groups-rings-fields

[thorn jay]

eh i guess its pretty direct by currying

is that how you spell it?

[glad osprey]

Yeah, I think that's a better way to phrase it, use the term scalar multiplication in both C1 and C2

Currying? Yep 👍

[velvet hull]

why do they call it currying when you curry in the scalar curry out multiply scalar vector the multiply

[thorn jay]

exactly

[glad osprey]

Curry on my wayward son

[white oxide]

Why is this highlighted line true

[rocky cloak]

It is exactly point (iv)

[white oxide]

Oh

I think you’re right

[twilit wraith]

I have to prove that an element has order 2 in Sn iff its decomposition is the product of commuting 2-cycles

the forward direction seems to be handled by noting that if a cycle has order 2 then given its disjoint cycle presentation, the lengths of the cycles in that presentation has to be 2 since 2 is prime

but how can i know that the backwards direction consists only of disjoint 2-cycles

maybe thats just the definition of decomposition of cycles

[rocky cloak]

I mean, you can determine the order of an element from its disjoint cycle decomposition

So you may want to think about what that relationship is

[twilit wraith]

is it right that the cycle decomposition of an element in Sn consists of only disjoint cycles?

because in that case this problem becomes much easier

but the thing thats tripping me up is that commuting cycles are not necessarily disjoint

[rocky cloak]

The decomposition into disjoint cycles consists of disjoint cycles. So unless you mean something else by cycle decomposition yes.

It is correct that cycles can commute without being disjoint, but maybe you can determine when two 2-cycles commute

[twilit wraith]

im pretty sure in that case it is only when theyre disjoint

because i think the case two cycles commute iff either theyre disjoint or if they can both be given by some power of another element of Sn

but of course that second case isnt an issue because the only nonidentity power of an element x of order 2 is x

[proud vigil]

I think for the reverse direction you can just say that if x is the product of commuting 2 cycles x1 x2… xn, then (x1 x2… xn)^2 = (x1^2) (x2^2)…(xn^2) = e since they’re all 2-cycles

Then you don’t really have to think about disjointness at all

[wicked patio]

unfortunately the identity is an empty product of commuting 2-cycles 🤓

[dim wagon]

I am having trouble with this exercise for the second part on finding basis for the second set of homomorphisms.

$U_3=\operatorname{span} {v_1,w_2}$

[proud vigil]

omg wait @twilit wraith hi haven’t we vaguely talked before in other servers

[cloud walrus]

somethingwrong

[dim wagon]

i have this proof which is what w used to find basis of first set (claim is the $\phi_i$ will form a basis). I don't see why defining $\phi_i$ similarly won't give the basis of the second set of homomorphisms

[cloud walrus]

somethingwrong

[twilit wraith]

I dont think so we only share physics and mathematics

[twilit wraith]

Oh ofc

[proud vigil]

thats bc i left chaos to be productive LOL

i think ur in chaos

im not sure

[twilit wraith]

Yeah i am in chaos lol

[proud vigil]

okay then yes we definitely have talked before then

ur really into philosophy right

[twilit wraith]

Oh I see that ur newish to this server

Yeah I study philosophy secondarily

Math is my main thing tho

[proud vigil]

yaa makes sense

[twilit wraith]

Actually tho for this how do we know then that the product of x1 ... xn isnt the identity

[proud vigil]

we dont but it anyway isnt true if we consider the identity right

or "empty products"

[twilit wraith]

Oh well I suppose if we assume x1 ... xn is the decomposition then we assume it isnt order 1

[proud vigil]

that works too ya

[twilit wraith]

Alright cool

it hasnt been that long since i took undergrad algebra but i still gotta remember the kind of thinking it takes

[twilit wraith]

alright so i just finished a question determining all the decompositions of elements of order 4 in S4

i could do it, but im curious about something now

to determine the number of such elements, could i just look at the size of the kernel of a surjective homomorphism from S4 to S3

wait not quite

[kind temple]

they are only the four cycles, so you want the number of circular permutations on 4 symbols

this is just (4 - 1)! = 3! = 6

the reason you want circular permutations is because the cycles represented by (1 2 3 4), (2 3 4 1), (3 4 1 2), (4 1 2 3) all represent the same 4-cycle

[twilit wraith]

fortunately i was able to deduce this

i suppose what im going for is recalling how we can use the first isomorphism theorem to find how many elements of order k are in some arbitrary group

[kind temple]

i don't think there is way to do this for an arbitrary group...

i wouldn't be surprised if there is some NP hard or NP complete problem involving automorphisms of graphs or something like that, which would prove P = NP if you could determine the number of group elements of order k for arbitrary k and arbitrary group

or maybe there is a simpler explanation...

but the intuition should be that there just isn't enough information in general

the thing that you are leveraging here is that S_4 has a "free" cycle decomposition

[cedar vault]

Automorphism group of graph will be finite(unless you are talking about infinite graphs), and in finite groups cant you just go generating cyclic groups generated by single elements by picking elements one by one and you should be done with this in polynomial time . And then calculate all divisors of the cardinalities of the cyclic groups etc. and it looks like you will be finished in finding all elements of order k in polynomial time in |G| (assuming multiplication is atmost polynomial time in |G| or whatever). maybe for some infinite groups with elements of finite orders this will be harder? I don't know much cs or complexity stuff so I might be completely off track here

[kind temple]

oh, so true

i don't think you can do anything leveraging the first iso theorem tho

my first idea is like, look at n |-> x^n, but that is only useful if the groups are abelian

anyways, thanks for catching that

[cedar vault]

but I guess even non constant polynomial time won't be good if you are finding order k elements by hand

[kind temple]

yea, seems like it

there is another idea i have, where you can use the first isomorphism theorem for sets in this particular case:

there is a function f : S_4 -> S_4 that takes a permutation s and maps it to the 4-cycle (s(1) s(2) s(3) s(4)).
call two permutations equivalent if they are the same under f.
the image is the 4-cycles, and 4! / #(of eq. classes) should be the number of 4-cycles

but counting the number of eq classes seems hard

without already knowing the 4-cycles

[quiet pelican]

I feel like that’ll be no easier than the usual combinatorial approach

[kind temple]

yea

[limber sequoia]

Im a bit confused about vector spaces as modules
So $\mathbb{R} \to End(\mathbb{R} )$ are vector fields of the reals over the reals. But there are some strange group morphisms of the reals, like the Hamel functions.

So what promises us that we get the "normal" vector field?

[cloud walrus]

tomer_k

[limber sequoia]

For starters, I know that 1r = r, but it's not clear to me that r1 = r, with r in the LHS being the endomorphism corresponding to r.

https://math.stackexchange.com/questions/1562845/number-of-ring-homomorphism-from-the-ring-of-real-numbers-to-itself

can adapt this monotonicity argument

[rocky cloak]

So if I'm understanding your question:

You've understood that a vector space is exactly an abelian group A together with a ring homomorphism
R -> End(A).

Then you pick your abelian group to be R and want to prove that the only vector space structure is the 1 dimensional one? This is simply not true.

R is for example isomorphic to R^2, so it would be very strange for the vector space structure to be unique.

The one dimensional vector space
R -> End(R)
is however defined to be
r |-> (x |-> rx)

[signal wadi]

I'm curious if anyone has references to a setup where we have multiple groups acting on possibly non-disjoint sets.

Like groups G, H acting on S, T respectively with a nontrivial intersection between S and T. Like it's possible for an s \in n S to be carried to a u \in S \intersect T by some g \in G and then u to be carried to some other t \in T by an h \in H.

There is a possibly interesting notion of joint orbit here... Curious what I might search for.

[rotund aurora]

what if S=T? It's not strange that the action of G has some property, and if H is related to G in some manner, then H also has that property, or a similar one. For example, if G is a Fuchsian group of the first kind and H is a finite index subgroup of G then H is also a Fuchsian group of the first kind

[delicate orchid]

it sounds like you're trying to define a (G, H^op)-biset structure on S \cup T with some specifics on which elements are fixed by the left/right actions

[signal wadi]

It could be... Although, while I did describe the situation for two groups, I think the situation I'm particularly interested in would be for a finite number of groups acting on just as many sets and having a more complex intersections story. I don't know that I am thinking about left and right actions so much as just many left actions that have the possibility of conditionally composing...

[delicate orchid]

I do know of a situation where this situation appears in the wild but it's quite niche

[signal wadi]

I'm considering symmetries of many functions simultaneously and where the inputs come from some common "pool" of inputs if that helps

[delicate orchid]

and it's not in this level of generality

is it possible to just extend your functions to the entire pool of inputs? So f(x_1, x_2) extends to f(x_1, x_2, x_3) but f(x_1, x_2, x_3) = f(x_1, x_2, y_3) for all x_3, y_3 outside of the inital "pool", for example

then your "mixed" action is just a group action

if not, you might want to look into partial group actions

[thorn jay]

I believe you can just extend it to an action of their free product right

like an action on X with X \sub Y extends to an action on Y by just fixing all elements not in X

and then use the universal property of free products

[delicate orchid]

that's what I'm saying here

but you worded it better lol

[thorn jay]

ts

[signal wadi]

I suppose so, yes. However, I think here is value for computing these groups on fewer variables -- you can see structure in the symmetries emerge and you can say some. F and some G are "the same", but I guess in the expanded pool of inputs you could just say their symmetry groups are isomorphic.

