groups-rings-fields

[bleak abyss]

GL_n acts on flags, the stabilizer of this flag is the set of upper triangular matrices in GL_n, in particular our Sylow subgroup is the subgroup of this stabilizer which acts trivially on consecutive quotients

I think in fact this property characterizes Sylow subgroups of GL_n (in general algebraic group shtick the relevant term is "maximal unipotent")

I also think it's useful if you're doing certain number theory things? Group cohomology, mod p representation theory

[fading ibex]

Interesting. I don't think I understand the significance, but perhaps I will read up on it in time. In my leisure time I've been working through the subjects that I learned but did not master from when I was an undergraduate. My very first course as an undergraduate (as in, literally it was the first lecture on the first day as a student) was group theory and I did not understand very much of it besides the basic definitions.

[formal ermine]

if $\varphi : R \to S$ is a homo and $I$ is an ideal of $R$. what does $\varphi(I) = 0$ mean? specifically the 0?

[cloud walrus]

illuminator3

[thorn delta]

It just means the direct image of I under phi is {0}. i.e. I is a subset of the kernel of phi

i.e. phi maps everything in I to 0

[formal ermine]

0 is the additive neutral element?

[thorn delta]

yeah

[formal ermine]

okay

thanks

[thorn delta]

npnp

[wind steeple]

Yes but this fact in general isn't that obvious but for the case = 2 it's obvious

[delicate orchid]

Think about the kernel of characters

[long narwhal]

heyo- just wanted to ask a quick algebra question:
if you have two univariate polynomials, compute a sylvester matrix, then find the resultant of the two is zero, i know this means that these polynomials have a shared factor.
i'm trying to find the gcd of two polynomials that have such a resultant. is this shared factor always the gcd, or do i have to be a bit more rigorous and calculate the gcd with polynomial division?

[delicate orchid]

ker$(\chi) = {g \in G \colon \chi(g) = \chi(1)}$

[cloud walrus]

Wew

[delicate orchid]

this turns out to be equal to the kernel of the corresponding representation (I'm assuming you're working over C)

and the representation is a homomorphism, so you may see where this is going

that sounds right to me

[median pawn]

i'm just trying to show that this map is surjective

why is (c,d,N) = 1?

[chilly ocean]

Because ad-cb = 1+Nk for some k

No?

If p divides c, d, N then from above it divides 1

[median pawn]

Ahh yeah makes sense

Thank you!

[formal ermine]

what is an example of a polynomial ring over a non integral domain that is also not an integral domain

[unreal crag]

Hi

Question abt the advanced section of this server

Does A level further mathematics count as advanced

[elder wave]

remember that your Ring is embedded in your polynomial ring as constant polynomials

[formal ermine]

depends on the topic

[unreal crag]

Oh lol oki

[elder wave]

so any zero divisors in your ring will also be zero divisors in your polynomial ring

[formal ermine]

ah wait I phrased my question wrong, sorry

what is an example of a polynomial ring over an integral domain that is not an integral domain

that way

[elder wave]

every (advanced) channel has the topics that are supposed to be discussed in them in the info

[formal ermine]

can a polynomial ring introduce additional zero divisors?

no, right?

[elder wave]

try proving it

[formal ermine]

yes I'll try that later

[elder wave]

but yes you're right

it's not hard to arrive at a contradiction when assuming you have p*q=0 for two polynomials

[formal ermine]

wait so D[x] is always an id for an id D

[elder wave]

yes

[formal ermine]

ah I was like really confused lol

in the lecture he said K[x] is an id for a field K, so I wasn't sure why he didn't just let K to be an id lol

[elder wave]

dunno

i'd assume that it's because people are maybe more familiar with K[X] from lin alg

[formal ermine]

ah fair

I didn't take lina

[elder wave]

taking AA without lin alg can be tricky

[lunar ledge]

I found that what I learned in linalg (matrix theory) didn't really intersect with what I learned in abstract algebra

[chilly ocean]

At the basic level, it's more of a "mathematical maturity" thing than anything.

Linear algebra doesn't really come up in introductory abstract algebra.

[lunar ledge]

yeah

[elder wave]

fair enough i guess

[chilly ocean]

Yeah. But a dive into module theory gives similiar things

[elder wave]

my prof teaches basic abstract algebra in the preceeding lin alg course which makes the AA course a lot more interesting

yeah when studying modules it's especially helpful to have studied vector spaces extensively before

[celest cairn]

Have I done this correctly?

[elder wave]

there's a lot of context missing

[celest cairn]

I’m working with the Quaternions

[woeful sage]

you can use a quaternion multiplication table to check quaternion multiplication

[celest cairn]

okie

[woeful sage]

from wikipedia

[celest cairn]

yay it’s correct

[woeful sage]

D[x] is the total derivative of x

[formal ermine]

if K is a field then what does the notation $K^n$ mean?

[cloud walrus]

illuminator3

[chilly ocean]

Most likely the n-fold direct product of K with itself.

[cloud walrus]

sebbb

[formal ermine]

what does "n fold direct product" mean?

[pastel cliff]

.

[chilly ocean]

What the person above wrote.

They are trying to snipe me.

[formal ermine]

\times is the cartesian product?

[pastel cliff]

ignore the latex

[chilly ocean]

The set of all n-tuples of elements of K.

[formal ermine]

what's the difference between that and the cartesian product

[chilly ocean]

Nothing.

[formal ermine]

ah

oke thanks

I wasn't sure if there's a special definition for fields

[chilly ocean]

You can equip it with the structure of a vector space over K, as well.

[pastel cliff]

dont overthink it

[chilly radish]

implicit ring structure

'cartesian product' refers to just the set-theoretic product

in this case K^n is probably not equipped with a ring structure tho

[delicate orchid]

not quite

but for every normal subgroup N you can find a set of irreducibles such that the intersection of their kernels is N

the key thing to consider is lifting characters

don't worry about it then

[formal ermine]

what's a trivial nilpotent element (rings)? just 0?

[wispy pivot]

I'm trying to understand this statement. ``Let $P \in Syl_p (G)$ and assume $N \trianglelefteq G$. Use the conjugacy part of Sylow's Theorem to prove that $P \cap N$ is a Sylow p-subgroup of $N$"

So for e.g. let $G = C_6 $ (cyclic group of order 6), $P = \langle 2 \rangle$ (order 3), and $H = \langle 3 \rangle$ (order 2). $P$ is a Sylow 3-subgroup and $H$ is normal since $C_6$ is abelian. So $P \cap H = {0}$ is a Sylow 3-subgrp in $H$. Am i missing something or is this correct?

[cloud walrus]

HimmyHow

[agile burrow]

That's correct. Since H has order 2, the largest power of 3 which divides it is 1. Thus, any 3-Sylow subgroup will have order 1, hence it will be trivial.

[wispy pivot]

Gotcha that makes sense, thank you very much.

[formal ermine]

bump

[pine patio]

why do we need K to be the splitting field of some separable polynomial? can't it just be any polynomial?

[rustic crown]

nope, separability is important

[pine patio]

i guess it has something to do with characteristic of F

[rustic crown]

the standard non-separable example is F = F_p(x), and the polynomial is t^p - x

[pine patio]

when it says D(f) is not 0 does it mean it's not identically 0?

[rustic crown]

yep

[pine patio]

ah that makes sense now

[rustic crown]

the problem in this case is that the splitting field is F_p(x^(1/p)) and the polynomial splits as (t - x^(1/p))^p

so the roots don't separate

which is why you get fewer automorphisms than the degree of the extension

[pine patio]

yeah i see now thanks

[main needle]

Does the symmetry group of the octahedron act faithfully on the set of opposite faces ?

[elder wave]

i guess?

