groups-rings-fields

[sweet echo]

yeah i see that its being twisted by the involution

but like i could take omega to be identity and i just get a regular derivation

[delicate orchid]

If that involution is -1…

Then it actually is a derivative but backwards

Perhaps it is a generalisation?

[sweet echo]

i couldnt quite match it up to this, it does seem more general yeah

[delicate orchid]

Oh I see - yeah

Perhaps it would be more accurate to call this a F_2-graded derivation or whatever but that’s a mouthful

[rocky cloak]

I mean usually something is called anticommutative if

xy = -yx

Here you have that the "commutativity" of x and Omega is governed by omega. So at least if omega was just multiplication by -1 I think it would make sense to call this antiderivation

[void cosmos]

k = R

and let A be any algebraic set in A^2(R)

then A = V(F) for some F in R[x,y]

[delicate orchid]

true!

[void cosmos]

does the proof just consider the product of all polynomials

in the ideal

and its f.g

cuz its noetherian?

thats what i thought about

like A = V(I) for some ideal in R[x.y]

but this is finitely generated so ?

or maybe i can do it by like

that classification thoerem

cuz ik the irreducibles

of A^2(k) in general if k is any infinite field

[rocky cloak]

So fg is 0 if one of f and g is 0, thus V(fg) is the union of V(f) and V(g), while V(<f,g>) is the intersection of V(f) and V(g)

Are you able to find a polynomial function such that h(f, g) is 0 iff both f and g are 0?

[void cosmos]

yeah

the sum

f+g

[rocky cloak]

Not quite

You can have f = -g without them being 0

[void cosmos]

fuck ur right

okay but ig this si the

only proble

so

just

square them

lmfao

shuld fix it

right?

so like f^2+g^2

[rocky cloak]

Yes

[void cosmos]

yea got it

im so stupid hahaa

tysm

idk why i immediately went to the product

[rocky cloak]

Note that it's important that we're working over the real numbers here

[void cosmos]

idk why

tho

cuz

[rocky cloak]

(or at least an ordered field)

[void cosmos]

it states at the end of the problem "

thats why we consider an algebrically closed field "

idk what closure has to do with anything here

or non closure of R

[rocky cloak]

So if f^2 = -g^2 it could still be that f = ig, without them being 0

[void cosmos]

yeah ur right

damn

these little things just went over my head

haha

ur right

so ig the main point of the problem was to show how like

algebraic sets of diff fields differ for the same polynomial

(which is obvious? or idk)

[rocky cloak]

Well, it's showing that the Nullstellensatz doesn't hold

An alternative approach, instead of limiting yourself to algebraically closed fields, could be to replace A^2(k) with the spectrum of k[x, y].

That basically amounts to evaluating your polynomials on the elements of the algebraic closure of k. (Up to some Galois automorphisms anyway)

[void cosmos]

idk whats that yet anyways

as u can see its still the next section

but cool

tysm

[void cosmos]

yo

bad question

the fact that rad (I) = intersection of all prime ideals that contain I

is it analogous to like

the decomposition of algebraic sets into irreducibles?

[echo path]

can someone help at hellp 20

[void cosmos]

yea sure

[south patrol]

Well primary decomposition is

Yes

[void cosmos]

yea

i kinda felt that it is just primiary decomposition

but idk the proof just felt instant

where with primiary decomposition is much harder

to show that it works for noetherian rings

[south patrol]

Hm the proof I remember seeing on the decomposition of algebraic sets is just an appeal to Lasker-Noether lol

[void cosmos]

points in C^n

😄 HAHAHAHA im just memeing i literally just learnt this like 2 hours ago

its probably wrong

[south patrol]

C^n lol

would be fine for finitely many variables

[void cosmos]

wait

yeah he said inifnitley many variables

which isnt even noetherian so

AG fails

[south patrol]

Anyway you can show that C[x_1,x_2,...] is Jacobson

[void cosmos]

lol

[south patrol]

which is at least a partial answer

[south patrol]

Yeah okay i checked and it seems you should still get the same answer essentially using this

Lemme think of the argument lol

Ah okay lol here goes lol i love this stuff

Basically uh idk if you've seen the "cheap" proof of Nullstellensatz for C

Basically uh

Let $\mathfrak m$ be a maximal ideal of $K = \mathbb C[{x_i}_{i \in I}]/\mathfrak m$ where $I$ is a countable set. We claim $K/\mathbb C$ is algebraic - if $t \in K$ were transcendental then $K/\mathbb C$ would be an uncountable extension (since the elements $\frac{1}{t-a}$ would be independent) whilst it's clear this is impossible

[cloud walrus]

potato

[south patrol]

And then uh

That implies K is isomorphic to C, so we have an isomorphism K -> C which we can think of as coming from a map C [ {x_i} ] -> C with kernel m

Now any map C[ {x_i} ] -> C is just given by sending x_i -> a_i for some a_i in C

And the kernel contains all the elements x_i - a_i

So m contains ( x_i - a_i | i in I)

However, the latter is a maximal ideal

So we have equality

This idea is actually kinda funny, like there's a similar proof that uhhh

Any endomorphism of C^infty must have an eigenvalue i think

as if T had no eigenvalues then the operators 1/(T-a id) for a in C would be linearly independent in some sense etc

[white oxide]

is it true that $N\bigl(\frac{\epsilon}{c + di}\bigl) = \frac{N(\epsilon)}{N(c + di)}$ for $c + di \neq 0 \in \mathbb{C}$?

[cloud walrus]

okeyokay

[white oxide]

wait

wait oops

wait

[chilly ocean]

try to prove it

[south patrol]

That what is a lie group?

[lethal dune]

[coral shale]

<@&268886789983436800> see msg history of 979887687353896990

[lethal dune]

what did I miss

[white oxide]

for 6, wouldn't it be better to use the division algorithim

given 5) lol

i.e. given an ideal I select an element with minimal norm and use the analagous proof used for showing that Z is a PID

idek if that makes sense but it was my first thought

[crystal turtle]

:3c or just show all euclidean domains are PIDs

[delicate orchid]

the proof is basically the same cause Z is inital or whatever ;3

[south patrol]

yeah idk why they mention euclidean algorithm

[crystal turtle]

Like just do it all at once

[south patrol]

i would just use division

i imagine if anything the person meant division alg lol

[crystal turtle]

rather than for Z[i] and then again later for all EDs

[white oxide]

i'll try this then hopefully i don't get points off for not using the euclidean algorithim lol

[crystal turtle]

i don't remember how hard the proof is

[white oxide]

yea my prof has hella typos but this seems too big to be a typo anyways

[south patrol]

lol i would just say "I assume division alg is meant ?" but maybe i'm too snarky

had to do that in an exam recently lol

[delicate orchid]

I feel like this question is fine

[white oxide]

i mean

for showing that Z is a PID we were instructed to use the euclidean algorithim but i posted a question about it and the TA said just use the division algorithim lmfao

so hopefully it should be chill

[crystal turtle]

me correcting the prof's question on exams

[white oxide]

so i would have to take the infimum of the norms then to select a minimal element since they're real valued

nvm

i'm stpuid

gaussian INTEGERS

[uncut girder]

Whats the difference between division algorithm and Euclidean algorithm

[south patrol]

division alg is like

given a and q write a = qb + r etc

[crystal turtle]

division algorithm is such a dumb name for it. It's not an algorithm

[south patrol]

eucilidean algorithm is the annoying long one that does it loads of times

and gives you the gcd

[south patrol]

ig you can just say lol "subtract b"

yeah that is what is in Elements apparently

and then just say the algorithm terminates in finite time ig

[white oxide]

so i could choose the minimum of floor(x), ceil(x) as my integer coefficient

[crystal turtle]

one of those is always greater than or equal to the other

[white oxide]

right.

