groups-rings-fields

[crystal vale]

Yes

Thank you

Let P be a p-Sylow subgroup of G and Q is any p-subgroup of G, then is Q\intersect N_G(P) is Q?

| QN_G(P) | = |Q| | N_G(P) | / | Q intersect N_G(P)| , implies | Q intersect N_G(P) | = | Q |, right?

I know I am making mistake but where?

[delicate orchid]

you've assumed what you're trying to prove? this intersection is Q iff Q is a subgroup of N_G(P), and |QN_G(P)| = |N_G(P)| if and only if Q is a subgroup of N_G(P)?

[crystal vale]

Thanks i got it ❤️

[tardy hedge]

.

[next obsidian]

I < U ann(m_i) where m_i are generators. Each annihilator is contained in an associated prime by a Zorn’s argument and then prime avoidance

Wait this isn’t quite right

For each generator a_i of I you pick an m_i where a_i•m_i = 0, and then the argument works

[tardy hedge]

Thanks!

[tardy hedge]

Why doesnt this original way work

m_i being generators of M?

[next obsidian]

Because it isn’t true

lol

Afaict

[next obsidian]

Wait no what I said also doesn’t work

I think it needs to be the original argument

But I can’t figure out why an element of I has to kill some generator

The argument is something like this though lol

Oh wait okay this is dumb

Because each ann(m) is contained in an associated prime, the set of zero divisors of M is the union of all associated primes

There’s finitely many so you apply prime avoidance

[tardy hedge]

What was the problem with this?

[next obsidian]

Try to prove it

[tardy hedge]

Ok ill get paper later im at gym rn lol

[cinder onyx]

I feel so silly to not even know where to begin here

I presume the answer is no?

[coral spindle]

Hint: what does the orthogonality condition tell you about the length of the columns of a matrix in SO_2? Try with just R first. By length I mean euclidean norm.

[rocky cloak]

Since it's SO2 and not SU2, I guess there might be some weirdness going on

[knotty badger]

what does SO2(C) even mean

[rocky cloak]

Matrices with
A A^T = I
I would think (and determinant 1)

[knotty badger]

huh i see

i mean

i guess there's nothing stopping you from considering bilinear forms on C^2

instead of sesquilinear forms

just

why would you

[cinder onyx]

same as SO2(R) just with complex coefficients? so preserve the dot product on C^2

[knotty badger]

but that'd be SU(2) surely

[cinder onyx]

SU(2) preserves the hermitian form no?

[knotty badger]

what's the euclidean norm on C^2

genuinely don't know

[cinder onyx]

just the dot product I guess

idk this is the first and only time I've seen this group so

[knotty badger]

but that's not a norm right...?

it's not positive definite

heck it's not even real

[cinder onyx]

it is not

I think that might be the point of the question

[knotty badger]

i see

[rocky cloak]

I think it comes up as like a complexification of SO(n), so it's not necessarily about structures on C^n. But I don't really know about these things

[knotty badger]

ah i only know about complexifications of lie algebras

[rocky cloak]

Yeah, I have seen so(n, C). And I guess every lie algebra gotta have it's lie group

[rocky cloak]

I guess bonus exercise for you, you can prove both

[cinder onyx]

I guess the question I could just consider by construction just thinking about how <v,v>=1 doesn't impose a bound on the norm of the coefficients of v over C

[south patrol]

yeah

I guess a funny way to think about this is as follows: what is a nice "parametrisation" of SO(2, R)?

[twilit wraith]

Hey everyone, I was asked to find all subgroups of order 4 of Z/4Z \times Z/4Z

I know how to do this, but im wondering if there's a proper/formal way to write down these subgroups

Im pretty sure there's 4 cyclic subgroups and 1 klein-4 group

[south patrol]

The cyclic subgroups are fairly straightforward by thinking about which elements have order 4. For the klein 4 subgroups H, I'd recommend considering how many elements of order <= 2 are in Z/4 x Z/4

I wonder if there is any nice expression for, say, the number of subgroups of order p^n in Z/p^n x Z/p^n. I guess you can do this by induction on n

[next obsidian]

Joe mama

[twilit wraith]

Pretty sure its 3

We know the order of (a,b) is lcm(a,b) so both the orders of a and b are maximally 2

Sorry its 4

There's 2 elements in Z/4Z that are order at most 2 so there's 4 in Z/4Z

[rocky cloak]

I think there should be six cyclic subgroups of order 4.

[twilit wraith]

I think you get that two of em are equal to two other ones

[south patrol]

yes

and how many elements does the klein 4 group have

[twilit wraith]

So like (1,3) eventually generates (3,1)

[twilit wraith]

Four

[rocky cloak]

Should be
p^n + p^{n-1}
I think. For the cyclic subgroups of order p^n

[twilit wraith]

Its all ordered pairs such that at least one entry is order 4

[south patrol]

ye

[twilit wraith]

Which gives 4 cyclic subgroups

[south patrol]

better way of putting it lol

[rocky cloak]

||(1,0), (1,1), (1,2), (1,3) all have order 4||

[south patrol]

oh wait ye i'm silly lmao

ignore this stuffs

[rocky cloak]

Then there's ||(0, 1) and (2, 1)||

[twilit wraith]

Oh shoot it might be three actually

Yeah it is 3 cyclic subgroups

No nvm it is 4

[rocky cloak]

I'm still counting 6

[twilit wraith]

No nvm it is 3 cyclic subgroups

Its bc <(1,3)> = <(3,1)>

[south patrol]

but there's more from this description

[rocky cloak]

Like there's 2x2 = 4 elements of order 1 or 2.
So 16-4 = 12 elements of order 4.

Giving 12/2 = 6 cyclic subgroups

[south patrol]

divided by 2 right cause of generators

as in like Z/4 has only 2 elements exactly of order 4

ye okay

[rocky cloak]

You will have nontrivial intersections yeah

[south patrol]

ye like (1,2) and (1,0)

[twilit wraith]

Well no we have that (1,0) (1,1), (1,2), (1,3), (3,0), (3,1), (3,2), (3,3)

[clear ingot]

U can consider it as images of endomorphisms of Z/4xZ/4

[twilit wraith]

But notice that some match up as inverses

[rocky cloak]

And (2, 1) and (0, 1) etc

[twilit wraith]

Oh I see

Yeah it might be six

[south patrol]

i guess elements of order 4 in G = Z/4 x Z/4 correspond 4 to one to elements of order 2 in G/2G = K_4 so there are 12 elements of order 4 and then two things generate the same group iff they agree up to inverses ye

this is just jagr's argument lol

but i guess i mean this reduction argument probably lets you do induction

so i believe you that it just multiplies by p each time and then you reduce to counting elements of order p in Z/p x Z/p, which is easy

[rocky cloak]

The way I actually thought about it was that
(1, x) generates a subgroup, so that's p^n things.
Then (x, 1) also generates a subgroup, but it's already accounted for if x is relatively prime to p. So that's only p^n-1 new subgroups.

For a total of p^n + p^n-1 = 4 + 2 = 6

[south patrol]

ah nice

[twilit wraith]

Phew thus group theory stuff is a lot of counting

[south patrol]

i guess that doesn't work for n = 1 though

but maybe for n >= 2?

[twilit wraith]

At least it wasn't one of the heinous proofs I had to do yesterday tho

[rocky cloak]

It doesn't?

[south patrol]

oh wait no i was silly

ignore

i agree

there are p^2 - 1 elements of order p in Z/p x Z/p, but p-1 generators of Z/p

hence p+1 subgroups of order p

[rocky cloak]

Accounting for the noncyclic ones like this seems a little tedious though.