[pastel cliff]

why are you asking this?

not a diss just seems like it might not be terribly useful to ask

[thorn delta]

prob written in a book without explanation or something

[lavish sigil]

Let $k$ be a field, and consider the map $k[x,y]\to k[x,x^{-1}]=k(x)$ mapping $y$ to $x^{-1}$ (and leaving $x$ and elements of $k$ intact). Formally, how would I show that the kernel of this map is precisely the ideal $(xy-1)\subset k[x,y]$?

[cloud walrus]

gustavn64

[noble saddle]

FYI, k[x, x^{-1}] and k(x) are not the same thing

For instance, 1/(x+1) is only in one of those rings

But really, there's no trick. Write down an arbitrary element of k[x,y], sub in x^{-1} fpr y and think about what conditions there are on the powers to get 0

[lavish sigil]

I realized that now

[sharp peak]

Is there some significance as to why the prof chose this sequence? Or am I looking too far into things

The question itself is from an AG class and is straightforward

I guess its a projective resolution of Z/pZ in homological algebra but why would anyone use this resolution instead of a resolutions of Z's

[next obsidian]

I don’t think this is a projective resolution

Yeah it isn’t, this can’t embed into Z because this has torsion

Anyway, I thinks it’s just because it’s funny?

Like it’s just an infinite resolution of the same thing over and over

Sometimes actual free resolutions look like this

[sharp peak]

Can you elaborate on why it not being embedded into Z implies it isn't a resolution

[next obsidian]

Well it’s a resolution yes

[sharp peak]

and thanks

[next obsidian]

But not by projective modules

A projective module is a direct summand of a free module, but a free module hs no torsion

Or alternatively, projective modules are flat which have no torsion

[sharp peak]

ah yeah thats right; projective implies flat implies torsion free

[cloud walrus]

Raghuram

[next obsidian]

Whew

are you okay with a reference?

this is an entire section of Bourbaki Algebra 1

I'll take a photo

this is the statement for the specific case of the dual

[tough raven]

looool there's a section called “the homomorphism from V* (×) W to Hom(V, W)”?

[next obsidian]

Yes

[tough raven]

What module structures do E, F, G have? That seems to have been specified on the previous page.

I'm guessing E is left-A, G is left-A, right-B and F is left-B?

[next obsidian]

Yeah this is general

for non-comm rings, it's just what makes it work

[tough raven]

Ye I understand

[next obsidian]

G is actually a bimodule

to be able to tensor Hom_A(E,G) over B

[tough raven]

Yes by left-A,right-B I meant bimodule

[vestal snow]

What is the least painful way to show $\mathbb{C}[x]\otimes_\mathbb{C} \mathbb{C}[y]$ and $\mathbb{C}[x]\otimes_\mathbb{Z} \mathbb{C}[y]$ are not isomorphic? I was thinking of showing that there is some $\mathbb{C}$-module P such that $Bil_\mathbb{C}(\mathbb{C}[x] \times \mathbb{C}[y], P)$ is empty and $Bil_\mathbb{Z}(\mathbb{C}[x] \times \mathbb{C}[y], P)$ is non-empty.

[cloud walrus]

Finitely Many Bananas

[next obsidian]

The former is C[x,y]

[vestal snow]

Yeah

[next obsidian]

This should probably make it pretty easy tbh

[vestal snow]

I actually went the opposite direction lol

[next obsidian]

lol

[vestal snow]

I thought writing it as a tensor product might make it easier

[next obsidian]

in what category do you want to show they're not iso?

as modules or as rings

[vestal snow]

Rings

Though showing that they are not iso as Z-mods would imply that

[next obsidian]

yeah

I'll be honest

[tough raven]

You know that in the former ix (×) y = x (×) iy but would expect not in the latter.
So maybe find a Z-bilinear map out of C[x], C[y] which disagrees at (ix, y) and (x, iy)

[next obsidian]

I don't think the latter is an integral domain

[tough raven]

which would show they are different as abelian groups

[next obsidian]

or Noeth

it seems way too big

[tough raven]

(actually more like you can't find an iso that respects the bilinear map of C[x], C[y] into both, but who wants an isomorphism that doesn't)

Or do you want to rule out any isomorphism

[vestal snow]

any isomorphism

Because they goal is to show that they are not isomorphic as rings

[tough raven]

That is sad

[vestal snow]

This is the original formulation of this problem btw

[next obsidian]

consider this

[vestal snow]

Since we're doing stuff involving Spec(Z), I don't even have a structure morphism to Spec(k)

[next obsidian]

(i (x) 1) - (1 (x) i))((i (x) 1) + (1 (x) i))

do the computation and you see this becomes 0

I think it's very unlikely that either of these are 0 when you tensor over Z

basically since you shouldn't be able to move the i across I think

If you can show this, then you have demonstrated a zero divisor in C (x)_Z C

[vestal snow]

True that could work

Though showing that the two factors are not 0 is probably really messy

[next obsidian]

I think you can do it like this

[tough raven]

They're not equal in C (×)_R C I think

[next obsidian]

find a Z-linear map C x C -> A such that they disagree at i (x) 1 and 1 (x) i

[tough raven]

where you can show that very directly from bases

[next obsidian]

so actually what Raghuram originally suggested

[tough raven]

which is a quotient of C (×)_Z C

[next obsidian]

I can turn into a proof that works to disprove they're iso as rings

now admittedly I can't tink of a single Z-linear but not C-linear map

from C into anything

[vestal snow]

Yeah lol

[next obsidian]

maybe project down one of the factors?

like

the imaginary part

[tough raven]

R-linear but not C-linear is easier

because everything is free module

[next obsidian]

I mean

R-linear is also Z-linear

lol

so that is a strictly harder problem

[tough raven]

C (×)_R C has basis 1 (×) 1, 1 (×) i, i (×) 1, i (×) i

[next obsidian]

wait lol

he's right

this is true by the most basic field theory

[tough raven]

Stronger but not harder, because there's a basis
Is what I'm trying to say

[next obsidian]

and yeah C (x)_R C is a quotient of C (x)_Z C

which sends a (x) b to a (x) b

[next obsidian]

so this tells you that the things in my product aren't zero

if they were zero over Z, they'd be zero over R

[vestal snow]

Is it?

[next obsidian]

but these things form a basis

Yes

recall the construction of these

it's the free module on C x C

modded out by relations making things base ring linear

or uhh bilinear or whatever

[vestal snow]

Won't it be the free R-mod in (x)_R and free Z-mod in (x)_Z?

[next obsidian]

Uhhh

Okay well you can surject onto it right?

Like

The map C x C -> C x C -> C (x)_R C

[vestal snow]

I think so

Yeah

[next obsidian]

Is Z-bilinear

[vestal snow]

Ohhhh that makes sense

Yeah that works

[next obsidian]

It induces a surjection because you know on simple tensors it sends a (x) b to a (x) b

[formal ermine]

because my homework says: "show that R/N has no non trivial nilpotent elements" or something like that

[next obsidian]

So by linearity you’re good

[vestal snow]

Lmao we should have thought of that

[formal ermine]

but we never defined what that means

[next obsidian]

Raghuram is smart

[vestal snow]

The R trick was super cool

[next obsidian]

[tough raven]

Or even the universal property:
if f1, f2 are bilinear maps into tensor product over Z, R resp.
then f2 is Z-linear => factors through f1 by say π

You can show π is epi because if g1 π = g2 π, then g1 f2 = g1 π f1 = g2 π f1 = g2 f2 => g1 = g2 by universal property again
Then epi is surjective
or directly surjective since tensor product over R is generated by things like z1 (×) z2 all of which are in the range of π

[tough raven]

Hmm
I think this doesn't matter because C is an R-module
so r(z1 (×) z2) = (r z1) (×) z2 (in tensor over R) and the latter is in the tensor prod over Z

[vestal snow]

Thanks!

