groups-rings-fields

[alpine remnant]

isomorphic as sets

that's it

[dim widget]

No the point is that (much stronger) R \cong R^2 as abelian groups

[alpine remnant]

do I need them to be isomorphic as fields?

[dim widget]

No just as abelian groups

under addition.

[alpine remnant]

huh

[dim widget]

K^2 is not a field unless you add some extra choices.

Anyway "many fields" will have this property, the only real problems in characteristic 0 are Q and number fields, where these examples dont work.

But if the question was "is it possible to have such a map f on an infinite field" then yes most infinite fields admit such a map.

WHich is why I zeroed in on Q, because I think there it's very unclear whether such an f exists or not.

[alpine remnant]

Inma go have lunch and try to write this all down properly to interiorize it coz I'm struggling

thanks a lot

why wouldn't this work for a simple bijection as sets between K and KxK?

[dim widget]

Ah you’re right, even just a random bijection works. I was over complicating things.

[slim kayak]

1. Does there exist a neat characterisation of the kernel of a map from a free module that sends any elements to the sum of their coefficients for a given basis?

2. Similarly, does exist for a basis consisting of two subbasis B1 and B2 (each of the same size as the one of the previous question) such that the sum of coefficients of the vectors must be equal 0 for each subbasis respectively

Somehow those are meant to be isomorphic

[frigid lark]

The one cycle, (1) only means that the permutation sends 1 to 1

It could still send 2 to 3

yeah your correct

This is exactly the reason we ignore 1 cycles

Personally I prefer to denote the identity map by e or 1

[delicate orchid]

except it doesn't have any different ways, because we don't write down 1-cycles

[frigid lark]

So do most groups, we can write down the identity in G as aa^{-1} for any a in G

But in the end it's still the identity

[delicate orchid]

I think of it as analogous to why we don't consider 1 or -1 prime in Z, its so each integer has a unique prime factorisation and we can't just keep adding on 1s forever

[formal ermine]

hi wew

[delicate orchid]

who is this guy ^?

[south patrol]

So true

[formal ermine]

POTATO

[south patrol]

Hallöchenlein Illum was geht

[formal ermine]

I'm still trying to figure this variety thing out

[south patrol]

Which thing oof

[formal ermine]

ok so definition 1 is

[delicate orchid]

nur englisch bitte...................

[formal ermine]

Z(f) is irreducible iff f is irreducible

definition 2 is through noetherian topo space

how are they equal

the definitions

[slim kayak]

What's the second defo in more detail

[slim kayak]

This is equivalent to the definition of varieties as algebraic sets that can't be written as a sum.

[formal ermine]

cannot be written as union

[slim kayak]

Ah

[formal ermine]

what obvious thing am I missing

wait

[slim kayak]

We just that Convo on hallidays thread

[formal ermine]

Z(fg) = Z(f) cup Z(g) no?

I'm stupid

[slim kayak]

Take any point from your variety

[south patrol]

Hi stupid I'm dad

[slim kayak]

It is a zero of some polynomial, if it is the zero of an irreducible polynomial that polynomial is the minimal polynomial of that root

[formal ermine]

):

[dusty verge]

Oh hey that's me

[south patrol]

You're not stupid

[slim kayak]

And so any other equation you could add that has that root must have that minimal polynomial as a factor

[dusty verge]

Dad? But you died in the fire...

[frigid lark]

They have gained enough karma from answering silly math questions to be reborn

as not a slug

[formal ermine]

I have 501 rep on mse!

[south patrol]

But as a potato

[frigid lark]

potato > slug

[south patrol]

Fair enough. A slug can suffer existential dread but a potato cannot ig

[frigid lark]

potato feed all of Russia...

's vodka addiction

[formal ermine]

why is the definition of a scheme so... unintuitive

[dusty verge]

Existential dread is lame just make your own meaning

[dusty verge]

What's a scheme?

[formal ermine]

do you know what a ringed space is

[dusty verge]

Yeah

[ivory trail]

a scheme is like a manifold but instead of R^n as charts you have affine schemes

[formal ermine]

an affine scheme is a locally ringed space that is isomorphic to some spectrum

[dim widget]

Yes

[dusty verge]

So it's a spectrum?

[ivory trail]

yes

[dim widget]

Just like R^n is a manifold whose ring of smooth functions is isomorphic to C^infty(R^n)

[dusty verge]

Im of the philosophical opinion that isomorphic objects are different ways of writing down the same thing and will fight anyone that disagrees

[ivory trail]

if only there was a person versed in hott here

[formal ermine]

a morphism of ringed spaces is weird

[slim kayak]

That sounds kinda lukewarm of a take

[dusty verge]

I'm versed in hot, that the same?

[dim widget]

Isn’t that an extremely basic tenet of all mathematics for the last 150 years? It’s not super controversial

Maybe more like 100 years

[slim kayak]

Isomorphic is a very popular notion of "the same"

[ivory trail]

sometimes the isomorphism is interesting though

[dusty verge]

Idk people have disagreed with me before though

Mostly my advisor

[formal ermine]

like why the f^sharp : O_Y -> extension of X for X -> Y, like why is the domain O_Y

[ivory trail]

but when it is canonical it is more fair to say that it's "the same"

[frigid lark]

I think you have to argue it most of the time

[ivory trail]

you should think of the map on sheaves as "pulling back" functions on Y to functions on X

like precomposing in topology

[dim widget]

Nooooooo not this

[formal ermine]

the codomain is on Y tho

[dim widget]

It is equivalent by some adjunction

[dusty verge]

Yeah but like, why?

[ivory trail]

oh yeah lol

[dim widget]

But the way it’s written in modern mathematics is that you think of the sheaf of functions on X as defining a sheaf on Y by “pushforward”

[slim kayak]

Obviously

[dim widget]

Meaning functions on V\subset Y are functions on f^-1 V as a subset of X

[dusty verge]

What's x->y denote here?

[dim widget]

And the morphism intuitively assigns a function on Y it’s “pullback” which is a function on X

[slim kayak]

Probably some "initial" morphism from which you are building f sharp from

So f

[ivory trail]

the structure sheaf of a ringed space can be thought of as the "scalar functions" on it in the same way that continuous real-valued functions on a manifold are the scalar functions on it (which has a scaling action on the tangent bundle, for example)

smoothness or whatever

[dusty verge]

Scalar functions?

Are you drawing an analogy to vector spaces here?

[slim kayak]

Like a dual space sorta?

[dusty verge]

Oh neat

[slim kayak]

That was a question

[dusty verge]

Lol

[ivory trail]

like in physics, you can have a vector field and then scale the vector field by multiplying each vector by a scalar

but if the scalar varies continuously/smoothly ("scalar field") you can still do this

[slim kayak]

I apologize for not using mandatory the ", no?" Question format

[dusty verge]

Oh neat

[ivory trail]

so the locally ringed space keeps track of what "scalar functions" it considers to be admissible

i don't think anyone else uses the term "scalar-valued functions" that's just how i think about it

[slim kayak]

The fact that you aren't using the term dual space there is making me doubt myself

[ivory trail]

yeah no dual spaces are involved

[dusty verge]

Yeah I think of ringed spaces as zooming in on shit and stuff

It's a ring where you zoom in and get a quotient

[ivory trail]

people do use the phrase "admissible functions" though

[slim kayak]

Ringed spaces are clearly chain mail

[dusty verge]

A ringed space is that dark souls 3 dlc

[ivory trail]

the analogy to manifolds is pretty important though

https://mathoverflow.net/a/756 idk how relevant this is

but it provides a reason why we should do geometry with locally ringed spaces

[slim kayak]

Slave algebraic Geometer Gael was a cool fight

[south patrol]

Lmfao

Is that from the ringed city

[slim kayak]

Locally ringed city

[dusty verge]

Dark souls 1 is the greatest game of all time and I will fight anyone that disagrees

Or at least have a math off with them, since I can't punch through the internet yet

[slim kayak]

Where do you guys think would be a better place for a question used by cheap knockoff homology of 2-cells attached to (multi)graphs as used by my algebra course. Here or on algebraic topology

[dusty verge]

Here

[slim kayak]

Liked sekiro and Bloodborne most

[dusty verge]

Yeah but the map

Think about the map my guy

The dark souls one map is homeomorphic to perfection

[ivory trail]

oh sorry i thought this was #algebraic-geometry

i guess my explanation doesn't help much if you're not familiar with manifolds and stuff

[dusty verge]

A manifold is just a shape

So what's an atlas though? Or chart or whatever

Physics guys mentioned them to me before

[carmine fossil]

Isn't DS1 like really slow

[dusty verge]

Compared to what?