[crystal turtle]

So that's not quite it

[white oxide]

$\text{min}{|\floor{x} - x|, |\ceil{x} - x|}$

[cloud walrus]

okeyokay

[white oxide]

nvm

[crystal turtle]

closer

kinda

not realy

Here's how I would write it

Let $a_i = \begin{cases} \text{floor}(a_i) &: a_i-\text{floor}(a_i) \leq \frac{1}{2}\ \text{ceil}(a_i) &: \text{ otherwise} \end{cases}$

[cloud walrus]

Ryx (Home for flowers)

[white oxide]

ah right the 1/2 thingy

damn i'm not that smart

ok thanks

[white oxide]

can somebody explain to me the intuition/usefulness of projective/injective modules? i'm kinda struggling to see the importance of its defining property

[delicate orchid]

projective is like free modules but actually general enough to come up

injective is the dual

a useful fact to know about projectives (for me) is that if P is projective then there exists some Q such that P+Q is free

you also have nice things like projective modules are flat and torsion free

[white oxide]

oh yea i remember seeing that briefly

wait every free module is projective or every projective module is free

the former right

[delicate orchid]

being free is just being projective without requiring the map involved to be surjective (and the induced morphism is unique in the case of being free but who cares)

[white oxide]

i see

[white oxide]

how does proposition 5.2 follow from proposition 5.1? i understand that stab(H) is the normalizer of H, so the order of the orbit GH should be equal to the index of the normalizer of H in G. but i'm not seeing how every element in GH is a conjugate subgroup of H

[terse crystal]

S={subgroups of G having same cardinality with H}, G acts on S by g•K := gKg^-1 for g in G, K in S

[delicate orchid]

Nearly just said “apply orb stab” without realising how useless that would be here

[vagrant zinc]

guys could someone help me with these exercises

[chilly ocean]

what exactly does it mean to say "\bar{x} = e^\varepsilon x is a Lie group?"

half asking because you probably have a definition to work from and half asking because i have no clue what that could be

[vagrant zinc]

It's like in group theory only I don't make sense of the first exercise since it is a composition and assumes a binary operation.

As in this example, I don't know from which sleeve this was taken. \widetilde{x}

I have done this exercise in this way

Sorry if I send a lot of screenshots, I feel it's easier to understand the question I guess.

[vagrant zinc]

[rocky cloak]

This is the weirdest notation I've seen I think, but the point is that you have some subset of R or R^2 that forms a 1-parameter group

And then f: R -> G is a group homomorphism such that

f(s + epsilon) equals the right hand side, when f(s) = x (or (x,y))?

If so I guess you would just plug and chugg to show that it's a group homomorphism

[terse crystal]

P is projective, therefore a summand of a free module of finite rank. k[x] is pid, Recall that submodule of a free module of finite rank over pid is still free

[chilly ocean]

If G finite group and H proper subgroup of G and H abelian, ord(G)=2ord(H) and Z(G)={e}

The problem says to show that every 2 elements in G\H dont commute

The index of H in G is 2 so H normal

[terse crystal]

Pick any a not in H, Notice that any element of G\H is of the form ah for some h in H

A pair of commutative elements in G\H=aH will lead you to a belonging to Z(G)

[chilly ocean]

Look at my try

And explain me what I did wrong

If we pick x in H{e}

|C(x)|>=|G|/2

But C(x) it is a subgroup of G and x≠e and Z(G)={e}

So |C(x)|=|G|/2

For every x in H{e}

Now from class equation

We have |G|=|Z(G)| + 2(|H|-1) + A, where A is sum of index of centraliser of in G where |orb(x)|>=2 and x in G\H

We have from here A=1

Which is false

[chilly ocean]

What is wrong in this plz

[terse crystal]

You mean x in H - {e}?

[chilly ocean]

Yes

[chilly ocean]

Btw this is from the fact H is normal cuz has index 2 în G

in*

[terse crystal]

Actually I didn’t need the fact H is normal. Simply H of index 2. Anyway I am still reading yours

[chilly ocean]

Ok ok

[terse crystal]

No idea how you got 2, to me |G|=|Z(G)|+(|H|-1)+A so A=|G|/2

[chilly ocean]

The index of C(x) in G is 2 when x is in H and x≠e

[terse crystal]

For any x in H - {e} The orbit of x is {x,y} two elements yes, but x,y are both in H

You count x then you can’t count y

2(|H|-1) a lot of elements overlapped

So it’s 2(1/2)(|H|-1) still |H|-1

[chilly ocean]

How the orbit is 2

[terse crystal]

You said yourself, [G:C(x)]=2

So orbit of x under conjugation is of two elements

But those two elements are both in H

H is normal as you said

[chilly ocean]

Right

Oh

Right

Thanks

[terse crystal]

Np

[chilly ocean]

Yeah so the mistake I did is that in class eq we need to say sum when |orb(x^)|>=2

Where x^={x,y}

Right?

[terse crystal]

Yeah

[chilly ocean]

Thanks

.close

[terse crystal]

Oh it’s not help channel

[chilly ocean]

Hahha

[chilly ocean]

How we can finish from here

If we pick two elements in G\H the product is in H

[terse crystal]

ah1 commutes with ah2 gives you a commutes with h, where h=(h2)(h1^-1)

Then a commutes with all elements

[chilly ocean]

How

By a bijection?

[terse crystal]

I probably thought something wrong, rethinking

[terse crystal]

Oh nvm, I fix it like this:

Two elements x and y commute , any a in G-H we have ah1=x, ah2=y, so a commutes with some h(a)=h2(h1^-1)

Now any b in G-H, similarly bh1’=x, bh2’=y, so b commutes with h(b)=h2’(h1’^-1)

But h(b)=h2’(h1’^-1)=(b^-1ah2)(b^-1ah1)^-1=b^-1ah(a)a^-1b=b^-1h(a)b therefore h(b)=bh(b)b^-1=h(a)

h(b)=h(a)

So for all elements in G-H we can choose a common h that commutes with all of them

so h in Z(G) contradiction

[vagrant zinc]

I have to do a degree project based on this topic and apply it to the Galois group.

The worst part is that this is not my forte, but I really like abstract algebra.

[chilly ocean]

How I can show that on every finite set M with ord(M)=2^k, we can define a Boole ring (M,+,×).

[delicate orchid]

M can be identified with the power set of some set with size k - take the boolean algebra of this power set and then pass to the corresponding boolean ring in the standard way?

does that work?

by "the boolean algebra of this power set" I mean the boolean algebra arising from the subset lattice of the set of size k- I'm not sure what the standard terminology is sorry

[molten rivet]

I'm struggling with part b, why should a cycle not change the product here?

[delicate orchid]

hint: conjugate by x_p

[molten rivet]

OH

that helps alot

thank you!

[delicate orchid]

no worries!

[errant shadow]

Can a group of order 25 have five subgroups of order 5?

