groups-rings-fields

[next obsidian]

Lol

Basically I have a sheaf of modules, and I've defined them via colimits

So on a base I have defined F(U) to be an ideal of O_X(U), on a cofinal set of open U's (open affines of a scheme if you care)

Then I've defined F(V) arbitrarily to be a colimit of the F(U) for U a subset of V

And I want to show that this actually leads F(V) to be an ideal of O_X(V), and I have a condition about the maps which says the maps from like F(U) -> F(U') for U,U' in my base play nice

[vocal depot]

Anyone home?

I'll check back at normal person hours.

But just in case, I'm trying to get my thoughts organized enough to answer a question from a list of problems my professor suggested we look at.

I figure some part of it has to do with the relationship between reflections and rotations but something about how points at the vertices move I think has to be addressed here.

Because a reflection about a line of symmetry doesn't itself equate to a rotation but two reflections does so somehow I'm supposed to describe how the act of reflecting it twice moves the points out of the right alignment and then back into it.

So if anyone could potentially guide me toward the logic I might need to construct a coherent thought about that I would be greatly appreciative.

[jovial nexus]

@next obsidian are you sure you extended by colimits?
Because the Sheaf property literally states that F(V) = lim_U F(U) where U ranges over a covering of V

[next obsidian]

I think it’s by colimits... maybe it’s a limit

[jovial nexus]

I kinda doubt it's a colimit tbh

[next obsidian]

Oh my god

[jovial nexus]

A section over V should be given by a compatible family of sections over basis elements contained in V

[next obsidian]

Inverse limit

Is not a colimit

[jovial nexus]

Not an equivalence class of stuff

Yeah it's not

[next obsidian]

☠️

This makes a lot more sense

[jovial nexus]

yeah

Haha

So anyway

You've got a limit of rings

[next obsidian]

I actually think I have it then

[jovial nexus]

And take a limit of ideals over each of those rings

[next obsidian]

I realized if I have an injection into A considered an A-module

[jovial nexus]

And want to check that considered as a module over the limit of rings, it stays an ideak

You can think about that in this abstract setting

[next obsidian]

Yeah

[jovial nexus]

No need to get yourself confused with sheaf notation

[latent anvil]

lmao

owned

[next obsidian]

Sorry I guess by abstract setting

[jovial nexus]

hope it helps your problem, ping me if you still got questions

[next obsidian]

Yeah, will do

I think this does end up solving it

[latent anvil]

wait so did you have a colimit of schemes?

[next obsidian]

No

I mean

I thought it was

But it’s supposed to be a limit

So take a base for a topological space B

[latent anvil]

but don't you want a limit on the ring side?

[jovial nexus]

it wasn't even a colimit of schemes tho?

[next obsidian]

Yeah it wasn’t of schemes

[latent anvil]

Oh okay

[next obsidian]

But I think I knew what he meant

[jovial nexus]

Like there were no real schemes involved

[latent anvil]

I was misinformed earlier

[next obsidian]

I never claimed I took a colimit of schemes

[jovial nexus]

Chmonkey was going over some stuff on how sheaves are determined by a Sheaf in a basis for the topology

[latent anvil]

You totally did

[jovial nexus]

I reckon

[latent anvil]

Brb finding screenshot

[next obsidian]

No I did colimits over

A base of O_Y modules

But it should’ve been a limit

[jovial nexus]

colimits of schemes are generally a bad idea, very few colimits of schemes exist

finite limits of schemes always exist though

[latent anvil]

Okay you said "This is for a thing about like closed sub schemes"

[next obsidian]

Is that because you can glue schemes?

[latent anvil]

not that the colimit was of schemes

[next obsidian]

Also yes it is about closed subschemes

[latent anvil]

gluing schemes gives you colimits, right?

[jovial nexus]

closed subschemes correspond to Quasi-Coherent ideal sheaves @latent anvil

[latent anvil]

Just not all of them

[next obsidian]

Yeah so

[jovial nexus]

yeah, glueing schemes along open subschemes is one of the few examples of colimits that exist

[next obsidian]

I’m trying to do something from Vakil

Which is before anytbing about qcoh comes in

[jovial nexus]

Just not all of them
@latent anvil mhm?
Quasi-Coherent Ideal Sheaves give you all closed subschemes

ah

[next obsidian]

Given an ideal I(B) for all Spec B in a scheme

Such that the map

[latent anvil]

gluing schemes doesn't give you all colimits

is what I was saying

[jovial nexus]

yeah no I guess it doesn't

[next obsidian]

I(B)_f -> I(B_f) is an iso

You show that these ideals define a closed su scheme

The thing I first wanted to do was extend this to an ideal sheaf

Then form the quotient

[jovial nexus]

I(B)_f -> I(B_f) is an iso
@next obsidian you're not aware of this, but this is one of several equivalent conditions making sure I defines a Quasi-Coherent Sheaf of modules

[next obsidian]

Yeah so, I figure it is

[jovial nexus]

anyway

[next obsidian]

But I think the point is to get this without resorting to higher level stuff

And the big issue I had was that I didn’t get closed subschemes

[jovial nexus]

yeah ofc

[next obsidian]

Since I couldn’t define the scheme theoretic image

And most sources only do it for a qcqs morphism

[jovial nexus]

Well the underlying space that you want is Supp(O_X / I) if I'm not mistaken

[next obsidian]

So I wanted to try and get my hands dirtier

I think so too

To try and understand this stuff better without just going to stuff about quasi coherent sheaves

[jovial nexus]

But yeah, your construction of how to extend I to a proper sheaf of ideals works

[next obsidian]

Gotcha

[jovial nexus]

yeah go ahead then ✌️

[next obsidian]

Thx for the help

[jovial nexus]

always

[whole basalt]

I've been watching some commutative algebra lectures recently and Wikipedia-rabbit-hole-following

And I think I'm starting to grok the idea of a local ring, and specifically localization of a ring at a given point

So let me see if I have this intuition right

In algebraic geometry, we often want to be able to look at, say, real-valued functions on a given variety V near a particular point p, and we might want to do algebra-y things to them. That means we might want to divide by certain expressions. Those expressions might be zero at other points on V, but we don't care about that as long as they're not zero at p. So we "localize" at p and say, okay, any function f that isn't zero at p now has an inverse "locally", and now we can happily move terms about without worrying about dividing by zero.

In another example I've seen of a local ring — the fractions with odd denominators — this is the integers localized at 2, because we've essentially said "okay, we really care about 2 in this ring, so we'll allow ourselves to divide by anything that isn't 2 and doesn't contain a 2."

1. Is there anywhere I'm off base, or anything else I should consider?
2. How would this apply to the dual numbers over R, i.e. {a+bε: a,b∈R and ε²=0}? I can't easily visualize what we'd localize where to get R[ε].

[next obsidian]

the dual numbers is R[x]/(x^2), but this is local simply because any prime ideal containing (x^2) has to be (x) so there's only one maximal ideal

I think that case is more about the fact that it's almost R

but it's like only one level away, so you get (x) inside there too

Idk jack about varieties, I think that's sort of the idea about localizing at a point

[whole basalt]

Ah so it's more related to the fact that fields are already trivially local?

