groups-rings-fields

[ripe harbor]

My book defines a group action as transitive when there is only one G-orbit. Does the name indicate transitivity in some sense of something?

[copper kestrel]

does this work btw

bats my eyelashes

please say yes

[marble hinge]

I tend to think about it like this. If it’s just one orbit, all the set elements are connected with action, they are all in one equivalence class. So, in a way, you can transit the whole set using group actions :) So it is transitive

[ripe harbor]

oh

[marble hinge]

But I have no idea if it’s indeed related to the actual reason of why it is named like that

[ripe harbor]

Thats good, thanks!

[marble hinge]

If it’s not transitive: it will have two or more disjoint graphs (two orbits, or more)

So you can’t transit it in one go

[cloud walrus]

ILikeMathematics

ILikeMathematics

[ripe harbor]

Oh yes

[cloud walrus]

ILikeMathematics

[ripe harbor]

Yeah ok that works out, thanks!

[small yacht]

How is (R, +) isomorphic to (C, +)?

[proud vigil]

https://math.stackexchange.com/questions/337084/are-the-groups-mathbbc-and-mathbbr-isomorphic

[marble hinge]

That’s actually a fun problem to show that there always exists an element g of the group that does this cycle. I.e. that if you keep acting with just this element you will traverse the whole set!

It’s called fixed-point free element (because it doesn’t fix any elements of A: no fixed points)

[ripe harbor]

You mean for finite groups?

[next obsidian]

Both are vector spaces over Q of the same cardinality, the cardinality of the continuum

This makes them isomorphic as vector spaces which is strictly stronger than being isomorphic as abelian groups

[small yacht]

I don’t know this stuff

[next obsidian]

Vector spaces are isomorphic iff they have a basis of the same size

Cuz R is infinite it’s bijective to R^2 which is what C is as a set, and cuz these are both really big compared to Q this is enough to say the size of their basis has to be as large as |R|

This means there’s some isomorphism between them as vector spaces, and part of that isomorphism is a bijective map respecting +, so it also is a group isomorphism

[small yacht]

Uhm
I think I conceptually got it but still
My textbook asks this with exclusively group theory background so I think a group theory explanation would probably benefit me more

[next obsidian]

Linear algebra is gonna be assumed for group theory

I don’t think you can argue this very directly

[small yacht]

It didn’t really mention it at all up to this point

[next obsidian]

I mean usually if you’re studying group theory you’ve already done linear algebra

[small yacht]

alright
Thanks anyway

[rocky cloak]

Strictly stronger, but also equivalent

[next obsidian]

I’m pretty sure it’s gonna be almost impossible to produce an isomorphism that doesn’t exploit their vector space structure

[small yacht]

It didn’t really talk about vector spaces relating to homomorphisms either😭

[next obsidian]

I was thinking of qualifying my statement that I think maybe by some fuckery it’ll be equivalent but I wasn’t sure

[rocky cloak]

The magic of Z -> Q being epi

[small yacht]

Well, you may be right actually, it’s phrased in a way that I think allows for the possibility of it being undecidable

(As in, undecidable without using linear algebra)

[proud vigil]

is that aluffi

[small yacht]

Yeah…

[proud vigil]

i remember that on my homework, i dont think we were expected to come up with an answer

imo at least the main idea was that none of the tools from that chaoter work on (R, +) and (C, +)

so you cant determine from those tools and need a different approach

[next obsidian]

You should check Exercise VI.1.1

Also man this is a throwback for me

I think Gomez told me this fact ages ago with this proof and I was like how tf are they isomorphic lol

And it blew my mind, I was like this is fucked up

[small yacht]

Yeah I would totally say they’re not isomorphic if I didn’t see it say that they are later

[novel star]

does this require choice?

[chilly radish]

Yes

[novel star]

ew

[chilly radish]

I looked this up and apparently it is consistent with NOT choice that R and C are not isomorphic

@next obsidian @small yacht this might interest you

So in a strict sense you CANNOT prove isomorphism as abelian groups without appealing to the vector space structure

[novel star]

lowk choice always pmo when it shows up in algebra

[chilly radish]

Algebra has choice all over the place. You just don't really see it at first

[novel star]

oh i know

[chilly radish]

Some of the mods fundamental theorems in algebra, such as every ideal being contained in a maximal ideal are equivalent to choice

[novel star]

it just seems like the constructions are too strange

i remember being really annoyed when showing that projective modules over local rings are free bc it required choice

[chilly radish]

Non-constructively asserting the existence of stuff is a bit weird I'll concede but constructions using Zorn's lemma are generally pretty easy to understand

[fading field]

one thing that always made me feel better is just using zorn’s lemma instead of explicitly calling it choice

[next obsidian]

I kinda figured

[novel star]

true

but it always feels like trying to dig a hole by plucking atoms

[next obsidian]

Zorn’s Zorn’s Zorn’s Zorn’s

[chilly radish]

Where do you need choice in that proof (also you forgot f.g.)

[novel star]

not finitely generated

all

[chilly radish]

Wait is that true

[novel star]

kaplansky showed this

[chilly radish]

Ok that makes sense to use choice then

My bad

[next obsidian]

Yes its a transfinite induction

It’s in Matsumura if ur curious

[chilly radish]

I might look at that later, thanks for the reference

[next obsidian]

There’s some fascinating interactions with set theory and weird homological stiff

[chilly radish]

I should teach a commalg class so I can brush up on my commalg. Maybe some children here would like this

[chilly radish]

Wdym

[next obsidian]

There’s a paper called something like “big projectives are free”

I think the result is something like a projective module whose rank is higher than some bound involving the cardinality of the ring are all free

[fading field]

this is notes from the underground right

[chilly radish]

Oh I see

[fading field]

not chapter 0

[chilly radish]

God Aluffi has such cringe book names

[fading field]

lol

[chilly radish]

Too bad since he's a good writer

[fading field]

yeah i used to shill aluffi so much. i actually have managed to taper my opinion of his writing as time has gone on

[chilly radish]

That's really interesting

[chilly radish]

You're probably aware but apparently there's a page about this on stacks about one of the results from that paper

Of course it's a bass paper he does all the weird fucky stuff

I feel like whenever there's a weird fucky kinda set theoretic result in algebra it's always him

[swift root]

notes from the underground sounds like it would be some hiphop adjacent thing lowkey

[chilly radish]

It gives me kinda big

https://tenor.com/view/maximo-espectacular-dardano-gif-20543566

[swift root]

a kinda low quality non-genuine cash grab but still

[chilly radish]

vibes

[swift root]

yeah lmao

[swift root]

lords of da underground (lords of da underground)

[tardy hedge]

I heard you were new here, hi!

[swift root]

oh yeah im new no one knows me

never been here before, in fact

[tardy hedge]

Welcome! This is a friendly space to share all your curiosities and questions about Mathematics. I'm sure you'll fit right in here, champ. 😊

[swift root]

i genuinely cannot tell if your3 doing a bit Kian 😭

[tardy hedge]

🙃

[swift root]

coPOV: shielding myself from the sun

[small yacht]

Nope

[small yacht]

It’s a reference to a Dostoevsky novel😭

[swift root]

oh well my bad

i guess its fitting that this is what first came to my mind though lol

[next obsidian]

About big projectives being free?

[chilly radish]

One of such results in the paper from what it seems
https://stacks.math.columbia.edu/tag/0GVE

[next obsidian]

Okay I think I’ve seen this yeah

This stuff reminds me of this one thing Johan proved

You know enough AG right?

So take an ample invertible sheaf O(1) and then for any coherent sheaf F there’s some n where F(n) has no higher cohomology

Stupid lemma: if there was some n where F(m) had no higher cohomology for m > n, over all coherent F, then the same holds for quasi coherent F because such a sheaf is a colimit of coherent sheaves.