[south patrol]

yeahh

yeah i guess i had in mind like if G = Z/p^n x Z/p^n then G/pG = Z/p x Z/p, so there are |pG| ( |Z/p x Z/p| - 1) = p^{2n-2) ( p^2 - 1) elements of order p, which you then have to divide by p^{n} - p^{n-1} = p^{n-1} (1 -p) lol to get p^{n-1}(p+1). But agreed, that is tedious

and ultimately i rely on counting the # of generators which still uses smth similar to the argument you did with coprimality lol

though it feels like there i ssome symmetry

[cobalt heath]

Sad, a linear iso Z[zeta_n]^N -> Z[zeta_N]^n is way harder than it should be

[twilit wraith]

Prove that there is no homomorphism from Z/16Z × Z/2Z onto Z/4Z × Z/4Z.

ok so i have to somehow prove there isnt a surjection

i cant do the easy thing since the first group is larger than the second

there must be some invariant property im not recognizing

seems like i could use the first isomorphism theorem but i probably shouldnt

wait wait wait

the two groups have different sizes and are finite

therefore theres no bijective homomorphism between them

but since the groups are finite this means theres no surjective homomorphism between them

[cobalt heath]

You can have surjective homomorphisms without having bijective homomorphisms between finite groups tho

[twilit wraith]

really?

but if its surjective then its injective

[vast tangle]

when theyre the same size

[cobalt heath]

Z/2Z × Z/2Z -> Z/2Z, send (i, j) to i

[twilit wraith]

oh fuckin duh

wow im silly

welp i have to tackle this some other way then

[vast tangle]

G -> {e}

[twilit wraith]

yeah

silly me

ok but somehow i have to show theres no homomorphism onto Z/4Z \times Z/4Z

i really badly want to use the first isomorphism theorem but i dont think i can

[cobalt heath]

There is one such homomorphism, isn't there

Trivially, the zero map

[twilit wraith]

"onto" is the big word here

[cobalt heath]

Ah, surjective homomorphism

[twilit wraith]

yes

i know that the Z/16Z part cant map onto Z/4Z x Z/4Z twice

since cyclicity is invariant

[cobalt heath]

Yeah, so inspect image of the map

[twilit wraith]

alr

i just need to show that |im(f)| is never 16

[cobalt heath]

Z/16Z part can only map to a cyclic subgroup

[twilit wraith]

ye

oh of course

if any element gets mapped to something in Z/4Z x Z/4Z it has to be a part of a cyclic subgroup of order 16 right

[cobalt heath]

Can you elaborate?

[twilit wraith]

so lets take <(1,0), (1,1)> as the generating set of Z/16Z x Z/2Z

every element from this group has to get mapped to some cyclic subgroup of Z/4Z right

since phi(k,0) = kphi(1,0)

[vast tangle]

Z16 x Z2 is not cyclic

[twilit wraith]

well yeah but thats because its generated by two elements

well thats ambiguous

i think u know what i mean tho, maybe

[vast tangle]

which cyclic subgroup of order 16 then

[twilit wraith]

there is none, thats the thing

but Z16 x Z2 has two cyclic subgroups of order 16

so any element from those two subgroups need to get mapped to some cyclic subgroup as well

and im pretty sure that subgroup has to be order 16

but there is no cyclic subgroup of order 16 in Z4 x Z4

[vast tangle]

im not following this part

[twilit wraith]

i may have just extrapolated incorrect info

maybe im just not thinking about this the right way

oh wait i think i know it

if phi is a homomorphism then phi(4a) = 0 for some a in the first group

but clearly there are more than two multiples of 4 in the integers mod 16

so |ker phi| > 2 and thus the mapping cant be a surjection

[vast tangle]

i buy it!

i think i saw this exercise in gallian and this seems like roughly the approach intended for the section

[twilit wraith]

the only concern i have is we havent really gone over anything like this yet

[vast tangle]

wdym

which part exactly

[twilit wraith]

well its not clear that |ker phi| > 2 implies that theres no surjection given all the content we have covered thus far

today we did direct products and tomorrow is lagranges theorem

[vast tangle]

ah

[twilit wraith]

further, i am pretty sure that this is a result from the first isomorphism theorem

[vast tangle]

indeed

[twilit wraith]

since we know that Z16 \times Z2 /|ker phi| is isomorphic to a group of less than order 16 if |ker phi| > 2

but only by the first iso thm

[vast tangle]

im trying to think about something about deriving a contradiction considering the fact that ord(phi(g)) divides ord(g) but im blanking due to a skill issue

i dont think this would work

ah wait

@twilit wraith

recall that if H \leq G then phi(H) \leq K

[twilit wraith]

oh right

[vast tangle]

but there is no subgroup of order 16

[twilit wraith]

so if the kernel of phi is non trivial

oh thats even easier

[vast tangle]

damn that got the noggin joggin

[twilit wraith]

yeah all of these problems do

yesterdays problems were especially bad

im being sure to do the 5 hardest questions on the homeworks every day just because i wanna be good at algebra

and the prof is pulling the questions from D&F or the prelim he took in grad school

[vast tangle]

ah sweet

[vast tangle]

i do think it was a good exercise to look for a more "primitive" proof

[twilit wraith]

this was seriously hard

[twilit wraith]

yeah i agree

though i cant say i thought of that myself

i shouldve recalled that

whatever im still new to this stuff

[vast tangle]

i had to look in my anki deck for properties of subgroups under homomorphisms since im starting to build rust on group stuff

[twilit wraith]

ill have notes for these things once i get to grad algebra

for now though my algebra class isnt lecture style

so theres not really any notes to be taken

[twilit wraith]

wait

isnt there a subgroup of order 16

just the group itself

[vast tangle]

oh lol im dum

[twilit wraith]

but the kernel necessarily being non trivial means the subgroup must be order less than 16

at least i think thats right

[twilit wraith]

Oh wait

Doesn't |f(1,0)| <= 4 and f(0,1) <= 2 imply that |im(f)| <= 8

[velvet hull]

okay this question was kind of disgusting

thanks for nerd sniping 1 hour away from my final prep time

[twilit wraith]

lol

that only took u an hour?

without the hint is crazy

[velvet hull]

I had to take a hint

[twilit wraith]

ok thats fair

[velvet hull]

spoiler

[twilit wraith]

oh u right

[twilit wraith]

btw theres a question on my homework im supposed to do without the first iso thm

and its just stumping me

[velvet hull]

Sure what’s the question

[twilit wraith]

Prove that there is no homomorphism from Z/16Z × Z/2Z onto Z/4Z × Z/4Z.

my prof gave me the hint that " Z/16Z x Z/2Z is generated by (1, 0) and (0, 1). If f is a homomorphism, show that |f(1, 0)| <= 4 and |f(0, 1)| <= 8. Then use this to show that |im(f)| <= 8."

but im still not sure how these relate to |im(f)|

seems most of my struggles so far have been not realizing a property of homomorphisms

[velvet hull]

Oh by the 1st iso theorem, if such a homomorphism existed it would correspond to a kernel of size 2

[twilit wraith]

im also pretty sure he meant that |f(0,1)| <= 2

[velvet hull]

But then there are only 2 possible kernels, and the quotient group of either are not isomorphic to Z/4Z x Z/4Z

[twilit wraith]

yes but i cant use that

[velvet hull]

What do you mean

[twilit wraith]

im supposed to do this problem without the first isomorphism theorem

[velvet hull]

Oh sure

What’s the maximal image of Z/16Z under such a homomorphism

And so what’s the maximal image of Z/16Z x Z/2Z under such a homomorphism

[next obsidian]

It should be that |f(0,1)| <= 2

And from there it’s obvious that the image has size bounded by 8

[twilit wraith]

not to me

well i suppose its because any composition of the two elements inside the homomorphism have order 8

ah of course

i was just overlooking a fact i knew very well

[next obsidian]

That’s not exactly it

But if you’re generated by an element of order say 4 and one of order 2 and they commute

your elements look like x^ay^b

Where the only thing that matters is a mod 4 and b mod 2

Giving only at most 8 combinations

[twilit wraith]

well i should say

oh wait no i see what u mean

dont the elements not commute tho

wait no they do necessarily

alright i got it

thank you chmonkey

[crystal vale]

How?

[crystal vale]

Let P be a p-Sylow subgroup of G and Q is any p-subgroup of G, then is Q\intersect N_G(P) is Q?

| QN_G(P) | = |Q| | N_G(P) | / | Q intersect N_G(P)|

Since N_G(P) contains P so it has order p^a m, where a is the highest power of p, and Q is p-subgroup, Q intersection N_G(P) also subgroup of Q so it has order some power of p.

Q intersection N_G(P) must have order |Q| otherwise |QN_G(P)| has power of p greater than a, right?

[rocky cloak]

Think about S3 and p=2.

Specifically investigate what happens with your very last point.

[crystal vale]

I see there is no problem with | Q N_G(P) | has order greater than p^a

[quartz wind]

does the following work?$\$
Statement. $(\mathbb{Q},+)$ is not the direct product of two nontrivial groups.$\$
Proof. We start with the fact that $A_p:\mathbb{Q}\to\mathbb{Q}:q\mapsto pq$ for $p\in \mathbb{Z}$ is an automorphism (note that $pq$ is repeated addition of $\text{sgn}(p)q$ $p$ times, not multiplication). Assume $\mathbb{Q}= G\times H$ where $G,H$ are nontrivial. From the universal property of the product, we obtain the commutative diagram shown below. Every automorphism of $\mathbb{Q}$ admits a unique decomposition from $\mathbb{Q}$ into $G\times H$. Thus, for $\varphi:\mathbb{Q}\to G$ and $\psi:\mathbb{Q}\to H$, $\varphi = \pi_G(\varphi\times \psi)$ and $\psi = \pi_H(\varphi\times\psi)$. For $(\varphi\times\psi)(x) = A_px=px$, $\varphi((g,h))= pg$ and $\psi((g,h))=ph$. Thus, $\varphi\times\psi$ cannot possibly be surjective, for $(p\pi_G(x),q\pi_H(x))$ has empty preimage for $p\neq q$, a contradiction.$\qed$

[cloud walrus]

esca

[velvet hull]

it's a pretty good idea, but concluding that the maps cannot be surjective does not look right to me - why are you sure that pg != g?

