groups-rings-fields

[pastel cliff]

ok related but then if J isnt a domain then will R/J not be a domain either?

i feel like thats not necessarily true

ohhh wait

wait no

[next obsidian]

An ideal a domain

[coral spindle]

I mean, I guess it's typically a non-unital ring lmao

[ripe basalt]

well my midterm blew

also Z_p is standard notation for the integer ring btw

[chilly ocean]

Yep

[ripe basalt]

i also hse it for groups

so the grader is not only a dick

but they are a wrong dick

[chilly ocean]

It's a bit like with groupoids

[next obsidian]

hurb

[ripe basalt]

imagine writing "the cyclic group Z_6" and losing a point because u mightve been talking about p aidcs XD

hilarious

[next obsidian]

the p-adics at the prime number 6

[ripe basalt]

yes, youll get to that level of advancement some day

[coral spindle]

Two marks removed bc 6 isn't prime(!)

[ripe basalt]

hahahaha

thats even better

[chilly ocean]

Three because Z_p aren't a cyclic group

[ripe basalt]

anyways i am sad that my midterm went horribly

thanks

imagine failing a midterm in grad school

i will soon have that experience

[pastel cliff]

what is missing here

(given prev. discussion i feel like something is missing - prof said we didnt have to be as rigorous with this but still)

[coral spindle]

Consider the ideal 2Z x 3Z of the ring Z x Z

[pastel cliff]

hrm

[coral spindle]

Oh, my mistake, I thought this was sth else

I like the reasoning here

So this does show that the prime ideals are of the form R x J, I x S, but it doesn't show that these are exactly the prime ideals

[pastel cliff]

hesitating bc i didnt use a quotient or the iso chmonkey mentioned before

ahhh

[coral spindle]

Meh, the method chmonkey mentioned is lovely but this is fine

[pastel cliff]

so "technically" there could be different prime ideals

get meh'ed chmonkey

[pastel cliff]

what am i missing to show that they're all of this form is what im meaning to ask tho

[coral spindle]

You have shown "If X is a prime ideal then it is of the form blah blah blah"

but you have not shown that everything of that form is a prime ideal

[pastel cliff]

ah okie. that's what i thought

[coral spindle]

(Technically, you have not argued why the P in P x S must be prime either)

(But that's easy so who care)

[gusty thistle]

So fk insnane

Wishing you good luck with the rest of it

[ripe basalt]

no that wasnt me who lost the point

although i did separately bomb a midterm just now

[pastel cliff]

wait this is easy

i think

one sec

bit verbose but i think that's enough

[next obsidian]

[pastel cliff]

i asked this before but i didnt get it: is there a commutative ring with identity which has a unique non-zero prime ideal

i originally thought a field would be an easy trivial example but i think walter pointed out some silly defn jumbo as to why it was wrong

[agile burrow]

Not silly

[pastel cliff]

hi walter

ok so someone had said "something something polynomial ring"

but (x) isnt the only ideal of R[x] is it

[next obsidian]

yes

Consider R[x]/(x^2)

[south patrol]

Good ring

[pastel cliff]

i was never gonna consider that on my own lol

why is it a good ring

outside the context of this q

[next obsidian]

Maps out of it with image a specific point are the Zariski tangent space of a scheme at that point

[pastel cliff]

i know some of those words

not many but some

[pastel cliff]

[south patrol]

nah i thought you said like unique maximal ideal lol

and was gonna say like this is an entire class of rings lol like local rings

Anyway R[x]/x^2 is interesting, the dual numbers (for R the real numbers ig)

you can think of it as R with this 'small' element x lol

[next obsidian]

More interesting when R = F_37^42

[pastel cliff]

just to be clear, R[x]/x^2 is the set of polynomials with terms in x^0 or x^1 and that's it right

[next obsidian]

Yeah

And you multiply in a kinda obvious way

It’s dim 2 over R

(Really I should have used k, you want a field)

[south patrol]

stfu

[pastel cliff]

https://tenor.com/view/hardcore-potato-angry-rage-pissed-off-gif-17086436

potato rn

[toxic zephyr]

is there a name for elements of a module other than "vector"? just started learning them in class and i'm still learning the terminology

[next obsidian]

I just call em elements

[coral spindle]

We typically don't call elements of a module vectors, unless you're working over an algebra

[toxic zephyr]

fair enough. i guess there isn't really a name for elements of a group/ring/field either. why not groupies tho lol

[toxic zephyr]

oh okay. my professor was probably just easing us into the concept then i guess.

[coral spindle]

Elements of a field do have a name; they're typically called scalars.

Though this depends on the author a fair bit

[toxic zephyr]

oh right. never made the connection. but my prof called the ring elements scalars for the module though. is there a different word for modules then?

