groups-rings-fields

[rocky cloak]

And how many real numbers have that property?

[vast verge]

Well just pi by using the sandwich theorem

[elfin wraith]

Squeeze theorem? In my algebra? Heresy

[vast verge]

I know right? Certainly wasn't expecting it

Anyway I'm pretty much finished with the chapter on ring homomorphisms, next chapter will be on polynomial rings which should hopefully be a little less intense

I'm also about halfway through gallian's abstract algebra

[tribal moss]

Compare this to C, which has 2^continuum wildly discontinuous automorphisms in addition to the identity and conjugation.

[glad osprey]

@vast verge btw, the fact that an automorphism of R fixes Q is not unique to R; any field F of characteristic 0 contains Q as a subfield, and any automorphism of F fixes Q. One way to think about it is that there is a unique ring homomorphism from Z to any field F. This ring hom must have kernel (0) or (p) for a prime p. By the first isomorphism theorem this means that every field either contains Z/pZ or Z, depending on the characteristic. If it contains Z, then it must contain its field of fractions, which is Q. Q and Z/pZ are called prime fields, because they contain no other subfields

[vast verge]

I saw this in the few collaries following the "homomorphisms from Z to rings with unity theorem"

[glad osprey]

Good to see Gallian dare to mention rings with unity

[vast verge]

You'd be surprised to know that Gallian also mentions the rare and elusive integral domain and field /j

Oh wait, integral domains don't necessarily contains unity

Cowabummer

Wait? Do they or do they not?

They have unity

Commutativity and unity always trips me up with definitions

[glad osprey]

I think most people would define them to have unity, but I'm not sure, maybe Gallian is crazy enough to say integral domains don't have unity either

[vast verge]

No, no I checked the textbook

Integral domains are commutative and have unity there

It's just that when I think integral domain I only think 'no zero divisors' and when I think field, I think 'every non zero element is a unit'

So I forget the whole commutative ring with unity part

[rocky cloak]

The question is, do they require 1 =/= 0 in integral domains?

[south patrol]

Lol I still find the choice of writing a textbook where rings don't have unity a lil baffling

[vast verge]

I always assumed that the additive identity and multiplicative identity must be distinct

Or otherwise the entire ring collapses to the trivial ring

[vast verge]

What? In other textbooks, they define a ring already with unity?

[glad osprey]

The reason we want 1 != 0 for integral domains is that we want the definition of prime ideals to be nice, right? Otherwise, if R is commutative then R itself is a prime ideal?

[tribal moss]

Yes, so the question was whether the zero ring counts as an integral domain

[vast verge]

Okay so let's say that zero is unity

Then a0=0a=0 by 0 being additive identity and a0=0a=a by 0 being unity, and this is for any a in the ring

So that means that 0 is the only element in the ring

[karmic moat]

composition of injection/surjection is injection/surjection

so yes

[fading acorn]

yes of course it is an equivalence relation

[karmic moat]

yeah composition of homomorphisms is a homomorphism

if this isn't clear to you you should check this

if the composition of homomorphisms wasn't one then we'd be in huge trouble

[tribal moss]

Well, I mean, category theory would stop working ... but would anything important break?

[karmic moat]

true

[fading acorn]

no category theory 😭 💔🥀

[karmic moat]

mathematicians are always in trouble

if they weren't then they wouldn't have jobs

[south patrol]

i would say this is the norm

[twilit wraith]

Wouldn't it be incorrect to define a ring as including unity though

Maybe im thinking of a rng

[karmic moat]

The i in ring means unity

[fading acorn]

algebraic geometers: ring stands for commutative ring with unit always

[vast verge]

Then nZ for n>1 wouldn't be a ring since it doesn't have unity

[karmic moat]

It’s an ideal

[south patrol]

I feel it is much more common just to say ring and nonunital ring

at least in my experience

[vast verge]

And an ideal is a subring which in turn must be a ring

[fading acorn]

same

[karmic moat]

Say “rng” and I’ll automatically think there’s a typo

[karmic moat]

No

[south patrol]

Ideals are subrings if you don't have identity

[karmic moat]

Subgroup under addition that absorbs under multiplication

[karmic moat]

Okay true

[vast verge]

Sorry I don't understand why you mean

I'm too Gallian-brained

So which part is wrong? Ideals are subrings or subrings are rings?

[karmic moat]

Unfortunately I enjoy life and so I have that rings contain unity. If I’m feeling frisky I’ll throw in Noetherian too

[karmic moat]

In rings with unity, ideals are not subrings

Subrings are rings

[vast verge]

Gallian calls ideals a type of subring

So that's where the misunderstanding comes from

[twilit wraith]

I think its coming from the unital stuff then

[south patrol]

Woah now we disagree

[sonic coral]

don’t forget about local

[karmic moat]

what's the matter? envious of my simple awesome life?

[elfin wraith]

I bet you even assume commutativity… sickening

[thorn jay]

hey now hey now

assuming Noetherianness is completely okay

even I assume the classes i work with are some form of Noetherian

[elfin wraith]

Yeah I’ll be honest I don’t think I would even know where to begin studying non Noetherian rings, most of my ring theory knowledge comes from the class I took called “non commutative Noetherian rings”

And the rest is all commutative nonsense where you just add conditions until it’s easy so that doesn’t really count

[near idol]

Every time one assumes commutativity, a child dies.

[twilit wraith]

https://tenor.com/view/let-it-die-gif-16851571680202593785

[tall igloo]

a ring should always be commutative. wdym ab≠ba? we learned this in middle school bro

[barren sierra]

If I say "prime ring" do people know what I mean by that?

[covert cliff]

I have heard of prime ideals?

[barren sierra]

this is not that 😔

[tall igloo]

Do you mean prime fields?

[vivid tiger]

now i do

[barren sierra]

well I want to include the integers

not necessarily Q

[vivid tiger]

(arb=0 : for all r in R) => a or b = 0

[barren sierra]

so like Z in char zero case, Z/pZ in char p case

[barren sierra]

not that

[tall igloo]

What do you refer to as a prime ring then

[vivid tiger]

oh then do you mean

Z

[barren sierra]

the subring generated by 1

[vivid tiger]

(mod p passibly)

[barren sierra]

yea

[vivid tiger]

yeah that one!

[barren sierra]

I just wanted to gauge if this was a common term

and it is not

[vivid tiger]

it is i just forgot

i think i see "prime field" more often

[barren sierra]

but Z not a field

it'd be cool if it was

[vivid tiger]

i mean, for fields

[small tendon]

I just define rings to be fields

[swift tundra]

If you are talking about rings, wouldn’t you want to include Z/nZ for any positive n? Since we aren’t limited to just fields anymore, and rings can have any characteristic.

Or maybe you are just trying to distinguish between characteristic p and characteristic 0?

[cedar vault]

.

[thorn jay]

minimal subring

[long obsidian]

I guess the ring of n by n matrices doesn’t exist or at least we must never use it 👺

[rocky cloak]

I actually wouldn't be opposed to the idea of calling commutative rings "rings" and (noncommutative) rings "Z-algebras"

The algebra of nxn matrices rolls of the toung just as nicely

[tall igloo]

Yeah no one has ever multiplied two matrices together

[karmic moat]

I wish this were the case because every time I do I have to look up the formula

[elfin wraith]

We can start studying the Weyl algebra^2

[vast verge]

Can it real be factorised into linear factors?

[tribal moss]

It cannot. The only root of the original polynomial is 1, so if it factored into linear factors it would have to be (x-1)³, which it isn't.

[vast verge]

Bizarre

I assumed it could be factorised into linear factors like the question said

[cedar vault]

Such a phenomenon does not occur in all fields

You need the field to be algebraically closed

[vast verge]

Can someone explain this?

[rocky cloak]

So a finite abelian group is a product of cyclic groups.

If m and n are relatively prime Z/m x Z/n = Z/mn, so the only way for it to not be cyclic if there is a common prime q dividing both m and n

In which case Z/q x Z/q would be a subgroup

[crystal vale]

the following proof is of if M is R-module and N is submodule of M which is Noetherian module and M/N same, then M will be Noetherian.

To show M is Noetherian, it is enough to show every submodule of N is finitely generated.

say A is submodule of M, now take image of A, f(A) under natural mapping M to M/N. Then image of A will be submodule of M/N, since M/N is Noetherian so f(A) is finitely generated say generated by {m1 +N,..., m_k +N}.

so for each a in A there exists c_i such that a = \sum c_i m_i + n, for some n in N.
Now if i take the set of such n so it will make submodule of N, which is finitely generated by n1,.., n_s ( assumption ).

So A will be generated by {m1,..., m_k, n1,..., n_s}

is it correct?

[terse flint]

I am confused with a simple question. Someone please answer. Is it true that if every group where all elements except identity have order p for some prime p then it is abelian?
I think it's not true but can't find a concrete counterexample

[south patrol]

(Sorry, made a mistake in reply)

I believe a concrete counterexample is the Heisenberg group of order p^3

[karmic moat]

this just looks like the correspondence theorem in disguise

you can argue this directly from correspondence theorem

take an ascending chain in M

eventually some M_i in your ascending chain contains N (it may be the case that this is just M itself)

[south patrol]

Like this is the set of matrices of the form $\begin{pmatrix} 1 & a & b \ 0 & 1 & c \ 0 & 0 & 1 \end{pmatrix}$ where $a,b,c \in \mathbf F_p$ (which is a bit funny) under multiplication

[tall igloo]

(i think you want a b c no?)

