groups-rings-fields

[rustic crown]

another idea is to reduce it to the gauss sum.
define g = sum_{a in F_p} (a|p) zeta^a

since sum_{a in F_p} zeta^a = 0
notice that tau - 0 = g

essentially,
(0 + 2QR) - (0 + QR + QNR) = (QR - QNR)

it makes the exponent less annoying to deal with, and the coefficient is still multiplicative so not bad

[covert cliff]

here C_m is the cyclic group of order m

Am i misunderstanding something? If for example m = 5, then is this saying that {g \in C_5 | ord(g) = 5} has size 4?

[rustic crown]

yep

all non-identity elements have order 5

[covert cliff]

C_5 is isomorphic to Z/5Z, so can we say that all non-identity elements in Z/5Z have order 5?

[frigid lark]

Yes

[covert cliff]

hmm

ok but isn't the order of 2 = 3?

[frigid lark]

You can generalise to any prime (and any positive integer with a bit more effort)

[frigid lark]

2+2+2 = 6 and 6 is not equivalent to 0 mod 5

[covert cliff]

oh right, I was confusing it with the multiplicative group

sorry about that

but thank you

[dusk sand]

Can someone help me with sign of a permutation?

Q=(1627)(398)(45)
T=(1734256)

Q=(16)(62)(27)(39)(98)(45)
T=(17)(73)(34)(42)(25)(56)

sign(Q)=(-1)^6=1
sign(T)=(-1)^6=1
Is this correct?

[ebon heron]

Hi you can reach out in my inbox for help

[chilly ocean]

no

post your question here

[summer path]

#groups-rings-fields message reminds me of this

[dusk sand]

find the sign of Q and T

[chilly ocean]

not yours

i am not talking about yours

[dusk sand]

haha

[hidden haven]

Instead of decomposing the permutations into transpositions, you can first prove that the sign of an n-cycle is +1 if and only if n is odd and that signs multiply on composition. Then you can compute it directly from the cycle decomposition

[dusk sand]

you mean by the order of each permutation

[hidden haven]

What do you mean?

Oh τ is not given as a cycle decomposition

[dusk sand]

like o(T)=12 o(T)=7

[hidden haven]

What are you trying to do with the order? I didn't get it

[dusk sand]

i thought you said i can use the orger to find the sign

[hidden haven]

Only for cycles

The sign of an odd cycle is +1 and of an even cycle is -1

[dusk sand]

ok...

and n is the number of different circles?

[hidden haven]

No, n-cycle means a single cycle of n elements

[dusk sand]

like this:
Q=(1627)(398)(45)
T=(1734256)

[hidden haven]

If Q = AB, then sign Q = sign A * sign B

[dusk sand]

ok... so for Q=(1627)(398)(45) we do sign(1627) * sign(398) * sign(45) = -1 * 1 * -1 = 1

[hidden haven]

Yes

[dusk sand]

and T=(1734256) is sign(1734256) = 1

ok thx

[ivory trail]

check out my cool new discord username

i think 196884 is still up for grabs

[dusk sand]

and what do i do when i asked to find TQT^(-1)

[uncut cloud]

In general you have to compute this by hand

[dusk sand]

all 3 or is there shortcut

Q=(16)(62)(27)(39)(98)(45)
T=(17)(73)(34)(42)(25)(56)

[uncut cloud]

In general there is no short cut. Maybe here there is one

[dusk sand]

ok.. thx

[summer path]

i mean it's not too bad i think here

albeit a little annoying

[uncut cloud]

For instance if the commute it will be very easy. And you can see that with the decomposition in cycles.

The main point in the study of permutations is that decomposition in cycles is better than decomposition in product of transposition

[dusk sand]

how do i find the largest order for permutation symmetric group

like S9 or S13

[uncut cloud]

Do you remember how you did that for S_4 earlier ?

Its gonna be the same way

But a general formula is rather inoperable

[carmine fossil]

So you want to find a n and n integers $x_i$ such that $\sum x_i = k$ and lcm ${x_i}$ is maximum

[cloud walrus]

DraK(night)

[carmine fossil]

I think brute force is the best approach for this

[dusk sand]

ya its just checking a lot of numbers this way

[uncut cloud]

Yes it is exactly this absurd formula. For S_9 I find 20 for 9=5+4. But not sure it is the maximum

[summer path]

3+3+3=9, gives you 27

[carmine fossil]

lcm, not product

[hidden haven]

Only an asymptotic formula is known for this

[summer path]

oh wait

im bad at reading

[carmine fossil]

If it were product, you just do a variant of AM-GM

[hidden haven]

How? Even then it seems difficult because you only allow integral partitions

[slim kayak]

Thats just the landau function here, and apparently quite a few people already tried to attack it

[hidden haven]

Landau's is for the LCM thing right?

[carmine fossil]

Same thing

[slim kayak]

Yeah, that's what the want for finding elements of maximal order in S_n tho

[carmine fossil]

Yeah lcm here

https://en.m.wikipedia.org/wiki/Landau's_function

[hidden haven]

Ye

Was just curious about the product thing

Like if you decide on the number of parts in the partition then you just go as close to equal parts as you can

But how do you figure out what number of parts maximises the product?

For 8, 3+3+2 is better than 2+2+2+2 curious

Maybe we want the young tableau thing for the partition to be as square as possible lmao

[carmine fossil]

I think you can look at the maxima of (a/n)^n, as n varies

[hidden haven]

Yeah that would make sense

[carmine fossil]

And then use that as a heuristic to determine which n gives you the best product

[hidden haven]

Should give a good approximation at least

[dusk sand]

find the 4 last digits of 2^2016

2^2016 = x mod 10000

[summer path]

#elementary-number-theory

[dusk sand]

2^16 = x mod 10000

[dusk sand]

ya did for 1000. 1000 * (1-1/16)*(1-1/625) = 9360

[summer path]

,ti

[cloud walrus]

The current time for Tubular Cat is 04:53 AM (EDT) on Wed, 07/06/2023.

[summer path]

brain too tired to do math right now

[formal ermine]

,ti

[cloud walrus]

The current time for illuminator3 is 11:01 AM (CEST) on Wed, 07/06/2023.

[formal ermine]

I'm tired too

[dusk sand]

can i just write: 
2^2016 = 2^16 = x (mod 10000)
2^16 = (2^4)^4 = 16^4 = x (mod 10000)
16^2 = 256 (mod 10000)
16^4 = 256^2 = 5536 (mod 10000)

[formal ermine]

#elementary-number-theory

[dusk sand]

you mean to put this there?

[formal ermine]

yes

[fair quartz]

i think my question goes in here? (if not, please point me to the correct channel)

im currently looking at operators in a hilbert space V (with the operator A being element of B(V) )
and i just need to prove that the norm of said operator is equal to the norm of its adjoint operator A*

i tried a few things but im not quite getting there:
in our script there is a proof that ||A*y|| =< ||A|| ||y|| so i tried doing something with that
and using Cauchy-Schwarz i managed to show that |<x, A*y>| = |<Ax, y>| =< ||A*|| ||x|| ||y|| but im unsure if that accomplishes anything

(x and y being vectors in V)

could somebody give me a hint or point me a to a resource that goes over this proof?

[uncut cloud]

Recall that in a Hilbert Space ||A|| = sup |<x,Ax>| for all |x|=1

With this equality your assertion will be obvious

But to prove this equality there is Cauchy Schwartz

[fair quartz]

ah yeah, the definition was the first thing that i looked at but i didnt see anything to approach there at first 😅

[wraith cargo]

I think you can do it with || A || = inf{C : || Ax ||<= C || x ||}

I think that's what your prof was alluding to

[fair quartz]

ill give these ideas a shot, thanks!

[rustic crown]

did you mean sup ||Ax||? cause what if your hilbert space is R^2 and A is rotation by 90degrees?

[wraith cargo]

Another hint might be that you just have to show one half of the inequality because the other side follows trivially from symmetry

[uncut cloud]

Lol I wrote Bs indeed, sorry. I am going to fix this

[rustic crown]

oh Bs as in not manny B's but bs >.<

[south patrol]

Noice thank

[feral rock]

Is there any relationship in some groups of |a| |b| and |a+b|? For elements a and b

[agile burrow]

You can impose some nice conditions to get some relationship. For instance, if a and b commute then the order of ab is the least common multiple of the orders of a and b respectively. However, in general there is no good relationship. In fact, one can show that for any integers l, m, n there is a finite group G with elements a and b such that the order of a is l, the order of b is m, and the order of ab is n

[feral rock]

Tysm

[covert cliff]

Can we conclude that $(\mathbb{Z}/p\mathbb{Z})^{\times} \cong \mathbb{Z}/p\mathbb{Z}$ for prime $p$?

[cloud walrus]

ForJoke

[uncut cloud]

The cardinal are not the same so no

But if you take Z/(p-1)Z then yes

[covert cliff]

oh, that makes sense, I was tripping

thank you

And also Z/(p-1)Z is cyclic right? so can we conclude that (Z/pZ)* is also cyclic?

