groups-rings-fields

[tough raven]

I never really understood what exactly qualifies as "descent".

😅

[dull ginkgo]

https://people.math.harvard.edu/~elkies/M250.04/artinlemma.html#:~:text=Proof of Artin's Lemma,are linearly dependent over F.

[tough raven]

Oh I thought you were talking about linear independence of characters.

[dull ginkgo]

Chinese Remainder Theorem is my favorite proof of that

I'm fond of any wacky way to use chinese remainder theorem to prove stuff

[tough raven]

I used to like that, but now my favourite proof is to interpret it as linear independence of eigenvectors with different eigenvalues.

[dull ginkgo]

(e.g. commutative artinian rings are a product of finitely many local rings)

[dull ginkgo]

huh

[chilly radish]

You have an object X "over" an object Y. You have something over X, and a way to "lift" it to something over Y. Given something over X, when does it come from something over X via lifting, and what data do you need to find the thing over X it came from

As an example, suppose I have a galois extension of fields E/F and I have an algebra over F. I can lift it to an algebra over E via the tensor product, i.e. A(x)_F E. given an algebra A over E, is there an algebra A' over F such that
A'(x)E \cong A

The information we need to find A' in this case is related to the galois group. It turns out that if you have a semilinear G-action on A (whatever that means), then

A^G (x) E \cong A

As algebras via a natural isomorphism. There's actually a lot of structure this isomorphism preserves

[dull ginkgo]

I am now rabbit-holed into double centralizer theorems

[chilly radish]

The idea of étale descent generalises this

[dull ginkgo]

@chilly radish there seems to be a fuckton of different cases of double-centralizer theorems, is there any particular motivation why this property is saught after?

[dull ginkgo]

also is this not false

Right modules mean it's an anti-homomorphism from R to E no?

[chilly radish]

I don't have a good answer for this

[chilly radish]

They're only looking at end of the underlying group

Oh wait

Yea you're right

This is an anti-homomorphism

[chilly radish]

I was looking at the highlighted passage

[dull ginkgo]

I was confused about this for like 20 minutes lol

[junior badge]

I'm looking for a minimal primary decomposition of (6) in R = Z[√-5].

I thought (6) = (1 + √-5) ∩ (1 - √-5) would work: both (1 + √-5) and (1 - √-5) are prime, and hence primary. It's clear that 6 belongs to the ideal on the RHS, while any element that is a multiple of both (1 + √-5) and (1 - √-5) is also a multiple of 6. Minimality also seems to be OK.

Does this seem reasonable? I'm pretty sure I've missed something. In https://math.stackexchange.com/questions/3953873/find-minimal-primary-decompositions-of-ideals, they end up with a different decomposition...

[toxic zephyr]

if f is an onto ring homomorphism from R to S, and we have f(J)=S for J an ideal of R, then under what assumptions can we conclude that J=R?

[coral spindle]

This happens iff J + ker f = R

It's hard to really unpick this condition. There is likely to be a difficult interplay between J and ker f that you can't really say very much about for a general ring R.

But this is at least a common enough condition that it has a name: J and ker f are coprime ideals.

So you need the only ideal coprime to ker f to be R.

[coral spindle]

Sorry, to be clear: you have f(J) = S iff J + ker f = R

[toxic zephyr]

I is a maximal ideal of S, and I'm trying to prove finv(I) is maximal. I supposed the preimage of I was strictly contained in an ideal J. I was able to show f(J)=S since I is strictly contained in f(J).

I proved this with FHT years ago but I'm trying to do a direct proof. can we perhaps conclude it's coprime to ker f somehow or is there an easier way?

[coral spindle]

So it suffices to show that f^-1(I) is a maximal ideal of R/ker f, right? Indeed there is an isomorphism R/ker f → S so it should be immediate

[toxic zephyr]

would you mind explaining why it's sufficient to show it's a maximal ideal of R/ker f? that's not obvious to me atm. I'm pretty rusty on my algebra

[coral spindle]

For any ideal J of R, the ideals of R/J are in bijection with the ideals of R containing J

Moreover this bijection preserves the order

(This is sometimes known as the correspondence theorem)

Ergo, if you wish to show that J (an ideal containing ker f) is a maximal ideal of R, it suffices to show that J/ker f is a maximal ideal of R/ker f

This is an important picture you should have in mind when considering quotient rings

In the end the proof of this correspondence theorem will really be the same thing, but this just elucidates it more I think.

[coral spindle]

You actually left out this assumption in your first question up here.

Since J = f^-1(I), it contains ker f (= f^-1(0))

So J + ker f = J

:)

[toxic zephyr]

oh that's super interesting. yeah I remember my prof mentioning that. that's really cool.

[toxic zephyr]

sjshajdhdjjfnfjd
yep there we go

thank you @coral spindle ! (:

[coral spindle]

No worries taylor

[coral spindle]

It is a great result and it's super useful

[rocky cloak]

If you want it this to hold for all J, then it's equivalent to the kernel of f being in the radical of R.

[coral spindle]

Oh ofc, nice

[rain grove]

This is very clear with examples but idk how to make sense of it in general

gcd(m/d,m) = m/d

[knotty badger]

I think this is just because $m/d$ divides $m$, no?

[cloud walrus]

Pseudonium

[knotty badger]

Any common divisor of $m/d$ and $m$ must be $\leq m$ and $\leq m/d$

[cloud walrus]

Pseudonium

[knotty badger]

But $m/d$ itself is a common divisor - and that means it must automatically be the greatest one

[cloud walrus]

Pseudonium

[rain grove]

Ohh yeah nice

[outer gull]

I don't know if this is the place for this but I'm looking to get into research on Lattices and order theory, Boolean rings and whatnot to understand the attached paper. Can someone please help me build an intuition for this?
The most I've understood is the powerset example, from which I could understand inclusion. Then I was looking at given an ordered lists of elements, a radon arragement of n elements assuming elements form every single list falls into a new list- would that technically classify as partially ordered? I felt that this was a very primitive intuition though, and i really was not able to make head or tail of the rest.
Excuse my english, I had a hard time translating this stuff into words.

[shell pilot]

So using the paragraph under the Definition, we can write $x^2-2$ as $(x-\sqrt(2))(x+\sqrt(2))$ so then it would be considered irreducible over Q. But if you use the definition, then neither $(x-\sqrt(2))$ nor $(x+\sqrt(2))$ is a unit in Q so based on the definition (not the paragraph under the definition, why is $x^2-2$ irreducible over Q?

[cloud walrus]

Soap_Opera

[barren sierra]

I'm confused

[cloud walrus]

Spamakin🎷

[barren sierra]

Wait wait I see your confusion

[cloud walrus]

Spamakin🎷

[shell pilot]

Ah

Ok

Thank you for making me read it more closely haha

Neither of those are polynomials in Q[x]

[barren sierra]

bingo

[sonic coral]

i assume you already figured it out @barren sierra but for my own sake of understanding, can you not argue that a root r of p(x) with multiplicity m is a root of p’(x) with multiplicity m-1, so you can divide it out and remove the singularity

[barren sierra]

No I meant something even stronger

And you can

[south patrol]

Which question is this lol

[barren sierra]

I'll message back and repost the question I deleted once I get back to my laptop

[south patrol]

Ah ok

Have you seen the nice Conway (?) proof that sqrt(2) is irrational

Works to show N^1/m is only rational if N is an mth power of an integer lol

[tough raven]

There is no attached paper BTW.

[hidden wind]

ooh do you have some suggested introductory reading for this

[rotund aurora]

Yes, Szamuely's "Fundamental groups as Galois groups", for example. Volklein's "Groups as Galois groups" is also nice, and more concrete.

[south patrol]

Oh i didn't know that was meant to motivate AG lol

[hidden wind]

amazing, thank you very much

adding this to my pile of to-read things currently at 297 items

[rotund aurora]

The point is that by Hilbert's irreducibility theorem if you can construct a finite Galois extension with Galois group G of Q(X1,...,Xn) then you will also be able to construct such an extension for Q. It is not clear how to construct Galois extensions with prescribed Galois groups of Q(X1,...,Xn) but the situation for C(X1,...,Xn) is nicer, and you can try to pass from C(X1,...,Xn) to Q(X1,...,Xn). But idk much more

But the Galois theory of the extensions of C(X) is basically the theory of Riemann surfaces, and so on.

You can also consider Qp(X1,...,Xn), etc, and this will lead to p-adic kind of AG stuff

This lecture is also nice https://youtu.be/ImLmr1YHn24?si=lcClIiYd_frU-QCU

[hidden wind]

i’m pretty sure this is related to the topics of my bachelor thesis mwahaha

[rotund aurora]

What is the topic? I hear you sometimes talk about hyperbolic space and stuff

[hidden wind]

which was a hodge-podge of too many things so that really i did not get any understanding of most of them ehehe, but it set a landscape i am excited to get a better sense of

[hidden wind]

the starting point was the klein quartic, but i spent most of my time trying to get a sense of the history rather than the mathematics that lead up to that… which i realize was probably a rather bad move as history is such a mess but we rollin, albeit slowly

[rotund aurora]

mmmh maybe there are interesting connections

[hidden wind]

klein’s work certainly is important in the theory of riemann surfaces

[rotund aurora]

Iirc one of the main features of the Klein quartic is that it had large automorphism group, I think it achieves Hurwitz bound of 168 elements? Maybe that was PSL2(F7)? but not sure

[hidden wind]

yep

[rotund aurora]

Maybe there is some way to use that to construct PSL(2,7) Galois extensions of Q. But I have no idea

[hidden wind]

he also wrote a whole book on the quintic and the icosahedron

well, a small book tbh

he wrote a lot of books with fricke… many of them behemoths that i’ve been told foresaw many later developmenrs

but the icosahedron and its relation to the galois theory of the quintic (their symmetry groups are isomorphic) is clearly the main inspiration for much of it

[long obsidian]

The integers Z is a ring. In fact for any integer k the multiplication by k:Z->Z is an injective endomorphism. In particular image(k)=kZ. The inverse k^{-1}:kZ->Z is basically division by k.

I can also consider multiplication by k^2 that is k^2:Z->k^2Z.

Is it obvious that multiplication by k^2 and division by k commute?

That is
k^2 k^{-1}=k^{-1} k^2?