[next obsidian]

I think so

[whole basalt]

And so you haven't strayed too far

[next obsidian]

There's dimension theory stuff and polynomial reasons to justify it a lot

But in terms of schemes

If you know what the topology on Spec A is

[whole basalt]

All I know about schemes is "varieties with multiplicity"

[next obsidian]

The sets D(f) form a base

Spec A is the set of primes

and closed sets are just sets Z(I) which are primes which contain I

[whole basalt]

And I don't know MUCH about varieties yet, I'm taking my first AG course this semester

(First lecture is in 2.5 hours)

[next obsidian]

The important part

here is that On the set D(f)

you actually look like Spec A_f

As in like that patch

is isomorphic to Spec A_f

the same way a patch of a manifold is like R^n

[whole basalt]

Okay so ... I know Spec R is the set of prime ideals of R

[next obsidian]

Yeah don't worry about it too much I guess

[whole basalt]

There might be some base information I'm missing 😛

[next obsidian]

But the fact it's a base for a topology

means that any open

can be covered by sets like this

[whole basalt]

I'm sort of slowly piecing together the bits.

[next obsidian]

and that they're like arbitrarily small

If you get into scheme theory

you'll see that these sets are super duper special

they sort of glue stuff together

[whole basalt]

Interesting.

[next obsidian]

I guess this doesn't really give much about like

what a visualization is, and I'm definitely not the one to ask for that

[whole basalt]

Hah well that's okay, I appreciate it regardless

[next obsidian]

Another thing to consider

about A_p

where p is a prime

You know how often you solve stuff by quotienting out by an ideal?

And you can relate a lot of stuff about A/I with A

and vice versa

Like ideals containing I are in bijection with ideals of A/I

you get the same sorts of stuff with localizing at a prime, but only really with prime ideals

So primes of A_p are in bijection with primes of A contained in p

where as primes of A/p are in bijection with primes of A containing p

so the two together tell you everything about the primes in relation to p

but A_p being local means you can apply a lot of powerful commutative algebra stuff and then draw those results back into A

[whole basalt]

Does A_p mean A localized at p?

[next obsidian]

At A\p

[whole basalt]

Okay, didn't know that was a notation

[next obsidian]

Yeah

it sort of sucks because

let x be a prime element

then (x) is a prime ideal

Then A_x and A_(x) are different

[whole basalt]

Whoopsie

[next obsidian]

Which sucks but it's just the notation haha

There's a lot of properties that can be checked "locally" though

For an A-module

M

You can also form M_p, localizing M at p

then M = 0

if and only if M_p = 0 for all prime p

if and only if M_m = 0 for all maximal m

So in a sense

this is like saying M_p = 0 is I guess it being 0 close to p

but if you look 0 close to all points

then you're just 0

There's a surprising number of things you can check at localizations

In fact, you can make this better, if (f_1,...,f_n) = (1), so all these f_i generate A

then M is 0 if and only if M_{f_i} is 0 for all i

So I guess in a way it's almost like you bring in localness of analysis into the picture

but algebraically

(by this I mean like, usually you can show something globally by showing for arbitrarily small balls around every point that it holds, at least in analysis)

Honestly though localization is something you just work with and start to appreciate more and more and see how powerful it is

[whole basalt]

Okay I will keep that in mind. 🙂 I think I'm still working on grokking all of this and will need to piece things together myself.

[next obsidian]

I think if you pick up a commutative algebra book

[whole basalt]

I can't quite decipher all of it right now.

But I'm working on it.

[next obsidian]

and see them prove stuff you'll see how its powerful

I don't personally really get much by trying to figure out stuff by trying to visualize it or think of it in some general way, so I guess that's a differnce in how we both learn things

I just kind of learned how to form a localization, then saw how its applied

[whole basalt]

Yeah. I need intuition in order to deal with things like this, I can't just definition-juggle

I'm only now getting to the point where I'm somewhat comfortable with what modules are, even, and that's only after working with multiple examples and playing around with them.

[vocal depot]

Anyone around?

[jovial nexus]

1. Is there anywhere I'm off base, or anything else I should consider?
2. How would this apply to the dual numbers over R, i.e. {a+bε: a,b∈R and ε²=0}? I can't easily visualize what we'd localize where to get R[ε].
   @whole basalt take a polynomial f(T) in R and evaluate the expression f(T + eps) when considered as a polynomial in R[eps][T] to see what the dual numbers are useful for

[smoky cypress]

What does “every (r+1)-rowed minor of A is 0” here mean?

This is jacobson’s basic algebra page 184

[woven delta]

The determinant of the minor

@smoky cypress

[smoky cypress]

Oh

That makes sense thanks

[smoky cypress]

Does this theorem need to assume that D is commutative?

Because I don't remember it says that assume the rings are commutative, but it just suddenly uses it in the proof

[golden pasture]

pid=commutative

[smoky cypress]

Oh wait really?

Oh frick

[golden pasture]

imagine unique factorization when things aren’t commutative
slightly difficult

[smoky cypress]

Ah.....

Fantastic I think I need to go over the ring chapters again to see which parts depend on commutativity

Oh yeah

[golden pasture]

just assume all rings are commutative almost everything will hold

[smoky cypress]

If it's not commutative then you can't even cancel the factors

rings without identity can always be embedded into a ring with identity

[delicate bloom]

even numbers + 1

amidoinitrite??

[dusty oracle]

Wot

[smoky cypress]

The ring (without identity) of even numbers can be embedded into Z, which is a ring with an identity

[smoky cypress]

I don't see why "the i-rowed minors of QA are linear combinations of the i-rowed minors of A"

[smoky cypress]

Oof ok

[jovial nexus]

Rings without 1 are dumb

[elder valley]

perfect for me, who is also dumb

[golden pasture]

rngs :uninteresting:

rigs :uninteresting:

[delicate bloom]

rs and gs are pretty cool

[smoky cypress]

What are rigs

[golden pasture]

rings-negatives

[smoky cypress]

Oh what

So just two monoids?

[golden pasture]

yes

like say cardinals

[smoky cypress]

Ah

[golden pasture]

but its dumb

[next obsidian]

Is $(A_f)g$ isomorphic to $A{fg}$?

[cloud walrus]

Chmonkey:

[next obsidian]

answer: yes

Okay

I have an ideal $I$ of a ring A, and I have $f_1,\dots,f_n$ such that they generate the unit ideal of $A$.

[cloud walrus]

Chmonkey:

[next obsidian]

How do I show that the sequence

[0\to I\to \prod_i I_{f_i} \rightrightarrows \prod_{i,j}I_{f_if_j}\to 0]

[cloud walrus]

Chmonkey:

[next obsidian]

Is exact (interpret the two arrows as actually being the difference of the two maps there)'

If I let $B = \prod_i A_{f_i}$ then this is

[cloud walrus]

Chmonkey:

[next obsidian]

[0\to I\to I\otimes_A B\rightrightarrows I\otimes_A B\otimes_A B\to 0]

[cloud walrus]

Chmonkey:

[next obsidian]

And $B$ is faitfhully flat, I think this is really some whack Cech complex or something, so some cohomology stuff might actually prove this, but I want to show this with less firepower

[cloud walrus]

Chmonkey:

[latent anvil]

what's your problem

Oh I'm reading

Is this showing the thing is a sheaf on a base

[next obsidian]

yup

Basically I need to show that the ideal sheaf is a sheaf on the base

[latent anvil]

You're doing the max proof

[next obsidian]

but cofinal blablabla

Yes

Means I only need to look at a cover of the form

[latent anvil]

hmm

if I is flat you're good but it probably isn't

which is sad

[next obsidian]

$I(\text{Spec }B)\to \prod I(\text{Spec }B_{f_i})\rightrightarrows \prod\text{Spec }B_{f_i}\cap \text{Spec }B_{f_j})$

[latent anvil]

right

[cloud walrus]

Chmonkey:

[next obsidian]

But the thing you're assuming is quasicoherence (secretly)

[latent anvil]

oh

Isn't B (×) B flat

Since B is flat

[next obsidian]

which says the map $I(\text{Spec }B)f \to I(\text{Spec }B{f})$ is an iso

[cloud walrus]

Chmonkey:

[next obsidian]

so I can turn it into the better looking complex

[latent anvil]

you pick up a Tor1(I,B (×) B) term on the left when you tensor the SES for O_Spec A with I

Do you hear what I'm saying

[next obsidian]

Umm

the thing is

I don't see how I get this sequence

as tensoring a sequence

Like

[latent anvil]

[0\to A\to B\rightrightarrows B\otimes_A B\to 0]

[next obsidian]

Chmonkey:
@cloud walrus

[latent anvil]

is exact

[next obsidian]

Wait wut

why?