[coral spindle]

I don't know what you mean by that

[toxic zephyr]

like he called it "scalar multiplication" in the module

[coral spindle]

Oh, like they act on the module

sure

That too

[toxic zephyr]

i also wanted to ask about this. he showed us an interesting module. taking a vector space V over a field F and a linear operator T:V->V. Then V is an F[x]-module where p(x)*v is defined to be p(T)v. would F[T]-module also be appropriate notation?

[thorn delta]

I haven't seen that notation before but i don't dislike it

[next obsidian]

Wait no I don’t like this

If anything it’s like a T-module

Anyway you’re starting to see one of the beginning ideas in representation theory

[toxic zephyr]

yeah? i'd be interested if you could tell be a lil bit if you don't mind.

[next obsidian]

A representation of a group G is a group homomorphism G -> End(V) for some vector space V over say, C

If V is dim n then this is the same as GL_n(C)

[coral spindle]

No, calling it an F[T] module is not correct

[next obsidian]

Conversely in a manner similar to what you saw here, such a map is equivalent to giving V the structure of a C[G]-module

Where C[G] is a group ring

And you can take any module structure of V over C[G] and turn that back into a representation G -> End(V)

So you can talk about representations as maps or as modules, and this is important because a lot of results are proved by looking at the structure of modules over C[G]

Things like Artin-Wedderburn and Maschke’s theorem are about such modules

But then you can translate that to get theorems about the structure of representations

[toxic zephyr]

is there a more compact way to define it or do you have to explicitly define the V, F, T, and specific scalar multiplication every time?

[thorn delta]

you don't need to define T

[coral spindle]

Yes.

Part of the data that defines a module is the action of the ring

If you know about G-sets, it is entirely similar

You can't just have a set on its own – you must define the way the group acts on it, which is part of the data of the 'G-set'

[thorn delta]

also, what's really wrong with this notation? it seems pretty harmless to me

[toxic zephyr]

this is interesting, i'll look into it. thank you very much for taking the time to explain it 🙂

[coral spindle]

It is incorrect; there is a significant difference between F[x] and F[T] where T is some predefined linear transformation of an F-vector space

[thorn delta]

also what i meant by this nix, is that if you have a F[t] module structure on V, then you can think of the ring-action as a map F[t] -> End(V). The image of "t" under the ring action is your linear map "T"

conversely, if you have a linear map "T", it defines a unique F[t] module structure such that t maps to T

[toxic zephyr]

what would F[T] denote?

[coral spindle]

I would read that as a certain subring of whatever GL(V) it is that T is an element of

[thorn delta]

GL(V)

[coral spindle]

If you don't know what it denotes why'd you write it tho

[coral spindle]

This is typical basis free-notation :)

Perhaps you'd like GL_n(k)?

[thorn delta]

yea, but T doesn't need to be invertible

[coral spindle]

Good point!

End(V) then

my mistake

[thorn delta]

correct

[coral spindle]

:)

Next time, I'd like a more direct correction btw

[thorn delta]

but ur feeding my defense of the F[T] notation

[coral spindle]

I really don't care terribly much

[toxic zephyr]

because it seemed like there was little point in writing p(x)*v=p(T)v when i could just consider all the scalars to be polynomials over F in the "variable" T

just cut to the chase of my scalars being some p(T) basically

[thorn delta]

to be very precise, (the only sensible way to interpret F[T] is) F[T] = F[t]/(f) where f is the minimal polynomial of T, i.e. the kernel of the ring action F[t] --> End(V). So if you have an F[t]-module, you can always get an F[T] module by
considering F[t]/(f) --> End(V) induced by quotienting out the kernel of the F[t]-module action

[uncut girder]

someone please fill this in

draw edges between field extrensions

[next obsidian]

Bruh

[uncut girder]

chmonkey do it

if you dont do it you dont know galois theory and youre a loser L

[next obsidian]

I have better things to do

Like learn about CRYSTALLINE COHOMOLOGY

[lunar ledge]

Waltuh

[agile burrow]

Aured

[uncut girder]

cringe

[noble saddle]

Holy shit dude, they did the hard part for you

[pastel cliff]

#❓how-to-get-help

pwned

[pastel cliff]

does the existence of a multiplicative identity in a ring imply that some elements have inverses

[delicate bloom]

the identity 😎

[pastel cliff]

oh wait Z

it has a multiplicative identity but only 1 and -1 have inverses

though i guess -1 is a non-id element

[next obsidian]

F_2

Only has one

[pastel cliff]

tf is that

[next obsidian]

Field with 2 element

Z/2Z

[pastel cliff]

ah

simple but im trying to show that these have identity, where e is idempotent

R has identitty

[next obsidian]

Yeah

It’s an identity within that ring

It won’t be 1

Cuz that isn’t in them

[pastel cliff]

but 1e will still be mult. id?

oh shit yeah

re1e = r1ee = re

[next obsidian]

Yeah

🙂

[pastel cliff]

identitty

[delicate bloom]

don't you also have to show ere=re too?