[carmine fossil]

Sekiro

[dusty verge]

Yes

[carmine fossil]

And ig Blood borne

[ivory trail]

you know how an atlas of earth works

[dusty verge]

Now replace ds1 in those sentences with any game that's not sekiro or bloodborne

[ivory trail]

you can't draw the globe on a flat piece of paper

[dusty verge]

Yeah the little map goblin in the book asks his friend in the sky where I am

[ivory trail]

so you instead draw several patches

[slim kayak]

So we have a slightly modified definitions of graph so that the set of edges has twice as many elements by adding e^-1 for each e and we have some functions that spit out the starting vertex, end vertex and an involution sending any e to its opposite oriented version.

Defined the chain complexes like this:

[ivory trail]

also known as charts

[carmine fossil]

Ok damn most games are slow

[slim kayak]

And the chain map like this:

[ivory trail]

the charts have coordinates on them

[slim kayak]

no my question has a break

[dusty verge]

Let me go drink an unhealthy amount of coffee real quick

[ivory trail]

and then you have to bend the analogy a little and say that the atlas also says that if you're in a region that is mapped on two charts, there should be a "transition function" turning coordinates of the first chart into coordinates of the second

and then furthermore that if you have three charts and you're in the triple overlap, going from the first to the second then from the second to the third is the same as going from the first to the third

bit of a technical condition that just says that these transition functions are compatible

actually it kind of depends on how you want to define a manifold

sometimes you start with a bunch of charts (i guess "patches" in this situation) and transition functions and verify the above compatibility condition to construct your manifold

sometimes you start with a topological space and a bunch of charts that makes the topological space into a manifold

[slim kayak]

A complex is defined as a graph together with a set of closed paths, which act as gluing instructions for 2-cells onto the graph. What I am meant to show is that the homology isn't changed if I change some things, one of them is adding a new vertex "e" to the original set and splitting the closed paths from this to that leaves the homology groups unchanged.

That is, geometrically, instead of attaching one 2-cell to a closed path of a graph attaching 2 smaller 2-cells

[ivory trail]

this looks like stitching together a bunch of patches in a specified and compatible way that gives you a manifold. while the other one is just a way to make computations on manifolds easier by writing down charts where you can do local calculations

[slim kayak]

One thing after another, the second chain map looks weird, it would just be summing up the edges of a cell rather than the usual alternating sum

[dusty verge]

That sounds sheafy, but that's maybe just surface level

[dusty verge]

Involution?

[ivory trail]

it definitely is, because if you stitch together structure sheaves on R^n then you get the structure sheaf of the manifold

[dusty verge]

Also what's that notation Vgamme | emptyset?

[slim kayak]

self-inverse function

[dusty verge]

What's VGamma here?

[slim kayak]

vertices

[dusty verge]

Oh it's just the set of vertices?

[slim kayak]

EGama correspondingly the edges and I is the index sets of the closed path i.e. the 2-cells

yeah

[dusty verge]

What are the patches? Open subsets of a larger space?

What's the umlaut fur?

?

I'm sorry my keyboard doesn't write in German

@slim kayak what's that notation you're using with the empty set here?

[slim kayak]

?

[dusty verge]

Wait are those sets?

Do you smooth bracket your sets?

[slim kayak]

implicitely

[slim kayak]

where?

[ivory trail]

for a manifold, the patches are open subsets of R^n

[dusty verge]

Wait why wouldn't you generalize it to any space?

[ivory trail]

well if you instead use all affine schemes and appropriate kinds of transition functions you get schemes

if you use open subsets of R^n and continuous/smooth/differentiable/whatever transition functions you get the respective kinds of manifolds

[dusty verge]

I'm trying to figure out your notation of [VGamma|emptyset]_Z

[slim kayak]

Its free module over set with relations

[dusty verge]

Oh that's a nice notation

[slim kayak]

in this case empty set

probably you can write anything you want in the right column as long as it makes sense

[dusty verge]

So the empty set is the relations?

[slim kayak]

yeah

[dusty verge]

Cool cool

[slim kayak]

as you can see by the isomorphism there

[dusty verge]

Yeah that makes sense

This is giving me leavitt path algebra flashbacks

[dusty verge]

So people do use this construction in other contexts, they just don't call the things they get manifolds?

[slim kayak]

for

if u see any german words i left just read them as "for"

[dusty verge]

What's a chain map?

[slim kayak]

looks like thats a technical term already

in this case just the boundary map between the chain complexes C0, C1 and C2

[dusty verge]

Boundary map?

[slim kayak]

oh, and 0 so that it doesnt feel left otu

[dusty verge]

I.e. these are closed sets and it maps their boundaries?

[slim kayak]

For defining homology groups?

[formal ermine]

write ue for ü, oe for ö, ae for ä, and sz for ß

[dusty verge]

Uh give me a sec to go figure out homology

[delicate orchid]

sometimes u people just need to say "u don't have the prereqs sorry"

[formal ermine]

halliday is a machine

he'll figure it out

[dim widget]

Sz för ß? Low German degeneracy

[slim kayak]

u told me here would be fine, you traitor

[formal ermine]

says the person who ghosted my dms

[slim kayak]

Ich hasze sowas zu sehen

[dusty verge]

That's the only heart I could find

[formal ermine]

haße

[slim kayak]

gotta have to wash my hands after this is over

thats a crime

[delicate orchid]

I would do this also

[slim kayak]

Anyways, given these changed 2-cells and the definition for the boundary map here it seems that C2 goes from being Z^n to Z^2n

So delta(x) = $\sum a_{i} \partial(i) = \sum a_{i} (e_{1} + ... e_{n})$ should hold.

[cloud walrus]

Kerr

[slim kayak]

( I had to write "\partial" based on "partial derivates" on the abstract algebra channel 😔 )

[delicate orchid]

oh are we doing graph homology

[slim kayak]

is it really called that? Cuz googling that didnt find me stuff that looked like it

[delicate orchid]

one moment

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

[dusty verge]

Okay so a homology is like an exact sequence, but less strict?

[delicate orchid]

a chain complex is a less strict exact sequence

the homology of the chain complex measures how far it is from being exact

[dusty verge]

Woops my bad yeah that

Oh

[slim kayak]

oh cmon i was typing that 😦

[dusty verge]

And exact sequences are just first iso theorem

Neat

[delicate orchid]

you what

[slim kayak]

vine thud sound

[dusty verge]

So

[slim kayak]

my name changed

[delicate orchid]

no no exact sequences are not just the first iso theorem

[slim kayak]

I was called out for not touching grass enough

[dusty verge]

Well no but also yes. Or rather they're important cause of it

[delicate orchid]

sure you can construct 0 -> N -> G -> G/N -> 0 as an exact sequence? I think that's what you're getting at?