[tough raven]

Well, both map k x to k^2 x for any x in Z, so yes?

[long obsidian]

Oh okay so it's obvious just by checking each element on kZ. Thanks! I forgot the domain had to be kZ

[tough raven]

If you want to soup it up abstractly, you could say that since k is cancellative, we can embed Z in Z[k^{-1}] where the equality is obvious.

I'm not sure if this has any advantage over just comparing the two though.

[glad osprey]

It's not a ring endomorphism though, right? Since k * a * b != k * a * k * b

[barren sierra]

ok finally back from errands

part (b)

this is the solution for the bound:

relevant exercise which defines that value B

I guess I need to normalize by the leading coefficient of p(z)

[kind kindle]

Alright, we have F(2), which is the Algebreic Closure of GF(2)

We have Addition and multiplication defined on this field

[dull ginkgo]

strange algebra

[kind kindle]

Are you talking to me or the other person?

[dull ginkgo]

No just talking to myself

[marsh scaffold]

Guys is there a nice possible way to view R^2-0 as a coset space of SL_2(R)

I'm taking like the coset space of this being G-isomorphic to R^2-0 as G-sets

G being GL_2(R)

[agile burrow]

Yeah, SL_2(R) acts on R^2 - 0 just by usual matrix multiplication acting on vectors. Since elements of SL(2, R) are invertible, it has trivial kernel (hence sends non-zero things to non-zero things). The stabilizer of (1, 0) is the subgroup of strictly upper triangular matrices (with 1s on the diagonal) so G/N = R^2 - 0

[marsh scaffold]

Wait i didn't quite get that

I meant I want to identify action of GL_2(R) on GL_2(R)/SL_2(R) the same as action of GL_2(R) on R^2-{0}

[agile burrow]

Oh

I don't think that quite works, GL_2(R) / SL_2(R) is isomorphic to R - 0 via the determinant map

[marsh scaffold]

No like I don't mean they're isomorphic as groups

More like as G-sets

Wait let me get you a reference

[agile burrow]

Right but even as coset spaces, I don't think you can identify GL_2(R) / SL_2(R) with R^2 - 0

[marsh scaffold]

https://kconrad.math.uconn.edu/blurbs/grouptheory/transitive.pdf

It's on pg 5

[agile burrow]

Yeah, he's saying that R^2 - {0} is a coset space of SL_2(R), meaning it's identified with SL_2(R) / N where N is some subgroup of SL_2(R)

[marsh scaffold]

Ah i see

Then i misinterpreted

[agile burrow]

And then the description I gave above is the relevant one here. I can elaborate on the details of that if you'd like since I skipped a couple of steps

[marsh scaffold]

Wait let me try that first

I'll let you know

Thanks for the clarification

[marsh scaffold]

Yeah I think I understand this ( you were basically saying the sl_2(R) action is transitive and then took the stab of (0,1)

This is like how we prove stuff for transitive actions

So yeah

[agile burrow]

Yeah, that's right

Orbit stabilizer is one of my favorite things

[knotty badger]

It’s quite useful, yeah

I love proving the binomial and multinomial theorems with it

Also gets used in quantum field theory

[dull ginkgo]

This is going to seem extremely dumb

Actually I answered it in my head lol

I was going to ask if the centralizer of a ring R embedded by left multiplication into End_grp(R) is iso to it’s opposite ring

But I realized it’s just the map from R^op via right multiplication lmao

[rotund aurora]

how do you actually prove this? Idk anything about Clifford theory

[errant wedge]

Where's the highlighted bit coming from? We know $|P|=p^k: \big| : |N(P)|=i: \big| : |G|=p^ml$, but how do we know we dont just have $|N(P)|=p^kj$ with $\gcd(p, j)=1$? Like how do we know the quotient still has order a multiple of $p$?

[cloud walrus]

Sara

[rocky cloak]

P acts on G/P by left multiplication. The size of an orbit divides |P|, so is either 1 or a multiple of p.

Counting modulo p we can ignore those that are multiples of p, so only count those in orbits of size 1. I.e. those in N(P)

[barren sierra]

I think they're asking why it's = 0 (mod p)

which isn't immediate to me either hm

not familiar with the proof of this generalized statement

[rocky cloak]

Oh, I totally missed that 0

But isn't that just because |G/P| is a multiple of p

[dull ginkgo]

yes

I haven't seen the action on the cosets themselves before

i've usually seen it on subsets of a given size

[barren sierra]

oh bruh

I was totally overthinking what the size of N_G(P) was 😭

[dull ginkgo]

*unfathomable sad trumbone noise*

[barren sierra]

VERY much using the fact that k < m

[rocky cloak]

I guess the more general thing is just that if a p-group acts on a finite set X, then |X| is congruent to the number of fixed points mod p

[dull ginkgo]

Yes

Jacobson had this form of Sylow as guided proofs

I liked them

[delicate orchid]

b-b-b-buh?!?!

[dull ginkgo]

It’s easier to just, partition the sets of cardinality p^k by the p-adic valuation of their left-stabilizers and do it from there

Show the equivalence classes are congruent to eachother mod mp^(differences)

[errant wedge]

Ahh ty

[tidal lotus]

hey guys whats the best book to self learn algebra? I wanna be prepared before i have to take the first abstract algebra class

[barren sierra]

I like Artin's Algebra

[tidal lotus]

what do you think about foote and hungerford if you have gone of them before?

[crystal vale]

In Z define a•b = a+b-ab, then there is a isomorphism between monoid ( Z,×,1) and (Z,•,0) such that a-> 1-a.

Now for injectivity I want to show that 1-a = 1 -b => a = b.

Now I don't see that it is valid to do 1-a = 1-b => a = b because how can I cancel 1, is it because now we are working on Z so we can use its property? But on Z there is a different operation, right?

[rocky cloak]

Z has a ton of operations.

When you have any equation x = y, and you do the same thing on both sides you get
f(x) = f(y)

This is completely independent of what kind of operation f is. Be it one you've used up until know, one that's familiar from before, or something completely arbitrary.

[crystal vale]

So if I take f:(Z,×,1)->(Z,•,0) such that x->1-x.

Then if 1-x = 1-y then f(1-x)=f(1-y) => x = y

But to show f is a well defined function if I take x = y I need to show 1-x = 1-y

So here in (Z,•,0) can I use if x = y then 1-x = 1-y?

[rocky cloak]

Well, I think it's well established that subtraction is well defined

[crystal vale]

Yes but I want to define subtraction in (Z,•,0)

[rocky cloak]

If you want to prove that, I guess it depends on your fundations and how you define Z

[rocky cloak]

No, you don't want/need to do that

I'm not even really sure what that would mean

[crystal vale]

So when I take set Z, can I assume addition and subtraction properties here ?

I mean there is a different operation on Z • so if I have a=b can I say 1-a = 1-b

[rocky cloak]

Yes, you can use addition and subtraction. Or the absolute value function. Or cook up you're own function out of the blue

[crystal vale]

Okay thank you ❤️

[rocky cloak]

The operation doesn't dictate what integers are equal

It doesn't really address it really

[crystal vale]

So here I can use the addition property of Z

And in the definition of • they used addition also

[errant wedge]

Z^2 as a Z-module doesn’t have invariant basis number right?

Or well IBN is a property of the ring, I mean to say rank is not well defined right

Oh wait nvm

[rocky cloak]

The rank is 2

[errant wedge]

Commutative rings have IBN. So IBN doesn’t imply that for an R-module of rank n, a linearly independent subset of n elements is a basis, what do we need for that? My understanding is that if this is not true then there’s some element not in the span, and thus giving us a basis of size n + 1 which is a contradiction

I’m doing something very wrong lol

I’m looking at $\langle (3, 4), (-4, 3)\rangle \subseteq \Z^2$

[cloud walrus]

Sara

[errant wedge]

Is this not lin indep?

Sorry lemme do this on paper again I probably messed up checking independence 💀💀

[knotty badger]

Pretty sure this is linearly independent

A linear relation needs integers a and b such that 3 a - 4 b = 0, and 4a + 3b = 0

[topaz solar]

If you can divide, obviously it’s true, but divisibility failures are wacky

[knotty badger]

View this as a matrix system, and then the matrix has nonzero determinant

[topaz solar]

Is it linearly independent over Q? Over R?

[errant wedge]

Its not over Z

afaik

[topaz solar]

I mean it’s not exactly the same, but if there’s no rational way to scale to add to zero, then as for integers…

[errant wedge]

Yeah

[topaz solar]

They can be linearly independent of size 2 without being a basis for that rank 2 module though

[errant wedge]

Ah

[topaz solar]

You have proper submodules of Z iso to Z

Namely

2Z

What about (2Z)^2 < Z^2

Good luck getting (1,1) when every entry has even coordinates

But that’s isomorphic

[rocky cloak]

I think this might be true if R is artinian.

But you're reasoning that you can find a basis of size n+1 is wrong. Just because the n elements doesn't span the module doesn't mean you can find another linearly independent element.

2Z < Z being a very easy example.

[topaz solar]

Mega sniped on why it’s wrong, but I didn’t think of the artinian

[rocky cloak]

I'm thinking maybe it's an iff also...

[topaz solar]

If R is artinian, stack those n-independent vectors, iso submodule, iterate

[topaz solar]

And uhhh, principal ideals should be reachable reasonably? I dunno about, like, n-generated? But some silly R^n quotient things maybe?

[rocky cloak]

Yeah, you just set up the matrix mapping standard basis to the new one.

Then it's injective iff the determinant is non-zero divisor. Hence a unit

[topaz solar]

nice

[rocky cloak]

Not an iff though. Since same argument works in any ring without non-unit non-zerodivisors

So I guess that's the condition

[topaz solar]

I see I see, unfortunate but understandable

The tyranny of only having weird equivalent formulations

[coral spindle]

There are two main results being used: Clifford's correspondence theorem and Gallagher's theorem. I'll explain each one.

\bigskip
Suppose $N \unlhd G$ and let $\vartheta$ be an irreducible character of $N$. Noting that $G$ acts (by conjugation) on the irreducible characters of $N$, consider the fixed point subgroup $G_\vartheta$. Then the induction of characters forms a bijection $\operatorname{Irr}(G_\vartheta \mid \vartheta) \to \operatorname{Irr}(G \mid \vartheta)$, where by this notation I mean characters whose restriction contains a copy of $\vartheta$. So if I can understand the characters of $G_\vartheta$, I understand the characters of $G$. This is the Clifford correspondence.

