groups-rings-fields

[crystal vale]

is directed limit category object?

[frail shoal]

people often study the concept of finiteness abstractly using the concept of filtered colimits

[knotty badger]

The directed limit of a diagram is an object of your category together with some maps

From the diagram to your object

[frail shoal]

to elaborate on this, in a lot of nice categories, we have the slogan "finite limits commute with directed colimits" is true. it's not true of all categories though

[crystal vale]

What is filtered colimits

[swift root]

what's the condition? they have to exist?

[frail shoal]

i think it's the same? before posting this message i referenced nlab, and it defined a filtered colimit as a colimit over a "filtered" poset,

and they defined a filtered poset the same way

[swift root]

direct limits are filtered colimits over a small thin category

[frail shoal]

ah, that is the distinction, oops

[frail shoal]

*i should have said filtered then, my bad

[swift root]

(co)cofiltered just means every finite diagram has a (co)cone

[frail shoal]

the point is, same concept as what is described in exercise 14

[frail shoal]

no clue

haven't studied it

ok so i can say for sure that, at least in any category of algebras, filtered colimits commute with finite limits

[swift root]

yeah, the underlying sets are the same

[frail shoal]

for example, suppose you have two directed systems M and N. you can compute M × N as the direct limit of all the M_i × N_i

it's a good exercise

note that this only works for finite products

and it only works for direct limits, if you look at general colimits it also fails

so it's very nice

[crystal vale]

So 16 is just asking to prove that the directed limit is just co limit

[coral shale]

codirected

[swift root]

no this is directed

[coral shale]

i was portmanteauing

[swift root]

a large travelling bag

[crystal vale]

So by the hypothesis every x_i -mu_ij(x_i) gives 0 therefore the submodule generated by those elements is {0} therefore the direct limit is just the direct sum of M_i

Right?

[frail shoal]

not quite, we have to consider each M_i as having distinct elements here
that being said, modding out by each x_i - mu_ij(x_i) does have the effect of identifying everything in M like how you would expect

[crystal vale]

Why? They are submodules of some bigger module

[frail shoal]

just for the purposes of the construction, we then identify them as having equal elements using the mu maps

[crystal vale]

Sorry i don't get where I am wrong?

[chilly radish]

It won't be a direct sum though, just a sum

[crystal vale]

Yes my bad

[chilly radish]

Ok I get what keith is saying now

In the disjoint union/direct sum you do distinguish between these elements, since they are in different submodules

But once you quotient out you recover the original identification

[frail shoal]

yeah that is what i mean to say

[crystal vale]

I see

Thank you

[frail shoal]

i might have misinterpreted what you were saying, sorry about that

when you said "gives 0" i thought you meant in C, not in M

and i assumed by direct sum you meant direct sum, as opposed to sum

[crystal vale]

So is it something wrong in my argument?

[frigid epoch]

the direct limit is a certain quotient

it's just not a massive sum of everything, you quotient a lot of it out

for example, for a point $x$ in a topological space and two open neighbourhoods $U,V \ni x$

[cloud walrus]

PKThoron

[frigid epoch]

two functions $f_U: U \to \CC$ and $f_V: V \to \CC$ become equal if $f_U|{U \cap V} = f_V|{U\cap V}$

[cloud walrus]

PKThoron

[frigid epoch]

(as far as the point x is concerned, those two functions are in the same equivalence class)

[sudden edge]

what about inverse limit do we cut in same way,

[frail shoal]

i meant to say, i might have misinterpreted your correct argument

im not sure

[sudden edge]

in inverse limit p adic integers uncountable still

[crystal vale]

I think you were correct, I need to look at m_i as a different element

[frail shoal]

in practice, i think direct limits tend to have smaller cardinalitiy than projective limits, but this is just vibes

[frail shoal]

yeah, they are different in C, but get identified in M = C/D

is the ultimate idea

[swift root]

inverse limits are subsets/groups/rings wtv of a product

[sudden edge]

yes cut the whole countable product in case of p adics

[swift root]

inverse limits glue the elements together, while direct limits patch the structures themselves together, in a sense

[swift root]

wdym by "cut"

[frail shoal]

btw i will rant rn, the terminology surrounding inverse/projective limits (limits) and direct/inductive limits (colimits) is really confusing

(to be clear inverse and projective are synonyms here, so are direct and inductive)

[sudden edge]

filter out nice ones

[swift root]

right

[frail shoal]

it's really hard to know at a glance which one someone is referring to 😭

[frigid epoch]

colimits are usually quotients of sums, limits are usually subobjects of products

like in topology, pullbacks are usually some fibered product

while pushouts are some glued sum

[frail shoal]

inverse limits are \forall, direct limits are \exists

incomprehensible posting, this is not for the sake of pedagogy, i just need to speak my piece

actually,

products are "for all", equalisers are "such that", coproducts are "there exists", coequalisers are "up to"

[swift root]

yes

[frigid epoch]

nice

[frail shoal]

from this, we build all of math...

/shitposting tone tag

[frigid epoch]

binary products are "and", binary coproducts are "or", initial objects are "false", terminal objects are "true"?

[frail shoal]

this sentence is a terminal object nuke

[frigid epoch]

and exponentials are "implies"?

[frail shoal]

heyting nuke

[frigid epoch]

https://tenor.com/view/yesss-fire-mistazelf-excited-restaurant-gif-22088135

[swift root]

okay so, what about zero object

falsity?

[frail shoal]

(to be clear im sorta vibeposting, i dont have a rigorous idea in mind behind what im saying)

the issue with making it rigorous is that, you can always take the dual category, and then it flips everything around

[frigid epoch]

it's probably in one of the geometry-logic correspondences

[frail shoal]

so you have to restrict to something nice and atp you lost the vibe generality

[frail shoal]

yea i could like appeal to topos theory or something but idk enough of that to be confident in it

[frigid epoch]

yeah like in categories with zero object, biproduct etc.

I know that "there exists" is something like elements in a fiber bundle

and "for all" is something like sections of that fiber bundle

in some HoTT-adjacent framework

[frail shoal]

i do know that there exists and for all can be viewed as adjoints in a really brainrot way

[frigid epoch]

yeah

[frail shoal]

but like, im not thinking about that when i think about most stuff

[frigid epoch]

"there exists a base space element such that x is in the fiber"
"for all base space elements (something something you get a section)"

[frail shoal]

there is a sense in which i could argue zero objects tend to be false

so like, any possible equation you could write down in the language of, say, R-modules, or groups, is satisfied by the identity element

so in a sense, identity has a trivial theory, which makes it function like a contradition

a lot of theorems are phrased like "if x satisfies blah then x = 0"

so in a way, 0 is the failstate

[tough raven]

A preadditive category is not a topos.

[swift root]

trivial category?

[tough raven]

Yeah except that

[frail shoal]

i would argue that the trivial category is false

which is ironic, cuz it's final

ok i guess final objects are falsey when you deal with equational theories

(before you complain, enpeace, my justification is that categories form a GAT)

[swift root]

a gyat?

oh generalized algebraic theory

[swift root]

martlet would NEVER

i think I said this exact thing before

[frigid epoch]

my absolutely unfounded take is that categories are a "bare analytic structure"

[knotty badger]

what does that mean

[frail shoal]

yea im curious too

[frigid epoch]

I have NO idea

just that they somehow permit integrals, have a sesquilinear form with values in a base of enrichment, and vaguely have some notions of flow and tangency at a point, limits and colimits, continuity and cocontinuity, completeness and cocompleteness...

and also the funny "groupoid cardinality of FinSet is e"

[swift root]

whar

sum of 1/|Aut(X)|?

[knotty badger]

yes

$\sum_{S \in \text{Sk}(\mathbf{FinSet})} \frac{1}{|\text{Aut}(S)|} = \sum_{n=0}^\infty \frac{1}{n!} = e$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[swift root]

groupoid cardinality is crazy lol wtf

[vapid vale]

what are these integrals

[swift root]

co ends

[frigid epoch]

there's an nlab article...

[swift root]

yes ive just read it

generalization of the class formula

[frigid epoch]

I think there's a generalization which is something like $\frac{|\pi_0|}{|\pi_1|}\frac{|\pi_2|}{|\pi_3|}\frac{|\pi_4|}{|\pi_5|}\cdots$ (or the inverse of this)

[crystal vale]

I don't get how they are identified in M?

[cloud walrus]

PKThoron

[swift root]

(take the sum of these over all x ∈ π_0)

[swift root]

by the relations x - μ_ij(x)

[crystal vale]

Yes so it shows D = {0}

So C/D, isn't same as C?