I think this is false but I am not able to prove it. Can someone please give a hint?

or is this even actually false?

[coral shale]

hint: check Z25 first

[delicate orchid]

do you know anything about groups of order p^2

[rocky cloak]

It can have 6

[coral shale]

(no idea what goes on in here)

[rocky cloak]

There aren't that many groups of order 25, one of them has 1 the other has 6

[delicate orchid]

if you do, this is very easy to answer - if you don't then I'll let jagr explain lol I cba

[coral shale]

i think ik which one has 1

[errant shadow]

I did, it only has 1 right?

[delicate orchid]

correct

[errant shadow]

there's one which can have 6?

[delicate orchid]

now check C_5 x C_5

[errant shadow]

👀

[errant shadow]

woah thanks, I think I might be getting somewhere

[delicate orchid]

in fact you're done

those are the only two groups of order 25

don't know if you can use that fact or not

[coral shale]

sir, i request a proof

[errant shadow]

is there a specific reason why that's so?

[coral shale]

Have you done Sylow

[errant shadow]

I am following Dummit and foote, does it have theory related to it

[errant shadow]

nope

[rocky cloak]

Well, every group of order p^2 is abelian, and there is a classification of finite abelian groups

[delicate orchid]

jagr please stop sniping me on the internet I might start crying

[rocky cloak]

The second part is definitely something you should check out if you haven't. The first is a little more involved to prove

[delicate orchid]

here's my source nerds
https://books.google.co.uk/books?vid=ISBN9780198535485&redir_esc=y

[rocky cloak]

I was born a camper

Always have been

[coral shale]

infinity is a prime number, now what, sir

[errant shadow]

thanks

[errant shadow]

would do, thanks!

[rocky cloak]

Random question: is there a big focus on p-groups when you do this fusion system stuff, or will any group do just as well?

[delicate orchid]

They have to be on p-groups by definition

[rocky cloak]

I see

[delicate orchid]

Cause sylow is nice or whatever

[rocky cloak]

I have already forgotten the definition, but why doesn't it work with other groups?

[delicate orchid]

You could define the exact same concept on an arbitrary group I just don’t think the theory is developed or interest to nG

[rocky cloak]

I see

[delicate orchid]

interest to nG
Lmfao *interesting

Fuckin phone

[dim bane]

interesting to nGroupoid?

[delicate orchid]

Exactly

[rocky cloak]

Hmm, is there sylow theory for groupoids?

Does the question even make sense...

[delicate orchid]

Consider the p-completion

Only analogue that comes to mind

[rocky cloak]

https://arxiv.org/abs/2101.07420

Found this

[delicate orchid]

smooth operator

oh I see they use the yuck definition of a groupoid

[barren sierra]

I think I looked at this book for some stuff from my REU

[topaz solar]

There was that one text on finite Morley rank simple groups which suggested a sort of fusion thing for 2-Sylow kinda

[delicate orchid]

always banging on about the morely rank bro morely this morely that I can't take much morely of this

[crystal turtle]

Who actually thinks of groupoids as partially defined binary operationz

Awful

[delicate orchid]

pretty much nobody - I read the first sections of that paper completely thinking in categorical terms lmfao

the strange thing is I can't see what's made easier by the algebraic interpretation? perhaps it's in the section on sylow stuff

[topaz solar]

If only I knew enough to make a pun, but I’m simply not to that degree

[delicate orchid]

you couldn't sink sylow to make a pun

[south patrol]

We're morely and morely, woooooah

[delicate orchid]

Free if and only if iso to R^n for some n perhaps?

[terse crystal]

It’s a fact. A projective module is a direct summand of a free module

So it’s a submodule of a free module, therefore it’s free itself since k[x] is pid

[rocky cloak]

Structure theorem for PIDs

[flint crater]

Im a bit confused. Why is the affine n space over k written as A^n_k if it just means k^n? Atleast that is what I seem to be reading unless I'm misunderstanding it

[terse crystal]

Because A^n_k doesn’t mean k^n, it is spec(k[x1,…,xn])

[rocky cloak]

A space is more than just a set, it usually comes with other structure. In this case with a topology and a sheaf of functions

k^n also has structure that A^n doesn't have, because k^n is also a vector space. It is only by choosing a coordinate system you would get an isomorphism between A^n and k^n

[night onyx]

Just to differentiate it from k^n as a vector space, you think of A^n as a topological space (with the Zariski topology) where k^n is the underlying set, not as a vector space (even though k^n obviously is).

[south patrol]

Restrict scalars from k to F1

[delicate orchid]

MODS

[white oxide]

Can somebody explain to me how one would start on this problem? Let $M$ be the $\mathbb{Z}$-module of $\mathbb{Z} \oplus \mathbb{Z}$ generated by $m_1 = (1, 2)$, $m_2 = (2, 3)$ and $m_3 = (3, 4)$. Decompose M as a direct sum of cylic modules, and express $m_1$, $m_2$, and $m_3$ as linear combinations of generators you choose

[cloud walrus]

okeyokay

[white oxide]

Apparently you were suppose to come up with the generators for Z \oplus Z using a linear combinations of the m_is

So I lost all points on that question

[delicate orchid]

what do you mean by "the Z-module of"

[south patrol]

Z-submodule

[delicate orchid]

right ok

[white oxide]

oh yea

my fault

[south patrol]

or sub-Z-module

[delicate orchid]

uhhh smith normal? lol

[south patrol]

hmm

[white oxide]

yea

[south patrol]

[delicate orchid]

it is smith normal right

been a while :pack:

[white oxide]

i have no clue what that is lmfao

[south patrol]

Smith normal would work ye

i fink

[delicate orchid]

bezout's lemma 2345 times if it fails

anyway lets do this ad hoc

we can get (1,1) clearly (huzzah!)

[south patrol]

Basically you can view this module as like uhhh

[delicate orchid]

and (1,2)

so (0,1) is in there

[south patrol]

Take the map φ: Z^3 -> Z^2 given by mapping e_i |-> m_i basically and then M = Z^3/ker φ

@ okey

And then you can express the kernel in a nice way etc

But yeah uhhh

[white oxide]

wait I'm confused, so we had to know that M was Z \oplus Z at the start of the problem? because the answer was to show that (0, 1), (1, 0) was in M and was thus equal to Z \oplus Z

[white oxide]

oh damn

[delicate orchid]

yeah, that's what I was in the process of doing before I got distracted

[white oxide]

damn bro i wrote some really idiotic shit lmfao

[delicate orchid]

since (0,1) and (1,2) is in M then (1,2)-2(0,1) is in M so (1,0) is in M ergo M = Z \oplus Z

[white oxide]

sneaky tricky crazy

[delicate orchid]

what potato says is how you do this generally

third time's the charm

[south patrol]

oh is M just the whole thing lol

yeah fair

hm

i don't think it shoul be but maybe i am smoking

uhhhh

yes nvm ur right yes ignore me

[delicate orchid]

I've been told by the mark scheme that I am right

[south patrol]

lol

lol here is a dumb way

submodule of Z^2 is free of rank <= 2

jk

cause that doesn't give you the generators

[delicate orchid]

yeah it's a subgroup of a free abelian group and is thus free abelian

[south patrol]

prove that

[delicate orchid]

but we need da representaiton

[south patrol]

(idk the proof)

[delicate orchid]

https://tenor.com/view/no-i-dont-think-i-will-captain-america-old-capt-gif-17162888

[south patrol]

something with transfinite induction ig

iirc

[delicate orchid]

google nielsen-schrier nerd

[south patrol]

it's not nielsen-schrier though

[delicate orchid]

it's not it's way easier than NS

[south patrol]

Lol

I know how to prove NS though lol

i don't know this one

[topaz solar]

There’s a separate thing by Dedekind for free abelian

[delicate orchid]

structure theorem for f.g modules over a PID

[topaz solar]

Prove it without full choice

[rocky cloak]

The pressence of the word abelian helps a bit

[south patrol]

f.g.