[cloud walrus]

shamrock:

[next obsidian]

Umm

OH SHIT

[latent anvil]

because it's the equalizer condition ok a base for O_Spec A

right?

[next obsidian]

I'm not tenosring by B

I'm tensoring by A!

[latent anvil]

no

Still wrong

lol

[next obsidian]

No I mean

yes this is true

[latent anvil]

Oh I see

Yeah you're tensoring I by A

and then you pick up a tor term

[next obsidian]

I mean

by I

yes

[latent anvil]

But with B (×) B

Which is flat

right?

So the Tor vanishes

[next obsidian]

yeah i think so

dude holy shit

giga brain

[latent anvil]

man I forgot how much fun algebra is

I should do algebra

[next obsidian]

Yeah that makes sense

ughhh

I was tryna see how I got my sequence by tensoring with B

[latent anvil]

[next obsidian]

but no, the common thing is an I

lmfao

[latent anvil]

owned

[next obsidian]

honestly yeah

😔

Tfw

You do AG really hard

[latent anvil]

okay ty for the break from analysis

[next obsidian]

and Shamrock still

epic owns all ur problems

[latent anvil]

now I gotta get back to convergence in measure

[next obsidian]

f'ing

closed points

Some dumb coho stuf

[latent anvil]

love how we're both doing math at 2:40

you knew you could ask me for help

[next obsidian]

Yeah

okay literally

I saw a notification in analysis

and was like "that must be shamrock"

[latent anvil]

lmaoooo

[next obsidian]

then I confirmed and sent a msg

KEK

[latent anvil]

The argument I just made was omegabrain

[next obsidian]

That reminds me of like

[latent anvil]

I ended up with this weird bound at the end

And my proof wasn't working

[next obsidian]

https://discordapp.com/channels/268882317391429632/496784958430380033/737722835736330280

When I made this argument

I felt my brain expanding

[latent anvil]

Because I had chosen a sequence of $f_k$ such that
$m({x \in E : |f(x) - f_k(x)| > 1/k) < 1/k$

[next obsidian]

https://discordapp.com/channels/268882317391429632/496784958430380033/737727039347359774

This was the real argument I made

that I felt giga for

[cloud walrus]

shamrock:

[latent anvil]

And I ended up with $\lim_{K\to \infty} \sum_{k > K}m({x \in E : |f(x) - f_k(x)| > 1/k) < \lim_{K\to \infty} \sum_{k > K} 1/k$

[cloud walrus]

shamrock:

[latent anvil]

And I wanted that limit to be 0

but it isn't

It's infinity

By the divergence of the harmonic series

So I went to the top of my argument and said "let $a_k$ be the terms in a positive convergent series"

[cloud walrus]

shamrock:

[next obsidian]

lmfao

[latent anvil]

it was so good

[next obsidian]

Yes now we're bounded by infinity!

Dude I said that infinite integral converged

on the midterm that one time so

why didn't you just

do like 1/k^2

did you just want to flex

[latent anvil]

I did in my notebook

But it was messy so I needed to post on discord

and cmon how could I not

[next obsidian]

lmao

[latent anvil]

the tails of an infinite series goes to 0 and I think that's beautiful 😊

[next obsidian]

[latent anvil]

man people keep telling me this analysis class is gonna be shit

e.g. Andreas

and I don't believe

I have faith

It's gonna rock

[next obsidian]

That's the spirit!

I can tell you that

the two weeks I did it were cool

[real panther]

Hey, could someone help me with a problem?

[woven delta]

[cloud walrus]

Neil_P:

[woven delta]

so R-{0} under multiplication has an element of order 2

[real panther]

Or if my mapping is bad

[woven delta]

think about that

(order 2 means that that element squared is 1)

[real panther]

I'm sorry - I don't follow there.

ah, got it

[woven delta]

what map do you have?

[real panther]

ln(x) goes from the postive reals to the reals

[woven delta]

yes

[real panther]

I haven't figured out a way to get the nonzero reals to map to the positive reals, but ik a bijection exists

[woven delta]

a bijection does exist, but it does not preserve the multiplicative structure

[real panther]

oh. You're right.

Does order 2 mean that there are two unique elements a, b such that a^2 = b^2 = 1?

[woven delta]

no it means that (-1)^2 = 1

[real panther]

what would an order 3 look like?

[woven delta]

$(\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3}))^3=1$

[cloud walrus]

Liquid:

[real panther]

That's pretty elegant

Would you say the group of complex numbers has an order 3 element?

[woven delta]

more generally $e^\frac{2\pi i}{n}$ is of order n

[real panther]

and for that matter, an order n element for every integer n?

cool. that makes sense

[woven delta]

so this is the group of complex numbers-{0}

under multiplication

[real panther]

Lol we just covered that in complex

[cloud walrus]

Liquid:

[woven delta]

lol forgot the i

[real panther]

Thanks for the help. I get why I'm wrong on my original question

[woven delta]

think about the order 2 thing

[real panther]

but how would I prove that you cannot be 1-1?

sure.

[woven delta]

basically if there was an isomorphism, there would have to be a corresponding element of order 2 in R under addition

[real panther]

That makes sense

[woven delta]

which would mean there is an element $r\in \mathbb{R}$ with $2r=0$

[cloud walrus]

Liquid:

[real panther]

a=-1, b=-1 and x=1, y=1 all map to the same thing, despite being different. Therefore this mapping is noninjective

[woven delta]

but you can write $\mathbb{R}^\times = {1, -1} \oplus \mathbb{R}^+$

[real panther]

I've seen you around here a lot - are you a just in upper level classes or some type of phd student?

[woven delta]

I'm a PhD student

[cloud walrus]

Liquid:

[real panther]

I'm a second year undergrad

Sorry, what does that notation mean? (the X and oplus?)

[woven delta]

$R^\times= R-{0}$ (more generally the group of units of a ring, for a field it is what I just wrote). $R^+$ is just positive reals

[cloud walrus]

Liquid:

Neil_P:

[real panther]

I don't know either, sorry

But it seems interesting

Yes

Is that a mapping from one group to another?

Thank you both

[vale flint]

~~can we show that if a polynomial in $\mathbb{F}_p [x]$ takes on every value in $\mathbb{F}_p \cup { 0 }$, then its leading coefficient must be divisible by $p$?~~stupid

[cloud walrus]

deiknymi0:

[smoky cypress]

Is that true? If the leading coefficient is divisible by p then it’s 0 in Z/pZ

I’m not sure if I’m understanding the problem correctly

Also 0 is already in F_p

[vale flint]

yeah sorry about the F_p thing

[woven delta]

Yeah the statement doesn't make much sense

[vale flint]

what i'm asking is if we have a polynomial of degree k, with integer coefficients, and modulo p, it attains all the values in (0, 1, 2..., p-1), can we show that its leading coefficient is divisible by p?