[quartz wind]

good point

thanks

[velvet hull]

even if you prove that pg!=g I do not think that is sufficient to conclude that they are not surjective

try a different approach, using the UP is a good idea

[quartz wind]

UP?

[velvet hull]

universal property

[quartz wind]

oh universal property

got it thanks

[knotty badger]

👀

[south patrol]

I guess you can say that if Q = A (+) B then A and B are both Q-modules and hence we have a contradiction by lin alg lol

[quartz wind]

yeah but no modules

[south patrol]

Well it's just vector spaces

But I guess also yeah any map Q -> Q is determined by where you send 1 so are no (nontrivial) idempotent endomorphisms

But that is essentially the same argument lol

[shy fossil]

Since people are posting other proofs :p Non trivial subgroups of Q intersect non trivially...

[thorn jay]

||Let N, M be subgroups of Q, and a/b, c/d nonzero elements of N and M respectively with b, d nonnegative. Then cb * a/b = ac must be in N and ad * c/d = ac must be in M, so N and M have nontrivial intersection, as neither a nor c is 0. This means that Q can't be a direct product, as that would need subgroups with trivial intersections.||

[thorn jay]

Ye that's what I did lol

[south patrol]

Nice question tho

[quartz wind]

yeah but we dont have inner direct sum iso to direct prod. we also dont have the definition of subgroup at all

i think im meant to prove it from almost purely categorical notions

[south patrol]

Well you can just take subgroup along iso

[quartz wind]

ill read this once i have a real proof

[thorn jay]

It's just Nekorys argument but written down

[shy fossil]

Ooooo that's fun, I don't immediately see anything wrong with your proof, but I just skimmed it

[quartz wind]

yeah i mean like hchan said, theres no guarantee that elements of G and H have infinite order (yet)

[shy fossil]

Btw, follow up question. What groups satisfy the property that non-trivial subgroups intersect non trivially?

[quartz wind]

but we do have some basic facts about the order of groups and order of their elements so i think i can salvage it?

[shy fossil]

Q and prufer groups. Quaternions. Any infinite non abelian things?

[shy fossil]

Seems like it. Sounds doable honestly

[white oxide]

Any hints for i)?

[thorn jay]

For finite abelian groups it's equivalent to subdirect irreducibility

For infinite abelian groups it's implied by subdirect irreducibility

Infinite abelian groups no clue

[shy fossil]

(x+x)^2 two ways

[thorn jay]

But it's smt like "for every nonidentity g, h there are integers n, m such that g^n = h^m ≠ 1"

[velvet hull]

if you chase the diagram you should get that ||p^n = p^m for all integers n,m||

[quartz wind]

oh maybe i can exploit the up with the identity on G. since its an automorphism, pi_G preserves order, otherwise we lose information in the process of mapping to Q.

[quartz wind]

ah yeah makes sense

thanks

[velvet hull]

so very cool diagram chase, like it

[tardy hedge]

Forgive me if im being stupid but like. The set of all zero divisors of M is the union of all ann(x). So I is contained in union of ann(x). Each ann(x) is contained in an associated prime of M, so I is contained in a union of associated prime of M. Prime avoidance -> I is contained in some associated prime of M

[next obsidian]

You can’t do infinite prime avoidance

[tardy hedge]

Ohh

[next obsidian]

This is why you need finiteness of associated primes

Which is where the fg assumption is used

[white oxide]

Thanks lol

[tardy hedge]

The fact that there are finitely many associated primes isnt a trivial thing right

I just found the statement now in matsumura

[next obsidian]

It is not

I mean I don’t think it’s very difficult either

[tardy hedge]

Does M an A-module always have A/ann(x) for some x in M as a submodule

I saw a statement that said if P is an associated prime then M has A/P as a submodule

[next obsidian]

This is just the span of x lol

[tardy hedge]

Cool

So if there is an element of M that has zero annihilator then A is a submodule of M i guess. Idk just never thought of that before

I guess it makes sense

[tardy hedge]

Otherwise, ann(x) includes P but may be bigger, right?

[tardy hedge]

random: if M1->M2 with N submodule of M1 but is disjoint from kernel, N injects into M2? I guess n+k with k in kernel cant be equal to something else in N because its disjoint from kernel, so every element in N does get mapped to unique thing in M2

[next obsidian]

That isn’t really the issue, it’s just that A/P is an integral domain so it has no zero divisors

[next obsidian]

Yeah the kernel of the restrict to N is the intersection with N

[tardy hedge]

Yeah ann(x) = P because A/P has no zero divisors. But if P was not prime we can just say P is contained in ann(x) right

[next obsidian]

Yeah

[tardy hedge]

ty

[next obsidian]

But what primes can arise in this case

Is the associated primes of A/I

Which are related to primary decomposition of I

[tardy hedge]

yeah, i didnt really learn that connection yet

but im not really sure what you mean "in this case" what is I here?

[next obsidian]

P

But P is always prime cuz P

So I use I

[tardy hedge]

oh youre saying the primes that can arise in the case of ann(x) for nonzero x in A/I are the associated primes of A/I which is related to primary decomposition of I

[next obsidian]

Yes

Which as you observed

Are the ann(y) for y in Ax

[next obsidian]

Which are prime

[tardy hedge]

Hm sorry im not really sure what this is about

[next obsidian]

annihilators of A/I are the same as annihilates of y in Ax

[tardy hedge]

I = ann(x) right

Ok so for nonzero y in Ax (so y = ax, a not in ann(x)) then some r in ann(y) makes ry = (ra)x = 0 so r is in ann(x) for the x in quotient A/I

k im just gonna come back to this l8r

[white oxide]

I don't really understand why we need this corollary. G_{K, T} is the free group with basis all edges (p, q) in K which are not edges in T. Since the fundamental group of K with basepoint p is isomorphic to G_{K, T}, doesn't it immediately follow that it's a free group with basis in bijective correspondence with what's proposed?

[tardy hedge]

Im reading a proof that had A/P (P prime ideal) and then it localized at P. How do i interpret that?

And then apparently the result is a field

[delicate orchid]

it's giving field of fractions oomfie

localising the quotient by P at P is a lot like localising at 0 imo imo

[tardy hedge]

Ah

Should it technically say localize at the image of P in A/P lol

Cause i guess i was a bit confused since P is not a prime ideal in A/P but yea it makes sense now

[echo vine]

Let $K$ be an infinite field and let $f(x), g(x) \in K[x]$. Prove that if the functions $f, g : K \longrightarrow K$ induced by these polynomials are equal, then $f(x) = g(x)$.

\textit{Hint:} Use the result from theory that says a non-constant polynomial of degree $n$ has at most $n$ roots.

\textit{Note:} The function induced by a polynomial is the function $f : K \longrightarrow K$ which consists in evaluating the polynomial at each point, that is, $f(a)$ is the evaluation of the polynomial $f$ at the point $a$.
\

\textit{I don't understand how the polynomials and the functions could be different. I understand they might be slightly different mathematical objects, but I don't understand how something like this could even be proven, it seems like its basically just a definition!}

[cloud walrus]

Aguacate

[echo vine]

Like what does it mean for the polynomials to be identical if not that they are the same at every point

[shy fossil]

two polynomials (say they are elements of k[X]) are equal if they are equal in k[X].

here's a quick counterexample and equality of polynomials is not the same as evaluating the same on every point. if k is finite, say of size p^r, then look at x^{p^r}-x. This and 0 are not the same polynomial

but they evaluate to 0 everywhere

[echo vine]

why is x ^{p^r} - x = 0 everywhere as a function?

what does being finite have to do with it

[shy fossil]

fermat's little theorem

[barren sierra]

try a concrete example

r = 1, p = 5, so you're looking at integers mod 5

not too hard to compute f(x) = x^5 - x over this field for all possible inputs

and then this concrete computation extends to what Nekory said by Fermat's Little Theorem which states that for any integer $a$ and any prime $p$ that $a^p \equiv a \pmod{p}$.

[cloud walrus]

Spamakin🎷

[tardy hedge]

whats the gist of the proof for this? i forget it

[thorn jay]

Isn't it literally just localising at (0) in R/P lol

[delicate orchid]

oh my science

[tardy hedge]

yayy the classic

brings me so much joy when i see a wew do

[glad osprey]

a^(p-1) = 1 in Z/pZ by Lagrange, then just multiply by a (you get Euler's theorem by basically the same proof)

I mean (Z/pZ)^*

[tardy hedge]

oh lol

(Z/pZ)* is group of units and that is size p-1 right

[glad osprey]

I remember we proved it in elementary number theory, it's crazy how much easier it is with group theory

[tardy hedge]

could you say quotients satisfy the universal property that if f: A->B sends all of I to 0, then f factors uniquely through g: A/I -> B st g o pi = f

or does it make more sense to say f factors uniquely through pi?