Thanks!

[thorn jay]

yeah, probably for computing, using the groups themselves is better

but for proving stuff it's probably nice that you can see it as one big group action

(though maybe not if you want to infer something about sizes and generalise like orbit stabilizer like theorems)

[delicate orchid]

so close to partial group actions being relevant 💔

[thorn jay]

is that studied often?

[delicate orchid]

no lol

[thorn jay]

figured

[glad osprey]

Our lecture notes define the symmetric power $Sym^n V$ as the vector space of coinvariants of $S_n$ acting on $V^{\otimes n}$. Does this mean that $v \otimes w = w \otimes v$ in $Sym^2 V$ for all v, w? Or would that be the vector space of invariants of $S_n$ rather?

[cloud walrus]

sheddow

[delicate orchid]

the invariant space would be spanned by things that look like v(x)w + w(x)v etc.

you're correct, v (x) w = w (x) v in Sym^2 V

think of it as "forcing the tensors to commute"

[glad osprey]

I see, thanks

hmm, so we want Sym^n V to be this quotient space of V^{\otimes n} because we want any symmetric bilinear map from V^n to factor uniquely through Sym^n V So Sym^n V plays the kinda the same role as the tensor product, except for symmetric bilinear maps

[delicate orchid]

exactly, you can definite it using an analogous unviersal property, replacing bilinear maps with symmetric ones

I'll let you think about what the corresponding algebra is for alternating bilinear maps

(although I think you mean multilinear)

[glad osprey]

yeah, multilinear 👍

for alternating maps, I want to say just replace S_n by A_n, but that seems too simple

[delicate orchid]

at the risk of going off on a tangent, what's the definition of an alternating multilinear map

[glad osprey]

f(v1, ..., vn) = sgn(sigma) f(v_sigma(1), ..., v_sigma(n)) for any sigma in S_n

so I guess we act by S_n still, but multiply by sgn additionally?

[delicate orchid]

exactly!

[south patrol]

Any what?

[delicate orchid]

sigma balllssss

[crystal vale]

I want to understand the structure theorem of modules over PID, Jordan canonical stuff, any good resource and what I need before reading it?

[maiden heath]

Is there a way to recover to recover the structure from a quotient. I.e. If we have a ring $R$ and an ideal $I$, there's a map $R \twoheadrightarrow R/I$. Is there a map in the other direction, like by tensoring with $I$ or something? Thanks

[cloud walrus]

Funky_Funktor

[tropic spade]

I don't think this would work without more requirements.

Since you can quotient two distinct rings in such a way that you have the same quotient ring up to iso.

So, if I just give you a mystery ring R and I tell you it's a quotient without telling you the ring or the ideal I used to construct it there could be several choices of rings and ideals you can quotient by to produce R.

[rapid cave]

for a commutitive ring R, I think if I is principal this is possible

[velvet hull]

not big requirements, only basic knowledge of rings and modules (iso theorems and PIDs) shoud be enough

[rocky cloak]

If R = F2[x]/(x^2) and I = (x) or if
R = Z/4 and I = (2) you can't really tell the two rings appart from just looking at R/I and I.

That being said I'm not actually sure what Funky Functor is asking

[rapid cave]

I interpreted this as R/I being projective

[rocky cloak]

Alright, well R/I is projective iff I is generated by an idempotent

[thorn jay]

is R commutative?

[maiden heath]

Maybe I should add more info. In particular the ring $R$ is commutative and we know the ideal $I$, I just want to know if there's a way to recover $R$ from $R/I$

[cloud walrus]

Funky_Funktor

[maiden heath]

If more info is needed. $R$ is $R_n \otimes_k R_n$ the $n$-variable polynomial ring tensored with itself and $I$ is the ideal generated by elementary symmetric polynomials

[cloud walrus]

Funky_Funktor

[thorn jay]

how do we know I?

[rocky cloak]

What does it mean to recover R from R/I to you exactly?

Like you're given just the ring R/I and want to know it's coming from R? That's obviously not possible, and I'm not sure what else you would mean.

[thorn jay]

as an R-module?

[maiden heath]

Like how to extend $R/I$ to get $R$

[cloud walrus]

Funky_Funktor

[thorn jay]

many ways

[maiden heath]

Would tensoring with $I$ be a way?

[cloud walrus]

Funky_Funktor

[thorn jay]

what is I here

what ring are you tensoring over

is I given as a module? abelian group?

rng?

[rocky cloak]

I don't see any interpretation that would make it true anyway

So I guess, no. Tensoring with I wouldn't be a way

[thorn jay]

^ this is an interesting problem

[delicate orchid]

the fundamental problem with this is that quotienting destroys information

[maiden heath]

$I$ is the ideal generated by elementary symmetric polynomials, i.e. for n=2 $I = <x_1+x_2, x_1x_2>$

[cloud walrus]

Funky_Funktor

[delicate orchid]

we'd need assumptions on R, I, and/or the splitting of the surjection R -> R/I to do anything

[thorn jay]

so we know the ring already

[rocky cloak]

Anyway, in this example R/I and I are equal in any meaningful way to compare them, but the two Rs are not

[thorn jay]

pf, to get all the extensions just calculate the cohomology group

[rocky cloak]

If the ring R is fixed, what does it mean to recover it?

[maiden heath]

I want the functor for example $-\otimes I$ (I don't know if this works) so that $(R/I) \otimes I \cong R$.

[cloud walrus]

Funky_Funktor

[rocky cloak]

So presumably a functor between their module categories.
Such a functor would give a ring homomorphism
End(R/I) -> End(R)
So a ring map
R/I -> R

This you wouldn't expect to exist in general

But if there is a ring map R/I -> R, then you can do extension of scalars

[rapid cave]

and there is a map when R/I is projective right?

[rocky cloak]

No, that's not related

The map R -> R/I splits as a homomorphism of R-modules iff it's projective

[maiden heath]

I see, I have an embedding of $C_n \otimes_k C_n$ into $R_n \otimes_{R^{S_n}_n} R_n$ where $C_n$ is the coinvariant algebra under $S_n$

[cloud walrus]

Funky_Funktor

[hidden wind]

this def of irreducibility does not pass the vibe check

[south patrol]

It is the usual one but yeah

[hidden wind]

according to this definition, 4 is irreducible in Z/6

2 = -4 being associate to 4

but 2 is not a unit…

[south patrol]

It is a bit of a funny definition for rings that aren't UFDs or anything yeah

[hidden wind]

is it really?

[south patrol]

Yeah

Just very often in practice you have irreducible is equivalent to prime

[hidden wind]

here is wikipedia’s definition of irreducible elements

In algebra, an irreducible element of an integral domain is a non-zero element that is not invertible (that is, is not a unit), and is not the product of two non-invertible elements.

[thorn jay]

"integral domain"

[hidden wind]

ah true

[thorn jay]

associates meaning their generated ideals are the same (in the commutative case)

?

[south patrol]

Tbh I have never used the notion of associates much in my life other than very carefully checking some element wasn't irreducible in a non-domain for counterexamples

[hidden wind]

what do you mean “very often”

[south patrol]

Yes exactly

[south patrol]

UFD

[thorn jay]

other rings arent nice to work with

[south patrol]

Like where you have unique factorisations into irreducibles

Now non-UFDs do come up all the time lol, as do non-domains more generally, just I feel like usually irreducibility is thought about over some nice ring

E.g. irreducibility of polynomials over a field

This is all in my experience though like I am sure there are people interested in irreducibility in rings that aren't domains or something

[hidden wind]

no i mean two elements being associate is by definition being the same up to a unit

guh, yeah ik those two conditions are not equivalent, but in my head associate is defined on the level of elements

[thorn jay]

idk I just guessed

[south patrol]

No im wrong

[thorn jay]

I believed the "same up to a unit"

[south patrol]

Sorry lol

[hidden wind]

no worries there are a lot of inequivalent definitions floating around

[thorn jay]

that happens

especially when you go more general, a lot of things fall apart

so you've gotta be careful with keeping track of everythin

g

[hidden wind]

this just feels horrible to me

[thorn jay]

lmao

[south patrol]

Fortunately like

When was the last time you cared about factoring in Z/6

[thorn jay]

well, as 2 and 4 are associates, it makes 2 an idempotent-ish element

[south patrol]

I think it also makes sense if you think of Z/6 as Z/2 x Z/3

[thorn jay]

and 4 is literally just idempotent

[south patrol]

Like I guess 4 is (0,1) which "should" be irreducible as we just have a product of fields

[thorn jay]

and that

[hidden wind]

this is a good point thanks

[south patrol]

But it is a lil funny agreed

[rocky cloak]

Tbf irreducibility doesn't have much value outside integral domains.