[woven delta]

What about the identity?

[vale flint]

hmm true. what if we make an additional constraint that k > 1 and k |(p - 1)?

[woven delta]

x^p-x

Oh

k | p-1 might be good

But like a better statement might be that the only polynomial in F_p[x] of degree < p that is surjective is x

[smoky cypress]

I don’t think that’s the case. The set of functions $\bF_p\to\bF_p$ is isomorphic to $\bF_p[x]/(x^p-x)$, so every function can be realized as a polynomial with degree $<p$. Then, the linear polynomials in $\bF_p[x]$ are bijective, so this gives $p(p-1)$ bijective polynomials. But there are $p!$ bijective functions from $\bF_p\to\bF_p$, so there must be some bijective polynomial with degree greater than 1 and less than $p$, then you can just scale it to make the leading coefficient to be 1 in $\bF_p$

[cloud walrus]

Whoever:

[woven delta]

That's neat whoever

You can actually realize functions as polynomials explicitly using Lagrange interpolation which is cool

And then reduce using x^p = x mod p

[delicate bloom]

But like a better statement might be that the only polynomial in F_p[x] of degree < p that is surjective is x
@woven delta also x+1, x+2, ...

I guess then whoever went on to talk about that more in depth sorry for the ping lol

[woven delta]

Yeah

I wasn't thinking lol

Also I may have been implicitly discounting those functions in my head

[delicate bloom]

yeah to be fair making mistakes like that is easy to just lump all things of the same kind of class together or whatever lol

I do it all the time

[woven delta]

Yeah same

[smoky cypress]

@woven delta actually if you use lagrange interpolation you also don’t need to worry about degree, since by construction the polynomial has degree at most p-1

[woven delta]

That's true

Good point

[vale flint]

thanks for the discussion everyone

[steep hull]

@vale flint You also asked if the degree of such a polynomial must always divide $p-1$. The answer to that is also no for $p>3$. We can actually show that there are no surjective polynomials of degree $p-1$ over $\bF_p$ for $p>2$ by Lagrange interpolation. Then it is just a matter of bounding the other possibilities. There are at most $(p-1)p^{\frac{p-1}{2}}$ possible polynomials as the degree must be at most $\frac{p-1}{2}$, which is less than $p!$ for $p>3$ because $p!=p(p-1) (\prod_{k=2}^{\frac{p-1}{2}} k(p-k))>(p-1)p^{\frac{p-1}{2}}$ follows from $k(p-k)>p$ for $k>1$.

[cloud walrus]

tree3:

[smoky cypress]

@steep hull how do you show that there is no surjective polynomials of degree p-1?

[smoky cypress]

[delicate bloom]

my vague guess to try to prove it would be to think of uhh

$f_a(x) = 1-(x-a)^{p-1}$

[cloud walrus]

Merosity:

[delicate bloom]

it's 1 at x=a, 0 otherwise

try to play around with linear combinations of these and other junk ... + magic

[steep hull]

Alternatively, note that for any polynomial of degree $<p$, $\sum_{n=1}^{p-1} f(n) \equiv -a_{p-1} \mod p$, where $a_{p-1}$ denotes the $x^{p-1}$ coefficient.

[cloud walrus]

tree3:

[steep hull]

Use Lagrange interpolation to write the $x^{p-1}$ coefficient as $\sum_{k=1}^{p-1} f(k) \frac{\binom{p-1}{k} (-1)^k}{(p-1)!}$ which becomes $\sum -f(k) \mod p$, which is $0 \mod p$.

@smoky cypress

[smoky cypress]

Ugh I don't quite see how the sum comes from but lemme see

[steep hull]

I gave two slightly different proofs

[smoky cypress]

Well I don't get either one so 🙃

[steep hull]

For the first one, I used that $\sum_{n=1}^{p-1} n^k \equiv 0 \mod p$ for $k<p-1$

[cloud walrus]

tree3:

[smoky cypress]

Ah ok

I see now

[steep hull]

For the second, just plug it into Lagrange interpolation formula

And that’s what you get

[cloud walrus]

tree3:

[steep hull]

The first formula is the exact representation of the x^{p-1} coefficient of the polynomial of minimal degree going through p chosen points

Which when taken mod p turns out very nicely

[vale flint]

thank you, that makes a lot of sense!

[cloud walrus]

tree3:
Compile Error! Click the reaction for details. (You may edit your message)

[fallen bluff]

Is there a universal way to turn a magma into monoid ?

[woven delta]

Scrap the magma you have and output Z

[next obsidian]

Based

[fair obsidian]

I'd have to think about it, but if you can quotient magmas then you can probably quotient to make it associative

Then add an identity

[next obsidian]

I want to say no

Magmas have so little structure, and enforcing associativity and an identity seems like a large task to do universally to something which literally is just "lol I have a binary operation"

[fair obsidian]

I believe it, but the identity definitely isn't a problem

[uncut girder]

Ok

This was a problem on my Algebra qual @woven delta

For a finite group G, with G/Z(G) abelian of order pq where p and q are distinct prime, prove that G is abelian

I couldn't solve it, anyone know how to do this?

[woven delta]

Yeah I think I do

So in most cases a group of order pq is cyclic

And in that case it's a very classical problem

[uncut girder]

Sure

[woven delta]

But okay, another good thing to know is that G/Z(G) is isomorphic to the inner automorphism group of G

Hmm

[vital quail]

G/Z(G) abelian

[woven delta]

Dude S_3 is not abelian

Smh

[vital quail]

For a finite group G, with G/Z(G) abelian

[woven delta]

In general the other group is a semidirect product

Which is not abelian

Oh lol

Yeah this is trivial

I didn't read the G/Z(G) abelian thing

[vital quail]

really it just amounts to knowing G/Z(G) cyclic <=> G abelian

[woven delta]

Yeah

And knowing that the only 2 groups of order pq are the semidirect product of Z/p and Z/q and Z/pq

And that the semidirect product is abelian iff direct product

@uncut girder

[uncut girder]

OH SHIT

Z/p × Z/q = Z/pq by chinese remainer theorem, how did I forget that

[woven delta]

Yeah

[uncut girder]

:c

This was a very easy problem

[fair obsidian]

I forget, but out curiousity you know that some H of order pq is a semidirect product if p < q by Sylow theorems?

Something like the order q subgroup is normal

[woven delta]

Yeah

You get one of them is normal

[fair obsidian]

Cool

[woven delta]

And then you can explicitly make the action

By writing down Aut(Z/q) = Z/(q-1)

We need p divides q-1

So the action is multiplication by p in Z/(q-1)

That's the outer way of making the group

[fair obsidian]

Otherwise you only get the direct product

Cool

[woven delta]

Yeah

[fair obsidian]

Btw I went through to check the quotient works, it seems like there's a unique smallest quotient of a magma that makes it associative

If the relation is works on $M$ is $\sim$ then the associated monoid to $M$ should be $M/\sim \sqcup {e}$

[cloud walrus]

The_Vman:

[fair obsidian]

Which should have the property that any magma morphism f from the magma M to a monoid N factors through this monoid

[gritty hedge]

Could anybody explain to me how exactly an irreducible polynomial defines a Galois field?

i.e x^8 + x^4 + x^3 +x + 1 defines the Galois field GF(2^8), but what does that actually mean?