[south patrol]

Yes typo sorry

[cloud walrus]

Prismatic Potato

[karmic moat]

so then your ascending chain breaks into two parts: everything after M_i are submodules containing N (which correspond to submodules of M/N) and everything before M_i are submodules of N. Both M/N and N are noetherian, so this is just the combination of two finite chains, which gives you a finite chain

[terse flint]

I see.

[tall igloo]

you can probably also get some more fun examples with semidirect products of Cp's

[south patrol]

Honestly idk any other examples besides silly variations on this

Ah

[terse flint]

Okay. But we need to check if every non trivial element is of order p, right?

[south patrol]

Yes

[tall igloo]

(by this i mean, here is a construction one can use to sometimes get nonabelian groups out of abelian groups, and would probably spit out something with everything order p. i have not actually tried it myself)

[terse flint]

But in case of p= 2 the matrix with a= 1, b=0, c=1 does not have order 2. I may have done some calculation mistakes. Please verify 🙏🏻

[south patrol]

Maybe you need p > 2 sorry

But it definitely works for p > 2 at least

[tall igloo]

yeah the mod 2 one is D_4

[south patrol]

Sorry for the mistake

[south patrol]

controversial

[terse flint]

Oh. I appreciate the idea. But I have yet to study semi direct products. Sorry, I actually don't understand this method of creating non abelian groups out of abelian

[south patrol]

I have a weird monad

[tall igloo]

no worries! it's just a way to take a product of groups but kind of twisted by an action of one on the other

[tall igloo]

D_square

i will agree with D_{2n} the day we start saying S_{n!}

[south patrol]

Actually here is a funny question: say you have a map $A \to B$ of rings and a $B$-module $M$. There's a canonical map $B \otimes_A \mathrm{Hom}_B(M, B) \to \mathrm{Hom}_B(M \otimes_A B, B)$. What can we say about it lol

Like i guess this is an equivalence for field extensions

[velvet hull]

hell yeah Ill start saying that

amazing argument

[cloud walrus]

Prismatic Potato

[south patrol]

Maybe there are other cases it is an isomorphism though

[rocky cloak]

The Heisenberg group is a semidirect product of Cp acting on (Cp)^2.

Might be hard to predict if everything will have order p for bigger products

[karmic moat]

wikipedia denotes D_n as Dih_n

[south patrol]

Dih

[karmic moat]

which made me giggle because i'm basically 5 years old

[tall igloo]

horrible

[south patrol]

Oh

[karmic moat]

but i looked into it and i've literally never seen that anywhere else

[south patrol]

You have called me 5 years old implicitly

lol

[karmic moat]

you are 6

fine

[south patrol]

ok

[karmic moat]

can't wait for recess

[south patrol]

This reminds me I need to fix a wikipedia page or two

The stuff that is ""my area"" is often dodge

[karmic moat]

anyway i'd like to think someone on wikipedia deliberately went through and changed each instance of D_n to Dih_n

[tall igloo]

i will now call this group Iso_* (im(Z/n --> R^2))

[elfin wraith]

It seems like the kind of nonsense D&F would pull but I don’t even think they do

[south patrol]

*finite field extensions

Ah this is just some compactness type thing

[terse flint]

Okay thanks guys for showing me the way to build the counterexample. I really appreciate it

[rocky cloak]

$Hom_B(M\otimes_A B, B) = Hom_A(M, Hom_B(B, B)) = Hom_A(M, B)$

so I guess this is just the multiplication map

$B\otimes_A X \to X$

$b\otimes x \mapsto bx$

[cloud walrus]

Not jagr2808

[south patrol]

Wait am I cooked chat

[south patrol]

Oh for X = Hom_A(M,B)? Lol

Sure thanks

Wait not quite

[rocky cloak]

Oh wait one is Hom_B(M, B) and the other is Hom_A(M, B)

So you have to compose with the inclusion I guess

[south patrol]

Cause it is B (x)_A Hom_B(M,B) -> Hom_A(M,B)

Yeahh

It is a bit of a funny situation

Basically the point is that like given A -> B, you get this comonad - (x)_A B on B-modules

[rocky cloak]

Like without chasing through all the details, since there is a natural map between the two I'll just assume this is it

[south patrol]

And if you take duals before and after applying this you should get some monad on B-modules

But I am confused as to what it really is

¯_(ツ)_/¯

[rocky cloak]

Wait, what's the comultiplication?

[south patrol]

So like more abstractly this is just from the adjunction

But yeah the comultiplication is the map $B \otimes_{A} M \to B \otimes_{A} (B \otimes_{A} M)$

Confused why that did not render...

[elfin wraith]

Does it want curly brackets around the As?

[south patrol]

seems so (if it loads)

lol

$B \otimes_{A} M \to B \otimes_{A} (B \otimes_{A} M)$

Ragebaited

[elfin wraith]

Are we just having a texit moment

[south patrol]

Maybe

[cloud walrus]

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

[south patrol]

Oh that's amusing

[cloud walrus]

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

Prismatic Potato

[rocky cloak]

So just
$b \otimes m \mapsto b \otimes 1 \otimes m$
?

[cloud walrus]

Not jagr2808

[south patrol]

Yeah

[rocky cloak]

And then you want to dualize by applying Hom(-, B)

[south patrol]

Before and after yeah which is a bit funny

[crystal vale]

Any hint, I want an infinite group which is simple, I know A_n is simple for all n ≥ 5.

[south patrol]

But yeah I will think about this more dw, like the monad bit is more stuff I am familiar with, just this small lemma about what happens when you plug in a dual was what I wondered about

[south patrol]

Try gluing together the A_n

[rocky cloak]

You can try letting n "be infinite"

[crystal vale]

Yes i am thinking about it

[south patrol]

Actually lol this is a neat fact jagr apparently like

The double dual functor on Vect_k has a monad structure and the algebras are given by Vect_k^op lol

which is quite funny

Basically just cause (_)*: Vect_k^op -> Vect_kis (up to op) its own adjoint and it is monadic

[crystal vale]

We can prove that if G is an infinite abelian group which has a composition series then G has to be finite, right?

[rocky cloak]

But will that work.

Like Hom(Hom(-, B), B) is not the identity, so how are you dualizing here?

[rocky cloak]

At least assuming composition series are defined to be finite yes

[crystal vale]

Oh so composition series can be infinite

[south patrol]

The point is like you do the comultiplication on the dual, and then dualise that map

Like this is a case of Koszul duality heh

[rocky cloak]

But then you would need
(T^2 X)* = (T (T X)**)* no?

Otherwise I don't understand how you're doing that

[south patrol]

Oh I just mean you define a new monadU via like U(V) = T(V*)* and then the map U^2(V)) -> U(V) is uh

Wait lemme thinkies

[rocky cloak]

I guess you pass through the natural map
X -> X**

[south patrol]

Yeah

I mean tbh I have been a little imprecise - for my set up one would usually restrict attention to fg stuff and then extend formally

which simplifies some stuff

but i think using the natural map works in general anyway

Actually yeah maybe I am being silly

All this is really doing is composing one adjunction with (-)^* being "self-adjoint" I think

So then you get some big adjunction between Mod_A and Mod_B^op

[crystal sundial]

Hi, I am currently reading through this section in Dummit and Foote. I cannot see the significance of this theorem. I get that it tells us that the determinant is a special scalar that gives us something, but I don't see why this is important.

[elfin wraith]

It’s telling you what the determinant is, it is the unique alternating multilinear form that’s 1 on the identity

[knotty badger]

ofc you do have to check that this condition is nontrivial

i.e. that $\dim(\Lambda^n(V)) > 0$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

(and doing that is about as hard as constructing the determinant in the first place)

[long obsidian]

Some would say this is an intrinsic definition of a determinant because it doesn’t use coordinates to define the det I guess

[south patrol]

A nice thing is that since $\bigwedge\nolimits^n$ is compatible with composition, you recover $\det(AB) = \det(A)\det(B)$

[cloud walrus]

Prismatic Potato

[south patrol]

(Painlessly)

And the idea of "top exterior power" is geometric and can be generalised to other contexts

e.g. in manifolds

[crystal sundial]

I see, thank you all for the insights! I will take a closer look after I eat.

[south patrol]

Np

[south patrol]

I guess depends a little on how we are defining these exterior powers aha

But this is a good point

[karmic moat]

Didnt know \nolimits was a thing

Cool

[crystal vale]

say G is finitely generated group, is it imply every subgroup of G is finitely generated? give hint

[unborn terrace]

No. This fact is true of abelian groups (just use structure theorem), but there are noncommutative groups for which it does not hold.

LMK if you want a hint on how to construct one

[crystal vale]

you mean if G is abelian finitely generated group then every subgroup is finitely generated

[crystal vale]

yes

[unborn terrace]

Think about the commutator subgroup of the free group on 2 letters

[knotty badger]

how would you define them?

[warm ember]

why are normalizer groups subgroups?

i understand in the finite group case

gSg^-1=S because everything is finite and they have the same cardinality

but if G is infinite cant you have gSg^-1<S

[tribal moss]

What's your definition of "normaluzer"?

[warm ember]

set of all g such that gSg^-1<S

wait thats not the normalizer

is it?

idk

[tribal moss]

I would require equality not inclusion.

