[chilly radish]

Frac is usually the field of fractions

U can define a ring of fractions

Which generalises.the notion

But it won't be a field

[coral shale]

Yh i'm confused why it isn't.

How is it defined? I can't seem to find it

[chilly radish]

Zero divisors can't be inverted

You localise at the set of nonzero divisors

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

[coral shale]

ty

[chilly radish]

Np

The problem with trying to invert zero divisors is that, well, you can end up dividing by zero

Which is bad

Even in the field.of fractions 1/0 is never defined

So the best you can do is invert all non-zero divisors

[coral shale]

I either didn't pay attention or never came across this when I did rings

which is a mystery

[chilly radish]

This works because this is a multiplicatively closed set

You usually see this in commalg first I think

Idt you'd see this much in an intro rings class

[coral shale]

oh ok 😂 ty

[delicate orchid]

you also have to augment the equivalence class if it's not an integral domain I think

a/b ~ c/d <=> (ad-bc)u = 0 for some unit u iirc

it has been over a year though so I will double check that

I was mistaken

u is just in the multiplicative set

[chilly radish]

Yea this is for a general localisation

[delicate orchid]

should review these notes tbh

very good lecture series

[chilly radish]

Note that if S has a pair of zero divisors, then this automatically implies that the localisation is just the zero ring, since this implies 0\in S, and so all.elements are equal. This is why we need the extra condition

If it has a zero divisor with the other zero divisor in R it's fine since they never interact

[delicate orchid]

lemme think about that one

ok yup got it

[coral shale]

My linear algebra knowledge kinda stops shortly after eigenvectors/values. I vaguely know what diagonalization is, am ??? on Jordan-Normal form.

I want to have a basic understanding of what a Tensor is - can someone give me a basic outline/point me to a good defn?

I vaguely know some related ideas (generalised version of a matrix? levi-civita, kronecker-delta are rank 3, rank 2 examples?), maybe, but really, I don't know.

[latent anvil]

in the context of linear algebra, a ($k$-)tensor can mean one of two things. A $k$-tensor on $V$ could mean a multilinear function $f : V^k \to k$, multilinear meaning that $f(v_1,\ldots, x + y, \ldots, v_k) = f(v_1,\ldots, x, \ldots, v_k) + f(v_1,\ldots, y, \ldots, v_k)$ and $f(v_1,\ldots, a x, \ldots, v_k) = a f(v_1,\ldots, x, \ldots, v_k)$ for any $x,y,v_1,\ldots,v_n \in V$ and scalar $a$

[cloud walrus]

Maiden-having Shamrock

[latent anvil]

so if you fix all but one arguments you get a linear function in the remaining argument

[kindred jay]

i’ve got i, ii, iii, how would i do iv? (the second iii)

[latent anvil]

There's a related concept called a tensor product, which takes in two vector spaces and gives you a new space, and one might call elements of that tensor product tensors

[coral shale]

so the determinant of an nxn matrix is a tensor

an n tensor

[latent anvil]

yep!

[kindred jay]

also, any hints on this would be appreciated:

[latent anvil]

f(v1,...,vn) = determinant of matrix with columns v1,...,vn is an n-tensor of k^n

(k is a field here, could just be R or C)

[kindred jay]

(these are practice problems for a midterm)

[coral shale]

dual space?

[latent anvil]

not quite

sorry, one sec

[coral shale]

and yes thats what i was looking for more or less. I see terms like tensor product, tensor field floating around sometimes and I don't have any idea

[coral shale]

will reply in a bit

[latent anvil]

the relationship between the tensor product of spaces and the multilinear maps I talked about above is that there's a bijection between linear map $V \otimes V \otimes \ldots \otimes V \to k$ and multilinear maps $V^n \to k$

[cloud walrus]

Maiden-having Shamrock

[latent anvil]

where we have $n$ copies of V in the tensor product

[cloud walrus]

Maiden-having Shamrock

[latent anvil]

so the dual space of the tensor product is the space of multilinear maps

also, we don't need to look at the same space every time. it's totally fine to talk about a multilinear map $V \times W \to k$, and similarly there's a tensor product $V \otimes W$

[cloud walrus]

Maiden-having Shamrock

[coral shale]

Thanks, this is kinda what I rlly needed

[latent anvil]

np!

the elements of $V \otimes W$ are of the form $v \otimes w$ for $v \in V$ and $w \in W$, and if $a_1,\ldots,a_n$ is a basis for $V$ and $b_1,\ldots,b_m$ is a basis for $W$ then $V \otimes W$ has basis ${a_i \otimes b_j : 1 \leq i \leq n, 1 \leq j \leq m}$

[cloud walrus]

Maiden-having Shamrock

[latent anvil]

so $V \oplus W$ is like a vector space where we take the disjoint union of the bases for $V$ and $W$, but $V \otimes W$ is a vector space where we take the cartesian product

[coral shale]

is that circled cross the tensor product both times

[cloud walrus]

Maiden-having Shamrock

[distant elk]

hi im just posting here to ask about if anyone has any interest in or info about the 8-dimensional algebra over the reals which is (roughly speaking) the "superset" of the complex numbers (i^2 = -1), split-complex (aka hyperbolic) numbers (j^2 = 1) and dual numbers (k^2 = 0), aka the "superset" of all three 2-dimensional real algebras (each of which i individually find interesting); i also ask this because these 8D numbers, despite having zero-divisors, have a product which is commutative (not just associative) (quaternions for example are anticommutative not commutative) and because i cant seem to find any reading material about this specific 8 dimensional algebra, i believe these may be called tessarines?