[dusty verge]

It is, but that idea generalizes to any exact sequence

[delicate orchid]

does it now

[dusty verge]

If a -> b- > c is exact, then the three things are analogous to your N, G, and G/N

[delicate orchid]

yur fairs

I suppose I see what you're saying

[slim kayak]

short exact sequences be like

[dusty verge]

So it encodes the quotienty relationship between consecutive elements of the sequence

I wonder if you can express solvability of groups with exact sequences idk I don't wanna figure that out

[ivory trail]

nlab has a page or two on this kind of thing

[dusty verge]

But that expands to any three consecutive elements in a sequence

[dusty verge]

Nlab is scary

[delicate orchid]

not quite I don't think

long exact sequences are compositions of short exact sequences but it's more complicated than just putting them end to end

goes like this

[slim kayak]

That sounds like you are stating the definition of an exact sequence vaguely

[dusty verge]

Yeah exactly

That's my entire niche

[slim kayak]

What does that have to do with the first iso thm?

[dusty verge]

But the point is that that quotienty relationship is the first iso theorem

But harder

And written in a way to confuse ambitious undergrads

[slim kayak]

If you quotient out by "harder" then those are the same, true

[dusty verge]

That diagram is what you get if you don't restrict to images right?

But either way that structure makes sense if you go ahead and derive it yourself

It's not actually mean, it just has resting bitch face

Pre images? What the hell was I saying lmao

[slim kayak]

anyways it looks like that for the modified set of 2-cells one has the free module over paths whose image under the boundary map is the sum of their edges, so one then have:
for $x = \sum a_{i} \gamma^{1}{i} + b{i} \gamma^{2}{i}$ and so since edges are mostly disjoint for the split paths that means that the kernel consists out of the elements such that $\sum a{i} = \sum b_{i}$ = 0 holds.

[dusty verge]

Okay disregard the first sentence there, you can't restrict without losing a lot of generality

[cloud walrus]

Kerr

[agile burrow]

You can, sort of. It's easier to see with nilpotent groups, you can define a group to be nilpotent if it's built out of central extensions of abelian groups. So for example, G is nilpotent if it fits into an exact sequence 1 -> N -> G -> G/N -> 1 where N is central and G/N is nilpotent, where you define the trivial group to be nilpotent.

A similar thing holds for solvable groups, where you can inductively define a group G to be solvable if it can be built out of abelian extensions of abelian groups

[slim kayak]

But that makes the statement of the exercise false, since one could take the graph complex with a single 2-cell / closed path and split it into the two and then one has instead of kernel being trivial that it is generated by a single tuple (1,-1)

[dusty verge]

Oh cool. I'm not sure what a lot of that means, so id need to spend a bit on it, but still cool

Whoever wrote "abelian groups or modules" on Wikipedia has some explaining to do

[slim kayak]

Abelian groups can be equipped with Z-module structure

[dusty verge]

That's why they have explaining to do

Smh distinguishing between abelian groups and modules

[slim kayak]

Or are you asking whether there is a notion of solvability for modules over some ring

[dusty verge]

No, I'm going back to figuring out homologies

[slim kayak]

sure

[dusty verge]

Oh fuck you Lang

I just realized why he threw an iso theorem into groups and never mentioned it again

It's so he could use it 2000 pages later when he got to homologies

[slim kayak]

halliday whenever he sees the iso theorem anywhere

[agile burrow]

I'm fairly confident that the isomorphism theorems would be used throughout the book

[dusty verge]

It's almost like it's the central idea in groups

[slim kayak]

You'd often be hardpressed to know when he is using it though

[agile burrow]

Fair enough

[dusty verge]

Kerr I thought you were cool and then you called me a monkey :(

It's okay you're not wrong

[agile burrow]

Admittedly I've only used Lang as a reference once or twice so it really isn't my place to comment on it

[slim kayak]

no, I didnt mean it like that : (

I am sorry

[delicate orchid]

it's the central idea in many categories not just groups

the central idea of groups is conjugation :smoked:

[slim kayak]

Did lang remove his jordan-hölder section or sth

[dusty verge]

No clue I've only got through groups

I needed a year long break after that

[slim kayak]

it used to be in groups

[dusty verge]

Uh doesn't sound familiar then

[slim kayak]

that section i was my first into jordan-hölder and group composition series

and I still have nightmares

[dusty verge]

Btw if you wanna be impressed, I also did the exercises

[formal ermine]

what's the connection between rep theory and normal subgroups?

[slim kayak]

Of the group section?

[dusty verge]

It's almost definitely important

Okay yeah vector spaces have dimensionality

So probably it lets you use dimensionality arguments to derive stuff about the structure of groups

[slim kayak]

nvm its still in the group section, just somehow 3 pages long

[formal ermine]

[dusty verge]

Hmm, no clue then

[formal ermine]

okay and also

[dusty verge]

My rep theory is non existent

[formal ermine]

number of reps = number of conjugacy classes

[formal ermine]

mine is shit too lmao

I only know like one or two definitions

just google stuff right now lol

[dusty verge]

Right that's the finite simple group classification bit right?

[formal ermine]

so we needa use class formula or something to find the number of reps

if we work over the complex numbers

[agile burrow]

Bro it's first iso theorem

[slim kayak]

more or less, it says that each group has an unique tower to them

[dusty verge]

It always is

[formal ermine]

I see a homomorphism, I see a normal subgroup

MUST BE ISO THEOREM

[dusty verge]

It's the central idea of algebra

[agile burrow]

Ok but it literally is the first iso theorem. If you have a rep of G with kernel N, this yields a faithful rep of G/N

[dusty verge]

The complex plane is just the first iso theorem

[formal ermine]

yeah lmao

[dusty verge]

Galois theory and free algebras and topological gluing

[formal ermine]

oh

nvm

I see

me is stupid

[slim kayak]

the what?

[dusty verge]

Plain? Idk

[slim kayak]

Do you mean R[X]/(X^2+1)?

[formal ermine]

he's talking about the algebraic closure of R

[rotund aurora]

if you have something that is like an identity, you just call it an identity

so only the identity is the identity

and the action is faithful

[formal ermine]

yes I am stupid

I mentioned that already

[dusty verge]

The complex plane is clearly just ring set of polynomials of sets of sequences of pairs of sets recursively containing the empty set quotiented out by the equivalence relation of cross multiplication quotiented out by Cauchy sequences quotiented out by the by the smallest multiplicatively closed additive subgroup that contains x^2+1

Will that last one define an ideal? Feels like it should

[agile burrow]

...the ideal generated by x^2 + 1?

[dusty verge]

Yeah

Math is wacky lmao

We come up with all these complicated definitions for shit in order to make them match our intuition

Okay so I'm reading the definition of a homology on Wikipedia, but from what I can see they don't actually attach the homology to the top space in any way. What relationships do the chain complex C(X) have to satisfy with the space X?

[agile burrow]

You associate a chain complex to a topological space and then compute the homology of that space. If you've built up your theory correctly, the homology should be an invariant under whatever equivalence relation you're considering. So to compute homology of a topological space, you compute homology of say the singular chain complex, or the cellular chain complex if you're a CW complex, etc.