\bigskip
Suppose that there exists some $\vartheta'$ an (irreducible, necessarily) character of $G_\vartheta$ which restricts to $\vartheta$ on $N$. Then every character of $G_\vartheta$ above $\vartheta$ is of the form $\vartheta'\kappa$ where $\kappa$ is an inflated character of $G/N$. This is Gallagher's theorem, which allows us to understand the characters $\operatorname{Irr}(G_\vartheta \mid \vartheta)$.

\bigskip
In my situation we have a normal subgroup $G^p$ and quotient $C_p$. The characters of $G^p$ look like $p$-tuples of characters of $G$ and $C_p$ cyclically permutes them. The stabiliser is just $G^p$ iff some characters differ, and we can use Burnside to count them, and then just use Clifford's correspondence. On the other hand the stabiliser is the whole wreath product iff they're all the same, in which case we extend and use Gallagher.

[junior badge]

"Let O_P be the localization of K[x, y] at a point P. Let F and G be coprime polynomials in K[x, y]. Then O_P/(F, G) is zero-dimensional and local, and hence Artinian." Now I don't quite see why O_P/(F, G) is zero-dimensional...

We know an ideal I in O_P/(F, G) corresponds to an ideal I + (F, G) in O_P. But (F, G) = (1) = O_P, so I would be the entire thing. But this makes no sense.

[cloud walrus]

Boytjie

[rocky cloak]

OP/(F, G) is the localization of K[x, y]/(F, G) so it's enough to see that this is 0-dimensional.

F is nonzero, so K[x, y]/(F) is 1-dimensional. F and G are coprime, so G is not contained in any height 1 prime that contains F. Hence G is not contained in any minimal prime in K[x, y]/(F). So K[x, y]/(F, G) has smaller dimension, i.e. dimension 0.

(height 1 primes of K[x, y] is just of the form (f) with f irreducible, because K[x, y] is a UFD)

[toxic zephyr]

probably a really stupid question but what is a ring homomorphism on Z[x] with kernel (x^2)?
in general, how can I construct a homomorphism with kernel (p(x)) or (p1(x),...,pk(x))?

[knotty badger]

Are you requiring it to be an endomorphism…?

Cause if you are, no such endomorphism exists

And if you aren’t, then here’s a stupid answer - take the quotient map Z[x] -> Z[x]/(x^2)

[toxic zephyr]

just needs the domain to be Z[x]. and yeah I mean something nontrivial :p

like for (x) or (x-a) there's the map to p(a). or to (2,x) there's p(0) mod 2, etc.

[knotty badger]

I mean, by the first iso theorem, this is essentially the only example up to isomorphism

And combined with inclusion into a larger ring, I suppose

Since the first iso theorem tells you that every homomorphism takes the form of a quotient map, followed by an isomorphism, followed by an inclusion map

[toxic zephyr]

I'm sort of going the reverse direction. like rather than saying I have f:R->S is onto, then S is iso to R/ker f, I'm trying to find a ring S that R/I is iso (but not equal) to by constructing f such that ker f=I.

so I guess the question is if a generic ideal (p(x)) has something like an evaluation map to get it as the kernel

[knotty badger]

The universal property of Z[x] tells you that ring homomorphisms Z[x] -> R naturally correspond to a choice of element of R - namely, where to send x

So just - pick a ring R with an element r such that r^2 = 0, and where that’s the minimal polynomial

That’ll give you a ring hom whose kernel is (x^2)

And all such ring homs arise this way

[toxic zephyr]

interesting. I was representing the elements of $\bZ[x]/(x^2)$ as matrices
$$a+bx\sim \m{a&0\b&a}$$
so maybe that's just it haha

[cloud walrus]

inconspicuous old man & mime

[toxic zephyr]

the [0,0;1,0] matrix works as something that squares to zero

[knotty badger]

Mhm

[toxic zephyr]

L

[knotty badger]

This works pretty nicely with the universal property of quotient too

[toxic zephyr]

I wonder if we can always represent polynomial quotient rings as matrix rings

[knotty badger]

It’s like

Ring homs Z[x]/(p(x)) -> R

Naturally correspond to ring homs Z[x] -> R which send p(x) to 0, by the universal property of quotient

Which naturally correspond to a choice of element r in R such that p(r) = 0, by the universal property of Z[x]

So you need only pick a matrix whose minimal polynomial is p, for example

And this should always be possible (companion matrix works, I believe)

[knotty badger]

This will get you an isomorphism, not just a homomorphism

[toxic zephyr]

so what about ideals that need more than one generator? like (2,x) is one example, but idk of any others

[knotty badger]

So we can use universal properties again

Ring homs Z[x]/(2, x) -> R

Naturally correspond to ring homs Z[x] -> R such that 2 and x get sent to 0

So really you just need a ring of characteristic 2

Like e.g. Z/2Z

For it to be an isomorphism you’d have to choose something like Z/2Z

But if you just want a homomorphism, any ring of characteristic 2 works

[toxic zephyr]

yeah. I'm not sure of any other examples of nontrivial ideals of Z[x] that require two or more generators which might lead to something more interesting.

like what about Z[x]/(x^2,x^3-1) or something idk. nvm that doesn't work

[knotty badger]

Well, ring homs out of that naturally correspond to

Choice of an element r in R such that r^2 = 0 and r^3 = 1

[toxic zephyr]

yeah that was a bad choice because it means 0=1.

[knotty badger]

These only exist if R is the zero ring, yeah

[toxic zephyr]

but so basically in general you're saying if you want Z[x]/(p1,...,pk) then we need to send x somewhere where there's an r such that pi(r)=0 for i=1,...,k?

[knotty badger]

Yes! That’s exactly the universal property of the quotient (combined with that of Z[x])

That gives you ring homs out of Z[x]/(p1, …, pk) - for ring isomorphisms, you have to check there are no “extra” polynomials which send x to 0 other than those you can deduce solely from looking at p1, …, pk

E.g. you can define a ring hom out of Z[x]/(x^2) by sending x to 0, since 0 satisfies 0^2 = 0, but this isn’t a ring isomorphism

[crystal vale]

Any hint to show any finitely generated subgroup of the additive group of the rationals (Q,+,0) is cyclic.

Now let H is a subgroup generated by {x_1,...,x_n } now I don't understand what I can choose such that H is cyclic.

First I thought about taking the gcd of x_i but it seems it is not the correct choice.

Wait what if I take x_1...x_n ?

No

[still dew]

All your objects are general but Q is a specific object

What are the elements of Q and what properties do they have

Try to do it for two rationals first

[crystal vale]

Let H be generated by p/q and r/s then it can be generated by 1/(qs) because we can generate p/q and r/s in < 1/(qs)>, right?

[still dew]

Not really

[crystal vale]

Why?

p,q,r,s are integers

[still dew]

can you write 1/pq as a integer combination of p/q and r/s

Okay so first things first do u understand what we mean by a group generated by {p/q, r/s}

[crystal vale]

I see my mistake

[crystal vale]

Linear combination of p/q and r/s

Integer linear combination

[still dew]

*Integers

Okay

Just apply the definitions and use the property of intergers

You should be okay

[crystal vale]

I don't get it

[still dew]

What about integers, is every finitely generated group cyclic?

[silver kettle]

step 1: notice every finitely generated group in Q is isomorphic to some finitely generated group in Z.
step 2: apply bezouts identity
step 3: apply the reverse isomorphism

[coral spindle]

@rotund aurora did you have any questions about this btw?

This is like, my bread and butter

[junior badge]

thank you so much!

[outer gull]

my bad-https://www.cambridge.org/core/journals/journal-of-the-australian-mathematical-society/article/on-a-common-abstraction-of-boolean-rings-and-lattice-ordered-groups-ii/AC5FFFCEC2F92B59E4AE6E80BD77728E****

thought it sent

[outer gull]

what

[hidden wind]

look for a “share” button on your end, instead of just pasting the url you are at

[outer gull]

https://www.cambridge.org/core/journals/journal-of-the-australian-mathematical-society/article/on-a-common-abstraction-of-boolean-rings-and-lattice-ordered-groups-ii/AC5FFFCEC2F92B59E4AE6E80BD77728E?utm_campaign=shareaholic&utm_medium=copy_link&utm_source=bookmark

[hidden wind]

lovely, that link works

[outer gull]

thanks

[rotund aurora]

what is Irr(G_theta | theta)? I assume G_theta is the stabilizer of theta under the G action, which forms a group. So then you can look at the irreducible characters Irr(G_theta). If I interpreted it correctly, Irr(G|theta) are those irreducible characters of G that, when restricted to N, equal theta.

[coral spindle]

Let $H \leq G$ and let $\vartheta \in \operatorname{Irr}(H)$. A character $\chi$ of $G$ ``lies above'' $\vartheta$ if $\langle\operatorname{Res}^G_H \chi, \vartheta\rangle \neq 0$.

\bigskip
The set $\operatorname{Irr}(G \mid \vartheta)$ is the set of irreducible characters of $G$ lying above $\vartheta$.

[cloud walrus]

Boytjie

[coral spindle]

In other words, these note the representations of G that, upon restriction to H, have a constituent with character theta.

[rotund aurora]

and just to be clear, Res_H^G chi is given by composing the inclusion H-->G with chi, right

[coral spindle]

That's correct

It just forgets everything outside of H

[rotund aurora]

ok but how can an irreducible character of G_theta lie above theta? Does G_theta contain N?

[coral spindle]

G_theta must contain N, yes

Since theta by definition is a character of N, it is invariant on conjugacy classes of N

Ergo the action of N on theta stabilises theta

[rotund aurora]

ah, right

[coral spindle]

To be totally crystal clear, the action of $G$ on characters $\vartheta$ of $N$ is defined by $g \cdot \vartheta \colon n \mapsto \vartheta(g^{-1}ng)$ which is well-defined only due to normality of $N$.

[cloud walrus]

Boytjie

[coral spindle]

On the level of representations, it looks entirely similar.