[delicate orchid]

why tf was I not taught this about tensors

[latent anvil]

yes, one time it means the tensor product of elements and one time it means the tensor product of spaces

[coral shale]

ok i will look this up lel

[latent anvil]

np!

[coral shale]

@kindred jay do u wanna open a help channel

and ping me

[latent anvil]

the tricky thing is that differential geometers (and physicists) use tensor to mean tensor field. A tensor field is a smooth function which gives you a tensor on the tangent space at each point

[latent anvil]

you could start a thread

[kindred jay]

how do i do that?

[coral shale]

different thing?

[latent anvil]

Help

[coral shale]

ty

[kindred jay]

ty yea

[latent anvil]

Scorpio Question

[latent anvil]

usually it comes up in a differential topology/geometry course

sometimes it's introduced earlier

[delicate orchid]

I mean I've been taught about the tensor product

but I've never seen it that plainly and clearly

[latent anvil]

Ah okay

Well thank you and you're welcome :)

It's tricky because two different fields (geometry/algebra) use the tensor product to mean related but different things

Like, there's a tensor product of modules and then there's tensor fields on manifolds and they're sort of the same thing but there's a lot of ambient complexity

[chilly ocean]

@latent anvil I suppose you know well about the wedge product as well. It seems related in some ways to the tensor product, though I can't get the intuition really, what are we modding out by exactly when we define $V \wedge V = (V \otimes V)/I$ where $I = v \otimes v \forall v$ ?

[cloud walrus]

Entelechy

[latent anvil]

Oh and the last connection between multilinear maps and tensor products is that $(V_1 \otimes \ldots \otimes V_n)^* \cong V_1^* \otimes \ldots \otimes V_n^*$, so you can take the tensor product of elements of dual spaces to get a multilinear map
(this isomorphism is only for finite dim spaces Vi)

[delicate orchid]

oh so that's what the wedge product is

[cloud walrus]

Maiden-having Shamrock

[latent anvil]

Yeah so I like to think of it differently than this

[delicate orchid]

I think I knew this one at least

[latent anvil]

Remember my definition of tensors as multilinear maps?

[chilly ocean]

mhm

[latent anvil]

In that context an alternating tensor is an multilinear map $\omega$ such that if you swap any two arguments, you swap the sign of the result

[cloud walrus]

Maiden-having Shamrock

[latent anvil]

Then this implies ω(v, v,...) = 0 because swapping v and v doesn't change the input, and so ω(v, v,...,) = -ω(v, v,...)

But in characteristic 2, x = -x doesn't imply x = 0 so we define it like you did

But really the intuition comes from multilinear maps where swapping two arguments swaps the sign

Does that make sense?

And the reason we care about these is for differential forms

[chilly ocean]

oh so really v tensor v = 0 cause that's alternating

[latent anvil]

Right

[chilly ocean]

why char 2 though ?

[latent anvil]

Well the point is that the two conditions are equivalent over any other characteristic

That v /\ w = - (w /\ v) and that v /\ v = 0

Yeah?

[chilly ocean]

yea

[latent anvil]

But the first definition doesn't tell you anything in char 0

Er

Sorry, the first definition says v ^ w = w ^ v

Because - 1 = 1

[chilly ocean]

oh right

makes sense lol

[latent anvil]

So if you mod out by the I generated by that relation you'd get the symmetric power

Not the exterior one

[chilly ocean]

and how do you get the exterior one ?

[latent anvil]

You use the second relation

v ^ v = 0

So in char ≠ 2 your thing I is generated by v (×) w + w (×) v, but in char 2 you need to use the thing generated by v (×) v

Does that make sense?

[chilly ocean]

ah !

[latent anvil]

!

[chilly ocean]

I understand now

so v x v' = v' x v

[latent anvil]

No this is for S(V)

The symmetric power

[chilly ocean]

well yea that gives you the symmetric power

[latent anvil]

right

[chilly ocean]

that's if you quotient by S(V)

[latent anvil]

If you flip the sign you get the alternating power

So v × v' = - v' × v

Sorry I'm worried I've been confusing, all I'm trying to say is that we want the wedge product to satisfy the relation v ^ w = - w ^ v but this is bad in char 2 and so we replace it with v ^ v = 0 (and in all other characteristics the two identities are equivalent)

[chilly ocean]

when you say char 2, you mean char(V) = 2 ?

[latent anvil]

No, I mean char k = 2 where V is a vector space/module over the field k

This implies v + v = 0 for all v in V

[chilly ocean]

mhm I see

well all of this is confusing by default but you make it so much clearer aha

[chilly radish]

don't elements of the form $v\otimes w$ just generate the space? I think in general an element is not gonna be expressible as an elementary tensor, but as a linear combination of elementary tensors

[cloud walrus]

ShiN

[latent anvil]

You're right lmao sorry

[chilly radish]

just a small point, no bother

[kind jungle]

Are there any other examples that are Integral domains that is not a field that is not Z

[chilly radish]

For any field F, the ring of polynomials F[X] is an integral.domain

That's one family of examples

F[X] is never a field, since X is not invertible

[kind jungle]

thank you

[barren sierra]

should (4) not say "Show that N(R/N(R)) = {N(R)}"?

[chilly ocean]

0 can mean the 0 ring :P

I guess it's annoying if it's unclear within a text which conventions the author is using though

[kind jungle]

is 2Z a field? and is 2Z and integral domain?