It’s stupid because no such n exists by observing O(-k)

But if there was an “O(\inf)” so to speak, then for any quasi coherent F F(\inf) would have no higher cohomology because you take the union of coherent subsheaves and for each of those there’s a finite n where F(m) has no higher cohomology for m > n, and \inf is larger than all n

Well it turn out they showed such an “O(\inf)” exists on certain schemes (I think the hypotheses weren’t too strong)? And it’s super duper hella not finitely generated. And it’s something stupid where like, you make it via taking… direct sums(?) of O(n) for various n in some way that branches out like a tree or something. My memory is kinda hazy on what he said

But it reminded me of the eilenberg swindle and some of this sort of crap with big projectives and freeness and shit

[azure cairn]

what's a good intuition for S_3?

like i understand it's the set of bijections on a set of 3 elements (or the set of functions that shuffle a set of 3 elements, less abstractly speaking)

but in ch2 of artin (which defines groups, subgroups and additive subgroups of Z, cyclic subgroups, homomorphisms/isomorphisms, equivalence relations, cosets, etc.) he keeps using S_3 as an example for stuff and it feels opaque to me

like i can verify what he's saying is true but it offers no intuition

uhh for example here

[somber goblet]

it's used because it's the smallest nonabelian group

[azure cairn]

yeah i understand that

[somber goblet]

it's also the symmetries of an equilateral triangle if that helps

in fact i think that's the presentation he's using there

where x is a rotation 120 degrees and y is a reflection

[azure cairn]

he says x = (1 2 3) and y = (1 2)

[somber goblet]

yep that's it

[azure cairn]

wdym by symmetries

[somber goblet]

maps of the triangle onto itself?

like if you rotate a triangle 120 degrees you end up with the same set of points, just moved around

[azure cairn]

oh

[somber goblet]

likewise if you reflect along an axis of symmetry

[azure cairn]

im bad at Geometry.

but yes i understand

this is still like weird idk

i probably just need to spedn more time convincing myself of these theorems

[somber goblet]

in this case x is a rotation 120 degrees and y is a reflection

so you can see how $x^3 = 1$ and $y^2 = 1$

[cloud walrus]

lexi

[azure cairn]

ah

yes and then

x^2 is a 240 degree rotation so

ummm

lemme think

hm ok nvm that's hard to write out

it's like

yxy^{-1} or something

ty for the intuition using triangles that is helpful

[somber goblet]

it might also help to draw the cayley diagram out

this helped me a lot in understanding these things

[azure cairn]

the who

i think the geometric perspective of this will be discussed in later chapters

[somber goblet]

replace sigma with x and tau with y

or this one might be better

replace r with x, s with y, and e with 1

[azure cairn]

e is what, identity?

alr

[kind temple]

yea, it helps if you have edge colorings

[somber goblet]

the red arrows follow applying x

the blue arrows follow applying y

[kind temple]

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

[somber goblet]

i found that this helped me understand the group structure of S3 better

[azure cairn]

Idk I feel like integers are more intuitive especially if the geometry behind S_3 hasn't been explained yet

[somber goblet]

and its a good way to compute product and things

[azure cairn]

Like he could've very easily let G be the additive integers and H be the subgroup of evens

And then 1H is a coset of odd integers

If I am thinking about this right?

Then the equivalence classes for G are evens and odds

[somber goblet]

that would have worked as well

[azure cairn]

Ok great I'm using that intuition then

[somber goblet]

and i typically see that provided as an example as well

you can generalize this to multiples of any integer

[azure cairn]

Yeah he'll do that soon

[somber goblet]

then the cosets are the residue classes mod n

[azure cairn]

When we get to right cosets and modular arithmetic and quotient groups presumably

[somber goblet]

yep

[azure cairn]

What's a residue class?

Oh ok I googled

[somber goblet]

equivalence classes on the mod relation

yeah

[azure cairn]

It's just the equivalence class for that yea

[somber goblet]

group theory is fun :3

[azure cairn]

I feel bad I haven't touched my real analysis book in 2 weeks cuz I've been doing thie

This

[kind temple]

the cayley graph of the integers looks like this, right?

... -> -2 -> -1 -> 0 -> 1 -> 2 -> ...

okay, yea this is right

[somber goblet]

yep

[kind temple]

i like that lol

[somber goblet]

i did a paper on cayley graph automorphisms

they're a lot of fun

[kind temple]

isn't the automorphism group of the cayely graph of a group just the original group?

[somber goblet]

i took out directedness

then it gets interesting

[kind temple]

oh

[azure cairn]

can the order of a group, if infinite, be ascribed cardinals like in set theory

if so would this lemma be true for groups of infinite order

cuz you have a bijection H -> aH since its inverse is a^-1 H

and so the left cosets would be equinumerous

[kind temple]

yes, this is true of any group.

[somber goblet]

oh cool

we can have index cardinals

thats funny

[azure cairn]

thanks.

[kind temple]

yea, i think |H| [G:H] = |G| still holds in the infinite case

[kind temple]

why does removing the directedness make things interesting?

[somber goblet]

the group gets a lot bigger

and harder to predict

[kind temple]

okay, so like, what happens for an infinite stick extending in both directions, the undirected cayley graph of Z?

Z is still in there as a subgroup

[somber goblet]

in that case the automorphism group is $\mathbb Z \rtimes C_2$ i believe

[kind temple]

but now you can flip things

[somber goblet]

yeppp

[kind temple]

okay i buy that

[somber goblet]

or wait times other way

[cloud walrus]

lexi

[kind temple]

i always forget lmao

[azure cairn]

automorphisms are isomorphisms between the same group right

[kind temple]

i don't use semi-direct products enough and i always have to relearn them

[azure cairn]

hmm i didnt know there was a group for those

interesting

[somber goblet]

it gets kinda cool because for some geometric figures you can describe them as an undirected cayley graph

[kind temple]

and then like, how does the cayley graph of this compare to the undirected cayley graph of Z?

[somber goblet]

and then work on the automorphism group of that to get the symmetry group of the original figure

[somber goblet]

wdym

[kind temple]

im not sure lol. you can disregard that question

[somber goblet]

the cayley graph of $\mathbb Z \rtimes C_2$ looks like two copies of the cayley graph of Z laid parallel, where the arrows point in opposite directions, and with arrows across from one to the other like a ladder

i think

[cloud walrus]

lexi

[kind temple]

okay, neat

[kind temple]

i wonder if there is some sense in which this thing "lives over" the undirected version

[somber goblet]

well it certainly has it as a subgroup lol

[kind temple]

truth

yea, this does seem pretty cool

what were some of the main results of your paper?

[somber goblet]

it focused on the symmetries of the hypercube

that i demonstrated were $C_2^n \rtimes S_n$ iirc

[cloud walrus]

lexi

[somber goblet]

and then talked about relevance for hamming distance stuff

on binary strings

nothing too fancy

[kind temple]

very cool!

[somber goblet]

you can do other shapes too

you can use cayley graph automorphisms to show that $D_{2n}$ is in fact the symmetry group for the $n$-gon

[cloud walrus]

lexi

[somber goblet]

by taking the cayley graph of $C_n$

[cloud walrus]

lexi

[kind temple]

so Aut(Cayley(C_n)) = D_n? (i hate having the extra 2 in there lol)

[azure cairn]

ahh ok so

if we fix H <= G, all cosets of H partition G and all cosets of H are uniformly ordered. thus the order of G is simply the order of H times the number of cosets that exist

and by partition i mean "fit evenly" into G, all the cosets of H are disjoint and their unions equal G

the way i reasoned about the partition argument is that if H <= G is fixed, define any arbitrary homomorphism phi for which ker phi = H. then phi(a) = phi(b) <=> b in a(ker phi) = aH and so you phi determines the equivalence relation that forces cosets to be equivalence classes

[kind temple]

"uniformly ordered" is awkward phrasing ha

i would say that they "have the same order"

[kind temple]

you can't do this

[azure cairn]

why?

[kind temple]

a subgroup H of a group G is normal if and only if it is the kernel of a group homomorphism

this theorem applies even for subgroups that are not normal

[azure cairn]

oops

ah yes i see

a subgroup H <= G is normal if every conjugate of every element in H lives in H. then ker phi is a normal subgroup since if a in ker phi, then phi(gag^-1) = phi(a) = 1 for every g in G

so choosing phi s.t. ker phi = H implies H must be normal

[kind temple]

yea

[azure cairn]

why'd he use the congruence symbol then

he said earlier that congruence comes from an equivalence relation determined by a homomorphism

[kind temple]

can you show me the paragraph you are thinking of?

[kind temple]

right, so a group homomorphism (or more generally, a function between sets) determines an equivalence relation

[azure cairn]

yeah he uses ~ in general (including for maps of setS) but said congruence is specifically what is used for the equivalence relation by a homomorphism

[kind temple]

but this isn't the same equivalence relation

[azure cairn]

if i am reading right

wdym

[kind temple]

this equivalence relation doesn't come from a group homomorphism unless H is normal

[azure cairn]

indeed

so why is he using congruence

[kind temple]

its not the same congruence that he is referring to here

[kind temple]

what you said here is wrong

yes, that is a good example

[azure cairn]

he hasnt discussed those or modular arithmetic at all yet (its in a few subchapters)

[kind temple]

they are the congruence relations induced by multiplication by n : Z -> Z

[azure cairn]

so did he just drop the term congruence relation without explaining what they are

[kind temple]

congruence relations and equivalence relations are used interchangeably, they are synonymous

[azure cairn]

like i dont understand why he didn't write a ~ b if b = ah and then prove ~ was an equivalence relation

[kind temple]

i do think that what the author did here was confusing

[azure cairn]

why did he specifically use congruence

[azure cairn]

so then why does artin specifically say that group homomorphisms determine congruence and not just a general equivalence relation

like f: S -> T determines an equivalence relation on S, but for a homomorphism he says that the equivalence relation is specially denoted by congruence

[kind temple]

i guess if it comes from a homomorphism, he is free to use congruence, but its not the only way he gets to use the congruence notation

[azure cairn]

..