Also, for here, are we considering the solution to "C" to be the solution of this proposed universal mapping problem?

so is this universal mapping problem kind of combining universal property of quotients and universal property of localizations into one thing?

i havent really seen things proved in this way too much before

[knotty badger]

Yeah

[tardy hedge]

ty

[knotty badger]

Probably this

[tardy hedge]

yea

[knotty badger]

This is essentially using the yoneda lemma

[tardy hedge]

Oh ok, ideally i would like to understand it without that first

[knotty badger]

why?

[tardy hedge]

because the book isnt using any category theory

[knotty badger]

Ah I guess universal properties are very naturally category theory though

[tardy hedge]

yeah true

and Tbf i havent learned yoneda yet lol

[shy fossil]

draw the diagrams and chase diagrams

[tardy hedge]

is this the idea ?

[shy fossil]

you-need-a-lemma

[knotty badger]

Yes

[knotty badger]

To be fair I think it’s usually taught very badly

[tardy hedge]

yeah i didnt take a category theory class before, it was introduced in my galois theory class last term

[knotty badger]

Interesting, I’ve never taken Galois theory

[tardy hedge]

i know the basic stuff up to like adjunctions but i dont have a good intuition for what those are yet

Galois theory feels pretty unique

interesting combination of groups and commutative algebra (fields)

[shy fossil]

The galois correspondence is found in oh so many places

[tardy hedge]

yea ive heard

i think thats why my prof wanted to introduce category theory

he introduced the galois correspondence as an equivalence of poset categories or something?

or adjunction of them

[knotty badger]

Adjunctions are very very cool

[tardy hedge]

idk i didnt really understand it

[shy fossil]

yes

[tardy hedge]

its an equivalence of poset categories?

[shy fossil]

it is mostly clear when looking at galois connections in posets

[knotty badger]

An adjunction iirc

[shy fossil]

something that blew my mind was a prof saying "lie groups are like the galois groups for differential equations"

[tardy hedge]

ha sounds cool i havent learned lie theory tho

[knotty badger]

In my experience I found truly understanding the yoneda lemma was what led me to understanding basic category theory so much better

[tardy hedge]

yea i should give it some time to learn, i definitely know enough category theory basics and greater context to get a good feel for it i imagine

[shy fossil]

I explain yoneda exclusively as the mathematical statement of the philosophy that "You only truly know a person by how they interact with other people and how other people interact with them".

[knotty badger]

I’ve heard this before but I feel like that statement is just false for people

[tardy hedge]

maybe that includes how a person interacts with themselves too

[shy fossil]

it's a philosophy, what can be done

[tardy hedge]

in that case i would agree more

[shy fossil]

people are fickle, but it gets the point across

[knotty badger]

In the yoneda sense yes

But to me this isn’t really the way I think of yoneda

[glad osprey]

Imo, talking about poset categories in Galois theory is just unnecessary complexity. I like to think of the Galois correspondence as between preorders, ie. the preorder of subgroups and the preorder of intermediate fields. When the field extension is Galois this correspondence is an actual equivalence of preorders, but when it isn't you just get a Galois connection (an adjunction in cat theory lingo) - some of the intermediate fields map to the same subgroup

[tardy hedge]

Yeah he def mentioned preorders too

But it was all so abstract i didnt really get it

[glad osprey]

Yeah, I saw your problem set. That wasn't Galois theory

[long obsidian]

If I have a commutative k algebra A with a set of generators G={g1,…,gn} and B another commutative k algebra then under what conditions will a set map f:G-> B define a k algebra map F:A->B?say r in A is a relation. Is it only required that F(r)=0?

[thorn jay]

Calling it a preorder instead of a lattice is so cursed lol

Post's preorder ahh

[glad osprey]

Lol, I guess lattice is a preorder with meets and joins right? I'm just used to calling everything a preorder

[knotty badger]

Yeah I think so

You’re essentially expressing A as a quotient of a free commutative k-algebra

[shy fossil]

this is necessary and sufficient I believe

[thorn jay]

Let I be the ideal of the natural map k[x1, ..., xn] -> A induced by sending xi to gi. A set map f defines a map A -> B if and only if I is contained in the kernel of the map k[x1, ..., xn] -> sending xi to f(gi)

[knotty badger]

So you can apply the universal property for that

[shy fossil]

although you have a little hiccup, F should be extended to some other map to handle relations as inputs

[knotty badger]

Yeah here’s the “quotient of a free commutative k algebra” thing I was talking about

[thorn jay]

I'm not sure how a preorder with meets and joints would look like

Maybe it'd be forced to be an order

But usually it's defined as a poset with meets and joins

(and I don't really care abt preorders)

(Because I really only care about (semi)lattices)

[glad osprey]

Oh, posets have antisymmetry, I was thinking it was the other way around okay, then I also don't care about preorders

Every preorder I care about is actually a poset

[thorn jay]

Me when preorders have more conditions than orders

[tardy hedge]

Should this also say: and every map f: A->B that satisfies 1) and 2) makes g factor uniquely through it

[tardy hedge]

Also if S cap I is not 0 are both those rings just 0

Imma say probably true for both of what i said

If not then poo

[tardy hedge]

How do i know that if i have a description of an object by a universal property, then the object that satisfies it is unique? Is that a category theory kind of thing?

Yea im gonna read the wikipedia page

[knotty badger]

Yes

Up to unique isomorphism

[glad osprey]

I can't remember the proof in the general case, but I think the idea is that if A and B satisfy the same universal property, then there is a unique morphism f from A to B and g from B to A, then by some straightforward cat theory magic you find that f o g = id and g o f = id. I think Aluffi chapter 0 talks about it

[tardy hedge]

Yea i was thinking it was something like that too

[knotty badger]

it’s ✨ yoneda✨

[south patrol]

Shoukd maybe emphasis this means unique maps A to B etc subject to compatibility with the universal properties

[knotty badger]

oh yes ofc

[quartz wind]

is it true that if an R module M is nontrivial and has no proper submodules then its cyclic, equal to <e> for some e, and iso to R/ann(e)? for if it had a non singleton generating set then the span of any proper subset of the generating set would be a proper submodule?

[tardy hedge]

Yes but im not sure if what you said with if it had a non singleton generating set is relevant

[quartz wind]

okay thanks

[tardy hedge]

Also a subset of a generating set may still be a generating set

[quartz wind]

well the assumption to contradict is that it isnt cyclic so if one element in the set still generates it then its proven

[south patrol]

Ye nice

I guess you can rephrase it just as like: if M has no nontrivial proper submodules and m any nonzero element, then <m> is a nonzero submodule and hence equal to M

[quartz wind]

thank you

[cobalt heath]

Is there notion for automorphism on Z_q [X] / (X^N + 1) preserving Z_q?

For fields, Gal(K/Q) would be a related notion

But is there for Z_q?

[white oxide]

Can I have a hint for ii) please? I'm assuming that $\mathfrak{p}$ is not maximal, so that $\mathfrak{p} \subset J \subset A$. Then if $x \in J \setminus \mathfrak{p}$, $x^2 - x = x(x - 1) \in \mathfrak{p}$ so $x - 1 \in \mathfrak{p}$. However, I've been toying around a bit (adding $2x$ and stuff) and I can't get $x \in \mathfrak{p}$ or $1 \in \mathfrak{p}$ for a contradiction. Any hints

[cloud walrus]

okeyokay

[cobalt heath]

Maybe using x^2 + x in P would help?

[white oxide]

Yeah I've tried that too but maybe I'm just dumb

💀

We have x - 1, x + 1 in P

[next obsidian]

Preserving meaning what?

That it fixes Z_q?

Or that it sends Z_q into itself

[white oxide]

I was able to get 2 in P but I don't think that's sufficient

[cobalt heath]

Fixing Z_q

[next obsidian]

That’s exactly a Z_q-algebra automorphism

:)

[next obsidian]

If A is Boolean so is A/p

[cobalt heath]

Ah, dang. Yeah.

[next obsidian]

You’re welcome. 😒

(Jk)

[cobalt heath]

How do I characterize Z_q-algebra automorphisms in this case?