[chilly ocean]

it's not a field, you can check it easily

and it's an integral domain because it's commutative and it has no zero divisor

though it doesn't contain the multiplicative identity so be careful with your definition of integral domain

[kind jungle]

thank you

is there an example of a noncommutative ring with characteristic 4? or does that not exist? I was trying to use matrices but no luck

[tribal moss]

2x2 matrices with entries in Z/4Z?

[chilly ocean]

should work yea

[barren sierra]

I am stuck on showing closure under subtraction

so if x, y are in J(R)

then 1 + xr is in R^x and 1 + yr is in R^x for all r in R

I want to show that 1 + (x - y)r is in R^x for all r in R

1 + xr is in R^x but then that tells me nothing about -yr

and that's where I got stuck

I mean my only intuition is why 0 is in J(R)

otherwise it's not even clear why this should be an ideal

[tribal moss]

Why subtraction? Have you shown closure under addition already?

[barren sierra]

you only need to show subtraction

for subgroup

idk we can consider x + y if you want

still equally unintuitive for me

[tribal moss]

We know that 1+yr is a unit. Therefore we can construct q = r/(1+yr). Now 1+xq = (1+yr+xr)/(1+yr) is also a unit by assumption; multiply that by the known unit 1+yr and you still get a unit.

[barren sierra]

ah damn

that's clever

[kindred jay]

can i get a hint on this problem?

[next obsidian]

It’s true for arbitrary y, and when x is a square. Maybe you can show that everything is a square?

No this is dumb

Oh wait maybe it’s possible, idk

[chilly ocean]

problem has me pressed

yeah idk this one is either annoying or there is nice trick

xy=xy(yxyx)=x(yyx)yx=

ok

thats it

last two steps prove commutative

[hollow imp]

@chilly ocean here's how I'd do it:

• show that each element is its own inverse
• show that a group where each element is its own inverse must be commutative.

[kindred jay]

ok so like i got that x^2 = e for any x

[hollow imp]

Oh wait

[chilly ocean]

it isnt a group

[hollow imp]

Pinged wrong person

[kindred jay]

so like each element is necessarily its own inverse

[next obsidian]

How

This isn’t a group

[kindred jay]

the only thing is i cant figure out that its closed

[hollow imp]

Oh wait it ain't a group

[next obsidian]

There’s no identity

[kindred jay]

oh

if theres no identity

[chilly ocean]

continuing off this
=xxyx=yx

[kindred jay]

then what do you do lol

[chilly ocean]

ig step by step

[next obsidian]

How do any of these equalities hold lol

[chilly ocean]

start with xy

xy=xy(yx)^2

[next obsidian]

Oh

Hurb

[chilly ocean]

then you use associativity twice

[kindred jay]

how do we know xy = xy(yx)^2?

[next obsidian]

Apply the hypothesis

[kindred jay]

can we assume yx is in the group

"group"

[next obsidian]

You know ab^2 = a

[kindred jay]

bc this isnt acc a group

[next obsidian]

For any a and b

[chilly ocean]

yeah its closed associative operation

[next obsidian]

a = xy, b = yx

[chilly ocean]

so if y,x in G so its yx

there is a better way i think

that doesn’t require you to think

[kindred jay]

so if you let a = xy, b = yx

[next obsidian]

Yeah then xy(yx)^2 = xy

[kindred jay]

then xy(yx)^2 = xy

i see

yeah i was just like

not sure about closure

it didnt seem clear

[next obsidian]

Lmfao

[kindred jay]

nah this isnt for a hw

im studying for an exam

[chilly ocean]

nice

[next obsidian]

What do you mean closure?

There’s no closure you need to prove

[chilly ocean]

binary operations are closed

[kindred jay]

like, does x, y in G imply that xy in G

oh

yeah i see

by definition whoops

ok so

wait, how come x(yyx)yx = xxyx?

we know that y^2x = x, right?

ah sorry nvm

x^2y = y means y^2x = x

[chilly ocean]

ye

[chilly ocean]

these are both true

[kindred jay]

alternatively

wait, can you assume the existence of an identity element?

[chilly ocean]

nah

no more structure than associative and closed

[kindred jay]

gotcha

[chilly ocean]

an example of this structure thats natural idk

im not sure there is

or maybe but it is probably convoluted

[kindred jay]

yeah that's a weird problem

thank you for your help though

[chilly ocean]

any

[median pawn]

hey i need some help

[cloud walrus]

Hausdorff

[median pawn]

how would you do this

[hidden haven]

Take a chain of R-submodules of M

They are also F vector spaces

So is finite chain

[next obsidian]

I’m not sure that quite does it

I think you have to go via contradiction

[hidden haven]

I mean instead of finite chain

You say that has bounded length

bounded by dim M

[next obsidian]

Yeah

That would do it

[bronze junco]

Got a question about notation.. I'm on the group theory section of Artin's Algebra book (ch. 2), and he introduces a squiggly arrow when talking about compositions. I've never seen that arrow before. Is there a standard definition for the squiggly arrow, or is it just a notation Artin uses for showing a law of composition, and nothing more than that?

[next obsidian]

Can you send a photo?

[cold frigate]

It's artins funny notation for $\mapsto$

[cloud walrus]

Buncho Spheres

[median pawn]

my instructor gave a proof but i don't understand it, so let me send that

[next obsidian]

Lmfao

But yeah it’s what buncho spheres said

[bronze junco]

Oh. Wow. Glad I asked

Much appreciated!