[rotund aurora]

yeah

[coral spindle]

But yeah this works out as a very nice case, because my stabiliser here is either the normal subgroup or the whole group, and we have a canonical way to extend in the latter case.

The rest is counting.

[rotund aurora]

but when chi is itself irreducible this can only happen if the restriction of chi to H is theta?

I suppose what you did is sum |Irr(G wr C_p | theta)| where theta runs through the irreducible characters of G^p. I think I could get the right number, but it is not so clear why you are not overcounting

mmh but you had to divide by p

I believe the np comes from multiplying the number of irreducible characters of G (so the characters of G^p that have all "coordinates" equal) with the number of characters of C_p, which I assume is p. (By Gallagher.)

There are n^p-n characters of G^p whose coordinates are not all equal, in that case the stabilizer is G^p itself and Irr(G_p | theta)=theta, for theta irreducible character of G^p, no? But why did you have to divide by p?

[coral spindle]

No this isn't true

Let's take a simple example

The dihedral group with 8 elements has a normal Abelian subgroup of order 4

If what you claim were the case, then the restriction of every character of Dih(8) would be irreducible on this subgroup, hence would be degree 1. But in fact Dih(8) has a character of degree 2!

In the case of normal subgroups, we can describe the situation a bit. Clifford's theorem comes in and tells us that the restriction of an irreducible character is some multiple of a sum of a G-orbit of a character of the normal subgroup

That's a mouthful, so let me write it in symbols.

If $\chi \in \operatorname{Irr}(G)$ and $N \unlhd G$, then there exists some $\vartheta \in \operatorname{Irr}(N)$ and integer $m > 0$ such that $\operatorname{Res}^G_N \chi = m\sum_{\mathclap{gG_\vartheta \in G/G_\vartheta}} {}^g\vartheta$

[cloud walrus]

Boytjie

[coral spindle]

Clifford's theorem is more general, but no matter.

The point is that reps may split up, but they split up into G-conjugate reps at the very least.

[delicate orchid]

https://tenor.com/view/locked-in-monkey-monkey-meme-monkey-locked-in-monkey-face-gif-940796321824170934

[rotund aurora]

So, when the stabilizer is G^p, the irreducible characters above theta are conjugates of G^p wr C_p and hence conjugates of C_p? So there are p many of them?

[delicate orchid]

oh this is an interesting way of doing it

[rotund aurora]

wait

[rotund aurora]

I suppose it's the other way, an irreducible character of G^p wr C_p has p constituents?

lol

[coral spindle]

When the stabiliser of theta is G^p, we have the Clifford correspondence, which in particular means that the only character of the whole group lying above theta is exactly Ind theta, which looks exactly like conjugates of theta, yes

But there is only one

However, we can simply count how many characters theta are stabilised, and note that they never produce the same representation of the whole group. There are exactly n of these characters, corresponding to each of the representations of G. got this the wrong way round lol

[rotund aurora]

but there are n^p-n irreducible characters of G^p whose stabilizer is G^p, no?

[coral spindle]

Oh whoops I got it the wrong way round!

Yes sorry

Indeed. So we get a unique induced character for every G-orbit of these characters

So there are n^p - n such characters, but they come in orbits of size p. So in the end we get (n^p - n)/p irreducible characters of the wreath product produced in this way.

The ones that are stabilised by the whole group look different, sorry

[coral spindle]

N.b. this you can use Burnside to count

[rotund aurora]

wait to count what? When you know that there are n^p-n such characters and that they come in orbits of size p you are done, because of the sum |orbit|=|X| thingy (which is not called Burnside)

[coral spindle]

You can count the orbits via Burnside

But yes you are right, it is not necessary

[crystal vale]

Yes we take gcd of generators

[crystal vale]

Step 1: let H be generated by finite set S, such that S = { p_1/q_1,...,p_n/q_n } where gcd(p_i,q_i) = 1 for all i = 1,2,...,n.

So if I take S' = {p_1,...,p_n} then we can map H to <S'> such that

a_1p_1/q_1 +...+a_np_n/q_n -> a_1p_1 +...+ a_np_n, a_i are integers

Then my guess is it is isomorphic mapping.

Is it correct?

[silver kettle]

set all a_i to be equal and be equal to the product of all the q_i then the multiplication map is an isomorphism

[crystal vale]

Sorry, I don't get it

[silver kettle]

the map f(x) = q_1...q_nx is an isomorphism from S to a subgroup of Z

[crystal vale]

Yes to show it is one-one if q_1...q_nx = q_1q_2...q_ny => x = y, can I use here q_i's multiplicative inverse ?

[silver kettle]

sure

[crystal vale]

And in Z every subgroup is cyclic. And H is isomorphic to the subgroup of Z so it is cyclic.

Thank you ❤️

To show if n≥4, A_n is generated by 3-cycles (abc).

So there atleast 4 elements and if I showed we can generate (ab)(cd) then it is enough, right?

So we can generate (ab)(cd) = (abc)(bcd).

Is it correct?

[delicate orchid]

yeah this works

[crystal vale]

In question they ask for n≥3 but in n=3 I think I don't know how to show without computations so I tried for n=4

And in A_3 there are only 3 elements

And they are 3-cycles (123), (132)

Identity by (123)^3

[delicate orchid]

idk what you mean "I don't know how to show without computations". Computing things in proofs is fine

[crystal vale]

Okay thank you

[rotund aurora]

what is the lie algebra here

[rain grove]

In a group G, ord(a) = m and ord(b) = n. If ab = ba show that ord(ab) divides lcm(m,n).

So using what I learned last time I did: $(ab)^{mn} = 1 \iff ord(ab) | mn$

[cloud walrus]

OHHELLNAH

[coral spindle]

Hint: what is the defining property of lcm(m,n)? It's in the name.

[rain grove]

it is divisible by gcd(m,n)?

[coral spindle]

What does LCM stand for?

[rain grove]

least common multiple

[coral spindle]

Great.

So it is a multiple of m and n, and it is the smallest such thing.

Pseud don't

[knotty badger]

What might help here is

Hey why?

[coral spindle]

Now you need to spot one more thing

[knotty badger]

Ok hang on

I am allowed to talk here too

[coral spindle]

Is it always the case that (ab)^x = a^x . b^x? No, but in this case it is. You should use this special property.

[knotty badger]

You do want to use this additional property of order, but in a different way

What’s an equivalent way to say ord(ab) divides lcm(m, n)?

[rain grove]

ok ty for hints ill try to figure it out for a minute and msg again if i figure out smth

[knotty badger]

Cool

[knotty badger]

I still don’t understand why you said this

[coral spindle]

I don't want you distracting people by going on about how this is a universal property again, it feels like it is in every other thing you say and it gets old fast.

[knotty badger]

I mean, i actually helped this person solve the last problem

[coral spindle]

Great

[knotty badger]

I use them because they’re useful

I don’t think your preferences factor in

[coral spindle]

OK

[rain grove]

Hey its ok you are both very helpful and i thank you for helping me with this :))

[coral spindle]

I want to point out that everyone uses them because they're useful, and in fact considering this example is a definition, you can't really avoid it! But not everyone is mentioning it at every opportunity.

[kind kindle]

Has there been any research into the matrix representation of elements of the algebraic closure of GF(2)

(Yes, I know the matricies would be infinitely large)

[coral spindle]

I'm not sure what you mean by this, can you clarify?

[kind kindle]

At least i'm pretty sure they would be.

The Algebraic closure of GF(2) is a countably infinite field consisting of well... the algebraic closure of GF(2)

[knotty badger]

I mean, I mentioned it here cause it’s useful for this problem

[coral spindle]

Yes I'm aware of this, but what more specifically are you looking for?

[coral spindle]

The definition is indeed helpful.

[kind kindle]

A way to represent elements of the field reasonably so I can work with it a bit.

[coral spindle]

OK I see what you mean

The usual approach to this is actually by bounding the field you need.

[knotty badger]

Hmm, I don’t think the universal property is usually the definition of order, though.

[kind kindle]

I want to see if I can derive a nice distance function for the field. So I can define some equivalent of the real numbers for it.

[delicate orchid]

it's repackaging the data of the definition into an abstract mess

why do it

[coral spindle]

Since the algebraic closure of GF(2) is 'filtered' as you like by smaller fields (that is, GF(4), GF(8) etc) you can usually restrict just to the smaller fields

[delicate orchid]

it's pointless

it's just complete busy work

[coral spindle]

So maybe you could be more specific about what you're looking for Colo?

Like what kind of calculations do you want to do

[kind kindle]

I want to see if I can define calculus for fields very different than the real numbers.

[knotty badger]

Oh, interesting

Though why would algebraic completion be necessary for that?

[spice whale]

tangent vectors are linear derivations
for v,w tangent vectors, let [v,w] = vw - wv, where vw is the pointwise product of derivations v and w

you can prove that this is still a derivation

[delicate orchid]

oh so it is exactly like the matrix lie group case

[spice whale]

yes

i did just look this up on wikipedia

[kind kindle]

To eventually derive some equivalent of the real numbers using this algebraic closure.

[coral spindle]

While calculus per se is a little tricky, this has actually been done! In the field of algebraic geometry, there's a sensible notion of a tangent space of a variety, so you can define the derivative in that case. I believe that Gathmann's notes on algebraic geometry (you can google that and a pdf will come up) covers this.

[knotty badger]

Oh wow

[delicate orchid]

I wanted to say that ally but I couldn’t think of a good way of formalising multiplication

[knotty badger]

I didn’t know AG did calculus

[coral spindle]

It doesn't do 'calculus' so much as its own version of differential geometry.

[knotty badger]

Isn’t that like

[delicate orchid]

Diff geo is calculus

[knotty badger]

Coordinate-free calculus

[coral spindle]

You can see it like that, but my point is that you're not gonna start with limits.

[kind kindle]

I'll look into that sometime, might not understand it though.

[rotund aurora]

you can given an absolute value. Fields of finite characteristic over a prime field only have trivial absolute values tho

[coral spindle]

Limits are tricky in varieties.

[knotty badger]

Hm, right, no limits

[delicate orchid]

Derivation on a field 😭

[knotty badger]

Ohhhh wait

Is this the infinitesimal thing

[coral spindle]

Yes, unfortunately it is much harder than in the real case :)

[knotty badger]

Like dual numbers?