[warm ember]

ye thanks

[tribal moss]

Indeed you can -- for example, let G be the group of permutations of Z, and S be the subgroup that fixes all of the negative numbers. Then conjugating with a right shift gives you a strictly smaller subgroup.

[warm ember]

in that case the right shift would not be in the normalizer group right

[tribal moss]

Right.

[candid patrol]

You can think of it as the stabilizer of a certain group action

[tribal moss]

Namely the group acts by conjugation on the set of all its subgroups.

[empty kernel]

I need help with c
it asks with what group the quotientgroup is isomorph

G is a commutative group

I waas thinking that it is isomorph with Z_n but im unsure

where n is the dimension of g

and i for sure dont know how to prove

[rocky cloak]

You can try to think about what kind of map would (g,g) in its kernel

Like what can you do with (x,y) that gives the identity iff x=y?

[empty kernel]

yeah i did x-y

more general i should do xy^(-1)

wait is it isomorph with G again?

oh yeah

omg yeah that works lol

dank

[thorn jay]

andere nederlander🔥

[elfin wraith]

Silly language

[slate bramble]

Hi! A question on a practice mGRE set I was doing asked whether Z[sqrt 2] is a unique factorization domain. I realize that I don't know how to show that a ring is a UFD (although, after review, I understand how to show that some rings aren't UFDs with obvious counterexamples).

Is there a simple rule for this case, since Z[sqrt D] is a canonical ring extension, or would there be some more general strategy for this?

(I see that there are proofs online that it's a Euclidean domain as well, but I'm curious specifically for the intuition about being a UFD, although I understand that ED -> UFD.)

[elfin wraith]

I dont actually know many general ways to directly show a ring is a UFD, however wikipedia lists some equivialent conditions that may be of use. https://en.wikipedia.org/wiki/Unique_factorization_domain#Equivalent_conditions_for_a_ring_to_be_a_UFD.

IIRC for Z[sqrt d] its not a UFD for d<=-3 or d>=3, or something similar to that anyway (I dont have the time to reprove this for myself right now)

I think im wrong about the positive d actually, im pretty sure thats the right condition for negative d though

[slate bramble]

thanks! they say that (2) is useful for proving this, but it's not obvious to me how; in particular, if I take arbitrary prime ideal I \subseteq R and want to show that I is generated by an element p \in R, how would I go about this? My first thought is using a norm function and showing the existence of an element with minimal norm, but this probably wouldn't work for non euclidean domains?

[elfin wraith]

Sorry, slow moment there lol

[slate bramble]

haha no worries

oh it seems this is related to the class number problem

so there is a necessary and sufficient condition i think, but it's just not feasible to use for this type of exam

[elfin wraith]

IIRC for the specific case of sqrt 2 theres a number theortic style proof but I dont remeber it off the top of my head and im not at home just now to think about it properly, but im sure I did it in my number theory class

[elfin wraith]

Yeah I think in general this isnt such an easy thing to do, but theres people here who know a lot more comalg than I do how can hopefully be of more help

[slate bramble]

this is so interesting lol i've been thinking about hwether i should do a statistics phd lately since i like analysis so much but i've forgotten how much i enjoy algebra/nt and it's reminding me why i wanted to do a math phd

[slate bramble]

Oh also I'm realizing this is a more interesting result than I thought, for example in the case of a non PID like Z[x] where (2,x) is maximal -> prime but is not generated by a prime element

[rocky cloak]

So there are some properties of Z[sqrtD] in particular that can be used I guess.

It's 1 dimensional, so it's enough to prove that all primes are principal.

Z -> Z[sqrt D] is an integral extension, so every prime ideal lies over some prime (p) in Z.

If D does not have a square root in Z/p, then (p) is prime.

If a is a square root of D modulo p, then (p, a-sqrtD) is a prime ideal and the question reduces to where this ideal is principal.

Which should reduce to whether there is x and y with
x^2 - Dy^2 = ±p

which I guess might be hard to determine in general... But easy if D is negative

[slate bramble]

thank you for the explanation! i think maybe i'm not quite all the way in understanding this 😭 i gather that dimension is krull dimension but i haven't studied this yet

[south patrol]

I would just show the norm defines a Euclidean function. It is quite subtle beyond that - the general way I would show a number ring is a PID (equivalent to a UFD for number rings) is by using the Minkowski bound which boils down to only checking whether a few prime ideals are principal. But the Minkowski bound will be "too good" for this problem

[slate bramble]

thank you!

i think maybe i'll wait until i take grad ring theory or nt and revisit the problem in its fullest generality 😭

[south patrol]

Of course

On a related note, note that whether a number ring is a ED is also pretty subtle

(because to prove something isn't a ED you have to show that no possible function could work, and there isn't an obvious obstruction if ur ring is actually a PID)

[slate bramble]

gotcha

yeah nonexistence results seem hard in general lol

[unborn terrace]

It's easy to do when D is negative because then the LHS just expresses the squared absolute value of an element of the number field, so the problem reduces to geometry

[toxic sapphire]

is there a simple way to show that the group of rationals under addition is not finitely generated without the fundamental theorem of finitely generated Abelian groups?

[thorn jay]

look at the primes which can appear in the denominator of elements in the subgroup generated by { a1/b1, ..., an/bn }

[south patrol]

You can even show Q isn't a finitely generated ring by the same argument

[thorn jay]

if Q were finitely generated, that would contradict a certain popular theorem

unless.. you're a finitist...

[rocky cloak]

Write Q as the union of Gn := (1/n) for all n.

If Q is finitely generated there is an N such that all generators are contained in the union up N. This is contained in G(N!), but Q is not equal to G(N!)

[dull ginkgo]

Heegner nightmare

[dim wagon]

im having trouble following the proof here

[knotty badger]

universal properties :D

[dim wagon]

how are we concluding that $\Psi\Phi=\operatorname{Id}_{F_1}$

[cloud walrus]

somethingwrong

[knotty badger]

it's cause of uniqueness

the composite makes the diagram commute

but so does the identity

therefore they have to be equal

[dim wagon]

wouldn't only $\Phi$ and $\Psi$ have to be unique, why would their composite have to be unique?

[cloud walrus]

somethingwrong

[knotty badger]

their composite is a morphism from $F_1 \to F_1$ making a triangle commute, right?

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[dim wagon]

ahh okay

[knotty badger]

but the identity is also such a morphism

hence, by uniqueness, it must equal the identity

(it is a bit of a dumb proof to be fair)

[dim wagon]

so if i were to look at this definition, im choosing $\varphi=i$ and $M=F_1$ right?

[cloud walrus]

somethingwrong

[knotty badger]

yep!

[dim wagon]

ahh okay it makes sense to me now

[knotty badger]

it's maybe a little non-obvious from the notation that M can also be chosen to be F_1 itself

because it's often phrased as "for any other R-module M, ..."

[thorn jay]

thats really poor phrasing 😦

[dim wagon]

is there any reason, my prof wrote this down? this doesnt seem to help me

[knotty badger]

idk

[dim wagon]

its copied by me, i will change it

[knotty badger]

i prefer doing it via naturality and yoneda

but that requires a bit more familiarity with cat theory

[dim wagon]

okayy thanks for the help, i will check the purpose of this part with my prof tmr

[knotty badger]

[knotty badger]

honestly the "exists a unique morphism such that the diagram commutes" is such an unfortunate way to phrase universal properties 😭

[elfin wraith]

Why so?

[knotty badger]

ok so i have a good analogy for this

suppose $X$ and $Y$ are sets

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

we say that a function $f : X \to Y$ is a $\textit{bijection}$ iff for every $y \in Y$ there exists a unique $x \in X$ such that $f(x) = y$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

we say that a function $f' : X \to Y$ is an $\textit{isomorphism}$ iff there exists a function $g : Y \to X$ such that $g \circ f' = \text{id}_X$ and $f' \circ g = \text{id}_Y$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

as you're likely well aware, these are equivalent definitions, right?

[thorn jay]

probably comes from algebra

because for algebraic structures it is equivalent to being generated by a set, if the function is uniquely determined by where it sends the set

[elfin wraith]

Yep, bijections are isomorphisms in set

[knotty badger]

the thing is, despite being logically equivalent, these definitions are not necessarily practically equivalent

for example, it's really easy to turn a function $h : X \to Z$ into a function $h' : Y \to Z$ using the isomorphism definition - you just define $h' = h \circ g$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

but with the bijection definition, you have to work "in coordinates" by saying "for every y in Y, there exists a unique x in X with f(x) = y, we then define h'(y) = h(x)"

if X and Y had more structure (say, were groups), you'd have to do more work in terms of showing h' is a homomorphism as well

whereas with the isomorphism def, this just follows from composite of homomorphisms is a homomorphism

also, if you had a chain of bijections, proving the composite is a bijection is again a bit of an annoying "coordinate" proof

"for every z in Z there exists a unique y in Y such that f(y) = z, and for this y there exists a unique x in X such that g(x) = y, so we have an x such that (f o g)(x) = z. moreover this x is unique because..."

whereas with the isomorphism def, it just follows from $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

do you see what i mean?

in a sense, the bijection def is easier to prove, but the isomorphism def is easier to use

[elfin wraith]