[south patrol]

I mean usually the proof of this isomorphism comes from proving (Z/pZ)* is cyclic lol

[covert cliff]

oh

[south patrol]

So this seems kinda the wrong way round

Fun fact lol if F is a field then any finite subgroup of F^x is cyclic

[covert cliff]

I was trying to show that (Z/pZ)* is cyclic lol

[south patrol]

Ah

[covert cliff]

So ig I will keep trying then

Thank you for the heads up tho

[south patrol]

Not the easiest problem to do by yourself

[uncut cloud]

It seems you have misunderstood something. Z/(p-1)Z is cyclic by definition, you see that ?

[covert cliff]

yes

it is generated by 1

<1> = Z/(p-1)Z

[covert cliff]

is there any condition between d and m?

because in Z/5Z there are no elements of order 3, for example

[frigid lark]

Well d | m

By Luigi

https://en.m.wikipedia.org/wiki/Joseph-Louis_Lagrange

[grand cliff]

what a fun question

[lusty marlin]

This is an interesting result. How does one go about proving it, if at all it is provable with a standard undergraduate level of understanding of group theory?

[south patrol]

The first bit or the last bit lol

[dim widget]

@covert cliff you ought to use the following lemma: a group is cyclic of order n if and only if there are exactly \phi(d) elements of order d for each d|n. Then use what you know about how many roots a polynomial can have over a field applied to x^{q-1} - 1 = 0

[lusty marlin]

Haha the last bit, of course

The first bit is incorrect. The order of the product of commuting elements divides the least common multiple of their orders. It isn't necessarily equal to it.

[south patrol]

Agreed

[covert cliff]

well the forward direction is the image I sent

[dim widget]

@covert cliff I am telling you how to prove that that isomorphism exists in the first place

Or rather the corrected version

[covert cliff]

oh I see I see, my bad

[dim widget]

@lusty marlin with very little work you can also show that lcm/gcd divides the order of the product and that these bounds are sharp.

[covert cliff]

I will try to prove this lemma first, I am doing all this for a number theory class and I have literally no algebra background, I love this uni

[dim widget]

@covert cliff alternatively you could try to just show that the q-1’st cyclotomic polynomial has a root over F_q always. But the argument will end up being similar.

@covert cliff also the last thing is that it might be easier to prove G is finite cyclic iff G has a unique subgroup of order d for each d|n.

And use that to conclude

[covert cliff]

Ok lemme give these a shot

[lusty marlin]

Oh. Silly of me to assume that this wasn't the case. Thanks.

[agile burrow]

I don't quite recall the details, this result is proven in Milne's notes on group theory. My understanding is that one considers a sufficiently large number q and shows that PSL_2(F_q) contains elements of the prescribed order by using some basic facts about multiplicative groups of finite fields and some linear algebra. Nothing is terribly involved though

[delicate orchid]

step 1: G = <a, b : a^l, b^m ,(ab)^n>
step 2: is G finite? no? ADD MORE RELATORS
repeat step 2

[rustic crown]

in general order of ab would only divide the lcm. if the orders are coprime then it works, but if not you can only say that there exists an element whose order is the lcm that follows by writing lcm as a product of two coprimes and reducing to the previous case.

for example a and a^-1 have the same order, but the product has order 1.

so yea only other precise statement you can make is order of a^d is n/(d, n) where n = |a|

[coral spindle]

I really like the example of the infinite dihedral group.

Or I guess dihedral groups in general

two generators of order two, yet their product is of arbitrary order...

[delicate orchid]

... wtf!!!

[coral spindle]

so true

[rustic crown]

Dn = <x, y|x^2 = y^2 = (xy)^n = 1>

me like this presentation more

so symmetric and cute

[hidden haven]

You just assumed the axiom of empty set

[rustic crown]

[coral spindle]

I like the semidirect product version

becuase idk, I'm evil?

[rustic crown]

moldi is bacc

[hidden haven]

[coral spindle]

It's easier for rep theory to know about the normal subgroups

[rustic crown]

true >.<

[delicate orchid]

fuck the semidirect product we out here C_n \wr C_2

[coral spindle]

Babe

that is a semidirect product !!!!

[delicate orchid]

ermmmm but it's clearly a wreath product, babes

[rustic crown]

det never studied wreath >.<

why we care about it

[coral spindle]

Wreath products idk they're nice

Like to be clear

[delicate orchid]

name a funny group

[coral spindle]

OK

[delicate orchid]

it's sylow subgroups will be wreath products

[south patrol]

Sylow

[rustic crown]

oooh

[coral spindle]

Let $X$ be a $G$-set. Then $G$ acts by automorphisms on $H^X$, which is to say the group of functions $X \to H$. Then $H \wr G$ is defined as $H^X \rtimes G$.

[cloud walrus]

Boytjie (never-to-be-glomed)

[delicate orchid]

S_n, definitely A_n with n odd but also probably even, GL over finite fields

etc. etc.

all their sylow subgroups are wreath mfs

that's why I personally care about them

they come up in automorphisms of n-ary trees as well

[coral spindle]

I care because a bunch of Coxeter (Weyl) groups are wreath products

[delicate orchid]

we care for the same reason then

[coral spindle]

Can't remember, I think $B_n$ is $C_2 \wr S_n$

[cloud walrus]

Boytjie (never-to-be-glomed)

[delicate orchid]

it is

[coral spindle]

But this is coming up in my research

noice ty Wew

I cannot remember dynkin shit for the life of me

[delicate orchid]

le graph

[coral spindle]

that's french for the graph btw

[rustic crown]

[delicate orchid]

I care about complex reflection groups so I'm more like

C_m \wr S_n

[rustic crown]

wreath you mean?

[coral spindle]

Deeply fucked up 🙏

Joking

[delicate orchid]

there's topological construcitons using them as weyl groups which are entirely representation theory-based but are not lie groups

[coral spindle]

I don't understand complex reflection groups

[delicate orchid]

it is beyond fucked up

[coral spindle]

What how

[delicate orchid]

they're called spetses (like hte island) I know fuck all about them

[south patrol]

Who up fusing their systems

[delicate orchid]

who up pseudoing they reflections

[coral spindle]

Yeah I was gonna say, I assume this is coming from the Weyl group of a f.s.

[delicate orchid]

I got a pdf called "towards spetses I" and they don't even get around to defining them in 200+ pages why couldn't I have just done the funny fluids

much easier

[slim kayak]

Had a question. Can someone suggest me a neat way of showing whether or not the subgroup generated by subgroups H and K of a free product A x B with H = A and K = gBg^(-1) is free?

[agile burrow]

GOOD point

[chilly ocean]

Are these finitely presented or no?

[slim kayak]

A and B? Not specifically but I don't remember us ever working with infinitely many generators so for the moment let's say yes

[hot lake]

when you say free, are you asking if it's the free product of H and K

[slim kayak]

Isomorphic to the free product of H and K

<H,K> = H x K ( = being isomorphism symbol here)

I reduced it to showing whether a word with alternating letters in H or K is trivial in A x B, but it devolves into seemingly never ending nested cases

Can't wait for Cayley graphs, word problems are miserable

[rustic crown]

det bad at memes >.<

why is good excited >.<

[chilly ocean]

can anyone help me with math

need some help

[rustic crown]

#❓how-to-get-help maybe

[agile burrow]

idk lol, I just write like that 🙂

[slim kayak]

Maybe it absorbed a high-energy photon?

[white oxide]

does this example only apply to a free nonabelian group?

(cuz then each yk would just be y)

[south patrol]

Sure, free abelian groups aren't free groups anyway (in general)

[coral spindle]

You answered it yourself!

[white oxide]

just checking i don't trust myself lmfao

[white oxide]

huh ok

[coral spindle]

Maybe you should! You're smart!

[slim kayak]

Free abelian groups are free group with some relations sprinkled in, specifically just a bunch of commutators

[coral spindle]

Kerr that is literally any group

Literally any group is a free group plus some relations

[slim kayak]

Yes, so?

For the same reason free abelian groups aren't free just like almost any group, they have relations

[white oxide]

how will the prime power decomposition of the fundamental theorem of finitely generated abelian groups follow if we take d1 = 3, d2 = 18, ....? because then the proposed group will be isomorphic to Z_3 x Z_18 x ... and 18 cannot be written as a power of a prime

[south patrol]

You can write it in a different way

using CRT

[white oxide]

ewww CRT

[summer path]

CRT is very nice

[slim kayak]

CRT, my favourite change of basis theorem

[rustic crown]

CRT is uwu

[coral spindle]

I think of the CRT in the same way that I do matrices

Matrices are a wonderful computational tool. They tell you so much! But they are also a brain worm

So often I see people computing with matrices as if they don't correspond to linear transformations. Often it is easier to do certain proofs by looking at things abstractly, rather than as matrices.

It seems that the CRT is similar: people see it as just a statement about solutions, but in fact it's an enormously useful fact about elementary number theory

E.g. here is a proof that Euler's totient function is multiplicative. We define phi(n) = |(Z/nZ)*|. Then if n,m are coprime, then phi(nm) = |(Z/nmZ)*| = |(Z/nZ x Z/mZ)*| = |(Z/nZ)* x (Z/mZ)*| = phi(n)phi(m).