[frigid epoch]

if for $i \leq k \geq j$ and $m_i \in M_i$ and $m_j \in M_j$, you have $\mu_{ik}(m_i) = \mu_{jk}(m_j)$ they get identified

[frail shoal]

cuz we mod out by x_i - mu_{ij}(x_i) and all the mu's are subset inclusions

oh i didn't scroll down

[cloud walrus]

PKThoron

[frail shoal]

ah ok so yeah i do want to emphasize, indeed, that C consists of a direct sum of all the M_i, and as a result, we have that x_i ∈ M_i can be distinct from x_i ∈ M_j, which is confusing notationally, idk how to properly notate it

[velvet hull]

i thought my cool image would have explained this identification cleanly :(

[frail shoal]

yeah it is distinct

maybe someone else knows how to notate this better, idk

[crystal vale]

Okay they are different but how they are same in M?

[frail shoal]

it's like how when you take the disjoint union of two sets, two identical elements can become different

but it's also a notational nightmare there too

uhh

[frigid epoch]

example: $U \supseteq W \subseteq V$ and $f_U \in \mathcal{O}(U)$ and $f_V \in \mathcal{O}(V)$ with $f_U|_W = f_V|_W \in \mathcal{O}(W)$

[cloud walrus]

PKThoron

[frail shoal]

maybe we should notate the elements as ordered pairs

(i, x_i)

i encodes which structure we're in

and we require x_i ∈ M_i

and we take linear combinations of those, to define M

so we get that D is generated by (i, x) - (j, x) for all x ∈ M_i ⊆ M_j

[frigid epoch]

example of a directed set

[crystal vale]

I see

Thank you

[frigid epoch]

two elements earlier in the directed sets may "become the same"

like two functions being the same when restricted to a common open set

[crystal vale]

Oh

[crystal vale]

Any hint for the later one? Okay it says x_i in D but i don't get any further idea

[frigid epoch]

what's D

I guess it's the direct limit module

so $D = \bigoplus_{i \in I} M_i/\sim$ where $m_i \sim m_j$ if there is a $k \geq i,j$ with $\mu_{ik}(m_i) = \mu_{jk}(m_j)$ and $\mu_i: M_i \to D$ is the map sending $m_i$ to $[m_i]$

(more technically I should speak of something like $\iota_i(m_i)$ where $\iota_i$ is the inclusion of $M_i$ into $\bigoplus_{i \in I} M_i$)

[cloud walrus]

PKThoron

PKThoron

[frigid epoch]

ohhh my bad for conflation notation then

[crystal vale]

D is the submodule of C generated by relation

[frigid epoch]

perfect

so first we must write down what the 0 in M even looks like

since it's a quotient, 0 is the class 0+D

[crystal vale]

Yes

[frigid epoch]

in other words, if $\mu_i(x_i)$ is $0$, then $x_i$ is in $D$

(again it's actually iota_i of x_i but anyway)

[cloud walrus]

PKThoron

[frigid epoch]

now I wish we could say that makes it automatically of the form $x_i - \mu_{ij}(x_i)$

[cloud walrus]

PKThoron

[frigid epoch]

cause in that cause mu_ij(x_i) is 0, exactly what the exercise demands

but in general, $x_i \ in D$ means it's some finite linear combination of those differences, so in general $x_i = \sum_{r=1}^q \lambda_r(x_{i_r} - \mu_{i_r j_r}(x_{i_r}))$...

[cloud walrus]

PKThoron

[crystal vale]

Yes

[frigid epoch]

this is where I find it rather helpful to spell out the inclusion maps into the sum module

even if it looks ugly

so $\iota_i(x_i) = \sum_{r=1}^q \lambda_r [\iota_{i_r}(x_{i_r}) - \iota_{j_r}(\mu_{i_r j_r}(x_{i_r}))]$

[cloud walrus]

PKThoron

[frigid epoch]

ok

and then do some godawful coordinate checking in the direct sum module

$\iota_i(x_i)$ is concentrated in the $i$-th coordinate

[cloud walrus]

PKThoron

[frigid epoch]

meh tbh I don't know how to cleanly formulate it either

the idea is that somehow, the right side ends up just saying x_i as well and that the mu_ij term must vanish, which is what you want

[crystal vale]

How?

I think there are two cases

First when there is no r such that i_r = i, in that case it shows x_i = 0

So we are done

In second case it shows there is some r such that i_r = i

Then we get x_i = \lambda_r (x_i - u_ij(x_i) )

If j = i we are done

If not it gives x_i = \lambda_r(x_i)

[frigid epoch]

yes, and that means mu_ij(x_i) is 0

which is exactly what you wanted to show in exercise 15

[crystal vale]

i don't get it

let me think

[rocky cloak]

I guess probably not the intended method of the exercise, but the construction have actually works for arbitrary colimits.

There is a nicer construction specific to direct limits where the two properties you need proving are obvious. So one possible approach would be showing that these two are isomorphic (because both satisfy the universal property for example).

The construction is similar, but with a key difference.

Start with the disjoint union of the Mi, under the equivalence relation from mi ~ mu_ij(mi). Then you define addition mi + mj by first replacing mi and mj with equivalent things so they live in the same Mk and add there. This gives a well defined module structure and is the direct limit

[crystal vale]

i think you replied on different question

[rocky cloak]

What do you mean?

The second part of your exercise 15.

[crystal vale]

okay

why we are taking disjoint union, why not direct sum?

[rocky cloak]

Because union is easier than direct sum.

An element of a union is just an element in one of the things. An element in a direct sum is some horrible sum

[crystal vale]

okay

and to define m_i + m_j, since there exists k>= i,j, so by identification mi = mu_ik (mi) and mj = mu_jk(mj)

and both mu_ik(mi) and mu_jk(mj) is in Mj so we can add

okay

so i need to show it has universal property

yeah got it

in that case u_i is just inclusion mapping

not inclusion

i got that it is direct limit but still i don't get how this helps me to solve that problem?

[rocky cloak]

In this construction mi becomes 0 iff mi and 0 are equivalent i.e. if there is a k with mu_ik(mi) = mu_jk(0) = 0

There are no sums you have to worry about cancelling eachother

[crystal vale]

oh i see

ah great

so that equivalence relation is generated by m_i - mu_ij(mi)

[rocky cloak]

Well you can describe the relation explicitly as
mi ~ mj iff there exists k with
mu_ik(mi) = mu_jk(mj)

It's obviously symmetric and reflexive and transitivity follows from directedness

[crystal vale]

yes

thank you

[rocky cloak]

Another proof using the universal property could be:

Let Ni < Mi be the set of elements mapped to 0 by some mu. Then Ni is a submodule and mu takes Ni into Nj. Further Mi/Ni -> Mj/Nj is injective.

Then if xi in Mi is not in Ni, then you get a map Mi/Ni to an injective that doesn't vanish on xi. Using the lifting property of injectives and Zorn you can lift this to compatible maps from all Mj/Nj and hence from Mj.

Then by universal property there should be a map from the direct limit that doesn't take xi to 0, so xi can't be 0 in the direct limit

[supple pecan]

here, do i need to talk about conjugacy of h^i and g^i?
i get why it makes sense, since its the defintion of conjugacy, but can i just take g^i = e and then say
h^i = kek^-1 = e and then similarly h^i = e => g^i = e so
g^i = e <=> h^1 = e so the same powers of g and h are = e, so they have the same order
to me it seems like what im saying is equivelant to them being conjugate without saying it, but the proof uses the fact that any h conjugate to e is equal to e, but have i just shown that instead of stating it essentially?
sorry about the long text tldr: are these equivelant statements to h^i is conjugate to e so h^i = e? and stating that explicitly isnt required

[rapid cave]

Too me it looks like a you rephrased the proof

[cosmic lagoon]

yh you've just inlined the step that e is only conjugate to itself which is fine. valid

[supple pecan]

okay thats what i figured, thank you

[supple pecan]

thank you so much !

[crystal vale]

What do I have to show in 19?

[rocky cloak]

Is there any specific part of the exercise you don't understand?

[chilly radish]

You have an exact sequence

M_i->N_i->P_i for each i that is compatible with the transition maps. Show that the resulting sequence of direct limits M->N->P is also exact

[crystal vale]

Oh, so M_i -> N_i -> P_i is exact is given, okay

[crystal vale]

To show M -> N -> P is exact, I have to show each M_i -> N_i -> P_i is exact, right? But that is given already

[chilly radish]

No, you have to show that the maps induced by M_i->N_i->P_i satisfy exactness

Refer to the previous exercise for how homomorphisms between direct limits are constructed

[crystal vale]

I see

[copper kestrel]

is the \rho_{s}^{1} \chi a typo here? should it be \rho_{s}^{1} x? (second sentence of proof)

also how did they get the justification that W_1 is stable? like why does f \rho_{s}^{1}(x) = 0 imply \rho_{s}^{1}(x) = 0? im probably just over thinking it bc its true since f is not equal to 0 but i jst want to make sure i understand fully

[chilly radish]

chi is definitely a typo. They make the full argument right after though. The kernel of f is either 0 or V_1 since V_1 is irred. It can't be V_1 cuz that would mean f=0, so it's 0 and thus f is injective

[copper kestrel]

oh woops

thank you shiN!

wait why does the first case imply f = 0

sorry

[chilly radish]

If the kernel is the entire space

Then where does every element map to

[copper kestrel]

oh i see

it maps to 0

[chilly radish]

Yea so the map is identically 0 in that case

[copper kestrel]

and the only way thats possible is if f = 0

[chilly radish]

Yup

[copper kestrel]

i see, thank you!!