Yeah, with you so far

[knotty badger]

the same phenomenon happens with universal properties

"there exists a unique morphism making the diagram commute" is analogous to the "bijection"-style def

it's easier to prove, but means a lot of arguments have to be done "in coordinates" with careful manipulation of commutative diagrams

on the other hand, "is naturally isomorphic to a certain functor" is analogous to the "isomorphism"-style def

it's easier to use in a lot of scenarios, especially for things like considering multiple universal properties at once

if all you know of universal properties is the bijection-style def, then all your arguments involving them have to be bijection-style

and this can make the arguments... unnecessarily awkward

[knotty badger]

akin to this

imo a good balance is knowing both defs and also knowing why they're equivalent

that way you can establish a universal property with the bijection-style def, but then afterwards you can use the isomorphism-style def for further arguments

[elfin wraith]

Ok yeah I see what you mean, though im not sure I would go as far to say that the unique map such that the diagram commutes is an unfortunate way to put it, it's true and more approachable, which I think makes the idea much more concrete (because it can be a bit easy to get lost in the full definition), but I take your point, I think thats a nice anology

[knotty badger]

yeah i don't think one should eliminate the "unique map such that the diagram commutes"-style def

any more than one should eliminate the term "bijection"

it's just that, if it's the only definition you know, it can make your life a lot harder than it needs to be

(source: me suffering through my cat theory course during my masters 😭 )

[elfin wraith]

Yeah no I agree with that for sure lol

I guess in my head ive always treated the unique map thing as the idea rather than the definition, but I will admit that I do tend to over rely on the unique map style thing, which has likley made my life harder at points

[knotty badger]

so have i!

[glad osprey]

But you can't always formulate a universal property in the isomorphism-style, can you? For example the universal property of the product?

[knotty badger]

sure you can!

they're equivalent formulations, after all

[glad osprey]

What would it look like for the product? I'm thinking of the definition where you have objects X, Y and projections pi1, pi2, then X x Y is a product if for every object H and f : H -> X and g : H -> Y there is a unique morphism such that f and g factors through X x Y (this was kinda poorly written, sorry)

[knotty badger]

ah, so

pairs of morphisms Z -> A, Z -> B naturally correspond to morphisms Z -> A x B

in that you can interconvert between the two

so what the product "does" is let you represent pairs of morphisms by a single morphism, "packaging" them together, in a reversible way

more formally, $\text{Hom}(-, A \times B) \cong \text{Hom}(-, A) \times \text{Hom}(-, B)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[wraith cargo]

I feel like this works for small diagrams but for more complicated things I feel like this is a very slippery slope to go down

[knotty badger]

it works for all limits and colimits

as well as adjoints

[rocky cloak]

Given the diagram

AxB -> B
v ^
A <- H

There is a unique map from H to AxB making it commute

[knotty badger]

i don't really know what you mean by "slippery slope"

tons of things can be (and are) expressed in terms of representability

[wraith cargo]

eh well maybe I'm wrong
I just felt that as a way of thinking this could lead you to weird results when combining diagrams
Like I can't really see it for something like the inverse/direct limit
How exactly this type of reasoning would go
But realistically I'm probably just unfamiliar with this type of reasoning

[knotty badger]

so for the direct limit

[kind temple]

this follows from the fact that the yoneda embedding preserves limits, i.e.,
C(-, lim D) = lim C(_, D-), where D : J -> C is a diagram and C(_, D-) : J -> [C^op, Set] is a diagram in the category of presheaves.

[glad osprey]

I see

[wraith cargo]

do these ideas still work if your category isn't locally small?

[glad osprey]

Noo, you can't say unique map, that's the bad definition 😭

[knotty badger]

so morphisms $(\lim_{n \to \infty} A_n) \to Z$ naturally correspond to compatible families of morphisms $f_i : A_i \to Z$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

yes

[wraith cargo]

hmhmhm this is pretty cool

[knotty badger]

:D

representability-style definitions can be formulated "pointwise" in a way that sidesteps any size issues

so again, you can view the direct limit as a "packager"

[rocky cloak]

@knotty badger do you think about the universal property of tensor product with bilinear maps in some representability way?

[knotty badger]

it "packages" a compatible family of morphisms into a single one

in a reversible way

[knotty badger]

yes definitely

[rocky cloak]

How so?

[kind temple]

not sure…

[knotty badger]

bilinear maps $V \times W \to Z$ naturally correspond to linear maps $V \otimes W \to Z$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

so what the tensor product "does" is let you represent bilinear maps via ordinary linear maps

in general representability-style defs feel more suited to viewing through the lens of "is-does duality"

[rocky cloak]

Then I guess I don't really understand the difference

[knotty badger]

I did go to great lengths to outline the difference between these styles…

[wraith cargo]

I think it's a language thing mainly

how you frame these things in your head

[knotty badger]

.

[rocky cloak]

So the "diagram" perspective is that there is a bilinear map
VxW -> V(x)W
such that every bilinear map is given by the composition
VxW -> V(x)W -> Z

and the "representability" perspective is that for every bilinear map there is a map V(x)W -> Z?

And then the key difference is in surpressing / not thinking about this bilinear map VxW -> V(x)W ?

[knotty badger]

Yes, you can derive the latter bilinear map from the representability perspective

In my experience it’s often cleaner

It’s like the difference between talking about a linear map V -> W, and representing it by a specific matrix

[covert cliff]

1 -> N -> G -> H -> 1 where G -> H is f: (n, h) |-> h gives f(i(n)) = f(n, 1) = 1 so this is an SES and boom H \cong G/N

does this argument work?

[kind temple]

you don't need the language of SES's, but yea. this is just the first isomorphism theorem applied to your projection f : G -> H

[untold torrent]

c^2aca=e

c^2ac=a^-1

if i multiply by a both right side then

What should I do next?

[proud vigil]

maybe plug them all in and see which ones work

[untold torrent]

Could you tell me how can I plug option A?@proud vigil

[tribal moss]

This looks like a trick question. From c(ca)²=1, if we set b=ca we get
c = b^-2
a = c^-1 b = b^3
which implies that a and c commute, and both answer options (1) and (2) are right.

On the other hand (3) and (4) can be eliminated by considering (Z,+) with c=-2 and a=3.

[How did I discovered this: First I eliminated 3 and 4 by thinking "what if the group is abelian"; then it's just a linear equation to find examples of a and c. Then, in the hope of finding a counterexample for one of 1 and 2, I thought about how one might go about constructing a non-abelian example by brute force -- say, in a small permutation group -- and found they always ended up commuting.]

[tribal moss]

"Plugging in" doesn't feel like a promising strategy here; since (1), (2), (4) all contain a, after we plug it in we still have a mystery equation in both c and a, which seems to leave us none the wiser.
And even plugging in (3) would just yield c^9=e, which could perfectly well be true in some group -- in fact, one example of c²ac=a^-1 would be Z/9Z with c=1 and a=3.

[untold torrent]

a, c commute how did you check?

[tribal moss]

That's what the post you reply to explains.

[glad osprey]

oh, because a and c are both powers of b took me a while to get it

[languid trellis]

What is meant by 'quadratic character' here? It's the second time this phrase is used in the book and the first one has just as little explanation.

I'm guessing 'quadratic character of 2' means the value of the legendre symbol (2/p) or (2/q) but I have little idea

[steady linden]

Has anyone here ever played with finitely presented groups on sage?

[peak condor]

Can anyone check a proof for me?

[cursive spindle]

Yes it means exactly what you're thinking

[cursive spindle]

[cursive spindle]

Yes

[peak condor]

Basically, I’ve been labouring over this proof because I didn’t use one of the conditions so I felt it was incorrect, but I don’t see any logic errors and I just need reassurance that it’s correct

[cursive spindle]

Skimming through looks fine to me

I haven't checked in detail

[peak condor]

Thanks I appreciate you time seriously I have been stressing for whatever reason

[cursive spindle]

Yeah np

I feel like I recognize you from somewhere

Oh yeah of course lmao

[glad osprey]

Yeah, looks correct 👍 I feel like I have seen a generalisation of this, where you have (ab)^p = a^p b^p and (ab)^q = a^q b^q for coprime p and q, then conclude G is abelian. Is this true or am I misremembering?

[hexed cargo]

This is not true - take $G = (\mathbb{Z}, +)$, $N = 2\mathbb{Z} \leqslant \mathbb{Z}$, $H$ trivial

[cloud walrus]

Adayah

[wraith cargo]

it is very much true!

the issue is that you're think the map 2Z -> Z is the inclusion, but in fact it's the isomorphism 2Z = Z!

the problem statement is very subtle

and actually maybe even wrong lol

I don't know if H should be a subgroup a priori

[hexed cargo]

which assumption do you think my counterexample fails to satisfy?

[hexed cargo]

the problem statement doesn't mention any "map" from N to G, it's just a normal subgroup

[rocky cloak]

It should probably say that G equals (the internal) semidirect product instead of saying it's isomorphic.

[wraith cargo]

I think it should probably say that G is the internal direct product of N and H, not that it's isomorphic

so yeah ig you're right I was reading the problem wrong

[rocky cloak]

Well you were reading it correctly, it was just written wrong

[wraith cargo]

😭

[rocky cloak]

The point of communication is to understand what the author meant after all. Not to nitpick their typos

[wraith cargo]

I always forget that the 5 lemma fails for Grp

so these facts are so weird to me

tho actually

Z is abelian

so why does this fail lol

[rocky cloak]

It does?