[delicate orchid]

Quotienting out by x^2 you mean?

Cause yeah they do that

[coral spindle]

You can indeed use the dual numbers to obtain tangents of varieties

There are other ways

[knotty badger]

Oh neat!

[coral spindle]

Many.... many........... mannynnyytyy other ways

[knotty badger]

I kinda use those in physics anyway

[delicate orchid]

This is the one case where I think your obsession with categorical aspects will be very useful pseudo

[agile burrow]

I want to comment that there's a second way to define the Lie algebra structure and it's not immediately equivalent to the first but it's a useful perspective. G acts on itself by conjugation, and this action fixes the identity e. So given an element g in G, by differentiating the action of conjugation by g, I obtain a map G -> Aut(T_e G). One can check that this action is smooth, hence by again differentiating at the identity, I obtain a map T_e G -> End(T_e G). By currying, I can view this as a bilinear map T_e G x T_e G -> T_e G. This precisely recovers the Lie bracket as Ally defined it, but it's not at all obvious why this is the same, or why it's even a Lie algebra structure (I don't know of a direct way to see that this satisfies the Jacobi identity)

[kind kindle]

But can we define some equivalent of the reals for this field? Like how R is the continuous version of N, and shares some of the same properties?
I know this isn't very rigorous haha.

[south patrol]

If ur worried about limits in varieties it's a good sign you should move onto schemes

[coral spindle]

No

[knotty badger]

Yeahhh this is how I remember it being done in my Lie algebra course

[coral spindle]

There is no sensible ordering in a field of positive characteristic

[coral spindle]

You may be interested, though, in looking at the p-adic numbers

[rain grove]

Is this ok:

If x = ord(ab) then because 1 = (ab)^x = a^x . b^x so a^x = 1 and b^x = 1, now applying that a^x = 1 \iff m | x and same for b it means that x is a multiple of m and n. And because x is the smallest such number that is a multiple of both m and n it is by definition lcm(m,n).

But im not sure about the part where i say a^x = 1 and b^x = 1 cause they could be inverses no?

[kind kindle]

Who says the field has to be ordered?

[delicate orchid]

Interesting. Almost like you’re using the adjoint rep to get a Lie algebra structure. Does this work with other reps?

[coral spindle]

These are 'cousins' of R which have interesting properties, and people indeed do analysis on the p-adics. But they are characterstic 0 fields.

[kind kindle]

but that makes sense. I wouldn't expect it to have an ordering per-se.

[knotty badger]

Hmm, I’m not sure how you conclude that a^x = 1 from a^x b^x = 1

In general this is not true

[kind kindle]

So I presume that means no nice equivalent of calculus or continuity.

[knotty badger]

And yeah, they could totally be inverses

[kind kindle]

But, I do have an idea for a natural extention of this field.

[coral spindle]

No there is a topology on the algebraic closure of GF(2), but no good notion of distance.

Topologies define what it means to be continuous

[knotty badger]

Oh huh there are topologies on algebraic closures

[coral spindle]

Yes of course, they are profinite.

[agile burrow]

This is exactly what's happening. I don't know if something happens if you instead consider other representations because we're distinctly using the fact that the adjoint action of G on itself preserves the identity in order to obtain a characterization of group representations purely in terms of the Lie algebra. This is essentially the basis for the key principle relating group representations to Lie algebra representations

[delicate orchid]

I could metricise it. I'm just built different

[rotund aurora]

what does pointwise product mean here? why isn't vw=wv?

[kind kindle]

Ah

[knotty badger]

Uhhh what does that mean

[kind kindle]

Again, forgive my lack of knowledge here, i've mostly been looking into stuff like this on my own out of curiosity.

[rotund aurora]

shit, there are like 3 convos going on

[spice whale]

i think i might be incorrect on the internet

[kind kindle]

I have no formal education on this topic

[coral spindle]

inverse limit of a nice enough sequence of gadgets

[rain grove]

But one case is if a^k = b and then it should be easy, the other case is if a^k != b, but then they can't be inverses to they must be 1

[delicate orchid]

all reps preserve the identity tho 😹 Maybe we need a rep arising from a semidirect product action for it to work

[knotty badger]

Oh, and each of these has a topology? That’s cool

[delicate orchid]

yeah the profinite topology

the inverse limit of a bunch of discrete topologies essentially

[knotty badger]

Well, I know that a^x = b^(-x), but I’m not sure you can conclude much further from there

[kind kindle]

But, I do have an idea in my mind for how to extend this field.

[knotty badger]

Though hmm, I suppose your argument amounts to checking the intersection of the subgroups generated by a and b

[spice whale]

ok no it's literally just composition @rotund aurora

[knotty badger]

Maybe that could actually work

[spice whale]

compose the derivations

[kind kindle]

Basically, think of elements of this field as elements of the vector space {0, 1}^N

[coral spindle]

Doing this is hard.

[kind kindle]

but only the ones with a finite number of nonzero elements

[knotty badger]

It’s not how I would approach it but that doesn’t make it wrong per se

[kind kindle]

Addition would just be normal vector addition over GF(2)

and multiplication would be some sort of convolution i think

[knotty badger]

Ok I think I can believe a version of this

[delicate orchid]

it's how I think about all profinite things

[rotund aurora]

how does composition make sense? Isn't a derivation some function from the germs at the point to k?

[delicate orchid]

I don't think it's that bad

[coral spindle]

Nah I mean finding a basis

Of the algebraic closure of GF(2)

That's really frikkin hard

[delicate orchid]

2nd root of unity... uhhh cube root of uh... 1+x...

[south patrol]

Oh

[kind kindle]

[1, 0, 0, 0, 0, 0.....], [0, 1, 0, 0, 0, 0, 0...]. [0, 0, 1, 0, 0, 0...]....

[south patrol]

I've been reading GF(2) as GL_2(F)

[delicate orchid]

cooked

[south patrol]

This happens every time

[kind kindle]

1, x, x^2, x^3....

[south patrol]

Who writes GF(2)

[coral spindle]

You're not seeing what I mean. Let x = [0, 1, ....]. What's x^15 - x^2 + 2?

[delicate orchid]

undergrads

[coral spindle]

Do you see what I'm saying?

[south patrol]

Tbf never seen it in my ug

[kind kindle]

this is GF(2)

2 = 1+1

1+1 = 0 mod 2?

[delicate orchid]

ok so you don't see what he's saying

[coral spindle]

Good point yes lol but you're not seeing my point still

[delicate orchid]

the multiplication in that basis is not the natural multiplication you'd assume

[kind kindle]

I might be totally wrong, again.

[delicate orchid]

it's incredibly difficult to figure out what the product of those two mfs should be

[kind kindle]

Sorry if I am.

[coral spindle]

A basis of the closure of GF(2) gives us an identification with these lists
BUT WHAT IDENTIFICATION???????

[agile burrow]

Maybe to add one step that's missing from ally's explanation: to each vector field at the identity, you obtain a left invariant vector field on G by pushing it around by left-multiplication. Given two vector fields X and Y on your manifold, the bracket XY - YX is another derivation on smooth functions, where (XY)(f) = X(Y(f)) (here Y(f) is another smooth function on the manifold). If X and Y are left invariant, then the bracket [X, Y] is again left-invariant (exercise). Take the value of the bracket at the identity, and you recover a Lie algebra structure on the tangent space at the identity.

[coral spindle]

You have to choose what the identification is first

and that's fucking hard!

It's really really hard!

[rotund aurora]

I have seen it. It might be slightly common in more combinatorial contexts. GF probably stands for "Galois field"

[kind kindle]

What do you mean by this?

[south patrol]

It does

[kind kindle]

You can do this for GF(2) and GF(4) easily I think

[delicate orchid]

it's what's used for finite fields in GAP

[south patrol]

I have seen it just only on this server lol

[knotty badger]

That’s a nice coordinate-free way to do it

[delicate orchid]

they're finitely generated, of COURSE you can

[south patrol]

Or progrms

[knotty badger]

Our course just did matrix commutator

[delicate orchid]

now do it for all of them at once

[coral spindle]

Yes, for tiny fields it's easy. But as the fields get bigger, computing them gets much harder. Can you compute GF(128)?

[kind kindle]

Sorry if i'm making you angry, like i'm really really sorry. Thanks for putting up with my math knowledge.

[coral spindle]

For the infinite case, it's very very hard.

[delicate orchid]

I'm not angry lmfao

I just type very emphatically

dw boss

[agile burrow]

Yeah, there are kind of slick ways to do this for general Lie groups which is nice if you don't want to restrict to linear ones

[kind kindle]

Why can't you just represent it with the basis [1, 0, 0, 0, 0...], [0, 1, 0, 0, 0...].... (128 vectors later) [0, 0, 0, 0, 0.... 1]

[knotty badger]

Right, for us we were mostly concerned with applications of Lie groups to physics

[kind kindle]

Yes, multiplication would NOT be pointwise.

[delicate orchid]

ok, then what's the first basis multiplied by the second

[knotty badger]

So standard model stuff

[coral spindle]

This does nothing! This tells us only the addition!

[kind kindle]

It would be a convolution

[delicate orchid]

yeah exactly

[coral spindle]

So you still need to work out what the multiplication is

And this

again

is really hard!

[delicate orchid]

also not 128 vectors, just 7

[agile burrow]

Fair enough, most Lie groups I've thought about are linear anyway. Maybe the one exception that's cropped up frequently enough in my life is the universal cover of SL(2, R)

[knotty badger]

Ooh, where does this crop up?

[agile burrow]

Mostly when I think about hyperbolic geometry and braid groups. It's one of the eight model Thurston geometries for three dimensional manifolds

[knotty badger]

Ooh, I think I’ve heard that braid groups get used in condensed matter

For anyons and stuff, you’ve gotta look at representations of braid groups

[rotund aurora]

Could you explain how are you composing X and Y? Isn't the tangent space the set of derivations (germs at the origin)-->k?

[knotty badger]

You’re composing the vector fields I think

Which work as endomorphisms of C^infty(G)

[agile burrow]

I've extended X and Y to be vector fields on the manifold, so now given a smooth function f on G, Xf is a smooth function on G where at each point p, (Xf)(p) is defined by applying the the vector X_p to the germ of f at p. In particular, I can apply Y to Xf (and in a parallel way I can apply X to Yf). The bracket [X, Y] applied to f is by definition X(Yf) - Y(Xf)

[tribal moss]

I'm not exactly sure what's going on here, but it doesn't really look like the kind of introductory algebra #groups-rings-fields was created for. Are you sure it shouldn't be happening in a more advanced channel?