[delicate orchid]

u draw they triangles

then consider they continuous maps from they triangles

[agile burrow]

Yeah, the general procedure is to start with some object, associate some chain complex, and then take the homology to get an invariant which is hopefully useful

[untold cloud]

Hi, guys, in these assumptions, let pR be the ideal in R generated by p, and m be the maximal ideal of pR in R, then why R/m is the splitting field of f_p

[dusty verge]

I'm R[x] you mean?

Or please just be more specific with your definitions, cause I think you're only giving like 80% of the info here and then missing 20% seems critical

[untold cloud]

suppose F is the splitting field of f_p, then i want to show that R/m is isomorphic to this F

sorry about the confusion

[dusty verge]

What's R?

Chunky or not chunky?

[untold cloud]

R= Z[\lambda_1,...,\lambda_n], where Z is integer and \lambda_i is root of f in K

[dusty verge]

Okay, so there's no initial relationship, it's just that the structure of the space will force one, which tells you stuff?

[dusty verge]

Oh gotcha. It's just the thing in the thing

[agile burrow]

Well you have to be smart about how you build a complex out of an object, I wouldn't say it's "forced"

[dusty verge]

You have to be smart? Oh no

Okay, so it's a "I guess this will be useful, hey look it is" kinda deal?

[agile burrow]

Take cellular homology for example. There are infinitely many CW structures you could attach to a space, yet they all yield the same homology

[dusty verge]

Do you have fundy thm yet?

[dusty verge]

Sorry, CW?

[agile burrow]

Probably another thing for you to look up

[dusty verge]

Oh lmao it stands for CW

[formal ermine]

CW stands for CW obviously

[dusty verge]

You can probably do this with the iso theorems. Define the splitting field of f_p and try and find it as a subfield of F

[slim kayak]

Hi polynomial ring over R

[slim kayak]

R/m is F?

Being a splitting field means that your polynomials dissolves into linear factors, since f_p splits in F_p any field containing is hence also a splitting field.

In case of p a prime pR is already maximal so in your case the statement says F_p is a splitting field and F=R/pR = F_p is a splitting field

[chilly ocean]

Would this be the right channel to ask about stuff with Complex numbers? Or is that calculus? (Sketching subsets with it)

[slim kayak]

depends a bit on the question

Does it involve complex numbers or is it about them?

[chilly ocean]

it involves them ( i.e |z-2-3i|)

[delicate orchid]

probably one of the lower uni or hs channels then

[slim kayak]

Have you taken abstract algebra or a proof-based course linear algebra course?

Yeah, probs a level too high here

[formal ermine]

the entrance bar here is knowing what a group is

[chilly ocean]

Ah well.. Thought it may be in here cause our topics are also rings n stuff rn (feels like its all over math)

[delicate orchid]

what

ok nvm yeah this is the right place

[chilly ocean]

I mean last week we had stuff like an abelian group.. this week its complex numbers .. to fast to understand

[formal ermine]

is this some hs summer camp lol

[slim kayak]

sounds like my first semester analysis course

[dusty verge]

Complex numbers are an abelian geoup

[formal ermine]

without 0

🤓

[dusty verge]

Groups are just things like addition

[chilly ocean]

nah its just 2nd Semester maths for Computer Science ;-;

[formal ermine]

oh

that explains it

it depends on the question really

[dusty verge]

Groups are just things that look like addition

[formal ermine]

if you state it we can tell you where it belongs

[dusty verge]

Abelian groups even moreso

[slim kayak]

free group starring blankly

[chilly ocean]

How would you sketch something like {z ∈ C:0≤arg(z+3-4i)< π/4} ? I know how it would look like if it just says arg z but this confuses me.. I already have done it but idk if its right

[slim kayak]

Anyways, if you can combine things and cancel things and also do nothing at all, thats a group

z : 0≤arg(z+3-4i)< π/4 ?

[chilly ocean]

ah yeh sry

[formal ermine]

ok so

a scheme is something that locally looks like an affine scheme

yes?

[slim kayak]

Did you already cover using the exponential function to depict complex numbers?

[formal ermine]

@delicate orchid

[delicate orchid]

firstly #algebraic-geometry

[formal ermine]

scary place

[delicate orchid]

who gives one stop being a baby

you might actually get an answer there

[agile burrow]

Lol Wew told you to grow up

[delicate orchid]

I don't know alg geo beyond basic affine varieties and results relating to them

[chilly ocean]

uhh idk

[formal ermine]

I am learning classical ag and scheme theory at the same time rn

after I'm through this and the next chapter I'm gonna switch to diff geo for a couple of weeks

[delicate orchid]

le wedge has arrived

[slim kayak]

Nevermind, just think for what z arg(z) lies between 0 and pi/4.

Then shift those values accordingly

[agile burrow]

Yes

[formal ermine]

epic

thanks walter

[chilly ocean]

ahh k .. thanks @slim kayak

[formal ermine]

it's like a manifold innit

[agile burrow]

Yes, that's the analogy

[formal ermine]

I still need intuition for an affine scheme though lmao

[agile burrow]

[slim kayak]

what do the values z look like that have arg pi/4?

[chilly ocean]

Its just the area between the "x axis" ("" cause its the real axis or whatever you would call it) and x=y

[slim kayak]

Yes, now you have everything to know the values z+3-4i, now you only have to translate that to z.

[chilly ocean]

And with (z+3-4i) the point which would be (0,0) moves to (-3,4)? right?

[slim kayak]

yes

[chilly ocean]

good.. then I fully understood it now .. Many thanks

[daring hinge]

So according to a YouTube video the answer is C although in answer key the given answer is A. Can anyone confirm?

[formal ermine]

how should 2 be true lol

[dusty verge]

Free group looks like addition

[daring hinge]

Well the one who explained on video did something like f(a+1) = (a+1)³ = a³+1+3a²+3a = f(a)+f(1)+3a²+3a
So 3a²+3a = 0 implies 3a(a+1)=0 and since a is arbitrary 3=1+1+1=0

[chilly ocean]

hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

[wraith cargo]

Doesn't factoring out like that work only in an integral domain?

[slim kayak]

Well, f(1+1)=(1+1)^2 = 1+1+1+1 and f(1+1)=f(1)+f(1)=1+1

so 1+1=0

[daring hinge]

Is this for first part?

[slim kayak]

yeah

[daring hinge]

Ah ok

[delicate orchid]

le frobenius endomorphism has arrived

[daring hinge]

Yeah I'm doubtful the way he did it

[wraith cargo]

Tho i think 2 actually is correct lol

[daring hinge]

Oh

Damn

[delicate orchid]

yeah you can't automatically conclude from 3a(a+1) = 0 that a+1 = 0 or 3a = 0

[daring hinge]

Alright, do you think part 2 is wrong then?

[delicate orchid]

not sure tbh

[slim kayak]

I mean, it sure is valid that expressing the abelian group as a Z-module, that 6.1=0.

but something we have to break that down further

[wraith cargo]

Yeah i think that you can immediately conclude that 3x1=0
I think you can only conclude that 6x1=0

So uhhh

[slim kayak]

is there any reason the second ring can't be Z/6Z

[delicate orchid]

ok everybody, lets name our favourite char 6 rings!

[delicate orchid]

the map isn't a ring homomorphism

[formal ermine]

uhhhh

[wraith cargo]

Tru

[delicate orchid]

or is it

[formal ermine]

R/6Z

[delicate orchid]

nope, it is a ring homomorphism

Z/6Z works

[slim kayak]

petition to declare Z/6Z favourite char 6 ring

[wraith cargo]

Aye

[formal ermine]

nay

[south patrol]

F_6 is my favourite char 6 field

[slim kayak]

aaaaaa

My field is cooler than yours

F_1

[slim kayak]

so hey

counterexample is Z/6Z which has char 6 i.e. 1+1+1 is not 0 but instead 3

[daring hinge]

Alright I'll check it

[slim kayak]

just gotta have to check if cubing a number is equal to itself reduced mod 3.