[fair obsidian]

You take the quotient of the polynomial ring $F_2[x]$ by the ideal generated by your polynomial, in this case x^8 + x^4 + x^3 +x + 1

[cloud walrus]

The_Vman:

You take the quotient of the polynomial ring $F_2[x]$ by the ideal generated by your polynomial, in this case x^8 + x^4 + x^3 +x + 1
```Compile error! Output:

! Missing $ inserted.
<inserted text>
$
l.54 ...erated by your polynomial, in this case x^
8 + x^4 + x^3 +x + 1
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

[smoky cypress]

Let $f\in F[x]$, $u=x+(f(x))\in F[x]/(f(x))$, then is it true that $F[u]\cong F[x]/(f(x))$?

[cloud walrus]

Whoever:

[smoky cypress]

Ah yes of course

Lol direct application of first isomorphism theorem

Nvm

[smoky cypress]

I don’t see what he did to show that you can do row operations such that b_{11} | c_{kl} for every k,l

And the “equivalence” here means that A is equivalence to A’ if A’=PAQ where P,Q invertible, P is mxm and Q is nxn

[fair obsidian]

I think the point is that if b_{11} doesn't divide some c_{kl} then you can decrease the degree of the minimal entry of A, and you can only decrease the degree of the minimal entry so many times

[smoky cypress]

I kinda get that point, but I don’t really see what happened in the line “Repetition of the first process replaces c_{kl} by ...”

Ah I think we can come up with an alternative proof then

Let $m(A)$ be the smallest $\delta$ of all the nonzero entries, then we can find $\min m(A’)$ where $A’$ goes through all the matrices equivalent to $A$. Let $A’$ be one such matrices and by row operation we can move the entry with smallest $\delta$ to $a_{11}$. If $a_{11}$ doesn’t divide all other nonzero entries then we can find matrix that has a smaller $m(A)$, then we can just use row operation again to make the first row and column to be in the form (24), and proceed with induction

[cloud walrus]

Whoever:

[fair obsidian]

Yeah that makes sense

Definitely a lot clearer

[smoky cypress]

Alright thanks I guess I can stick to this

[fair obsidian]

No problem

[pure crest]

if a,b are in a field extension B of Q, and an arbitary element, can i say that if a+b+c in B then c is in B?

[bleak abyss]

c = (a+b+c)-(a)-(b)

[pure crest]

right thats what i was thinking

thank you

[smoky cypress]

I have asked this already, but I don’t see “Hence the i-rowed minors of QA are linear combinations of the i-rowed minors of A and ...”

[smoky cypress]

<@&681260374879633482>

[timid hull]

what does i-rowed minors mean?

[smoky cypress]

The determinant of some smaller ixi matrix

[next obsidian]

Determinant of an i by i submatrix

[smoky cypress]

I guess you can find a ixm submatrix Q’ of Q and a mxi submatrix A’ of A such that the minor of QA is det(Q’A’), then by this link https://en.m.wikipedia.org/wiki/Cauchy–Binet_formula?wprov=sfti1,
the minor is a linear combination of the minors of A’, which are minors of A

But that’s kinda nontrivial

[next obsidian]

Yeah dude this is non-obvious to me

[fair obsidian]

Are you familiar with the exterior algebra? If so there's a not too terrible way of seeing it

[smoky cypress]

I think I know the very basics

[fair obsidian]

Do you know how the determinant is related to the exterior algebra?

[smoky cypress]

Ah I do not

[cloud walrus]

The_Vman:

[fair obsidian]

Where the wedge operation is like multiplication, being linear in each entry and anti-commutative

[cloud walrus]

The_Vman:

[smoky cypress]

Ah yes I do know that they are anti symmetric and linear

[fair obsidian]

Awesome

[cloud walrus]

The_Vman:

[fair obsidian]

In general the kth exterior power will have dimension n choose k

[smoky cypress]

Yes

[cloud walrus]

The_Vman:

[fair obsidian]

In the case of V = W, since the nth exterior power is 1-dimensional, the induced map is simply scaling by some scalar

You can take that scalar to be the definition of the determinant of T

(I should note that in general having $\wedge$ be anti-commutative is equivalent to $v \wedge v = 0$ for all $v$ but in characteristic 2 you need to define the exterior algebra using the latter)

[cloud walrus]

The_Vman:

[fair obsidian]

Does that all make sense? I can give an example

[smoky cypress]

Kind of

I learned \wedge as a way to construct an alternating multilinear map from two multilinear maps

But sorry it’s 4am here and I can’t quite think clearly

I’ll reply tmr

[fair obsidian]

Yeah this is pretty much the same, except you define it as formal products/wedges instead of as maps

Cool, it's pretty late here too we can keep talking about it tomorrow of you want

[smoky cypress]

Alright

[steady axle]

is it true that if H is subgroup of G then [G,G] ∩ H=[H,H]

certainly [H,H] ⊆ [G,G] ∩H but i am having hard time proving the other way

[next obsidian]

This is not true

I'll define something here, a group is called "Perfect" if [G,G] = G (this is a real term)

If you search, you'll find that the smallest non-trivial perfect group is A_5 which has 60 elements.

Now suppose it were true that [G,G] ∩ H=[H,H] and G is a perfect group, then for any subgroup H of G, then H is a perfect group because
[G,G] ∩ H=[H,H] turns into G ∩ H=[H,H] which clearly says H = [H,H]

Note that [G,G] = {e} if and only if G is abelian, so take any perfect group, like say A_5, then take any non-trivial abelian subgroup of it, the abelian subgroup is not perfect so by contradiction the statement is false

@steady axle

Nice problem to think about

[steady axle]

right

[next obsidian]

As a bonus, any non-abelian simple group is perfect so there's a wealth of examples of perfect groups if you care

[pure crest]

Hey guys, how can i prove that $\sqrt{2} + \sqrt[3]{5}$ is an element in $\mathbb{Q}(\sqrt{2}, \sqrt[3]{5} )$?

is it enough to just say that $\sqrt{2}$ and $\sqrt[3]{5}$ are in $\mathbb{Q}(\sqrt{2}, \sqrt[3]{5} )$ so their sum must be as well?

[cloud walrus]

boat:

[solemn rain]

ye

[cloud walrus]

The_Vman:

[pure crest]

i was thinking of going by the definitions by saying $sqrt{2} = a_0 \sqrt[3]{5} )^0$ where $a_0 \in \mathbb{Q}(\sqrt{2})$ but i dont think thats necessary

[cloud walrus]

boat:

[solemn rain]

yea just say what you did

nothing wrong with that

[pure crest]

whcih one, the latter one?

[wispy urchin]

regardless of your definition, closure of addition holds anyways

[solemn rain]

yes

the first one

first argument

nothing wrong with it

[pure crest]

ohh alright thank you!

[fair obsidian]

I was just trying to say that when you write $\mathbb{Q}(\alpha, \beta)$ for some real/complex alpha and beta it means the smallest subfield containing Q, alpha, and beta. So in particular it should be closed under addition

[cloud walrus]

The_Vman:

[pure crest]

ohh yeah i see what you mean

i wasnt sure the level of detail one should go to prove this

[wispy urchin]

its personal taste

of your professor ofc you have no say in it

[pure crest]

:/ thats true

thank you guys!

[neat ginkgo]

what's the operation in a quotient group

i dont get it like

Z4/Z2 = {0+z2, 1+z2, 2+z2, 3+z2} = {{0, 2}, {1,3}, {2, 0}, {3, 1}}

right

the elements of the group

are sets

but its a group of sets, what's the operation

[fair obsidian]

So in a quotient group, the sets are cosets by the normal subgroup you quotiented by

I'm not quite sure I get the quotient you wrote down

[neat ginkgo]

wait did i fuck up

is z2 not a normal subgroup of z4

[fair obsidian]

It is

[cloud walrus]

The_Vman:

[neat ginkgo]