[kind temple]

which is silly and confusing

i would say that it isn't even worth paying attention to the distinction

[azure cairn]

with my misconception earlier, it happened because i thought he used congruence because i thought he implicitly meant H could be the kernel of some homomorphism

[kind temple]

yea, idk

ppl make mistakes

writing books and being consistent is hard

[vapid vale]

my understanding is that when he calls it congruence in 2.7 he never says thats specific to being defined by a homomorphism. and by that point, he has not introduced cosets yet, where he also wants to use the term congruence

[azure cairn]

i dont understand why you would make the distinction for a homomorphism if it wasnt meant to be used specifically for them

idk whatever, should i just treat congruence and equivalence relations synonymously?

[glass sinew]

what did he use it for last time? if it's about homomorphisms, the cosets are the fibers of the natural projection homomorphism

[vapid vale]

i imagine if he had introduced congruences by the time you had covered cosets and homomorphisms then he'd use the term for both

if anything the part where he uses it for cosets should just be understood as a generalization of when he does in 2.7, once the new concept has been introduced

[azure cairn]

what did he use what for last time

[glass sinew]

the "congruence relation" idea

[vapid vale]

yes

it would be pretty irresponsible to say something was "usually denoted" as something and actually have it entail a specific differentiating definition

[kind temple]

what is artin's definition of congruence, btw?

[azure cairn]

um he didnt define it

[kind temple]

oh nice

[glass sinew]

oh he really just says it like that

[kind temple]

im not the biggest fan of artin's text

d+f is better imo

[vapid vale]

he wants to communicate that its just a synonym that is preferred in the context of fibers/cosets

[azure cairn]

hmm okay

[kind temple]

we need to teach setoids lowkey lmao

[glass sinew]

an setoid

[azure cairn]

ok sure i can work with that idea

[vapid vale]

well there is a subtle difference i guess since congruence is specifically an algebraic notion

[azure cairn]

for some b in the image of a map on sets, the fiber is just the set of elements that map to b right

[glass sinew]

i liked the d&f illust. for fibers under a homomorphism, solidified my intuition pretty well

let me get it out

[azure cairn]

[kind temple]

yes this is my favorite example

[azure cairn]

this makes sense though yeah

each fiber contains complex numbers of the same modulus

and so the circle's radius represents the magnitude/modulus of that complex number right

[azure cairn]

like if you think of a complex number as a vector and then swivel it around, all those vectors are in the same equivalence class

[kind temple]

i was trying to explain this to somebody early today with a similar example

[azure cairn]

and the horizontal line is the set of points that maps to the single point on H?

[kind temple]

the horizontal line is like the "H-axis"

[glass sinew]

it's the codomain H

[kind temple]

the points are where the lines get mapped to

those are the fibers living above the points

[azure cairn]

yeah, the fiber for each element in the image of a map is the set of elements in the domain that map to that element

if i had to sum it up Concisely

i just think of it as a sort of generalization of a bijection, where instead of a unique inverse you have a "set" of inverses

[glass sinew]

yeah!

the fibers are in a one-to-one correspondence with each point in the image

[kind temple]

thinking in terms of fibers/bundles becomes important later, say, when you study diff geometry or sheaves

[novel star]

bro got the isomorphism thoerems

[glass sinew]

agreed, very broad and important idea

[azure cairn]

interesting

[kind temple]

which one? the modulus map?

ah

[azure cairn]

the ordering of this chapter feels awkward though, i feel like he couldve introduced quotient groups in tandem with cosets and used integers as good examples

[kind temple]

https://pseudonium.github.io/2026/01/10/Indexed_Fibred_Duality.html

you may find this interesting

[azure cairn]

cuz quotient groups is a few subchapters ahead

[novel star]

naturality is a beautiful thing ig

[azure cairn]

https://tenor.com/view/nerd-emoji-nerd-meme-radar-gif-26497624

[glass sinew]

diagram so good i just explained to my dad what fibers are

[kind temple]

why are they called jets?

im blaming the... physicists?

is that accurate?

woosh

oh nice

how were you thinking of them as subobject classifiers?

okay, maybe i will try to meditate on this too. im not the most familiar with subobject classifiers

okay, cool. will play around with these

[crystal vale]

For b),

Let r be the primitive l-th root of unity over F_p.

Now if r in F_p^k, then r has to be order l in F_p^k^{×}. It implies l should divide p^k -1 , implies f≤k.

So it implies n≤f. Now I will show r in F_p^f.

Since l divides p^f -1 hence F_p^f has an element e such that e has multiplicative order {l}. Hence from that e I can generate r.

Is it correct?

And any hint for part c, latter part?

[rocky cloak]

So you know phi factors into irreducible polynomials, and those factors will be the minimal polynomials of its roots.

So then what you need to determine is, what are the degrees of these polynomials and do you have multiple roots?

[crystal vale]

Roots are distinct

Because l-1 is relatively prime to l

So f' and f has gcd 1

So all irreducible polynomials are distinct

So I have to show if I pick any irreducible factor then one of the primitive root should be the root of that irreducible polynomial

[rocky cloak]

The clue is in the first part of the exercise

[rocky cloak]

Not sure I follow this reasoning. But an easy way to show the roots are distinct is just to list all the roots

[crystal vale]

If they have repeated roots the gcd of f and f' is polynomial of degree ≥ 1

[rocky cloak]

But how is that related to l-1 being relatively prime to l (which is just something that would be true for any polynomial f)?

[crystal vale]

because i am showing f = x^l -1 has distinct roots, because \phi(x) is a factor of f so if f has distinct roots then \phi too.

now f' = lx^l-1, since l \neq p so l^-1 exists, hence xl^-1f' - f = 1.

sorry not l-1, i think now it is clear

how do i find the number of primitive roots ?

i know over Q

oh it is same

ah it depends

[maiden crater]

So here the quotient is ${(a)+b \mid b \in (b)}$ for instance, right

[cloud walrus]

wai

[crystal vale]

ah i don't get it

[rocky cloak]

Do you know the relationship between the degree of F(a) and the minimal polynomial of a?

[supple ice]

in a PID a prime ideal has basically two moods: either it is 0 in which case the quotient is just the original PID again or it is nonzero hence maximal so the quotient is a field and a field is the ultimate lazy PID since it has no nontrivial ideals to worry about

[maiden crater]

yea, that's what I thought]

cool

thanks

[crystal vale]

yes both are equal

[rocky cloak]

So for an lth root of unity a what is the degree of the minimal polynomial of
a^n?

[crystal vale]

a is primitive root ?

btw i got it

[rocky cloak]

Sure

[crystal vale]

l is prime and all roots generate by a, so for F[a^n], the degree is same as f

where f defined in the exercise

n is not divsible by l

[rocky cloak]

So then all the irreducible factors have degree f, and then there must be (l-1)/f of them

[crystal vale]

yes

so when p not divide l there is a primitive l-th root of unity, but when p divide l then there is no primitive l-th root of unity

can i change l by any n ?

[maiden crater]

could I have help with (a)

There does exist a common multiple trivially yes

but how do I prove it's the least

[crystal vale]

have you proved ED implies PID?

so try to find the lcm of a and b in terms of ideals

do you how to characterize the gcd(a,b) in terms of ideals?

[maiden crater]

the interesction is generted by the gcd

[crystal vale]

no

[maiden crater]

hmm?

[crystal vale]

how did u defined the gcd?

[maiden crater]

is this the defn you're talking about

[crystal vale]

yes it is <a,b> not intersection of (a) and (b)

both definition is equivalent

so <gcd(a,b)> = <a,b>

but find if (a) \intersection (b) = (c) then what property c have

[maiden crater]

a is genertaed by both b and c

so it's a multiple of both

[crystal vale]

(a) \cap (b) = ( c )

so how a is generated by b and c?

[knotty badger]

oh yay categorical product

[maiden crater]

oh right

I'll think a bir more I guess

[azure cairn]

I appreciate this diagram a lot more after learning $[G:\ker T] = |\im \phi|$. Tyvm for sharing

[cloud walrus]

Altanis

[azure cairn]

Cuz each fiber can represent a coset of ker phi

[glass sinew]

i owe it to d&f, it's their visual aid for cosets as well

[azure cairn]

Not sure why artin didn't include it tbh. Moreover he randomly dropped the term isomorphism class without defining it just now while reading 😭

[glass sinew]

algebraists ☕

[delicate orchid]

heavily depends on what the action is

[cursive spindle]

lmao

I do like the stuff Artin did in math though

[ripe harbor]

Is it often used to compute |X| though or is it more something you use in proofs?