[chilly radish]

Glad to help

[copper kestrel]

rep theory is so strange

its fun but strange

[copper kestrel]

one more question, why does the ground field of complex numbers imply that there exists at least one eigenvalue?

sorry my linear algebra knowledge is lacking bc i havent done it since 2024

[quiet pelican]

Eigenvalues are roots of the characteristic polynomial in the ground field

[copper kestrel]

oh true

duh

thank you :]]]

[twilit citrus]

Okay so I am looking at F_16 as an artin schreirer extension of F_4; I want to explicitly compute the generators of the multiplcative group of F_16 explicitly, how do I go about it? I mean I can look at divisors of 15 and check the root of my irreducible polynomial (more than one if I look case-wise) and what its order is and if its order is 15 or not but, I feel like there's something very basic I am missing and this process seems very very tedious

is there a better way?

And am I doing it right?

[rocky cloak]

So the root will be a generator, so I guess you're just asking for the quickest way to prove this.

I feel like computing x^3 and x^5 and noting they are not equal to 1 is fairly quick.

I guess you can notice F4 already has all the cuberoots, so you really only need to check x^5.

So
x^2 = x + a
gives
x^4 = x^2 + a^2 = x + a + a^2 = x + 1
x^5 = x^2 + x = x + a + x = a

[twilit citrus]

yep, basically

[twilit citrus]

oh wow, thanks

gosh

a here is?

Like, I got 6 possibilities for irreducible polynomials

and the coefficients for x is not always 1

and if it is then a must not be in F_2

in order to be irreducible over F_4

[twilit citrus]

but this does doesn't consider the cases where coefficient of x is not 1?

I guess this would give me two? Generators? but not all of them?

orr can I find the rest using a similar method is what you are saying?

👀

[candid patrol]

I don't get it

How is it sufficient to assume H' < H ?

[rocky cloak]

I'm not sure I follow.

Like that you're thinking of an artin schreier extension means you're considering
x^2 - x - a
right?

And then a can't be 0 or 1 because then it's not irreducible

[rocky cloak]

Once you have one generator it's trivial to find all.

They will just be x^i for i relatively prime to 15

[twilit citrus]

but there are a total of 6 irreducible polynomials and this doesn't give us all of them

[rocky cloak]

What do you mean? Which irreducible polynomials are you talking about?

[twilit citrus]

of the artin-schreirer form

[rocky cloak]

I can just see two

[twilit citrus]

hmmm that's okay

but

to find just the one

[rocky cloak]

x^2 - x - a
and
x^2 - x - a + 1

[twilit citrus]

I wanted something more concrete than brute force ig

idk

I am not sure

[twilit citrus]

there's the case where coefficient of x is not in F_2 but a is in F_2

that gives us some polynomials

(I got)

and the case where a and coefficients of x both don't belong to F_2

[rocky cloak]

It gives a polynomial, but not on the artin schreier form
X^p - X - a

[twilit citrus]

hmmmm

fair

makes sense

but then so does artin schreirer always give us

or wait, like, does one of the roots

or by frobenius

does one of the irreducible polynomials' (both roots of it in this case or in general in case of quadratic irreducibles)

always give us generators?

both being generators?

[rocky cloak]

So you're asking if you always have a multiplicative generator that is the root of an artin schreier polynomial in general? That I don't know

[rocky cloak]

They said before that if
Aut_H = Aut_H', then also Aut_H' = Aut_H+H'.

So conversely if Aut_H' is different from Aut_H+H' then also Aut_H' and Aut_H are different. Hence we may set H = H+H'

[twilit citrus]

yep that's what I was asking

alsooo

[twilit citrus]

does this give me two generators?

[rocky cloak]

Conjugate of a generator is a generator so sure.

(It also just gives you all the generators though)

[crystal vale]

I got it, in this exercise I used exercise 15

[twilit citrus]

gah, I wrote wrong silly me

I meant, it gave me two irreducibles, so it gives me

4 generators

but the totient function gives me 8 possibilities for generators

so if I am explicitly looking for the minimal polynomials in a systematic way, how do I do that

to get all the generators in terms of the minimal polynomial

[crystal vale]

and this is well defined

Jagr, i think by doing the same construction of disjoint union and equivalence as before, then we can define multiplication in A.

m_i •m _j = mu_ik(m_i)mu_jk(m_j) and it is well defined operation

[crystal vale]

under this ring structure, identity is any 1_i, where 1_i is the identity of A_i, 1_i is identified with 1_j for each i,j

[crystal vale]

I have one doubt, so in the universal property of tensor product of family {B_\lambda}, the multi linear maps goes from \oplus {B_lambda} or direct product of family {B_lambda}?

What is Tor_n?

[chilly radish]

The product

[chilly radish]

If you're not familiar with Tor you probably shouldn't do this exercise. The definition takes a while to get to terms with, unless you are already familiar with derived functors, homology, and/or resolutions of modules.

[crystal vale]

No, I am not familiar with those concepts yet

[chilly radish]

Then you should skip it for now

[crystal vale]

Then I am trying to show that A-algebra B has universal property of tensor

[chilly radish]

Yes wrt the product and multilinear maps

[crystal vale]

But if it is a product, how do I construct a map from product to B ?

Multilinear map

[chilly radish]

Sorry I got my arrows backwards

[chilly radish]

I think I'm wrong and it should be the direct sum and not the direct product for this exercise. In general you can define an infinite tensor product by using the product as well but this construction is much harder to describe

[crystal vale]

I see

[upbeat tendon]

you would be working with direct sums here, i think you could generalize this to most tensor product construction processes tbh

definition for pure tensors can be extended linerally to the whole tensor product as you said

[maiden crater]

(I)=I[x] is just abuse of notation right

here

[rocky cloak]

I'm not really sure I would call it abuse.

Do you mean because R[x] should only be defined for rings?

[next obsidian]

I think it’s jus a definition of notation

There’s a notion of M[x] for modules and I think this will end up being the same

[maiden crater]

I'm used to (I) being the ideal generated by a set

[swift root]

well it is

(I) is the ideal of R[x] generated by the set I

[south patrol]

Yes

[calm fern]

Can someone please gimme a hint on how middle cancellation implies that a group G is abelian

[wraith cargo]

What do you mean by middle cancellation

[calm fern]

this

[wraith cargo]

Take x=b^-1a^-1

Or actually

[calm fern]

but what would that do ?

[wraith cargo]

Well you wanna prove that ab=ba

[calm fern]

yes

[wraith cargo]

So you need to find some x such that axb=bxa

[calm fern]

right

ohkayyy got it, thanksss

[wraith cargo]

What did you choose for x?

[calm fern]

this only that you suggested, that way I am getting a(b^-1) = b^-1(a)

[wraith cargo]

Yeah but idt what I took is the right idea

I thought it was at first but idt that's the correct choice

[calm fern]

if we take x = a^-1 or b^-1 , it will hold the given property

[wraith cargo]

Yeah that's a better choice

[calm fern]

alrightyy

[maiden crater]

I was thinking of proceeding by induction here

[vocal pebble]

Not necessary, you easier would be to just show that the map R[x1,..,xn] ->R[xpi(1),..,xpi(n)] has an inverse

(But I'm kind of confused why they're asking for isomorphism, when the rings are infact equal.)

[maiden crater]

I was just thinking of saying this by using commutativity

[vocal pebble]

Even if R wasn't commutative, they would be equal, unless you mean something else by R[x1,..,xn]

[quiet pelican]

It depends on how exactly you defined the notation R[x1, …, xn]
I can think of definitions where they are not set-theoretically equal

[vocal pebble]

Yeah I had said this before but I edited it out because the definition I know defines R[S] for a set S, and since pi is a bijection, S stays the same(which is clearly the superior one vs a definition that depends on the order of the variables)

[maiden crater]

mhm

So here I'll probably want to show $R/(x)$ is an integral domain. I also know that $(x)$ generates $\Q[x]$( correct me if I'm wrong pls)

[cloud walrus]

wai

[maiden crater]

so Let $[p(x,y)+(x)] \cdot q(x,y) + (x)$ where both $p,q $ can perhaps be polynomials of exclusively $y$. The product of two such polynomials cannot be a function of exlclusively $x$ so their product can't be $Q(x)$ we're done

[cloud walrus]

wai

[maiden crater]

This works I suppose

[vocal pebble]

Oh here if you mean x generates Q[x] IN Q[x,y] then that's not correct, since for example (x) has xy not in Q[x]

[maiden crater]

No on it's own

[vocal pebble]

Sure, not sure how that's relevant though

[vocal pebble]

I didn't really understand this, all you have to do is show that Q[x,y]/(x) is an integral donain

[maiden crater]

Ye, so I have to show it has no zero divisors right

Which is what I am trying to do

[vocal pebble]

Yeah but like (if I understood what you're trying to say) you seem to be assuming that the image of an element is 0 if it is only exclusively a polynomial in x, which is not true, eg : [xy]=[0]

[maiden crater]

Well I'm talking about Q(xy)/(x)

[vocal pebble]

Yeah, and I'm saying even in Q[x,y]/(x), you have for example xy=0 and xy is not exclusively a polynomial in x

one way to do this simply would be to show that Q[x,y]/(x)=Q[y] which is an integral domain

[maiden crater]

Got it

[coral steeple]

Does this look right for the implication a -> b?