did i fix it?

[fair obsidian]

Not quite

[neat ginkgo]

:/

[fair obsidian]

While {0, 2} is fine, the other subset shouldn't be {0}

The quotient group consists of elements like x + {0, 2}

[neat ginkgo]

oh damn, i totally messed up the operation

i was doing *

[fair obsidian]

Yeah happens, especially when you are used to multiplying everything and suddenly you start working with +

Looks good now

[neat ginkgo]

yeah makes sense that it contains all of Z4 as elements of the cosets

[fair obsidian]

So the quotient is {{0, 2}, {1, 3}}

Yup

[neat ginkgo]

yea

so the thing i dont understand is, why is this a group? what's the operation? what's the neutral element?

[fair obsidian]

So if you write the cosets like x + H and y + H, then the operation would be (x + H) + (y + H) = (x + y) + H

[neat ginkgo]

oh

oh that's it

everything makes sense now

damn

[fair obsidian]

If you think of them as sets, you just take x and y to be elements of the respective sets, and then the sum of the cosets is whatever coset corresponds to x + y

And you only quotient by normal subgroups so that this operation is well-defined, doesn't depend on these choices

[neat ginkgo]

yeah i get it now

thank you so much !

[fair obsidian]

No problem

[smoky cypress]

@fair obsidian i read over what you said, and now it makes sense

[fair obsidian]

Awesome

Do you see how you can use it for the minors?

[smoky cypress]

No

[fair obsidian]

Ok, and did you get the relationship to the determinant?

[smoky cypress]

Yeah the nth exterior product of V (dimV=n) has 1 dimension, so for a linear operator in V we have an induced map on n vectors, and since the exterior product has 1 dimension this induced map is multiplication by scalar, and we can define this scalar to be the determinant of the linear operator

I guess the n vectors can be chosen to be the standard basis

Or any linearly independent set

[fair obsidian]

Yeah exactly and by like linearity and such the choice doesn't matter

Ok so I think the best way to see the minors thing is to go through an example

[smoky cypress]

Alright

[cloud walrus]

The_Vman:

The_Vman:

[fair obsidian]

I'm missing one term I'm too lazy to write down

[cloud walrus]

The_Vman:

[fair obsidian]

Anyways the point is that if you use the v's and w's as basis and write down the matrix corresponding to T, then factors on the wedges are exactly the determinants of the 2x2 minors

[smoky cypress]

Ah

Yeah I think I kinda see it

From an informal view point ofc xD

[fair obsidian]

Lol yeah its all just symbols

But now to compute the 2x2 minors of some composition ST, you just need need to apply S

Which would be something like $(a_{11}a_{22} \cdots) S^(w_1 \wedge w_2) + (a_{11}a_{23} \cdots) S^(w_1 \wedge w_3) + \cdots$

[cloud walrus]

The_Vman:

[smoky cypress]

Yeah

So it is going to be a sum where each term is $\det(\text{some minor})\cdot S^*(w_i\wedge w_j)$

[fair obsidian]

Yeah exactly

[cloud walrus]

Whoever:

[smoky cypress]

And then $S^*(w_i\wedge w_j)$ will also be a linear combination where the coefficients are minors of the second operator

[cloud walrus]

Whoever:

[fair obsidian]

Yup

[smoky cypress]

Yeah that makes sense

And also I see the first equation now

https://discordapp.com/channels/268882317391429632/496784958430380033/746753773338099823

This one

You have $T^*(v_1\wedge v_2)=\br{\sum_ja_{j1}w_j}\wedge \br{\sum_ja_{j2}w_j}$

[cloud walrus]

Whoever:

[smoky cypress]

And then you can just expand the wedge using linearity, the fact that wedge is antisymmetric

And I’m gonna place my trust on your algebra that this will simplify to the expression you wrote

[fair obsidian]

You can trust

It is nice to go through a map from 2d to 2d. Like if T only mapped into the span of w_1 and w_2 then you should get the usual ad - bc

[smoky cypress]

Yeah I can see that without writing it down

[fair obsidian]

Awesome

Oh btw, this space of k-wedges is related to alternating k-tensors

[smoky cypress]

Yeah that’s how I learned wedge product, as a way to produce alternating k tensor from smaller multilinear maps

I read it in Spivak’s calculus on manifold ch4

[fair obsidian]

Oh nice I also used Spivak

Although I should say, that construction is better mirrored not in this space but in an isomorphic space which is embedded in the general tensor product

Here you don't have to deal with the coefficients consisting of a bunch of factorials

[smoky cypress]

Oh ok, I read another construction that involved some k shuffle or sort of stuff, and I found the construction in spivak to be much nicer

[fair obsidian]

He used like an alternating map right? You sum over the permutations with sign

[smoky cypress]

Yeah

Then you multiply by a weird coefficient

That makes wedge associative

[fair obsidian]

Yeah

Yeah that space of alternating k-tensors is (isomorphic to) the dual space of the kth exterior power

[smoky cypress]

Oh ok

[fair obsidian]

It's some sort of universal property. The kth exterior power has any alternating k-tensor factor through it

[smoky cypress]

Alright

Where can I read more on exterior algebra?

[fair obsidian]

Honestly not entirely sure. You can find the condensed definitions and construction on Wikipedia, I kinda learned them from some TAs and by thinking about them on my own, applying what I knew about it while doing the alternating k-tensors in Spivak and such

If I do find a good resource I'll link it to you

[smoky cypress]

Alright thanks

[neat ginkgo]

let G be a group such that G/Z(G) = <aZ(G)> where a is from G, prove that a must be from Z(G)

[fair obsidian]

np

[neat ginkgo]

do you guys read this and be like

"oh yeah of course it is"

[fair obsidian]

Lol no, I still need to think about it

But with practice you maybe remember similar things and make connections

Ok so, do you have any ideas so far?

[neat ginkgo]

so the whole conversation we had earlier was just so i can wrap my head around what this even means

i understand now that

there is a specific coset aZ(G) which generates all the other ones

i need to prove that a is an element of Z(G)

is this right so far

[fair obsidian]

Yeah exactly right

[neat ginkgo]

i dont really see a path towards the proof, i know that the quotient groups cyclic therefore any coset mZ(G) is just a^i * Z(G)

[fair obsidian]

That's a great start

So what do you need to show to get a in Z(G)?

[neat ginkgo]

don't really know honestly

should i do it by contradiction

[fair obsidian]

Well what is Z(G)?

[neat ginkgo]

its the subgroup of G consisting of only elements that commute with G

don't know if that sounds right, trying to translate from the language im taking the course in lol

[fair obsidian]

No its fine, I would say the subgroup consisting of elements x that commute with every element in G

So to show that a is an element of Z(G), you need to show that a commutes with all elements of G right?

[neat ginkgo]

yeah

m is a generic element from G

[fair obsidian]

Cool, so to answer your question, I think you should do it directly

But go ahead

Sorry, I wasn't sure if you wanted to say more or not

[neat ginkgo]

im trying to do something with just algebraic manipulation

but i think its not working lol

[fair obsidian]

Ok, I'll say this, you started by explicitly writing down the elements of G/Z(G)

Can you somehow use that to write down what elements of G look like?

[neat ginkgo]

you mean

m = a^i

[fair obsidian]