[sharp peak]

Super stuck on these questions:

I am sure knowing 12 you can do 13 and vice versa

but err... hmm. Is this a well known result or something?

Right yeah 12-2 is Lazard's lemma

[next obsidian]

(1) virtually amounts to the fact that any of these are set theoretically a union of its finitely generated sub-things

(2) is harder because you no longer can use sub-things

[sharp peak]

I think the intuition is pretty clear. But what I am stuck on is the actual details of the proof

Or even a sketch of what a proof like this should look like

[next obsidian]

think about how to prove that a map from a set X -> Y is exactly the same as a collection of maps X_i -> Y which agree on intersections where {X_i} is some cover of X

The proof will be virtually the same for the algebraic case

[pastel cliff]

[next obsidian]

And will exhibit your module M as a universal cocone of the system of finitely generated submodules

[tough raven]

BTW is remark (1) here correct?
(Proposition 12 says E is projective iff for any generating set {α_i}_{i in I}, there is a corresponding family of linear form {β_i} such that for any x, x = Σ_i β_i(x) α_i, this sum being finite for each x. This would seem to say (Σ_i β_i (×) α_i)(x) = x, except that the latter is not a well-defined element in the tensor product.)

In the case of vector spaces over a field, I know that 1_E is in the range when E is finite-dimensional but is not when E is infinite-dimensional.

Should the remark say this characterises being a finitely generated projective module? Or that it is projective iff 1_E is in the closure of the range under pointwise convergence?

[next obsidian]

Idk

[elder wave]

Then yes, show that your only nilpotent element is 0, N is the nilradical or just some radical ideal i guess

These rings are called reduced in case you're curious

[cloud walrus]

bwpvbzz

[gusty thistle]

how do i delete this embarssement hehe ($\mathbb{Q}(\zeta_10) = \mathbb{Q}(\zeta_5)$ lol

[chilly ocean]

you can't

[quiet pelican]

Proof that that V^{\chi_i} is a subset of that
Suppose that there is some v, g with gv =/= chi_i(g)v. Then span(v) is an irreducible sub rep not isomorphic to chi_i

And we clearly see that that is a subset of V^{\chi_i}

So they are equal

[white yoke]

Does anyone have a pdf of Introduction to Algebraic K-Theory by John Milnor?

[wintry pasture]

I've working on this one for a bit now and come to the conclusion that $|G_\gamma| = k_1|\Delta|$ and $|G_\delta|=k_2|\Gamma|$ for some $k_1,k_2\in N$ as well as that if I manage to prove that $G_\gamma$ acts transitively on $\Delta$ and vice versa I'm done. Could someone here give me a hint or tell me that i am totally on lost tracks?

[cloud walrus]

Philip

[pine patio]

is this correct: Q in Q(alpha) in Q(alpha, beta) where alpha, beta are distinct roots of t^3 - 2

do i only need to show that Q(alpha) isnt the splitting field of any polynomial in Q[t]

[next obsidian]

It’s a little bit easier though, you just need to find a polynomial with a root in Q(alpha) but not every root lives there

[pine patio]

hmm how does that show there isnt another polynomial that splits over Q(alpha)

are you saying that if an arbitrary g has a root in Q(alpha) then it doesnt contain all of its roots?

[thorn delta]

finite galois extensions are normal and separable, so in other words u just need to show that Q(alpha)/Q is not normal: that there exists a polynomial f in Q[x] which has a root in Q(alpha) but does not split in Q(alpha)

[pine patio]

im using this definition

[next obsidian]

One can show any splitting field is normal

i.e. if f is polynomial with a root in it, all its roots lie in it

well, irreducible polynomial

[woeful sand]

is bourbaki (algebra books) a good read?
my taste: i don't like extensive and computational exercises/proof

[next obsidian]

no

I like them because I'm insane, and already know a lot of algebra

it's not a textbook you read to learn something

[woeful sand]

can you give me a couple of recomendations

im always looking for new books to read

[next obsidian]

there's plenty of them pinned in #book-recommendations

[woeful sand]

oh, sorry i didn't remember

[next obsidian]

Dami made a big post

if you want Bourbaki-lite then try Lang

[woeful sand]

i keep jumping from book to book, i can't finish a single one lol

i understand what you're saying

[next obsidian]

if that's the case it sounds like the book isn't the issue

to be honest

I think you jsut have to commit to working through a book

[woeful sand]

no, there's no issue, i like to do it that way

[pastel cliff]

what the fuck is it for then

[next obsidian]

references

or when you need to know really really general theorems about a specific topic

[pastel cliff]

[coarse forge]

so I'm trying to prove QR using Gauss sums and I've come to a roadblock; part of the way to proving the result is to show that [\sigma(G) = \left(\frac{p^}{q} \right) G,] where $G = \sum_{a = 0}^{p - 1} \left(\frac{a}{p}\right) \zeta^a$, $\sigma : K \to K$ is the $q$th power Frobenius (automorphism) endomorphism over a field extension $K / \mathbb F_q$, and $p^ = \left(\frac{-1}{p}\right) p$. The problem asked us to use the fact that the fixed field of $\sigma$ is $\mathbb F_q$ and the fact that $G^2 = p^*$ to prove the result, but I haven't really used the fact that the fixed field of $\sigma$ is $\mathbb F_q$ to prove the result, was wondering how I'd go about it

[chilly ocean]

why is this ..

[pastel cliff]

can someone explain the ascending chain condition to me like im 5

(for rings)

is it just a restatement of the fact that there is always a maximal ideal in a ring with identity

[next obsidian]

no

that holds always

the ascending chain condition does not

prove that it's equivalent to every ideal being finitely generated in the ring

[pastel cliff]

i saw that they're equivalent (haven't fully understood proof yet) but what does ACC give that that ^this^ doesnt

[next obsidian]

That every ideal is finitely generated

Really it just tells you that any collection of ideals contains a maximal element

[pastel cliff]

[next obsidian]

This is true via Zorn’s when your poset is “all ideals”

ie maximal ideals exist

[pastel cliff]

okay this is still the same dumb question but - there is always a maximal ideal in a ring with identity \neq any collection of ideals contains a maximal element???

oh i guess yeah

[toxic zephyr]

we learned today that the rational numbers as a Z-module cannot have any linearly independent set of two or more elements. won't that always be true for any ring R as an R-module? since {r,s} has the R linear comb. (s)r+(-r)s=0?
does this generalize to if I is an ideal of R, the same will be true of R as an I-module? just let a be any element of I and then (as)r+(-ar)s=0?

i guess R would have to be commutative

[pastel cliff]

also how do yall remember the diff between prime vs. irreducible elements in an integral domain

[delicate bloom]

all primes are irreducible, but not all irreducibles are primes

[toxic zephyr]

except in a PID

[delicate bloom]

just speaking generally here

[toxic zephyr]

it'd probably be good for me to have an example of an irreducible elt that isn't prime off the top of my head, but i'm coming up blank. probably something in Z[x]

[pastel cliff]

2 in Z[sqrt(-3)]

but i only know that cuz it's in my lecture notes

[toxic zephyr]

oh yeah totally. can't believe i forgot about that, i had an assignment on that last week lol

[toxic zephyr]

<@&286206848099549185>

[thorn delta]

won't that always be true for any ring R as an R-module
yes, if R is commutative, R is a rank 1 free R-module, so any set containing more than 1 element would be linearly dependent just like in linear algebra. However, this example is a little different than Q as a Z-module because Q is not even free (i.e. a direct sum of copies of Z)

does this generalize to if I is an ideal of R, the same will be true of R as an I-module? just let a be any element of I and then (as)r+(-ar)s=0?
you at least need as != 0 and ar != 0. Also, when you start talking about modules over (noncommutative) rings without identity and weird stuff like that, certain stuff may become less obvious/false. For example if {r,s} are linearly dependent, i believe there still could be a set of three elements for which are linearly independent over R as an I-module.