(Of course, the largest subfield of L contained in E' and E'' is their intersection)

Also, the point is that you use the same trick with alpha and beta to ensure that these are square roots of elements of K. I only said to do it with alpha, but this might not have been totally clear

... Another note: I could have referred to L^H' instead of E', in case that causes confusion

[maiden crater]

how would I do this

Like isn't (x) = Q[x,y]

[coral steeple]

Then how do you write 1 as an element of (x)?

@maiden crater

[maiden crater]

oh riht

got it

thanks

[velvet hull]

a little verbose, but the proof idea is correct

[ripe harbor]

Let $L/K$ be a Galois extension of degree $45$. I want to show this has at most $12$ subfields using Galois theory. So, we know the Galois group of this also has cardinality $45$ and every subgroup must divide it, so possibilities are the orders $1, 3, 5, 9, 15$ and $45$. Exactly one for orders $1$ and $45$, for $5$ we can use the third Sylow theorem: $$n_5 \mid 45, \qquad n_5 \equiv 1 \pmod{5}.$$ But how is that possible?

[cloud walrus]

ILikeMathematics

[supple pecan]

when they say = in the example, do they mean isomorphic?

is it fair to say theyre equal for groups, rings, etc if theyre isomorphic or is the ≅ notation needed

[knotty badger]

Yes most likely

Them being strictly equal depends on how you implement them

But they’re certainly isomorphic

[merry harness]

Depends how you define Z5 as a ring

I mean you could have Z5 = Z/(5)

If it's something else then yeah just isomorphism

I think defining Z5 as the quotient ring is common

[knotty badger]

Yes though you can also define it manually

[twilit wraith]

youll often see algebra textbooks eventually just use = for isomorphism

usually its when the isomorphism is obvious just by the structure of both objects

[valid fox]

I've seen things like a vector space and it's dual are isomorphic but not "canonically" so (whatever canonical means). But when we have an inner product they are canonically isomorphic (by using transpose iirc). So is a vector space with inner product somehow "more isomorphic" to its dual than a vector space without additional structure?

Are there other examples of this?

[supple pecan]

oh? does it make sense to define rings as a ring theyre isomorphic to

[supple pecan]

ah i see, we've just been pretty consistent on the notation so this confused me. thank you !

[supple pecan]

how does that work?

[merry harness]

If you define Z5 = Z/(5) then it's equal not just isomorphic

You also could define Z5 more manually like pseudo said

In the grand scheme of things, it doesn't matter how you construct Z5 specifically, since it works the same anyway (Cat theorists salivating)

[knotty badger]

What would say is that the inner product picks out or induces a particular isomorphism between V and V*

[rocky cloak]

How it's possible?

You just list the factors of 45 and see which are 1 mod 5

[valid fox]

If we think of the vector space as lists of scalars, then the map $\phi: V\to V^*, v\mapsto v^T$ where $v^T$ acts on V by matrix multiplication

[cloud walrus]

Tiessie

[knotty badger]

Welllll this is not quite what you want to do

[valid fox]

Or something like $v\mapsto \langle v, .\rangle$?

[cloud walrus]

Tiessie

[knotty badger]

Yes

For the former, you need to choose a basis of V to represent it as lists of scalars

[valid fox]

Right

[knotty badger]

And it’s possible to have an inner product on V without a particular basis being specified

However, the inner product still gives you enough information to define a specific isomorphism V -> V*

[knotty badger]

Which is this

[valid fox]

I'm more so curious about this word canonical

[neon jewel]

Third Sylow theorem says that $n_5 \mid 9$ not $45$.

[cloud walrus]

Lin Xia

[valid fox]

I've seen it in another places too but I never quite understood the significance

[noble nexus]

One explanation of the word canonical is that you can choose the isomorphism uniformly across different spaces

you can formalize that with category theory but you don't really need to know all the details to understand the idea

the non-canonical isomorphism of a space with its dual requires picking a basis

so if you had two vector spaces, there's no way to ensure that the way you picked a basis in both cases is compatible

[knotty badger]

One theme in higher mathematics is that it doesn’t just matter that two things are isomorphic, but how they are isomorphic

If V is finite dimensional, then it is isomorphic to V*

[noble nexus]

but with the double dual isomorphism, it's the "same" isomorphism regardless of the vector space

[knotty badger]

But there are many possible isomorphisms you could choose, and it’s not clear if one really stands out from the rest

However, if you’re given an inner product on V, then there is a standard way to choose an isomorphism between V and V*, using this inner product

You could then say this is a “canonical” isomorphism

[knotty badger]

This is kind of true though not sure if I like thinking of naturality this way these days

[valid fox]

Is there some category theory language for this?

[knotty badger]

Yes, it’s to do with the concept of a natural transformation

[valid fox]

Ahhh

I have heard of that

Are there other good examples of this I could look into?

[knotty badger]

Yes loads

One of my favourite examples is the group operation

For every group $G$ you have a function $*_G : G \times G \to G$ called the group operation

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[chilly radish]

The canonical counterpart to this is the fact that you can define a map V->(V*)* without referencing a basis, and this will be an isomorphism for any finite dimensional V

[knotty badger]

It’s what you use to multiply elements of your group

This is a natural transformation $U \times U \to U$ where $U : \mathbf{Grp} \to \mathbf{Set}$ is the forgetful functor

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

More generally, any kind of algebraic operation (group inversion, ring multiplication, a ternary thing like “a(b + c)”) defines a natural transformation

[valid fox]

FYI I'm familiar with abstract algebra up to and inclduing module theory

[knotty badger]

Ah then I can give a good motivation for naturality

I think a good word to keep in mind when thinking about naturality is orthogonality

Two things can be geometrically orthogonal, meaning at right angles to each other

But there’s also a notion of two things being conceptually orthogonal - this means they refer to separate, independent ideas

As an example I was booking travel to a conference recently, and figuring out how to get to the airport was mostly independent of figuring out what flight I needed to take

[valid fox]

Right

[knotty badger]

They’re independent legs of my journey

How good you are at swimming is also mostly independent of how good you are at mental maths

In that sense they’re orthogonal skills

One insight from cat theory is that there’s often a connection between these two senses of orthogonality

For example, with the travel example, it doesn’t really matter which order I book the legs of my journey

I could book travel to the airport first and then book the flight, or do the opposite order, and end up with the same result

[knotty badger]

You can actually represent this with a kind of commutative diagram

Note that the independent legs of my journey have now become geometrically orthogonal processes

In the most literal terms, the two arrows are at right angles to each other

This sort of thing happens quite often when you have two independent processes - you can run them independently such that the order in which they finish doesn’t matter

(One thing that’s confusing is another name for this is running the processes in parallel, but obviously parallel and orthogonal are very different geometrically)

Does that make sense?

[valid fox]

For sure

[knotty badger]

This actually happens a lot in algebra

You have internal operations that happen within your algebraic structure

Like the group operation, or vector addition, or scalar multiplication, or ring multiplication, or field division

And then you have external operations that happen between two algebraic structures - the homomorphisms

The way you define homomorphisms is to ensure they are “orthogonal” to the internal operations

In the sense that it doesn’t matter what order you apply the internal and external operations in - you arrive at the same result

For group homomorphisms, that’s $\phi(ab) = \phi(a) \phi(b)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[valid fox]

These would just be called morphisms right

Is a natural transformation then a specific kind of functor?

[knotty badger]

Again, you can represent this with a commutative diagram

In the most literal sense, the external operations are orthogonal to the internal ones

[knotty badger]

Well, this is actually what naturality is about

Naturality is a categorification of the concept of orthogonality/independence

The way it manifests within category theory

Because the internal operations are always orthogonal to the external ones, the internal operations define natural transformations!

Another good example of this are operations on lists in programming

You can have “data-changing” operations such as squaring each element of a list, and “shape-changing” operations like reversing the list

These are orthogonal to each other as well

So the shape-changing operations end up defining natural transformations

[knotty badger]

It can be seen as one yes - more often it’s viewed as an arrow between functors

But you can also view it as an arrow-valued functor, as well as a bifunctor - each point of view has its merits

The double dual example earlier is saying that the evaluation map V -> V** is orthogonal to all linear maps V -> W

[valid fox]

What do you mean by arrow? Morphism?

[knotty badger]

Yes, I use the terms interchangeably

[valid fox]

Okay

[knotty badger]

One thing I like to emphasise is the geometric side of category theory

“Morphisms generalise functions” is a decent intuition, but an equally important one is “morphisms generalise paths”

Not all morphisms take the form of structure-preserving maps, and getting too attached to this can hinder understanding

[valid fox]

Paths in a kind of graph theory sense?