[next obsidian]

Uhhh

Idk what you mean

It’s determined by where you send x

So you need to send it to something which will be 0 mod x^N+1

To descend it to the quotient

And I’m not really sure about making it invertible, at least off the top of my head

[cobalt heath]

Ah I see, I was overcomplicating things

[next obsidian]

Wait no sorry

You need to send it to something which has x^N = -1

Because you need x^N + 1 sent to 0

[next obsidian]

Okay this is also false sorry

Okay, send x to f

Then f^N + 1 must be in (x^N + 1)

lol

[cobalt heath]

X^N + 1 | f^N + 1
X^N + 1 | f^N - X^N = (f - X) ..

I see, I need to think of the cases where f isn't of X^i form. thanks!

[next obsidian]

I guess?

¯_(ツ)_/¯

[tardy hedge]

Yoo waddup gang

Its poppin in here finally

Earlier today was weird vibes

But, the lack of jagr makes me sad

[next obsidian]

Lol

[tardy hedge]

That read like poetry^

[twilit wraith]

is this right? it feels too simple to be true

[next obsidian]

Yeh

It’s easy af

[twilit wraith]

wow

why is this a level 3 question on my homework

[next obsidian]

I mean

You know that the order of the image under a homomorphism divides the original order right

[twilit wraith]

yeah

[next obsidian]

Apply that to G -> G/H

And there u go

That’s what u did

[twilit wraith]

oh

yeah i forgot that theres a homomorphism from G to G/H

i mean thats part of the first iso thm

[torn heron]

guys who wanna create a group just to solve unsolved problems?

msg me in dm's if you do

[twilit wraith]

Nobody wanna join ts

[crystal vale]

Why do you think that?

Group of positive rational numbers under multiplication can be generated by { 1, 2,3,5,..}, right?

And a group of positive rational numbers p/q such that both p and q are odd number can be generated by {1,3,5,...}, right?

[twilit wraith]

Bc if someone comes to this server wanting to solve an unsolved problem, they are highly likely a crank

[ashen hill]

huh 😭

[crystal vale]

I don't think so

[lusty marlin]

Yes (if you're assuming that 'generated by' includes the inverses of those elements)

[crystal vale]

Yes

[lusty marlin]

This is also correct

[crystal vale]

So can I say these two groups are isomorphic?

Maps generator to generator?

[lusty marlin]

Yeah I think mapping the nth prime number to the nth odd prime number should give an isomorphism

[crystal vale]

I mean they will follow the relation, which is ab = ba

[lusty marlin]

Due to prime factorisation both groups can be viewed as a direct sum of countably many copies of ℤ

[crystal vale]

How?

[lusty marlin]

Each positive rational number can be written as a product of prime powers (powers may be positive or negative)

[crystal vale]

Yes

So we will map to their exponents?

[lusty marlin]

So we get 'coordinates' corresponding to the powers of each prime

[lusty marlin]

Yes

[crystal vale]

Got it

Thank you

[shy fossil]

Interesting question

Shifting the prime factorization by shifting the primes by 1 should also do the trick, right?

[lusty marlin]

Yes

[shy fossil]

Nice!

[crystal vale]

But I like the idea of the direct sum of Z copies

[shy fossil]

Structure theorem is a little too strong here, isn't it?

This can be given as an exercise to people in group theory after like a couple of weeks lol

[crystal vale]

Yeah

[lusty marlin]

We aren't invoking the structure theorem though, just saying that both groups are isomorphic to the direct sum of countably many copies of ℤ

[crystal vale]

And there is no relation between generators, only ab = ba holds

[lusty marlin]

Besides, isn't the structure theorem for finitely generated abelian groups?

[crystal vale]

Yeah you are right

[lusty marlin]

Here we're working with a group that isn't finitely generated.

Hence the structure theorem isn't relevant.

[crystal vale]

Yes

[shy fossil]

Fair

[torn heron]

if we were numbers you'd be a negative one

[balmy gyro]

I did not know that group theory a. Existed and b. Is used in chemistry

Can someone give me a brief rundown on the rigour(coming from someone who knows zero group theory)

[knotty badger]

oh yeah something something point groups of molecules

idk anything more than that honestly

[shy fossil]

Group theory is mainly an abstraction of symmetries. Chemical structures have a lot of symmetries so it is a nice tool to use to understand those structures

[balmy gyro]

I’m assuming there are matrices involved

[shy fossil]

They are what we call representations. You have symmetries, and you want to see if you can represent those symmetries using matrices that respect the way the symmetries interact with each other

[balmy gyro]

Ah ok

[shy fossil]

And since some matrices have physical interpretations as translations, rotations, reflections etc, you get the physical picture from the representations

The keywords for you are point groups, representation theory of groups etc

[arctic prawn]

The matrices end up being permutation matrices, and their traces are used to count fixed points.

[balmy gyro]

I’m a little interested in group theory now thanks to my chem textbook, you guys know any good way to start learning?

[lusty marlin]

Dummit and Foote

[arctic prawn]

Artin's algebra is a good book imo

[balmy gyro]

Ok so im assuming prerequisites are like linear algebra, set theory and proving

?

[lusty marlin]

You don't really need linear algebra for the initial part.
Knowing how to write and understand proofs is needed though, obviously.

Although one could theoretically pick that up as one reads the book

[balmy gyro]

Now ideal

Not

[shy fossil]

DnF for starters in group theory is like lawful evil

[lusty marlin]

I'm only recommending what I did myself. My friend started it right out of grade 12 and he was able to pick the initial parts up.

[shy fossil]

Personally, I feel DnF is very dry. Like it's an encyclopedia you go to when you need something

[lusty marlin]

I do agree with this sentiment though. But somehow it gave better 'vibes' than Artin for me personally.

Although Artin is definitely a good choice too

[arctic prawn]

Really, for Chem applications all you need is character tables and Maschke's theorem. The first chapter of Serre's linear representations of finite groups was written with chemists in mind. So perhaps read that after brushing up on group basics 🙂

[lusty marlin]

Serre wrote that chapter for chemists? Damn

That's an interesting fact

[shy fossil]

For me, I started with Gallian. It had good intuitions

[arctic prawn]

He took it from a Chem resource with some editing, at least I think

[lusty marlin]

Oh ok

[shy fossil]

True, but is mostly a representation theory book. Should one study that before studying a little bit of abstract group theory?

Genuinely asking because I never studied group theory because I needed to apply it

[arctic prawn]

Here's a screenshot from Serre's book

[lusty marlin]

What stuff in group theory, specifically? That determines how important representation theory would be beforehand

[shy fossil]

Say like a first course in group theory?

[arctic prawn]

Picture won't send. It says "the first part was originally written for quantum chemists"

[lusty marlin]

You're out there explaining what a bezout domain is, and you're asking about a first course in group theory?

I'm confused now

Since you seem well-versed enough already

[lusty marlin]

In any case, I don't think one needs to know representation theory for a first course in group theory.

Standard practice is for representation theory to follow group theory rather than precede it.

[shy fossil]

No no, I am asking "should one study rep theory before some intro to group theory if one is only interested in applications like in chemistry/physics?"

[lusty marlin]

Ah lol

[shy fossil]

I added this because I came from the pure side, so the applied pedagogy is pretty much unknown to me

[lusty marlin]

Then I think one can study things concurrently.

[lusty marlin]

Yeah I misinterpreted the phrasing on that message

[lusty marlin]

That's what some physics people do, I've heard

[arctic prawn]

Yeah, learn it from a Chem/physics resource though

Because a lot of the theoretical underpinnings are unnecessary in practice

In particular, the whole representation as a module thing is wholly useless in science applications in my experience

[shy fossil]

I did study it from Griffiths during COVID, it was... fun(ny)

[arctic prawn]

👍

[cyan skiff]

I have a question. It would be great if anyone can help me with it. That how many subgroups of order 6 does D6 have? And generalise to Dn where n is a positive integer divisible by 6?

[shy fossil]

Is D6 the one with 6 elements, or the one with 12 elements? I believe 12 or else the question makes very little sense

[cyan skiff]

My reasoning is for D6 we have C3 and two permutations of D3 as subgroups(S3's) . similarly for D12 we have C6 and two permutations of D6 all of which correspond to C3 and two permutations of D3

[cyan skiff]

The one with 12 elements

[cyan skiff]

So for any Dn divisible by 6 we have exactly 3 subgroups of order 6

Is it correct or am I missing something?

[candid patrol]

You are wrong

We have 5 subgroups of order 6 in D12

$\langle r^4, r^2s \rangle \simeq \langle r^4,s \rangle \simeq \langle r^4, r^3s \rangle \simeq \langle r^4,rs \rangle \simeq D_3$ and $\langle r^2 \rangle \simeq C_6$

[cloud walrus]

UGOBEL

[crystal vale]

I think < r^4, rs > = <r^4, r^3s>

Because you can make rs in <r^4, r^3s>

[vast stump]

there are multiple errors in this classification

[candid patrol]

you right

< r^4, rs > = <r^4, r^3s> and < r^4, r^2s > = < r^4, s>

So we get 3 subgroups again

[velvet cave]

i would highly recommend these group theory notes someone sent me

(not sure where they're originally from)

they're absolutely gorgeous

but also extremely in depth, so a bunch of it might be obsolete for your purposes

and it doesn't have many exercises, so to make sure you get adequate practice you could use another textbook like others have recommended

[chilly ocean]

most imp excersises are there

[cyan skiff]

So is it also true for all Dn where n is divisible by 6?