This is infinitely more insightful than that horrible counting argument one typically sees when one first encounters phi

Anyway this is all to say that if you say "ewww CRT" you're a bad person and you should feel bad and I hate you and I'm never talking to you again >:((((((((

[summer path]

I think crt makes a lot better sense when one sees it in the context of rings

Or at least it was for me

[coral spindle]

I agree. It's really a statement about rings.

[summer path]

Learning it in high school as purely a computational tool (with no proof, no surprise) for number of solutions was just like bruh

[coral spindle]

Just like how in high school you learn how to multiply matrices, right?

[summer path]

Kinda yeah

Tbh I don't really understand why matrices aren't just introduced in the context of linear algebra

Instead of stand alone as a computational tool

Because you can reasonably do linalg as a class pretty much immediately after precalc anyway

But I guess crt is understandable somewhat since you probably can't do abstract algebra that quickly, though one of the discrete math sections last year tried to give intuition of ideals on a pset and people were rather confused lol

[chilly ocean]

<@&268886789983436800> delete

[coral spindle]

wow I thought tubular cat was spitting facts

[summer path]

Nami

[rustic crown]

maybe because it also appears in other ares of math? like say graph theory. the incidence matrix and other nice matrix you associate to a graph dont' have a deep meaning via linear maps, but the matrix multiplication still manages to capture some nice combinatorics in it.

[summer path]

Hmm maybe

[rustic crown]

det likes matrices

[formal ermine]

the field of rational functions is the fraction field of the coordinate ring or?

[rustic crown]

yee

[formal ermine]

thanks

my prof uses Q(...) for fraction field

confused me

[dawn stump]

I'm gonna assume you guys aren't talking about critical race theory or cathode ray tubes

[formal ermine]

?

[coral spindle]

Chinese Remainder Theorem

[chilly ocean]

what

[coral spindle]

Some people nowadays call it Sunzi's theorem, though it can't really be attributed to any one person imo.

[dawn stump]

Thanks

[south patrol]

Nice

[south patrol]

lol nice

Isn't CRT more of like a vibe at this point lol

Like refers to slightly different things depending on how general you want lol

[formal ermine]

crt is a trivial observation

[south patrol]

once you already have the set up of rings, sure lol

[coral spindle]

The sea rises, illum

[formal ermine]

[south patrol]

It's like easy once you have rings and first iso but those are like lol thousands of years after CRT was found innit but yes

Tbh all of ug maths is just a trivial observation

[rustic crown]

[summer path]

much of undergrad algebra sequence is like just

[rustic crown]

[summer path]

stuff after the main sequence was more uwu though

[south patrol]

why are we talking about astronomy now...

[summer path]

huh

[south patrol]

"main sequence" lol

[summer path]

oh

i have forgotten

[rustic crown]

(det never knew)

[formal ermine]

I'm starting to like commutative diagrams

last alg nt lecture we proved dedekind's criterion with a single commutative diagram and 🐍 🍋

[rustic crown]

illu finally cat pilled

[formal ermine]

we also proved this other thing about lattices with a HUGE diagram

[formal ermine]

nooo only diagram chasing

[rustic crown]

whuts dedekind's criterion

[summer path]

diagram chasing

[formal ermine]

stating it takes half a page

[south patrol]

Oh do you mean like

equivalently Dedekind-Kummer?

like in algebraic number theory

[formal ermine]

yes dedekind kummer

[south patrol]

for how primes split up

Noice

[summer path]

from a google search, it seems so

[formal ermine]

my prof just called it dedekind

[south patrol]

We did too ye

I think Dedekind-Kummer is over more general extensions without the ground being Q necessarily (?)

[formal ermine]

[south patrol]

Didn't realise you could use big diagrams to do it lmao

[formal ermine]

yeah it's just that diagram and then commuting quotients

[south patrol]

Oh is that just an arrow before Schlangenlemma

I was trying to decipher the word lmao

[formal ermine]

loool

[summer path]

ours called it kummer-dedekind

[formal ermine]

DEDEKINDS KRITERIUM

[south patrol]

I always find it interesting when people swap round names or whatever

[summer path]

i can't wait for
dedekind

kummer

[wraith cargo]

Diagrams are great

better than anything else

I would marry a diagram if I could

(I am horribly addicted to algebra)

[formal ermine]

hi irony

[wraith cargo]

hello illum

How is life
I'm learning abt Pontyargin duality rn

[formal ermine]

scary words

I'm doing toric geometry at the moment

[coral spindle]

Hey did you know Pontryagin was a dick?

[rotund aurora]

Factually false

[wraith cargo]

aaa toric varieties I still don't understand how they relate to a torus

[rotund aurora]

I think its a variety with a doughnut inside

[wraith cargo]

I wouldn't be surprised lol
Only thing I know abt him was that he was totally blind

and his mother read math to him

[rotund aurora]

Wow thats interesting

[coral spindle]

Pontryagin duality cool tho

I should actually learn topological groups

[formal ermine]

you let the funny n algebraic torus act on your variety

[wraith cargo]

I liked the Gelfand transform too tho
and how it's an isomorphism in the case where you're transforming a closed unital self-adjoint commutative subalgebra

[summer path]

the only math history i know is: galois died young because he lost a duel and some guy who's name starts with a B was a nazi

[formal ermine]

Badolf Bitler

[wraith cargo]

name any german mathematician from the 20th century and there's a good chance he was nazi affiliated lol

[coral spindle]

One of Noether's students collaborated with the nazis to expel the Jewish professors from Goettingen, which of course included Noether herself.

It's truly heartbreaking.

[summer path]

maybe it was brouwer

idk

[coral spindle]

Brouwer was Dutch haha

[white oxide]

ewww CRT

[coral spindle]

blocked and reported

[dawn stump]

That goddamn Gödel

[rotund aurora]

what are algorithms for classifying groups up to isomorphism of order <=n for a given n

[formal ermine]

yoo I wrote one like that at an internship earlier this year

it was trash tho lol

[rotund aurora]

I mean, algorithms should exist right? cuz finite problem and then you only have to decide when two groups of the same (finite) order are isomorphic, again a finite problem

like you could actually run through all the |G|^{|G|^2} functions from GxG to G and check the axioms

[slim kayak]

There aren't any

[white oxide]

okay idk if this is the right channel or not but does anybody know any good articles/papers to read about the p-adic numbers (preferably not wikipedia lol)

[slim kayak]

We don't know, say, how many groups of order 2048 there are up to isomorphism

[summer path]

Gouvea has a text on them, but I personally have not used it that much

[coral spindle]

There are loads of books on valuation theory, you could check one of them

[white oxide]

oh i was looking for like a paper or article or something brief not really a textbook (just want to gain some basic knowledge)

i'll check that out later tho thx

[coral spindle]

Oh idk. I usually just use wikipedia or nlab

the nlab page for the p-adic integers (read it first) is p good

p un intended

[summer path]

i learned about them originally when i was trying to learn some arithmetic dynamics, so i kinda just read about them from silverman's text on them

but i think generally most intro texts on something using them will have a section to catch you up on them

[slim kayak]

There are a lot of ways to enumerate all groups of a given set of length n, but the problem arises once you want to figure out when two groups are isomorphic. This is related to the word problem of groups, since any isomorphism must have trivial kernel and any map of mapping generators to combinations of generators will come down to verifying whether all non-trivial elements map to non-trivial elements. Since the group is finite and the aforementioned being equivalent to checking if the map is injective, this is automatically a criterion by which to see whether a given homomorphism is an isomorphism between finite groups/ their presentations

[white oxide]

ok bet

[white oxide]

yea that makes sense

what field (pun intended?) are they used most in

[coral spindle]

yeah you should bet (much money to be won)

It's a good page ok

[summer path]

im not sure, perhaps someone else can answer this

i just study whatever looks fun

[coral spindle]

So something I learned recently about the p-adics is they can be used to construct galois groups

If you want to study the galois group of some polynomial, that can be really quite hard

[ivory trail]

bieberbach

but yeah also a bunch of other germans

[coral spindle]

so one way of doing it is finding some prime p over which your polynomial splits, whence it is fairly easy (i.e., we have algorithms) to find explicit solutions in the p-adics

And they're relatively easy to work with

But fwiw the p-adics are just a general area of interest. There's a family of results wherein properties of the integers are reflected in the completions, which is to say R and all the p-adics

This is known as the Langlands programme, basically

It's a huge area of research.

My current research project is related to this. I have to deal with the p-adics and when group representations are realisable over them.

This is all in hope that we'll be able to show a correspondence between certain 'local' characters of a group and some global characters

This 'local-global' correspondence is precisely what we're looking for when I talked about these properties of the integers (the 'global') being reflected in the p-adics (the 'local')

I rambled for a bit there, I hope that broadly motivated the p-adics.