[maiden crater]

p[retty silly question, I know

my idea was $( \Z/ (n_{1}\Z \times\dots \times n_{k} \Z) \cong \left(\Z \ n_1 \Z \times. \dots \times \Z \ n_{k}\Z \right)$

[cloud walrus]

wai

[rocky cloak]

Typically |X| is easy to compute / given.

The places I would see this come up would for example be in proving that one of [G:G_x] is not divisible by p.

For example if |X| is a multiple of p, but |X^G| is not, then you know at least one [G:G_x] is not as well

Or the other way, say G is a p-group, so that each [G:G_x] is a multiple of p. Then |X| = |X^G| modulo p.

[ripe harbor]

Thanks!

[small yacht]

How can a function that takes you to a function be a homomorphism?

[maiden crater]

Groups can be groups of functions no?

so Aut(G)is a group of functions on G

[small yacht]

I guess. What would the operation be, then?

[maiden crater]

function composition afaik

[small yacht]

I guess that makes sense

[maiden crater]

I have a question for you

What is S_3( this is key)

[small yacht]

Permutations of 1 2 3?

[maiden crater]

what are permutations really ( they are bijective functions, think about this)

like (132) isn't that a function from (1,2,3) to (1,2,3)

[small yacht]

Hm it is

Oh, and operations in S_3 are just compositions of those

[maiden crater]

yay!

yes

[small yacht]

I probably would’ve figured this out quicker if I thought of groups as automorphisms of an element of a groupoid

[maiden crater]

,w groupoid

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by math_rocks

[maiden crater]

is that cat theory lingo

[small yacht]

Yeah, sry 😭

Category where every morphism is an isomorphism

[maiden crater]

https://tenor.com/view/say-what-now-excuse-me-gif-13138805

[copper kestrel]

sorry for the blurry picture, i'm guessing that for Z[x], it factors into 2 * 2 * (x^2 - x + 2), and then i have to show that x^2 - x + 2 is irreducible?

i'm not quite sure how to do it for Q[x] and for Z_11[x]

and irreducible means every factorization has a unit, but Q[x] is also a field iirc? so then how would we have an irreducible element in Q[x]

ah wait i did smth wrong

[valid fox]

Q[x] is definitely not a field

[copper kestrel]

i had it as x^2 - x + 4 instead of x^2 - x + 2

[valid fox]

x has no inverse

[copper kestrel]

oh shoot youre right

[valid fox]

But indeed in Q[x] you can divide out the 4 and make the polynomial monic

Same thing for Z/11Z

[copper kestrel]

monic?

[valid fox]

Leading coefficient is 1

[copper kestrel]

ah

[wraith cargo]

it's neither

you use it to study the properties of either [G:G_x] or |X^G|

so for example you'd use this formula to prove the Sylow's 3rd theorem

[copper kestrel]

i ended up doing that and got it to factor into irreducibles but i dont think i'm doing it right

bc i dont think i can just divide by 4 out of nowhere

[small yacht]

Uhmmm a collection of sets where every function between them has an inverse

[maiden crater]

no,no, it's just that I know no cat theory

[valid fox]

4 is a unit in Q and Z/11Z so it doesn't really influence the factorization and irreduciblility of the factors

[copper kestrel]

ah

[valid fox]

It's different in Z though

[ripe harbor]

Thanks, yeah, I guess the most important thing you follow out of it is that |X^U| = |X| mod p for U a p-group, and then you use that everywhere

[supple pecan]

hi new to rings, but just wondering. do other rings like say M_n(F) have an analogue for primes?
like are there prime matrices. i assume this doesnt rlly work for matrices since the multiplicative inverse doesnt exist for all matrices (or is there a different reason?) but are there non standard analogues for primes in other rings or groups or fields or whatever

[karmic moat]

i'm not sure what you mean by "non standard analogue" but there's just a general notion of a prime element in rings

[rocky cloak]

The usual convention is that an ideal of R is prime if it is proper and if aRb is contained in the ideal then either a or b is in the ideal.

Then an element is prime if it generates a prime ideal.

In Mn(F) the only ideals are (0) and Mn(F) so you don't get any nonzero primes

[supple pecan]

I see, we havent gotten to that notion yet, its only been in the context of Z

[supple pecan]

ooh okay, i dont know what most of these words mean they only now defined a ring. But thank you ! i shall look into this

also do we have left/right cancellation in rings?

like if c(ab) = c(ba) can we say ab=ba or no

[rocky cloak]

Not in general no

For example, not in Mn(F)

[wraith cargo]

noncommutative rings have analogues of primes but they're define differently
Jagr gave a definition but an equivalent definition is that an ideal P is prime if for any two ideals, I and J, IJ \subseteq P implies I \subseteq P or J \subseteq P

[supple pecan]

Oh right so is this because there isnt always multiplicative inverse or is it because multiplication doesnt always commute? as in which property implies which i forget

[wraith cargo]

only in rings with no zero divisors does this hold

they're usually called domains (or integral domains if they're commutative)

[supple pecan]

so an ideal is a subset of the ring? do we not define primes in the integers as elements, and not subsets? this generalization confused me with how it would work wrt to integer primes

though idk what an ideal is so maybe i just need to follow the course a bit more 😅 so when we say 5 is a prime in the integers, what ideal are we refering to? or have i confused it

[wraith cargo]

primes and prime ideals are the same think basically in Z

[supple pecan]

so is an ideal not always a subset then?

[wraith cargo]

the prime ideals in Z are sets of the form {px | x in Z}

i.e. pZ

where p is a prime number

[supple pecan]

so how does 5Z represent the prime number 5 in Z?

[rocky cloak]

The ideals of Z are exactly of the form nZ for n a non-negative integer. And then nZ is a prime ideal if and only if n is prime

This is actually the historically origin of the name ideal.

Since statement about non-negative integers can be made into statements about ideals in Z, it turned out ideals better generalized certain statements, so they are what numbers "should" have been. They are the "ideal numbers"

[karmic moat]

that's interesting

[supple pecan]

i see we learn about ideals at some point so this will probably make more sense later !

thank you so much !

[rocky cloak]

Specifically, it's that unique factorization might fail in ring of integers of a number field, but factorization of ideals into prime ideals work.

So Kummer made a definition for "ideal numbers", and then dedekind extended this definition to arbitrary rings (naming them just ideals)

[karmic moat]

oh nice

yeah before i fell asleep last night the thought of "why are ideals called ideals" crossed my mind

[wraith cargo]

The original reason he defined these is because he wanted to prove FLT

Since when you have a statement like x^p + y^p = z^p you can factorize this in Z[ζ_p] and through unique factorization get a contradiction

But cyclotomic fields are very rarely UFDs

So to get around this he introduced his theory of ideal numbers which today are a predecessor to ideals to rephrase the statement in terms of ideals

[karmic moat]

i see

[wraith cargo]

He only managed to prove FLT for regular primes but it was still a huge accomplishment!

He was the first to notice that in dedekind domains even though we don't have unique factorization of elements we do have unique factorization of ideals

So his proof went along these lines

The idea is quite clever

You write the factorization of Fermat's equation in terms of principal ideals generated by elements like (x+ζ^k y) and their product equals some principal ideal to the p-th power

Now by unique factorization you have that each of these (x+yζ^k) ideal is equal to A^p_k where A_k is some ideal

i.e. it's an ideal that to the p-th power is principal

But p doesn't divide the class number

So clearly A_k has to be principal as well

And now by some case checking (I forget the details off the top of my head) you get a contradiction

[karmic moat]

yeah that's slick

[wraith cargo]

Classical Cyclotomic methods don't really get you much farther than this

[karmic moat]

maybe i'll look at that proof tonight before bed

[wraith cargo]

There are some generalizations and stuff

But if you wanna prove the full FLT you need two very difficult hypotheses

One is Vanderviers conjecture

And another is a technical hypothesis we don't really know much about lol

[karmic moat]

lol primes less than 2^31

[elfin wraith]

Can’t wait for 2^31+1 to provide a counter example

[karmic moat]

,w 2^(31)+1 is prime

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by aanzhuu

[karmic moat]

L

[elfin wraith]

That’s basically prime

[crude kiln]

,w 2^(31)+3 is prime

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by thefullestcircle

[crude kiln]

damn

[copper kestrel]

hi everyone i'm back, i'm trying to express the gcd of 49349 and 15555 in the form a(49349) + b(15555) = 61, but i'm not quite sure how to find a and b for some reason

here is the euclidean algorithm i did

i think i vaguely remember needing to go backwards in it?

or smth?