[bleak abyss]

This channel has gone without p-adic spherical functions for too long

@tropic spade proud?

So G is a locally compact Hausdorff group

C(G) is continuous functions G->C

L(G) is compactly supported shit (I'm sorry it's the notes)

If K is a compact subgroup, C(G,K) is the space of K bi-invariant functions on G

L(G,K) is compactly supported shit in that guy

Fix a Haar measure dg on G

And then L(G,K) becomes an associative C-algebra under convolution

[cloud walrus]

Sloth King Daminark

[tropic spade]

I retract my comment about you being a nerd and a coward. I now live in fear of your wrath.

[bleak abyss]

So obv if you replace g_0 with kg_0, then f_1 being left bi-invariant means nothing changes

On the other side, do a change of variables h = kg

Since we're Haar measure this just becomes

[cloud walrus]

Sloth King Daminark

[bleak abyss]

But then f_2 is right K-invariant so yea

Now already if one of these guys has compact support, then for any g_0 this integral is finite

f_2 is supported on E then we just integrate over E^{-1}, f_1 is supported on E then this is integrating over g_0E

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Now if K is open

Things get interesting

Let's see why this is all true

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Okay I mainly need to think about this normalization

I guess let's try

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Chalkboard time probably brb

Okay change of variables is the answer

So resetting

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Good

That shouldn't have taken me this long lol

Now for the projection

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Dude idk what this symbol is I'll replace it with a p

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Let's say phi is supported on the set E, well k_1gk_2 in E when g in KEK (lol)

[wooden ember]

What is the dami rant on today

[bleak abyss]

And that's compact so phi^p has compact support

And then we're integrating on compact shit

Point is

[cloud walrus]

Sloth King Daminark

[bleak abyss]

So first equality is change of variables I think?

Wait we only have a left Haar measure this is sus

Okay one at a time

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Yeah k_2 on the right is a bit sus ngl

Next equality feels like cheeky Fubini

And then phi^p (yeah that's what I'm calling p the screwy thing in the line above "by hypothesis" deal with it) is in L(G,K) so yea

Now the point is that given any continuous function, including omega circ inverse

If its integral against all continuous functions of compact support is 0 it's gotta be 0

I guess I'll spell it out in the complex setting

If it's non-zero it's non-zero on some open compact neighborhood

\overline{omega(g^{-1})} \chi_{some compact translate of K}

gg

Now spherical measures/functions

Yeah okay Macdonald is full of shit

Maybe in the first couple definitions he didn't assume unimodularity

But no shot he's not assuming it now

The thing I just did needs that

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

And then this is a spherical measure

So let's see

If G is unimodular I see why the first point holds

As for the second

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Do a Fubini swap and we win

So now

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

So now

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

The key point was that if you're already in L(G,K), f^p = f

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Now Fubini swap pull out f(h_1) and you get their thing

Ah okay so we are assuming G is locally compact, Hausdorff, and second countable

So it is sigma-compact

And these guys are taking only measures on G for which compact sets have finite measure (this includes Haar measure, and this is the "continuity hypothesis" they speak of earlier)

[quiet pelican]

If it’s not solved when I get back from lectures, I’ll prob help with this

[bleak abyss]

Obv f is compactly supported so yea Fubini is gucci here

So now

He meant mu(f) = \int_G f(g_1)omega(g_1^{-1}) dg_1 in the next line

Now omega is left K-invariant because f_0 is and is right K-invariant because mu is spherical. And continuity I won't think about it's some dumb bs

Oh another relevant point is

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Should be Fubini again

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Inner integral is just mu(f) because mu is spherical, then use that Haar measure is normalized

Anyway back to this

[cloud walrus]

Sloth King Daminark

[bleak abyss]

And the point is that K is compact, hence unimodular, so change of variables + maybe thinking of Fubini = I can swallow the k_1 and k_2 in omega and use bi-invariance, then yeah normalized and shit

So

[cloud walrus]

Sloth King Daminark

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Now

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Now

[cloud walrus]

Sloth King Daminark

Sloth King Daminark

[bleak abyss]

Whoops trailed

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Last equality is change of variables and Fubini

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Alright yeah I'm at this point gonna probably pause the rant since I'm better off not trying to go through shit systematically lol

[elder wave]

[bleak abyss]

Rather I'll just use my scratch work server for scratch work

What's up Timo

[elder wave]

Nothing

I just checked this channel during my lecture 4 hours ago

Now I checked again and was like damn dami is still at it

[bleak abyss]

Oh I meant not what's up = what's wrong but "how you doing fam"

[elder wave]

Im good mate wbu

Had a really interesting lecture about intro AG earlier

[bleak abyss]

Nice, in algebra class?

[elder wave]

Yeah in my computer algebra class

What are you trying to figure out btw

[bleak abyss]

Oh just working through some notes on spherical functions

[elder wave]

I see I see

Good luck with that

[bleak abyss]

Thanks fam

[wooden ember]

This is a dumb question but do irreducible elements in a UFD generate a prime ideal (I mean multiple such elements)

Of course it’s true for a single one

[bleak abyss]

Hmm

k[x]

x-1 and x-2

[wooden ember]

They’re coprime so it’s fine no?

[bleak abyss]

That means they together generate the whole ring

[wooden ember]

Yeah exactly

[bleak abyss]

Which isn't a prime ideal

[wooden ember]

Wut am I being dumb

How is the whole ring not prime

Isn’t it vacuously prime

[bleak abyss]

Convention is to manually exclude it

"A prime ideal is a proper ideal such that..."

[wooden ember]

Huh how did I not remember that

Well that won’t be an issue for my case, do you know of an example where we get a proper non prime ideal?

[bleak abyss]

Hmm, so in a PID prime ideals are maximal

Meaning two irreducible things already generate the whole ring

[wooden ember]

(If I’m more specific to my case I’m looking for the case of a multi variate polynomial ring)

[bleak abyss]

Can you specify the full context here actually?

[elder wave]

Oh I saw an example of that a while ago I think

When I was browsing for something else

Let me see if I can find it!

[wooden ember]

Yeah o just have an ideal in C[x_1,…,x_4]

And I’d like to get its radical

But it looks very prime to me

[bleak abyss]

Which?

[wooden ember]

And it’s generated by irreducibles

(x_1x_4-x_2x_3, x_1x_3-x_2^2, x_2x_4-x_3^2)

[elder wave]

https://math.stackexchange.com/questions/1807526/do-prime-elements-generate-a-prime-ideal/1807596#1807596

[cloud walrus]

Sloth King Daminark

[bleak abyss]

For readability lol

[wooden ember]

Gimme a few minutes I have to walk to the hospital I’ll be back in 10m

[bleak abyss]

So thinking about this in an AG way

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Hmm let me play around here

[cloud walrus]

Sloth King Daminark

[wooden ember]

Ok im back

So the vanishing locus is (u^3, u^2v, uv^2, v^3)

This is how I got this ideal in the first place

And now I’m trying to compute I(vanishing locus) which Id like to think is just the ideal I already have

[bleak abyss]

Oh

[wooden ember]

Thats a shame

[cloud walrus]

Sloth King Daminark

[wooden ember]

Ah and we’d like to show our ideal is its kernel?