[wraith cargo]

I heard this somewhere once but I forgot where maybe it's wrong

[hexed cargo]

The point of mathematical rigour is to express yourself clearly, not have others guess what you have in mind

[tidal schooner]

It actually does work in Grp, and more generally in any homological category

[wraith cargo]

ohhh okay very interesting

then I misremembered haha

[wraith cargo]

lol

I wish papers were written this way

🙏

[hexed cargo]

The discussed problem is a perfectly correct formulation of a false theorem

How is the reader supposed to know that something else was intended?

[hexed cargo]

I wish too, but sloppines of papers or maybe incompetence of their authors should not be a reason to lower your own standards and promote it to others

[wraith cargo]

I think that here the issue was that people sometimes really like to conflate isomorphisms with equality and it's very important not to

and this is kind of an epidemic issue in math

[rocky cloak]

Context for one.

And the fact that one of the interpretations is clearly wrong as per your example

[wraith cargo]

if you ever ask kevin buzzard about this he'll talk your ear off lmao

[hexed cargo]

I think your speech could use a little more verbs

Otherwise I honestly have no idea what you just said

[glad osprey]

I'm pretty sure he meant to reply to the message above

[hexed cargo]

That's a reasonable guess

[glad osprey]

I knew that because I read jagr's message with context in mind

[rocky cloak]

And indeed it was correct

Like the problem is a little sloppily written, so I'm not saying it's bad to point out what they meant or whatever

[hexed cargo]

I don't think context indicates that a different statement was intended

The fact that the interpretation is wrong should make me think the author had something different in mind?

So first of all, people never really mean to formulate actually false theorem?

And second of all, I have to look at every possible way to change the assumptions to make the theorem true, just because the author was not careful enough to write it down correctly?

[rocky cloak]

Unless they're trying to trick you, or are asking for it's validity. No that's not my experience

[hexed cargo]

Plenty of times I had people ask me to prove flat out false theorems

[rocky cloak]

No, you don't have to look at every possible way. It's okay to be confused. Miscommunication happens, and you can blame the speaker if you like

[hexed cargo]

I don't necessarily blame the speaker, it's just that I don't think I am to blame for giving a correct counterexample rather than spending my evening guessing what was intended

[wraith cargo]

I don't think it's about lowering standards it's more that math communication is more complicated than just pure rigor and a lot of stuff can go wrong
Secretly math isn't about rigor at all, rigor is just a final check to make sure that the ideas you have are correct so it's more of a check and balances thing than the actual essence of mathematics
And when you're senior enough most mathematicians switch over to a more conceptual type of thinking in my experience

[hexed cargo]

And that response:

The point of communication is to understand what the author meant after all. Not to nitpick their typos

certainly had that vibe

[hexed cargo]

When you're senior enough, is when you truly value the clarity of speech

[rocky cloak]

Your not to blame. It is good to clear up potential miscommunication. So that would make you the opposite of to blame I guess (to honor?)

[hexed cargo]

It's not about rigour just for the sake of it, it's about making sure that others understand you

I've seen plenty of senior mathematicians throgouhly confused and displeased by sloppy delivery

[rocky cloak]

Then I'm sorry for giving bad vibes I guess

[hexed cargo]

Already forgot about it 🙂

[wraith cargo]

that's a very personal thing
Every mathematician I've met has a totally different things they like and you can't please everyone
And then at some point I think when you try to communicate things you have a style that pleases you personally and that can lead to errors like so

this doesn't excuse wrong statements ofc

it's more that there's a rhyme and a reason as to why these things happen and there isn't really blame to pin on just sloppy work

[hexed cargo]

I'm just saying that against a false theorem you have every right to just state a counterexample

If the problem author wants to continue the discussion, it is on them to correct the statement as to make the counterexample invalid

From some level upwards, that is

[rocky cloak]

I mean, exercises in a book aren't really something you're solving for the author. It's not like an advirsarial relationship

I feel like asking, could they have meant something else? Should be the reasonable next step.

But if there isn't another reasonable interpretation then you'll just have to move on

[hexed cargo]

Fair

But strong emphasis on why the theorem in its current form is false can be really englightening

If you just casually assume the corrected interpretation, the asker might assume there was nothing wrong with the original

That it's just a different phrasing

In this case, for one, the difference between just being isomorphic and being isomorphic through the most "obvious" isomorphism is fundamental

[tribal moss]

Ah, sorry, perhaps I should have stressed that more.

[rapid cave]

Hi, does anyone have any hints to this exercise? Also by closure here they mean a topological closure or just the derived subgroup of G?

[tribal moss]

Hmm, I suspect as soon as the word "profinite" occurs we're at least in #advanced-algebra territory. That’s Scholze stuff, isn't it?

[rapid cave]

neukirch ANT

[wraith cargo]

profinite means inverse limit of finite groups

[coral spindle]

The commutator subgroup is another name for the derived subgroup, so indeed (unless I am making a terrible mistake, which I am sure is possible) they mean the closure of the derived subgroup.

You are probably aware that for plain ordinary groups, any map G → A for an Abelian group A factors through G/[G,G] since the kernel of the map G → A always contains the commutator. What you need to show is that it also contains the closure of the commutator, which should be an argument involving closedness.

I feel like if I lay this out too clearly I will solve the whole thing for you, so I won't say more.

[rapid cave]

thank you man

[woeful sail]

Does anyone have any hint on this? Let p be a nonnegative integer. What group is determined by the group presentation with generators {x, y, t} subject to the relations {x^2=y^2 = 1 = (xy)^2, t^p = x}?

It comes from doing Van Kampen, I can clearly erase x from the presentation but I would like something more familiar, I don't know if it is a familiar group I just hope

[rocky cloak]

It doesn't seem to me like for example (yt) would have finite order, so probably not a very familiar group.

If you mod out by x, then you get the free product of C2 and Cp, so you can think of it as a central extension of this by C2 if that helps

[rocky cloak]

What is the space you're computing fundamental group for?

[glad osprey]

What is the additive group of GF(p^q)? Is it just (Z/pZ)^q, by the classification theorem of f.g. abelian groups and the fact that each element has order <= p?

[sonic coral]

yes

[south patrol]

Yes, but here is an easier way to see it: the canonical map F_p -> F_(p^q) gives the latter the structure of an F_p-vector space.

(And then the dimension has to be q by counting)

[glad osprey]

Ah right, that's a nicer proof

[tribal moss]

Alternatively, the construction of F_(p^q) as Fp[x]/<some irreducible degree-q polynomial> gives you the Fp-vector space structure (almost) explicitly.

[swift tundra]

Iirc in Dummit and Foote they define F_(p^q) as the splitting field of x^(p^q)-1

Over F_p of course

[tribal moss]

Yeah, it takes a bit additional work from there to see that it can also be made as a simple quotient of Fp[x].

[rocky cloak]

Whether you write is as a quotient of Fp[x] or Fp[x1, x2, ...] It's still a vector space though

[tribal moss]

True.

I had just internalized the "quotient of Fp[x]" construction enough that it was the first that sprung to mind.

[south patrol]

I just want to emphasise that this is a very general thing: fields admit a unique map from their prime field (Fp or Q) and hence become vector spaces over those. I don't know how helpful referring to explicit constructions is for that, since the Fp vector space structure on those comes from them being Fp-algebras anyway

Note the same argument shows any finite field has cardinality F_q with q a power of a prime

[steady linden]

||Sorry for changing account but I did not realize i was logged in another one on the phone|| I don't really know haha I was given by my advisor and I guess it come from some Van Kampen given the structure and what he does

I don't even know what he really wants, just played with a bit for two days and only got your same conclusion, I don't really see how to think about the problem if not by some random Tietze moves

I would say that it in some sense it is similar to this, but I don't really believe it, I think it is just the need to find something haha

[rocky cloak]

Your advisor just gave you a group presentation, and asked you to do something with it, but not saying what that something was?

I'm confused

[steady linden]

Yes. We have a meeting scheduled next week, hopefully if will become more clear but it is literally what happened

[warm needle]

Maybe I'm missing something obvious, but I'm not sure how I can derive anything else from this table just by knowing the given entries and that G is a group.

[stark mist]

each row and column will have an element only once, so can you eliminate some possibilities and figure out what bb = dd should be?

[coral spindle]

Well here's an example: is it possible that b, c, or d are the identity? If not, what is the identity?

[warm needle]

Ah. I guess not because (let # be the operation symbol)
If b was identity element, b#c=b, but b#c=d
If c was identity element, b#c=c, but b#c=d
If d was identity element, d#b=d, but d#b=c
So then we say that because an identity element must exist, by cases, a is an identity?

Then at that point we basically fill in top row and left column.

From there, we still have two empty spaces in each row. Maybe there's a way to solve that with logic?

[coral spindle]

There is indeed a way to solve that with logic

[warm needle]

Like I can leverage my b#b=d#d and a string of disjunctions.

[coral spindle]

and indeed, there you go, a is the identity so you've just got 7 entries for free

[warm needle]

Thanks

[warm needle]

What is this property called by the way?

That x,y,z \in G, *:<x,y> \to z is surjective?

[glad osprey]

you can think of it as left multiplication by g being a bijection, so it permutes the elements a, b, c, d

[stark mist]

well, if you had a row with two same entries, then you'd have something like b . c = a and b . d = a. Can you see why that's not possible? nvm I guess you were asking for the name: bijective.