[rocky cloak]

Notice
61 = 549 - 488
and
488 = 2135 - 3*549
so
61 = 549 - (2135 - 3*549) = 4*549 - 2135

Etc

[azure cairn]

does the fact that, for homomorphism $\phi: G \to H$, $|G| = |\ker \phi| \times |\im \phi|$ have any connection whatsoever to rank nullity in linalg

[cloud walrus]

Altanis

[rocky cloak]

Yes, they are both consequences of the first isomorphism theorem:
Im phi = G/ker phi

[azure cairn]

i can see that

[azure cairn]

my book puts this fact before that thm tho

but yes i see it ty

[azure cairn]

well i mean i could basically prove the first isomorphism thm rn, i know there's a bijection from the left cosets of ker phi to im phi

[copper kestrel]

ah that'll do it. thank you!

[rocky cloak]

If you're working over a finite field (say with q elements) then dim V = log_q |V|, so that would be another way to connect them

[azure cairn]

im very excited to get into quotients because it was a really cool topic in my linalg book but didn't have much use

[azure cairn]

ummm where is this from

that's very interesting

and |V| denotes the order of the vsp as a group?

wait what

ah yes finite field

i was confused cuz |V| < infty but thats cuz i worked over R/C

[rocky cloak]

And n-dim vector space is isomorphic to k^n and |k^n| = |k|^n = q^n which you can rewrite as
log_q |k^n| = n

[azure cairn]

what is k

the field?

[rocky cloak]

The field

k for körper

[azure cairn]

this swould be covered in my abstract alg book right lol

i understand though

[azure cairn]

    \begin{theorem}[Chaining Property of the Index]{Let $K \le H \le G$ be subgroups of $G$. Then $[G: K] = [G: H][H: K]$.}
        Let $[G: H] = m, [H: K] = n$. Then choose $g_1, \dots, g_n \in G$ and $h_1, \dots, h_m \in H$ for which $g_1H, \dots, g_nH$ are distinct left cosets of $H$ and $h_1K, \dots, h_nK$ are distinct left cosets of $K$. Then note their unions partition $G$ and $H$ respectively. Fixing some $1 \le k \le m$, note $g_kH$ is partitioned by $g_kh_1K \cup \cdots g_kh_mK$. Doing this for each $k$ yields $mn$ distinct cosets of $K$ that partition $G$, completing the proof. 
    \end{theorem}

    \begin{remark}[Chaining Property for Multiple Subgroups]
        By induction, we can generalize this theorem for subgroups $G_1 \le \cdots \le G_n \le G$ of $G$.
    \end{remark}

is there an analogue for groups of infinite order perhaps?

[cloud walrus]

Altanis

[azure cairn]

could you use cardinals and just redefine what m * n means for cardinals

[charred iris]

I think so?

although (assuming axiom of choice) m * n = max(m, n) if one of m, n is infinite (and both are non-zero, which we get for subgroups)

So really you only get something interesting if K is the same cardinality as G

[lusty marlin]

Yes, essentially the same proof works for infinite groups too.

[azure cairn]

Idk how cardinal math works so :p was just wondering

[azure cairn]

Alr nice

[rocky cloak]

The proof is basically more important than the statement here.

If {gi} are coset representatives for G/H and {hj} representatives for H/K, then {gi * hj} give you representatives for G/K. This is true infinite or not

And so you get a bijection between G/K and G/H x H/K.

[azure cairn]

What is the group theoretic definition of G/H

Is it the set of all cosets of H?

I haven't gotten to quotient spaces yet but I feel like I could make that guess

Alr

Wait what about left and right cosets

[azure cairn]

Does the quotient group or wtvr comprise both

[rocky cloak]

G/H = left cosets
H\G = right cosets

[azure cairn]

Oh interesting

[rocky cloak]

For a quotient group you would have the left cosets equal to the right cosets

[azure cairn]

This is equivalent for normal subgroup H right?

[rocky cloak]

Yes, for normal subgroup the left coset gH equals the right coset Hg

[azure cairn]

Cuz normal subgroups are invariant under conjugation so it shouldn't matter where the group element is appended to

[azure cairn]

Ah ok

[tepid light]

(problem 4 part b) How does one construct an isomorphism? I don't want the solution to the problem, I just need to know how to solve it lmao. My professor insists on teaching through examples so we haven't done really any definitions or methods yet. The full problem is for context, since most of what I've seen otherwhere is isomorphisms over vector spaces.

[cursive spindle]

You can explicitly write the elements for each one if you want

then match

[tepid light]

Is there an order they should be in, or does it not really matter.

And how would I do it generally

[cursive spindle]

Identity maps to identity, transpositions map to transpositions etc

[rocky cloak]

Note that GL2(F2) and S3 are both defined by an action. The former acting on (F2)^2 and the latter on {1, 2, 3}.

So for example if you can cook up three things that GL2(F2) permutes you would get a homomorphism to S3

[tepid light]

Like this?
Also, what does 'defined by an action' mean?

[rocky cloak]

Just that they are some set of functions where the group operation is composition.

For S3 they are bijections on {1,2,3} and for GL2(F2) they are invertible linear maps on (F2)^2

[noble nexus]

yeah if you can show that GL2(F2) acts faithfully on a set of 3 points then you are done p much

[tepid light]

So in a general case, I would look for things that GLn(Fp) permutes in a similar way to Sk?

GL3(F2) would have an isomorphism with S8 since 2^3 different 3x1 column vectors can be made, each aligning with one element in S8?

[noble nexus]

not quite, because you wouldn't get all of the permutations

its a special thing for GL2(F2)

basically because all possible permutations of the nonzero elements can be reached

[tepid light]

interesting

kk, thank you

[tepid light]

nonzero = nonzero vectors?

[noble nexus]

but yeah in general you can get a lot of nice isomorphisms or relations between symmetric groups using this method of just finding a nice set that the group acts on

yeah

another example is that the general affine group GA2(F2) (i.e., transformations of the form x\mapsto Ax+b where A is in GL and b is in F2)

is isomorphic to S4

because GA2(F2) acts on (F2)^2 in the obvious way, so each transformation gives you a permutation, and you can show that (1) all permutations can be reached and (2) a permutation fixing every point must be trivial (this part is obvious)

[azure cairn]

hm i guess this belongs in another channel actually cuz it's not really abstract alg related

[tepid light]

It doesn't?

[azure cairn]

no sorry i was talking about something i sent that i deleted

[tepid light]

ah kk

[azure cairn]

#advanced-lounge message

[cursive spindle]

For non-commutative rings what structure does the set of torsion element enjoy? I know for commutative ring it is a submodule but in the non-commutative case the property of being a subgroup fails

[ripe harbor]

"As p-Sylow groups of G, P and Q are also p-Sylow groups of N_G(Q)" when can we make such a statement? I didn't send more context yet, maybe it's possible to answer this in general so that it's not necessary

Don't we need the power of the prime p in |G| being the same as in |N_G(Q)|?

N_G(Q) is the normalizer of Q in G, why is that the case then?

[rocky cloak]

Note that Q is a subgroup of N(Q), and that the power of p dividing |N(Q)| can't be bigger than the one dividing |G|.

So Q will be the p-sylow subgroup of N(Q). Whether P is depends on whether P is a subgroup of N(Q) at all.

[cursive spindle]

nvm they would just be a set. nothing interesting

[crystal vale]

I don't get the c part, how the class equation is, the center can contain more than q-1 elements, right?

[quiet pelican]

q is defined as the order of Z

[crystal vale]

Ah my bad

[quiet pelican]

(And hence the intersection of Z with non-zero elements is a-1)

[crystal vale]

Yes

Thank you

[crystal vale]

Okay | q - r | > q-1, where r is any root of unity, how does it lead to contradiction?

Okay, q-1 = \phi_n(q) × t

But it implies |t| < 1

[quiet pelican]

But t must be an integer, because everything in sight is

(It would help to see (d))

[crystal vale]

Why? Yes other things are in Z

[quiet pelican]

phi_n is an integer polynomial
So phi_n(q) is an integer
Which divides the integer q-1 an integer number of times
And that number is t

[crystal vale]

I think i have to prove phi_n divides x^n-1 / (x^m_i -1) in Z[x]

[quiet pelican]

Gauss’ Lemma

[crystal vale]

How?

[quiet pelican]

They divide in Q basicallt by definition of phi_n

[crystal vale]

Yes

I got it, thank you

[fleet cairn]

Is this not true for bijective functions (it looks like it claims for every function). Example a' related to a'' but not equal.