Almost

There's an element a^i in each coset, but a generic element in the coset a^i * Z(G) might be just slightly different

So in the previous example we had cosets like 1 + {0, 2}, which had elements 1 + 0 and 1 + 2

What are elements of the coset a^i * Z(G)?

[neat ginkgo]

{a^i * z1, a^i * z2, ...}

[fair obsidian]

Exactly

So a general element in G is in some coset, so it has the form a^i * z for some z in Z(G)

Right?

[neat ginkgo]

oh god

i completely forgot

the the elements in the cosets

are actually just elements of G

...

im so stupid

hahahahaha

i think i get it now

god

[fair obsidian]

Lol no, it happens

That took me a while too actually. When you look at a quotient G/N it gives you a description for the elements in G.

It takes a little bit to get that you can use that to your advantage. That if you want to know something about elements in G, you can sometimes use what you know about the cosets

Ok but back to the question

Now you just need to show a is in Z(G)

[neat ginkgo]

nah it aint coming to me

lmao

every element in G is just some power of a * an element from the center

[fair obsidian]

Right

And you want to show that the element a commutes with all such elements a^i * z

[neat ginkgo]

m = a^i * z
ma = a^i *z * a
ma = a^(i+1) * z

[fair obsidian]

Yup

[neat ginkgo]

but here z just commuted with a

doesn't that just show that

[fair obsidian]

Well what is a*m?

[neat ginkgo]

(z or a) is from Z(G)

oh

you're right

damn

this was weird as hell

[fair obsidian]

That's math lol

[neat ginkgo]

why did it take me so long just to realize i need to show ma = am

lol

thank u again ! ! ! 🙏

[fair obsidian]

Really I would say that's the hardest part. You might be able to do every step, but if you don't know which steps to take it gets a lot harder

You're welcome

[neat ginkgo]

took me a while to get my head around what im even working with

like what actual mathematical objects

[smoky cypress]

Here Jacobson defines the $p$-component $M_p$ of a module (over pid $D$) to be the set of $y\in M$ such that $p^ky=0$ for some $k$. I have two questions: first is $p^k=\underbrace{1+1+\cdots+1}_{p^k\text{ times}}$; second, it says that $M_p\subseteq\operatorname{tor}M$, but part of the definition of torsion submodule is that $p^k\neq0$, which is not guaranteed in a ring with characteristic $p$, so is it still true?

[cloud walrus]

Whoever:

[fair obsidian]

I'm not too familiar with modules but I would say that p^k * y is the repeated addition of y, so yeah p^k is 1 added to itself p^k times

You always have a map from Z to a ring with identity by sending an integer n to 1 added to itself n times, and I think if not specified you think of integers as being in the ring in this natural way

[carmine fossil]

@neat ginkgo Are you familiar with the viewpoint, that all cosets are equivalence classes,such that the relation respects certain conditions?

[smoky cypress]

Yeah @fair obsidian, and by first iso theorem this map shows that R contains Z/(n) or Z, and if R is an integral domain then n=p is prime

[fair obsidian]

Yeah makes sense

Just reading some definitions, you define the torsion submodule when the ring is an integral domain, are you assuming that here?

Wait if R is has characteristic p, then isn't any q^k a unit if q is some other prime?

[smoky cypress]

Not sure

But you do define torsion submodule when the ring is pid

[fair obsidian]

Ok good to know

But my point was that if the p.i.d $D$ was characteristic $p$, then any prime power $q^k$ is a unit as it is a non-zero element of $Z/(p)$ inside $D$. So if $D$ has prime characteristic any torsion element $y$ with $ay = 0$ for $a \neq 0$ and $y \neq 0$ would have $a$ not be some $q^k$

[cloud walrus]

The_Vman:

[fair obsidian]

So if you used the above definition, the q-components would all be zero anyways

So it only really makes sense to talk about this if D isn't characteristic p

[smoky cypress]

It would only make sense if D has characteristics 0 I guess

[smoky cypress]

https://discordapp.com/channels/268882317391429632/496784958430380033/746766403675226303

https://discordapp.com/channels/268882317391429632/496784958430380033/746772276724105227
<@&681260374879633482> can someone these few questions xD

[woven delta]

The answer to the first one is yes

Z acts on all modules in that way

And that's always how we interpret multiplying by an integer

[smoky cypress]

Alright

[woven delta]

I'm not seeing where part of the definition of torsion submodule is what you said

Can you post the definition you are referring to?

[smoky cypress]

https://discordapp.com/channels/268882317391429632/496784958430380033/746770653977247786

[woven delta]

What's your point then whoever

[smoky cypress]

p^k must not be 0

But it could be

[woven delta]

If p^k is 0 in the ring R, then p^k m = 0

[smoky cypress]

Then M_p will be the whole module

[woven delta]

Hmm

[smoky cypress]

So M_p is not a subset of torM anymore

[woven delta]

Okay so this can only happen if our ring is of characteristic p

[smoky cypress]

Yeah

[woven delta]

But yeah you're right

This definition needs a few more words

To make it make sense

[smoky cypress]

Alright thx

Also I tried to google it

But nothing really showed up

Eh

[woven delta]

There's a definition in D&F

One sec

[smoky cypress]

I will assume that the pid has characteristic 0 ig

Ok

[woven delta]

Just let p not be the characteristic

[smoky cypress]

Then what v man said applies

No element will be in M_p

Except for 0

Here https://discordapp.com/channels/268882317391429632/496784958430380033/746772276724105227

[woven delta]

Yeah I guess the characteristic 0 case is the interesting one

[smoky cypress]

Yeah

I found the definition in D&F, it’s not a direct definition

Oh

The primes are not primes in Z, but primes in the pid

Bruh

@woven delta @fair obsidian I think it might be that by primes they meant primes in the pid

[fair obsidian]

Bruh

Well that makes sense

[smoky cypress]

Yeah

Bruh moment

[woven delta]

Lmao

[woven delta]

Apparently there's a generalization of sylows theorem that states that the number of subgroups of order p^e for any integer e is congruent to 1 mod p

Does anyone have an idea for how to prove this?

[smoky cypress]

There is an outline of that proof in the last section of chapter 1 of Jacobson

It's in the exercises

@woven delta

[woven delta]

Awesome thanks

[latent anvil]

We have this on a problem set in my algebra reading group thing

It's awesome

It's on the homework the week before the sylow theorems are in lecture

[neat ginkgo]

TIL what quaternions are

used for a brief second in my algebra book

what field of mathematics is this studied by

[bleak abyss]

For fuck's sake

[half rapids]

pretty trivial

[latent anvil]

dami I think there's a typo

you should go look for it

[bleak abyss]

ur the typo here

[nova plank]

I got to the first box, then i was tired and stopped

[latent anvil]

sharmock

[fair obsidian]

what field of mathematics is this studied by
algebraists? Not sure if there's a specific branch that specializes in quaternions. Probably people that study real associative algebras or something like that.

[bleak abyss]

Quaternions I don't think are specifically studied but they kick in elsewhere

Mostly in algebra

The group Q8, quaternion algebras

[latent anvil]

they show up in topology, when you're looking at classical (matrix) groups