[white oxide]

cool! thanks for typing that up

i actually i remember that my freshman professor was doing some research involving the langlands programme

My research is in number theory and geometry. In the late 1960s Robert Langlands proposed a series of deep and elegant conjectures linking arithmetic, geometry, analysis and representation theory. Since its inception this program has, thanks to the work of mathematicians such as Pierre Deligne, Vladimir Drinfeld, Alexander Beilinson and Robert Langlands himself, evolved in many directions and now permeates much of mathematics. My work predominantly explores the p-adic and geometric aspects of the Langlands program and is highly interdisciplinary, involving number theory, arithmetic geometry, algebraic geometry, p-adic analytic geometry, D-module theory, p-adic Hodge theory, motive theory and higher category theory.

yeah

[slim kayak]

I know p-adics as funny infinite number and number theory thing or as "try out categorical limits exercise no. 12"

What makes p-adics "local" ?

[coral spindle]

I couldn't tell you exactly. In some ways it's just a name

I'm sure a good number theorist could explain it, but alas I am neither a number theorist nor good

[echo gull]

this is very cool research

I love how there is like two paradigms for research writeups on faculty websites, it is either an entire history of their discipline or its like "I do harmonic analysis, particularly the very harmonicy sort"

[slim kayak]

If there is almost ring theory maybe there is almost harmonic analysis, how harmonicy the analysis is seems very important to state IMO

[echo gull]

definitely

[glossy crag]

Haven't seen such a proof before, could you share the relevant pages from (what i assume are) the professor's notes (statement and proof)?

[rotund aurora]

Lol veritasium just uploaded a video on p-adics

[rotund aurora]

https://youtu.be/3gyHKCDq1YA maybe you can check this

[south patrol]

10 adics

[wraith cargo]

wdym local

[rotund aurora]

If you are solving diophantine equations or related things then the name "local" is obvious

If some polynomial in integers has an integer root, then it has a root mod p for every prime p, etc

[wraith cargo]

obvious

[coral spindle]

obvious

[white oxide]

there's no algorithmic way to find a generator for this group right

not tryna do all that work sadly

that's some chinese remainder theorem shit

[coral spindle]

Sure there's an algorithmic way! You search through every element 🥰

At a glance I can't see any obvious way to find one.

Actually come to think of it

Wait, no. Nvm.

You may find something by inspecting powers of x, but this is just a hunch.

[rotund aurora]

So there are algos? Like I wasnt exactly asking for an efficient explicit algorithm, ig more of a decidable thing and something that works for most 2 digit numbers and some numbers with a low number of prime factors.

And btw, the word problem is decidable on finite groups no? Maybe im saying nonsense rn, I dont remember how it went

[coral spindle]

It is really not saying much that the word problem is decidable

There is a LOT of variation of ease-of-computation in the class of decidable problems.

Like we are dealing with a finite problem, so ofc it's decidable. But this is far from saying it's easy.

[echo gull]

there's no ontology to anything in math being obvious

[rotund aurora]

I agree, Boytjie

[rustic crown]

if you're doing geometry, to get an open subset you can look at the non-vanishing set of a function. now you can think of an integer f in Z as a function on the set of prime numbers, namely if p is a prime, then f(p) := image of f in Z --> Frac(Z/p)
ofc you would want these functions to be continuous, and the natural topology to put on the set Spec Z which makes all these things true is the zariski top.
now you can look at germs of functions at the point p in Spec Z. that is, functions defined in an arbitrarily small nbhd of p. in particular, if a function f doesn't vanish at p, you should be able to define 1/f. so if you work this out, the germs of functions is exactly Z localized at the prime p. i.e. rational numbers where the denominator isn't divisible by p. this is a local pid aka DVR, and to understand a few things better you can take its completion. that's exactly the ring of p-adic integers... so in this sense it's "local".
similarly to understand an abelian groups which are Z-modules, you can reduce your work to understanding what happens with respect to each prime looking at Z_{(p)} modules and completing to get modules over Z_p, and hope that these primes capture enough information and tell you about your original question.

[coral spindle]

Hey det what about the reals?

How come they're involved?

[rotund aurora]

Ill be honest, I dont get why people call these things geometry

[coral spindle]

Something to do with the 'prime at infinity'?

[rustic crown]

maybe comparing it to a different example would be helpful.

consider complex numbers and polynomial functions. a (non-zero) prime ideal in C[x] looks like p = (x-a) and if f in C[x] then f(p) is the image of f --> C[x]/(x-a) = C

let's focus at the prime p = (x-0) = (x). a germ of a function at p would be a rational function C(x) such that the denominator doesn't vanish at x = 0. this is again a local ring with the maximal ideal generated by x. sadly, the zariski topology is super coarse, so even this local ring "sees" a lot of global structure and is thus harder to analyze. if you complete it, that can be thought of as making the element x an infinitessimal and the completed ring is C[[x]], the ring of formal power series in x.

[rustic crown]

right, i'm not very sure about this relates to algebraic geo, but a natural question is to ask what are the absolute values on Q (or number rings in general). there is one that you get from the embedding Q --> R. and by a theorem of ostrowski, if you define two absolute values to be equivalent if they induce the same topology, then up to equivalence only absolute values are the euclidean absolute value and the p-adic absolute values. so in some sense, you can think of this absolute value because of R as a prime which you didn't see before. this makes a few things really pretty, like remember given a natural n, we define |n|_p = p^-(v_p(n)) and so one way to package the fundamental theorem of arithmetic is by saying (prod_p |n|_p) * |n| = 1

det wanna learn more math

[coral spindle]

ur cool det

thanks

[rotund aurora]

Isnt that like saying sum_(p in X) ord_p(f)=0 in a variety or something

So that in particular it is not equivavalent to unique factorization lol, although it is impled by

[slim kayak]

Makes sense, but how is are the p-adics the completion of the localizations of Z at a prime? I thought they were the completion/limit of the Z/p^nZ or is it both?

[chilly ocean]

Both descriptions are correct

If we localize Z at p, then every element of that ring can be written as a unique power of p times a unit. Then we can complete it with respect to the -adic topology by considering the space of all formal power series in powers of p

[slim kayak]

Is it also still the limit sort of completion? If so, what is the diagram?

[wraith cargo]

it's the inverse limit of Z/p^nZ

[next obsidian]

If you take an m-adic completion for a maximal ideal m, it doesn’t matter if you localize at m first

The completion at an ideal I is defined the inverse limit of R/I^n

If you do this at m a maximal ideal, because R_m/m^nR_m = R/m^n the completions are the same

[slim kayak]

I said as much, I am just trying to understand in what sense it is also the completion of localizations at p also

[slim kayak]

Oh, so like the usual inverse limit but instead with the image of I in the localization?

Well, the set of fractions with numerator in in the ideal I

[next obsidian]

Well you’re completing at mR_m

If you do it after localization

But it doesn’t matter because the quotients you look at are the same either way

It won’t be true if you take a completion at some other prime after localizing

Only when you do m-adic completion and localize at m

But in the case of p-adics this is the case since (p) is maximal in Z

[slim kayak]

Geometrically, is it sorta like poking a hole and then excising a set having contained that hole?

[next obsidian]

¯_(ツ)_/¯

I think not

These operations are what happens zooming into the point

You aren’t taking a punctured neighborhood

[pastel cliff]

chmonkey

[small bramble]

Why is it the case that if G is a cyclic group then if d divides |G| then G contains d elements of order dividing d?

[uncut cloud]

you can see this very explicitly

[delicate orchid]

Consider the subgroup generated by an element of order d

[small bramble]

That has order (n, d) right, and then adding all the divisors would give d cus number theory says so

Kool

Ty

Oops order n/(n, d)

[delicate orchid]

It has order d

[small bramble]

Oh lol I was thinking power

[uncut cloud]

If you take G=Z/nZ, then the elements of order d will be the k such that n | kd. But you also know that d must divide n. If we write na= kd and n= bd you then have ab=k. So k is a multiple of b=n/d, and must be <= n So you have d choices possibles. But it gives you the number of order a divisor of d, not exactly d

[small bramble]

Yeah so I got D&F as a reference and it says with proof in 2.3 that for any divisor a that divides n, there are phi(a) elements of order a (where phi is the Euler totient function), and I knew that the sum of phi(k) for k a divisor of d is d. Where I was failing to make the jump to the conclusion was cus I was thinking "but this is an arbitrary group of order n, not d" but I just realized it doesn't matter cus any divisor of d divides n still 🤦‍♂️. I guess I should be good now as long as I internalize the proof of the statement I paraphrased, ty.

[formal ermine]

what is an open orbit?

[oblique river]

An orbit which is open

[formal ermine]

ah

[valid night]

If a group $G$ is generated by a (not necessarily finite) set $S$, is every group homomorphism $\phi : G \rightarrow H$ determined by $\phi [S]$?

Like how a linear transformation is determined by where it sends the basis vectors?

[cloud walrus]

Kroros

[uncut cloud]

Absolutely

[lusty marlin]

Yes

[valid night]

Alright thank you very much

[uncut cloud]

But unlike the linear case you can not in general send S on whatever you want

[rustic crown]

more precisely, it's determined by the set function S --> H. and not just by its image phi[S].

[lusty marlin]

My bad, read the question as having ϕ|S, which I interpreted to be ϕ restricted to S.