Squiggly arrow is non-standard..

[cloud walrus]

Hausdorff

[chilly ocean]

Does anyone have any good books on Galois theory?

[small bison]

i think you can just pull back N_k via the quotient map M -> M/M_1

[cloud walrus]

Hausdorff

[small bison]

yeah

[next obsidian]

The later facts follow by the third isomorphism theorem

[median pawn]

agreed

[median pawn]

what all do i need to check to know that this construction actually works?

[next obsidian]

Nothing really

This all follows from the third isomorphism theorem

And the fourth isomorphism theorem or the correspondence theorem or the lattice isomorphism theorem blah blah blah it has a ton of names

But the one that relates submodules of M/N to submodules of M containing N

[chilly ocean]

how come i never heard of G-modules before?

a module is a ring action on an abelian group.. so seems natural to think about group actions on a abelian group too

i did see group actions on different stuffs

but never heard them called G-modules

i did hear of group rings tho

why is G-Mod an abelian category?

what category of modules over a ring is G-Mod subcategory of?

[chilly ocean]

satisfies the definitions

G-modules sorta useful depending what you looking at

duh

they come up in group cohomology

[chilly ocean]

u asked?

[cloud walrus]

Hausdorff

Hausdorff

[chilly ocean]

but freyd mitchell embedding

seems unintuitive that G-mod is some subcategory of R-mod for some ring R

is that what freyd mitchell embedding states?

because maybe you just look at endomorphism ring of G

and then it corresponds to the R

maybe

i feel like G-modules are cool

tell us more

I dont know much besides that they generalize group actions on groups

how is it a generalization tho

isnt a a G-module literally just a group action on an abelian group

yeah but i mean as a category

wym as a category

its a category and for any fixed group its a way to talk about how other groups interact with it

i feel like thats cool

er

[hidden haven]

G modules are ℤ[G] modules

[chilly ocean]

its just a functor category?

you can do that for group actions on any other category

[chilly ocean]

yeah

the definition group homology is just tensor by Z[G] iirc and look at homology

[hidden haven]

Yes

@median pawn

[chilly ocean]

how to learn homology and cohomology stuff without already knowing it?

you might be able to pick up from wikipedia lol?

[hidden haven]

[median pawn]

okay, cool! thanks

[cloud walrus]

Hausdorff

[median pawn]

what are all the simple modules

[hidden haven]

You quotient by a maximal ideal

[cloud walrus]

Hausdorff

[median pawn]

right?

and are there any others? @hidden haven

[hidden haven]

Maximal ideals in F[x] are the ideals generated by an irreducible polynomial

[cloud walrus]

Hausdorff

[median pawn]

on top of that we have x^2 + m for all m > 0, right?

wait this way we would also have x^4 + m for all m > 0

there is no way to write them all down

[hidden haven]

F

[median pawn]

am i missing something lol

[hidden haven]

No

Alg closed field good

Not alg closed field not so good

[chilly ocean]

wikipedia page for chain complexes only defines them for abelian categories. Isnt this done also over more general stuff like semi-abelian categories or homological categories?

[next obsidian]

Who cares

Tbh

[chilly ocean]

i care thats why im asking.

[chilly ocean]

two questions

are there kernels and cokernels

there are

then why care

everything should work out from that

huh?

why not care?

this is semi abelian

https://ncatlab.org/nlab/show/semi-abelian+category

[chilly ocean]

time i wont get back from my life

i cant think of any examples

Grp

and it doesnt say kernels and cokernels exist

[chilly ocean]

strong examples

oh yeah maybe not idk

in Grp exist tho

yeah but look at kernels and cokernels of non abelian groups

well snake lemma holds or at least some version of it

cokernels arent always contained

bruh this isnt good use of my time qq

noone is forcing you to be here

idk who actually cares or studies the topics on nlab like that

is just logician stuff?

[chilly ocean]

why would anyone care about anything at all?

good question

i dont know why people would care about abstruse topics on nlab all the time besides curiosity

and im starting to think curiosity is becoming a weaker reason as i get older

well i do have a good reason

let me in

im on a undergrad research program and this is the topic they want me to work on lol

any applications to cool math?

like whats the upshot

of research program

money

thats all the motivation i need baby

im interested too now

nah but frlt

nine million percent of topics on nlab idk why i should care

i try and read articles and link follow

and then i feel like i wasted time

other articles are rly good tho

what do you care about

[chilly ocean]

not sure for sure

i like topics in topology though

are you a pro in using homology to study topological spaces?

[chilly ocean]

hell no

i am noob

please someone talk about homology im begging you

this world doesnt have enough people talking about homology

[delicate orchid]

non-discrete spaces

[hidden haven]

Clerk talks about homology half the time

[hidden haven]

Yes

As long as you have a 0 object you can define a chain complex

[chilly ocean]

pointed categories

[coral shale]

Is it me or is 2.5.8 and 2.5.10 just wrong

If L was algebraically closed I wouldn't disagree

ohhhh wait nvm about 2.5.10 embeddings are defined in C

But 2.5.8?

The splitting field of z^5 - 2 is a finite extension, but surely what they wrote is untrue

[chilly ocean]

Its in a book so it must be true

[chilly ocean]

Is splitting field generated by single element?

[weak oriole]

Sure?

you can actually even show that K(a,b) = K(a+kb) for most choices of k \in K

[coral shale]

no, 5th root of 2 and a 5th root of unity

[coral shale]

huh?