But yeah, I would instead say that something is irreducible if it can't be written as a product of two non-units

So just 4 = 4*4 should make it reducible

[thorn jay]

that's the definition that seems more useful too

[rocky cloak]

It's funny though, because it makes 4 prime, but not irreducible

[hidden wind]

horrible

[gentle crow]

i was reading a couple of solutions to D&F exercise 1.2.17 and i don't understand how n = 3k implies there are no further relations satisfied by powers of x in the first solution or how n = 3k implies x, x^2 are distinct nonidentity elements of X_{2n} in the second solution. would anyone mind explaining?

[rapid cave]

well by the group presentation if x^2 = e then X2n must be trivial

but its order is divisable by 3

[rocky cloak]

The fact that D6 satisfies those relations, means that any forced relations would also need to be satisfied by D6

[rapid cave]

D3?

[rocky cloak]

Sure, with more reasonable notation you would call it D3

[elfin wraith]

D&F moment

[rapid cave]

@rocky cloak may I ask what is your math background

you are just so knowledgeable

[gentle crow]

it's supposedly more common in group theory literature for the subscript for dihedral groups to be its order rather than the number of sides of the n-gon to which it refers

[rocky cloak]

I do rep theory of finite dimensional algebra stuff

[rapid cave]

but you seem to understand a lot of things in algebra at a very high level

like I feel you outdo me on every answer

which makes me respect you tbh

[rocky cloak]

I know some stuff and know less about other stuff

[rapid cave]

are you a phd student?

[elfin wraith]

They better also write S_n! then

[rocky cloak]

Not anymore 😢

[rapid cave]

wdym 😱

[elfin wraith]

Got kicked out for spending too much time on discord

[rocky cloak]

PhDs don't last forever

[rapid cave]

so you finished yours?

[rocky cloak]

Yeah, now I'm out in the world

[south patrol]

Actually what like rep theory phd students r there here

Hmm

[rapid cave]

representation theory

[rocky cloak]

The good kind I hope

[south patrol]

Yeah

Wdym by good kind

Or just good at rep theory

[rocky cloak]

The kind that spends most of their time on discord

[south patrol]

Nice

[elfin wraith]

Wew boytjie and ragharum I think

[south patrol]

Ah good point

[rapid cave]

tbh after I took a rep theory course this year I really can't appriciate the field

[south patrol]

I didn't rly know what ragharum does

[rapid cave]

I still don't understand why do we care

fr

[elfin wraith]

Something neiche I guess, he seems to just come out with some crazy specific questions then dip again lol

Ive no idea what he studies because I did see you changed it from alg to rep theory so

[rapid cave]

@rocky cloak would you mind to enlighten me?

[south patrol]

Ah lol

Sorry

Yeah I mean I just feel like "algebra" is a very vague thing to say you research

[elfin wraith]

My basic understanding is, group rings are hard, direct sums of vector spaces are not so much, work out what those are and study those instead

[south patrol]

My university has an algebra department but it seems they could rename it to "rep theory" without any problems

[rocky cloak]

I mean, what do you care about?

Rep theory has connections with physics, algebraic geometry, algebraic topology, combinatorics.

But to me it's more like rep theory is a reason to care about algebra / algebraic structures.

[rapid cave]

I am a number theorist

the only use I know is with Artin L-functions

[south patrol]

Lol

Isn't representation theory totally ubiquitous in NT lol

That is motivation already

[elfin wraith]

To be honest the majority of the rep theory ive formally studied has also been for L functions in my analytic NT class lol

[thorn jay]

you can even research algebra while technically being a logician

[thorn jay]

based

[rocky cloak]

I don't know that much number theory, but I thinklike group cohomology and representation theory of Galois groups of number fields is kind of a big thing there

[elfin wraith]

My noncom class made us work out the representations of some group rings but never realy told us anything about it

[south patrol]

Totally lol jagr

[rapid cave]

this was kind of my experience

[elfin wraith]

Basically just a you know Maschke and Artin Wederburn and so go crazy

[south patrol]

Kinda funny to me how Wedderburn is like a Scottish name

[rapid cave]

the class was about the theory itself and it didn't even mention once why?

[south patrol]

I always thought it should be Dutch or smth for whatever reason

[thorn jay]

giving us too much credit

[elfin wraith]

It feels very quinisentially Scottish to me lol

[thorn jay]

also wedderburn is not a dutch name 😭

[rapid cave]

I think we are in the wrong channel xD

[thorn jay]

"group" representations

smh

[rocky cloak]

burn seems fairly British, wedder is a bit harder to place for me

[south patrol]

Yeah

[south patrol]

It does now

[south patrol]

Yeah I mean definitely not lol

Idek how I got this impression lmao

You know one thing that grinds my gears

[elfin wraith]

You can tell its not dutch because you can spell it

[south patrol]

Why is Noetherian often/usually "mispronounced" smh

But then again, so is Artinian

But at least with Artinian it is consistent with Artin

[rapid cave]

in my language its translated

[south patrol]

Whereas I think the default is to pronoune Noether and Noetherian differently

[south patrol]

Wdym lol

[elfin wraith]

Me too, I feel like such a prentious dick for insiting on prononcuing it correctly but idk it just feels respectful

[rapid cave]

like there is a term for these in my languege

[thorn jay]

hey my name is very spellable

I always have to correct people because the spelling is weird though

[south patrol]

For me it is cause I like heard of Noether whilst studying German in high school so I never like pronounced it otherwise idk

[south patrol]

Ah a more specific one

[hidden wind]

i’m joining this rep theory class a bit late i have to catch up somehow, guh

[south patrol]

I was just confused what you mean by translating a name

[thorn jay]

Martijn Garritsen

[elfin wraith]

You would be the first dutch person ive met with a spellable name

[thorn jay]

How is "Daan" not spellable

[south patrol]

Dan

[elfin wraith]

Everytime I want to see the name of the prof from my quantum computing course I need to look up his book

[rocky cloak]

Wait, how are people pronouncing it?

[south patrol]

As in English th

Instead of hard t

[elfin wraith]

No-thee-reian

[rapid cave]

Ne te rian

[hidden wind]

Nøter

[south patrol]

Cute

[rocky cloak]

Ah, yeah. I do that to when speaking English

[south patrol]

I also don't expect a serious answer cause I know why it is pronounced this way

I just find it funny Noether as a name seems usually pronounced correctly but then ruined by noetherian

[elfin wraith]

It is interesting that Noether always spelled her name as Noether rather than Nöther

[south patrol]

Think this is fairly common w names

[elfin wraith]

huh

[rocky cloak]

I mean it makes sense, when you make it into Noetherian it ceases to be a name, so then you just pronounce it according to English rules

[hidden wind]

names tend not to follow standard orthography

[elfin wraith]

but you still capitalise it like a name

[south patrol]

Especially since German has had many relatively recent orthography reforms

[rocky cloak]

That's English for you

[elfin wraith]

Ive noticed people tend to capitalise Noetherian and Artinian but not abelian which I find odd

I usually capitalise them all, but I think im the weird one

[south patrol]

I imagine this is one kf things where there were interchangeable spellings and then a reform depreciated one of them but names are preserved

[rapid cave]

abelian doens't sound like a name

[south patrol]

My surname is a "mispelled" version of an actual word aha

[elfin wraith]

Yeah honestly through learning german im becoming convinced that nouns should just have an all or nothing capitalisation, I appriciate the consitency

[rocky cloak]

I mean, Abel both sounds like and is a name

[thorn jay]

my last name is literally just "from [village]"

[regal mango]

Please help in q 16

[hidden wind]

in norwegian we just use the lower-case version for all of these

[thorn jay]

I lived pretty close to that village even

[elfin wraith]

Lore accurate

[south patrol]

Classic

[rapid cave]

abelian doesn't feel like it to me even though I know it comes from a name

[south patrol]

Average Dutch name tbh

[thorn jay]

it's like the most dutch thing about me

[south patrol]

You know a good Dutch name

Gijs

[thorn jay]

Hertog-Jan Schoenmaker

[glad osprey]

How would you pronounce Artinian correctly? 🤔 Art-æ-ian?