[delicate orchid]

IT'S RANK 2 TF U MEANNNN

[white oxide]

ye but my question is how do we know that the action of conjugation by all elements of G on H yields H again bc isn't the orbit GH just equal to xHx^{-1} for all x in G

[south patrol]

I mean in general

😭

[south patrol]

Lol

[topaz solar]

make sure you email me the proof so that I definitely won’t write an article on the failure of NS to imply choice

[south patrol]

Actually where does choice come into the topological proof like just when constructing the "correct" covering space?

[terse crystal]

I didn’t say they are all H, the orbit is exactly {gHg^-1: g} as you said

[rocky cloak]

According to Wikipedia:

In the proof based on fundamental groups of bouquets, for instance, the axiom of choice appears in the guise of the statement that every connected graph has a spanning tree.

[crystal turtle]

Idk what the context is but if you take an infinite product it's probably there

[south patrol]

Oh yes sure

Yeah thanks jagr

[delicate orchid]

explicitly, (0,1) = 2(1,2) - (2,3) and then (1,0) = (1,2)-2(0,1) for the fans in the back

[white oxide]

ye but the proposition says the the orbit of GH is equal to the index of (G: Gs) which is the index of the normalizer of H in G which I understand, but if I substituted what they said in prop 5.1 for H then I get the orbit of GH is equal to {gHg^{-1} | g in G} = (G: N_H) so i guess i'm confused as to how {gHg^{-1} | g in G} consists of all conjugate subgroups to H (don't we require gHg^{-1} = H for it to be conjugate to H)?

i'm sorry this is prob a fundamental misunderstanding lmfao

[delicate orchid]

lemme take a look

cause I bet this is either orb-stab or burnside's

don't we require gHg^{-1} = H for it to be conjugate to H
absolutely not

do we require y = gxg^{-1} = x for y to be conjugate to x? no

by the very nature of being gHg^{-1} it is conjugate to H

[white oxide]

lol i should look at the def of conjugate again rip

[delicate orchid]

y conjugate to x means there is a g such that gxg^{-1} = y

replace y and x with subgroups it's the same

[terse crystal]

No it doesn’t have any thing to do with gHg^-1=H
G acts on X , the bijection G/Fix(x) -> orbit containing x is mapping gFix(x) to gx

[delicate orchid]

so the set of conjugates is (hopefully) clearly the orbit under the conjugation action

[terse crystal]

So G/Fix(H)->orbit containing H
gFix(H) |-> gHg^-1
Where Fix(H)= N_G (H)={g: gHg^-1=H}

So [G: N_G (H)]=|orbit containing H|=number of subgroups conjugate to H, I don’t see any question in it

[white oxide]

huh okay

i see

thanks

[south patrol]

Group theory is so hot

[terse crystal]

By the way your new question. You can calculate its smith form. That three rows two columns matrix

Oh they already said that

[white oxide]

damn i've never heard of that nor have we went over that in class

but thanks i'll ask in OH about that

[delicate orchid]

a temptress....

[white oxide]

also

so many questions

first of all what is M(x) 💀

as you can see i was copmletey fucking lost

[coral shale]

field adjoin i would guess?

[white oxide]

nah we haven't gone over fields yet lol

[rocky cloak]

Multiplying by the ideal (x) I guess

[delicate orchid]

gross notation lmfao

wtf is wrong with Mx

[rocky cloak]

Or xM

[delicate orchid]

I think I'm gonna be sick...

[rocky cloak]

#leftmodulegang

[delicate orchid]

that module is awful

[terse crystal]

Here k[x] is pid. So again smith form shows up, like considering the smith form of diag{((x-1)^2)(x^3), (x-2)(x^2)}

[delicate orchid]

x^4 = 2x^3+1
x^3 = 2x^2
scrumptious

oh ffs yeah of course u can just smith normal again

ignore me

[summer path]

M(x) looks so awful compared to (x)M

[white oxide]

dude how am i suppose to know how to do that when i've never encountered smith form in my life

damn that's crazy'

[delicate orchid]

basically by doing smith normal

but without knowing it

[white oxide]

wait is this related to jordan normal

[delicate orchid]

e.g. these are actually elementary row operation on a secret matrix

[white oxide]

are they the same thing

[delicate orchid]

no they are not the same thing

[white oxide]

damn

[delicate orchid]

different decomps

you can do smith normal on non-square matrices

which is why it's so good here

inb4 some nerd explains that ermm technically there's a generanlsation of JN form...

[white oxide]

is there anyway to do it w/o smith form?

smith normal

[rocky cloak]

You can just use Chinese remainder theorem

[white oxide]

fuck

[delicate orchid]

fuck

how did I not spot that LOL

[rocky cloak]

You don't know that one either?

[white oxide]

nah im not really in touch with my culture

[delicate orchid]

wait you don't know CRT

the ol R/I_1...I_n \cong R/I_1 ... R/I_n for disjoint blah blah blah

[terse crystal]

smith form of diag{f,g}=diag{gcd(f,g), lcm(f,g)} btw

[white oxide]

ok so i took a discrete math class about a year ago and basically goofed the entire class, saw it recently and read over it once but note nough to internalize it

[delicate orchid]

I dunno what you cover in "discrete math" I don't know what that is

did you do ring theory

realising this sounds like I'm drilling you

I'm just confused is all

[terse crystal]

Yeah discrete math sounds not very math to me, according to what other describe it, it seems to be a course gathering elementary results in different fields…

[white oxide]

yeah it's just the shit in #discrete-math lmfaoi

you know

covers proof-writing for 2 weeks

number theory for 2 weeks

probability theory for 2 weeks

etc.

then some graph theory

not enough time to internalize shit at all

[terse crystal]

Yeah graph theory, some simple Boolean algebra…

No idea what is the purpose of this course…

[white oxide]

a little dash here a little dash there

me neither

ig it's like

to expose you briefly to different concepts you might encounter in upper div classes but even then the premise is dumb

[rocky cloak]

I thought the point of discrete math was "here's the math you'll need for your computer science degree"

[south patrol]

isn't that notation defining M(x)

[white oxide]

yea that's true but im not a cs major

and it's also required for math majors

[terse crystal]

Ah no wonder some people mention things like “calculus 3” I have no idea what it is, but according to them it’s a course mixing calculus and linear algebra. I guess it’s for engineering, just like discrete math for CS

[white oxide]

wym

[south patrol]

sorry annotation i mean

though i don't really understand that annotation either since it has m^r

[terse crystal]

Oh yeah your picture

It isn’t full. Apparently you wrote the definition of M(x) near it.

We can’t see the full context of it…

(If it’s not (x)M…)

[white oxide]

yeah i thought that def was wrong tho... these were just my thoughts

i remembered some notation lang used and guessed it was that

[white oxide]

im confused, what does my prof mean by the group of automorphisms being a set of integers coprime to n? how are integers automorphisms? by left multiplication or whatever?