Let a = 2^{1/5}, b = e^{i2pi/5}
K(a, b) = K(a+b)

?

[waxen hedge]

If the base field is perfect, the theorem tells you that you can find a single element generating the whole splitting field

[weak oriole]

That probably works, ya

[waxen hedge]

Here the base field is a finite extension of the rationnals so it works

[coral shale]

whats this theorem called

[waxen hedge]

Probably some explicit variation of the primitive element theorem

[coral shale]

oh nvm it is that what am i reading

in my notes

[weak oriole]

No idea
Just take minimal polynomials of a,b and then try to be clever

[coral shale]

z^5-2
1+z+z^2+z^3+z^4

what clever stuff can I do to make this into 1 min polynomial that contains a and b 🤔

[weak oriole]

You just need to show one of a,b lie in K(a+kb)

[coral shale]

Is my example too hard numerically?

z^3-2

1+z+z^2

hmmm

[weak oriole]

Its prolly a better idea to do it more abstractly
The actual construction computation will look bad

[coral shale]

I'm curious what the actual polynomial will end up looking like

[tribal moss]

You don't need to assume an identity element: The equation x²y=y=yx² for all y tells you exactly that x² is an identity element. This element can't depend on a choice of x because x² = x²y² = y². So we do have an identity element, and then we can say that every element has an inverse, so G is indeed a group even if it's not explicitly stated to be.

[broken stirrup]

hello friens

anyone took a elliptic curves?

[cloud walrus]

Hausdorff

[median pawn]

How should I prove this?

[cursive temple]

How can we deduce that $[\prod B/\mathfrak{B}_i^{e_i} : A/ \mathfrak{p}] = \sum [B/\mathfrak{B}_i^{e_i} : A/\mathfrak{p}]$

texit isnt working so what im basically asking is that how we can turn the dimension of a product into a sum of the dimensions of the factors

[coral shale]

If I give an algebraic complex number in cartesian form is there an algorithm to find its minimal polynomial?

Or say I tell you it is a specific linear combination of the root of this polynomial and that polynomial

[lethal dune]

in can get very complicated very fast

[rustic crown]

how will you give me the algebraic number?

[coral shale]

I just want to see a simple example 😅

say cube root of 2

and a cube root of unity

sum them

[lethal dune]

use sagemath

[rustic crown]

there are ways to find one polynomial with that as a root

[coral shale]

I was shown sometime ago the method has to do with tensor product?

[rustic crown]

but irred will be hard

[lethal dune]

I already tried this one

it's complicated nevertheless

[coral shale]

uhh why im doing this.......

I just want an explicit example for

[lethal dune]

@rustic crown is there?

[coral shale]

K(a, b) = K(alpha)

where I can see this

and im curious what are the other roots of the minimal polynomial of alpha would look like

[rustic crown]

i don't know 🙈

[coral shale]

I understand the proof of the primitive element theorem, mostly

[lethal dune]

you can start by $x^3-2 = 0$ then substitute $x \mapsto x-\omega$ then try to eliminate $\omega$ by squaring and blah blah

[lethal dune]

I had my hopes high😔

[coral shale]

im going to have to lookup in chat history ig

i asked something related before

[lethal dune]

not sure if this is correct tho, I don't have energy to check

try doing something similar

[coral shale]

thanks i will look/try

I think I found it

What I was referring to

except I don't know that tensor stuff lol, so was hoping for an explicit construction with an example

the context - so I was initially trying to show if alpha and beta are algebraic, alpha+beta is also by constructing an explicit polynomial (and not realising that was a bad idea)

[lethal dune]

looks interesting

[coral shale]

Well how do I put this. I find this interesting because

K(a) isn't normal. Only adjoins 1 root
K(a, b) is normal
K(a+kb) is normal for 'most' k (finitely many aren't)

a being cube root of 2
b being 3rd root of unity

[coral shale]

Ah, the minimal polynomial of a+kb needs to be degree 6 in this case, didnt realise earlier

(no wonder its not easy to 'guess')

I get the feeling its roots form a basis

[chilly ocean]

In fact you have $\alpha , \beta$ algebraic over $K\subseteq L \iff \alpha + \beta$ and $ \alpha \beta$ are algebraic over $K$. It's an equivalence, can you not use vieta's formula ?

[cloud walrus]

Entelechy

[coral shale]

To show this? Don't think so?

Not that I know of

[chilly ocean]

well if S is the sum and P is the product, alpha and beta are roots of X^2 - SX + P no ?

but that's the other implication anyways

[coral shale]

a, b algebraic over K does not mean (x-a)(x-b) is a polynomial living in K[x]

it means there exists polynomials in K[x] which have a and b as roots?

[chilly ocean]

well it just says that if a, b are algebraic over K then their product and sum are algebraic over K too, and the other way around

[coral shale]

Yes, but how is this quadratic and vieta relevant?

(x - a)(x - b) = xx - (a+b)x + ab

[chilly ocean]

that's for the converse implication so that's irrelevant, sorry

[coral shale]

I dont think its relevant for either implication?

[chilly ocean]

if you have a+b and ab that are roots of a polynomial in K then X^2 - SX + P has a and b as roots

[coral shale]

If you have already proven the forward implication, then you can use what you said

is that what you mean

[chilly ocean]

the => is a lot longer

[coral shale]

im still not sure how this

a^2 + b^2 = (a+b)^2 - 2ab

[coral shale]

I agree

But x^2 - Sx + P doesn't live in K[x]

[chilly ocean]

why not ?