[coral spindle]

People sometimes call it the Latin Square Property

but in particular, it's actually a bijection

This could be seen as something called Cayley's theorem, but that's a story for another time.

[warm needle]

hmm, I guess I just don't recall (and don't see in my notes) my professor mentioning this when defining groups. The latin square thing

But I mean I definitely see how I can fill in the remaining rows under the assumption that each element must appear once. Like the sudoku property

[coral spindle]

Well, it's a consequence of the fact that every element has a two-sided inverse. It's not part of the definition of a group because all groups already have this property.

it's a bit of a shame your professor didn't mention it though.

Note: not all latin squares are groups.

[warm needle]

Well, she mentioned the two-sided inverse. I just don't think she showed that such a result guarantees that each row/column has an entry appear exactly once. Kinda feels similar though to what we were talking about with cyclic groups.

[thorn jay]

(Im people)

[coral spindle]

I should hope so

[thorn jay]

that would have as disastrous consequence that the trivial group can have more than one element

[tribal moss]

Did you figure out why this is true? If not it's a good exercise to try to construct a proof of it.

[white oxide]

Why is $\sqrt{(x, y)^2} = (x, y) \subset k[x, y]$

[cloud walrus]

okeyokay

[warm needle]

Not yet, but I decided I would just come back to that problem a bit later. I feel confident that I can probably prove it but if I get stuck I’ll come back here with my progress.

[tribal moss]

I suppose that's a radical ideal? Is there something else than <x,y> you'd rather want it to be?

[velvet hull]

hint - ||(x,y) is a maximal ideal||

[cedar vault]

In general you can prove that ||The radical of a monomial ideal is a monomial ideal and is generated by the radicals of the minimal monomial generators of the ideal||

[cedar vault]

Where the ||radical of a monomial is just the squarefree part of the monomial||

[tardy hedge]

What up mathletes

[thorn jay]

hello hello

[elfin wraith]

Jürgen Herzog?!

[rocky cloak]

Posting on discord from beyond the grave

[elfin wraith]

The man’s just that dedicated to monomial ideals

[tardy hedge]

Y is multiplying by g a homeomorphism in a topological group? Is it continuous because its just a restriction of G x G -> G map to {g} x G -> G

[thorn jay]

yes, both it and its inverse are continuous cuz theyre restrictions of the multiplication to {g} x G and {g^-1} x G

[static mauve]

I've spent a couple hours trying to prove the following, but if anyone wants to give me a tiny hint, that would be appreciated.

If we have a cyclic group G, and a group homomorphism f: G -> G, show that f(x) = x^n for some n when x is an element of G.

It's obvious that this doesnt hold when f is not a homomorphism. Since homomorphisms have some very strict requirements, I was trying to see if any of the things that are preserved under homomorphism would lead me down a good path, but alas I havent had success there.

Maybe something about subgroups being preserved under homomorphism is the trick.

[rocky cloak]

So the fact that G is cyclic is very important. Maybe fix some generator g and think about f(g)

[elfin wraith]

Use the fact that the group is cyclic to give you another way to write x would be my advice, there isn’t a huge amount to do in this problem

[static mauve]

Yeah, Im assuming Im missing something blatantly obvious.

Since G is cyclic, G = <a> which means any element x = a^n for some n.

f(x) = f(a^n) = f(a)^n

and I know that f(a) is an element of G as well... I want to say that f(a) is the generator for f(G), but I havent proven that.

[velvet hull]

ignore my previous message, oh wait you said f(G)

it is true, but I don't think it's necessary for you to prove the question

[coral spindle]

element x = a^nelement x = a^nelement x = a^nwhich means any element x = a^n for some n.which means any element x = a^n for some n.which means any element x = a^n which means any element x = a^

HUH

[elfin wraith]

Discord moment

[coral spindle]

Discord moment indeed

Let me try again

[coral spindle]

which means any element x = a^n for some n.

Correct, and f(a) is also an element of G. So...?

That was very strange

[thorn jay]

digital stroke

call the bondulance

[coral spindle]

It was very strange, the message I thought I was sending looked literally nothing like the one that arrived

I shall keep it up for the preservation of curiosities

[thorn jay]

maybe an outerspace particle flipped a bit

[coral spindle]

I think more likely discord has a bug :P

[static mauve]

Since f(a) is an element of G -> f(a) = a^m... but m and n are seemingly unrelated

[rocky cloak]

If f(a) = a^m what does f(a^n) equal?

[thorn jay]

did someone talk about an insect collection?

[static mauve]

f(a^n) = f(a)^n = (a^m)^n = a^mn = (a^n)^m => f(x) = x^m

[coral spindle]

They are indeed unrelated

Well done, you've provided a proof!

[static mauve]

Idk why that took me so long to see, but thanks for all the help everyone. I now feel very dumb 😛

[velvet hull]

the good part about spending a day on a proof is that you'll never forget the question ever again

and it sounds like a joke but I'm being completely honest, its the most permanent way of learning something you'll ever get

[static mauve]

Yeah, ive been trying really hard to not look up answers too quickly... always feels like im cheating myself

[coral spindle]

Gonna steal this line for the future

[elfin wraith]

You underestimate me

[knotty badger]

This is true

I remember spending 2 weeks on trying to solve an analysis Q

[south patrol]

Spending only a day on a proof sounds amazing to me now

[uneven bobcat]

Do not underestimate my ability to forget things.

[elfin wraith]

I’m very good generally at remembering I have done a problem, it’s more so the how I tend to struggle with

[unborn terrace]

Is there a notion analogous to projective modules for groups in the sense of a group which is always normal whenever it is a subgroup?

I guess 0 group does this, but are there others?

[thorn jay]

if you mean that the image of every injection must be normal, then the group must be abelian, as { (g, g) } < G x G is normal iff G is abelian

[elfin wraith]

A group which is always normal whenever it’s a subgroup? Assuming you mean the other way around this is just abelian groups

I ’m possibly misunderstanding what you mean though

[tribal moss]

No -- embed your group in a "more than sufficiently large" symmetric group and it won't be normal there. (Unless it is trivial).

[thorn jay]

oh, yeah that's way easier lol

[elfin wraith]

Ah ok yeah that makes more sense to me lol, I wasn’t sure what he meant

[thorn jay]

a perhaps more interesting example is the holomorph, the semidirect product of a group with its automorphism group using the natural action

[unborn terrace]

Do you have a more specific construction? I'm not sure how to do this?

[elfin wraith]

Actually this isn’t even true, all subgroups of the quaternions are normal and that isn’t abelian

[thorn jay]

this action won't be trivial; the only group with a trivial automorphism group is the trivial group, so we do not have that G is a normal subgroup of G \rtimes Aut(G)

[thorn jay]

there's a weird ass classification theorem for those groups

[tribal moss]

For example, let G act on G cup {*} by left multiplication for the G part, and g.* = * always.
The injects G as a subgroup of S(G cup {*}).
However, now every nontrivial element of G has a conjugate that doesn't fix *, so it's not in the original subgroup, so the subgroup is not normal.

[elfin wraith]

Really? I guess that’s quite a strong requirement for a group to have but that still feels crazy

[thorn jay]

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

[elfin wraith]

Do you know what they’re called?

[thorn jay]

it's not very convoluted like with simple groups, but it just feels very arbitrary that it's either abelian or the quaternions x elementary abelian 2-group x torsion abelian group with only elements of odd order

[elfin wraith]

That is a whacky ass classification

[thorn jay]

it is perhaps a very strange thing to ask to force every subgroup to be normal

it's a miracle in itself that congruences correspond to some kind of subgroup at all

[elfin wraith]

Yeah this is why I could believe there was a classification in the first place because this is quite a strong condition, so there’s likely something you can say, but it is an odd feeling classification

I’m guessing it makes some amount of sense if you see the proof though, I imagine that in quite a logical way you have to build them like that

I mean yes that’s how proofs work but you get what I mean haha

[thorn jay]

mhm

it's still weird

math is full of weird stuff

tomorrow my study officially starts

it's crazy

[elfin wraith]

Insane that you haven’t even started uni yet

I start my masters in a couple of weeks and I’m pretty confident you’re more knowledgeable than me

You’ll enjoy it though, it’s nice getting to immerse yourself in it

[tribal moss]

There's also this weird classification result for locally compact Hausdorff non-discrete topological fields ...

[elfin wraith]

That sounds sufficiently terrible

[south patrol]

Local fields?

Lol

Funny way to phrase it

[thorn jay]

(this) uni is great in that I can basically just pick and choose what courses I pick so long as I complete some necessary subjects

[south patrol]

I mean that is the definition just saying it like that makes it sound bizarre aha

[south patrol]

That is cool

[tribal moss]

Yeah, that was my point.

[elfin wraith]

That’s actually less insane than I was expecting lol

[south patrol]

These are used all the time in NT

[elfin wraith]

That sounds lovely, pretty perfect if you already know as much as you do

[thorn jay]

current idea is to do some necessary subjects of the next year, so in third year I can spend most of my time on the bachelor's thesis which is probably going to end up being much more than is asked from me lmao

[elfin wraith]

Yeah I’ve heard of them lol, the definition is adjective soup enough that it sounded as though it could’ve been horrible and weird but I guess it’s a field and compact haussdorff so you can’t really get a much nicer structure

[south patrol]

Lmao

[thorn jay]

when is the field totally disconnected though 🤨

[south patrol]

One thing I find funny is how "almost perfect" is now a technical term

[south patrol]

Often

[thorn jay]

\o/

[south patrol]

Fun thing I learnt about relatively recently is this notion of being extremally disconnected

closure of open is open

[thorn jay]

heard of it

where is it used?