[rapid cave]

They probably makes some great shoes

[south patrol]

That looks wrong to me

[thorn jay]

maybe someone used to

[elfin wraith]

Whathaveyoutriedalready?

huh

[south patrol]

I mean how it is the nasal French -in, at least for the older Artin aha

[thorn jay]

How would you pronounce jan otherwise

ian has a different inflection from jan

[south patrol]

Idk if the younger Artin pronounces it in a non-French way

[hidden wind]

generally we try to do everything in norwegian but if there’s anyone who dont speak norwegian in the class then we do english instead, more often than not (especially in the more advanced courses fsr) there is someone who dont speak norwegian

[regal mango]

Using euler theorem

[south patrol]

Tbf could have hard j

[regal mango]

But cant move forward from it

[south patrol]

In English

[elfin wraith]

I dont know why my space bar wasnt working there that was odd

[south patrol]

Indeed this is a women's name, Jan

[thorn jay]

dutch doesn't have hard j's :3c

[hidden wind]

also not all the profs speak norwegian particularily well in which case they just teach in english

[regal mango]

No problem

[south patrol]

Yeah I assumed so

In German it funny innit

[rapid cave]

Are you all german or something

[south patrol]

Dschungel

[south patrol]

Are any of us German

[rapid cave]

no

[south patrol]

Saw this in a thesis intro today

[thorn jay]

no Im dutch

[thorn jay]

ugh

ai

[south patrol]

I'm british

[hidden wind]

guh “generelly” is such a norwegian misspelling im embarassed

[rapid cave]

same thing

[south patrol]

Yeahh

[elfin wraith]

As a first hint, what is the smallest m such that a^m \equiv 1 mod a^n -1? Understanding that might give you a hint as to how to apply another nice result from group theory

[south patrol]

Immediately dates the thesis

[thorn jay]

you think the additional captions are ai generated to?

[south patrol]

But nice-looking paper otherwise

[thorn jay]

I dont know enough math to know if theyre related to the names of the desserts

[regal mango]

Getting a little point where are you going for

Let me see about it

[south patrol]

Idk

[elfin wraith]

Honestly dont most like younger Dutch and Norwegian people speak better English than half the people in the UK though?

Im conviced the average Norwegian would have an easier time communicating with people in London than I do

[thorn jay]

no but what the fuck is this actually

[glad osprey]

Jes, ve du 💪

[thorn jay]

1 to 3-5 to 2 to section 2.3???

[south patrol]

Yeah lol

I didn't even notice that until I shared it here

[south patrol]

Not hard

[elfin wraith]

AI is going to change the world bro, dont get left behind

[thorn jay]

I like dead fields anyways I'm not gonna miss much money

[south patrol]

If the area of maths ur in is dead then you'll probably make more money anyway

(By leaving math)

[thorn jay]

b

but

universal algebra :(

[elfin wraith]

Im being fully serious though, I have on multiple ocassions found it legitmatley hard to communicate with english people in the UK

What do you mean by this?

[rapid cave]

reflect-rotate-reflect is rotation in the other direction

[elfin wraith]

The motivation is that it works

like thats what you do

[thorn jay]

coxeter group moment

[rocky cloak]

If you rotate in front of a mirror it looks like you're rotating the opposite way. So
yxy^-1 = x^-1

[thorn jay]

because

basically

[south patrol]

In what way oop

Exercise to the reader

[elfin wraith]

They just cannot understand me

[rapid cave]

well all elements can be shown to be xy^a or y^a

[south patrol]

No but I mean like you are basically specifying what xy is in terms of other stuff so you can do this yeah

[rapid cave]

do you want me to stay? (I will got to sleep otherwise)

[thorn jay]

dialect momento

[hidden wind]

i get this is probably a joke but uhm no

[south patrol]

Up to you lol

[rapid cave]

ok

was nice to talk with you all

[south patrol]

Aight cuz

[rapid cave]

gn

[thorn jay]

cyaaaa

[hidden wind]

sure many young norwegians are very good at english, but very few, certainly less than half of all young norwegians have english as their mother tongue, whereas more than half of people in the uk do have english as their mother tongue

[elfin wraith]

Its somewhat of a joke, potentially not true for norwegians, but iirc like 96% of dutch people are proficent in english

[thorn jay]

many people here find the english cambridge easier than regular english because it's less learning

lol

I may be biased though

I'm pretty sure it's different at different school levels

[elfin wraith]

Yeah no of course, my joke here was honestly mostly about dialects, like English is my native and only langauge and it can be very hard to talk to people for whom that is also the case

[hidden wind]

right

[rocky cloak]

I mean UK has a pretty wide family of dialects.

I don't know how exposed brits are to other dialects, but it's kind of a big thing in Norwegian media I feel

[rocky cloak]

And also Danish and Swedish language (and English) is very present

[glad osprey]

It's interesting how varied the english proficiency is among norwegians, I feel like it's almost bimodal. There are some who grew up on internet speaking english and have an almost native understanding, and then there are some who struggle just reading a sentence

[elfin wraith]

Yeah I think I remeber hearing that like a significant portion of Norwegian media is in English which some think really helped with peoples understanding

But I think some people, especially in the south of England, quite easily grow up without much exposure to any of the stronger dialects from elsewhere in the country

[south patrol]

fairly exposed to others, just if someone has a v strong accent it can be harder to understand

actually yeah i am speaking as a londoner

[elfin wraith]

Yeah this is true I dont even think the dialect argument applies because I very intentionally speak clearly and ""properly"" for you fuckers and still get blank stares

[hidden wind]

this is true, there are many norwegians, even young ones, who are incredibly bad at english

i learnt english quite late and my pronunciation for the longest time was nothing short of a comedy sketch

it’s mostly fine now, which i have come to find unfortunate

accents are charming

[elfin wraith]

I think ive lost a lot of my accent since moving away from home, but people keep telling me its still very strong. I do feel kinda bad about losing some of it though, it feels like part of my identity, but it is rather impractical

[glad osprey]

I don't think you're speaking as clearly as you think, I found a clip of you trying to talk to some police officers: https://www.youtube.com/watch?v=Cun-LZvOTdw

[thorn jay]

are the guns also cannon?

canon

fdhsk

[glad osprey]

yes, Nope is armed to the teeth

[toxic sapphire]

I think I know the axis, but I'm not sure how to justify it

[elfin wraith]

(FYI im from litterally the opposite side of the UK, hot fuz is set about as far south as you can go lol)

[rocky cloak]

You're more of the billy Elliot type then?

Just because I like bellet don't mean I'm a puffah

[toxic sapphire]

I'm pretty confident that r^a h is equivalent to reflection across the line going through the origin with angle a2pi/n from the origin, I also found a description of this line via labeled vertices on P_n, but I don't know how to prove that this reflection is equivalent to r^ah

[elfin wraith]

further still, more of the trainspotting direction

[rocky cloak]

I guess the reasonable thing would be to see how this reflection acts on the vertecies and compare with his r^a h acts

[rocky cloak]

Shiet! You couldn't even find a decent civilization to be colonized by

[hidden wind]

i love billy elliot

[rocky cloak]

I love trainspotting

[toxic sapphire]

right, I just don't know how to do this rigorously. I'm not sure how to compute where this new reflection sends a vertex algebraically

[rocky cloak]

Well you just see where h sends it, then where r^a sends that

[elfin wraith]

Have you tried explictly working it out for some small cases?

[toxic sapphire]

yeah I got that part, but now I should show that the reflection across the line I'm claiming is equivalent by showing where that reflection sends a vertex, but I'm struggling to compute that

[south patrol]

Lol

[thorn jay]

what 😭

[south patrol]

Lol

I mean I should've looked up "Ind-Artinian"

[thorn jay]

of course

[south patrol]

Maybe I should learn more commutative algebra

[thorn jay]

join us

in struggling

[toxic sapphire]

for this problem, it seems like you can just define the map as sending a/2 + k to a/2 - k (mod n) when a is even, since this line of reflection passes through a/2

[south patrol]

Lol

[toxic sapphire]

but I'm not sure what to do when a is odd and n is even, since the line doesn't pass through any vertices

[rocky cloak]

It passes in the middle between two though

[toxic sapphire]

yeah it does, I see that. in particular it passes through the midpoint of (a+1)/2 and (a-1)/2. is it correct to say that it maps (a+1)/2 + k to (a-1)/2 - k?

yes I think that does it!

[limber sequoia]

huh

I was trying to ask whether more bizzare structures like R as a vector field over Q are ruled out by this definition (maybe there is some isomorphism), but I didn't know that as well

How are you building that isomorphism?

[rocky cloak]

As an abelian group R is isomorphic to Q^(|R|) which can be seen using the axiom of choice.

So then the isomorphism would just come from a bijection between 2|R| and |R|.

[rocky cloak]

If you want to think of R as a vector space over Q then you would be thinking of a map
Q -> End(R).

R -> End(R) would be realizing it as a real vector space

But they can get as bizarre as you like

[limber sequoia]

So in year 1 linear algebra we said two things:

1. a vector field is .... (all the axioms)

2. R is a vector field over R with the standard multiplication

And I implicitly assumed that this is the only way to take R as a vector field over R.

and you're saying that the implicit assumption is plain false.

[rocky cloak]

Yes

[south patrol]

Nothing special about the reals here either

Just that is the "standard" A-module structure on a ring A but there may be others, but this is what one kinda should suspect given you only are taking into account the underlying group of A (on one side)

[limber sequoia]

this makes a lot of my professor's comments from year 1 less cryptic and their emphasis on parts of the theory i dismissed...

thank you, im much obliged.

[candid patrol]

Hi, does anyone have a PDF or online link about the automorphisms of Z/2Z×Z/4Z, and more generally the automorphisms of a non-elementary abelian p-group ?