[rustic crown]

ah okie >.<

[long nebula]

You can express this as a universal property

[valid night]

Ah alright, thank you

[formal ermine]

how do we define a group structure on an n dimensional algebraic torus

[long nebula]

A function from G to S can be extended to a homomorphism from G to F(S) (the free group on S) and then compose that with the quotient from F(S) to H

🥺

[rustic crown]

the other way right

F(S) --> H, and then G is by definition quotient of F(S) by the relations, so factors as F(S) --> G --> H

[formal ermine]

ah

R^n/Z^n lol

wait nvm

that confused me even more

[dim widget]

Z^n is a normal subgroup of R^n you are welcome

[formal ermine]

because we defined T_n to be (C^*)^n and I doubt that's isomorphic to (R/Z)^n

[dim widget]

Ah that's a different torus

an algebraic torus

[formal ermine]

yeah

[dim widget]

But as a topological group it's just $(R^*_{>0})^n \times T$ where $T$ is $R^n/Z^n$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

so it's similar

Anyway C^* is a group in the obvious way, and a product of groups is a group soooo....

[formal ermine]

oh lmao I forgot that I knew the definition of T_n for a second

[dim widget]

i'm going to sully you now

in 3

2

1

new meta is slow-rolling the sully

[formal ermine]

that makes it even more painful

[rustic crown]

[south patrol]

[long nebula]

yea sorry lol

That was silly of me

[rustic crown]

https://media.discordapp.net/attachments/796976320067272704/818150000482582528/612933479730249728.gif

[lethal dune]

that just gives them some time to contemplate life decisions

[formal ermine]

this doesn't make sense

in the next sentence he says

"a cone in V = Q^n is a subset of the form blabla. If V = N_Q for a lattice N then we call blabla a lattice cone"

cuz like

if N_Q = Q ⊗_Z N = Q^n and V = Q^n then when will V not be N_Q

[wet zodiac]

sad how she died of cancer too

[slim kayak]

Given an HNN extension with monomorphism that arent surjective, why does it then contain a free group of rank 2 by Britons lemma? If one chooses say an element a not in the image of of phi, then the words of a and t can only be reduced if a is of the form t psi(x) t^-1, which is then reduced if phi(x) isn't equal to some power of a. But Idk how to show that, or whether I even chose the right elements as generators

[glossy crag]

makes perfect sense to me (in general finite-dimensional Q-vector spaces call X a cone, call Y a lattice cone if it's a cone in the finite-dimensional Q-vector space N_Q)

[formal ermine]

how?

X is a cone in Q^n if it's a subset of the form ...

a cone is a lattice cone if Q^n = Q^n

so shouldn't all cones be lattice cones?

[formal ermine]

@dim widget this is literally what my prof wrote

"A lattice N is a Z-mod isomorphic to Z^n for some n. Define N_Q as Q tensor_Z N iso Q^n. A cone in V iso Q^n is a subset of the sigma = ... If V = N_Q for a lattice N then we call (sigma, N) a lattice cone"

but V is iso to Q^n and N_Q is also iso to Q^n

[dim widget]

No condition on \sigma???

[formal ermine]

yeah

that's why I'm confused

[dim widget]

Yeah that is confusing, I assume in that case the v_i should come from N.

But I guess this makes sense (although it is kinda weird) if $N \subset V$ is a lattice then for any $v \in V$ there exists $n \in \mathbb{N}_{>0}$ such that $nv \in N$, so $cone(v_1, v_2, \dots, v_r) = cone(n_1, n_2, \dots, n_r)$ regardless of what the $v_i$ are.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

Idk lattice cone just doesn't seem like a very useful piece of terminology when defined this way, it would've been useful to add a remark about this

[formal ermine]

wait

so what is the definition of a lattice cone lol

[jaunty forum]

Can anyone explain what exactly the fixed field of a subgroup means, I don't really get how it's different to the Galois group of a field extension

[formal ermine]

galois group is a group

fixed field is a field

[dim widget]

I think the definition that your professor gave is fine. It is equivalent to another definition that you might make

[formal ermine]

they are inverse to each other

[formal ermine]

when is a cone not a lattice cone? cuz like

if V = N_Q for a lattice N

then our sigma is a lattice cone

but V = Q^n

and N_Q = Q^n

[dim widget]

When is a raven not a writing desk?

[formal ermine]

not a lattice cone

🗿

[dim widget]

All cones $\sigma$ in $N_Q$ are canonically lattice cones via $\sigma \to (\sigma, N)$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

But a lattice cone is the data of both $\sigma$ and the choice of lattice $N$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

So for an abstract $V/\mathbb{Q}$ not all cones are lattice cones.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

However they can be enriched to lattice cones for any choice of lattice in $V$.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

Essentially a lattice cone is a pair of a cone and a lattice, and they don't have to satisfy any compatibility

Just come from the same V

[formal ermine]

what does V = N_Q have to do with sigma though

[dim widget]

Nothing except that $\sigma \subset V$.

[cloud walrus]

Topos_Theory_E-Girl

[jaunty forum]

I'm doing a commutative algebra course basically and I'm having trouble understanding the difference

[dim widget]

If K/F is a galois extension with galois group G then for any $H \subset G$ a subgroup we can look at the subset K/L/F where L is the set of $k \in K$ such that $h(k) = k$ for all $h \in H$.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

That's the definition.

In this case K/L will be Galois with galois group $H$, and if $H$ is normal $L/F$ will be Galois with galois group $G/H$.

[cloud walrus]

Topos_Theory_E-Girl

[jaunty forum]

Yeah I think I largely get it, at first I just thought okay it's just the identity, but obviously in C/R the complex conjugate is there too

[dim widget]

I'm assuming here that all of your extensions are finite extensions.

[jaunty forum]

Um idk, it's not in our definition, he covered it before he covered finite extensions too

His definition was:

[formal ermine]

What

how do you do galois theory without knowing what a finite extension is

[jaunty forum]

"Let L:K be a field extension. Then Gal(L/K) denotes the subgroup of Aut(L) of field automorphisms $\phi \in$ Aut(L) such that $\phi(a) = a \forall \a \in K$"

[cloud walrus]

LeftySam
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[jaunty forum]

It's a fairly introductory course we've basically just covered UFDs, Euclidean Rings, symmetric polynomials, what a field extension is, kronecker's theorem, algebraic numbers over C, splitting fields, finite fields, cyclotomic fields, galois group of a field extension, separable field extensions, primitive element theorem, normal field extensions, galois field extensions and galois' dream. Basically, I know what it is but in the definition we're given it's not specified that L:K is a finite field extension in the definition we're given for Gal(L/K)

[formal ermine]

huh

how does one do finite galois theory with infinite extensions lmao

[jaunty forum]

My impression was just that Gal(L/K) was just a set of automorphisms, since it wasn't specified I thought there might exist some example between infinite extensions

[formal ermine]

infinite extensions happen to be more complicated

you use topology for that

[jaunty forum]

I still don't really get what the fixed field is saying

[formal ermine]

do you know the fundamental theorem of galois theory

[jaunty forum]

Galois' dream?

[formal ermine]

wat

no the fundamental theorem

[jaunty forum]

Let L:K be a finite Galois extension and G:= Gal(L/K). There is a natural bijection between the subgroups H <= G and the intermediate fields F which is given by the following two maps that are inverse to each other a: F -> Gal(L/F) and b: H -> Fix(H)

[formal ermine]

yeah

[rustic crown]

fixed field of H is the subfield of L fixed by H. i.e. it consists of elements x in L such that phi(x) = x for each phi in H.

[jaunty forum]

Like I know it, but idk if I understand that because idk what the Fix is doing

[jaunty forum]

Since L:H is a field extension surely this is just an element of the Galois group, since it fixes elements over the subfield

[rustic crown]

it's weird talking about finite galois extensions but not other finite extensions

[jaunty forum]

We've spoken briefly about finite extensions

[rustic crown]

L is a field, H is a group, L:H is a whut

[jaunty forum]

Oh

So if you take a Galois extension that is finite, let's say L/K, that means it's normal and separable. I.e. it's algebraic, and every irreducible polynomial in K[t] that has a root in L splits into linear factors in L[t], and every element in L is separable over K, i.e. the minimal polynomial of every element in L, call it g, has gcd(g,g') = 1 in K. Then if you take some intermediate field of L/K, say, F, then you do the Gal(L/F).

I guess I'm not sure from that point how you go from Gal(L/F) back to F by Fix

That is Fix(H) for some subgroup of Gal(L/F)

[dim widget]

There is no "minimal polynomial of every element in L"

You can take the minimal polynomial of a generator for L if L/K is finite separable

[jaunty forum]

Sorry I meant to say for each element take the corresponding minimal polynomial

[dim widget]

Yeah that's the point of galois theory is that for a finite extension L/K intermediate extensions F are in bijective correspondence with subgroups H \subset Gal(L/K)

[jaunty forum]

Yeah I get they're in bijection but I don't get what the Fix function is actually doing

All I know is it's acting upon a subgroup of Gal(L/K)

[dim widget]

Fix(H) takes H to the set of $k \in L$ such that $h(k) = k$ for all $h \in H$ as I already wrote...