Which it is, so f is just the identity map

[daring hinge]

Yeah I was thinking about that

You mean mod 6?

[slim kayak]

Yes mb

[daring hinge]

Alright thanks

[white oxide]

why does x^6 - 2 not split in the given field? i thought the given field also included -2^1/6?

and those are all the roots of x^6 - 2 right

unless im trippin

oh it's some shit involving the primitive nth roots of unity

[dim widget]

Good to keep in mind that there couldn’t be only 2 roots because it’s degree 6

And irreducible

[white oxide]

wdym

i thought a polynomial of degree n has at most n roots

or am i trippin again

[dim widget]

Over the splitting field it has exactly n roots

[delicate orchid]

Not over it’s splitting field

[white oxide]

ah right

gg

thx

[dusty verge]

Yeah, the roots form a circle around the origin in the complex plane, meaning some of them are negatives of each other, i.e. integer multiples of the other, edit I misread your issue, ignore this

[white oxide]

ah ok

wait holup

[dusty verge]

Basically your issue was in the opposite direction right?

[white oxide]

why are they all zeros if they're not equal to 1

for k = 1.... n - 1

[south patrol]

A degree n polynomial has <= n roots

They should add in 1 as well lol

but ig by complex they just mean C \ R lol

[white oxide]

wait but like w^k \neq 1 and so w^k - 1 \neq 0

confused

for 1 <= k < n

[south patrol]

But it's x^n - 1 that it's a root of

[dusty verge]

Think of a complex number of module one as a thing that rotates. The roots are the things that rotate n times, and get to one, but don't get to one rotating less than n times. Another way of looking at is as generators of z/nz^*

[south patrol]

not x-1

[delicate orchid]

These notes are very worrying lol

[white oxide]

ahhhh shit my fault

hahaha

[south patrol]

Why are they using degrees...

[delicate orchid]

Both degrees and not using e^ix

[south patrol]

Tbf Wew we did this before Euler for some reason lol

Idk why, seems a bit pointless delaying Euler's formulae but ye

[dusty verge]

Degrees are king because I learned them in highschool and I refuse to adapt

[delicate orchid]

You did not do splitting fields in A level

[south patrol]

What lol

[delicate orchid]

We learnt euler’s in year 12/13

[south patrol]

yh

lol

[dusty verge]

It's why miles > meters

[south patrol]

Bleak

[delicate orchid]

???

[delicate orchid]

Also ???

“Oh yeah mate I’m 0.006miles tall”

[south patrol]

I was just confused why you brought up splitting fields lol

[delicate orchid]

The question is about the splitting field of x^6-2

[south patrol]

oh i didn't see that bit above lol

I was talking about the roots of unity ting

Okay

[dusty verge]

Jesus Christ you're 31 feet tall?

Make sure you don't knock down any buildings on the way home

[delicate orchid]

I am a nephilim this is known

[dusty verge]

Is your dating pool restricted to the statue of liberty?

[delicate orchid]

Jealous

[white oxide]

boys gentlemen, is Q(2^1/6) not a splitting field for x^6 - 2 because it doesn't include the complex roots included in the primitive 6th roots of unity?

[dusty verge]

Lady liberty's got back

[white oxide]

<@&286206848099549185>

[south patrol]

Why did you sully your own message

[delicate orchid]

Pinging helpers… lol

[dusty verge]

Because it don't include imaginary numbers

[white oxide]

idk what you're talking about

that was Wew

[south patrol]

lol and why gentlemen etc

anyway

[white oxide]

is that sane?

[south patrol]

That is yes a way to reason it

[delicate orchid]

Yeah stick an e^ipi/3 on that thang

Also it just doesn’t factor into linears over that field

[dusty verge]

No complex numbers in thy field, but there are complex sixth roots of two

Thy lmao

[south patrol]

Yeah like Q(2^1/6) is a subset of R, and x^6 - 2 only has 2 real roots

[delicate orchid]

Think theyve got it by now lads

[south patrol]

Lol

No I mean I was giving another way without appealing to the preexisting work lol

[dusty verge]

Yes

To sum up everything else here

[white oxide]

okay thank you lads!

[formal ermine]

@south patrol you should join the best thread out there

#1100503863586455632

[dusty verge]

What about halliday asks questions?

That's my least favorite thread because it makes me feel like I'm sectioned off :(

[near star]

confusion

oh

it switches them

I thought it was the exact same thing xd

nvm

ignore

[south patrol]

Ignore what 😎

[long nebula]

Ngl I did too when I saw what you posted

[jaunty glacier]

How would I show that, if a group has at least one reflexion, then this group is a dihedral group

[delicate orchid]

you don't, because that's not true

[jaunty glacier]

I have that written in my notes 😮

[delicate orchid]

There are many reflection groups which aren’t dihedral

[jaunty glacier]

Oh wait I didN't write it in the righ way

Let G be a finite symmetry group, then if G contains at least one reflexion, then this group is a dihedral group (there exists a n for which G=D_n)

Actually I think I found how to do it 🙂

[white granite]

poooooooooooooooooooooooooooooooooooo

[white oxide]

why do they have to show that the common divisor is of common degree? isn't x - a already of positive degree?

also by the last theorem how is this true

that the only divisor of positive degree of f(x) is f(x) itself

bc correct me if i'm wrong but for (x - 1)^2 in F[x], 1 is a multiple zero, but (x - 1) is a divisor of (x - 1)^2 which is not equal to (x - 1)^2

lol is that just not true

f(x) = (x - 2)^2(x + 1), f'(x) = 3x(x - 2) so they share a common factor of (x - 2), and x - 2 divides f(x) and is of positive degree?????

or rather what do they mean by up to associates

<@&286206848099549185>

[lapis trail]

If they differ by the product of a unit AKA if f(x)=ap(x) for constant a, then f and p are called associates

[white oxide]

ah i see thank you, so then a(x - 2) = (x - 2)^2(x + 1) for some a nonzero in F?

[lapis trail]

Hold up let me read and see if I understand your question

[white oxide]

than ku

[lapis trail]

Start the proof by assuming f is not reducible

[white oxide]

ahhhh i didn't see that

i think i got it now

[lapis trail]

Yeah they left that out

[white oxide]

got it got it

[lapis trail]

But they say it in the statement of the theorem

[white oxide]

right i kinda glazed over it

well yeah i got it now

tys

m

[lapis trail]

Np np

[white oxide]

what do they mean by minimal degree?

irreducible over R?

[lusty marlin]

There are no polynomials of degree 1 in Ker ϕ. Hence a polynomial of degree 2, like x²+1, is of minimum degree in Ker ϕ

[warm ember]

im dumb lmao can i get a hint for cayley's theorem

[solar shore]

do you remember cayley’s theorem

[warm ember]

umm

it wasnt in a box

the book never proved it

[solar shore]

welp now you will

[warm ember]

ye

[solar shore]

so what should g be mapped to?

what kind of element is it being mapped to* i should say

[warm ember]

o

idk

like

obviously 1->1

[carmine fossil]

Hint: See left multiplication as a group action

[warm ember]

the previous problem was this which i solved

[warm ember]

oh

thanks

ill try

[white oxide]

ohhh ok thank u

[warm ember]

wait whats a group action

like

define f(g_i)=gg_i?