That’s smort

Does it actually make things easier though

[bleak abyss]

Image is subring of C[u,v] which is integral domain

[wooden ember]

Yeah I agree but I’m not sure computing the kernel is going to be easier

https://math.stackexchange.com/questions/95217/methods-to-check-if-an-ideal-of-a-polynomial-ring-is-prime-or-at-least-radical

This post has a similar question (very similar)

But the answers either use dimension theory or localization which we haven’t covered yet

And it honestly seems super painful

[quiet pelican]

Figuring it out

[quiet pelican]

Here’s b, c is seeming much harder

Intuitively for c, it’s averaging over the action of the component of rho_i, and when we average again, that can’t change it, but I just can’t seem to get the constants to work out

[quiet pelican]

Suppose that v is in some irreducible subrepresentation isomorphic to rho k

yes

because it gives the inner product of two distinct irreducible characters

[quiet pelican]

It’s basically identical to the proof for G^1, but you just have to deal with more constants

[chilly ocean]

Let <g> be a cyclic group of order n. Let i and j both divide n. Prove that <g^i> subset <g^j> iff j | i

hint please

[maiden heath]

Hi, i'm a bit stuck on this question. I don't see how self dual helps

[glossy crag]

<= direction is clear, for the converse divide i by j with remainder

Checking to see if the following proof of "for any n, Q/Z has only 1 finite subgroup of order n" is correct:
Let x in Q/Z be arbitrary, pick a representative in [0,1) and write it in lowest terms, then the order of x is the denominator. Because of this, if (x) is a cyclic subgroup, then it must be of the form (1/n) where n is the denominator of x. Now let G be an arbitrary finite subgroup of order n, bring all its elements to the same denominator m, then G is a subgroup of (1/m) and thus itself cyclic. But we've shown the cyclic subgroups are unique, so G is of the form (1/n).

[chilly ocean]

Hey, I don't see how one can conclude that G/N is generated by x bar here. Couldn't some relations get in the way?
(This is part of a non-rigorous solution given out by my professor)

[next obsidian]

Anything in G can be written like some combination of x’s and y’s (definition of being generated by x and y)

In the quotient you can replace those y’s with e

Since you quotiented out by y

So anything in the quotient can be written as a bunch of x’s

[chilly ocean]

Not sure I understand. Firstly, G is generated by x,y, and z - can the z's be replaced somehow?

[next obsidian]

Oh sure

But z is also now e in the quotient

Since you modded out by N which was <y,z>

Sorry, I misread slightly

But the same idea holds

[chilly ocean]

But then wouldn't one need to get the z's out the the right to make them e?

[next obsidian]

I’m not sure what you mean

So you know z-bar = e-bar

And y-bar = e-bar

So say you had some expression like say

xz^2yzy^2

This is some arbitrary element in G

When you go to the quotient

Every instance of z becoems a z-bar

So you can replace it with e-bar

Similar for y

So this becomes x-bar(e-bar)^6

= x-bar

[chilly ocean]

Ah, I see. Because the map is a homomorphism, so we can just distribute.

[next obsidian]

Yeah

Exactly

[chilly ocean]

Ah, makes total sense now. Thanks so much.

[pastel cliff]

just wanna check this proof pls, that last sentence feels eh but i think my logic is mostly sound?

also im being asked to prove that for any ideal $I \subseteq \Z[i]$, the quotient ring $\Z[i]/I$ is finite - im gonna assumed that's i = sqrt{-1} right?

[cloud walrus]

sebbb

[formal ermine]

what was the theorem called again that says if $|G| = pq$ for primes p,q then G is solvable

[cloud walrus]

illuminator3 (I/you)

[formal ermine]

I remember its name starting with a b(?)

[rotund aurora]

https://en.m.wikipedia.org/wiki/Burnside's_theorem

Literally copied and pasted what you said in google

[pastel cliff]

actually a hint for this would be cool and good

mmmm methinks correspondence theorem

[formal ermine]

aaa thanks

[formal ermine]

google search results are personalized lol

I'm getting this

[rotund aurora]

I think it is suficient to show that the ideal I will contain an integer

Then the result is easy

In fact, I think that holds for any ring of quadratic integers

[pastel cliff]

im not sure i see how sorry

as in, why that would be sufficient

[rotund aurora]

So, suppose your ideal is (2). Evaluate Z[i]/(2)

btw, the correspondence theorem is actually helpful if you don't see it immediately

but maybe is more confusing.

[pastel cliff]

yeah correspondence thm confuses me lol, i get the statement but i have a hard time using it

i'll be back tho, i had started another q in the mean time, wanna finish that one

[rotund aurora]

I think Artin does a very good job at conveying the importance of the Correspondence theorem

One nice corollary is that $R/(a,b)\approx (R/(a))/(\overline b)$ where $a,b\in R$ and $\overline b$ is the reduction of $b$ modulo $a$

[cloud walrus]

Croqueta

[rotund aurora]

So esentially, the order of the quotients doesn't matter

[glossy crag]

Yello?

[maiden heath]

Hiya, still struggling on this problem. I'm not sure how self dual helps

[hollow mica]

We say that V ⊗ W is naturally isomorphic to V ⊗ W** because of the natural isomorphism between W and W**, but can the same be said about V ⊗ W* ?

I know the answer is no, but why not?

Sure, the isomorphism between W and W* requires a choice of basis, but the tensor product doesn't require the choice of a basis

[chilly ocean]

If V is the base field, then you'd just be asking if there's a natural isomorphism of W with its dual.

As you know, this is not the case.

[chilly ocean]

If W is any vector space over k, then k ⊗ W = W. To prove this, take the map k ⊗ W -> W given by c ⊗ w -> cw and show that it's an isomorphism.

The simplest way is to simply write down the inverse W -> k ⊗ W.

[hollow mica]

(a_1 w_1 + ... + a_n w_n) -> ((a_1 ⊗ w_1) + ... + (a_n ⊗ w_n))

sure

how does this connect to my original question

[chilly ocean]

You do not need to choose a basis to write down the inverse.

[hollow mica]

oh

uh

[chilly ocean]

...

The inverse is w -> 1 ⊗ w.

So k ⊗ W = W.

[hollow mica]

oh I see

[chilly ocean]

W = k ⊗ W is not naturally isomorphic to W* = k ⊗ W*.

Take V = k and you have your exact original question.

This is how W = k ⊗ W is related to your original question.

[hollow mica]

oh I see

you're saying the same argument can be repeated if k is no longer the base field but any vector space V over k

[chilly ocean]

I am saying that V ⊗ W is not naturally isomorphic to V ⊗ W* in the case that V is simply the field that it is a vector space over.

For then these tensor products are W and W*, respectively, between which you know there is no natural isomorphism.

[hollow mica]

I wanted to see why it's not naturally isomorphic for an arbitrary V

but I guess there is no reason for why it should be

[chilly ocean]

It is for V = {0}.

For the rather silly reason that V ⊗ W and V ⊗ W* are then both {0}.

So you've got an example where they are naturally isomorphic and an example where they aren't.

[hollow mica]

I am at peace

[chilly ocean]

You might ask if for all non-trivial V the two are not naturally isomorphic. That, I do not know.

[chilly ocean]

Teach me your ways.

[chilly ocean]

But I suspect it is the case. You might get something like k^n ⊗ W = W^n and k^n ⊗ W* = (W*)^n = (W^n)*, so you'd then be asking the same question of whether there's a natural isomorphism between W^n and its dual. Same thing.

(I haven't checked the isomorphisms here, don't hold me to them just yet.)

[hollow mica]

I am sort of not at peace with V ⊗ W = V ⊗ W** though. Is it just "obvious" because W = W**

[chilly ocean]

Tensor the identity map of V with the isomorphism W -> W**.

This is the desired isomorphism.

[hollow mica]

I want a (bijective) linear map V ⊗ W -> V ⊗ W**. Tensoring, this is the same as asking for a (bijective) bilinear map V x W -> V ⊗ W**,

[chilly ocean]

(v, w) -> v ⊗ ev_w does the job.

ev_w taking an element of W* to its value at w.

[chilly ocean]

This is what I meant by this message.