[south patrol]

I wouldn't say "homomorphic to" is a standard term. What do you mean by it

There is a homomorphism between any two groups given by sending everything to 1

[rocky cloak]

$A \cong B$

A is isomorphic to B

[cloud walrus]

Not jagr2808

mq in his algebra/stats arc

mq in his algebra/stats arc

mq in his algebra/stats arc

[rocky cloak]

If there is a homomorphism
A -> B, then you can compose this with an isomorphism B -> C to get a homomorphism
A -> C
Sure

You can still compose them. But then the kernel might change

Well the kernel is what you're computing in the proof above

If you just write down the definition of kernel it should be appearant.

If g(f(a)) = 1, it doesn't mean f(a)=1 unless you have some assumptions on g

[elfin wraith]

The kernel of an isomorphism is trivial

[rocky cloak]

The kernel of
A -> B
might be different from the kernel of the composition
A -> B -> C

That's right

[elfin wraith]

Theyre just functions at the end of the day, it probably has been just maybe implicitly. It can get confusing though, I do usually need to sit down and work out where stuff ends up on a peice of paper

[south patrol]

Kinda funny cause aren't like G and H kinda irrelevant here

[elfin wraith]

Yeah this is sorta just a more general and fairly obvious fact that I think they kinda obfuscate by mentioning G and H

[south patrol]

And then you can like quotient out by kernel and do first iso

First iso is kinda funny to me like

Ye

What is p

[glad osprey]

Have you learned Cayley's theorem yet? Or haven't they introduced that at this point?

[south patrol]

Probably not since this is a special case aha

Which one lol

Presumably you mean G is acting on G/H and inducing this

[glad osprey]

lol, I feel like the general case is easier than this special case

[south patrol]

Yeah that's what I mean by this

[elfin wraith]

Yeah isnt the proof like 2 lines lol

[south patrol]

Like you are letting G acting on G/H but that action already factors through H (by normality) to give you the thing in Cayley's theorem anyway

[glad osprey]

yeah, this corollary is just Cayley's theorem applied to G/H

[south patrol]

Yeah

[elfin wraith]

Take the obvious group action, trivial kernel, first iso

[south patrol]

Real.

Cayley is yoneda

Joking

Cause I find this kinda cringe

[elfin wraith]

@ moderators, obvious troll

Yes, every group is isomorphic to a subgroup of the symmetric group

You should make sure you understand the intution behind that, because what I was saying above about G and H just kinda confusing stuff is true, the reasoning behind Cayleys theorem should be pretty clear

G/H is a group, Cayleys theorem

what is the map p here?

I think so yeah, but also youre kinda just overcomplicating it, you know that its isomorphic to a subgroup by Cayley, and you know it has index n (so that |G|/|N| = n because everything is finite) and so it must be a subgroup of S_n

Thats sorta how I would argue it anyway, maybe just in slightly more precise words

[thorn jay]

perish

[elfin wraith]

Pretty much yeah

[thorn jay]

cayley's theorem is an easy corollary of the fact that the permutation representation of a reduced Latin square forms a group isomorphic to its stabilizer under the action of permutating rows and columns, when said Latin square is the multiplication table of that group

[south patrol]

i don't like "is isomorphic to a subgroup of". Just say "embeds into"

:chad:

[elfin wraith]

Exactly what I had in mind

[thorn jay]

trivial, really

its all trivial once youve internalised it

[elfin wraith]

They’re not mutually exclusive, I have spent hours on problems before only to realise they’re trivial

That’s because I am not a smart man

[rocky cloak]

Towards the end of a lecture a student interrupts
"Why does that follow from the lemma, I don't see it"
"That's trivial", the professor says
"It doesn't seem trivial to me"
"you see... hmm..."
The professor stops and stares at the problem for a bit.

The whole class waits in silence, before the professor breaks out of thought.
"Look at the time, I will have to get back to you next week"

At the beginning of next class the professor says
"I have carefully considered your question over the weekend and have concluded that I was right. The problem is in fact trivial"
"Now, onto our next subject"

[thorn jay]

its trivial once you see the trick™

unfortunately, the trick™ is nontrivial and requires a fundamental shift in perspective

[elfin wraith]

I had a topology lecture in a vain similar to this, the lecturer wasn’t actually supposed to be teaching the class, and he generally doesn’t believe in lectures, but someone asked him a question about some theorem which he remarked “ah that ones easy let’s prove” and proceeded to spend the hour working out how on earth to prove it

He did succeed and it was honestly really cool to see how such a talented mathematician (the man is cracked) works through a problem but it was amusing none the less

[rocky cloak]

I mean historically they come from people teaching to people who can't read / can't afford books.

Typically consisting of one person just reading aloud to a large hall

[elfin wraith]

I think lectures can be useful, but if they’re literally just reading out the notes they’re certainly not

If the lecturer provides a different perspective or goes really in depth on some technical details of part of the reading I think that’s good

Or as that lecturer believed in, just treating it as a Q&A session (and if no one asked anything he’d just kinda rant about whatever was on his mind, we did a lot of scheme theory in that topology class)

I will say though I was very bad about actually ever attending lectures in my UG and even if I went I was pretty bad at paying attention rather than just doing other work (or doing puzzles on my phone)

[rocky cloak]

Not directly related, but funny. I was in a lecture where the professor wrote things like
x in X

Someone asked if he could do more to distinguish capital from lower case x in his handwriting.

He just answered "No" then continued the lecture

[thorn jay]

because hearing it explained and seeing them write the stuff on a chalkboard is much more parseable than a book usually

[elfin wraith]

That’s brilliant I love that, gave me a good chuckle

[thorn jay]

no but i can imagine it

and the couple lectures ive had so far were still kinda interesting

except calculus, that made me want to smash my head into the table at how boring it was

[elfin wraith]

My analysis lecturer sounded like gru with made those lectures far more interesting

[thorn jay]

i guess in general calculus is not my cup of tea

[elfin wraith]

Very glad to have just done that in highschool

[rocky cloak]

I've generally gotten a lot out of lectures. More engaging than books

[thorn jay]

yes same we also did integration and derivatives and optimization n stuff

but we also have it in uni

i guess as an intro to analysis

[elfin wraith]

Not really, the Germans and Italians have it worse than we do lol, their unis are notoriously hard for maths, there’s like a 50% drop out rate in first year (granted it’s free and there’s no consequences but still)

[elfin wraith]

Yeah my UG just made you do a test when you started and if you do poorly you have to take a like foundations of maths course

[thorn jay]

ugh lucky

[elfin wraith]

A level maths is pretty similar to the level of Abitur no?

Ouch lol

[thorn jay]

im probably just gonna skip most lectures unless smt interesting is being explained. Ive found a cool book in the library im gonna work through so ill just do that

good, teach em young

[elfin wraith]

This was my general approach, I’m going to try to be better in my masters though both because I’m paying silly money for it but also because I think the work is just going to be hard and I’ll take everything I can get

[thorn jay]

this uni allows higher year courses but unfortunately the interesting ones start in the later semesters

so ive gotta sit it out for a bit

wdym letdown

i also went to Nijmegen cuz the city is cool

all 3 years

there are interesting masters courses im just not at that level yet

and multivar analysis is for second year, which is cool, but i dont know enough analysis

[elfin wraith]

Bros 3 days into his UG and can’t even take masters courses yet, smh

[velvet hull]

oh, mq is in their UG now?

good luck!

you've got a decent head start on algebra already

lol

okay, you do have a decent head start on algebra then

[thorn jay]

it would be cool to take noncomm geometry some time

cuz apparently they just have that

[elfin wraith]

Algebraic or functional analysis?

[thorn jay]

hmm, id have to look

and they also have schemes but i think thats mainly manifold-pilled

[elfin wraith]

I’d guess it’s functional analysis and C* stuff

[thorn jay]

sadge

[elfin wraith]

I think all the algebraic stuff is pretty K theory pilled and just

That being said I’m possibly going to see if I can write my thesis on that kinda stuff lol

[thorn jay]

K theory gets involved? cool!

[thorn jay]

ooo

[elfin wraith]

Something something non commutative motives

[thorn jay]

ah

ive heard of motives somewhere

[elfin wraith]

It’s something to do with generalising cohomology theories I think, or at least relating them in some way

But yeah one of the, possibly the, biggest names in this kinda noncom AG specifically through noncom motives is at my uni so I may speak to him about supervision

But I also may just do some rep theory stuff because uhhh K theory hard

[thorn jay]

all math hard

until you find a book actually properly explaining it McKenzie

[elfin wraith]

But more seriously I just don’t think I really have the background to approach anything like that, but of course he would potentially know an avenue I could explore, time will tell

Like I know some AG and AT but I don’t think even close to enough

[thorn jay]

i suppose its finding the middle ground between really hard and doable

[elfin wraith]

https://arxiv.org/pdf/1108.3787 this is essentially where I know what I know (very little) of noncom AG and motives etc from

[elfin wraith]

Yeah which I guess is part of the job of a supervisor, we shall see, I still don’t start for another like 3 weeks anyway

[thorn jay]

notation mess lmao

[elfin wraith]

There is also other areas of noncom algebraic geometry, I know my non com lecturer did some stuff like that which wasn’t as topological but I don’t really know any of the details

[thorn jay]

plenty of time to mentally prepare yourself

[glad osprey]

Let f : V -> Z be a module homomorphism. Is it true that f factors uniquely through V/H iff ker(f) <= H?