[solar shore]

ye for some g in G

[warm ember]

and how f maps g_i to g_j

and stuff

idk

how indices move

[solar shore]

basically u wanna map group elements to functions right

[warm ember]

ye

[solar shore]

cuz its a subgroup of Sn

using his hint, left multiplication can be considered a function

[warm ember]

oh

[solar shore]

like $f_g(x) = gx$

[cloud walrus]

blanket

[warm ember]

isnt Sn permutations tho

wait

oh

i get it

[solar shore]

there u go

[warm ember]

thanks

[white oxide]

are we allowed to write (x - alpha)(x + alpha) because E is an extension of Q and therefore E[x] is an extension of Q[x]? so x^2 + 1 can be both viewed as a member of E[x] and Q[x], but in this particular instance we're viewing it as a member of E[x] because alpha is in E but not in Q?

and by identify do they mean "shifting the perspective" (F is isomorphic to a subfield of E, so we can view E as "containing" F)?

[warm ember]

does abstract alg use multivar

i only know single var calculus

[chilly ocean]

no it doesn't

[formal ermine]

that's analysis...

you only need the numbers from 1 to 10

maybe occasionally a big number like 57

[warm ember]

o

i was scared for a sec

[frigid lark]

Yeah to create fields

[warm ember]

cuz some metric space used integration

[past path]

Is anyone awake at this hour? I'm stuck showing that the following is a proper subgroup of the automorphism group.

I've proven everything but the word "proper" from part (c).

I first tried an (incorrect) argument by considering the algebraic closure of Z_p as the direct sum of infinitely many copies of Z_p.

And then I remembered the appropriate definition of F was not the direct sum and am puzzled.

That approach was one of "uncountably infinite number of subfields" until I realized there wasn't.

I tried appealing to ChatGPT for inspiration, but was not able to glean anything helpful from the conversation. It made some arguments that were woefully easy to find fault with.

[dim widget]

The earth is round there is always someone awake "at this hour"

[past path]

Fair enough. I don't know the scope of this server.

[dim widget]

I assume in this notation Z_p = Z/(p)?

[past path]

Do you have any idea how to approach showing that H is a proper subgroup of the Galois group? And yes, Z/pZ

I've proven everything but "proper".

[dim widget]

Yes, that's the hardest bit.

[past path]

Aye.

[dim widget]

So how are you describing the automorphisms of \bar{F}_p/F_p right?

[past path]

Z_p-automorphisms of F.

The Galois group of F over Z_p

[dim widget]

You are describing how the automorphism acts on F_{p^n}/F_p for every n (is what I was saying)

[past path]

Automorphisms of F that fix Z_p.

Ah, gotcha. Yes.

So are you saying

we could take one element a and map it to a^p

and another element b mapped to b^{p^2}?

I don't know how much is determined by the map.

[dim widget]

So one option is to just write a single integer m and be like "on F_{p^n} we map a \to a^{p^m} for all n".

[past path]

Since the structure of the algebraic closure is a bit puzzling to me. I get that it's the splitting field of...everything.

I don't know if that makes sense.

Couldn't we just compose phi with itself that many times?

[dim widget]

Yeah you shouldn't need to know anything specific about the structure of the algebraic closure as a field. All you need to know

[past path]

Or am I misunderstanding your point?

[dim widget]

Is that $\bar{F}p = \cup_n F{p^n}$ and $F_{p^n} \cap F_{p^m} = F_{p^{(n, m)}}$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

What is \phi? \phi is just Frobenius?

a \to a^p?

[past path]

I suppose that makes sense. Yeah, I think so. We haven't used that word I don't think but it's that map.

[dim widget]

Anyway so the thing to notice is that raising a \to $a^{p^{m + kn}}$ and $a \to a^{p^m}$ on $F_{p^n}$ are the same automorphism.

[cloud walrus]

Topos_Theory_E-Girl

Anyway so the thing to notice is that raising a \to $a^{p^{m + kn}}$ and $a \to a^{p^m}$ on $F_{p^n}$ are the same automorphism.
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.57 ... the thing to notice is that raising a \to
                                                   $a^{p^{m + kn}}$ and $a \...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

LaTeX Font Info:    Calculating math sizes for size <14> on input line 57.
LaTeX Font Info:    Trying to load font information for U+msa on input line 57.```

[slim kayak]

How can the subgroup be proper if it's fixed field is Z_p, by the galois correspondence E^H is contained in E^G, so G is contained in H?

[dim widget]

That's not how infinite galois theory works.

[past path]

unfortunately, yeah.

We'd just need to show the existence of an automorphism of $F$ that is not a power of $\varphi$.

[cloud walrus]

Amizar

[dim widget]

Yes, so what I'm suggesting above is that we make it different powers of p on each of the finite subextensions F_{p^n}, subject to some compatibility conditions on the intersections.

[past path]

But wouldn't $a\mapsto a^{p^n}$ be in $H$ for every $n$?

[cloud walrus]

Amizar

[past path]

How would we be able to define such different powers of $p$ while preserving the nature of the automorphism?

[cloud walrus]

Amizar

[dim widget]

So for instance on F_{p^2} we could make it $\phi$, and on $F_{p^3}$ we could make it $\phi^2$, these are both automorphisms to to stitch them together we just need to make sure that what power of $\phi$ that we describe on $F_{p^6}$ is compatible with the inclusion of these two fields (for which we can use the chinese remainder theorem). If we keep doing this for $F_{p^q}$ for infinitely many primes $q$ we get an automorphism which is not a power of $\phi$, because it gives a different power of $\phi$ on each $F_{p^q}$

[cloud walrus]

Topos_Theory_E-Girl
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dim widget]

I could tell you exactly what all of the automorphisms are but I think you'll understand the structure better if you come up with it yourself

[past path]

So, would it be possible to just change the exponent on a single piece?

Like, $\psi = \varphi$ everywhere but...

[cloud walrus]

Amizar

[past path]

Or would we need to have an infinitely-constructed automorphism?

[slim kayak]

What happens if you apply the it to a product that isn't in the "but" set but one element is

[past path]

Oh wait

I think I get what you're saying.

[dim widget]

If you don't use infinitely many disjoint finite fields to construct it there will always be some power of $\phi$ which agrees with it by the chinese remainder theorem.

[cloud walrus]

Topos_Theory_E-Girl

[past path]

Ohhhhh, I think I see what you mean. Wait, are you sure?

Wouldn't it have to be an infinite power of phi?

in order to be 1 mod every prime number?

except for one

[dim widget]

It won't even be "a power of \phi" that's the whole point!

[slim kayak]

Let phi(a)=a^n and phi(b)=psi(b)=b^m, with phi(ab) unequal psi(ab)

[past path]

That's what I'm saying

The Chinese remainder theorem thing you're talking about

Like, what's the problem with $\psi=\varphi$ everywhere but $F_{p^3}$, on which $\psi=\varphi^2$?

[cloud walrus]

Amizar

[past path]

Is it that it's not an automorphism, or that it's in $\langle\varphi\rangle$?

[cloud walrus]

Amizar

[dim widget]

That would work! But you need to be careful when you say "everywhere"

for instance on $F_{p^{2n}}$ for $n$ arbitrary it cannot be $\phi$, since it has to agree with what you've described on $F_{p^2}$

[cloud walrus]

Topos_Theory_E-Girl

[past path]

Wait, do you mean $F_{p^{3n}}$?

[cloud walrus]

Amizar

[past path]

Because that's the one that's been singled out.

But are you saying that on $F_{p^{3n}}$ it needs to be...surely not equal to $\varphi^2$... That would have some interesting consequences for factors of $n$. How would we combine them?

[cloud walrus]

Amizar

[dim widget]

Ah yes, I mean on $F_{p^{3n}}$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

You combine them using the chinese remainder theorem.