[agile burrow]

Yeah you're right, sorry I should've moved this to something like #diff-geo-diff-top

[kind kindle]

alright messing around with this whole multiplication thing a bunch, i don't see what the issue would be extending it.

We only need to define multiplication between our basis elements here I believe.

[delicate orchid]

yes that's true

so uh... good luck!

[kind kindle]

Yeah, that's not super hard. Define the basis as 1, x, x^2, x^3....

from there you should be able to represent every element of the closure of GF(2) right?

[tribal moss]

Is the issue here only to find an explicit way to calculate in the algebraic closure of F2, or is there something further after that point?

[delicate orchid]

ok and it's representable as a quotient of a polynomial ring because...?

we want a basis for F_2 bar basically

just so we can do multiplication

[delicate orchid]

and if so, we need the exact quotient to define the multiplication

these are the steps which must be completed in our quest

[tribal moss]

In an algebraically closed field, your "x" basis element should have a square root -- what is that?

[kind kindle]

x+1

this is gf(2)

or the extension of it.

meaning it contains GF(4) as a subfield.

ohhhh wait

yeah yeah, i see the problem

but, this is the closure of GF(2), so it has to exist.

I mean, this representation I think would contain a sqaure root of x

but the vector would have an infinite number of nonzero elements which... complicates things.

[spice whale]

ok so apparently, the smooth action of G on itself induces an action on the space of vector fields, which means there's only one vector field associated to each tangent vector at 1
and a whole vector field treated as directional derivatives takes a smooth function M → R to another smooth function M → R
which means you can just compose the associated vector fields and then project back down to the tangent at the identity

[tribal moss]

By definition, if you have a basis for a vector space, each element is a combination of finitely many basis vectors.

[kind kindle]

This is an infinite dimensional vector space.

[tribal moss]

Yes. That doesn't change the fact that only finitely many basis elements can be used in a linear combination.

[kind kindle]

Isn't R a vector space over Q

[tribal moss]

Yes.

[kind kindle]

yet the sum of any finite number of rational numbers will be rational.

[delicate orchid]

yeah

[tribal moss]

In a basis for R over Q, there will be at most one rational element.

[delicate orchid]

which is why R isn't finite dimensional over Q

also Q[i] is a vector space over Q and your statement still holds so actually your point is just inconsequential

I think you're thinking that basis vectors have to be in the underlying field for some reason

[kind kindle]

Alright, according to wikipedia that definition holds, but why?

Why can't we have an infinite linear combination?

[tribal moss]

Because making infinite sums mean something would require us to have a concept of limits, and we don't have that for a general vector space.

[delicate orchid]

it's also like... by definition of a basis?

but then you'd be right to ask why we define a basis like that ig

[kind kindle]

I have a hard time making sense of this still.

For finite dimensional vector spaces, that makes perfect sense.

For for say, the vector space of all functions.

[tribal moss]

If we do happen to have a topology on our vector space, we can speak about "Schauder bases" which must generate each element uniquely as a possible-infinite linear combination. But that is not the kind of basis we're talking about here.

[coral spindle]

Yes it is unintuitive.

[kind kindle]

How the heck would you define a basis for that, where each element could be written as a finite linear combination

[coral spindle]

You cannot write down a basis for this space.

It is impossible.

But it exists due to the axiom of choice.

I'm not just saying it's impossible because it's infinite. It's impossible in a much more mathematical way: without the axiom of choice, we cannot guarantee its existence, and the axiom of choice doesn't tell us how to construct something.

[kind kindle]

What if we do generalize the notion to allow for infinite linear combinations?

In that case, writting out a basis for the space of all functions becomes trivial actually.

[coral spindle]

It is a bit more complicated than it might at first seem

I'm hoping Tropo can comment on this

[tribal moss]

(By the way, I think one can show that the algebraic closure of F2 at least does have a basis where multiplication is computable. But implementing that proof straightforwardly would lead to absurdly inefficient multiplication algorithms).

[kind kindle]

According to wikipedia there is a viable algorithm for it.https://en.wikipedia.org/wiki/GF(2)#Algebraic_closure

[coral spindle]

This is the method I pointed out to you first!

This is just saying you can compute in the algebraic closure by computing in the finite fields.

But fwiw, this says nothing about finding a basis, which is harder.

[delicate orchid]

thank the heavens I never need to worry about anything other than adjoining roots of unity

[south patrol]

In which situations is it useful to do these explicit multjplications/ find bases

[tribal moss]

You can take a countably infinite set X and consider the vector space of functions X -> F2 with pointwise addition just fine. However that certainly cannot be the algebraic closure of F2, because the algebraic closure is countable and F2^X is uncountable!

[coral spindle]

Potat are you asking where finite fields are useful

[rotund aurora]

do you mean over the algebraic closure?

[south patrol]

No

[delicate orchid]

yes.

[coral spindle]

🙏

[kind kindle]

But if one were to allow for infinite linear combinations, you could just define the basis for the space of all functions with the set of functions f(x) = 0 unless x = some number in the reals, in which case it would be 1.

[south patrol]

I mean I do stuff w finite fields but haven't had to like write out a basis for F_q ig or pick some explicit model

[kind kindle]

the union of all of those functions in 1 set.

[south patrol]

Maybe this is more a computational math thing idk

[delicate orchid]

it is

[coral spindle]

OK so let f be the function with f(0) = 1 and f(x) = 0 otherwise. What's the infinite linear combination f + f + f + f + f + ... evaluated at 0?

[rotund aurora]

How would you calculate the number of solutions of an elliptic curve over F_q?

[delicate orchid]

but it's still nice to have a concrete picture of the elements of these things

[south patrol]

Idk it's hard

Weil conjectures

[kind kindle]

Fair

[coral spindle]

Yeah it is actually hard.

[south patrol]

But yeah true in that case you should do this and it'll be a pain sure

[kind kindle]

You could just say that's the driac delta function.

But that would mean we're now suddenly working with distributions not functions.

[coral spindle]

That would be an ad-hoc solution to a problem that remains.

Well, not even a real solution.

[kind kindle]

Which I have no idea about other than the fact that they are not functions.

[rotund aurora]

can you calculate the zeta function without counting solutions? Idk but I would have thought that the other direction is more common

[south patrol]

Owo

[coral spindle]

Anyway, it's useful now at least to keep in mind that vector spaces only support finite linear combinations.

[knotty badger]

Mhm mhm, that’s all you can really get algebraically

[kind kindle]

Still, I have a feeling that a basis might be possible to construct.

But uh.. not gonna try with my current level of math experience :P

[rotund aurora]

but in practice, would the faster method be to restrict the finitely many elements that you are dealing with to a finite field and use the computational methods available there? I think for finite fields there are efficient algorithms, but idk how they relate to one another when you change fields (of the same char) or pass to extensions/subextensions

[kind kindle]

thinking about it a bit more, the set of algebreic numbers is countable.

but uh.. that's not doing us much good.

the vector space would also have to contain the solution to x^5+x+1, which isn't solvable yay.

[tribal moss]

F2^X was definitely not something I was suggesting as a solution to this problem; my aim was to explain to C48 why his idea of just allowing arbitrary formal linear combinations (with possibly infinite support) wouldn't lead to one.

[kind kindle]

I overdosed on abstract nonsense and I need a break

[coral spindle]

You're confusing C and the closure of GF(2)

[kind kindle]

i met the solution in the vector space.

[coral spindle]

The closure of GF(2) is not contained in C in any meaningful way

[kind kindle]

Like how sqrt(x) would be in GF(2)

where x is one of our basis elements.

and sqrt(s) is the solution to s^2-s = 0

[rotund aurora]

Sorry, I meant to answer to this message. @tribal moss

[coral spindle]

That doesn't matter, because that's not how we work out the values of the Galois fields

In fact, every finite extension of a finite field is solvable.

So this is simply false.

[kind kindle]

my head hurts

[tribal moss]

Oh, right then. The part that I can't immediately see a way do do efficiently is if we want to add or multiply two elements from different fields GF(2^a) and GF(2^b), well need to promote them both to live in a larger field such as GF(2^lcm(a,b)). That means we have to remember how each of the fields sit inside the others, but there are infinitely many such inclusions to deal with, so we can't just have a table of them.

Of course, as long as you're only working with {+,-,×,÷} you can choose a large enough field once and for all.
But given that we want to have the algebraic closure in the first place, I suspect we'll want to at least take roots too, which can take us higher in the hierarchy.

[hidden wind]

if a group homorphism is bijective it's an isomorphism; if injectiive it's a monomorphism; if surjective it's a epimorpism; is there a name for a homomorphism that is neither surjective nor injective?

[knotty badger]

Probably just “homomorphism”

[tribal moss]

Not that I immediately recall.

[hidden wind]

yeah but that is less specific :S

[knotty badger]

The first isomorphism theorem does tell you that every group homomorphism takes the form of

A quotient map, followed by an isomorphism, followed by an inclusion

[tribal moss]

"inferiomorphism"?

[coral spindle]

"A homomorphism, that is neither surjective nor injective"

[hidden wind]

[dull ginkgo]

If A and B are rings, and B has finite rank N as a left-A module, then End_B(B) = B_r (right translations, isomorphic to B^op) embedding into End_A(B) ~= M_n(End_A(A)) ~= M_n(A_r).

Does that mean End_A(B) has finite rank?

[tardy hedge]

Oh HELLL nah he’s back at it again

[tribal moss]

What?

[knotty frigate]

dont we need banach spaces for infinite linear combinations to be behave nicely?

[rain grove]

So i was thinking a bit and now I think this is not even true:

so ord$(a) = m$ and ord$(b) = n$
but if $b = a^k$ then ord$(b) = $ord$(a^k)$ = $\frac{m}{\text{gcd}(k,m)}$
ord$(ab)$ = ord$(a^{k+1}) $= $\frac{m}{\text{gcd}(k+1,m)}$

But for example if $m = 6$ and $k = 3$ then lcm$(m,n) = m$ but ord$(ab) = \frac{6}{\text{gcd}(4,6)} = 3$

and then lcm$(m,n) \neq $ord$(ab)$

[cloud walrus]

OHHELLNAH

[knotty badger]

Would you like a suggestion?