[hollow mica]

ok I understand this isomorphism

but I don't understand how that's the same as "tensoring the identity map of V"

I thought you could only tensor bilinear maps into linear maps and vice versa

like a bilinear map V x W -> U gives rise to a linear map V ⊗ W -> U (universal property)

how can you do anything with the identity map V -> V

[chilly ocean]

I'm talking about considering the map id ⊗ phi (where I use phi to be the isomorphism w -> ev_w).

I'll explain the general idea.

[hollow mica]

that tensor is an element of Hom(V, V) ⊗ Hom(W, W*) ?

[chilly ocean]

If $T_1\colon V_1 \to W_1$ and $T_2\colon V_2 \to W_2$ are linear maps of vector spaces all over the same field, then the map $$V_1 \times V_2 \to W_1 \otimes W_2, \qquad (v_1, v_2) \mapsto T_1(v_1) \otimes T_2(v_2)$$ is bilinear. By the universal property, it induces a linear map $V_1\otimes V_2 \to W_1 \otimes W_2$, which is usually denoted by $T_1 \otimes T_2$.

[cloud walrus]

rakko

[chilly ocean]

What I meant by "tensor the isomorphism of W and W** with the identity map of V" was to apply this construction to id: V -> V and phi: W -> W** to get the map id ⊗ phi: V ⊗ W -> V ⊗ W**.

[hollow mica]

ohhhhhhhhhhhhhhhhhhhh

i didn't know the tensor product of linear maps was a thing

[chilly ocean]

It's easy to check that if T_1 and T_2 are invertible, then so is their tensor product as defined above.

The inverse should be obvious.

[hollow mica]

is this the same as the tensor product when we treat T_1 and T_2 as elements of Hom(V_1, W_1) and Hom(V_2, W_2)

wait T_1 ⊗ T_2 is a linear map between tensor product spaces

that's completely different, so nvm

[chilly ocean]

I think it's a good question to ask.

Yes.

(Up to natural isomorphism, of course.)

[hollow mica]

oh

[chilly ocean]

Let me write it out.

[cloud walrus]

rakko

[chilly ocean]

I'm sure if you actually write out the maps, you'll see that the tensor product of linear maps I defined above coincides with their tensor product as elements of vector spaces.

(The tensor product I defined above lives in the last vector space in this chain of equalities, and the actual tensor product lives in the very first.)

So maybe I should add the disclaimer that this isn't really a proof, but it makes you think it should be true.

[hollow mica]

wait why isn't this a proof

[chilly ocean]

I never actually proved that the map T_1 ⊗ T_2 that I defined above really is the tensor product of T_1 and T_2.

I just showed that the spaces they live in are isomorphic naturally.

[hollow mica]

oh I see

[chilly ocean]

This is actually me coping because I wrote something down too quickly and realized it didn't actually answer the question.

[hollow mica]

Ok if it wasn't true then the definition of T_1 ⊗ T_2 would just be a big scam

[chilly ocean]

It's true because it would really suck if it were false.

This kind of reasoning has gotten me pretty far.

All you really have to do is show that the map $\mathrm{Hom}(V_1, W_1) \otimes \mathrm{Hom}(V_2, W_2) \to \mathrm{Hom}(V_1 \otimes V_2, W_1 \otimes W_2)$ given by $$T_1 \otimes T_2 \mapsto \underbrace{((v_1 \otimes v_2) \mapsto T_1(v_1) \otimes T_2(v_2))}_{\text{the map I defined before!}}$$ is an isomorphism.

[cloud walrus]

rakko

[hollow mica]

right

[hollow mica]

wait you write "all you really have to do", but that's just you converting this in equation form right

like that is what you were saying basically

[chilly ocean]

Kinda.

I just wanted to write down the explicit isomorphism.

[hollow mica]

I see

Last question

I don't understand the definition of tensor algebra

and I'll write down why:

If $V$ is a vector space, we define the \emph{tensor algebra} of $V$ to be
$$ V = \bigoplus_{d = 0}^{\infty} V^{\otimes d} , $$
where $V^{\otimes d} = \underbrace{V \otimes \dots \otimes V}_{d~\text{times}}$.

[chilly ocean]

Do you mean otimes?

[hollow mica]

oh yeah

[chilly ocean]

It should be the direct sum of all the V^{otimes d}.

[hollow mica]

that's a scary looking otimes

ok so anyway

[chilly ocean]

Forget the bigotimes thing, my bad.

I meant bigoplus.

[hollow mica]

oh

right it's the direct sum

[chilly ocean]

I got confused about what you were trying to do for a moment, my bad.

Underscore before the {d = 0}.

[cloud walrus]

mrean

[hollow mica]

what the fuck is wrong with me

ok

so

[chilly ocean]

There we go.

[hollow mica]

an element of the tensor algebra

[pine patio]

quick qu: is Gal(K/F) defined for arbitrary field extensions K/F or does the extension need to be Galois (ie K is the splitting field of a separable polynomial in F[t]) ?

[hollow mica]

looks like

$(v_1, v_2, \dots)$ where $v_d \in V^{\otimes d}$

[cloud walrus]

mrean

[chilly ocean]

For the infinite direct sum, you need all but finitely many entries to be zero.

[hollow mica]

and v_x = 0 for all x > y for some y

yeah

[chilly ocean]

Yeah.

You're exactly right.

[hollow mica]

so

if we have another (w_1, w_2, ...) a part of the tensor algebra

people claim this to be the "operation" on the tensor algebra

but I don't get why this is relevant

this is just something that lets us find the tensor product of a v_i and a v_j (two components of an element of the tensor algebra)

how at all does it define (v_1, v_2, ...) + (w_1, w_2, ...)

where I use + loosely to mean the operation on the tensor algebra

[chilly ocean]

This doesn't define the sum, though.

The sum (v_1, v_2, ...) + (w_1, w_2, ...) would just be (v_1 + w_1, v_2 + w_2, ...).

[hollow mica]

ok yeah

[hollow mica]

so why is this relevant

[chilly ocean]

This is supposed to help you define (v_1, v_2, ...) ⊗ (w_1, w_2, ...).

[hollow mica]

oh, so you can take the tensor product of the tensor algebra with itself?

[chilly ocean]

I could have two elements $$v_1 + v_2 \otimes v_3, w_1 + w_2 \otimes w_3 \otimes w_4$$ of the tensor algebra, and I might hope to multiply them by doing something like
\begin{align*}
(v_1 + v_2 \otimes v_3)\otimes(w_1 + w_2 \otimes w_3 \otimes w_4) &= v_1 \otimes w_1 \
&+ v_1 \otimes w_2 \otimes w_3 \otimes w_4 \
&+ v_2 \otimes v_3 \otimes w_1 \
&+ v_2 \otimes v_3 \otimes w_2 \otimes w_3 \otimes w_4.
\end{align*}
Basically just expanding it out like you would if these were elements of any other algebra where you can multiply elements.

Christ that looks terrible.

One sec.

[cloud walrus]

rakko

[chilly ocean]

This would give a multiplication operation on the tensor algebra given by, well, the tensor product.

[chilly ocean]

This is relevant because it's defining the kind of multiplication I'm talking about by starting on the summands by which the tensor algebra is defined.

Define it on the components, and then build it up into a full thing on the tensor algebra.

[hollow mica]

I don’t understand “I could have two elements” of that form, because aren’t elements of the tensor algebra just infinite tuples ?

[chilly ocean]

Ah.

That's a common notation abuse.

I'm pretending that (v_1, v_2, ...) is a sum v_1 + v_2 + ...

Obviously this makes no sense. v_1 and v_2 and so on all live in different vector spaces.

[hollow mica]

ohh, you’re allowed to do that for direct sums, right

[chilly ocean]

As long as you remember what the undefined sums you're writing actually represent.

The idea is that if you treat these as actual sums, you ought to be able to multiply them accordingly using the tensor product to get another thing in the tensor algebra.