[past path]

How could we define it on $F_{p^{3n}}$? It's been so long since I've used CRT that I honestly am not sure I remember how it works.

[cloud walrus]

Amizar

[past path]

I know it has something to do with GCDs but beyond that I'm not sure.

Which is kind of ironic, as a number theorist.

[dim widget]

So $F_{p^{3n}} = F_{p^{3^d}} \cdot F_{p^{m}}$ where d is such that $3^{d-1}$ exactly divides $n$ and $m$ is coprime to 3.

[cloud walrus]

Topos_Theory_E-Girl

[past path]

Is the dot meant to be a straight up product?

Does that behave the way direct sums used to behave?

[dim widget]

It's the field generated by all sums of products of elements in either field.

Kind of like the ideal IJ for I, J two ideals in a ring.

[past path]

Ok, that makes sense.

Wait

no

wouldn't that be I+J

IJ is strictly contained in both I and J.

Maybe like $H\lor K$ for groups

[cloud walrus]

Amizar

[dim widget]

But the point is that if you describe an automorphism on $F_{p^{3^d}}$ for all $d$ and then on $F_{p^{m}}$ for $m$ coprime to $3$ you describe it on all extensions in an essentially free way.

[cloud walrus]

Topos_Theory_E-Girl

[past path]

Ohh okay.

[dim widget]

No

I'm talking about the ideal of all elements $\sum_k i_k j_k$ with $i_k \in I, j_k \in J$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

Anyway it's just an analogy.

[past path]

Maybe we learned ideals differently. Hungerford's definition has $IJ$ strictly contained in both. $I+J$ is our name for what you're describing.

[cloud walrus]

Amizar

[dim widget]

that's definitely not the case

The ideal I described is pretty clearly contained in the intersection of I and J.

[past path]

Wait, you said sums of products.

My bad

Yes, sorry.

I forget that products make ideals smaller and fields larger.

[dim widget]

Yes, because fields have identities in them!

[past path]

Aye.

Anyway, would it suffice to say $\psi=\varphi^{2d}$ on $F_{p^{3^d}}$ and $\varphi$ on $m$ coprime to 3?

[cloud walrus]

Amizar

[past path]

Or would it need to be $\varphi^{2^d}$?

[cloud walrus]

Amizar

[past path]

Or wait, could we just define it as $\varphi^2$?

[cloud walrus]

Amizar

[past path]

That last option definitely feels the easiest if it's safe to do so.

It's an automorphism on the whole thing so it should still extend to the subfields

[dim widget]

It should just be $\phi^2$, the other examples are not compatible with the inclusion $F_{p^3} \to F_{p^{3^d}}$

[cloud walrus]

Topos_Theory_E-Girl

[past path]

Right, that makes sense.

And we can extend the isomorphism for free because the fields are disjoint and we define them on the pieces?

[dim widget]

Yep

It's a nice exercise to show that if $K, L$ are disjoint galois extensions then the Galois group of $K\cdot L$ is a product of the two galois groups

[cloud walrus]

Topos_Theory_E-Girl

[past path]

I don't think we've covered that yet, but it seems reasonable.

You just...extend twice, right?

order notwithstanding?

[dim widget]

Yeah I mean in one direction you just restrict the automorphisms to the subextensions, this is clearly injective. If K, L are finite then all you have to do is show that K \cdot L is galois and you're done.

[past path]

normal+separable, right?

Or is it more complicated for the infinite case?

[dim widget]

Let's just do K, L finite since that's enough for us.

And we can use that an injection between finite groups of the same size is an isomorphism.

[past path]

So long as it's still a homomorphism.

[dim widget]

Well restriction to a Galois subextension is always a homomorphism

[past path]

I suppose that's fair.

[dim widget]

Anyway once you show all of this the only thing left is to show that there is no integer which is congruent to $2$ mod $3^d$ for all $d$ but congruent to 1 mod another prime.

[cloud walrus]

Topos_Theory_E-Girl

[past path]

I mean, I doubt it takes a ton of convincing to say that there are no natural numbers congruent to 2 mod every power of 3 and congruent to 1 mod every other prime.

The only integer congruent to 1 mod every prime but 3 is... 1.

And that one's congruent to 1 mod 3.

[dim widget]

Yep. I think your proof is fine, you might want to give more justification about why that is an automorphism depending on what level you're at/how strict your grader is.

[past path]

It's a 2nd-semester grad course, and she's not terribly strict but she's pretty thorough.

[dim widget]

Yeah it's up to you.

[past path]

But I've already passed the prelim in the subject and have basically full marks for both semesters so I'm not really that worried about grades.

Most of this is just for the sake of making sure I know what I'm doing.

Also, the underlying reality that grades are irrelevant calls to me.

[dim widget]

Anyway similar arguments show you that $Gal(\bar{F}p/F_p)$ is the set of families $(m_n){n \in \mathbb{N}}$ of integers such that when $d|n$ $d|(m_n - m_d)$ i.e. it is the inverse limit of the family of groups $Z/nZ$ with the obvious morphisms $Z/nZ \to Z/dZ$ when $d|n$.

[cloud walrus]

Topos_Theory_E-Girl

[past path]

I've been seeing that phrase a lot lately

What does inverse limit mean?

[dim widget]

Since you are a budding number theorist it is extremely important for you to know that this is $\prod_{p \text{ a prime}} Z_p$ where $Z_p$ is the p-adic integers.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

(although the topology is also important for infinite galois theory, and I haven't described it for you).

[past path]

The product of...all of them?

[dim widget]

An inverse limit of groups, rings, etc is something which you can associate to a system of maps of objects $R_i$ indexed by some partially ordered index set $I$ such that for any $i_1, i_2$ there is some $i_{1, 2} \in I$ such that $i_{1, 2} \geq i_1, i_2$. Given such a collections of groups/rings/etc. $R_i$ with maps $f_{i, j}: R_i \to R_j$ when $j \leq i$ one can form the collection of products $(r_i){i \in I}$ such that $f{i, j}(r_i) = r_j$ for all $j \leq i$.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

That can easily be verified to also be a ring, group, whatever.

Common examples of I are the natural numbers partially ordered by divisibility (which we're using here) and the natural numbers with the usual ordering.

[past path]

Perhaps this will make more sense to me when I'm not on the tail end of 10 hours of math at 6 in the morning.

Thanks for helping me understand the automorphism from earlier, though!

[dim widget]

Sure, feel free to read it later.

[past path]

I'll look at it when I'm more well-rested.

Meanwhile, I have a class to teach in a few hours 😂

[dim widget]

But it is really important for number theory (and math in general!) to eventually know what inverse limits of algebraic objects look like 🙂

Good luck with your class! Hope you do well. And have fun teaching.

[past path]

Sounds good! Thanks again

[formal ermine]

profinite completion of the integers my beloved

[frigid lark]

The one thing Lang mentioned in chapter 1 and ignored for at least 5 chapters

[dim widget]

?

What question?

[south patrol]

That isnt a question

[formal ermine]

beat me to it

[south patrol]

Lol

[dim widget]

I don't know what you mean, the thing you are referring to (as people are pointing out) is a definition. If you have a specific question about it you can ask!