[lusty marlin]

This holds for all functions. It should be easy to verify that the given relation is an equivalence relation.

[kind temple]

in the case that f is injective, then a ~ b if and only if f(a) = f(b) if and only if a = b

[fleet cairn]

I think i misunderstood, my real question is if f is bijective then doesn't this equivalence relation require every element of A to have their own unique equivalence class (also meaning the equivalence class has only one element). Because if not, then let's say a certain equivalence class C has two or more elements, and we let x and y \in C and let x != y (because there exist two distinct elements). Then x ~ y, yet f(x) != f(y) from the bijective property, but this is a contradiction that x ~ y iff f(x) = f(y).

[vapid vale]

yes

[marble hinge]

where F(S) is a free group generated by set S

I was somewhat surprised that it is just unique up to isomorphism, and not fully unique

does it come from the fact that we can choose any fresh elements for the sets to represent S^-1 and empty word?

i.e. here:

so we can have different isomorphic free groups on S, that differ only on S^-1 and on empty word, but are exactly the same on S

[candid patrol]

Wdym fully unique ?

[marble hinge]

that just one object exists

[wise socket]

You could take S = {a} and then S = {b}, both F(S) are isomorphic, but are different since a and b are different as a letter

[marble hinge]

mm, but the corollary says that the set S is the same, we fix it

"F(S) is unique up to isomorphism which is the identity map on the set S", that's for a fixed set S, so your example is not applicable

so my question is why it doesn't say "There is a unique F(S) for every set S"

[marble hinge]

I think the answer comes from that screenshot that I quoted. I.e. from our ability to choose any elements for S^-1 set and for the empty word. So our F(S) is not unique, because we can choose say 384 to represent the empty word, but in another universe we can choose \pi to represent the empty word. The result would be technically the same (because we are not using pi or 384 as actual numbers, just some symbols for "empty word"), but the construction is still slightly different. Hence those free groups on S would only be isomorphic, but not the same.

but wanted to double check

[kind temple]

something a bit stronger can be said, the free group is unique up to unique isomorphism

[marble hinge]

Mmm, I don't really see the subtle difference induced by that additional claim that isomorphism is unique... Can we have an example. Say, I have a set S={a}. And I generate some free groups out of it. Can you show me two different free groups on this set that are isomorphic to each other and that unique isomorphism between them?

[rocky cloak]

The difference would be as you described, for example a^-1 was something different.

Then the point is that there is a unique isomorphism fixing S

Once S is fixed everything else is determined

[marble hinge]

right. So the uniqueness of the isomorphism there is saying that once we determined what symbols we use for a^-1 and empty word, the mapping {a -> symbol for a, a^-1 -> symbol for a^-1, 1 -> symbol for empty word} is fully determined by "a |-> symbol for a" (in this example a is the only one member of S)

[desert verge]

you could also label the elements of S as (s,0) and their inverses as (s,1).
this way there's no need to choose S^{-1} or the bijection, and F(S) is well defined

[marble hinge]

yeah, but effectively it is choosing S^{-1} in a way

you choose that elements of S^-1 are to be tuples of elements of S combined with 1

[desert verge]

yes and also making it clear how to distinguish elements of Sx{0} and S^{-1} ( := Sx{1})

grr the notation for this is confusing

[marble hinge]

Mmm, combing back to the uniquness of isomorphism. I think it is saying that we can't "permute" this mapping: {a -> symbol for a, a^-1 -> symbol for a^-1, 1 -> symbol for empty word} in any way. Once we fixed correspondence between a and symbol for a and we chose other symbols.

i.e. we can't say map "a^-1" to symbol for empty word and claim that this is isomorphism

anyway, this is all a little bit "meta", and confusing indeed

[desert verge]

what's "symbol for a"

[marble hinge]

ah, right, we have no "symbol for a", we just use "a" itself

but by "symbol for a^-1" I meant an element from S^{-1} that we chose to be the inverse for a

I think I was also confused by what exactly "unique isomorphism" means

but now I think I get it. Once we fix those two free groups and their elements, the fact that isomorphism between them is unique means that there is just one bijective function that preserves the group property

and this isomorphism takes "symbol for empty word in free group 1" to "symbol for empty word in free group 2" and so on

[knotty badger]

I like to think of it as the isomorphism induced by the universal property

[marble hinge]

Right. That’s what they use in the proof of that corollary

[jaunty berry]

hi guys just joined this channel because i need help with mathematics

ill iform yall if i need help

[marble hinge]

haha, but there is a person with exactly the same pfp already @crystal vale

[jaunty berry]

lol guess ill change it then

[azure cairn]

is the phrase "reversing the roles of g and g^-1" valid

[quiet pelican]

Yeah

[marble hinge]

if you want more fomality, then you can introduce, say, $k$, as $g^{-1}$. And then you have $khk^{-1} \in H$, or substituting back: $g^{-1}hg \in H$

[cloud walrus]

dying_sphynx

[ripe harbor]

Oh, the power of p in |Q| has to be less than the one in N(Q), and even if they're not the same, it's still a Sylow p-subgroup of N(Q). Thanks! For some reason I was thinking for a moment that we want the powers to match, which they don't necessarily, but that's fine

[rocky cloak]

I mean, they will match.

Like say p^k is the largest power of p dividing |G|. Then |Q| = p^k. As Q is a subgroup of N(Q) we must have that p^k divides |N(Q)|. Since |N(Q)| divides |G|, this must in fact be the largest power of p dividing |N(Q)|

[ripe harbor]

Oh, say the power is k as Sylow-subgroup of G and q as ... of N(Q). Then k <= q from Q <= N(Q) and q <= k from N(Q) <= G. So k = q. Does that look right?

[copper kestrel]

not quite sure how to do this with polynomials, i know for regular numbers you just do the euclidean algorithm but this polynomial is super weird

[cursive spindle]

You do the same thing

[copper kestrel]

what

[cursive spindle]

But holy giant polynomial

Sounds miserable to compute

[copper kestrel]

:')

[azure cairn]

wait can't i simplify this proof by saying |gHg^-1| = |H|?

[cursive spindle]

Do you know polynomial division?

Use this

[copper kestrel]

i do

[copper kestrel]

but i was told that wont give us a solution in Q[x]

or is it just multiple polynomial divisions to get the gcd

[supple ice]

don’t do the exercise skip it

[cursive spindle]

Unironically this

[copper kestrel]

its homework

i cannot skip it

[azure cairn]

skip it anyways

[cursive spindle]

Have fun doing long division then

[copper kestrel]

chat 💔

[supple ice]

what’s the point of finding the damn gcd of two degree-10 polynomials lmao?

[cursive spindle]

The point is to go back idk how many years to remember how to do long division

Annoying exercise

[copper kestrel]

computationslop

[wraith cargo]

maybe there's some easy factorization lol

[cursive spindle]

Of course there might be one but it's not obvious at first glance

[azure cairn]

this proof is valid, right? if so, which proof is preferable

[wraith cargo]

but but but students gotta learn how euclid's algo works 🥺

[desert verge]

wait k[x] is a PID right

[wraith cargo]

yes

[cursive spindle]

Yeah it's an ED

[supple ice]

I didn’t even properly read the exercise lmao I just saw a polynomial of degree 10 and gave up

[desert verge]

ohh

ED implies PID

[cursive spindle]

Yup that's why you can do euclidean algorithm

[desert verge]

in k[x] yes

[copper kestrel]

whats k[x]

is k a field

[desert verge]

yes

[wraith cargo]

poylnomials with coefficients in k

[copper kestrel]

ohhhhh

i see

why is k/K notation for a field

[cursive spindle]

use wolfram and see what u get

[copper kestrel]

LMAOOOOO

[wraith cargo]

it means field extension

[copper kestrel]

ohhhh

[wraith cargo]

oh

[copper kestrel]

i have not learned that

[desert verge]

i don't think that's what they meant...

[wraith cargo]

I might have misunderstood what you meant

It comes from german

[copper kestrel]

ah

[cursive spindle]

Yeah Korper

[copper kestrel]

oh interesting!

my book uses F for field like a NORMY

[wraith cargo]

Körper

[desert verge]

F is only for finite fields

[copper kestrel]

ooh

[wraith cargo]

haha yes good one

[cursive spindle]

Hold my F_(1+i)

[supple ice]

I just messed around by hand for fun since I was too lazy to do the full algorithm properly I tried to write your little polynomial as (x^3 + 2x - 1) times something then I divided the big polynomial by x^3 + 2x - 1 and checked that the two resulting quotient polynomials are coprime so if I didn’t screw up the gcd is x^3 + 2x - 1

[copper kestrel]

🫡 i'll see if i can verify it

[cursive spindle]

big time

I almost plugged into wolfram

[ripe harbor]

But perhaps the isomorphism I used at the beginning is unnecessarily strong

[cloud walrus]

ILikeMathematics

ILikeMathematics

[ripe harbor]

From that, we get that there is only one q-Sylow-group of Z/(q(p - 1)Z) up to isomorphism

[cloud walrus]

ILikeMathematics

[ripe harbor]

But then, why is this useful

So is the key really the isomorphism at the beginning?