[knotty badger]

Yes exactly

In fact any graph automatically defines a category

Where the morphisms between two vertices A and B are precisely the paths in the graph that start at A and end at B

Called the “path category” or “free category” of the graph

Much like any group is a quotient of a free group, any category is a quotient of a free category

So it can be specified by supplying a graph, plus a rule about when two paths should be considered “equal”

This is the “generators and relations” PoV on categories

[valid fox]

Kind of like group presentations

[knotty badger]

Yes exactly

[valid fox]

Where you just impose the appropriate structure

That's nice

[knotty badger]

Things like “finitely presented” categories are important objects of study

[ripe harbor]

None..

[ripe harbor]

No, my book says its | |G|

[rocky cloak]

Well, you already wrote down the divisors, so let's go over them one by one.

The first one is 1, is that equal to 1 modulo 5?

[ripe harbor]

Oh crap I forgot about 1

Yes

[rocky cloak]

Alright, so it could be that n5 = 1

Could it be 3, 5, 9, 15 or 45?

[ripe harbor]

no

not 1 mod 5

[rocky cloak]

So then it must be 1

That simplifies things

[ripe harbor]

Right. Thanks!

[ripe harbor]

Now it follows that np = 1, 6, 11, 16, 21, 26, 31, 36 or 41

So np = 1

Because none of those divide 45

[rocky cloak]

One might notice that any divisor of 45 that is not a multiple of 5 will be a divisor of 9 (and vice versa)

[frail shoal]

when i think about it, the data of an inner product, a linear map V ⊗ V -> R, uncurries into a linear map V -> V*. so in a literal sense, an inner product space is literally a vector space equipped with a map from itself to its dual

i forgot the other axioms of inner product spaces though

oh, i honestly have no idea how to interpret the other axioms in the uncurried way

i mean i could do it very literally but i mean, like, a useful interpretation

[rocky cloak]

An isomorphism V -> V* is exactly a non-degenerate bilinear form.

The axiom
<v, v> >= 0
is maybe harder to uncurry

[frail shoal]

hell yeah

maybe symmetry and positivity could be viewed as making the operation resemble multiplication

dunno

[neon jewel]

sure mb

[maiden crater]

Here can I simpy say each $\mathbb{Z} [x_1 \dots ] / (x_i)$ is prime?

[cloud walrus]

wai

[maiden crater]

Like suppose p/(x) * q/(x) = (x) then one of them would have to have x so would simplify to (x), which is the 0 in this rong

ring

[wraith cargo]

But in Z[x_1,x_2,...]/(x1x2,x3x4,...) when you quotient by (x_i) you still have x_j x_k = 0 in your ring...

so the quotient isn't an integral domain

so (x_i) can't be prime

[maiden crater]

hmm

[maiden crater]

wdym

[wraith cargo]

well take R/(x1) right

this means x1=0

but you still have x3, x4 != 0

but x3x4 = 0 in your quotient ring

[maiden crater]

got it

how do I do it then

[rocky cloak]

Notice that whenever you have
xy = 0
a prime ideal would need to contain either x or y.

So if you're looking for minimal primes you might try to force it to only contain one of them

[maiden crater]

so I want the prime ideal to only contain x and its multiples and sims

[wraith cargo]

0 has to be in your ideal

[noble nexus]

I think the best is just a literal way, I've seen that in literature a bit

[wraith cargo]

and also x_i x_i+1 = 0,

so either x_i or x_i+1 is gonna be in your ideal

[rocky cloak]

So let's call our prime p.

Then
x1 x2 = 0

What does that tell you?

[noble nexus]

i.e. a map phi from V to V^* is positive semi-definite if phi(v)v\geq 0 for all v

[maiden crater]

One of them is a zero divisor

[rocky cloak]

Well they are both zero divisors by definition

[noble nexus]

for inner product spaces you have the notion of a positive operator which gives you positive semi-definite forms

[maiden crater]

I'm going for dinner now will be back in 15

[coral steeple]

oh... I thought it was much too concise thanks!

[ripe harbor]

When exactly is Gal(L/K) abelian? Is there some criterion?

[wraith cargo]

These are called abelian extensions

And characterizing these is a big thing in math

In number theory specifically this is known as class field theory

[ripe harbor]

Oh

[ripe harbor]

And one more thing: When we consider Z/20Z and want to know the number of group homomorphisms Z/20Z -> Z/20Z, do we consider these w.r.t addition or multiplication?

[wraith cargo]

Addition

[ripe harbor]

But then isnt there only one group homomorphism?

Because g(1) = 1

so g(2) = g(1 + 1) = 1 + 1 = 2

[wraith cargo]

This isn't true

[ripe harbor]

any group homomorphism maps 1 to 1

no?

[wraith cargo]

Shouldn't it map 1 to any generator?

Because 1 isn't the identity

It maps 0 to 0

[ripe harbor]

oh

Thanks!!

[ripe harbor]

So there will be as many homomorphisms as generators

Arent all of these also automorphisms?

[chilly radish]

It doesn't have to map 1 to a generator

Any mapping of 1 to a generator will be an automorphism, but 1 can be mapped to things of lower order as well

The only condition is that f(20)=20* f(1)= 0, so f(1) has order dividing 20, but that's every element in Z/20

So you can map 1 to any element

You should check this is always well-defined

[ripe harbor]

but that's every element in Z/20

Is there a quick way to see that?

[chilly radish]

Lagrange's theorem

[ripe harbor]

oh

Whats the problem if our only justification is that we map 1 -> a for any a in Z/20Z

And that will give us the rest of the homomorphism

[chilly radish]

Well you're explicitly defining this map via representatives, so you need to make sure it's well-defined.

So first showing it's independent of choice of representative. The next step would making sure that the image of the generator satisfies the proper relations, but here as we said this just means being an element with order dividing 20

So for example if you were looking for maps Z/2->Z/3, then sending 1 to 1 is not well-defined, since if I choose 3 as a representative of the equivalence class 1+2Z, then that representative will get mapped to 0

Formally the easiest way to do this is to define a map

Z->Z/20

By mapping 1 into any element (since Z is free abelian) and then using the first isomorphism theorem to get a map

Z/20->Z/20

[ripe harbor]

Thanks!

[chilly radish]

When I say first iso here I just mean that if you have a map f:A->B and a subgroup N of A contained in the kernel, then there's an induced map

f:A/N->B, not necessarily injective or anything

(changed containing to contained in)

[crystal vale]

So if G is finitely generated abelian group then each subgroup is also finitely generated, right?

[candid patrol]

Yes

[ripe harbor]

Why is $\text{Gal}(\mathbb Q(\sqrt[4] 2)/\mathbb Q) \cong \mathbb Z/2\mathbb Z$? Why cant we also have $(\sqrt 2 \mapsto \sqrt 2, \sqrt[4] 2 \mapsto -\sqrt[4] 2)$?

[cloud walrus]

ILikeMathematics

[coral steeple]

What is the other element you found in the group?

[swift root]

well i guess you can intuit that the Galois group cannot have order 4, as the extension is not Galois

[coral steeple]

The Galois group permutes the roots of x^4-2, and an automorphism in the group is determined by where it sends the fourth root of 2. The only purely real roots are plus or minus the positive root, so at most you have the identity and what you wrote (so I don't know why you're alluding to something else being in there)

Maybe I'm making a mistake

[swift root]

yeah this is correct

[swift root]

the automorphism you are describing is in fact the only nontrivial element in this Galois group

[ripe harbor]

Ah, we cant have sqrt(2) -> -sqrt(2)

Thanks

[candid patrol]

Guys, is that true that f € Aut(G) iff \bar(f) € Aut(G/Phi(G)) ?

I know there is a similar thing with p'-automorphism wich induces identity on G/Phi(G)

[merry harness]

euro is a new one

[azure cairn]

i think ive seen ppl. use euro

[chilly radish]

I've definitely used euro before on whatsapp

[candid patrol]

ahahha it's cause i'm french

[chilly radish]

What is Phi(G) here

[candid patrol]

frattini subgroup

[rocky cloak]

So if G is finite, then all that's needed to verify is wether f is surjective.

Note that if bar(f) is surjective, then G = Im(f)Phi(G).

If Im(f) was not all of G you could pick a maximal subgroup containing it. Then it would also contain Phi(G), contradiction. So Im(f) = G

In the infinite case I guess G = Q has G = Phi(G), so there it wouldn't work

[lilac mango]

Does anyone know of a computer algebra system that does localization?