[delicate orchid]

yur

[white oxide]

no way that's crazy

thx

[delicate orchid]

multiplying on the left/right by any element of a ring yields an endomorphism of that ring

[white oxide]

in this sense then every single left multiplication is an auto right

oh yeah

endo imeant

so i dont have to check that, i think this becomes very easy if you know that stuff which aren't zero divisors are the integers coprime to n, which satisfies injectivity i believe

or maybe he wants us to show that integers coprime to n are not zero divisors to satisfy injectivity

oh wait

is there no multiplication structure since we're only considering the group

Zn

oh wait no i'm dumb

[delicate orchid]

ok no that's a fair point

[white oxide]

ah Zn as a left multiplicative group of integers coprime to n module

?

[delicate orchid]

no clue what that means

[white oxide]

me neither

[delicate orchid]

I'm just looking at Z/(n) as a ring

[white oxide]

what about Zn as a Zn module lmfao

[delicate orchid]

which then gives the action of Aut(Z/(n)) on it actual meaning

[terse crystal]

The author wants you to discuss U(Z/nZ)

[delicate orchid]

I was trying to dance around outright stating that multiplication is an automorphism iff ur multiplying by a unit but meh

[terse crystal]

Oh

[delicate orchid]

tbh they realised it couldn't be a zero divisor which for Z/(n) is good enough

[white oxide]

lol maybe he meant ring but after all

🤯

[terse crystal]

Group

Where binary map is composition

[white oxide]

i'm glad he gave us #1 that was a fun warm up

[white oxide]

wdym

oh

wait why couldn't we justinterpret it as Zn as a Zn-module lmfao

am i high

[terse crystal]

He wasn’t talking about Z/nZ

[white oxide]

oh

[delicate orchid]

??? ok now I'm confused

[white oxide]

wait huh

?

[terse crystal]

He was talking about Aut(Z/nZ), group binary operation defined by composition

[white oxide]

OHH

i see what u mea

n

wait

yeah that's how i interpreted the question

[terse crystal]

Aut(Z/nZ) is isomorphic to U(Z/nZ), group of integers coprime to n mod n under multiplication

[white oxide]

so yea he's saying that Aut(Z/nZ) are integers coprime to n

[terse crystal]

So multiplication

[white oxide]

ohh so he wants to show the isomorphism basically

[delicate orchid]

yes

[terse crystal]

Yeah

And he wants you to look into cases when n=p and n=9

[white oxide]

so i just send a to left multiplication by a i'm assuming....

[terse crystal]

A homomorphism f from G to H, where G is cyclic

having a generator x

f is determined by f(x)

[white oxide]

what's the standard notation for the units in Zn again?

Zn^X or smt

[terse crystal]

U(Z/nZ)

[white oxide]

cogwheels you say U(Z/nZ) right

hm i've never seen that b4

kinda trips me up the U is like a union

[terse crystal]

Or Z/nZ *

Or Z/nZ ^ multiplication symbol

Never mind just U(Z/nZ)

[delicate orchid]

(-)* is morally correct (funny adjoint to R[-] or whatever)

[white oxide]

bruh

[white oxide]

this one is based

[terse crystal]

A* or A^ ❌ are usually for when A is a field , I guess

[delicate orchid]

I've literally only ever seen U(-) used for Z/nZ

it's so the little undergrads don't CRYYYYYYYYY WAHHHHH

[terse crystal]

I see Jacobson use U(A) to denote units of A

Don’t know what notation Lang uses though

[delicate orchid]

actual beast of the beneath

[white oxide]

damn bro i'm tearing up rn

[terse crystal]

U(Z/nZ) is fully discussed in 4.11 in Jacobson btw if you want to know all the cases

[white oxide]

alas

my downfall

surjectivity requires some knowledge of elementary number theory

of which i have none

it's equivalent to stating the existence of an x such that ax \equiv y mod n for y in Zn given and a coprime to n

[delicate orchid]

it's clearly invertible

[white oxide]

and this

oh

[delicate orchid]

you're multiplying by a unit

[white oxide]

AYOO THAT'S FACTS

damn bro i see you in #advanced-number-theory i see you

[delicate orchid]

you most certainly will NOT! that place STINKS!!

[white oxide]

damn

[rotund aurora]

does every field K of char 0 admit a proper valuation ring R

so for every x in K either x in R or x^{-1} in R, but K!=R

[next obsidian]

As long as K admits a subring of dimension > 0, the answer is yes

Take R such a subring, consider R_m for m not height 0, and then we have a local subring of K

Then K is dominated by a valuation ring

[next obsidian]

This is gonna be equivalent to having a valuation ring because other direction is trivial

Oh

It’s char 0 so Z < K

Lol

[rotund aurora]

I thought char 0 because in char p you can take Z/pZ

[next obsidian]

Wait no

Not true no

Because Frac(Z) ≠ K necessarily

[rotund aurora]

yeah right

I was also thinking about localizing, but that's an obstruction

[next obsidian]

So what I said was wrong, you need a subring of dim > 0 with field of fractions K

[white oxide]

can somebody explain to me how r1, \dots, rk are relatively prime? r4 and r6 will have a common divisor of p_1^{n_1}, so that they aren't relatively prime? what am i misinterpreting

[next obsidian]

They mean the entire collection

So no prime divides all of them

[white oxide]

oh i see

thx

[rotund aurora]

what did dim>0 meant isn't this the same as saying that R is not a field?

[next obsidian]

Yeah I guess so

[rotund aurora]

ok

and you can definitely find a non-field R such that K is the field of fractions of R right

[next obsidian]

I think so

I think if you took the integral closure of Z inside of K that might work

But also, idk

Or uh

Do like, take K as an extension of Q

Grab generators

Adjoin them to Z

[rotund aurora]

isnt the integral closure of Z in Q(x) just Z lmao

[next obsidian]

That probably works

Yeah so adjoin variables so that like K/Q(shit) is algebraic

And then adjoin as many variables to Z as needed

[rotund aurora]

maybe look at K^x as a group, and find generators

[next obsidian]

No that’s complicateder

[rotund aurora]

then take the ring generated by those, but dont take inverses

but works

xd

[chilly ocean]

Hi

[next obsidian]

Take a transcendence basis S for K over Q

[rotund aurora]

yeah right

[next obsidian]

Then Frac(Z[S]) = Q(S)

So K/Q(S) is algebraic

[rotund aurora]

yeah

[chilly ocean]

Hello my Name is tom

[next obsidian]

Now you should be able to like take an integral closure

[chilly ocean]

Wat

[next obsidian]

Or just even take generators for K over Q(S)

And adjoint tbem to Z[S]

[rotund aurora]

ye

thanks

[quiet pelican]

Are there any simple examples where $H \times \bZ \cong K \times \bZ$ but $H$ and $K$ are not isomorphic, where $H$ and $K$ are groups?
The only one I can find online leaves me thinking “Ok. How was I meant to think of that?”

[cloud walrus]

Micose

[white oxide]

how does p^nc in p^nA - 0 imply that there is a least positive integer j such that p^jc \in Ra? is it just saying that the set of positive integers such that p^kc are in Ra is nonempty and thus by the well-ordering principle there exists a least positive integer with that property?