[coral shale]

S = a+b
P = ab

Did I read you right?

If yes, then you only know these 2 are algebraic

Not that they are members of K

[chilly ocean]

hmmm

then I don't really know about the converse

are you still trying to prove the => implication ?

[coral shale]

No no

I was referring to what someone else said before for this particular question (the question is done already)

The process of constructing polynomials with roots a+b and ab (v. non-trivial, which is why the proof should be approached another way)

[chilly ocean]

yea I assumed a and b were in the field but they're in the extension

[pastel cliff]

someone post first iso im on mobile

[next obsidian]

You can do it on mobile

Smh

[lethal dune]

,tex \begin{tikzcd}
G \ar[r, "\varphi"]\ar[d, "\pi"{left}] & H \
G/\ker(\varphi) \ar[ur, dashed, "\tilde{\varphi}"{below}]
\end{tikzcd}

[lethal dune]

I always keep this in my clipboard

[pastel cliff]

yes

we "learned" kernels today in lecture

stupid ignorant guess from the picture but is this saying that there's a bijective map from G/ker(phi) to H

[coral shale]

homomorphism f : G -> H
G / ker f is isomorphic to im f

[pastel cliff]

:O

[cloud walrus]

ShiN

[chilly radish]

wait

[cloud walrus]

ShiN

[chilly radish]

that still doesn't explain it, since I could just omit 2 variables and do the same type of thing, just putting in I_n twice

so what's wrong here??

[chilly radish]

I'm trying to find a polynomial identity of degree <2n

Yea basically. Shouldn't this be the same as assigning the identity to the last variable?

Hmm ok wait yea, it's not gonna be exactly the same polynomial

Right

Yes I understand now that's my mistake

Yea if you remove one of the variables but keep the signs the same I bet it'll vanish

Or

Just not be an identity

Either way it's not gonna be a counterexample

Ok, time to try and prove there aren't any of lower degree then

Can I have a hint

I'm thinking Smth with the matrix standard basis

Alright i'll try playing around w/ the 2x2 case to build up intuition

Then do the general case

Thanks!

Oh and if u just remove the last variable in retrospect it obviously vanishes identically cuz all the permutations where x_2n would be moved around cancel with themselves

[chilly radish]

I think I have the sketch but I'm a bit lazy to write it out, but the idea is that when k<2n and the polynomial is multilinear, then if it's nonzero, you can write it as a sum over Sk. Choose some permutation where the sum doesn't vanish, I think you can show (Maybe by induction or directly) that there is a combination of basis vectors (That is 1 in one place and 0 in the rest) such that they will give a nonzero basis vector for the permutation AND that is the only occurance of that basis vector with that assignment in the polynomial.
When you get to 2n this breaks because you don't have enough indices. I think that's the big idea, I might write down a formal proof later, seems a bit tedious

I'll try and write it down better later. Btw a very basic result in multilinearisation says that this implies there aren't ANY polynomial identities of degree less than 2n that aren't 0

Since a polynomial identity implies a multilinear identity of lesser or equal degree

No problem, have fun!

[kindred jay]

any ideas on this? i thought that it followed from the fundamental theorem of cyclic groups, but my professor said that was "almost correct" but not exactly accurate - i've spent a while and am not really sure as to where to go. i know that if G = <a> and |a| = n, then all solutions will have the form a^(kn/d) for k between 0 and d-1, but i don't really know how to go about formalizing that argument.

[chilly radish]

I think it's as easy as taking, when 2k <2n (Assuming n>2 here cuz I did that case already:
E11 E12 E22 E23 ... E_(k-1)k E_k1

And when 2k+1<2n
E11 E12 E22 E23 ... E(k-1)(k-1) E(k-1)k Ek1

Up.to taking suitable permutation
And when u get to 2n this fails because you have 2 ways to get Ekk (it's not this exact setup cuz you run out of indices) whereas here you only have 1 way (namely Ek1 moves to the start)

That wasn't so hard (Given this is correct)

[coral shale]

Can't you write those d solutions explicitly and show anything else cannot be a solution?
ie. consider powers mod (n/d). Only powers === 0 mod (n/d) would solve the equation

[kindred jay]

so like, any power that is a multiple of n/d

[coral shale]

Explicitly in terms of a pre-chosen generator a that is (G is cyclic, so you can do this)

Well this might not work out, but is how I would approach at least.

[next obsidian]

You can classify the order of an arbitrary element g^k

In terms of the order of g, and k

It turns into a statement about some divisibility stuff

I think that this should probably completely finish it

[kindred jay]

ok so

if g^k = e

then |g| divides k

lemme try and play around with that a bit

nvm im stuck

[next obsidian]

If |g| and k are coprime, what’s the order of g^k?

[kindred jay]

|g^k| = |g|

bc |g^k| = |g|/gcd(k, |g|) = |g|

[next obsidian]

Oh yo

You have your formula

[kindred jay]

yeah

[next obsidian]

Okay so now like

What you wrote down about the kn/d

[kindred jay]

k is a divisor of |g| though

[next obsidian]

Not necessarily

That formula |g^k| = |g|/gcd(k,|g|) always holds

[kindred jay]

oh yeah

im talking about for the sake of my problem

[next obsidian]

Oh sure

Actually not necessarily

[kindred jay]

oh?