[south patrol]

But usually you'd want compact + Hausdorff too

[thorn jay]

I'm getting the idea that compact Hausdorff spaces are somewhat nice..

[tribal moss]

Up to a few months ago, the Wikipedia article didn't even state that as a definition but just essentially said "local field means one of these three possibilities".

[south patrol]

Well the important thing is that these are like the projective objects in compact Hausdorff spaces, so they appear in some things for that reason

E.g. they appear in this CONDENSED MATHS

[thorn jay]

that's why we invent other names for it

[south patrol]

I guess like locally compact Hausdorff spaces are a very large class of "reasonable" spaces beyond manifolds

[thorn jay]

Im starting to feel like you're trying to push some message onto me...
can't quite figure out what though...

[elfin wraith]

I’m guessing not related to perfect spaces in topology? Because how could those be “almost” lol

My intuition says it should have some measure 0 set of isolated points but that seems silly

[karmic moat]

I’m going to introduce the opposite notion. I’m gonna call it “not at all disconnected”

[south patrol]

I still wonder why these are called perfect lol

[thorn jay]

aren't smooth manifolds always assumed to be locally compact Hausdorff?

or manifolds in general

[thorn jay]

hmm, it would be very nice if we had an opposite of the word "disconnected"

[south patrol]

Sorry, by "beyond" I mean like "more general than"

[karmic moat]

Not quite sure. Could be huge

[thorn jay]

codisconnected?

[south patrol]

Isn't this actually a thing that like

[karmic moat]

Ohhhh wait that’s awesome

[south patrol]

I forget the term

When you care about law of excluded middle and have a constructive version of this

[thorn jay]

A codisconnected space is a topological space X such that, for any disjoint open sets U and V, their union is never the whole space

[karmic moat]

Ure a genius omfg

Annals is calling u

Pick up the phone

[thorn jay]

thank you thank you

[elfin wraith]

Yeah I truly have no idea, I’m guessing someone thought they were lovely to work with

What is almost perfect though?

[thorn jay]

ofc it's the constructivists...

[south patrol]

Actually idk what I mean w constructiveness

It is a slightly different point

The point is often one defines connected as basically "not disconnected" I guess but there are nice ways of phrasing it

[thorn jay]

almost perfect is when it's perfect for some finite number of elements removed

[elfin wraith]

See this was my guess based on how we usually name shit but this seems silly

[south patrol]

So like if you have a chain complex over a ring R you can ask for it to be quasi-iso to something which is a bounded complex of finitely generated projectives — this is called being perfect

[thorn jay]

in the context of topology that would be a pretty useless/nonsensical definition lol

given how huge the "nice" spaces are

[south patrol]

And then for almost perfect (better known as pseudocoherent) you can ask for the same but with that complex (in homological grading) bounded below

[elfin wraith]

Ok yes so I am right, you’re just trolling lol, I was doubting my sanity there

[thorn jay]

I am indeed trolling

you may call me, the master baiter, even

okokok im sorry im sorry

I'll be good

[south patrol]

Jk jk

[elfin wraith]

I understand all of the words in this definition, but not well enough to understand what it really means lol

[south patrol]

Lol

[elfin wraith]

Pseudocoherant and quasicoherant feels like dangerously similar terms

For what are, I’m guessing, quite different things

[south patrol]

So like the idea is finitely presented modules are the "small" modules

Like just finite free stuff and summands of them lol

You can make this precise like categorically that they behave in a special way (they are the "compact" modules)

Then perfect is the version of this for the derived category

Like they should be smol things

[thorn jay]

what does compact mean?

or is that just the analogy

[elfin wraith]

Hmm ok I see, around derived categories is kinda where my homalg runs out so

Gimme a year I’ll hopefully have more to say about that

[south patrol]

Means Hom(X, -) preserves filtered colimits

[thorn jay]

very intuitive

😭

[south patrol]

Unfortunately compact topological spaces are not compact in general

But uh

[south patrol]

this is basically saying that if Y is a union of some Y1 <= Y2 <= ..., then any map X -> Y has to land in some Y_n

So for example you can see what the "compact sets" are

[thorn jay]

oh okay

[south patrol]

Which is very nice!

[thorn jay]

is it a categorification of compactness in posets?

[south patrol]

Uhh what is that aha

Or is it just the analogue here like

[thorn jay]

gah, you youngins without any knowledge of order theory

[south patrol]

If X < sup(S) for some increasing sequence S then X < Y for some Y in S

[thorn jay]

yes

[south patrol]

Something like this

[thorn jay]

no

[karmic moat]

This is, frankly, disgusting. I am throwing up

[south patrol]

Oh lol I messed it up

[thorn jay]

X < sup(S_0) for a finite S_0 \subset S

[south patrol]

Ok so this

Ah

Ok

Makes sense

Is that equivalent to the sequences thing

[thorn jay]

(think every cover has a finite subcover, that is where the term comes from at least)

[south patrol]

I imagine so

[thorn jay]

smt smt Zorn

probably

[south patrol]

Yeah fair

Heh

I think the compact objects in topological spaces are finite discrete spaces which is sad

But kinda makes sense cause Sets "sits inside" Top in a nice way

[elfin wraith]

This just reminded my I have a D in both metric spaces and general topology lol

[thorn jay]

fun fact: in a subobject lattice, the compact elements are precisely the finitely generated subobjects

[south patrol]

Good

[karmic moat]

I got a B- in graph theory

I hate graph theory

I’m surprised the prof didnt fail me

[thorn jay]

so for example in the lattice of intermediate field extensions of F < E, the compact objects are all algebraic over F

[south patrol]

Scientists are good at graph theory

[elfin wraith]

I somehow managed to be ill for both my metric spaces and gen top exams despite them being a year apart

[south patrol]

Nice

Thank u for enlightening me

[karmic moat]

I’m not interested in it. Too applicable

[thorn jay]

(if I am right in remembering that this lattice is algebraic, I am much too lazy to check)

[south patrol]

Compact objects are very important for some things in cat theory which is nice

[elfin wraith]

Don’t even talk to me about graph theory, I got a 92 on course work and an 86 in the exam that got scaled down to a 74

I got stuck on one single problem, still upset about it

[south patrol]

Like often a bit category can be controlled by the compact things of which there aren't too many

[karmic moat]

Scaling down is crazy

[south patrol]

I'm sorry

[karmic moat]

I have never heard of a negative curving

[elfin wraith]

Edinburgh hates to see you winning

[south patrol]

Easy exams are the hardest when it gets curved tf down

[karmic moat]

That professor is malice incarnate

[thorn jay]

oo, just like how in a nice (algebraic) lattice, every element is the join of compact elements

[south patrol]

EdinBRUH moment

[elfin wraith]

In fairness it was definitely like way too easy but that’s not my fault

[thorn jay]

wtf curving down is diabolical

actual devil

[south patrol]

Oh nice

I guess they like the join of all smaller compact things lol

[thorn jay]

(think every subobject is the union of it's finitely generated subobjects)

(oh and this condition is exactly what makes a lattice the lattice of subalgebras for some algebraic structure)

[thorn jay]

can you elaborate on this btw?

[elfin wraith]

Anyway the thing that makes this more funny is that I got an A in algtop and an even better A in my thesis titled Betti Numbers in Algebra and Topology

Possibly the first person who’s reasonably good at algtop and ostensibly doesn’t know more than the definition of a topology

[thorn jay]

Ive heard somewhere that algebraic topologists either really want to act as if their topology is algebra, or that their algebra is topology

[karmic moat]

Life is so good when u pretend everything topological space has every nice property

[thorn jay]

I suppose you are the former

[thorn jay]

"assume X to be indiscrete Hausdorff"

[elfin wraith]

Just throw adjectives at it until it complies

[elfin wraith]

I definitely am, but also I can actually do topology I was just coincidently genuinely ill for both of my topology exams so

I had 100% in my metric spaces course work and I think I got 20% in the final exam

I was just straight up delirious

[thorn jay]

ouch

[south patrol]

Lol sounds fair

[thorn jay]

people who study quandles are knot theorists who pretend their topology is algebra so much that they've become algebraists

[south patrol]

Sure, what I had in mind was that there are the notions of (local) presentability and compact generation which make categories really nice (like make it easy to apply the adjoint functor theorem, for example)

And these guys have all small colimits and limits and things

[south patrol]

Have you seen higher / "brave new" algebra aha

[thorn jay]

like what I'm looking into now are generalised knot diagrams which, truthfully, are more combinatorial objects than anything?? 😭

[south patrol]

Algebra is combinatotics

[karmic moat]

Aldous Huxley algebra

[thorn jay]

it becomes algebra again when you just attach a quandle to it in the obvious way and hope that an isomorphism on the level of algebra is enough for all the nice properties to carry over

[thorn jay]

I've heard of higher algebra relating to infty categories

[thorn jay]

(this is not going well so far)

[white oxide]

Sanity check: If N is a flat A-module then tensoring A-modules with N preserves isomorphisms

[thorn jay]

this is true by virtue of tensoring by N being a functor

[white oxide]