[copper kestrel]

i definitely did smth wrong, my answer was your answer times -59

[marble hinge]

I think you have to be a little bit careful here about infinite groups (then using |A| = |B| when those both are infinities is kinda suspicious). I think I liked the idea of the previous proof more. You just need to show that H is a subset of gHg^-1. For each h from H you find a corresponding h' from H such that h = gh'g^{-1} and that's it. It will be h' = g^{-1} h g . And we know that this is an element of H, because h is from H, and H is normal (and thus closed with respect to conjugates).

[copper kestrel]

well actually since Q is a field cant we simplify by multiplying by -1/59?

bc units

[ripe harbor]

Maybe like this: Instead, consider a general group G of order pq(p-1). Then |Syl_p(G)| in {1, p, q, (divisors of p - 1)}. But also |Syl_p(G)| = 1 mod p so it can only be 1 or q. But q is not possible, otherwise p | q - 1 even though p > q. So |Syl_p(G)| = 1. Now |Syl_q(G)| in {1, p, q, (divisors of p - 1)}. But also |Syl_p(G)| = 1 mod q so it can only be 1, p or a divisor of p - 1. It cant be p otherwise q | p - 1. It cant be a divisor of p - 1 otherwise q | d - 1 and so q | p - 2, (now im still trying to find a contradiction here, why is q | p - 2 not possible...)

$q \nmid (p - 1)$ but $q \mid (p - 2)$ where $p > q$ are prime seems fine..

[cloud walrus]

ILikeMathematics

[wraith cargo]

here how you should think about it

take a group G of order pq

then clearly it has a sylow p-subgroup of size p and a sylow q-subgroup of size q

Now using Sylow's third theorem you should show the number of p-subgroups is 1 and the number of q-subgroups is also 1

so clearly they're normal and their intersection is trivial

as well you should show that they generate the whole group

then you can conclude that your groups is exactly Z/pZ x Z/qZ

[ripe harbor]

It seems this didnt use q nmid p - 1?

[ripe harbor]

Ah probably here

Yeah ok this is basically my way just that I considered G of order pq(p - 1) and then used the third Sylow theorem, but yeah, then that seems unnecessarily complicated

Thanks!

[wraith cargo]

yeah I'm not really sure what you were doing haha

[marble hinge]

I think I've seen this exercise in D&F, before Sylow theorems chapter

are you specifically interested in solving it using Sylow's theorems, and not other means? I suppose so

Ex. 4.4.2 in section on Automorphisms (they show in the text that this group is Abelian, and then in the exercise additionally show that it's cyclic)

[ripe harbor]

Yes but I think I almost did that now, just thinking about the last step. What do you mean by other means? How else could we solve it

they show in the text that this group is Abelian, and then in the exercise additionally show that it's cyclic
this?

[rocky cloak]

What's your definition of normal here (since you say gHg^- < H by definition)?

Because the usual definitions of normality I can think of would be

• gHg^-1 = H
• gH = Hg
• H is the kernel of a group homomorphism

[ripe harbor]

Now whats left to show is that P cong Z/pZ and Q cong Z/qZ

[mossy zenith]

since there is an ordering for finite presentations (for fp groups), did anybody try going through them one by one see what groups come up? I understand that the isomorphism problem for finite presentations is not decidable, but one could just abort after some time and write down unknown. Also let me know if this is the wrong channel

[ripe harbor]

I want to show P and Q are cyclic. Take a in P. Assume 2 <= ord(a) < p. Then ord(a) | (...) [to be completed] but p is prime, contradiction!

[vocal pebble]

Well showing that PQ is the whole group is the "hard part". Their orders are p and q respectively, so they are automatically cyclic.

[ripe harbor]

Why?

Why are they cyclic

We just know they have p and q elements

[vocal pebble]

Groups of prime order p are cyclic (particularly they are Z/(p)), as you seem to have also proved.

[ripe harbor]

How does the proof go?

[vocal pebble]

Take any non identity element x. Then ord(x)>1 and also ord(x)|p by lagranges theorem. This must mean ord(x)=p because p is prime

[ripe harbor]

Oh, of course...

|G : <x>| * |<x>| = |G| is Lagrange

[versed flower]

showing lagrange's theorem now

[ripe harbor]

So |<x>| | |G|

[ripe harbor]

Why the thinking emote? That should seem right

[ripe harbor]

Thanks!

[candid patrol]

It is

[ripe harbor]

But also, I havent established yet that Z/pZ x Z/qZ cong Z/pqZ

Do we construct the isomorphism explicitly?

[azure cairn]

normal means that, for fixed g in G, ghg^-1 in G for every h in H

gHg^-1 = {ghg^-1: h in H} btw, the textbook uses that

[proper barn]

hi, here it states in a division ring every non zero element has a left and right inverse, and seems to say that they are always equal.
My prof said they can sometimes not be equal, and then corrected himself to say one inverse may exist while the other doesnt. He also said that whether theyre the same doesnt matter if its commutative (i.e they will be the same) but it seems theyre the same regardless. Would then commutative imply that both inverses always exist and are thus the same?

[ripe harbor]

Also, what we have until now is QP = G

[candid patrol]

See chinese lemma

[ripe harbor]

And Q cong Z/qZ and P cong Z/pZ

[ripe harbor]

I dont have that available

Is (Z/qZ) * (Z/pZ) actually the same as (Z/qZ) x (Z/pZ)?

[vocal pebble]

Not sure what you mean by *

[rocky cloak]

I see, then I guess dying sphinx gave you a good hint

[azure cairn]

what hint

also is there some alternative formulation of normality?

[vocal pebble]

In a commutative ring, one sided inverse exists implies two sided inverse exists yes (and they will be equal by what you said)

[rocky cloak]

If you're in a commutative setting them ab = ba

So if ab = 1, then you would also have ba = 1.

[ripe harbor]

UV = {uv | u in U, v in V}

[proper barn]

I see, so there are cases where only one exists?

[rocky cloak]

Yes

[proper barn]

ah okay. thank you <3

[rocky cloak]

What is true is that if
ab = 1 and bc = 1, then a=c

But it could be that ab = 1 and there's no such c

[ripe harbor]

What I have right now is: $$G = PQ$$ with $$P \cong \mathbb Z/p\mathbb Z \quad \text{and} \quad Q \cong \mathbb Z/ q \mathbb Z$$ so $$G \cong (\mathbb Z / p\mathbb Z) \cdot (\mathbb Z / q \mathbb Z),$$ no?

And not the cartesian product yet

[cloud walrus]

ILikeMathematics

[vocal pebble]

And to show that G=PxQ, you need P intersection Q to be trivial, and every element of P commutes with every element of Q

[marble hinge]

Yes

[stone sky]

Which would be the case since p and q are (presumably) distinct primes, so the intersection is trivial

[vocal pebble]

Yes, that was their exercise.

[ripe harbor]

Yes, I showed the intersection is trivial

[marble hinge]

Other means = without Sylow

[ripe harbor]

But why does that help

How does that allow us to turn multiplication into a cartesian product

[stone sky]

I’ll show you

I won’t prove it here by in general if G is a group with subgroups H,K with HK=G, the intersection of H and K is trivial and hk=kh then G is isomorphic to H external product with K

Try the homomorphism φ(h,k)->hk and check its an isomorphism

[ripe harbor]

Ok but actually, I think we dont need cartesian products here. We can just show directly that (Z/pZ)(Z/qZ) cong Z/pqZ, no?

Since we cant use CRT anyways

[ripe harbor]

Im just not sure about the isomorphism yet

[ripe harbor]

Ah, alright

[vocal pebble]

external product with K* yes

[stone sky]

Thank you lol

[ripe harbor]

where this is the normal product UV = {uv | u in U, v in V}

[marble hinge]

I don’t know how exactly, I haven’t yet tried to solve that exercise, but I can try and share with you later, if you are interested

[ripe harbor]

Oh, sure

[stone sky]

What can you tell me about P and Q

[ripe harbor]

They are cyclic, of order p and q

So we get $G \cong \langle a, b \rangle$ with $P = \langle a \rangle$ and $Q = \langle b \rangle$

[cloud walrus]

ILikeMathematics

[stone sky]

Nice, what do you know about cyclic groups and groups whose orders are distinct primes?