[cloud walrus]

Topos_Theory_E-Girl

[jaunty forum]

but this is already true? I don't understand. H is a subgroup of the group of Gal(L/K) such that phi(a) = a over K.

[dim widget]

Yes but i didn't write $k \in K$ i wrote $k \in L$.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

Fix(H) is a field which contains K but it is larger.

[jaunty forum]

Can you write it in more layman terms, so we have a subgroup of Gal(L/K), Fix(H) is all the k in the larger field such that h(k) for all h in H

[dim widget]

It's the set of elements of L which are fixed by H

[dim widget]

Meaning they are fixed by all the elements of H

[jaunty forum]

And H is a set of automorphisms?

[dim widget]

It is a subgroup of the group Gal(L/K)

So yes it is a set of automorphisms, but it also has a group structure.

[jaunty forum]

In one of my past papers, the question is:

True or False: Each unique factorisation domain has infinitely many prime elements.

Answer is given as false, since fields are UFDs but have no prime elements

[glossy crag]

Maybe I'm missing something, but I still think this makes perfect sense. If V is finite-dimensional over Q, define cones to be something. A lattice cone is a cone in the finite-dimensional Q space produced from a lattice.

[jaunty forum]

But if fields are ufd but have no primes, because they have no zero divisors, wouldn't it mean that all integral domains have no primes as they all don't have zero divisors? But this is obviously not true

[formal ermine]

what

a field has no primes because all elements are units

except 0

[formal ermine]

yeah but every such V is produced from a lattice, no?

[dim widget]

That's not what it is.

[glossy crag]

Guess I'm missing something, then.

[dim widget]

Yes this is False.

[dim widget]

It is a pair consisting of a cone and a lattice in V

[glossy crag]

That's not at all obvious from the way the definition is worded imo.

The way it's worded is "cones are this, lattice cones are cones in lattice vector spaces".

[dim widget]

They define it to be a pair $(\sigma, N)$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

see here

[glossy crag]

Yes, the cone sigma in the space N_Q, what'st he problem?

[jaunty forum]

what I don't get is if field implies integral domain, and fields have no primes; why isn't it the case that integral domains have no primes?

[formal ermine]

because not every integral domain is a field

[dim widget]

It's clear that N is part of the data is all...

[glossy crag]

Absolutely. Is that a point of contention?

[formal ermine]

.

[glossy crag]

Oh you mean that different lattices may produce the same space Q^n?

[dim widget]

apparently it is?

[dim widget]

Yes infinitely many

[glossy crag]

Your point being that the datum N in (\sigma,N) is sort of meaningless?

[dim widget]

No I think it's an important part of the datum which is why I think the ordered pair is important. Because at some point they presumably want to talk about the "integral points of the cone"

[formal ermine]

uwu?

[dim widget]

yes

[formal ermine]

so what's the point in differentiating between the two

[jaunty forum]

Is the following true or false

[dim widget]

the lattice is part of the data the lattice is part of the data the lattice is part of the data the lattice is part of the data the lattice is part of the data the lattice is part of the data

[jaunty forum]

"Let $\mathbb{Q}_n = \mathbb{Q}(e^{\frac{2i\pi}{n}})$ be a cyclotomic field. Then there exists n > 6 such that $\mathbb{Q}_n$ intersection $\mathbb{R}$ = $\mathbb{Q}$

[cloud walrus]

LeftySam

[dim widget]

you should use galois theory

[jaunty forum]

My lecturer used sigma: Q_n -> Q_n as the complex conjugation to start with but no idea how he got this idea

[dim widget]

what is the fixed field of the group generated by sigma?

[jaunty forum]

he says Q_n and R but no idea why

Also isn't the intersection of Q and R just Q? And if so sinc Qn is Q with primitve roots of n added why isn't Q_n and R intersection just Q again?

[dim widget]

Well what would Galois theory tell you about the degree [K:Q] where $K$ is the fixed field of complex conjugation on $Q_n$

[cloud walrus]

Topos_Theory_E-Girl

[jaunty forum]

I don't know frankly

[dim widget]

Okay well maybe look up the fundamental theorem of Galois theory

[jaunty forum]

so the group generated by sigma is {id, conjugate} right?

[dim widget]

yes

[jaunty forum]

I'm not entirely sure how fundamental galois theorem or whatever applies here

[dim widget]

Well it tells you what the degree of the fixed field is over the base field

If L/K is galois and H is a subgroup of automorphisms and F = Fix(H) then you can compute the degree of F/K by knowing the degree of L/K and the size of H somehow...

[jaunty forum]

I know that Fix(<sigma>) = Q_n and R from the solutions I just don't know why

Not sure how we can even talk about Fix anyway wrt to other fields

[dim widget]

No $Fix(\sigma)$ is $Q_n \cap R$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

Oh that's what you said yes

[jaunty forum]

Why is this the case though

[dim widget]

The fixed points of complex conjugation on all of C is just R

so the same is true for any number field considered as a subfield of C

[jaunty forum]

Why do we automatically take C as the larger field? Aren't there field extensions of C?

[dim widget]

You just get the elements of the field which are real numbers

<@&286206848099549185>

[jaunty forum]

I'm confused as to why it's Q_n intersection. We have <sigma} = id and conjugate, and Fix(<sigma>) is the field of all elements that map to themselves, so why is it not just R?

Oh nevermind

Because every element sigma is actually an automorphism of Q_n

So now I need to see why [Fix(<sigma>): Q] = [Q_n:Q]/2

[dim widget]

I believe in you

[jaunty forum]

So it's about demonstrating that [Q_n: Fix(<sigma>)] = 2

[dim widget]

yes

[jaunty forum]

And is that because the extra dimension is all the irrational Q_n?

[dim widget]

it's the part that is not Real numbers

The extra dimension is entirely irrational, but all of [Q(zeta_n):Q] comes from irrational numbers except for one dimension

[jaunty forum]

Q_n and (R\Q)?

[dim widget]

no

[jaunty forum]

Not entirely sure why, surely all elements of Q_n are just Q_n and R or Q_n are R\Q

So we're looking for [Q_n: Q_n and R]

[dim widget]

Yes try to show that that is 2-dimensional unless n is 2

[jaunty forum]

I don't get why you can't just say it's 2 dimensional as you can take Q_n and R\Q

Sorry C\R

[white oxide]

what do they mean by viewing the equation modulo 7?

[rustic crown]

just go mod 7

apply the ring homomorphism Z --> Z/7Z

[white oxide]

confusion

so would it just reduce to a^2 \equiv 5b^2 mod 7

number theory bad

[void cosmos]

yea and now u can brute force easily

[white oxide]

ok thx

[white oxide]

the second qualification for a UFD is "the factorization into irreducibles is unique up to associates and the order in which the factors appear" does this just mean that there can be multiple representations of an element in an UFD with there being different associates present in the product? im not entirely sure on what it means

i would assume "up to" in mathematics just means virtually the same in this lense

like there's only one field of order p a prime up to isomorphism

[wraith cargo]

You can have two factorizations but they differ by a unit

[white oxide]

would 6 = (2)(3) = (-2)(-3) be an example in the integers?

[south patrol]

ye

if a_1 ... a_k and b_1 ... b_l are two factorisations of the same ting then there are units u_1,...,u_k and there's a permutation σ : {1,...,k} -> {1,...,l} such that b_i = u_i a_σ(i)

[paper aurora]

can i please have help understanding this part of a proof? this is notes from a course on elliptic curves, but i think all i need to understand the highlighted parts are basic facts on group homomorphisms

i feel like it should be simple but i am having trouble understanding the last phrase of the highlighted part

[south patrol]

Well the way I'd think about it is that if φ: G -> H is a group homomorphism then the preimage of each element of H is of cardinality |ker(φ)| (or empty as it may be)

[paper aurora]

oh i see. so applying that here , we get the preimage of each thing in ker(beta) under gamma has size ker(gamma)

[south patrol]

Ye

That's what they mean w cosets

I'm just specialising to what matters here

[tropic bear]

Hi folks 🙂 I'd like to get a formal understanding of the relationship between the 2-adics and two's complement arithmetic. While the 2-adics represent a number $n = \sum_{j >= 0}^\infty \alpha_j 2^j$, in two's complement we truncate the sum to some length $K$ (say 32, or 64, commonly in computers). How does this affect the number system? Some results still seem to translate just fine. So for example if we have $p(x) = 1 + x + x^2 + ... = \frac{1}{1-x}$, then by $x = 2$ we get the 2-adic representation of -1 as ...1111, and that matches the two's complement representation of -1, as 111....111, there being $K$ ones. This then yields a common trick to compute the representation of negative numbers: Since -1 is all-ones, then subtracting n from both sides yields $-n - 1 = ...1111 - n$, and adding one to that yields $-n = ...111 - n + 1$, where subtracting $n$ from ...111 corresponds to "flipping all the bits" in $n$.