[balmy gyro]

initial

Well ik how to prove stuff

But my proving ability is like y=asinx

Depends on the day

Ik some linear algebra like axioms of vector spaces, and study real analysis already

So would you say im fine

[lusty marlin]

He just stopped after a short while.

[lusty marlin]

Taking your claims of knowing to be proper, I think you should be fine.

[toxic zephyr]

we can put Q(cos(6pi/23)) as an intermediate field of Q(e^(6pi i/23)), right? if we call it zeta then it's fine since (zeta+zeta^-1)/2 is in the field so there's containment?
and also that is a primitive root of unity right? since (6,23)=1 (or should it be because (3,23)=1?)

[delicate orchid]

yes and yes

[south patrol]

In fact this is precisely the fixed points of complex conjugation I.e. intersection with reals

[quartz wind]

Statement. The order of $\text{Aut}_{\mathsf{Grp}}(C_n)$ is the number of natural numbers $r<n$ relatively prime to $n$.$\$
Proof. Every automorphism of $C_n$ can be identified with $pC_n$ for some natural number $p$. For if not, then there exists some automorphism $\varphi$ and some distinct $r,s\in C_n$ such that $\varphi(r)= pr, \varphi(s)=qs$ for distinct naturals $p,q$. Then $(\text{lcm}(p,q)/p)\varphi(r)=(\text{lcm}(p,q)/q)\varphi(s)$ and hence $(\text{lcm}(p,q)/p)r= (\text{lcm}(p,q)/q)s$, which violates bijectivity. Now for a given automorphism $pC_n$, let $m\coloneq\text{gcd}(p,n)>1$. Choose an element $r\in C_n$ coprime to $n$. Then $\vert r\vert = n$, but $\vert pr\vert = n/m$, violating that group isomorphisms must preserve order. $\qed$

does this work?

[cloud walrus]

esca

[velvet hull]

you don't need to justify why every automorphism is of the form pC_n

instead observe that C_n is generated by 1, by definition, and any homomorphism is determined by the image of any generating set

[quartz wind]

ah good point

thanks

[velvet hull]

also, although it is true that group isomorphisms must preserve order it doesn't mean that you can always send an element to another element of the same order

why are you always allowed to send 1 to some number coprime to n, and that induces a well-defined homomorphism?

[tardy hedge]

How can I consider rings of fractions to be a functor? If I am localiizing wrt to some subset S then I can only consider rings that have that S as a subset right

[quartz wind]

i think it was given without proof that if both groups are abelian and its order preserving, they are isomorphic

i supposedly wouldnt know how to prove this

[tardy hedge]

or does it just make sense to consider it as a functor from A-modules to As-modules

[velvet hull]

taking the field of fractions is not localizing to a specific multi closed set

[tardy hedge]

yea i mean the latter

[velvet hull]

so it's a functor from integral domains to fields

and if you fix a base integral domain A it induces a functor from A modules to Q(A) modules (vector spaces)

[velvet hull]

could you explain a little more what you mean here

[velvet hull]

I don't see how that helps your proof

[quartz wind]

since for p coprime to n, Cn and pCn have the same number of elements of any given order, are finite, and are commutative, they are isomorphic

[velvet hull]

since you havent proved that sending 1 to p is even a well-defined group homomorphism we don't even know if pCn is a group

[quartz wind]

ah okay i see

thanks

i will verify that

[velvet hull]

also I will say that you should not need that theorem

it's an extremely powerful theorem, and cyclic groups are a lot more simple to work with than the average finite abelian group

[quartz wind]

so i should think of a simpler way

[velvet hull]

if you add the additional assumption that you are working with a finite cyclic group

then the theorem becomes provable, for where I think you're at

but maybe try and think of a different way, yes

that can be an additional exercise

[quartz wind]

got it, thanks again

[spice wagon]

is it really useful to study the chapter on rings, ideals, A-algebras, A-modules etc.. in the case where the ring is not commutative?

or is it sufficient to do so only if the ring is commutative

[white oxide]

Here is
[\text{Spec } A = \bigsqcup_{j = 1}^n \pi^*_j(\text{Spec } A_j)]

[cloud walrus]

okeyokay

[next obsidian]

Depends on what you end up needing ring theory for

[spice wagon]

algebraic geometry, homology theory, representation theory and maybe i would say algebraic topology

[next obsidian]

Rep theory might need it

[cinder onyx]

So my thinking is, we can parametrise any great circle into the form $G_\theta=\cos\theta X + \sin\theta Y, X,Y\in SU_2$, where $\langle X, Y \rangle = 0$. \ Any longitude can be similarly parametrised in the form $L_\theta=\cos\theta I + \sin\theta A$ with $A \in \mathbb{E}$. So then for G to be a left coset of L where must be some $M\in SU_2$ such that $MX=I, MY=A$ and the former condition clearly indicates $M=X^{-1}$. \I'm stuck on how to show that $X^{-1}Y=X^{*}Y=A\in \mathbb{E}$ follows, which would surely be sufficient (then $G=M^{-1}L$).

[cloud walrus]

mati

[cinder onyx]

for some unholy reason I'm missing it because it seems almost too obvious that the action of SU_2 on the 3-sphere should preserve orthogonality which'd make it easy but my brain is missing it

I think I'm overthinking it lmao

[tardy hedge]

if ur tryna test if a map A/I -> B defined by mapping representatives of A/I is well-defined, could you also just test if I is contained in the kernel? (I guess you need A->B hom first, and I is contained in the kernel of that map)

thats the same thing right

[south patrol]

Yes this is the standard way to do it

[coral steeple]

Is this correct? Extend to the field of fractions. There exist integers x,y so that xm+yn=1. Write a=a^(xm+yn)=(a^m)^x(a^n)^y=(b^m)^x(b^n)^y=b^(xm+yn)=b

(The case a,b not invertible in the FOF is trivial)

[cobalt heath]

Yeah but I don't think you necessarily need field of fractions, a * a^(xm) = a^(yn) is enough

BTW I am so lazy, would anyone please find an Z_q algebra automorphism of Z_q [X] / (X^N + 1) that is not in the form of X -> X^h

[cobalt heath]

When does X^k + 1 divides f^k + 1

Ah better idea to ask this in helper channels

[blissful crystal]

did you end up getting help on this?

[cobalt heath]

No, it did not work out well.

[shy fossil]

try looking at the k-th roots of -1. f^k+1 must also have those as roots.

[cobalt heath]

Ah, yeah that should be the easiest. I dunno why I did not think of that, while always working with the roots.

[blissful crystal]

i was going to go on but thats probably better than any hand waves i could have made

[cobalt heath]

In my case, N is power of 2. So it should just be the root of unity.

[shy fossil]

I believe you can classify what f has to look like using this

[cobalt heath]

So basically, f should map roots of X^N + 1 to another roots of X^N + 1, right?

[shy fossil]

yee

[cobalt heath]

I guess there is a map sending
zeta -> zeta^3
zeta^3 -> zeta
zeta^i -> zeta^i

[lean sail]

I am struggling a bit with the following theorem and proof from "A first course in abstract algebra" by John Fraleigh. Here is the theorem and proof:

\subsection*{6.15 Theorem}
Let $G$ be a cyclic group with $n$ elements and generated by $a$. Let $b \in G$ and let $b = a^s$. Then $b$ generates a cyclic subgroup $H$ of $G$ containing $n/d$ elements, where $d$ is the greatest common divisor of $n$ and $s$. Also, $\langle a^s \rangle = \langle a^t \rangle$ if and only if $\gcd(s,n) = \gcd(t,n)$.