[cloud walrus]

rakko

[chilly ocean]

Don't actually read this, just look at it and understand why someone might prefer to write these things as sums.

One usually calls this kind of notation abuse a formal sum.

[chilly ocean]

The fact that solvable groups are not the same thing as abelian groups is weirding me out.

Is there an intuitive way of thinking about how something non-abelian could be made of abelian composition factors?

[pastel cliff]

reading some lecture notes an my prof says that an important principle (ha) of the study of PID's is that containment of ideals is analogous to division of elements - why is this important?

i see that the idea of gcd's becomes possible because of this but idk i feel like im missing something

[hollow mica]

isn't it even abuse to say that "multiplication in the tensor algebra is the tensor product"

because really you are applying the tensor product to the components (not component-wise, but in a fashion similar to polynomial multiplication, not exactly sure if it has a name)

[chilly ocean]

Perhaps.

[hollow mica]

I guess my gripe is with this

[pastel cliff]

ssmol bump

[hollow mica]

can we really extend it linearly

sorry wifi is glitchy

[hollow mica]

this was where my confusion began, how in the hell does this alone define the multiplication operation

[hollow mica]

now even after I know what multiplication is, I still don't see it

[chilly ocean]

If this isn't answered in about an hour, I'll look at it.

[hollow mica]

Ok, thanks

[pastel cliff]

if I and J are comaximal ideals of a ring, do they have to be maximal?

maybee that's silly but i feel like there could be a counterexampl

[south patrol]

Not true in, say, Z

[south patrol]

In fact, if this were true then the notion of comaximal would be a bit useless I guess

If (a),(b) are ideals of Z, what does it mean for them to be comaximal?

[pastel cliff]

something something CRT

well

you probably mean that they're prime

[south patrol]

well not prime

[pastel cliff]

coprime*

[south patrol]

yes

a and b are coprime

because (a) + (b) is (d) where d is the gcd of a,b

But a and b can be coprime without (a) and (b) being maximal

so for eexample (4) and (49) are comaximal ideals, but neither is maximal

[pastel cliff]

[pastel cliff]

is this proof valid? not entirely sure about what i say for the gcd in the backwards part

[hollow mica]

bump

[chilly ocean]

Not home yet.

Soon.

[pastel cliff]

https://tenor.com/view/dog-husky-soon-gif-7801078

[hollow mica]

👍

[pastel cliff]

concern here is that it might only actually work in euclidean domains but idk

[chilly ocean]

I think the claim is that the \emph{composition} $$T^k V \times T^\ell V \to T^k V \otimes T^\ell V \to T^{k + \ell}V \hookrightarrow T(V)$$ extends by the universal property of the direct sum to $$T(V)\times T(V) = \bigoplus_{c = 0}^\infty \left(\bigoplus_{a + b = c} T^aV \times T^bV\right).$$

[cloud walrus]

rakko

[chilly ocean]

So you have a map T(V) x T(V) -> T(V) which should do the thing I wrote earlier.

I'm not really sure about this. It seems like a needlessly abstract way to talk about this when you can literally just write down what the multiplication should be.

@hollow mica Sorry for the delayed response. I was out getting some food.

Some things are indeed better left not explicitly defined. This is not one of them.

Wikipedia has failed here.

[pastel cliff]

sorry to interject but im a lil stuck on an easy question - Let $R$ be an integral domain. \claim $r$ is prime in $R$ if and only if $\ang{r}$ is prime in $R$. - any pointers

[cloud walrus]

sebbb

[chilly ocean]

Actually, no, I'm not happy with this explanation.

It is literally JUST too late to delete it.

[chilly ocean]

Ignore the deleted message. This should be more straightforward than what I originally wrote.

Where are you getting stuck? What did you try?

[hollow mica]

unpacking this, T^k v x T^l v -> T^k V ⊗ T^l v is the usual tensor product map (v, w) -> v ⊗ w. The second map is the natural isomorphism you get because of the associativity of ⊗, and the third map is the injection v -> (0, 0, ..., v, 0, 0, ...) where there are k + l - 1 0s before the v ?

[chilly ocean]

Yes, yes, and yes.

To be frank, I did not check that this gives the "obvious multiplication" that you can just define right off the bat.

My gut just tells me it works out.

I'm not happy with the way Wikipedia defined this.

[hollow mica]

i swear it's how every other source I could find defines it as

[chilly ocean]

https://math.stackexchange.com/questions/2245702/pointwise-multiplication-on-tensor-algebra

I think I just got spoiled by the way differential geometers tend to treat these things.

[pastel cliff]

trying to use the combindation of facts that rx + sy = u, rx = a, ry = a, and that r and s generate comaximal ideals

it does feel like it should be very straightforwards though

[chilly ocean]

It's literally just $$(v_1 \otimes\cdots\otimes v_k) \cdot (w_1 \otimes \cdots \otimes w_\ell) = v_1 \otimes \cdots \otimes v_k \otimes w_1 \otimes \cdots \otimes w_\ell$$ extended to the whole thing.

[cloud walrus]

rakko

[chilly ocean]

I wanted to delete the original explanation and replace it with just this.

[hollow mica]

the universal property you are referring to is:

given the natural embeddings x_k: T^k V -> T(V), for any arbitrary vector space M and linear maps f_k: T^k V -> M, there exists a unique linear map f: T(V) -> M such that f ∘ x_k = f_k for all k

[chilly ocean]

I think you should ignore my original explanation.

[chilly ocean]

This is extremely overkill.

You can argue just from the definitions of "prime element" and "prime ideal" without anything else.

In fact, I don't think you even need the fact that R is an integral domain.

[pastel cliff]

where do we have prime elements tho? we just have that theyre coprime no?

[chilly ocean]

"r is prime in R"

[chilly ocean]

Are we talking about the same problem? This is what I'm referring to.

[pastel cliff]

i posted the wrong question

[chilly ocean]

It happens.

[pastel cliff]

Suppose that $R$ is a PID, and that $r,s \in R$ have the greatest common divisor equal to a unit. \claim If $r \mathrel{|} a$ and $s \mathrel{|} a$ then $rs \mathrel{|} a$

[cloud walrus]

sebbb

[chilly ocean]

Extended by some kind of universal property, I ought to clarify.

[pastel cliff]

i was using this

[chilly ocean]

I think I've fumbled enough on this and should let someone else take over.

[pastel cliff]

to consider this

[long nebula]

what's the purpose of requiring a discrete valuation on a field to be surjective? what results break if the discrete valuation isn't surjective?

[hollow mica]

oh it extends because the multiplication operation should be distributive, right

[pastel cliff]

[next obsidian]

What do you mean? Can you expand your definitions?

[hollow mica]

I'd like to understand the connection to universal properties though, because that's supposed to be the good stuff

[pastel cliff]

someone shined the chmonkey-signal

[pastel cliff]

chomkey

[long nebula]

a discrete valuation on F is a function nu : F^\times -> Z satisfying

• nu(ab) = nu(a) + nu(b) for a,b \in F^\times
• nu is surjective
• nu(a+b) ≥ min(nu(a), nu(b)) for a,b \in F^\times if a+b≠0

[next obsidian]

Ah okay

So the image is a subgroup of Z

But these are all isomorphic to Z

[long nebula]

oh hmmm

[next obsidian]

So for technical reasons it’s just nice to pretend the codomain is the copy of Z which is the image

[long nebula]

so there's no additional strength to saying it's surjective?

[next obsidian]

Like now you know you can always say for any n, you can grab a such that nu(a) = n

[long nebula]

bc you can just redefine the valuation so that it's surjective? assuming it doesn't just send anything to be 0

[next obsidian]

Yeah basically

[long nebula]

ahh gotcha

ty!