[rocky cloak]

Macaulay2

[lilac mango]

Ok then I don't know how to read documentation hahaha

[rocky cloak]

https://macaulay2.com/doc/Macaulay2/share/doc/Macaulay2/LocalRings/html/_local__Ring.html

[lilac mango]

Oh ok

The thing is I want to localize not a polynomial ring

But like Z/nZ by some prime ideal

[vapid vale]

fwiw i just tried to do this in M2 for a bit and i couldn't figure it out

[rocky cloak]

I mean, you don't need a computer to localize Z/n

[lilac mango]

I wanted to verify something

So I think localization in Z/nZ is just the quotient ring

And I kind of did some calculations but I'm not sure so I wanted some computational intuition

[chilly radish]

You can post your work here

[rocky cloak]

The prime ideals of Z/n are just (p) for primes dividing n, and the localization is Z/p^k where p^k is the largest power of p dividing n.

Which is just a quotient ring of Z/n yes

[fickle dirge]

I don't understand where is the axiom of choice needed/used here

don't we construct the chain of ideals and show that each inclusion is strict so the chain is infinite

This is in dummit and foote proving that every PID is a UFD

[velvet hull]

that statement looks a little strange

[fickle dirge]

is it just the fact that i choose r11 , r111 ?

[velvet hull]

where is this from?

[fickle dirge]

Dummit and foote

[rocky cloak]

Well you do need to make an infinite number of dependent choices to construct the chain

[fickle dirge]

so this, right?

I am choosing from the set of divisors one of them

[rocky cloak]

yeah

[fickle dirge]

Okay, got it thanks

[wild jasper]

As (\mathbb{Z})-modules let (B = \mathbb{Z}[1/p]) and if (A = \mathbb{Q}/B)[p],) do need anyting else to show that (A = 0) than that (a/b +B = (pa)/(pb) + B = 0)?

[cloud walrus]

mh_le

[wild jasper]

ah yeah, I now realize why it doesn't make sense.

[candid patrol]

I see thx you

[candid patrol]

What can you say about Sylow subgroups ?

[ripe harbor]

Can we be sure there arent any more?

[candid patrol]

and what about the 5-Sylow ?

[ripe harbor]

Why do we care?

We want to quotient by order 27

[candid patrol]

Why not

But yes you can show that 3-Sylow subset ker(f) for all f : G -> Z/5

||(There is also one 5-Sylow, so G is nilpotent)||

[ripe harbor]

Oh wait, got it

Thanks

[candid patrol]

np

[solar kernel]

Hello!

I’ve got a question - this is for an assignment, so don’t help me too much. I’m largely just checking i have the right idea.

I need to prove that there doesn’t exist a simple group with order 132.

I’ve split 132 into its prime factorisation, $2^2311$.

Having a look at the Sylow 11 groups; the number of these 11-groups must either be 1 or 12. If it is 1, we are done(said 11-group is normal, and hence G is not simple). Suppose then there are 12 Sylow 11-groups.

Then there are 120 elements of order 11(intersection of sylow groups are the identity, and order 11 groups elements are all order 11 bar the identity).

Therefore we have 11 elements that are not in the sylow 11 groups. Now we consider the Sylow 3 groups. Either there is 1,4 or 22 of these groups. Can’t be 22, to few elements to work with. If it is 1, we are again done.
Suppose it is 4, then we have 3 elements left for the sylow 2 groups; there are either 1,3,11 or 33 sylow 2 groups, but there actually is exactly 1 sylow 2 group due to the number of elements left.

Therefore for any “choice” we have made, there is always a Sylow p-group that is a normal subgroup of G, and hence G is not a simple group.

I haven’t properly typed it up, so it isn’t in the formal wording just yet, but is my thinking correct? Cheers!

[cloud walrus]

Makon W

[quiet pelican]

That works
Another way to deal with the “4 Sylow 3 subgroups” case is to note that acting on them by conjugation gives a non trivial map G -> S_4 which must have non trivial kernel by looking at the order

[solar kernel]

Oh, and then the kernel of a homomorphism is a normal subgroup! I like that lol.

[lilac mango]

I'm trying to see that A/I tensor_A A/I is isomorphic to A/I but I'm struggling to prove injectivity

Mostly the problem is I don't know how to prove that some tensors aren't zero

For example if I is a proper ideal of A how could I prove that in (A/I tensor A/I) we have that (1+I) tensor (1+I) isn't zero

[rocky cloak]

The general method to proving that an element in M(x)N is to define a bilinear map from MxN where it doesn't vanish.

Hence it can't be 0 by the universal property of the tensor product

An alternative aproach to solve your exercise is to use right exactness to prove that
A/I (x) M = M/IM

[stone sky]

Quick question. I have the following statement that my professor asked me to prove. That $\mathbb{C}^{}/S \cong \mathbb{R}^{}$; however this is not true is it? Shouldn’t the result be $\mathbb{C}^{*}/S \cong \mathbb{R}_{0>}$ ?

[cloud walrus]

Surf

[south patrol]

By S do you mean S^1?

But yeah the usual thing would be R_{> 0}

[stone sky]

Yes S^1 sorry

[south patrol]

And indeed R^x and R_{>0} are not isomorphic groups because the former has the element -1 of order 2

[stone sky]

Correct

[south patrol]

So yee well noticed

[stone sky]

Ok cool stuff

[crystal vale]

I think the order of H is irrelevant here

I mean H has to be p-group

But not necessarily order p

[south patrol]

Yeah agreed

H is contained in one of the Sylows and all the Sylows are conjugate but conjugation doesn't change H

[crystal vale]

Yes

So I am trying to prove one fact which I think is true, if G is p-group then every normal subgroup intersect with center

[quiet pelican]

How do you prove that the centre of a p-group is non-trivial?
Can you use that proof technique here?

[crystal vale]

Oh

Let me try

I got it thanks

G acts on H by conjugation

[patent girder]

If $q\in \mathrm{Sym}^2(V^*)$ we can define $\mathfrak{so}(q)$ to be the trace zero matrices that preserve $q$. For dimension $3$ in the reals, I was able to show that the classification of quadratic forms gave the isomorphism classes of $\mathfrak{so}(q)$, using invariance of the Killing form. Is the same true for arbitrary $V/k$?

[cloud walrus]

Zander

[azure cairn]

is this notation used for any field

i.e. $\bF[x] = {a + bx: a, b \in \bF$, idk

[south patrol]

No. Usually the situation is that you have two fields F and K (with F contained in K) and an element a of K. Then F[a] denotes the smallest subring of K containing F and F(a) similarly for smallest field. Or you can view this as what you get from messing around with a applying ring operations.

Unfortunately there is also the minor conflict with the notation F[x] for a variable/indeterminate x, meaning a polynomial ring

Here you only need stuff of the form a + bsqrt(2) because sqrt(2)^2 = 2, so you can simplify any expression involving powers of sqrt(2) to something of that form

[azure cairn]

Interesting. Idk why the author dropped that notation randomly with no clarificatiob

[rocky cloak]

It's very standard notation, so they might have assumed the reader was familiar with it.

Or I guess it's just the standard name for Q[sqrt(2)], you don't need the general pattern of this notation to be told that

[tribal moss]

I think is is actually consistent with F[x] for a polynomial ring. If we imagine that, say, polynomial or rational functions already exist "out there" somehow, and already know how they react to + and ×, then then F[x] is indeed the smallest subring of "out there" that contains F as well as x.

[rocky cloak]

Or you can think of F[x] as the value of all formulas in the theory of rings with values in F and x. Then when x is an indeterminate the formulas just return formal expressions (that are considered equal if they're equal for all values of x)

[south patrol]

It is consistent but also kinda different

[cursive spindle]

This notation is common for number fields

[tribal moss]

Though you need to be careful with "if they're equal for all values of x" if F happens to be finite. Perhaps amend with something like "even if the value you choose for x is from some extension of F", but then can begin feeling very fuzzy until you know enough theory to know what the possibilities are -- which supposes you already understand polynomial rings to begin with.

[rocky cloak]

x does not take values in F

[tribal moss]

Clearly it doesn't because otherwise your proposed intuition wouldn't work. :-)

[cursive spindle]

btw if you follow this logic then Q(sqrt(2),sqrt(3)) would be a+ b sqrt(2)+c sqrt(3) which is almost correct

Intuitively it should make sense that this field should have dimension 4 over Q

but in this one you have 3

[azure cairn]

apologies for no reply im running on very little sleep and can barely comprehend anything rn lmao

[cursive spindle]

Type of shit numerical analysis would do to me

Take a rest pookie

[azure cairn]

unfortunately im unable to fall asleep in school for whatever reason

its too uncomfortable to sleep

soon tho :>

[azure cairn]

could you define dimension?

[swift root]

vector space

[rocky cloak]

Cardinality of any basis

[azure cairn]

Right, Q is a vsp over itself

[rocky cloak]

Yes, but more relevant: any field extension of Q is a vector space over Q

[azure cairn]

I dunno what that is so I'll wait till i see in my book :p

[cursive spindle]

Think of Q(i). This has a basis {1,i} over Q correct?

[azure cairn]

Yes

[cursive spindle]

Yes that's an example of what I mean by a vector space

[azure cairn]

I see

So then Q(sqrt2, sqrt3)

You said that this would have dimension 4..?