Also, you only need to show the order divides the lcm, not that it’s equal

[rain grove]

Ohhhh

[tribal moss]

Yeah, at least for some values of "nicely". But I think the raw definition at least makes sense without completeness.

[rain grove]

Omg

[knotty frigate]

are there any other topology that can be defined on vector spaces that make them behave that well?

[rain grove]

Sure if you already thought of one, but I think i got it now

[knotty badger]

I was going to suggest using the “universal property” of ord(ab)

If you’ve got it then that’s fine, though

[knotty frigate]

wait what is ord(a)? the only place ive seen ord is in set theory for ordinal

[knotty badger]

Oh here it’s just the order of a group element

The definition is $\text{ord}(g) = \min_{\leq} { n > 0 \text{ such that } g^n = e }$

[cloud walrus]

Pseudonium

[knotty badger]

The universal property is $\text{ord}(g) | n \iff g^n = e$

[cloud walrus]

Pseudonium

[knotty frigate]

isnt it also defined as |<g>|?

[knotty badger]

Yes, that’s equivalent to the definition I presented

[knotty frigate]

figured

sorry its been a while since ive touched AA

[knotty badger]

Mhm no problem

[knotty badger]

…perhaps with a caveat about elements of infinite order

[knotty frigate]

wait what happens with infinite order elements?

oh wait your definition would lean towards ordinals

[knotty badger]

Well then in the definition I presented, you’re taking $\min \emptyset$

[knotty frigate]

whereas mine would spit out cardinals

[cloud walrus]

Pseudonium

[knotty frigate]

right?

[knotty badger]

And that’s not really a nice thing to do

[knotty badger]

Your definition would give $\infty$

[knotty frigate]

oh let me reread this

[cloud walrus]

Pseudonium

[knotty frigate]

speaking of abstract algebra. are there any elementary implications of the abelian ring axioms involving multiplication. Im working within a system of axioms and just managed to prove that the structure this induces would be a ring but i dont know what implications it would have that i havent proven already (ive already proven that 0 * n=0; -(n) * m=n * -(m)=-(n * m) )

i wanted to see if i could prove the cancellation rule for multiplication (which is true for the structure i want to be able to induce but i dont know if its true just based on this theory and) it certainly doesnt follow from the ring axioms

[marsh scaffold]

Guys so I've been reading on doubly transitive actions of G on a set X and I was introduced to a nice characterisation that G acts doubly transitive on X iff it acts transitive on X and the stab_x_o acts transitively on X-{x_0}

Now I ve been looking for an example where we have stab_x_o acting on X-{x_0} transitively

But the action on X is not transitive

Like I would want to know why the G acting transitively on X condition is necessary

[rain grove]

Is this ok?

I think i went the longest path possible

But like also idk about (2) cause my result is that in (2) lcm(m,n) = ord(ab)

problem: In a group G, ord(a) = m and ord(b) = n. If ab = ba show that ord(ab) divides lcm(m,n).

[knotty badger]

I think this works, but you can considerably shorten the proof

[rain grove]

How would you do it?

[knotty badger]

So I’d start with the universal property of $\text{ord}(ab)$

[cloud walrus]

Pseudonium

[knotty badger]

Namely, $\text{ord}(ab) | n \iff (ab)^n = 1$

I assume this is familiar?

[cloud walrus]

Pseudonium

[rain grove]

yeah

[knotty badger]

Cool

We want to show $\text{ord}(ab) | \text{lcm}(m, n)$

[cloud walrus]

Pseudonium

[knotty badger]

By the universal property, this is equivalent to showing $(ab)^{\text{lcm}(m, n)} = 1$, right?

[cloud walrus]

Pseudonium

[rain grove]

yeah

[knotty badger]

Now, since $a$ and $b$ commute, $(ab)^{\text{lcm}(m, n)} = a^{\text{lcm}(m, n)} b^{\text{lcm}(m, n)}$

[cloud walrus]

Pseudonium

[knotty badger]

But both factors there are 1 - the lcm is a multiple of ord(a) and of ord(b)

So their product is 1

And you’re done

[rain grove]

ok wow

im so fkn sad now

this is so nice

[knotty badger]

Aw, don’t be!

You solved the problem, after all

[rain grove]

yep at least something haha

[knotty badger]

Adding this to the bank of problems where the universal property of ord(g) is useful

[marsh scaffold]

Oh so like I've got down that if stab_x_0 is a proper subgroup of G the we don't need G is transitive on X cause its implied

And like S_{n-1} acting on {1,2,...,n} works imo

As a counter to the problem not being true when the stab_n =S_n-1 then it wouldn't be doubly transitive

[long nebula]

I mean, I think any problem where the definition of order is useful would be a problem where the universal property of order is useful, no? They are basically the same

Since the definition of order is the least number such that...

[knotty badger]

Different sense of least

[long nebula]

Yeah I guess so

[knotty badger]

.

[knotty badger]

.

These are not exactly the same

[long nebula]

Yeah they're not exactly the same, it's nice that orders play nicely with the divisibility structure. The two ways of thinking about it are so equivalent to me though that I struggle to think of a situation where one would be useful and the other wouldn't haha.

[knotty badger]

Mhm I get what you mean

Cat theory helps make differences like this visible to me

Also there was a problem a while back where the universal property + yoneda was helpful

[long nebula]

Well yeah I mean I think it's obvious there is a difference, hence why I said "basically the same" and not "the same", I just found your statement funny because to me it more or less reads like "Adding this to the bank of problems where the definition of ord(g) is useful"

(modulo a small lemma about how the ≤ structure and | structure of N relate)

[knotty badger]

Mhm, but if someone is new to group theory

This isn’t the definition of order

[long nebula]

Yes I know

[knotty badger]

So I think it’s useful to have a name for it

Makes it more concrete/explicit in students’ minds

[long nebula]

I didn't say anything about that, I just said that it'd be funny to have a bank of problems specifically where the universal property of ord is useful, just like it'd be funny to have a bank of problems specifically where the definition of ord is useful

[knotty badger]

.

[tribal moss]

Forgive me, what is the universal property of order?

[knotty badger]

This

[tribal moss]

I see.

[knotty badger]

Usually one proves this soon after you state the definition of order

Which is either the one I gave, or sometimes “size of subgroup generated by g”

At least, I don’t think it’s common to define order by the universal property

[long nebula]

Yeah, it's not

[rain grove]

Ok so a similar problem but still idk how to finish this:

$a \in G, \text{ord}(a) = m$ and $b \in h, \text{ord}(b) = n$.
Show that for $(a,b) \in G \times H$ $\text{ord}((a,b)) = \text{lcm}(m,n)$

[cloud walrus]

OHHELLNAH

[rain grove]

With universal proprety i get "one side" ord((a,b)) | lcm(m,n)

im trying to get lcm(m,n) | ord((a,b)) now but no luck

[vapid vale]

a very direct way to solve this is argue that (a, b)^lcm(m,n) = (1, 1), and that for any value less than lcm(m, n), that isn't true

[long nebula]

It suffices to prove that ord((a,b)) is a common multiple of m and n

[knotty badger]

It might be needed to use the other definition of order for this, yeah

I suppose there is a universal property for lcm which helps

But that may be, uh, too many universal properties at once

[knotty badger]

This is likely a better approach

It’s like - you don’t have to choose between the universal property or the usual def, you’re allowed to use whichever is more convenient

[knotty badger]

Like in my case I’d solve this entire problem with just universal properties, but that’s just cause I’m very familiar with them

[dull ginkgo]

Showing R[X] is a graded ring using universal property alone (without knowing it’s a polynomial ring a priori”) is kinda a pain

[knotty badger]

Yeah I don’t think universal properties should be presented as the One True Perspective

Sometimes you really want to work with the actual construction

[dull ginkgo]

You have to show R[X]/(X) via equating it with the extension of the zero map, matching the kernels, then show the maps (X)^n/(X)^(n + 1) are module-isomorphic to R via factoring out the x due to it being principal

[knotty badger]

The way I usually phrase it is

A universal property tells you what something “does”, but you often also want to focus on what something “is”

[dull ginkgo]

It’s a good exercise with universal properties imo, but it sucks

[knotty badger]

Both are important

[dull ginkgo]

“constructing” the tensor algebra off of it’s universal property in a similar vain can be done but is it worth it? No :3

[knotty badger]

Mhm sure

[dull ginkgo]

Just nice to use the tensor product construction that already exists

[knotty badger]

Yep yep

For me, what’s nice is that you’re allowed to freely move between looking at what something “is” and what something “does”

That’s the essence of the yoneda lemma

[dull ginkgo]

Which in its own right, can also be constructed instantly off it’s universal property much like abelianization, quotienting out by a kernel that represent the relations of bilinear maps

[knotty badger]

Yep yep

[barren sierra]

Doesn't this just show it's a common multiple of m, n

Not necessarily least?

[knotty badger]

It’s to do with the universal property of lcm

Lcm(m, n) | k iff m | k and n | k

[dull ginkgo]

I want to learn some deeper category theory to understand natural maps and adjoints / derived functors better

[barren sierra]

i.e. you may have that 2lcm(m, n) is the order

or idk how far you're stretching the word "suffices" lol

[knotty badger]

I’ve found it very fun, certainly!

[knotty badger]

If you look back at the message

[barren sierra]

I still think there's a little to write after you show common multiple

[knotty badger]

This is what they’re trying to show

They’ve already shown the other direction

[barren sierra]

Oh

I see I see

[dull ginkgo]

The only derived functors i know off the top of my head are the two for the adjoint endofunctors Hom and Tensor product with their Ext and Tor ones

[knotty badger]

Yeah I haven’t actually done much homological algebra

[dull ginkgo]

I wonder if there is some connection between derived functors of adjoints

[knotty badger]

There’s a pretty short proof if you use the universal property of lcm, but I wonder if that may be too much at this stage

[dull ginkgo]

I’ll get there at some point

Another thing i want to learn is Morita Connections / Dualities

I already have enough shit i want to learn but I have no time constraint really so i can just cram what i want into my noggin

[long nebula]

They just asked how to prove that lcm | ord

They already proved the other direction

[barren sierra]

Ya I missed that while scrolling

lo siento

[tribal moss]

So the viewpoint here is that when something is a supremum (or infimum) in a poset, it makes sense to call this fact "the universal property of such-and-such"?