[cloud walrus]

proofman

[lean sail]

\begin{proof}
That $b$ generates a cyclic subgroup $H$ of $G$ is known from Theorem 5.19. We need show only that $H$ has $n/d$ elements. Following the argument of Case II of Theorem 6.10, we see that $H$ has as many elements as the smallest positive power $m$ of $b$ that gives the identity. Now $b = a^s$, and $b^m = e$ if and only if $(a^s)^m = e$, or if and only if $n$ divides $ms$. What is the smallest positive integer $m$ such that $n$ divides $ms$? Let $d$ be the gcd of $n$ and $s$. Then there exist integers $u$ and $v$ such that
\begin{align*}
d &= un + vs.
\end{align*}
Since $d$ divides both $n$ and $s$, we may write
\begin{align*}
1 &= u(n/d) + v(s/d)
\end{align*}
where both $n/d$ and $s/d$ are integers. This last equation shows that $n/d$ and $s/d$ are relatively prime, for any integer dividing both of them must also divide 1. We wish to find the smallest positive $m$ such that
\begin{align*}
\frac{ms}{n} &= \frac{m(s/d)}{(n/d)} \quad \text{is an integer.}
\end{align*}
From the division property (1) following Example 6.9, we conclude that $n/d$ must divide $m$, so the smallest such $m$ is $n/d$. Thus the order of $H$ is $n/d$. \medbreak

    Taking for the moment $\mathbb{Z}_n$ as a model for a cyclic group of order $n$, we see that if $d$ is a divisor of $n$, then the cyclic subgroup $\langle d \rangle$ of $\mathbb{Z}_n$ has $n/d$ elements, and contains all the positive integers $m$ less than $n$ such that $\gcd(m,n)=d$. Thus there is only one subgroup of $\mathbb{Z}_n$ of order $n/d$. Taken with the preceding paragraph, this shows at once that if $a$ is a generator of the cyclic group $G$, then $\langle a^s \rangle = \langle a^t \rangle$ if and only if $\gcd(s,n)=\gcd(t,n)$.
\end{proof}

[cloud walrus]

proofman

[lean sail]

So, where I am struggling is, why does $m = n/d$ have to be the smallest such integer? How do we know that there is not one smaller?

[cloud walrus]

proofman

[next obsidian]

Look I ain’t gonna read all that, but it should be because the gcd is the minimal number such that you can write it as a linear combination of n and s

If you had a smaller order you’d be able to produce a smaller such d

[lean sail]

thank you for your response! yeah, understand. don't blame you for not reading it... it is kind of a wall of text. just wanted to make sure i provide full context, but i don't think reading the entire thing is necessary

[lean sail]

just trying to think about this ...

[next obsidian]

If a^s has order k then a^ks = e, now divide ks by n with remainder r then you get that a^r = e. If r ≠ 0 then n isnt the order of a so r = 0 meaning n divides ks. This means n/gcd(n,s) divides k so k >= n/gcd(n,s)

There’s a direct proof

[lean sail]

this is an exact answer to my question lol

[lean sail]

just to be 100% clear, you're using the property of cyclic groups where every element a^s of a finite cyclic group G can be written uniquely as a^r where s = nq + r for 0 \le r <n?

like, you're using the division algorithm?

[next obsidian]

Uhh, no?

What I did proves that the order of a^s is gcd(s,ord(a)) in full generality

When you do the division algorithm to write s = pn + r you just use the fact that a^s = (a^n)^p•a^r = e^p•a^r = a^r

I mean I guess this is the exact same thing as what you said (by considering <a>) but I don’t think it’s that fancy

[quartz wind]

are annihilators necessarily cyclic? otherwise i dont see why we can talk about orders as if theyre elements of the ring

[tardy hedge]

Annihilator of a module?

[next obsidian]

No

Consider like Z_2^3 and the element (1,0,0)

It’s annihilated by 0 x Z_2 x Z_2

Or just replace Z_2 with any ring lol

[quartz wind]

yeah right

okay so

why should "relatively prime orders" make sense at all

or like o(v) = prod alpha_i

[tardy hedge]

M is an abelian group so its elements have orders

[quartz wind]

but what about for the A_i, where M = sum A_i?

[tardy hedge]

Doesnt that just mean the size of A_i

[quartz wind]

order here refers to any generator of the annihilator, which i suspect coincides with the group theoretic definition for elements

i mean its just multiples of the usual order from group theory

for a single element that is

[next obsidian]

Have they defined order differently?

[next obsidian]

R is a PID dude

You can talk about relatively prime because you have prime factorization

[quartz wind]

yeah but who says the order is just one element

[next obsidian]

Wdym?

[quartz wind]

like the annihilator must be cyclic for the order to make sense

or am i wrong

[next obsidian]

It’s a PID

Every ideal is principal

[quartz wind]

ah

i forgot what it stood for thats embarassing

thanks

[tardy hedge]

Tru

[crystal vale]

How can I classify the ring of order 35?

So the underlying additive structure is Z/35Z

And we want unity in our ring

It cannot be an integral domain

[tribal moss]

Perhaps split by characteristic first?

[shy fossil]

chinese remainder theorem?

[crystal vale]

Its characteristic 35, right? Because its additive structure is Z/35Z

[crystal vale]

How does the Chinese Remainder help here?

[tribal moss]

A priori you don't know where 1 is in the additive cycle; the characteristic might be 5 or 7 too.

[crystal vale]

Oh unity is not necessarily to be \bar 1

Right

[velvet hull]

I don't think the characteristic even has to be prime

[tribal moss]

It has to divide 35, though.

[crystal vale]

Yes

[tribal moss]

And if the characteristic is 1, then it's the zero ring, whose size is wrong.

[crystal vale]

Yes

[shy fossil]

Oh... show that if p\mid n implies Z/pZ and R/pR are isomorphic. Then note that in your case, no prime divides 35 twice, so any pR and qR are coprime, so you can use CRT on R/35R which is just R (in fact this goes through for all square-free n)

[tardy hedge]

Why do we want to consider characteristic first? How does that help us?

Wdym by “split by” characteristic

[crystal vale]

Let's find out

[crystal vale]

Cases

[tardy hedge]

Ok yea i guess it makes sense how that could help classify then

It lets us know some subrings doesnt it

[tribal moss]

In particular, if the caracteristic is n, then there's a subring isomorphic to Z/nZ.

[tardy hedge]

Yea

[crystal vale]

That subring generated by unity element 1?

[tardy hedge]

Ya

[tribal moss]

(And ad-hoc reasoning from there gives us a very short list of candidate rings in each case).

[crystal vale]

Isn't characteristic is equivalent to taking the LCM of additive order of all elements ?

Characteristic define as the least positive number n such that nx = 0 for all x in R, so this nx is additive x+x+..+ x or n•x ?

[tardy hedge]

Additive

[tribal moss]

Hmm, that sounds promising. (I had a more convoluted argument in mind myself).

[crystal vale]

So then it cannot be possible that Z/35Z has characteristic 5 and 7

[tardy hedge]

I thought it was least int so that n*1 = 0

Is it the same thing?

[tribal moss]

Yes: If n×1 = 0 then n×a = 0 for all a, so 1 has the largest possible additive order anyway.

[crystal vale]

Yes

[crystal vale]

That's why I said this, but then I got confused

[tribal moss]

Sorry for confusing you. You were just faster than me ...

[crystal vale]

So now characteristic 35, how will we proceed now?

[tribal moss]

Here :-)

Within the additive subgroup generated by 1 (which we now know is the entire ring), everything can be written as a finite sum of 1's, so by the distributive law that fixes how multiplication works there.

[south patrol]

You can argue structurally by considering the map Z -> R and its kernel

I guess that is more an aesthetic thing

[tribal moss]

Hmm, from that direction how do you quickly exclude the possibility of the kernel being (5) or (7)?

[crystal vale]

n -> n•1, so characteristic 35 implies Z/35Z is isomorphic to subring of R, and since R has order 35, so R is isomorphic to Z/35Z

[tribal moss]

Yes.

[south patrol]

What I mean is you can also just do the characteristic arguments thinking of this, like char 7 or 5 would mean you're a vector space over a field and have prime power cardinality

[tribal moss]

Yes, that's what I was steering towards at first too.

[south patrol]

Kinda nice how we don't have to worry about commutativity

[crystal vale]

Yes

[crystal vale]

Btw I don't understand the distributive argument here

[tribal moss]

E.g, (1+1+1)(1+1) = 1(1+1) + 1(1+1) + 1(1+1) = 1+1 + 1+1 + 1+1

[crystal vale]

So here 1 is the unity element

[tribal moss]

Yes.

And we get (n·1)×(m·1) = (nm)·1 where the · stands for repeated addition and × multiplication in the ring (and nm is multiplication in Z).

[crystal vale]

I see

Got it

And we can write all elements in terms of n•1

Thank you @tribal moss and @south patrol

[cinder fox]

Do equivalence classes preserve algebraic structure? For example, if we were to consider the following equivalence between $\bR^{n+1}$ and $\bR_n[X]$: $(a_k){1\leq k\leq n+1} \equiv \sum{k=0}^n a_{k+1}X^k$; then define, on the set of equivalence classes, addition and multiplication such that they correspond to the usual polynomial addition and multiplication, does that form a ring?

[cloud walrus]

struct ∅ {};

[crystal vale]

So we can extend this to the ring of order n, where n is squarefree number

[tribal moss]

What does R_n[X] mean here? Your equivalence suggests it's the set of polynomials of degree at most n, but that set is not a ring (it fails to be closed under multiplication).