[rotund aurora]

so reading about valuations I find some people dont care too much where the value group lives, so I was wondering about the following (see image for clarity). So then the question would be to classify the fields K such that rank(K)=1, etc. Is this something known?

EDIT: uh wait I messed up the definition of rank(K), I meant the smallest Omega that works for all R lol

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

more context:

so my main question is when does G(R) embed into \mathbb R for all valuation rings R of a given field K

This is the same as saying that for all x, y in R such that 1/x,1/y not in R there exists an n in N such that x^n/y in R, and this should be true for all valuation rings R of K

[rotund aurora]

uh wait I messed up the definition of rank(K), I meant the smallest Omega that works for all R lol

[quiet pelican]

How would you motivate the solution that I’ve found to the problem?
It very much feels like something you’d never even think to think of

[south patrol]

I am curious too cause this was presented to me as like an olympiad style question

And apparently it is true if G, H are abelian

[rocky cloak]

I guess what's happening is you can think of G as a semidirect product between C1023 (there's a typo in the answer it appears) and Z^2.

And while (1, 0), (0,1) and (2, 5), (1, 2) are both free bases for Z^2 the action that each basis vector has on C1023 is quite different.

But yeah, it's a kinda crazy example

[quiet pelican]

It’s trivially true if they’re fg abelian

Well
Not too trivial

But like
I can prove with little effort

Then there’s kinda my implicit question
Is there a nicer example?

[south patrol]

Yes

[rocky cloak]

This example is very minimalistic though. Like I'm not sure what a simpler group could be that isn't abelian

[quiet pelican]

It’s also true if they’re finite not necessarily abelian I’ve seen

[rocky cloak]

I think maybe you can bring that 1024 down to 32, unless there is something I'm missing

[terse crystal]

Yeah since the only finite subgroups of H product Z are H’ product {0} I think

[cursive spindle]

Can anybody give me an example and a counter example of prime ideals in ring theory?

[chilly ocean]

(2) is prime, (4) is not

[cursive spindle]

What does () denote for again?

Cyclic?

[chilly ocean]

ideal generated by the guy inside

[cursive spindle]

So you must have a commutative ring right

[chilly ocean]

these are ideals in Z

[cursive spindle]

You assume R is commutative with 1 then P is prime ideal in R iff R/P is an integral domain

[chilly ocean]

yes

[cursive spindle]

I'm having difficulties understanding the quotient

[chilly ocean]

what about it

[cursive spindle]

I don't really get it

[coral spindle]

Have you seen quotient groups?

[night onyx]

Remember that a + P = 0 iff a in P. So somehow "belonging to P" translates to being zero in R / P

[rocky cloak]

You can generalize the notion to noncommutative ring, but generally when people talk about prime ideals they mean for commutative rings

[night onyx]

so the idea of being prime involves ab belonging to P, and being an integral domain involves ab = 0

which is why P being prime (meaning ab in P iff a or b in P) translates directly to R/P being an ID (ab + P = 0 iff a + P = 0 or b + P = 0)

[cursive spindle]

Yes Z/6Z

[coral spindle]

OK, so ideals are like normal subgroups. If you ignore the multiplication on the ring for a moment, quotient rings behave exactly like quotient groups

The additional requirement that they be closed under multiplication from the ring is exactly what is needed to make this a ring, and not merely a group.

The way to work with quotient rings is therefore exactly the same as one would with quotient groups. Think of the things that are 0 in the quotient ring R/J as exactly those things that are in the ideal J.

[cursive spindle]

A normal subgroup is like having a group then you divide in into n groups right?

[coral spindle]

I do not know what you mean by that.

[cursive spindle]

I'm trying to think of it in terms of diagrams

Or maybe I haven't familiarized myself with examples enough

Anyhow I'm learning about these because I'm starting alg geo soon

[crystal turtle]

Hope this helps 👍

[night onyx]

that's more the idea of a quotient G/H, you take a group G and cut it up into copies of H. The reason why we care about normal subgroups is because the set G/H becomes a group itself when H is normal

[cursive spindle]

What does it mean when H is normal

[coral spindle]

Hold on a moment

Have you seen the definition of a normal subgroup before? Are you just asking for a reminder?

[night onyx]

normal means gH = Hg for all g. But really this definition makes not a lot of sense unless you think about what's required for G/H to be a group

[crystal turtle]

If you're not comfortable with normal subgroups and quotient groups, it would probably be best to review that before working with ideals and whatnot

[night onyx]

it's a really important exercise to take a subgroup H, consider G/H, and ask yourself "well I can multiply cosets aH * bH = (ab) H, but does this even make sense? Is it well defined?"

That's really how you arrive at the idea of normal subgroups

otherwise being normal is a REALLY weird definition

[coral spindle]

This wall is so talkative

[night onyx]

I really like the idea of classifying different types of ideals in terms of the properties of quotients, like prime corresponding to integral domains, maximality corresponding to being a field, even just the quotient having the same structure corresponding to being a kernel, it's cool

[coral spindle]

Me too :) another cute one is that an ideal is radical iff the quotient is reduced

[cursive spindle]

It's probably best to review quotient groups and normal subgroup before going into rings

[night onyx]

yeah, and you can ask so many questions like that, "if R/P is an {insert type of structure}, then P is a {insert type of ideal}"

[coral spindle]

Yes.

[coral spindle]

I hope you understand that this question was trying to gauge exactly that

[delicate orchid]

if R/P is a ring then I is an ideal :3

[night onyx]

I remember thinking about the first isomorphism theorem just starting from regular set functions f : X -> Y. Like you consider the equivalence a ~ b iff f(a) = f(b), realize that f(X) ~= X / ~ as sets, then ask what happens when f is a group homomorphism, realize the equivalence classes are all copies of the kernel, blah blah

that was kindof an "oooh shit I get it!" moment

[coral spindle]

Right, and this is exactly what the first iso in universal algebra says in this case

[rotund aurora]

@cursive spindle what is your motivation/interest for studying alg geo? Just curious

[cursive spindle]

Complex geometry, Diophantine equations and enumerative geometry

And much more

I think I like complex geometry more than real geometry

[rotund aurora]

cringe

so cringe

honestly I dont even know what complex geometry is

like ik what Riemann surfaces are and so on, but where's the geometry

[peak root]

A nice question

[rotund aurora]

I should be more open minded probably

[peak root]

I am starting to questioning the algebraic geometry. Where is the geometry?

Why am I always trying to transform a question in algebraic geometry into a question of algebra or cohomology

then i am just doing algebra

[rotund aurora]

ok if you pick Euclid and show him algebraic geometry or even just modern geometry for the first time he will be like

but something similar would happen with modern number theory

[peak root]

I mean anything trying to touch algebra will eventually be taken over by algebra.

look at algebraic topology. you start with some nice pictures and imagination, then you open a book on homotopy theory, just loads of diagrams and sequences.

OK. maybe need to move the channel to vent. sorry for that

[rotund aurora]

is there any notation to denote some lifting of an element of some quotient?

[delicate orchid]

I've seen putting a little tilde over it

or a '

[south patrol]

I like hat

Lol

[rotund aurora]

what about when you are lifting a function or a polynomial

it may be confused with the derivative

[white oxide]

by find a decomposition of this group into cyclic groups for n = 9 do they mean give an isomorphism with a direct sum of Z_p^is or whatever

ykwim

fundamental theorem of finitely generated abelian groups ya

[delicate orchid]

HIP HIP HURRAY YOU HAVE WON THE DAY!!!