In particular, I'd like to understand how division works. Say we have some odd number $n$. The representation of $-n$ is as we saw above. What's the representation of $\frac{1}{n}$? Let's say this is in the 2-adics, and I may be able to translate it to two's complement from there 🙂

[cloud walrus]

flebron

[glossy crag]

Is this a valid proof (it sounds almost tautological, but I want to make sure):
a) From a previous lemma we know that the ideals of A are precisely the (direct) sums of A's isogenous components. Therefore a component is a minimal ideal (by the lemma, a component is an ideal and any ideal it contains is also a component and components are independent) and a minimal ideal is a component (an ideal is a a sum of components and every component is an ideal, therefore a minimal ideal is a sum of 1 component). The simple submodules of A are the minimal left ideals, therefore every ideal is a sum of these.
b) A is a direct sum of its components and only finitely many are necessary, since A is finitely-generated. From a previous lemma, if M=\sum_i N_i where the N_i are simple, then every simple submodule of M is isomorphic to one of the N_i. Since A is the sum of finitely many components, its every simple submodule has the same isomorphism type as one of these => finitely many components in general.

[rustic crown]

truncating a 2-adic means you're essentially working in Z/2^KZ. in this ring, 2^K = 0, which is why -n = 2^K - n = (2^K - 1) - n + 1 like you noticed.
to find the multiplicative inverse of n, you're solving the equation n * m = 1 mod 2^K, which is same as finding integer solutions to n * m + 2^K * x = 1. there many ways to do this algorithmically, reverse euclidean algorithm, using convergents of continued fractions, etc.

another way to think about this would be looking at 1/(1 - (1-n)) and as 1-n is an even number, it's 2-adic norm is smaller than 1, so you can write 1/n = p(1-n), but this also looks computationally heavy.

[cloud walrus]

flebron

flebron

[tropic bear]

#include <stdint.h>
#include <stdio.h>
int main() {
    uint32_t n = 43;
    uint32_t m = (~n + 1) + 1;
    uint32_t x = 1;
    int i;
    for (i = 0; i < 31; ++i) {
      x = x * m + 1;
    }
    printf("n=%d, m=%d, x=%d, n*x=%d", n, m, x, n*x);
}

Which I realize is the same thing you said in your last thing 🙂

@rustic crown What's a way to make that "essentially" rigurous?

[cloud walrus]

flebron

[rustic crown]

i think you mean x^j instead of 2^x

[rustic crown]

it's a ring homomorphism

infact that's one way to construct p-adic numbers.

[tropic bear]

You're right of course 🙂

[rustic crown]

instead of thinking of p-adic numbers as formal power series in p and then manually defining how to add and multiply with messy carry overs, you define it via a sequence of partial sums.

what you can do is, look at the infinite product of the rings Z/p^kZ as k varies over N

so an element of this is just an infinite tuple (x_k) for x_k in Z/p^kZ

now you look at the subring of this consisting of elements such that x_k reduces to x_l mod p^l whenever k > l.

[tropic bear]

k here is >= 1, or >= 0?

[rustic crown]

doesn't matter, hehe

k can vary over any infinite subset of N :p

[tropic bear]

Oh right, Z/p^0Z = Z/1Z = Z/Z = 1

And there's only one thing there, so it's adding no data to the infinite tuple

[rustic crown]

with this definition it's easier to see why it's inherently a ring

[tropic bear]

If x_k reduces to x_l mod p^l whenever k > l, this means you''ve got projection maps "down the chain", right? That are just taking modulo the lower power of p

[rustic crown]

if you know some basic cat theory, we're looking at the inverse limit of Z/p^kZ

[tropic bear]

So e.g. if the thing has a 3 in the mod 2^2 position, then it must have a 1 in the mod 2^1 position

[rustic crown]

yep

if you know a number mod 64, you also know if it's even or odd.

[tropic bear]

Right, that makes sense, and it jives with "making more and more precise notions of 'how close to a power of p are you'"

[rustic crown]

you have natural map Z --> Z/p^lZ and p^k lies in the kernel, so you obtain natural maps Z/p^kZ --> Z/p^lZ

[tropic bear]

a.k.a. the valuation thing

So addition here is what, componentwise?

[rustic crown]

yep

[tropic bear]

The inverse limit thing is precisely this notion of projection maps, right?

[rustic crown]

you can think of these components as the partial sums of the power series in p

[tropic bear]

Right

Multiplication is more involved however, than componentwise

[rustic crown]

oh it is component wise

[tropic bear]

😮

[chilly ocean]

Bilinear form is a map AFAIK..
Then how dot product can be bilinear form <@&286206848099549185>

[summer path]

have you looked at the definition of a bilinear form?

but yes it is a map from $\mathbb{R}^m \times \mathbb{R}^m \to \mathbb{R}$ in this case

[cloud walrus]

Tubular Cat

[summer path]

such that it is linear in each argument

you can pretty easily check that the usual dot product satisfies this

detuwu

[rustic crown]

tubuwu

det tired

[tropic bear]

btw @rustic crown newton-raphson converges after 5 iterations, though it's not clear how to derive the "5" for me:

    int32_t z = 1;
    for (i = 0; i < 5; ++i) {
        z = z * (2 - n * z);
    }
}

using f(x) = 1/x-n, f'(x) = -1/x^2

[chilly ocean]

It says "on R^m"

[tropic bear]

@gilded stream the dot product "on R^m" is understood as the dot product between pairs of vectors, each in R^m 🙂

[rustic crown]

newton's method converges at a quadratic rate, so that makes sense. in each iteration, you would double the precision. we only care about 2^32, so that's why you need 5ish iterations

[tropic bear]

mfw 2^5 == 32

And precision here would be according to the valuation?

Is the damn thing that generalizable?

[rustic crown]

yeep

[rustic crown]

it's because you can describe p-adics in a very analytic way. the normal description of the real numbers is to start from rationals and complete in the missing holes by looking at cauchy sequences.
if you use the p-adic norm, you could do the same thing and get yourself a nice complete non-archimedian field Q_p, now p-adics integers are elements with non-negative p-adic norm. the estimate for convergence of newton's method is a simple consequence of taylor's theorem, so if you can make sense of all these things, it has to be true :p

[long nebula]

What do Ext and Tor stand for?

[rustic crown]

extensions and torsion i think

[dim widget]

no correct answers

[long nebula]

Wtf

[dim widget]

Extanea and Torsten

the two founders of homological algebra

[long nebula]

Why are there no correct answers who named these

[rustic crown]

(i think tteg is memeing >.<)

[long nebula]

😭

[summer path]

exterior?

[long nebula]

Followup question, why

[dim widget]

Have you computed what Ext and Tor are for abelian groups?

[dim widget]

see above

[long nebula]

I have not

[dim widget]

I think that's a good exercise then, what is Ext(G, G') parameterizing and what is Tor(Z/n, Z/m)?

Where G, G' are two arbitrary abelian groups

[long nebula]

Okay I will come back to this when I care about homological algebra

I'm not ready

[rustic crown]

Tor_1(R/r, M) = {m in M : rm = 0} is the r-torsion elements. (r needs to be a non-zero divisor so that 0 --> R --> R --> R/r --> 0 is a resolution)

[dim widget]

And Ext(M, M') parameterizes equivalence classes of sequences 0 \to M' \to N \to M \to 0

[long nebula]

That's very neat

[dim widget]

So like Ext(Z/p, Z/p) has two elements

one is just Z/p \times Z/p and the other is Z/p^2

[long nebula]

Ohh

That's nice

[dim widget]

morally this is what happens but there is a complication about the quivalence relation

Unfortunately the equivalence relation is not isomorphism of N, but an isomorphism of the sequence, so there are p-1 extensions corresponding to Z/p^2

so Ext^1(Z/p, Z/p) = Z/p and only the trivial element corresponds to the "stupid" extension

[long nebula]

ahhh okay

cool! :)

on an unrelated note, algebraic topology has actually made me happy about the snake lemma

before this, I was just happy about it because snakes are cool

[tropic bear]

so you can build differentiation, manifolds, that sort of stuff, over Z_p?

(the diffgeo i've done always mentions R as a sort of "privileged" domain and codomain)

[dim widget]

Yes

There are many ways to do this, but usually over Z_p you do this on analytic functions.

So you build p-adic analytic manifolds and p-adic analytic spaces

[chilly ocean]

what about (p, q)-adic manifolds

[dim widget]

probably zomeone elze know about the topic, zo i prefer to wait for zomeone elze that know about it

[chilly ocean]

i wonder how [redacted] would handle this

[rustic crown]

s = z

[somber sleet]

Can somebody explain to me what a free R-module of degree n is?

we defined it as an R-module, which is isomorph to R^n

[next obsidian]

That’s a perfectly fine definition

The point is that it has a basis, which lets you define R-linear maps by saying the basis element e_i goes to an element m_i in M

And then you have to send Sum r_ie_i to Sum r_im_i

They’re kind of like vector spaces in that sense

[somber sleet]

does it equivalently work as saying that the dimension of a finite vectorspace is n?

[next obsidian]

If you’re over a field, then yes

[somber sleet]

which means that if V is over R, then it is isomorph to R^n?

[next obsidian]

Using this language, you could say this

Any vector space is a free module, because all vector spaces have a basis

[somber sleet]

how does it generally work then?

is this intuition wrong? Do you have something, whihc makes more sense?