[rocky cloak]

H <= ker(f)

[glad osprey]

Ah, that makes sense, thanks

[thorn jay]

universal property!

[glad osprey]

And this is true for groups and rings too, right?

[rocky cloak]

And factors through V/H should mean through the canonical projection V -> V/H

[thorn jay]

any algebra

algebraic structure

using the right notion of "normal" "object"

[glad osprey]

And the right notion of "true"

[thorn jay]

it will be an object of the category but often not in the obvious way

something something categorical congruence relation

[south patrol]

Schemes.

[thorn jay]

jumpscare

[velvet hull]

I think about it in terms of the other direction, namely that ideals and submodules are defined as whatever the fuck they need to be for the "quotient thing" to work

[delicate orchid]

just in case you don't already know, these "whatever the fuck"s are called congruences and you can get very general with them

I don't mean to mansplain ts to u just in case u wanted to learn more

[velvet hull]

found the universal algebraist

[delicate orchid]

no no no no no for you see, quotient objects are exactly the coequalisers of congruences

so it was actually category theory

[velvet hull]

"corporate needs you to find the differences between this picture and this picture"

[thorn jay]

I FUCKING LOVE CONGRUENCES RAHHHHH

[thorn jay]

coping so hard

[delicate orchid]

category theory is better because I can put "\infty-" in front of everything

[twilit wraith]

Fuck my finitepilled bounded chud life

[south patrol]

Yeah

[elfin wraith]

An absolute all timer of a statement

[south patrol]

What is bounded

[delicate orchid]

STOP SAYING YEAH

[south patrol]

I'm tired bro

[karmic moat]

This guy so damn contrarian he hates agreements

[south patrol]

Yeah

[karmic moat]

Real shit

[south patrol]

Sup anamono

[karmic moat]

Nm

Waiting for class to start

Lie groups and Lie algebras

U?

[twilit wraith]

S is bounded if there exists a positive M such that |x| < M for all x in S

[delicate orchid]

U is one yeah

[south patrol]

Nice I hope that is funnn

[karmic moat]

It’s good yeah

[south patrol]

I have just been writing some stuff... tiring

Stayed up late thinking about maths rip

[south patrol]

M is bounded if H_k(M) = 0 for |k| >> 0

[karmic moat]

Real shit

[south patrol]

Sybau

[delicate orchid]

https://cdn.discordapp.com/attachments/1141450550425165875/1409950242639708311/lyzg1pltg2gf1.png?ex=68b9c9fd&is=68b8787d&hm=e147dc5a3ccf6661496be75e94b3c89f691bbfeb21fbd4cd76fe5a86d3b8314d&

[south patrol]

I have been thinking about dual vector spaces a little

Unironically

[delicate orchid]

in what sense

[south patrol]

Like Koszul duality uses these a lot and wondering how stuff generalises tbh

One thing I found amusing was I've always seen "Hom(V, W) = V* (x) W" stated only for finite dim V and W

But it (fairly trivially) holds as long as W is fd

Some generalisation of this fact came in handy for me

[thorn jay]

holy based

[south patrol]

Similarly for (V (x) W)* = V* (x) W*

In both you check they work when W = k and then both commute with finite biproducts in the W slot so all is good

Lol

[delicate orchid]

let me guess. Knowing u. We have a closed sym. monoidal (infty, 1)-cat and ur using the analogous statement with the internal hom

[south patrol]

I mean this is actually the case

[delicate orchid]

LMAOOOO

[south patrol]

But it is the derived category of a ring

So not that different morally

[delicate orchid]

just work with a ring spectra u coward (or I guess a field spectra since these are vector spaces)

[south patrol]

In this case I think stuff should go through ye

But also like duals behave v differently not over a field so eh

Eepington.

[delicate orchid]

so what exactly are you doing in this so called "derived" category

if that even IS it's name

[delicate orchid]

doing this?

but without the infinity, 1 part

[south patrol]

Eepy peepy

Oh I mean don't wanna spoil it tbh lol

[delicate orchid]

look foward to reading it in the next issue of the annals then ig

[elfin wraith]

Holy adjective soup

[delicate orchid]

it's like 50-50 noun-adjective

[elfin wraith]

Get a load of grammar boy over here

[thorn jay]

I can parse sym. monoidal (infty, 1)-cat somewhat

what does closed mean?

[delicate orchid]

it has an internal hom, so it's kind of implicit by the following clause

[thorn jay]

oh that was it

baller

[elfin wraith]

I just don’t know the (infty,1) part, I know wew has explained (m,n) categories to be before though

Or at least pasted multiple nlab links at me

[delicate orchid]

I know of no other definition of the word "explaination"

[thorn jay]

does (infty, 1) just mean a category but with some notion of isomorphisms between morphisms?

[south patrol]

All my symmetric monoidal things are closed

[thorn jay]

if I've parsed the definition correctly

[south patrol]

And isos between isos etc etc

[thorn jay]

right okay

ultrahomotopy

[south patrol]

Wew I am just thinking about some Koszul duals and computing basically but ye

I don't think I have anything new rly tbh

Or like

Ideas of things to do which would be new

[knotty badger]

why is that the case?

[south patrol]

By this

[knotty badger]

oh neat

yeah that tracks

Hom(V, W) cong Hom(V, k) tensor W is the idea you mean

hm and what if V is finite-dimensional

then it should also work right?

Hom(k, W) is iso to Hom(k, k) tensor W

[south patrol]

Oh I just mean is W = k then you just have the statement Hom(V,W) = V* (x)_k k

[knotty badger]

yes yes i'm aware

i think this should work if V or W is finite-dim

[south patrol]

Yeah same argument

[knotty badger]

it's essentially continuity of Hom

but in an enriched setting

[south patrol]

ig what i meant was like, this wasn't hte idea i had

[knotty badger]

oh

[south patrol]

It's just yeah Hom commuting with limits

as you say

[knotty badger]

hm then how does this fail when W is infinite-dimensional

ah i know why

[south patrol]

biproducts innit

[knotty badger]

yeah it's basically because

given a basis, a vector space is a big direct sum of its subspaces

but it does work when W is infinite-dim and V is finite-dim

unless i'm mistaken about that

[south patrol]

yes it does

but then you can do a similar thing in the other slot

[knotty badger]

so what fails in the generic case

that's what i'm trying to figure out

[south patrol]

Oh lol

well like

You get the comparison map from sum of product to product of sum

i think

[knotty badger]

right ok

[south patrol]

Maybe not quite that sorry i'm silly

i haven't thought about counterexamples aha

Well a nice thing to see is that in the case V = W is infinite dimensional, then like

Okay backing up, like there's the canonical map V* (x) W -> Hom(V,W) that induces the isos when stuff is good

which is just like sending (f, w) |-> (v |-> f(v)w) and linearly extension

if you take V = W infinite dimensional, then this doesn't hit the identity

Indeed this factors as an iso V* (x) W -> {V -> W with finite range} followed by inclusion ig

[knotty badger]

Yep

[south patrol]

If V or W has finite dimension then any map V ->W has finite dim range so you're good

[knotty badger]

Ah yes this tracks

thanks!

[south patrol]

But a different question is why they aren't like "abstractly" isomorphisc

like can you compute dimensions i mean

Maybe they are actually abstractly isomorphic, lol. like

[knotty badger]

wait really

[south patrol]

Actually yeah I mean suppose $V = \bigoplus_I k$, then $V*^* \simeq \prod_{I} k$. which you can write as $\bigoplus_J k$ for some (potentially bigger) $J$, then you get like $V^* \otimes W \simeq \bigoplus_J W$, whilst $\mathrm{Hom}(V,W) \simeq \prod_I W$, and like

these are gonna be different if W is massive

[cloud walrus]

Prismatic Potato

[south patrol]

Like I mean J is bigger than I, but now just wack up the size of W until the product becomes bigger than the sum i think

hm

[rocky cloak]

So it's true that for any infinite set I the dimension of W^I is equal to its cardinality |W|^I. And that |W| = |k|*dim W.

So comparing dimensions you have
|k|^I * dim W
and
|W|^I

If dim W <= |k| then both are just equal to |k|^I.

If dim W >= |k|^I, then both are equal to dim W.

If |k| < dim W <= I, then both are |k|^I.

If |k| <= I < dim W, then the bottom is just |W|.

So if the generalized continuum hypothesis is false you should be able to find dim W between I and |k|^I.

Otherwise it seems to me they will always be abstractly isomorphic.

[south patrol]

Yeahh this is kinda what I noticed, like if you wack up this seems an issue

So do you mean it may be false given like not gch?