[candid patrol]

Just use the second isomorphism theorem ^^

[azure cairn]

is a "congruence class" the same thing as a residue class

[ripe harbor]

They are isomorphic to Z/pZ

[ripe harbor]

How?

[stone sky]

Yes

[azure cairn]

and a residue is the representative of that class?

[ripe harbor]

(H n N)/N cong HN/N

[stone sky]

Yes

[ripe harbor]

(P n Q)/Q cong PQ/Q

Like that?

[stone sky]

namely that cyclic groups are abelian, and that groups of distinct prime order intersect trivially. Then you can use the isomorphism I showed you to prove that you indeed get the result you’re looking for

[candid patrol]

Not really sorry, I didn’t notice that you already said G = PQ

[ripe harbor]

Ah, alright

[stone sky]

Using the assumptions I outlined above, you can prove that you have an isomorphism between G and the product of the two quotient groups you have

[ripe harbor]

Can we directly show that $$(\mathbb Z/p \mathbb Z) \cdot (\mathbb Z/q \mathbb Z) \cong (\mathbb Z/pq \mathbb Z)$$ without dealing with any cartesian products at all?

[cloud walrus]

ILikeMathematics

[ripe harbor]

Since the isomorphism you showed at the beginning was the one with the cartesian product..?

[candid patrol]

I’m lagging

[stone sky]

Yes, just use a generator argument

[candid patrol]

Yes by finding an element of order pq

[ripe harbor]

Ok yeah thats what I want to do, we have Z/pZ = <a> and Z/qZ = <b>

[stone sky]

Okay perfect. Then what is the order of a and b respectively

[ripe harbor]

p and q

[stone sky]

Nice, now what would the order of ab be?

[ripe harbor]

pq

Since p and q are coprime

[stone sky]

So what can you say about G

[ripe harbor]

So now what we need is showing that ab is in G, right?

Then we are done since <ab> c G and they are of the same order (G is of order pq) => <ab> = G

[stone sky]

Well, what’s the order of G

Done

ab is in G because you already stated it’s the product of two groups. So a in P with b in Q is ab in G

Done

[ripe harbor]

Oh thats great! Thanks a lot!

[stone sky]

No problem

[azure cairn]

...but it is, right?? Zn is the identity coset 0 + Zn of the additive group (Zn, +)

[chilly radish]

I think they just want to distinguish between Zn with a subgroup structure and 0+Zn the coset

This is, like, incredibly stupid and confusing though

Oh

[azure cairn]

[chilly radish]

I think what they're trying to say is it's not a multiplicative coset?

Since you are using yhe same notation as in the multiplicative case

[azure cairn]

oh

ummmm

[chilly radish]

Still incredibly dumb and confusing way to put that

[azure cairn]

in general a subgroup is always one of its own cosets though right

[chilly radish]

Like Zn is not a coset in the sense the {Zn: n \in Z} does not partition Z

[chilly radish]

A subgroup, yea

[azure cairn]

yes

[chilly radish]

Again I think this is just trying to make some sort of semantic/notational point but failing spectacularly

[azure cairn]

i wrote notes on that page as such

but i was very confused reading that sentence on a first pass lmao so i asked about it here

[glass sinew]

always saw the integer multiples of n written as nZ, justifying the Z/nZ notation for residue classes mod n
-# you made the top 10 communist countries video?????

[raw delta]

hello. i have some questions. Could you explain the concepts of 'group', 'ring', and 'field' in abstract algebra? I am specifically looking to understand the underlying logic behind them and why they are so fundamentally important

[supple ice]

that’s not a question that’s several semesters

[elfin wraith]

They’re so important because they make precise the different ways that’s you can combine objects and their general behaviour.

Fields are the stuff you know from highschool, the real and complex numbers, you can multiply and divide, and everything commutes

A group is the information behind symmetries. Things associate, have inverses, but they might not commute and you only have one operation (think about rotating and reflecting an n-gon)

Rings are stuff that look like the integers or polynomials, you can add and multiply but probably not divide and they need not commute (all fields are rings, and all rings are groups under addition)

As to anything more specific, you’re just going to want to pick up a book like Artin Algebra

[quiet pelican]

A field is something you can do high school algebra in
A group is something you do geometry with
A ring is something you can do number theory on
/hj

Like I say /hj, but that’s a good low level motivation actually

[elfin wraith]

A very succinct summary of what I said lol

Though condemning rings to number theory is unfair, rings are actually interesting

[quiet pelican]

Commutative rings are boring
Non-commutative rings are fun

Speaking of which I should catch up on the NNR lecture I missed

[elfin wraith]

What are you doing right now? Like localisation?

[quiet pelican]

Yeah

[wraith cargo]

Noncommutative things shouldn't exist

[elfin wraith]

My course was in bijection with the Oxford one so I presume the Cambridge one is pretty much the same

[wraith cargo]

matrices aren't real

[elfin wraith]

Localisation is cool, hard though

[quiet pelican]

We just finished the regular (ie no left or right zero divisors) case, and now we’re doing the zero divisors case

[elfin wraith]

Oh we avoided zero divisors in ours, that sounds fun

[desert verge]

i know no one asked me but that sounds boring af 💀 😭

[glass sinew]

a semigroup (X, *) is a set X with an associative binary operation under which X is closed
a monoid (M, *) is a semigroup with an identity element 1 such that for each x in M, then x * 1 = x = 1 * x
a group (G, *) is a monoid where each element x has an inverse x^-1 such that x * x^-1 = 1 = x^-1 * x
a ring (R, +, *) is a triple where (R, +) is a group where + is commutative, for which the identity is called 0, and where (R, *) is a semigroup. if (R, *) is a monoid, the ring is unital. if (R, *) is a commutative monoid, the monoid identity is called 1, and the ring is commutative. furthermore, * distributes over +.
a division ring (R, +, *) is a unital ring where each element has an inverse under *; ie that (R - {0}, *) is a group
a field (K, +, *) is just a commutative division ring

hope this helps

[wraith cargo]

wdym lol

[desert verge]

localization

[wraith cargo]

oh well

it's a simple idea in principle but it's hugely important

probably one of the most important tools for constructing new rings with very interesting properties

[desert verge]

yeah local rings

rings with exactly 1 maximal ideal

omg so interesting

[elfin wraith]

It’s just like formally inverting things, I.e. adding inverses or letting you divide, and this is really easy in commutative rings, you just kinda make fractions in the obvious way. But for noncommutative rings it’s really not clear how you should do this

And in fact you can’t always do it

[wraith cargo]

well they're called local for a reason lol

In the case of commutative rings the original motivation is geometric

[elfin wraith]

Also yeah local rings are interesting, it’s a simple definition but like, that’s kinda what algebraic geometry is all about

[wraith cargo]

your ring of rational functions that behave well around a given point is in fact a local ring

[elfin wraith]

Not that I find algebraic geometry interesting but it doesn’t mean the theory is easy

[frigid epoch]

they show up in function germs

[wraith cargo]

this is why it's called "local" to begin with

[frigid epoch]

not just algebraic geometry but also complex and differential geometry

[raw delta]

i took abstract algebra lesson already some years ago. but i didnot understand the main meaning of them i think

[desert verge]

nice

[raw delta]

but i will think your message

thank you so much

i dont want formel definitions

formel definitions are everywhere

[frigid epoch]

pick a space (like R^n)

pick the ring of real-valued functions on that space (like C⁰(R^n, R))

in reality, this is not just a ring, but an R-algebra

check a point x in that space

[raw delta]

hey, your answer is realllyy gooodd

[frigid epoch]

then the continuous functions around that point x are either invertible in some small neighborhood of x

or f(x) = 0

in other words, locally around x, continuous functions are either invertible or they satisfy f(x) = 0

the functions f such that f(x) = 0 form a maximal ideal in that "local ring"

[desert verge]

omg that makes sense

because maximal ideal = set of non-units

[frigid epoch]

only in local rings

[desert verge]

if f(x) != 0 it's a unit

[frigid epoch]

in Z, (2) is a maximal ideal but that doesn't mean everything else is invertible haha

[frigid epoch]

yeah that's true!

sorry I misunderstood

[desert verge]

makes sense

[frigid epoch]

usually local rings show up in manifolds and functions

cause they all satisfy that local invertibility thing

[desert verge]

tru

inverse function theorem

for smooth mfds

[frigid epoch]

continuous functions on general manifolds
differentiable fcts on diff manifolds
smooth fcts on smooth manifolds
holomorphic functions on complex manifolds
regular functions on algebraic varieties

local rings like (Z localized at everything but p) are a bit stranger to understand

I don't fully understand them either