[novel parrot]

how can i do part b?

so we know that the kerf^n = kerf^n + 1 for large enough n

[wraith cargo]

Ok so what does it mean for a ring to be noetherian

[novel parrot]

chains of ideals become stationairy

[novel parrot]

this

[wraith cargo]

Ye perfect

[novel parrot]

i want to show that f is injective

f(a) = f(b) => a = b

[wraith cargo]

Ye
And what does that imply

[novel parrot]

kernel is 0

[wraith cargo]

Cool that's what we're aiming for

[novel parrot]

so taking a in Kerf

f(a) = 0

f^n(a) = 0

probably need to use surjectivity between f^n+1 and f^n ?

but i dont really know how its done

[chilly ocean]

assuming f(a) = 0 and ker(f^n) = ker(f^{n+1}), try writing a = f^n(b)

i've basically solved it for you by writing it out like this

[novel parrot]

so b is in kerf^n too

so a = 0

cool

does smith normal form always exist over a noetherian ring?

or need PID ?

my instinct says pid

[south patrol]

Apparently Noetherian + SNF always exists over the ring <=> PID

Actually yeah so you can kinda use the structure theorem for this lol

Like in reverse sort of

[delicate orchid]

yeah, SNF gives you the decomposition of the ring that appears in the structure theorem for PIDs

therefore, if it's Noetherian and you can do SNF, you get something isomorphic to a PID

[south patrol]

owo

[ebon pine]

Can all finite groups be generated by a generating set of 2 elements?

[formal ermine]

no

[ebon pine]

I think this is wrong but can't think of any counterexample. Dn is clearly generated by 2 elements

[formal ermine]

take like

a 3d cube

and its discrete rotations

axes

yk what I mean

[ebon pine]

Yes

What group does it form?

[next obsidian]

Take (Z/2Z)^3

there's only 8 elements, if you really want you can enumerate every 2-element set and see it can't span, or you can just see that there's 3 components and why this means it can't be 2-generated

[agile burrow]

interestingly, I believe it's true that every finite simple group is generated by 2 elements

which, as you might expect, depends on classification

[next obsidian]

no

take the walter-chmonkey group

[delicate orchid]

I swear I had a convo about this a week or so ago

[agile burrow]

they found the 27th sporadic group

or is it the 28th? i can't keep count

[delicate orchid]

yeah it's the tits group

[ebon pine]

Thanks

[ebon pine]

I see, maybe this was being asked. Is it something that an undergrad is supposed to know?

[next obsidian]

Absolutely not

[delicate orchid]

no

and definitely not prove my god no

[next obsidian]

You just apply the classification 4head

[delicate orchid]

yeah I'm aware

now prove the classification

[summer path]

Isn't the proof for classification of finite simple groups like tens of thousands of pages long scattered all over the place?

[delicate orchid]

yes

[summer path]

[delicate orchid]

the finite simple groups are: A_n, C_p, GL_3(2) and that's it

[summer path]

Reminds me of a homework problem I had in rep theory class where it was trivial by feit Thompson but obviously wasn't intended to be done that way

I forgot the original question though

[next obsidian]

The problem: show that every odd ordered group is solvable

[dim widget]

Was it showing that a 2-dimensional representation of a finite nonabelian simple group is trivial?

I remember someone asking in a rep theory exam if they could use Feit-Thompson for this question and not understanding at all what they were thinking.

[dim widget]

Was this not advanced enough for #advanced-number-theory (Edit: apparently it was advanced enough)

[summer path]

I think something along those lines

[void knot]

Given $A$ a square matrix with $\lambda_i$ eigen-values, show that [\sum_i |\lambda_i|^2\le \tr (A^*A).]
I forgot how to show that 😦

[cloud walrus]

RaD0N

[formal ermine]

#linear-algebra

[void knot]

this is related to singular values. I don't think it is "early university" 😄

[formal ermine]

you can pretty much ignore the category names

there are occasional advanced linear algebra questions in that channel

[agile burrow]

I'm guessing Schur decomposition or something would help

[void knot]

Maybe... let me check that out

[rapid junco]

Orthonormal eigenbasis

Is my guess

[void knot]

A is not necessarily normal

[rapid junco]

Yeah its self adjoint

[void knot]

@agile burrow yeah, I guess the Schur decomposition is the way I did it some time ago

[rapid junco]

Therefore its simile#ar to a diagonal matrix with respect to an orthonormal basis

Oh nvm its not self adjoint

[agile burrow]

matrix decompositions are awesome

[rapid junco]

True

[void knot]

A^*A is

[rapid junco]

So then take A*A and it will be similar to A diagonal matrix an orthonormal basis

And I may be wrong but im pretty sure this diagonal matrix has the eigenvalues across the diagonal

Then theres a property for a set of orthonomal vectors that the norm of any vector with those vectors as their basis is the scalars summed

Which looks a whole lot like your left side

[void knot]

take $Q$ unitary such that $A=Q^*TQ$, where $T$ is triangular. Then
[\tr(A^*A)=\tr(T^*T)=\sum_{i,j}|t_{ij}|^2\ge \sum_i |t_{ii}|^2=\sum_i |\lambda_i|^2.]

[cloud walrus]

RaD0N

[void knot]

@agile burrow actually working out 😄
thx!

[agile burrow]

happy to help

[void knot]

@agile burrow u don't see the help that u gave xd
I spent 2h on this crap that I did years before

[agile burrow]

Well a good review never hurt anyone

Unless they were like allergic to reviewing or something, idk

[formal ermine]

is there any easy way of counting conjugacy classes of a finite group other than counting normal subgroups

[agile burrow]

Certainly not in general, I don't think. There are probably various orbit-stabilizer tricks you could come up with for specific cases though.

[coral spindle]

Just compute the character table

Unironically though, sometimes this can be quite straightforward

If you have a nice enough normal subgroup, you can use Clifford theory to count the number of irreducible characters without necessarily actually writing them down

Apart from that, as walter says, finding conjugacy classes is very hard. In fact the conjugacy problem is undecidable in general.

[delicate orchid]

a word problem is undecidable? imagine my shock

[craggy lichen]

If G is a group of order 385 show that its 7-Sylow subgroup is in the center of G.

i understand everything up till the last line

can someone explain the last part to me

[agile burrow]

The natural map is the map induced by conjugation, i.e. the map taking an element g to the automorphism of G_7 given by conjugation by g. Then it's clear that G_7 is in the center iff this map acts trivially. The image of a group homomorphism is a subgroup, so by isomorphism theorems its order divides the order of both the domain and the codomain. Since the domain and codomain have coprime orders, the image must be trivial.

[solar shore]

so i remember someone asking about cayley's theorem, and it got me thinking when my professor was just mentioning how important it was

is there a similar theorem/idea for rings?

can rings also be isomorphic to a group of permutations? or does that question even make sense

[agile burrow]

There is an analogue of Cayley's theorem for rings. The idea is that a ring acts on its underlying abelian group by left multiplication and this is a group homomorphism. In particular, R is a subring of the endomorphism ring of its underlying additive group

[agile burrow]

I'm not really sure how to interpret this question, I don't think it really makes sense as stated

[solar shore]

this was what i was looking for, my bad

i was just thinking if there was something similar to cayley's theorem for rings

[craggy lichen]

in part c, why is G_q characteristic in N?

am i missing something obvious

[lethal dune]

see that G_q∩ G_r is {e}

so the product is a direct product and hence unique subgroup of order q

@craggy lichen

[static needle]

Anyone here know any groups isomorphic to Z^n aside the complex version?

[tribal moss]

What is "the complex version"?

[low void]

Why is f+(f)=0 where f is the ideal generated by the polynomial f?

[formal ermine]

f = 0 mod f

[low void]

How does this follow from the definition of ideals?

(f)=K[X]*f

[formal ermine]

gH = H iff g in H

[low void]

Where's the mod comin from?

[formal ermine]

f + (f) is just a permutation of (f)