Oh ya

[thorn jay]

okay okay I've just been hitting walls so far and basically all of them have consisted of me not having a good grasp of when two arcs might be identified

[tall igloo]

disneccted

[tall igloo]

kid named balmer spectrum

[hidden wind]

X = 0 or X = 1

[fading acorn]

fundamental theorem of physics: every group is lie group

[chilly ocean]

I just started watching House M.D. and thought this was a reference to that at first lol

Then I saw “physics”

[woeful sage]

It's never lupus

[dim wagon]

im having trouble following this proof

$|G|u=u(\sum g)$ makes sense to me

[cloud walrus]

somethingwrong

[dim wagon]

but im not sure where the next equality comes from

[rocky cloak]

Guess it just comes from V being in the center of CG

For any x in CG, x times (Sum g) just multiplies it by the sum of the coefficients of x, independent of if it's from the left or right

[dim wagon]

ah okay, so idea is just that $\sum ug=\sum gu$

[cloud walrus]

somethingwrong

[dim wagon]

and this holds since we are summing over the whole of G and uG=Gu

[rocky cloak]

Yeah

[dim wagon]

would this make sense?

ah okay

thanks for the help

[thorn jay]

i wouldnt say that the second equality is necessary at all in the proof

as V is a sub-G-module, its basically by definition that u(∑g) ∈ V

(i.e. V is an ideal of ℂG)

oh wait i see that your guys action is on the right for some reason

[dim wagon]

ah yes the book i am using does everything on the right

[thorn jay]

janky

[rocky cloak]

I mean, that's pretty common

[dim wagon]

im doing a separate module on R-modules and tensors, where everything is done on the left. quite painful to swap my mind between the two

[thorn jay]

ive seen people mixing left and right group actions on the same set

truly a sight to behold

[tidal schooner]

though as a quandle theorist you're probably used to people not agreeing on right vs left actions

[thorn jay]

I am

grrr

they dont even use the notation of the quandle operation properly

its a triangle because one arc goes UNDER the other

MY GOD

Anyways

[tough raven]

Is anything non-trivial known about the number of conjugacy classes of pairs of Sylow p-subgroups of G, for a finite group G and prime number p?

[spark veldt]

If Aut(G) is the set of all automorphisms on G and A(G) the set of all bijective mappings from G to G, then Aut(G) is a subset of A(G). I don't really understand this.

I think I need an example of where a mapping on A(G) is not necessarily an automorphism? Because aren't they both bijective?

[velvet hull]

is G a group?

[spark veldt]

yes

I'm having a hard time trying to constuct an example

[velvet hull]

the key is homomorphism

Aut(G) has to consists of homomorphisms that happen to also be bijective

A(G) consists of any bijective mapping, not necessarily homomorphisms

[spark veldt]

Ooh I see yeah I kinda missed that i think

Do you have a concrete example

[velvet hull]

any nontrivial group

the identity has to go to the identity

so find a bijection that doesn't send the identity to the identity

[crystal vale]

If I have to find an injective homomorphism Q8 to S8, how do I find it?

I found one injective homomorphism, where I mapped i to (1234)(5678) and j to (1638)(2547) and i verified that it holds the relation ij = -ji.

Is there any better way to do that? I am sure there is

[velvet hull]

quaternion group has 8 elements

[crystal vale]

Yes

I mean yeah I have to find some faithful action on {1,2,...,8} by Q8

[velvet hull]

do you remember cayley's theorem

[crystal vale]

Yeah I got

Thank you ❤️

If I have to find some morphism between S_n to G, then I try to find the appropriate image of (12) and (123...n), but which relation should follow the image, I mean if I have to find morphism Dn -> G then image of r and s, say x and y respectively, so xy = yx^-1 should holds there

So which relation should I need to verify when I have an S_n group?

[velvet hull]

just look up group presentations of S_n

although if you want to use that as proof the presentation of Sn may or may be allowed to be used without proof

[snow knoll]

Im reading a book of abstract algebra by pinter and this problem in the subgroups chapter is giving me a little trouble

I feel like the definition of K here is odd

I understood it to mean that any x in G is in K iff for any a in H, xax^-1 in H

This worked with no issues until I tried to prove K has inverses, and the solution in the textbook doesnt make a lot of sense to me

Oh whoops

[snow knoll]

I have a physical copy of the book and I took these screenshots from the pdf I just found online
This is NOT what my textbook says

The solution makes a lot more sense now lol

[rapid cave]

Another way to see this definition is that K is every x in G such that its left coset and right coset are the same

so if xH=Hx then of course x^-1H = Hx^-1

[distant summit]

Apologies for replying to such an old message, but it would be alright to phrase (C) as follows?

There is a bilinear product (-,-): FxV -> V defined by (a,x)=ax

[thorn jay]

you cant really define bilinear without scalar multiplication

but i guess it's true in that its a bimorphism of abelian groups? seems better to just state the axioms outright at that point

[glad osprey]

Yeah, using the term "bilinear" in the definition of a vector space seems circular. You need to define vector spaces before you can define bilinear

[distant summit]

The issue I have with Halmos' terminology is that when we talk of commutative multiplication, we want "multiplication of a by x" to be synonymous with "multiplication of x by a", and similarly we want "multiplication of something in F by something in V" to be synonymous with "multiplication of something in V by something in F"

So idk if there is a way of characterising (C) without making (what I believe to be) a false distinction between "multiplication (of a vector) by scalars" and "multiplication (of a scalar) by vectors"

I understand that these are different things

But the terminology doesn't seem to describe the difference adequately

[thorn jay]

commutativity is not inherently a "multiplicative" property

[distant summit]

I know

I'm saying that if multiplication is commutative, then xy=yx, so we should be able to call xy "multiplication of x by y" as well as "multiplication of y by x"

[thorn jay]

but scalar multiplication isnt commutative?

[distant summit]

by scalar multiplication do you mean FxV or FxF?

[thorn jay]

the former

the latter is just multiplication in the field

[distant summit]

Ok well as far as I know FxV is treated as commutative

ax=xa

[thorn jay]

??

i mean i guess because fields are commutative, left and right modules coincide but i certainly dont see people writing v ⋅ a

[distant summit]

If you're vector is (1,2,3), 4(1,2,3)=(4,8,12) (everyone agrees), but in my experience you would also have (1,2,3)4=(4,8,12)

[thorn jay]

what experience

can you point me to a textbook which treats scalar multiplication as something which can be done from the right?

like sure, it looks alright in that specific form but its not formally defined nor is it really useful to look at it that way

[distant summit]

Actually I think this is besides the point

My point is that both distributive properties concern multiplications that are FxV, so calling one "multiplication of a vector by a scalar" but another "multiplication of a scalar by a vector" seems weird

[thorn jay]

theyre formally the same thing, it just depends on what youre focusing

this discussion is just a repeat of the former

[glad osprey]

I don't understand your point, do you want multiplication of a vector by a scalar to be a map from VxF while multiplication of a scalar by a vector to be a map from FxV?

[rapid cave]

Its just the same thing

[distant summit]

And is the issue with my suggestion about bilinear the use of the word "bilinear"?

If I said

There is a bilinear product (-,-): FxV -> V defined by (a,x)=ax, with the properties that (a+b,x)=(a,x)+(b,x) and (a,x+y)=(a,x)+(a,y)

[thorn jay]

thats just how its usually defined lmao

[distant summit]

Not necessarily

My point is more that "multiplication of a vector (scalar) by a scalar (vector)" is being used to refer to the same operation, so the distinction seems unnecessary or inadequate at conveying the desired point

[thorn jay]

well id change the terminology from "defined by" to "where we write"

[delicate orchid]

is your issue the implict switching between left and right vector spaces?

[glad osprey]

But what distinction do you want to convey? Do you want some terminology to convey whether we're holding the scalar fixed or the vector fixed?

[delicate orchid]

because you're right - they're not the same operation. But every vector space is naturally isomorphic to it's opposite

so we don't really care

[distant summit]

Maybe?

The way he's written it kinda makes it sound like the multiplication in C1 is different to the multiplication in C2, but they are the same

[south patrol]

:cereal_2:

[delicate orchid]

I'm not particuarly fond of this phrasing either

[delicate orchid]

it wasn't this, then

[glad osprey]

Oh, yeah, "multiplication by vectors" is maybe not ideal, but I'm not sure there's any more precise way to say it in english other than just writing the equation

[distant summit]

There are two distributive properties, a(x+y)=ax+ay and (a+b)x=ax+bx

The former concerns distribution over vector addition, whereas the latter concerns distribution over scalar (or field?) addition

[distant summit]

Yes, this is my issue

The distinction lies in the kinds of addition (vector v. scalar), but the multiplication is always FxV

[thorn jay]

why is that an issue exactly?

[distant summit]

Because the phrasing makes it sound like the distinction lies in the kind of multiplication being used, but it doesn't, it lies in the kind of addition being used

Actually

That's not quite right

I think it's more that his phrasing makes it sound like there are two distinctions to make (between multiplications and additions), but there is only one (between additions)

Or maybe call both "multiplication by scalars"

[glad osprey]

A more precise way to say it is that scalar multiplication is a ring homomorphism from the scalar field to the endomorphism ring of V. But I'm not sure if that helps

[distant summit]

I haven't studied rings so unfortunately it doesn't

[thorn jay]

its induced by*