Iirc SU(2) is isomorphic to the unit quaternion is diffeomorphic to S^3?

the unit quaternion are a 3d sphere in a 4 dimensional space so you can use them to put a group structure on S^3

[bleak abyss]

Sounds right I think

The one I know offhand is that SO(3) is RP^3

So if SU(2) double covers SO(3) then yes that works

[latent anvil]

Yeah it does

In fact this is how Lee (outlines in an exercise) the proof that SO(3) ≈ RP^3

you like look at how a unit quaternion acts by conjugation on the imaginary quaternions

This is a linear automorphism of a 3d vector space

if you look at it in the basis {i, j, k} you actually land in SO(3)

and this is surjective with kernel {-1, 1}

[bleak abyss]

Oh my proof of that was much more direct lol

Like you just come up with a map S^3 -> SO(3) by being like

Err

No no

From the 3-disk

[latent anvil]

?

[bleak abyss]

You can realize RP^3 as D^3/antipodal points on the boundary

[latent anvil]

oh yeah sure

[bleak abyss]

So you map D^3 -> SO(3) by just being like

Aight for a point on the disk x, map it to the rotation about the axis given by x, by angle |x|\pi

(We're considering the axis given by x and -x to be flipped, so rotating the opposite direction)

Two things map to the same guy if they're antipodal points on the boundary

So boom it factors through RP^3 and is a homeomorphism (in fact it should be a diffeo?)

[latent anvil]

Nice

[lament dawn]

Does anyone know where I might find an algebraic proof (or something useful for constructing a proof) of the following claim?

$A, B \subseteq E \implies \mathbb{F}(E) \cong \mathbb{F}(A) \underset{\mathbb{F}(A\cap B)}\ast \mathbb{F}(B)$

Where F(X) is the free group on a set X and the RHS of that isomorphism is the free product with amalgamation

I can think of a method from a topological perspective using Seifert van kampen, but wondering if there exists an algebraic proof already before reinventing the wheel myself

[fair obsidian]

So A and B are sets, do you mean to say that the product with amalgamation gives F(A union B)?

[lament dawn]

Yes sorry, here $E = A \cup B$

[cloud walrus]

Migillope:

[lament dawn]

oh I forgot to copy paste the fixed version of that original latex

one moment

[fair obsidian]

The easiest way I can think of proving this I think is by using universal properties

[cloud walrus]

Migillope:

[lament dawn]

that was what I meant to type up

[fair obsidian]

Cool

I think you might be able to construct an explicit isomorphism, although that is similar to the universal property stuff

Do you know what I am referring to when I say the universal properties (of the free group, amalgamation, ...)?

[lament dawn]

Yeah, I tried some cursory attempt at that, but I'm unsure how I would go about constructing the diagram

Starting from A U B -> F(AUB) and AUB -> F(A) * F(B) amalgamated over F(A ^ B) would do it, but that last bit is kinda the entire problem (and not particularly well defined)

I've already simplified the problem to a reasonable algebraic problem, I just want to make sure I'm not working on it pointlessly before committing the time to it

[fair obsidian]

I'm not an expert but I think it should work out

Ok so lets let H = F(A n B) and G = F(A) * F(B) amalgamated over H

Using the inclusion AuB into F(AuB) you want to construct a map from G to F(AuB)

So G has the property that whenever you have homomorphisms f1: F(A) to K and f2: F(B) to K such that the restriction of f1 and f2 to H give the same map, you get a unique homomorphism h: G to K

Such that h matches up with f1 and f2 by restricting appropriately

Right?

[lament dawn]

such that the restriction of f1 and f2 to H give the same map
meaning K subgroup F(AnB)?

[fair obsidian]

No I mean that H is a subgroup of F(A) and F(B), so you can restrict the domain of f1 to H and the domain of f2 to H

And $f_1 |_H = f_2 |_H$

[cloud walrus]

The_Vman:

[fair obsidian]

Maybe you think of there being inclusion homomorphisms $\eta_A: H = F(A \cap B) \to F(A)$ and $\eta_B: H = F(A \cap B) \to F(B)$ instead, and then I mean $f_1 \circ \eta_A = f_2 \circ \eta_B$

[cloud walrus]

The_Vman:

[fair obsidian]

Maybe I should ask, how have you defined the free product with amalgamation? Are you ok using this property or should we be talking about quotients too?

I don't know how clear that is, but that's the kind of diagram I'm thinking about

[lament dawn]

that is more clear to me

I'm (ideally) defining G (free product w amalgamation) to be to be F(A) * F(B) / <<F(AnB)>>, so the free product of the free groups of A and B quotient out the normal closure of the free product of A and B intersection. From there it's pretty easy to use the first isomorphism theorem to complete the proof

I'm not sure if that diagram is true in all cases, what is that property? Pushout?

[fair obsidian]

Ok yeah that makes sense, but you mean the normal closure of elements $\eta_A(g)\eta_B(g)^{-1}$ for $g \in F(A \cap B)$?

[cloud walrus]

The_Vman:

[fair obsidian]

Yeah it is a pushout diagram

[lament dawn]

yes

[fair obsidian]

And yeah the pushout diagram corresponds to a free product with amalgamation

[lament dawn]

yeah my category theory game is not weak, I wasn't sure if that applied to free groups of free products w/ amalgamation

I had similar diagrams, just wasn't sure on the theory of that. This is generalized a lot from my application

[fair obsidian]

I mean I'm pretty sure it does, but I think my category theory game is probably weak lol

[lament dawn]

I figure I could avoid category theory if I could prove that for $f: F(A) * F(B) \to F(E)$ and $g: F(A\cap B) \to F(A) * F(B)$ we have $ker(f) = <<im(g)>>$ then the original statement would follow

[cloud walrus]

Migillope:

[lament dawn]

but that might be more convoluted than just figuring out the cat theory

[fair obsidian]

Hmm

Yeah I can't think of a good way of showing $ker(f) \subseteq <<im(g)>>$ other than going through some diagrams

[cloud walrus]

The_Vman:

[lament dawn]

I mean, I don't see any reason why pushout wouldn't work with free groups, but at the same time I don't want to write something relying on that fundamentally without really knowing for certain it does

clearly pushout applies to free products with amalgams, as it is literally the same, wiki even defines them as the same. I just don't know if it applies to free products of free groups with amalgams. I suppose that would just be a subcase, so why not?

[fair obsidian]

Yeah it's just a subcase

But that's not what's going on with the above f:F(A) * F(B) to F(E) and g:F(A n B) to F(A) * F(B)

[lament dawn]

yeah that's a different approach

since I was a little sketch about pushout

[fair obsidian]

Yeah I think the best to solve your problem is to show that the pushout diagram with F(AnB), F(A), F(B), and F(E) instead of the free product with amalgam, has the universal property

And so F(E) must be the free product with amalgam through some natural isomorphism

And using the property of the free group (and the pushout property of AnB, A, B and AuB) it shouldn't be that bad at all

[lament dawn]

I guess my question would be what conditions we need upon the homomorphisms such that the commutation in your diagram is an isomorphism

does F(A) -> K need to be an isomorphism?

or just homo? or surjective?

[fair obsidian]

Sorry so that which one becomes an isomorphism? The map from F(E) to the amalgam?

[lament dawn]

the map from F(A) *_H F(B) to K

[fair obsidian]

Ohh