[cursive spindle]

Yes

[azure cairn]

What's the definition of this

All I know is Q[x] = a + bx for every a, b in C (treating Q as a subfield of C)

Is Q[x, y] = a + bx + cy?

[cursive spindle]

Let's build it one at a time

Remove sqrt(3) for now

What's Q(sqrt2)?

[umbral ginkgo]

[ℚ(√2,√3) : ℚ] = [ℚ(√2,√3) : ℚ(√2)] · [ℚ(√2) : ℚ] = 2 · 2 = 4.
Une base est donnée par :
{1, √2, √3, √6}.
alors dim_ℚ ℚ(√2,√3) = 4.

[swift root]

Q[x] = a0 + a1x + a2x^2 + ... + anx^n for ai ∈ ℚ

[azure cairn]

a + bsqrt2 for every a,b in C

[cursive spindle]

tfw you try work out things step by step then you get someone spoiling it

[cursive spindle]

Right

[swift root]

bwa

[swift root]

Q

[umbral ginkgo]

oh sorry didn't mean to hahahaha.

[swift root]

not C

[cursive spindle]

And you can easily check this is a field

ah yes good catch enpeace

tfw i read a+bsqrt(2) and I assumed everything afterwards is correct

[swift root]

and it happens that if x^2 ∈ ℚ, then this collapses to what you said

[cursive spindle]

So now with Q(sqrt2,sqrt3) you want to maintain the property of it being a field

[merry harness]

@azure cairn be careful to note that e.g. (\bQ[x] \subsetneq \bQ(x)), since they are ring and field extensions, respectively

[azure cairn]

I have never seen this

[swift root]

(more generally if x satisfies a polynonial equation of degree 2)

[azure cairn]

I don't know what any of this means 😭

[swift root]

this is not true

[azure cairn]

Artin just dropped Q[sqrt2] as an example of a subfield of C

[umbral ginkgo]

x² is a polynôme how is it in Q.

[merry harness]

I thought brackets are ring and parens are field

[swift root]

they can coincide

[merry harness]

Oh you mean sqrt 2 specifically

Yeah I should put more general oops

[cursive spindle]

but anyways you want to multiply things. In particular you need sqrt(6)

[swift root]

you also specifically wrote down sqrt 2 vro

[azure cairn]

And said Q[sqrt2] is the set of complex numbers a + bx for every a, b in Q. Idk what Q[x] means in terms of being a polynomial ring

[merry harness]

Yeah I was looking at sqrt 2

I meant in general

[swift root]

x is an element from an overlying field

[umbral ginkgo]

I wanna explain but math terms with me are in french help🥲😂.

[swift root]

then what you wrote is still wrong

[cloud walrus]

qiu

[merry harness]

Correct example now

[azure cairn]

I don't understand where sqrt(6) comes from. Again, the only thing i know so far is that Q[sqrt2] = {a + bsqrt2: a, b in Q}. Idk what the notation Q[sqrt2, sqrt3] means

[cursive spindle]

Is there a way to get sqrt(6) by using a+sqrt(2)+csqrt(3) where a,b,c in Q?

[azure cairn]

It just feels like he introduced notation too early

[azure cairn]

Right ok so

,w a + sqrt(2) + csqrt(3) = sqrt(6)

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by altanis_

[azure cairn]

Ok let's see

[merry harness]

Means you can have any combination of sqrt 2, sqrt 3

[azure cairn]

Did you forget a b

[cursive spindle]

yes

mb

[azure cairn]

a + bsqrt2 + csqrt3 = sqrt6

[cursive spindle]

right

[azure cairn]

Is this possible

Idts?

[cursive spindle]

yes that's the point

[umbral ginkgo]

ℚ(√2, √3) is the smallest subfield of ℝ containing ℚ, √2, and √3.

In other words, it is the set of all combinations obtained from √2 and √3 using +, −, ×, ÷. that's why you found √6.

I hope the translation is correct.

[azure cairn]

Right but where did sqrt6 come form

[azure cairn]

Oh

[cursive spindle]

Remember we need a field

[azure cairn]

I didn't know that's what Q[sqrt2] meant, the smallest subring of C containing Q and sqrt2

[umbral ginkgo]

Because ℚ(√2, √3) is closed under multiplication.

Now:
√2 · √3 = √6
Therefore, √6 ∈ ℚ(√2, √3).

[azure cairn]

Ohh I see

[cursive spindle]

sqrt(2) times sqrt(3) is not in the set (yet)

[azure cairn]

So yes we need a fourth parameter

For sqrt6

[umbral ginkgo]

Oui ça forme la base.

[azure cairn]

Q[sqrt2, sqrt3] = {a + bsqrt2 + csqrt3 + dsqrt6 : a, b, c, d in Q}

[cursive spindle]

another way to view this as follows: think of Q(sqrt3,sqrt2) over Q(sqrt2)

[azure cairn]

And yes if you treat this as a vsp the basis is 1,sqrt2,sqrt3,sqrr6

[cursive spindle]

So a+b sqrt(3) where a,b are in Q(sqrt2)

[swift root]

sqrt4 also know as 2

[azure cairn]

Typo

[kind temple]

or Q[sqrt 2, sqrt 3] = Q[sqrt 2][sqrt 3]

[umbral ginkgo]

Non, ℚ(√2, √3) = {a + b√2 + c√3 + d√6 | a,b,c,d ∈ ℚ}..

[cursive spindle]

Bro turned on French mode

[azure cairn]

That's what I said no

[umbral ginkgo]

I'm still adapting to English math hahaha.

[azure cairn]

Anyways I see. For posterity this was just dropped out of nowhere. I didn't know Q[x] meant the smallest subring of C containing Q, x

[cursive spindle]

yeah no worries

And the beauty about this is that Q(sqrt2,sqrt3) is the smallest such field which contains the roots sqrt2 and sqrt3 (and obviously their combinations)

[azure cairn]

This is all I got about Q[x]

[cursive spindle]

Have you gone to rings?

[umbral ginkgo]

but that's still not the right definition.

Q[x] is the set of polynomials in the variable x with coefficients in Q,
Q[x] = { a₀ + a₁x + a₂x² + … + aₙxⁿ | n ∈ ℕ, aᵢ ∈ Q }.

it's the smallest subring of ℂ containing Q and x.

[cursive spindle]

I assume yes cuz ur already in fields

[azure cairn]

No

[cursive spindle]

oh hmm

[azure cairn]

I'm in a chapter about vector spaces

[cursive spindle]

Then I probably shouldn't talk more about this

[azure cairn]

SO I think he's just introducing basic stuff about fields

And will discuss it more in depth later

[umbral ginkgo]

but how they did a field without defining a ring??

[cursive spindle]

It's okay imo

[azure cairn]

It's a set with addition and multiplicatjon operations that is an abelian additive and multiplicative group

[cursive spindle]

unless you are doing field extensions and whatnot

[azure cairn]

That obeys associativity and distrivutivity

[cursive spindle]

anyways what I wanted to say is that for example you can think of Q(i) as a quotient of Q[x] by some ideal

If what I said makes no sense dw about it

[azure cairn]

I don't know

Ok

[umbral ginkgo]

a field is a ring so that's lack of structure in her course.

[cursive spindle]

right yeah

[karmic moat]

vector spaces are modules but you can do linear algebra without knowing what a module is

[azure cairn]

Again, it's in the future

[cursive spindle]

some courses still introduce fields (but don't do anything advanced with it)

It's the same flavor as "oh but a group is a groupoid"

[azure cairn]

Like artins book just introduces a field I'm assuming to define a vector space properly but isn't doing anything with fields yet except what a linalg book would do (like defining characteristic of a field) and connecting it back to group theoreitc stuff

[umbral ginkgo]

it should've been groups to rings to fields to vector spaces.

[slim kayak]

I vaguely remembering our introductory linalg course mentioning PIDs to talk about minimal polynomials

[umbral ginkgo]

a vector space over a ring so it doesn't matter but unlike not defining the structure of a ring.

[swift root]

well it really does matter whether youre working over a field or not

[azure cairn]

Well idk what to say @french person (I'm typing in the forward message box so I can't ping you directly). Here is the table of contents

[umbral ginkgo]

vector spaces are generally defined over a field.

[swift root]

you said "vector space over a ring so it doesnt matter" idk what you meant by that then vro

[umbral ginkgo]

I said it doesn't matter defining a structure of a module cause it's just a vector space over a ring, it won't change anything about vsp.
but not defining rings can make it difficult later for polynomials for example, or even studying fields without knowing they're rings.

[swift root]

yes and then i said that the structure of modules really depends on whether youre working over a field or nor

[umbral ginkgo]

yes I see, little bit messy if you don't organize your thoughts.