[white oxide]

DUDE I'M A FUCKING GENIUS

[delicate orchid]

genius you may be, but an answer I do not see

[white oxide]

$\mathbb{Z}_3 \oplus \mathbb{Z}_2$ i hope.

[cloud walrus]

okeyokay

[delicate orchid]

or as I like to call it, Z_6

[white oxide]

ah right they're iso because (3, 2) = 1

holy shit i'm on fire today

[south patrol]

gotem

[coral shale]

3 + 2 is not 5

[white oxide]

Why is $(\pi(\tau)f)(x_{\sigma(1)}, \dots, x_{\sigma(n)}) \neq f(x_{\tau\sigma(1)}, \dots, x_{\tau\sigma(2)})$?

[cloud walrus]

okeyokay

[white oxide]

also am i wrong for thinking this is cursed notation

[topaz solar]

It’s plugging in the sigma shuffled indices to the already tau shuffled function?

Maybe

[delicate orchid]

so true bestie...

[coral shale]

in groups, rings, whatever, you can consider the substructure generated by a set.
You either define it to be the intersection of all substructures containing that set, or you construct it by generation using the structure axioms.

What exactly do we call this phenomenon where the 2 are equal. Like some sort of compactness?

[delicate orchid]

not sure if it has a name

it just means that the underlying set was already a subgroup/subobject

[coral shale]

nuh, i mean like <S>

[delicate orchid]

oh

[coral shale]

<S> = intersection of all subgroups with S in it
= {words formed by S}

[south patrol]

Hm I mean not compactness but like

[delicate orchid]

this smells like model theory

[south patrol]

This is basically just one fact right

[coral shale]

well do we see it outside algebra?

[south patrol]

The smallest thing containing a set is what you get by iterating the stuff it has to be closed under

[delicate orchid]

the smallest dude is the one contained in all the other dudes

[coral shale]

why am i not finding this obvious
ill have a think

[south patrol]

Topology generated by a set of subsets of a set ig too

[coral shale]

right basis things

or subbasis in that case

[topaz solar]

Inductive sorta shenanigans

[coral shale]

uhhhh

I guess it boils down to

{words formed by S}

this thing

[coral shale]

has to be a substructure for this to work

[topaz solar]

This kind of induction

Ye, if it’s a substructure, then that immediately gets you what you want

And if that’s all words formed in your language + S, then any operations = words made out of elements of it are in it

[coral shale]

😵‍💫 troo

[topaz solar]

This isn’t always a substructure if you have extra conditions though

Like some existential shenanigans and all

[coral shale]

so if it isnt, at worst, this must be... a subset of your goal

itll be smaller

[topaz solar]

In particular, there’s some statement about how this sort of substructure thing works iff it’s universally quantified stuff

I forget the exact one

But think how substructures of fields need not be fields

Universal algebra has entered the chat

[coral shale]

bro wut

[topaz solar]

Boytjie probably knows what I’m referring to here

[topaz solar]

There’s no inversion function symbol

That kind of thing

Only that an inverse exists for all nonzero

[coral shale]

ill have a thonk, i think i recall this

[topaz solar]

So no words with inversion

And if you have relational symbols, words with them are kinda

[coral shale]

i mean for fields

say your set is {a, b, c}

cant u have like

words like this (a + bc - a^-1)^-1, and then chuck out the non well-defined ones

[topaz solar]

Well there’s no ^-1 symbol

[coral shale]

ok i think i get u

[topaz solar]

Also for finite things consider Wedderburn’s theorem

[coral shale]

theres none in the categorical sense or something right

because this symbols has to apply to every element

[topaz solar]

There’s no symbol in the universal algebraic sense, or model theoretically

Since we want M -> M

[coral shale]

yh sure

[topaz solar]

This can be turned into a categorical sense too

Since ya know

Commutative diagrams for the axioms

which works kinda iff it’s universally quantified stuff since haha LEM?

[delicate orchid]

I knew it was model theory

[coral shale]

oh yh i was kinda thinking about this the other day
rings are endomorphisms of abelian groups
but fields are not quite automorphisms of them...

[topaz solar]

I remember it now

Substructures are also models iff universal formulas

[crystal turtle]

Uhh assuming they're actually subobjects in the category, the statement that each subobject of X is the intersection of all subobjects larger than it would be something about the poset (lattice) of subobjects of X

[topaz solar]

Consider fields as comm rings + existential for inverses and all that

That doesn’t preserve into subring s

But the universal commutativity does

[crystal turtle]

altho that might not precisely correspond to generation from a subset since that might be icky to state from a category pov

[topaz solar]

That lattice thing is pretty related here, and intersection is product in the subobjects

Thing is: predicates are also subobjects

And if your thing isn’t universal then you’ll run into issues

*this is a subobject in a slightly different sense, and passing back and forth between the structures and sets, but it still more or less holds there too in a sort of internalized sense?

[crystal turtle]

yeah okay this might not work as well to state as i thought at first lol. But you might be able to state something about lifting subsets from Set (subsets as subobjects of UR, under th eforgetful functor U) to Ring or whatever through intersections blah blah blah idk

?

[topaz solar]

Think how ideals are predicates of divisibility

[topaz solar]

Pretty sure some wackyness like this is related ye

[crystal turtle]

Since we know that intersection of subobjects as sets is a subobject again in the category (and the intersection in the categorical sense), this is kinda the idea? Since then intersection commutes with the forgetful functor I guess (up to some isomorphism or something)

[topaz solar]

Something something HSP theorem

Subobjects of fields

Term shenanigans (the words made out of applying operations to a set) working is quite tricky for if you have existentials or relations

But universal & function only stuff would have it be a substructure and hence a submodel?

[sonic coral]

I am trying to show that D12 (order 12) is isomorphic to D6 cross Z/2Z. I have elements a,b in the direct product group that satisfy the presentation for D12, so i guess i have a homomorphism. I’m just not sure how to argue that this is in fact an isomorphism, i only need to show one of injective or surjective.

[delicate orchid]

I don't believe this isomorphism, the centre of D16 is order 2 the centre of D8 x Z/2Z is order 4

[sonic coral]

oh i’m sorry. you’re right.

i have shown that they weren’t isomorphic by counting elements of order 2

[topaz solar]

Btfo’d instantly

[sonic coral]

let me fix it

[coral shale]

ummm

[sonic coral]

i think they are isomorphic, right?

[delicate orchid]

it's a semidirect product with C_2 I believe

[coral shale]

have u been given this question

or r u making this up

[delicate orchid]

definitely not direct - the centre argument holds again

[coral shale]

like im not sure any dihedral apart from small ones is a non trivial direct product

[sonic coral]

i was given this

Show D12 is isomorphic to D6 x Z/2Z

[coral shale]

is this a prove or disprove

ok

[sonic coral]

is x not cross?