[next obsidian]

I don’t know what you mean

The intuition is that they behave like a vector space, but they can be defined over any ring

Vector spaces only exist over a field

They’re the modules which have a basis

[somber sleet]

Okay, I get that

I just don't want to ignore something important

[slim kayak]

Maybe that the degree isn't always invariant, it is for, say, commutative rings and some weird left or right special rings, but not in general.

[minor dragon]

I’m struggling with the exercise 1.7, here P designate the Poincaré Series and we are working with graded vector spaces over a field (commutative). I think that there is an issue, I tried to prove it and I found that it is (1+1/t)P(B*,t) and not (1+t)P(B*,t). Am I right ?

[glossy crag]

Why is the set of such N non-empty (we need to know this before we can apply Artiniannes)? Equivalently, if M is any module, why does it have a simple quotient?

[glossy crag]

NVM, here M is assumed noetherian => set of proper submodules has a maximal member => simple quotient.

[south patrol]

Can't you also take like N = M

[delicate orchid]

yeah but that's boring

M < M < M < ... < M nice chain stinky

[south patrol]

If M = 0 we have no choice lol

And then there are no proper submodules

Too picky ig

[meager fractal]

here L1[X] is the ring of polynomials with coefficients being polynomials modulo f

how do i know we can write f as (X-a)g(X)

??

f = X^3 - n , letting a = X + (f) becomes a^3 - n == 0 so X is a root of L1[X] in L1?

so we must be able to write it as (X-a)g(X) where a = X+(f)

why is this construction so confusing 😭

[lethal dune]

ℚ³?

[dim widget]

The cubes in $\mathbb{Q}$

[cloud walrus]

Topos_Theory_E-Girl

[lethal dune]

I see, weird notation

[dim widget]

It's pretty normal for this kind of thing, but out of context it can look strange.

[dim widget]

Yeah I think the way they're doing this is silly. Just say: $X^3 - n$ has 3 roots over $\overline{\mathbb{Q}}$ which are the three cuberoots of $n$. The ratio of any two cuberoots is a 3rd root of unity by an easy calculation, so the splitting field is of degree 6: $\mathbb{Q}(\operatorname{cuberoot}(n), \zeta_3)$. The galois group can be checked to be $S_3$.

[cloud walrus]

Topos_Theory_E-Girl

[south patrol]

here's a bit of a dumb question lol

is there any nice way to find the formula for the discriminant of a cubic/quartic or is it necessarily just a slightly annoying slog with symmetric polynomials lol

[delicate orchid]

a simple and intuitive way, if you will

[south patrol]

I guess the nicest I saw was if you use the stuff in terms of norm but eh

Lol

[dim widget]

There is a general way for all n just using simple and intuitive linear algebra. However for degree n you need to compute a 2n + 1 x 2n + 1 determinant.

[lethal dune]

that ain't bad

[south patrol]

Wait how come 2n+1 x 2n+1

7x7 for cubic lol

[dim widget]

Because the linear system you are trying to solve for an f of degree n is for a pair p, q such that pf + qf' = 0, so you want an expression which finds out when the resultant matrix is invertible. So actually its 2n-1 x 2n-1 which is better

So 5x5 for cubic

[glossy crag]

For a very specific kind of cubic/quartic (or more generally a trinomial) it's nice and simple. Source is Samuel's ANT.

And iirc every cubic/quartic is reducible to this kind of trinomial (which, iirc, is how Cardano originally found the formulas for the roots in terms of the coefficients).

[south patrol]

Ah ok

[south patrol]

Yeah that's the nice method I had in mind hehe

[chilly radish]

Qubes*

[meager fractal]

the way they do it gets a LOT more confusing

\

horrible

and it doesnt help with this really

here I can say that x^p - 2 has p roots over its splitting field

[cloud walrus]

The

[meager fractal]

so we have indeed p roots

and then say Q(pthroot(2),zeta_p) is the smallest field containing all roots?

i also assume that the splitting field is over Q

bc they dont state this

[long nebula]

Yeah that's right, you just have to say that you can get all the roots using multiplication on pthroot(2) and zeta_p and you can get pthroot(2) and zeta_p by multiplying/dividing the roots

Which proves inclusion in both directions

[meager fractal]

and inclusion would be showing Q(proot(2))(proot(2) x zetap)..(proot(2) x zetap^(p-1)) = Q(pthroot(2),zeta_p)

[ebon pine]

I was thinking of starting an abstract algebra study group using knapp. Where should I tell about this? Should I message modmail?

[long nebula]

Yeah

[meager fractal]

ok ty

[long nebula]

I think you just advertise around and see who's interested, you can prob message modmail to be allowed to post in #events about it

And then you can either start a thread in this channel for it or start another server and invite everyone interested

(This is not official advice, I'm not a mod lol)

[ebon pine]

fair enough, thanks for the info tao

[meager fractal]

quick question

[wet zodiac]

id be intetested if we use artin

and like the measure theory group theres some pressure/motivation

[boreal inlet]

[glossy crag]

You mean Basic/Advanced Algebra? Can confirm, he's very good.

[ebon pine]

Well, ngl I am not very fond of Artin plus I wanted to try something new

[ebon pine]

Yes indeed

[glossy crag]

For the bits on field & Galois theory look into Isaacs, his exposition is what made things finally click for me.

[ebon pine]

I'll also be taking wisdom from Rotman and other books. Isaacs is a new name for me.

[glossy crag]

Knapp's Advanced Algebra is pretty wild, does all sorts of things.

[ebon pine]

We'll stick to just parts of Basic Algebra. Advanced algebra is well... advanced

[glossy crag]

I think Isaacs is a little less known than the others, but I consider his book the absolute creme de la creme of basic algebra.

[glossy crag]

Also look into Jacobson's Basic Algebra I, also one of the best.

[ebon pine]

Oh yeah, I even have a physical copy of it

The only big issue is if people would be willing to commit

Knapp is definitely not for first timers

[glossy crag]

Purely hypothetically I think it is, if the person had a proper European-style LA 1,2 course.

[ebon pine]

Same for Asian universities, not sure about us and canada

[long nebula]

👀 what is this Knapp book

[glossy crag]

Knapp's Basic/Advanced Algebra two-parter.

[dim widget]

discussion of knapp book makes me sleepy

[long nebula]

k'dnap

[glossy crag]

I've got something to wake you up, gimme a sec to write up the question.

I don't get how the general case reduces to the case of a single isomorphism type via "grouping". Say we partition the index sets as $I=\bigcup I_\tau$ and $J=\bigcup J_\tau$ based on isomorphism type, how from $\bigoplus_{i\in I}N_i\cong\bigoplus_{j\in J}N'j$ can we follow that $\bigoplus{i\in I_\tau}N_i\cong\bigoplus_{j\in J_\tau}N'_j$ for every $\tau$?

[cloud walrus]

Ocean Man

[ebon pine]

Don't make me question my decisions

[dim widget]

knapp = nap

I had another pun in my native language but its niche

[rustic crown]

[ebon pine]

det no

[rustic crown]

but that was a cute pun

[dim widget]

I have never heard of the book, i have no opinion on it. Could be good!

[ebon pine]

Pls don't det-onate

Ah, cool

[south patrol]

knapp is more negative in german

Hm is the lattice of subgroups of (Z/nZ)^x hard to visualise

[dim widget]

it follows because there is no zero map from N_1 \to N_2 if N_1, N_2 are two nonisomorphic simple modules.

[dim widget]

does it mean something like "hardly"?

or is it cognate with knave?

[glossy crag]

More like "terse" or "short", but it has a variety of meanings afaik.

[south patrol]

Yeah like if you're saying you only just managed to do smth

[glossy crag]

"just under" is the DeepL translation and the most fitting, probably

[south patrol]

this is funny because i haven't spoken german in a while so sometimes when i go to say stuff i'm worried i'm just making it up cause i go by feel but no like it can indeed also mean like

smth is knapp = in short supply / nearly out

[dim widget]

I mean it is a direct sum of cyclic groups, it's about as easy as it gets.

[south patrol]

Okay sorry that was dumb I meant like uh

[dim widget]

And you can calculate the factors explicitly if you can factor n

[south patrol]

Okay actually I have just made my question more specific and now i can google probs lol

[glossy crag]

Ah, ok, so I can just apply Hom_A(N,--) to M and single out those N_i that are isomorphic to N.

[south patrol]

Basically it's like I was looking at particularly nice subfields of Q(zeta_n) like

in particular Q(zeta_m) for m a multiple of n

But that corresponds to a particularly nice subgroup i guess

[dim widget]

yep! suffices to work one $N$-isotypic piece at a time, for $N$ a simple module, because if you restrict the isomorphism to $N^m$ the image is isomorphic to $N^m$ as well.

[cloud walrus]

Topos_Theory_E-Girl

[south patrol]

well the k with (k,n) = 1 and k = 1 mod m

But that feels funky to state lol

[dim widget]

you mean a divisor

[glossy crag]

And I am correct in understanding that for N finitely generated Hom(N,\bigoplus_iM_i) is isomorphic to \bigoplus_iHom(N,M_i), right (there's always an injective map and finiteness takes care of surjectivity).