[delicate orchid]

if R_n[X] is the degree up to n polynomials then this doesn't form a ring - it's not closed under multiplication.

wow I see how it is

[tribal moss]

Jinx.

[cinder fox]

oh right

[delicate orchid]

but I think your question is still interesting

[south patrol]

Any if you ignore this sort of thing then it's just relabelling elements

So the answer is ye

[tribal moss]

And you don't really get "equivalence classes" when you have a relation between two different sets A and B.

[delicate orchid]

it turns out, for rings and groups, that the equivalence classes that allow you to form quotient structures like these correspond to substructures in a nice way (ideals and normal subgroups respectively). But this isn't always the case (for example, topological spaces and semigroups)

[cinder fox]

oh right

my abstract algebra is a bit rusty

[tribal moss]

We could take R[X]/<X^(n+1)> on the right, though -- that is a ring, but your correspondence doesn't preserve the multiplicative structure.

[earnest birch]

Well, I think not some arbitrary equivalence class. The equivalence class of “preserving algebraic structure” is when there is an isomorphism between the two algebraic structures

I hope I understood your question right

[thorn jay]

what you are talking about in your example is a function, lol, not an equivalence relation. But, to answer your question, for any algebraic structure (using a very general definition with n-ary operations), there is the notion of congruence relation, which is an equivalence relation which preserves the algebraic structure

these are necessary as general algebraic structures do not permit the use of things like ideals or normal subgroups (some algebraic structures do not even have constants)

[coral steeple]

Thanks!

[rapid gyro]

guys one question only tell yes or no "Every element of an abelian group has finite or infinite order" is rigth true¿

[kind temple]

thats true independent of the group being abelian

[chilly ocean]

hi AG hater

[thorn jay]

Every number is either finite or infinite

[chilly ocean]

i know a number which doesnt even exist

[shy fossil]

yes?

[chilly ocean]

my highschool crush's mobile number

[shy fossil]

not the place for jokes

[stone elbow]

#rant

[spice wagon]

when we say that a group G operates on a set X if there exists $\varphi : G \times X \to X, \ (g,x) \mapsto g \cdot x$ which respects certain conditions, by $\cdot$ we are talking about the law of G right ?

[cloud walrus]

tm

[delicate orchid]

no

[spice wagon]

ah it’s the action then

[delicate orchid]

indeed

[spice wagon]

okay thanks I hadn’t clarified that in my notes

[knotty badger]

The notation is there to make some of the laws of the group action seem similar to usual group laws

[spice wagon]

yeah sometimes it’s confusing

[knotty badger]

how do you mean?

[thorn jay]

If you see an action as a "left multiplication" of G on the set X, it should be more clear

[knotty badger]

You can also get “right multiplications” of G on X

These turn out to be group actions involving G^op

[shy fossil]

I let my sets act on my groups

[knotty badger]

I mean you can think of a “set action” as just a map B x A -> A

An algebra for the B x (-) endofunctor

This endofunctor is a monad iff B has a monoid structure

[shy fossil]

Are you kidding me

I don't know if I should be disappointed or amazed.

[delicate orchid]

https://en.wikipedia.org/wiki/Group_with_operators

[thorn jay]

Thanks Noether

[thorn jay]

A monad is just an abstraction of algebraic theories

[thorn jay]

I let my group on set acts

[shy fossil]

So what would be an abstraction of monads? And their abstraction? Where do we stop?

[thorn jay]

Uhh idk that's some higher category theory shiz

[shy fossil]

My group lets me act on sets

[delicate orchid]

you can extend monads to n-monads and keep going until infinity-monads but the real question is why would you

[thorn jay]

Funny

[shy fossil]

counter question : Why shouldn't I?

[delicate orchid]

although calling vertical categorification"generalisation" makes me a little uneasy. It's really just the same idea higher up

[thorn jay]

Isn't the whole idea of abstraction being the same idea but higher up

[shy fossil]

time to invent horizontal categorification, and then some other direction and then in some other direction, ad infinitum

The directions of categorification will then themselves form a category with morphisms as rotations

[delicate orchid]

horizontal categorification already exists

[formal cave]

al

[shy fossil]

little joke I was making... although I don't really know the general theory

[delicate orchid]

not really, it's taking the absolute key essential ideas of whatever you're working with and allowing it to be applied to more objects. This is just putting a "n-" suffix in front of everything

the vibes are off

[shy fossil]

I only know that groups to groupoids and monoids to categories are examples of horizontal categorification

[thorn jay]

Mm, I see what you're getting at

Abstraction actively removes the clutter, so to speak, so we can focus on the underlying idea

While this is literally the same idea just more

[delicate orchid]

that's a much better way of putting it

[thorn jay]

Thank you

Anyways my point stands, putting n- in front of everything is funny

[fading acorn]

we can bypass it tho
a monad is just a monoid///but in the category of endofunctors

[delicate orchid]

agreed

[shy fossil]

But the just more kind of lets us do a lot of things. yes it is not a generalization per se, but somehow forgetting certain information brings us down the ladder

[delicate orchid]

sure but you can always just pretend any 1-category is an n-category will all higher morphisms just being the identity

[thorn jay]

So studying infinity categories is also studying normal category theory guys!!

[shy fossil]

me when I pretend my groups are just sets

[fading acorn]

groups are group objects in sets

[thorn jay]

Groups are objects in the Eilenberg-Moore category of the group monad

[delicate orchid]

sets are 0-categories

and truth values are (-1)-categories

[delicate orchid]

ts guy needs to be stopped

[shy fossil]

what would be (-2)-categories?

[fading acorn]

$-\infty$-categories where

[cloud walrus]

0mem

[thorn jay]

When are we getting higher cardinal/ordinal categories

[delicate orchid]

"Always True" and then it stabilises for lower n

[thorn jay]

Periodic table of categories

[shy fossil]

so just the one object category?

[delicate orchid]

well no becaue it's not a category, it's a (-2)-category.

[shy fossil]

fair enough

and I guess we stabilize beyond this point

[shy fossil]

not actually a bad idea

[thorn jay]

It exists

[shy fossil]

I know

I just went to find the picture

I read it first in a paper by baez

absolute chad

[thorn jay]

I was about to get that qwq

https://ncatlab.org/nlab/show/periodic+table
Here's the link anyways

[delicate orchid]

here's how the idea works in very basic terms: in a strict 2-category you have a 2-category of objects, a 1-category of morphisms, a set (0-category) of morphisms of morphisms. So the question "what is a (-1)-category" is asking what the next layer of morphisms should be. Which comes back to what I said earlier where you can embed any n-category into a higher category by pretending all of the higher morphisms are the identity. So the 3-morphisms of a 2-category (embedded in a >3-cat as described) is basically just asking "is this morphism the same as the other", i.e. the 3-morphism set is either empty (false) or contains a single object (true)

[shy fossil]

This is why I love mathematics

now if only we knew what a (3/2)-category is

[thorn jay]

It's like backwards recurrence relations

[delicate orchid]

I'm leaving that one as an exercise for the reader

[fading acorn]

$\Gamma(z)$-category

[cloud walrus]

0mem

[fading acorn]

we need this

[thorn jay]

Why stop there

r-category for r an arbitrary element of a ring

[delicate orchid]

ok ts is silly 🥀

[shy fossil]

why ring? Go for set valued

[thorn jay]

why stop there

[fading acorn]

🥀

[thorn jay]

a-category for a in an arbitrary algebraic structure

[shy fossil]

Wait, we can make sense of this I think

[delicate orchid]

notation has already been used for enriched categories. You're out of luck buster

[shy fossil]

we allow 1-morphisms, and restrict 2-morphisms to be invertible

[thorn jay]

Fuckkkk you're right

[delicate orchid]

that's a (2,1)-category!

I think

[thorn jay]

Homotopy type shit?

[shy fossil]

YES!

[delicate orchid]

which way round do they go again

yeah I was right

[shy fossil]

2,1 is fine

so all (n+1/2)-categories make sense

[delicate orchid]

https://ncatlab.org/nlab/show/(0,1)-category I got the brain worms

[shy fossil]

sure

that is a name

[thorn jay]

😭

I quit

I'm going back to logic

[delicate orchid]

bad news

[thorn jay]

NOT categorical logic

..

FINE if you wanna do it this way IM TURNING TO ANALYSIS

[shy fossil]

Lmao, I just presented a report in my logic class last month

[delicate orchid]

bad news

[thorn jay]

You're kidding

[shy fossil]

non-standard analysis

which is essentially also categories

[thorn jay]

That's amazing

Get owned, category theory deniers

[shy fossil]

i mean, if database management and collaborative text-editing can be brought under the fold of category theory, there really is no escape

I am not joking btw https://bosker.wordpress.com/2012/05/10/on-editing-text/