[coral shale]

well wew at least seemed to think ofherwise

I certainly wasnt rlly aware of this kinda stuff tbh

[sonic coral]

the hint for the problem was to find elements in the direct product group that satisfy the presentation for D12, which i have done

a = (r^2,1) and b = (s,1)

where b^2 = 1

[delicate orchid]

oh wait sorry, D_6 is order 6

so it's just S_3

right I believe this could work

so D_12 = <a, b | a^6 = b^2 = 1, bab^-1 = a^-1>

[sonic coral]

yeah this book is using D2n but i’m used to Dn so i tried to specify

[delicate orchid]

D_6 x C_2 = <a, b, c | (ac)^6 = b^2 = [b, c] = 1, b(ac)b = (ac)^-1>

naively

yeah ok so the C_2 goes "into" the copy of C_3 inside D_6 to turn it into a C_6, then you still have the semidirect C_2

I completely believe this now

[coral shale]

how does this generalise?

[sonic coral]

Is Cp the cyclic group of order p

[delicate orchid]

yus

[sonic coral]

now that you believe me 😂 how do i argue that the kernel is trivial

[coral shale]

oh
D2k where k is odd should work im guessing

[delicate orchid]

I don't think that's the right map

[sonic coral]

you think the statement isn’t true?

[delicate orchid]

from the top
D_6 x C_2 = <a,b,c| a^3 = b^2 = c^2 = [a.c] = [b,c] = 1, bab = a^-1>
= <a,b,c| (ac)^6 = b^2 = 1 = [a,c] = [b,c] = 1, babc = a^-1c>
= <a,b,c| (ac)^6 = b^2 = 1 = [a,c] = [b,c] = 1, b(ac)b = (ac)^-1>
= <ac = w, b|w^6 = b^2 = 1, bwb^-1 = w^-1>
D_12 = <x,y| x^6 = b^2 = 1, bxb^-1 = x^-1>
not very rigorous (why must ac be order 6, what to do about [b,c] before applying the tiestze transformation in step 4? whatevs.)
going off of this I think the map you want is x -> (a, c), y -> (b, 1)

surjectivity of this map shouldn't be too hard

[sonic coral]

what are you doing in all the presentations, like what’s the purpose

trying to think of a map?

[delicate orchid]

yeah

anyway try that map and see what happens

[sonic coral]

just so i’m following, i map x,y in D12 to (r,1) and (s,0) respectively

[delicate orchid]

uhh yeah

maybe (s,1) would have worked actually

I thought you were using 1 to denote the identity

[sonic coral]

i’m C2 i’m using 0,1

[delicate orchid]

then sorry for going on this massive diatribe

[sonic coral]

dummit and foote says something useful about this maybe

,rotate

[sonic coral]

from what i’m gathering, since the elements i have found satisfy the presentation it is surjective?

[delicate orchid]

yeah by the substitution theorem

*test

not theorem

[sonic coral]

i’m not sure what that is

[delicate orchid]

swag

[sonic coral]

or do i need to show that the elements i’ve picked out of the direct product group that happen to satisfy the D12 presentation generate the direct product group

[delicate orchid]

that is a one liner if you do

[sonic coral]

yeah we havnt learned the substitution test

[delicate orchid]

that's alright

[sonic coral]

so do this you think?

[delicate orchid]

it's very obvious that they do, but sure go show it anyway

you don't need my permission

to quote genesis 4: "Do I look like my brother's keeper?"

[topaz solar]

Ew reverse 🇵🇱

That quantifier in parenthesis hurts

The way the () slant

[delicate orchid]

stfu nerd

[crystal turtle]

bad

[delicate orchid]

please see my previous message for my response! Ok ttyl hugs and kisses xoxoxoxoxoxo

[crystal turtle]

[coral shale]

whats theta^n, is this n just a scripting letter

[timber blaze]

Is it true that a group is the union of its cyclic subgroups?

[delicate orchid]

Yeah, although I’ve never used that fact lol

[timber blaze]

thank you

[coral shale]

what do we call the general structure of $(\bR_{\geq 0}, +, \cdot)$

[coral shale]

like a semifield?

[delicate orchid]

yeah

[coral shale]

(for no particular reason) i want to say the vector norm is a morphism of one of these kinda. but not quite

like kernel is 0

but u have subadditivity

[delicate orchid]

the vector norm doesn't map out of a semifield, and need not preserve addition

[coral shale]

oh right, we dont even have f(v) = 0 <=> v = 0 making any of these things necessarily an injection

inner product, norm

[delicate orchid]

that's pretty much the only property a norm does have that we want

[errant shadow]

Some texts mention that 'Such and such groups have such and such subgroups up to isomorphism'. I get what they mean by that, but what else is there to consider apart from isomorphism when we are trying to find the number of subgroups? And can that change the number?

[night onyx]

if they say something holds in a group G "up to isomorphism" it means it's the same between all groups which are isomorphic to G, and the number of unique subgroups is one such property, yeah

[errant shadow]

I get that but I am asking why 'upto isomorphism'? Why do you have to specifically mention that? Is there anything else out there that, when taken into consideration can change the number of subgroups?

[delicate orchid]

could you give the actual exerpt

because this could either be what kryojyn is saying or that it contains an isomorphic copy of some other group as a subgroup

[night onyx]

I mean as an example, there are many 'different' groups of order 2

so you can't say there's only 1 group of order 2

[errant shadow]

yea that's the context

[night onyx]

but they're all isomorphic

[patent girder]

In group theory, one of the things we wanna do is find every group. This is quite literally impossible, but we can find the interesting ones. In doing this we restrict to isomorphism

[delicate orchid]

it's less that they're "not interesting" and more that we just literally don't care about what the elements specifically are, just the structure

[night onyx]

if you want the mathematical precise definition, you can look at isomorphism as an equivalence relation on groups, and a property P(G) holds up to ismorphism if it's well defined on the equivalence class of G

[patent girder]

well yea, this is what i mean by interesting. It's not interesting to find out {1,-1} exists if i know about {1,a | a^2 = 1}

[delicate orchid]

I can say Q_8 is a subgroup of SL(2,3), but this is annoying to work with if you don't also have an embedding of Q_8 into SL(2,3)?

[void cosmos]

let M be a quasi-projective variety

jk jk

[delicate orchid]

tbh I don't really understand what you're asking at all

what's the actual thing you're not understanding

[night onyx]

for a broader context, almost all of group theory is concerned with properties of groups which hold "up to isomorphism", like those are specifically group-theoretic-properties

[errant shadow]

No I mean I get the point that you want to study the structure and I get why you'd say upto isomorphism. But I was asking if embeddings, automorphisms also affect that number? I am in the middle of an introductory group theory course so yea, taking some time to convince myself.

[delicate orchid]

ah so you want the number of embeddings, then yeah it does

take my Q_8 example, precomposing any embedding with an automorphism of Q_8 gets you another embedding

[errant shadow]

ahhh

[delicate orchid]

as does post-composition with any automorphism of SL(2,3)

but that's not general I don't think

actually no I'm not sure that's even true here - it is for inner automorphisms (by Sylow) but again that might not happen generally

so just ignore

[errant shadow]

oh well I'll just check out cases where it holds

atleast now I have some more context

[patent girder]

I, presonally, like to think about (Z/2Z)^n (mostly because i'm lazy at times). Maybe this'll help you

[errant shadow]

👍

[coral shale]

can we have an evaluation map for K(x)

or just nah

like howd it be defined

[delicate orchid]

x -> whatever

[crystal turtle]

I don't think so, but you could maybe do like a "partially defined" homomorphism or soemthing

[delicate orchid]

oh right you want one for each ring

[coral shale]

looks ugly

and sad.