[next obsidian]

For more general valuations (not valued in Z) there’s a reason too

We want to talk about valuations being equivalent

But for this it’s easiest to consider them always being surjective

[long nebula]

oh interesting

[next obsidian]

So like imagine if I said it takes value in any totally ordered abelian group

or even just subgroups of R

well so consider you now have a discret valuation okay?

nu:K^x -> Z

Well we can “modify” nu by considering it avtually mapping to R

via Z -> R

The inclusion

[long nebula]

yea

[next obsidian]

For all intents and purposes these are the same thing right?

[long nebula]

yea

[next obsidian]

So to phrase when they’re equivalent is annoying bexause what you want is an isomorphism of ordered groups from the images of our valuations

In this case, call the R-valued valuation nu’

It’s very easy to get the isomorphism on image of nu’ and nu

It’s the identity

But this whole “isomorphism on image” is really annoying

[long nebula]

right

[next obsidian]

So it’s nice to just always replace the codomain of a valuation with the image

Bexause now we only need an isomorphism of the codomain

[long nebula]

ahh interesting

[next obsidian]

This won’t work if it isn’t surjective bexause id be asking for an isomorphism of R and Z

Which obviously doesn’t exist lol

[long nebula]

right right

[next obsidian]

But for discrete valuations the image is always Z!

So it’s extra nice for discrete valuations

[long nebula]

that's cool!

[next obsidian]

[long nebula]

ty for the explanation

[next obsidian]

Okay this was very true

[pastel cliff]

bump to thiis

it's definitely easy and im definitely overthinking

but if you could talk a look at it mr chmonkey that would be cool and good

[next obsidian]

Okay so

What you’re looking for is basically the lcm

You know how in the integers the lcm of integers is mn/gcd(m,n)

So in particular if you were in the integers then what you want is true right?

[pastel cliff]

let me guess, this holds in a PID

[next obsidian]

Basically!

So you can’t really divide

But there’s a few ways to make sense of this

I think the easiest might be to interpret this in the field of fractions

[pastel cliff]

if gcd is a unit it has an inverse anyways so that's techincally well defined no?

[next obsidian]

Yeah

But you’d need to show it’s an lcm

Anyway, you know the gcd divides m right?

By definition

So suppose m = kgcd(m,n)

I think your lcm will be k•n

So maybe try to prove fhat

I think this will hold in any UFD actually!

The proof I have in mind is basically writing it as a prime decomposition

So maybe actually try to prove it in the integers

Then generalize tuat to a UFD

Sound doable?

[pastel cliff]

ufd's are my next assignment but yes

[next obsidian]

Kekw

[pastel cliff]

ok goober moment

what do we gain from an lcm here

[next obsidian]

By definition if a and b divide a common element c, the lcm divides it

[pastel cliff]

so showing the lcm of r,s is rs implies that rs divides a

[next obsidian]

If you can’t manage this, then maybe you can also try writing 1 = ur + ts which you can do by assumption of them being coprime

[pastel cliff]

that's what id been trying b4

[next obsidian]

From here idk, multiply by a, but tbh i don’t think it will follow directly from this

[pastel cliff]

yeah i dont think it does

[next obsidian]

Since you have to use more than just being an ID, you need at least UFD

[pastel cliff]

ohhhhh wait

r,s are coprime so (r), (s) are comaximal, which means that (a) \cap (b) = (ab) but since their gcd is 1 (ab) = (u) = R

so lcm of r and s is (u) = (rs) and the thing follows

[next obsidian]

Oh sure

[pastel cliff]

[next obsidian]

Well so

u is rs UP TO A UNIT

So WLOG it is rs

[pastel cliff]

you mean ideal generated by u and rs?

[next obsidian]

Or actually like

Your symbols don’t seem to match up

I think you can just do that (r)\cap(s) = (rs)

[pastel cliff]

oh yeah wait i messed up i think

[next obsidian]

So since a is in both (r) and (s) it’s in (rs)

So GG

[pastel cliff]

(r),(s) comaximal implies that (r) \cap (s) = (r)(s) is what i meant to say

[next obsidian]

Yeah

But that’s just (rs)

By like observation I guess

Haha

[pastel cliff]

yeah so it still holds i think

[next obsidian]

Yeh

[pastel cliff]

[next obsidian]

But my point is you don’t need to go through u

Swag

[toxic zephyr]

so in linear alg (vector spaces) if a set contains the zero vector then it is automatically LD. for modules, would the analogue be that if a set has a zero divisor then it is LD?

[chilly ocean]

Yes.

I got confused by you using "zero divisor" for an element of a module; if you saw the deleted message, ignore it. I know what you meant.

If m is an element of M with rm = 0 for some non-zero r in R, then this is your desired non-trivial linear combination.

If R was a field, then you could divide by r to get m = 0 and you reduce to your first case.

The term you're looking for is a "torsion element" of the module, I think. (At least when R is an integral domain.)

[toxic zephyr]

oh yeah i remember my professor talking about torsion now.

cool thank you @chilly ocean

[oak hill]

Is $\frac{\mathbb{Z}}{\langle n \rangle} \cong \mathbb{Z}_n$?

[cloud walrus]

F♯A♯ℵ0

[chilly ocean]

Typically one would define the right-hand side to be the left, although one should note that, at least when n is prime, the right-hand side typically denotes the n-adic integers (which are not the same as the left).

[blissful crystal]

i mean i usually write the group of integers mod n to be just $\mathbb{Z}/n\mathbb{Z}$

[cloud walrus]

Shell

[blissful crystal]

$n\mathbb{Z}$ is essentially just $\langle n \rangle$

[cloud walrus]

Shell

[proud bear]

hi @blissful crystal

[blissful crystal]

do i know you random guy?

hey lol

[proud bear]

lol

[pastel cliff]

why is this wrong

bc im assuming it's wrong

there's no way it can be right

[toxic zephyr]

why can't it be right?

[pastel cliff]

too easy

idk

[cloud walrus]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

User cancelled member selection.

Nov 10, 2022, 8:29 AM

[toxic zephyr]

wait yeah that doesn't make sense. i think they mean that <a,b>=<c>

[pastel cliff]

naw it's a cap

[toxic zephyr]

because <a,b> has elts of the form ax+by

and that's how you prove that elts in a PID have a gcd iirc

[toxic zephyr]

if you replace <a>cap<b> with <a,b> i think their proof works

[pastel cliff]

i mean it's my proof

but the assignment has a cap

so idk

[toxic zephyr]

oh lol. did you get marked off points for it or something?

[pastel cliff]

working on it rn i just like to get a second pair of eyes on my pfs

[toxic zephyr]

it's weird. if a is 4 and b is 10 for example in Z, then the intersection is the ideal generated by 20 which is the LCM not the gcd. but the LCM isn't too far off from the GCD so maybe

[pastel cliff]

on a different note, this look okay? only worried cuz i never used the fact that R is an int. domain

[toxic zephyr]

can you send the original question? or is it exactly how you wrote it

[toxic zephyr]

this looks good to me but I'm still learning this stuff too so I'm no authority lol

[pastel cliff]

that's exaactly it lol

[daring ermine]

agreed

[pastel cliff]

none of these feel like they exist, which makes me think they all exist

ok b) def exists

probably a) too

[quiet pelican]

For (b), you want to give two polynomials which have a gcd in Q[x] that isn’t in Q[x^2, x^3]

[quiet pelican]

(d) doesn’t exist, but the argument basically proves it’s a UFD

[pastel cliff]

a guess for b) - x^4 and x^8?

[quiet pelican]

Which may be going a bit far

[quiet pelican]

x^4 is the gcd of them

[pastel cliff]

but that's not in Q[x^2, x^3] is it?

ohhhh wait yes it is

[quiet pelican]

(a) can be done by just exhausting the elements of Z/12Z that aren’t units