[frail shoal]

i think an interesting part of module theory is learning why modules are nice in general and why vector spaces are really nice, even for modules

i think during the period i was ua-pilling myself, sometimes i would do things with modules again, and like, remember things that feel impossible. like, "wait, that shit is true???"

stuff like, how the submodule lattice is isomorphic to the quotient lattice

most recently this happened for me when i reproved to myself that
0 -> A -> B -> C -> 0
v v
0 -> D -> E -> F -> 0
induces a unique map of ses's

like i vaguely recalled it but it felt too nice to be true, so i had to check it

although i think this is actually true of groups so this is a bad example

yknow what this yap is kinda dumb. sorry for wasting everyone's time here

[novel scarab]

probably an elementary question, but for a fixed m, for which n and subgroup H of G = (Z/nZ)^{\times} do we have that the quotient G/H \cong (Z/mZ)^{\times}? equivalently, for which n can we get a surjection (Z/nZ)^{\times} → (Z/mZ)^{\times}? I guess I got a quite ugly/nonexplicit answer by considering the sylow decompositions but I'm looking for something a bit more explicit

[tulip otter]

let f(x)\in F[x] be an irreducible polynomial of degree 6 and let K/F be a quadratic extension of F. I am asked to check whether the following is true or false: f is either irreducible over K or factors as a product of 2 irreducible polynomials of degree 3 in K

any hints?

[quiet pelican]

Let a be a root of f and consider the degree of a over K

You know the degree of a over F is 6

[tulip otter]

F(a) isnt necessarily an extension of K tho right? But ig we can take a subfield K' of F(a) which is isomorphic to K and then everything is fine? In that case [F(a):K']=[F(a):F]/[K':F]=6/2=3

So any root of f has a minimal polynomial of degree 3 over K and hence if f is reducible over K, it must be a product of two degree 3 polynomials. Is that right?

[frail shoal]

well we dont need to use sylow cuz these are abelian groups, so we can use the classification of them instead

that being said, it should ultimately come down to number theory yea

there are explicit results on what Z/n's unit group looks like

they're somewhat nontrivial though

[rocky cloak]

I doubt there's a very "nice" answer to this.

Like the canonical examples would be when n is a multiple of m then the map Z/n -> Z/m gives you a surjective map on unit groups.

But there can be lots of "accidental" relationships.

Like
(Z/3)^x = (Z/4)^x
or
(Z/35)^x = (Z/39)^x

[frail shoal]

in particular, Z/(p^d) has a unit group that is isomorphic to [ignore this, i wrote the wrong thing]

for p > 2

for p = 2? different story

actually i feel like i just spread lies

but i forget the correct thing

but anyway from there you can use CRT and the fact products of rings preserves unit group

[rocky cloak]

The unit group of Z/(p^d) should be cyclic, so Z/((p-1)p^d-1)

[frail shoal]

ohhh that's it yea

[rich granite]

I appreciate the toki pona love in your bio

[frail shoal]

o moku e kala pona

[rocky cloak]

Ĉu iu ĉi-tie parolas Esperanton?

[frail shoal]

ala

[rich granite]

Toki Pina low-key does a better job at being an interlang than Esperanto I'm sorry 😭

[rocky cloak]

I guess both are crushed by English in that department, so not useful metric

[rich granite]

Crushed in the sense that English is already there. Not in the sense of actual practicality

[rocky cloak]

Being there is the practicality, no?

[frail shoal]

i like toki pona cuz i like minimalism and i think it's well-designed

when i was much younger i fantasized about building a minimal language

so when i grew up and realized that one exists already, i was like. wtf

it's similar to the reason i like algebra

[rich granite]

I used to make a bunch of conlangs with their own family trees and loanwords and shit

Don't have the time for that nowadays unfortunately

[frail shoal]

it feels like the most minimal math to me in terms of how many rules you need to establish first

[rich granite]

How did conlangs autocorrect to complaints

[frail shoal]

(tying this back into this channel)

[tulip otter]

Let $p$ be a prime, $u\in\mathbb C$ be such that $u^p=2$, let $\Phi_p(x)=x^{p-1}+x^{p-2}+\dots+1$ and $\zeta$ a root of $\Phi_p$. I want to find the degree of $\mathbb{Q}(u,\zeta)/\mathbb Q$. I am thinking that its $p$ but is that true?

[swift root]

universal algebra is then just making up your own rules lol

[cloud walrus]

ali yassine

[frail shoal]

admittedly there are rules to UA, which you learn very quickly when trying to explain to someone what a "term" is

[tulip otter]

Like at first I said that its p because the set {u,u\zeta,..,u\zeta^{p-1}} is a basis, but no element of Q can be obtained from a linear combination of these elements

[rocky cloak]

Well, definitely can't be p. Because zeta has degree p-1

[quiet pelican]

How do you get u^2 from those elements?

As a basis

[tulip otter]

you dont haha

[tulip otter]

ah right lol

ok let me rethink this

ok its p(p-1), is that right?

[rocky cloak]

Maybe you have an argument for why in mind

[tulip otter]

well I mean its Q(u)(zeta), u has minimal polynomial X^p-2 and zeta has minimal polynomial of degree p-1

[rocky cloak]

So then the degree is a multiple of both p and p-1 at least

And that seals the deal

[tulip otter]

right, but just saying this doesnt give the exact degree right?

I mean knowing that its a multiple of both means that it is np(p-1) for some positive integer n

[rocky cloak]

Well it would have to be some number between
lcm(p, p-1) and p(p-1)

[tulip otter]

ah true

I see, tysm jagr and mico

have a great day/night both of you!

[swift root]

so many options to choose from

[tulip otter]

one last thing, is it acceptable to just say that
[Q(u,zeta):Q]\leq deg(X^p-2)deg(Phi_p)=p(p-1) and that at the same time [Q(u,zeta):Q] is a multiple of p(p-1) since X^p-2 and Phi_p are the minimal polynomials of u and zeta respectively, or should I prove these statements/ justify in details

[rocky cloak]

I'd probably go with the
[Q(u, z):Q] = [Q(u, z):Q(u)][Q(u):Q] thing, but meh...

[tulip otter]

I see

[frail shoal]

why are yall sully face reacting

[swift root]

mathcord moment

[sudden condor]

yeah that looks sick

[frail shoal]

tbh i assumed i was just missing something

yall dont seem particularly toxic ?

[swift root]

idk lol

[mild parcel]

Hmm, maybe just no group visualization fans in here? Let me remove my original post...

[swift root]

have you posted it on eg r/math and the 3b1b subreddit?

[next obsidian]

https://tenor.com/view/klasik-clasik-clasic-gif-25398314

[azure cairn]

nvm dumb question

[dire wren]

On the second(1.21 b) qn at the bottom, am I missing something, or can I just take p = q = 1 and a = any element of h not 1.

[rocky cloak]

a is supposed to be taken from A not H, so no you can't just take that

It's also not given that 1 is in Q

[indigo vale]

what are some applications for groups in the real world?

[twilit wraith]

chemists use it a bit if i remember right

crystallography gets into some pretty hefty group theory in particular

[rocky cloak]

I guess cryptography is the big thing (see for example Diffie-Hellman key exchange)

In chemistry certain properties of molecules can be determined by the character table of their symmetry group.

Similarly in solid state physics there is some relationship between representations of the symmetry group of the Hamiltonian and observable states (I forget exactly how that works, but something like that)

[twilit wraith]

it wont help you do your taxes but its useful in the general STEM world

[rocky cloak]

I mean to do your taxes you may have to upload something over an https connection

So then you're using group theory

[twilit wraith]

very good point

[dire wren]

1 is a left factor of all words, so it is.

[dire wren]

Yeh, that was stupid mistake

[crystal vale]

what does it mean by M is divisble Z-module, i think it means for each m in M and for each n in Z, there exist m1 such nm1 = m

[quiet pelican]

Yes

[karmic moat]

From the Dynkin diagram of D_4
e4
|
e1 - e2 - e3
can I determine that e1 + 2 e2 + e3 + e4 is a positive root directly from the diagram? As in, without actually showing that <e1 + e2 + e3 + e4, e2> < 0, so adding the two gives me a positive root?

[rocky cloak]

I believe there's a theorem of Kac that the positive roots are exactly those positive integer vectors where the Tits form is <= 1.

So square of all nodes minus product of endpoints of edges, and indeed
1^2 + 2^2 + 1^2 + 1^2 - 1*2 - 2*1 - 2*1 = 1

(in the case of Dynkin diagrams it's when the Tits form equals 1, since it's never negative)

[karmic moat]

Oh interesting

[crystal vale]

So when we say R is a Noetherian ring, that means every ideal of R is finitely generated, as Z-modules?

[rocky cloak]

No, as an R-module / as an ideal

For example Q (the field of rational numbers) is Noetherian even though Q is not an fg Z-module

[crystal vale]

Okay

Thank you

[small yacht]

What is he referring to in the 2nd bullet point?
i don't think it's true for C_n since, for example in C_6, [4]_6 does not generate a subgroup.

[next obsidian]

Every element generates a subgroup

It’s what the definition of generate ensures

4 generates the subgroup {0,4,2}

[small yacht]

Why did I decide that it doesn’t
Ugh sorry my brain isn’t working rn

[next obsidian]

Joe mama