Maybe this is more interesting than I realised lol

Ah yeah I see what you mean now sorry

[rocky cloak]

Yeah if there exists sets I and J with
I < J < 2^I
(and your field is small |k| <= I)

You can pick J as the dimension of W and I the dimension of V

[south patrol]

Ah wait yeah I didn't realise the field size bound damn

I was gonna say just make |k| big but lol

So this statement is actually equivalent to GCH?

Peculiar

[rocky cloak]

I think so yeah.

I didn't properly check all the cases, but seems like it

Not that surprising I guess. The statement is mostly about comparing sizes of power sets once you translate into cardinality

[south patrol]

Yeah sure

I think I remember like

i think $(V \otimes W)^* \simeq V^* \otimes W^*$ abstractly always, but just like the canonical map is an iso iff one of the dims is finite

[cloud walrus]

Prismatic Potato

[south patrol]

I think this is much easier though lol

https://planetmath.org/tensorproductofdualspacesisadualspaceoftensorproduct

Yeah https://math.stackexchange.com/questions/573378/u-otimes-v-versus-lu-v-for-infinite-dimensional-spaces also shows that continuum hypothesis gives a counterexample, couldn't find much else tho

I believe in u

Wait actually lol this is a cointerexample given CH

and you have one given not CH lol

Odd

Which I guess shows it is independent of ZFC lol if both are corrrct

[rocky cloak]

I think I see some mistake in my thinking.

I was assuming
|X|^I = |X| for I < |X|, but this I guess is not the case judging from that mse

[south patrol]

Ah okay

[rocky cloak]

This is in fact a silly mistake, since of course
|X|^I >= 2^I

[south patrol]

Hm how does that give a problem unless 2^I > X > I

Lol keep having CH come up

I guess is this what you use later tho

[rocky cloak]

Well, I assumed I had that, then tried to claim
|X|^I = |X| < 2^I

which is silly

[south patrol]

Ah okay

[slate bramble]

is true that in a finite p group of order p^n, there exists a subgroup of size p^k for 0\leq k\leq n

[south patrol]

This is either the (first) Sylow theorem or a corollary of it (depending on what you mean by the Sylow theorems aha)

[slate bramble]

i think the first sylow theorem was presented to me as the weaker statement that there just exists a sylow subgroup for each p

this result rings a bell but i don't quite see how i get it from the first sylow theorem

[south patrol]

So one form of the first sylow theorem is that there exist groups of all p^k where p^k divides the group order aha but yeah I mean

In he p group case i suppose can be done directly

You can pick a non-trivial central element x of G, and a power of it has order p

Then <x> is normal so you can consider G/<x> and use induction + correspondence theorem to get everything you need

[earnest delta]

If D is Euclidean domain then D[x] would be Euclidean Domain?

[limber sequoia]

Can anyone help me understand this cryptic comment?

This refers to module homomorphs

what "some morphism"?

Given some morphism f I can construct cokerf, and then compose a function to cokerf and it's kernel will be everything, but this is always true and doesn't depend on monomorphisms.

I don't think he refers to qoutients, as he talks about ker/coker in themselves.

[rapid cave]

No, Z is euclidean but Z[x] is not

[earnest delta]

Which property it fails

[rapid cave]

Z[x] is not even a PID here

[earnest delta]

Can It be UFD

[rapid cave]

I think yes

[unborn terrace]

It is indeed

[swift tundra]

If R is a UFD then R[x] is a UFD by one of Gauss’ lemmas I think

[unborn terrace]

Stole my message as I was writing it!

[rocky cloak]

They are saying monomorphisms are exactly those morphisms that are kernels.

An explicit example could be that if f is a monomorphisms, then
f = ker(coker f)

So in the sequence
A -> B -> coker f
A is the kernel of B -> coker f if and only if f is a monomorphisms

[limber sequoia]

How are you treating a morphism as the same thing as an object here?

[rocky cloak]

Well I'm not actually.

A kernel consist of two things. An object and an inclusion map.

So the kernel is the incision of A into B by f

[limber sequoia]

oh i see

[rocky cloak]

Or if you want to think of it more set theoretically:
Since f is injective it identifies A with the image f(A).

And f(A) is exactly the things mapped to 0 by B -> cok f

[limber sequoia]

oh that makes a lot of sense

[knotty badger]

i think the idea is that you want a galois connection

[limber sequoia]

for a general function
f: A -> B, then f(A) is the kernel of B -> cokerf,
but f is not an incision from A to ker (B -> cokerf) ?

[knotty badger]

$S \leq \ker(Q) \iff Q \leq \text{coker}(S)$ i think?

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[south patrol]

Idk if this is really what you want other than a consequence

[limber sequoia]

Is there a categorical description of a inclusion function as opposed to a general injective function?

[knotty badger]

i don't think so

[limber sequoia]

because I know that two groups G,H can be isomorphic as a subset of a group T but the qoutient may not be isomorphic, so they must have something different categorically?

[knotty badger]

i think what you're looking for is the concept of subobjects

[limber sequoia]

I know about generalized elements if that's what you mean

[knotty badger]

a subobject is an equivalence class of monomorphisms

with fixed codomain

[limber sequoia]

i see so that works out

[languid trellis]

I'm struggling on (a). So far, I've deduced that as a is odd and positive, we can apply quadratic reciprocity for the jacobi symbol: (a/p)(p/a) = (-1)^(p-1/2)(a-1/2). As p = 1 (4), we know (a/p)(p/a) = 1. As such, either (a/p) = 1, (p/a) = 1 or (a/p) = -1, (p/a) = -1. Either way, (a/p) = (p/a). As p = a^2 + b^2, it is clear that p is a quadratic residue modulo a. From this, I want to conclude that the jacobi symbol (a/p) = 1, however I'm struggling to (also I know that the converse of this is false, if (a/b) = 1 that does not imply that a is a quadratic residue mod b).

Is this approach correct? (If so, can i have a hint on how to proceed), and if the approach is wrong, can someone tell me pls and maybe where to look instead. Thanks

[rocky cloak]

So assume q is a prime that divides a.

Is p a quadratic residue modulo q?

What is (p/q)? What is then (p/a)?

[languid trellis]

Can we just reduce p=a^2 + b^2 modulo q to get p = b^2 (q), then we know (p/q) = 1 for all primes q dividing a, then by definition of the legendre symbol, (p/a) = 1, and by my reasoning above (a/p) is thus 1

And we know q doesn't divide b because that would imply p = 0 (q), which implies p=q, which is a contradiction (after some details)

nice. I don't know why I found that so hard. Jagr you are a saviour

[white oxide]

Does this follow from the fact that $M_{\pi^{-1}(\overline{\mathfrak{m}})} \cong (M/\mathfrak{a}M)_{\overline{\mathfrak{m}}}$

[cloud walrus]

okeyokay

[south patrol]

What is pi

[white oxide]

The canonical projection A -> A/a

[south patrol]

And is m bar = m(A/a)

[white oxide]

Also does g o f flat mean that C is a flat A-algebra and g (faithfully) flat means C is a flat B-algebra

[south patrol]

Then like pi^-1(m bar) = m

[white oxide]

Yeah well just any maximal ideal of A/a srry

[south patrol]

Yes

Like an A-algebra C "is" a map A -> C

[south patrol]

Ye sure

You can simplify using correspondence theorem ig

[white oxide]

Let $\overline{\mathfrak{m}}$ be any maximal ideal of $A/\mathfrak{a}$. We claim that $M_{\pi^{-1}(\overline{\mathfrak{m}})} \cong (M/\mathfrak{a}M){\overline{\mathfrak{m}}}$. Define $f: M{\pi^{-1}(\overline{\mathfrak{m}})} \to (M/\mathfrak{a}M){\overline{\mathfrak{m}}}$ by $\frac{m}{s} \mapsto \frac{\overline{m}}{\overline{s}}$, where $\overline{m}$ [resp. $\overline{s}$] denotes the residue class mod $\mathfrak{a}M$ [resp. mod $\mathfrak{a}$]. Then $f$ is an isomorphism, so that
[(M/\mathfrak{a}M){\overline{\mathfrak{m}}} \cong M_{\pi^{-1}(\overline{\mathfrak{m}})} = 0] by assumption. By localization, $M/\mathfrak{a}M = 0$, as desired.

[cloud walrus]

okeyokay

[white oxide]

I was too lazy to check that it was an isomorphism and well-defined lol

I mean it seems canonical so hopefully it is...

[south patrol]

You should need to quotient on the left

If it probably easier if you start with m containing I

[So then pi^(-1)(m bar) = m]

But then this is a result that is AM I think anyway?

Like localisation behaves nicely with quotients

[white oxide]

Oh yeah localization commutes with quotients

[white oxide]

sorry wdym

[thorn jay]

grahhh I love correspondence theorem

[ruby forge]

some of questions in my assignments have things like "does it carry a magma structure" or "does it carry a group structure"
are they different things than magmas and groups or I am confusing myself?