[next obsidian]

This is true

Like your list of elements

[kindred jay]

yea, itd be smth like <g^n/d>

[next obsidian]

So like each of these will go to the identity after being raised to the d-th power

[kindred jay]

right yeah

[next obsidian]

So now you need to show that if you looked at

g^m for m not one of those

That like

|g^m| doesn’t divide d

Because if (g^m)^d = e

Then |g^m| divides d

Or actually

It might be easier to show that if (g^m)^d = e that it is of that form maybe

Cuz you get |g^m| divides d

[kindred jay]

yeah that makes sense

[next obsidian]

And then you know n/gcd(m,n) divides d

[kindred jay]

this seems like restating the fundamental theorem of cyclic groups in a way tho

[next obsidian]

And I think from this you probably can write m = kn/d

Uhh

[kindred jay]

bc we're essentially saying <a^n/d> has order d

[next obsidian]

What’s your fundamental theorem?

[kindred jay]

ok so given a cyclic group G = <a>

with |a| = n

then for each divisor n

the only subgroup of order d is <a^n/d>

[next obsidian]

It’s the generator

[kindred jay]

my professor said it

[coral shale]

misread the thm

[next obsidian]

Oh wtf

[kindred jay]

isnt exactly the same tho

[next obsidian]

I think it does follow from that

[kindred jay]

like

ofc any element in <a^(n/d)> will satisfy x^d = e

[next obsidian]

Right

[kindred jay]

(a^(kn/d))^d = (a^n)^d = e

and the order of that subgroup is d, so it has d elements

maybe i have to show that there are no other solutions?

[next obsidian]

Yes that’s the other part

But like

I think you can induct now probably

Like you can’t have another element of order d

Because if h was some element of order d not in that subgroup

Look at <h>

Woah contradiction

So such a thing would be of order < d

Say it’s order k (note that k divides d necessarily)

But inside of <a^n/d> you can just produce k things of order dividing k already

So if there was a thing of order k outside <a^n/d>

You have more than k things of order dividing k, and then you contradict the inductive hypothesis

Does that sketch make sense?

[kindred jay]

does the element necessarily have order d, or does it order divide d?

[next obsidian]

Why 6?

[kindred jay]

sorry i meant d 😅

[next obsidian]

Well

There is no element

But you do cases

If it were order d, then it has order d

Else it’s an h such that h^d = e

But |h| < d

From the first condition you have that |h| divides d, that’s the number I called k

[kindred jay]

mm

[next obsidian]

Does that make sense?

[kindred jay]

so, this means within that subgroup, you can get k elements whose order divides k?

[next obsidian]

Yeah

But you already did this!

This is a cyclic group of order n/d

And k divides n/d

[kindred jay]

is the cyclic's group order k/d? or is it d

[next obsidian]

And you showed earlier that in a cyclic group of order m, if k divides m, there’s at least k things of order dividing that

Oh whoops

d

Yeah you’re right haha

[kindred jay]

npnp

so k divides d here?

[next obsidian]

Right

Since k = |h| and h^d = e

First by definition of k

Second by assumption

[kindred jay]

there are at least k elements whose order divides m?

[next obsidian]

It’s this

Replace n with d

And the d there with k

We’re applying this to <a^n/d>

Which is the cyclic group of order d

[kindred jay]

right

wait so

we are showing that this has d elements ?

[next obsidian]

It has k elements of order dividing k is what we show

But then h is another element of order dividing k not in that subgroup

But this is a contradiction by induction

We already know we only have k things of order dividing k because k < d

[kindred jay]

wait we need to use induction here?

[next obsidian]

Yes

Or uh

Wait no we don’t need to do that I’m dumb

[kindred jay]

np

[next obsidian]

Okay easier idea lol

Suppose h is not in <a^n/d> but has order dividing d

Then let k = |h|, it divides d

[kindred jay]

right yea

[next obsidian]

Okay, now the element (a^n/d)^(d/k) = a^n/k generates the only subgroup of order k

The reason I wrote it that funny way is

this proves a^n/k is in <a^n/d>

[kindred jay]

bc d/k is an integer?

[next obsidian]

Yup

But then we know <a^n/k> is a subset of <a^n/d>, so that h is also not in <a^n/k>

But now <h> is a different subgroup of order k

Contradiction

[kindred jay]

<h> is a different subgroup of order k?

[next obsidian]

Right

Because h has order k

And h isn’t in <a^n/k>

So there’s no way it generates the same subgroup

[kindred jay]

and how is that a contradiction?

[next obsidian]

It contradicts the fundamental theorem of cyclic groups

[kindred jay]

yea ok that makes sense

i just emailed my professor to ask why we couldn't use the fundamental theorem of cyclic groups

[next obsidian]

I think their point is just that one direction is easy, showing that there’s >= d elements

But showing <= like we did here is harder

It isn’t immediate just from the statement of the theorem

[kindred jay]

ty for your explanation! also, if you have time, any help on this? i know that phi(x) = ax is definitely a automorphism, but how do you prove it's the only one?

(btw this is a practice midterm exam)

[coral shale]

You usually proceed constructively for these kinda things

In general, to construct a homomorphism, it is smart to look at where generators must map to

[next obsidian]

Yeah

[coral shale]

If you can't find a set of generators, a similar idea can still be used.