[knotty badger]

Well, a universal property for an object x in a poset under some relation <= is merely an alternative description for when x <= y or y <= x

Suprema and infima are examples of such universal properties

In general, a universal property is an alternative description for arrows into or out of your object

It’s just useful to have a common name to refer to this phenomenon, is all

Like, it’s a new set of words yes, but you only need to learn them once in a sense - then, you just see lots and lots of specific examples of them

[shell pilot]

Show that a nonconstant polynomial in Z[x] that is irreducible over Z is primitive.

Just looking for a hint. I know that the only units in Z are 1 and -1 and for a polynomial to be primitive means that the gcd of the coefficients of the polynomial is 1.

For the polynomial (call it f(x)) in Z[x] to be irreducible means that it can be written as the product of g(x) and h(x) where g(x) or h(x) are units of Z[x] and g(x) and h(x) are also elements of Z[x].

[glad osprey]

Do universal properties always uniquely identify objects?

[knotty badger]

Indeed, up to unique isomorphism! So long as you state them correctly

Of course, they don’t tell you such an object actually exists

That’s what you need constructions for

They tell you what an object “does”, how it relates to other objects, how to use it, but not what it “is”

Still, a universal property can give you hints for how to construct an object

[glad osprey]

And the universal property for ord(a) mentioned above says exactly which arrows go into and out of ord(a) in the poset of divisibility, right? It makes sense that's enough to show it is unique in that particular poset, but what would unique isomorphism mean in this case?

[knotty badger]

So, it tells you the arrows that go out of ord(a), not the ones that come in

It tells you $\text{ord}(a) | n \iff a^n = e$

[cloud walrus]

Pseudonium

[knotty badger]

Up to unique isomorphism here just means that - if $k$ and $k’$ both satisfy the universal property, then you always have $k | k’$ and $k’ | k$

[cloud walrus]

Pseudonium

[rain grove]

$G,H$ finite cyclic groups. Show: $G \times H$ cyclic $\iff |G|$ and $|H|$ coprime

This is the proof I wrote so far but im not sure how to argue the last part... The last part i wrote because i think its right but idk how to prove it yet

[knotty badger]

This is cause $k | n \iff a^n = e \iff k’ | n$, then substitute $n = k$ and $n = k’$ respectively.

[rain grove]

No other elements g', h' ...

[cloud walrus]

Pseudonium

[knotty badger]

This is essentially applying the yoneda lemma

[cloud walrus]

OHHELLNAH

[knotty badger]

Which is a very general result from category theory - no need to look it up or anything, just a fun aside

[knotty badger]

For the last part, you can actually just compute ord((g’, h’)) for arbitrary g’, h’

And show it’s always strictly smaller than the group size

It’ll help to use your ord(g^k) result

[glad osprey]

Aha, and since divisibility is antisymmetric, the unique isomorphism just turns into equality. So in this case there is just one object satisfying the universal property

[knotty badger]

Yep! Unless you allow for negative integers

Then -ord(g) works too

[glad osprey]

But you said the universal property. How can you be sure there isn't another universal property of ord(a) in another poset, or a completely different category?

[knotty badger]

I mean idk, this just seems to be a pretty obvious one for a group

[rain grove]

wdym compute?

[knotty badger]

Like there’s a formula for it

[knotty badger]

Like, if you find one, you can state it sure

I just haven’t found one lol

[glad osprey]

Yeah, seems unlikely there's two completely different universal properties for the same object 😅 Anyways, thanks for the info 🙂

[knotty badger]

I mean

It’s not uncommon to have multiple

The more the merrier I say

In this case technically $\min$ has a universal property, so you could say

[cloud walrus]

Pseudonium

[knotty badger]

$x \leq \text{ord}(g) \iff \forall n > 0 \text{ such that } g^n = e, x \leq n$

[cloud walrus]

Pseudonium

[knotty badger]

That’s a universal property in a different poset

It just ain’t as useful as the divisibility one

[knotty badger]

E.g. biproducts in abelian categories

[glad osprey]

I see. In that case though you would need to downgrade to the indefinite article, which is very sad

[knotty badger]

I mean I don’t mind

I’m happy to call it “a” universal property

After all I’m the only one who seems to call it this lol

[errant wedge]

Double checking an exercise, $\Z[i]/\langle(3+4i)\rangle\cong\Z_{25}$ right

[cloud walrus]

Sara

[valid rose]

yes, (\mathbb{Z}[i]/(a+bi)) is isomorphic to (\mathbb{Z}/a^2+b^2\mathbb{Z}) for coprime (a) and (b).

[cloud walrus]

Millie

[valid rose]

(and if a and b are not coprime, CRT applies)

[marsh scaffold]

Wait could you expound more on this

[valid rose]

(\mathbb{Z}[i]) is has unique factorisation (and in fact is a PID), so you have [\mathbb{Z}[i]/(z) \cong \prod \mathbb{Z}[i]/(\pi_{k}^{r_{k}})] where (z = \prod \pi_{k}^{r_{k}})

[cloud walrus]

Millie

[valid rose]

so then all you need to know are the primes of Z[i] and what happens for prime powers

[chilly ocean]

What do composition series/simple factors of a group tell us that (not necessarily maximal) normal series don’t?

[marsh scaffold]

I see but given arbitrary a,b we have to first find the irreducibles correct?

[valid rose]

yeah you have to be able to factor into gaussian primes

this can be done algorithmically though

[marsh scaffold]

Oh i would like to know how

Say I've been given 3+3i

[valid rose]

3+3i is particularly easy because it's 3(1+i)

but in particular you look at norms

[marsh scaffold]

Ah okay

I think I understand then check for factors

Then rinse and repeat

[valid rose]

yeah Z[i] is a euclidean domain with the norm a^2+b^2, so factorising is easy

[marsh scaffold]

Okay, thanks ❤️

[errant wedge]

can you explain how to do this sorry, i dont immediately see where norms come in

also what is norm in this context

[errant wedge]

I did this using the classification of free abelian groups

[valid rose]

the norm of a + bi is a^2 + b^2 (and also (a+bi)(a-bi), which is a very useful fact)

[errant wedge]

Mhm

Oh wait you just mean like norm on C

[valid rose]

ye

and it has the exceedingly useful number theoretic property that for y, z in Z[i] we have N(yz) = N(y)N(z)

[marsh scaffold]

Also a norm(in context of EDs) satisfies some properties i believe

[valid rose]

which means for factoring you only have to look at gaussian integers of smaller norm

[errant wedge]

I read this as norms being a property of EDs

[valid rose]

I mean it is, it's just in this case we're lucky that that coincides with the field norm

[errant wedge]

Oh ok, what is a norm for EDs?

Ty again

[valid rose]

so a norm for a euclidean domain is a function f: R\{0} -> N such that a and b in R, b != 0, you have q, r in R st a = bq + r and r=0 or f(r) < f(b) -- ie just enough to be able to the same style of divison arguments as we have for natural numbers

the field norm however, for an algebraic number field, is the product of all the conjugates of an element (like with Z[i], you have (a+bi)(a-bi), and you can do similar things in other algebraic number fields and their rings of integers)

but this isn't always a norm in the sense of euclidean division

[errant wedge]

Ah ok, thank you :)

[errant wedge]

So we'd have $\Z[i]/(3+3i)\cong\Z[i]/(3)\times\Z[i]/(1+i)$

[cloud walrus]

Sara

[errant wedge]

Ik $\Z[i]/(1+i)\cong \Z_2$, would I just use classification of free abelian groups for the former

[cloud walrus]

Sara

[valid rose]

you can also do these by drawing out the lattice -- though i don't have time to actually draw an example of this by hand

[errant wedge]

Is $\Z[i]/(3)\cong \Z_3\times\Z_3$?

[cloud walrus]

Sara

[valid rose]

it does indeed have 9 elements, but it's not quite that. An easy way to do it is with the isomorphism theorems for rings: [\mathbb{Z}[i]/(3) \cong \left[\mathbb{Z}[x]/(x^2 + 1) \right] / (3) \cong \mathbb{Z}[x]/(3, x^2+1) \cong \mathbb{Z}/3\mathbb{Z}[i]]Another way to see this is that since ((3)) is maximal in (\mathbb{Z}[i]), (\mathbb{Z}[i]/(3)) is a field, and it has 9 elements (write them down) so it must be the field with 9 elements.

[cloud walrus]

Millie

[errant wedge]

Ah ok. Is it true they're isomorphic as additive groups?

[valid rose]

yes

[errant wedge]

tyty

[errant wedge]

Sorry, just confirming, the last one is (Z/3Z)[i] and not Z/(3Z[i]) right

[upper pivot]

yeah

[errant wedge]

cool ty

[upper pivot]

been a while btw, how r u. U should apply to my WIM DRP project next term :3

[errant wedge]

omg john hi

im good

[errant wedge]

im doing a lot next term so idk if im gonna do drp, still on the fence, which is urs again?

[upper pivot]

Word hyperbolic groups

yeah fair

[errant wedge]

the one that mentioned boone novikov?

[upper pivot]

no the one like, below it

geometric group theory basically

[errant wedge]

Does look interesting

[upper pivot]

lol no pressure to apply to it ofc

[errant wedge]

hrmm I really wanna, need to decide if its a good idea by sunday

[errant wedge]

No I do wanna do drp again, just unsure if its a good idea

im taking
GER 101 - Elementary German 1
MATH 249 - Intro to Combinatorics (Advanced)
PMATH 367 - Topology
PMATH 446 - Intro to Commutative Algebra
PMATH 945 - Topics in Algebra (Rings and their Applications)
So my friends are telling me drp again is prolly a bad idea and im also on the fence hrmm, can you tell me what u think the time commitment might be 👀

[upper pivot]

hmm yeah i mean ur friend is probably right this seems like hard coursework

[errant wedge]

945 is with bell at least, its just biweekly assgns + take home final

[upper pivot]

oof yeah that will probably take a lot of ur time

u should do analysis courses too at some point!