[south patrol]

As in the fact quotients commute w localisation in this way means the formula you wrote shouldn't be true

[south patrol]

I can't say too much without knowing the question, but often you say "Does [say, a set] X carry a group structure" to mean "Can you turn X into a group?"

Without more context this is a bit contentless (any set can be given a group structure) but it probably means a "natural" group structure in some way

[ruby forge]

I am confused since there is a set T with 20 elements, a function f: T -> Z and it says does (T, f) carry a magma structure but f is not even a binary operation

[south patrol]

Lol

I don't really know what that means then sorry

[ruby forge]

no worries, instructor made a "beginner friendly" abstract algebra short course and questions are so weird haha

[south patrol]

Lmao fair

Slightly odd as personally like I don't know anyone who works with magmas and I don't think most unis would have a course that mentions them aha

[thorn jay]

wtf

maybe you need a bijection with T and ZxZ?

but that's impossible

[ruby forge]

I mean f is a uninary operation what even is the point of the question

[south patrol]

Lol

[ruby forge]

I have learnt fields and rings before in hs and they are teaching magmas in uni for beginners

kind of funny

[south patrol]

Just say "any set admits a group structure so in particular..."

[south patrol]

Yeah odd

[south patrol]

I do always wonder what happens if you do this in an exam where you answer their question "too" literally

"No, consider the empty set"

[thorn jay]

axiom of choice spotted

[south patrol]

Okay I mean in this case biject it to Z^20

[ruby forge]

once I got 30% marked off because I did not show a increasing monotonic function that gets a negative value at a point (given in question) can output value 0

[thorn jay]

30% 😦

[glad osprey]

but that's not true even for continuous functions. Or do you mean "can" as in "may or may not"?

[ruby forge]

it was a polynomial
to be fair I would understand his judgement in general but it was such a easy function and I was just thinking he was just trying to bully me or smth lol

whatever have good day everyone

[ruby forge]

as a evidence he showed me thomas calculus manual and said you did miss that step dont know, it was trivial in my eyes

[glad osprey]

I see 👍 30% seems like an overreaction, but early in university they are often extra stringent with stuff like that - you know it's trivial, but they don't know that you know that. A lot of other students might claim something is trivial without actually being able to prove it

I wrote that a bit ambiguously, but I meant that 30% is an overreaction IMO

[ruby forge]

there were also bonus points in his defence haha

[white oxide]

Can I have a hint to show that if f: A -> B and g: B -> C and g o f is flat and g is faithfully flat then f is flat

[rocky cloak]

Let
0 -> M -> N -> L -> 0
be a sequence of A-modules. Tensoring with B gives a sequence
0 -> K -> B(x)M -> B(x)N -> B(x)L -> 0

To show that f is flat you need to show that K is 0.

[glad osprey]

Is there a way to show that every element of U (x) V can be written as a finite sum of u (x) v with u in U and v in V, just using the universal property? It's easy to prove when constructing U (x) V as a quotient of the free vector space over UxV

[kind temple]

i mean, isn’t that enough? the quotient map q : U x V -> (U x V) / (tensor relations) is such that by universality, you get a linear isomorphism between U (x) V and (U x V) / (tensor relations)

so whatever representation of U (x) V that you have, you can just go through the usual one first, and transport everything over

[glad osprey]

I don't 100% understand what you mean, but you're saying that since U (x) V is uniquely characterized by its universal property, we can just use any specific construction of U (x) V to prove stuff, and it'll be true for any construction?

[kind temple]

yes. like, an element of (U x V) / (tensor relations) is a linear combination of things that look like (u,v) + R, right, where R is the kernel of quotient map?

the isomorphism to U (x) V takes (u,v) + R to u (x) v coordinate-wise, and since it is linear, then the linear combination transfers over

[glad osprey]

hmm yeah, I kinda get it, but I'm not sure if by U x V you mean the free vector space over U x V or just U x V as a set. If it's the latter, I don't see how we can talk about linear maps and linear combinations

[kind temple]

sorry, the U x V in (U x V) / (tensor relations) should be the free vector space generated the product U x V. everything should still work

[glad osprey]

ah yeah, that's the proof I'm familiar with 👍 the way I think about it is, the free vector space over UxV are by definition finite sums of the form (u, v), and after quotienting you can still write the elements (or cosets I guess) as finite sums of this form

[kind temple]

so yea, q is no longer a quotient, should be an embedding

[south patrol]

Let's say the universal property is that bilinrar maps U x V -> W correspond to maps U (x) V -> W. Take W = U(x) V and we get that there is a bilinear map U x V -> U (x) V corresponding to the identity. Suppose the image of this bilinear map does not span. Then we can find a proper subspace X < U (x) V and every bilinear map U x V -> Z factors uniquely through this inclusion X -> U (x) V, which is bad.

I think something like this works

[rocky cloak]

Sure.

The universal property is that any bilinear map factors through
UxV -> U(x)V

consider now the submodule generated by u(x)v. The map UxV to this submodule is clearly bilinear, so this must be a direct summand of U(x)V, and by the uniqueness property the complement must be 0

[south patrol]

Aha

Why does Jagr get catking and not me too

Ok phew

[glad osprey]

Nice I like potato's proof the best, because of finishing a proof by contradiction by "which is bad"

[rocky cloak]

There's enough to go around

[south patrol]

Jagr's is more complete though

I guess more informally U (x) V is the "initial recipient of a bilinear map from U x V", and clearly the span of the image of that universal bilinear map must also be the initial recipient

[merry sorrel]

Hello guys, i was having a chat with one of my proffessors and she came up with a question that made me wonder for a bit. Do all inverse functions generate an inverse element for a specific operation? This came up in the context of studying group theory.

[tall igloo]

i am a bit confused by the question, do you maybe have an example in mind?

[merry sorrel]

Gimme a minute, i'll grab a piece of paper and try to recreate what we wrote down in class

[tough raven]

Let X be the subspace of U (⨯) V generated by u (⨯) v for u in U and v in V. Let phi be the quotient map from U (⨯) V to (U (⨯) V) / X. Then phi corresponds to a bilinear map which vanishes on all pairs, i.e., the zero bilinear map, so phi = 0 (by uniqueness). This means that X = U (⨯) V.

[near pivot]

where could i learn the basics of groups rings and fields i am very interested in the subject but have very little knowledge

[rapid cave]

well they are mostly different subjects but they interact with each other a lot

[near pivot]

oh i see i didnt really know that

thats how little i know 😭

[rapid cave]

You can pick up a book that covers all three, you can ask on #book-recommendations

[karmic moat]

#book-recommendations message

[near pivot]

thank you very much

[ebon prairie]

if p and q belongs to a ring
assume p divides q, hence p= q.r
my question is will this r belong to the ring?

[velvet hull]

if p divides q in the ring, r has to exist in the ring by definition

so divisibility is a concept that is dependent on what ring you are talking inside

[ebon prairie]

how by definition??

qr belongs to the ring

[thorn jay]

because that is the definition of divisibility

[velvet hull]

if p divides q in a ring R, then there exists an r in R such that q = pr

[ebon prairie]

if ab belongs to the ring does it imply a and b belongs in the same ring?

[ebon prairie]

oh

[thorn jay]

you dont have the context of a larger ring

so the only way that ab makes sense is for a and b to already be in the ring

[ebon prairie]

will it imply on group under multiplication as well?

[thorn jay]

any algebraic structure

because an algebraic structure only has itself as a context, its not guaranteed that you can place it into some larger structure

because operations on algebraic structures are functions only defined on the underlying set

[ebon prairie]

a.b

multiplication

how many multiplication are there

normal multiplication

algebraic

yes

i forgot , when i say multiplication i mean the normal one only

yed

[rocky cloak]

So is your question only about integers, or are you asking about rings in general.

If a*b is a product of integers, then a and b are integers (because otherwise it wouldn't be a product of integers...)

2*(1/2) is a product of rational numbers that equals an integer

[remote zephyr]

2*(1/2)=
1

ab=a+b:a=1,b=2:1+2=3

[glad osprey]

<@&268886789983436800> spamming random stuff in lots of channels

[remote zephyr]

Err: Undefined

[fast kelp]

If you wish to continue being able to post in advanced channels, knock it off please

[fading field]

toooo late

[fast kelp]

Whoops

I got out smayed

[thorn jay]

mf really started spamming for no reason

imagine having that little to do with your life

[fathom viper]

bro seriously got caught in 4k ( yeah bro if you spam you're going to get cooked I would know from experience )

[tough raven]

Soooo difficult to imagine. Could never be any of us.

[pulsar idol]

ok im having trouble understanding the meaning of this question, surely i get different cyclic subgroups generated by $aba^{-1}b^{-1}$if i choose different $a,b \in G$ right

[cloud walrus]

lifelong dumbass

[pulsar idol]

cause looking at sources there doesnt seem to be any mention of the commutator subgroup being necessarily cyclic

[wraith cargo]

It's generated by all such products

So it's generated by such relations for all a,b

[pulsar idol]

uhh