[kindred jay]

generators have to map to generators

[next obsidian]

The idea is like

If x maps to a

2x maps to 2a

Because it’s additive

[kindred jay]

obviously ax works, because a(x+x) = ax + ax = a(x) + a(x)

but how would you show it's the only possibility

[next obsidian]

Well

That’s what I just said

[kindred jay]

mhm

[next obsidian]

You should think about it more

[coral shale]

Meanwhile I am suspicious the answer for (R, +) is non-trivial

[next obsidian]

It’s really easy to show that at least on Z, that you have phi(x) = ax for some rational number a

[kindred jay]

we know that phi(x+y) must be equal to phi(x) + phi(y)

[next obsidian]

I think you literally use analysis lol

And that Q is dense in R

[kindred jay]

showing it's onto and one-to-one is easy enough

i think i'm missing the part where you show it's necessarily a linear map

unless you don't have to do that

[next obsidian]

That’s by assumption my man

It’s a group homomorphism

Oh

[kindred jay]

wait lol so like

[next obsidian]

You mean that it’s Q-linear

[kindred jay]

you can just assume it's a linear map?

[next obsidian]

Yeah it’s clearly Z-linear

[kindred jay]

any group homomorphism is a linear map?

[next obsidian]

But you have to be a bit more clever to get it for other numbers

No, I thought by linear you meant Z-linear

Which is just additive

[kindred jay]

like when my professor went through it, he just showed a few examples

but didnt prove it

[next obsidian]

You should think about it more

I mean

What is a gonna be?

Like

For what number x is phi(x) = a

Assuming that this is bijective there’s a unique answer

[white jackal]

Can you guys help me with this?

[next obsidian]

Just use the first isomorphism theorem

[kindred jay]

yea i'll think about it a bit more

[next obsidian]

Anyway

My advice is just to figure out

1: what that x is

2: why phi(x/2) = a/2

Once you do that, the rest of it will fall out into place quite easily

[coral shale]

does this not follow from one of the isomorphism thms?

[white jackal]

is it the correspondence theorem?

[next obsidian]

.

[coral shale]

the correspondence thm is equivalent to the 3rd iso I think

But chm's hint is likely better

[next obsidian]

The correspondence like…

It goes hand in hand with it

That like gives the relations of the subgroups and you kinda need to know it in order to even state the third isomorphism theorem

Like to say (G/N)/(H/N) you need to know like what H/N is and that things appear in that form

[coral shale]

you kinda I read this as something else

[next obsidian]

[white jackal]

so i just use the first iso for this

[next obsidian]

Yeah

[white jackal]

okkk tyty

[next obsidian]

Protip

[white jackal]

what is it

[next obsidian]

If you’re ever asked to show that G/N is isomorphic to anything

You probably want to use the first isomorphism theorem

[white jackal]

ok

gotcha

[coral shale]

1/2 = 0.5 👀

[chilly ocean]

i love you Shuri

[lethal dune]

[median pawn]

hey could someone help me out here

[next obsidian]

That doesn’t really seem right to me

Those are in bijection with the maximal ideals that ring via the nullstellensatz, but the maximal ideals shouldn’t be simple

Oh simple modules

Uhhhh…

[chilly ocean]

Can someone help me solve this problem please?

I’m not sure how to even get started

[white jackal]

how do i get started with this proof?

[coral shale]

oi wtf

[coral shale]

.

Are u guys at the same uni/college lul

Readup par lul

Suffering alone is pain, but there is much to be enjoyed in suffering together

[white jackal]

i know i have to use the first iso theorem but idk how to start it

[chilly ocean]

LOL

[coral shale]

To show an isomorphism, you need to construct a map

and show it is isomorphic

👀

at least, this is the usual way.

[tender mist]

Huge coincidence lol

[chilly ocean]

Umm

[coral shale]

The map to choose... the 1st isomorphism theorem tells you which

On the other hand, I think you can show N x N~ is the kernel of a homomorphism G x G~ -> RHS ? Might be saying this wrongly

[white jackal]

hmmm ok

[chilly ocean]

Wait so I need to use first iso

[coral shale]

thats the plan

[chilly ocean]

And for that

I need G/ker(phi)

?

[coral shale]

You need to show N x N~ is the kernel

[chilly ocean]

Why

[coral shale]

for some smartly chosen homomorphism

uh

So we have this thing

And G/ker(phi) is isomorphic to im(phi)

So you can try to construct a surjective homomorphism phi from G to H instead

where ker(phi) is what you want it to be

[chilly ocean]

What would G and H in this case be

[coral shale]

There is only one relevant quotient to consider, really

on the left hand side

===
So you want
endomorphism f : G x G' -> (G/N) x (G'/N')
ker f = N x N'

Write out what the elements of (G/N) x (G'/N') look like.

[upper cape]

Does anyone have any general tips on constructing field extensions with a given Galois group?

[spice whale]

any groups

[coral shale]

this is generally unsolved I think

[dull root]

I have the following question. Let a be in C, and a- be its conjugate. In general what can I say about the degree [Q(a,a-):Q(a)]? If a is real, it is clearly 1, but it is not true for all c. My naive guess is that is degree is always either 1 or 2 (since if we look at the quadratic formula if a is a root of a quadratic, then a- is also another root), but does anyone know of a good way to see this or a counterexample

[coral shale]

define 'conjugate'

complex conjugate?

[dull root]

yes

[coral shale]

there's some theorem

which says what you wanted

why do you think its either 1 or 2

I don't see where you get that from your guess

[coral shale]

idk name so rip

But then again, I don't think you need to rely on it

I have the feeling you can show what you want explicitly but not totally sure atm

nah nvm, I think you need to state that theorem

Try googling 'roots come in conjugate pairs'