[pastel cliff]

e commutes with everything so we can make e1 -> 1e no?

then ee = e bc idempotent

oh

l o l

yeah you're right

is there anything we can say about the cartesian product of comaximal ideals

trying to prove that if $e$ is idempotent, $R \cong Re \times R(1-e)$

[cloud walrus]

sebbb

[pastel cliff]

actually Re probably isn't necessarily maximal

ok no hints yet brb

waiiiiiitttttt

i smell chinese remainder thm

[delicate bloom]

simply just ||extreme spoiler lmao||

[pastel cliff]

ok it's pretty easy to see that they're two sided ideals

they're comaximal

[next obsidian]

This is actually like

Even easier

You’re basically just writing it as an internal direct product

Which you probably saw for groups

So I’m pretty sure you can just emulate that

[pastel cliff]

bold of you to assume i retained it

ok but

[next obsidian]

I guess my point is

[pastel cliff]

Re and R(1-e) comaximal

[next obsidian]

Yeah

[pastel cliff]

their product is just 1

[next obsidian]

Yeah

Wait no?

The product is 0

[pastel cliff]

that's what i meant my b

[next obsidian]

Which is important you DEFINITELY don’t want the product to be 1 haha

Yeah so you can just CRT

But like

I think it’s worth thinking about this as an internal direct sum

Like your map sends x

To (ex, (1-e)x)

But you can interpret that isnide R

As ex + (1-e)x

And you can just see that multiplying x and y

Would correspond to multiplying ex + (1-e)x and ey + (1-e)y

[pastel cliff]

wait sorry does CRT work? what is R/Re

[next obsidian]

And the cross terms just die

[next obsidian]

It’s (1-e)R

[pastel cliff]

i knew that

i was just testing you chmonkey

[next obsidian]

Yeah I mean

I realized that

To show this

[uncut girder]

i also knew that

[next obsidian]

Requires writing it as a direct sum

Omegalol

[uncut girder]

because this was on my ag hw

[pastel cliff]

ok imma latex it brb

but imma do it my silly way chmonk

[next obsidian]

That’s okay too

But what I said is worth thinking about

[pastel cliff]

think i can get away with saying they are clearly ideals?

[next obsidian]

They’re defined to be

eR is shorthand for (e)

Or well

Idk if even shorthand lmfao

[pastel cliff]

probably missing something small but why is this true

ew

"Every prime ideal P in a Boolean ring R is maximal: the quotient ring R/P is an integral domain and also a Boolean ring, so it is isomorphic to the field F2, which shows the maximality of P. Since maximal ideals are always prime, prime ideals and maximal ideals coincide in Boolean rings."

R/P is an integral domain bc P is prime

but how does it get to the isomorphism to F2

[next obsidian]

Every element is an idempotent

Is there was one besides 1 or 0, then it’s a product of rings

But an integral domain can’t be a product of rings

[pastel cliff]

working on this

should i instead be looking to use CRT again?

that^ isn't done ofc

wait uhhhhh

the prime ideals of R will all be {0, r} for every r in R

they're all comaximal

(assuming comaximality can apply to more than two ideals of the same size)

then the product of all prime ideals is equal to the intersection of all prime ideals which is just 0

R \cong R/P x ... x R/P for all possible P

bit scuffed but i think this is good

[past path]

Why is the definition of proper subring distinct from the definition of proper subset and subgroup?

Is there a meaningful reason the identity group is a proper subgroup of every nontrivial group but the zero ring is never a proper subring?

[tough raven]

WDYM in the third-last sentence?

[past path]

Also, for a sufficiently small ring, is it possible to have the zero ideal be a maximal ideal?

[tough raven]

It doesn't contain 1?

[past path]

Even in rings without identity, if I'm not mistaken a proper subring is defined to be not the whole ring and not the zero ring.

[tough raven]

Yes, and “small” isn't quite the right word for it. (In a commutative ring,) this happens iff the ring is a field, and there are fields of arbitrarily large cardinality.

[tough raven]

Oh, I assumed ring meant with identity.

[past path]

I suppose that could also make sense.

I was thinking of a ring with 2 elements, but I guess that would make it a field.

[tough raven]

In a rng, that's … a weird convention IMO, which I disapprove of.
But people have their own conventions 🤷

[past path]

What do you mean weird convention?

The definition excluding the zero ring?

[tough raven]

It's upto convention whether you define proper subthing to allow trivial subthing

And I think not allowing it is weird.

A course I'm taking now does this though. 😫

Even for groups.

[past path]

I find it frustrating as well, though I believe it's standard convention.

[tough raven]

Is it?

[past path]

What's the verdict on redefining the word proper while I'm redefining notation?

[tough raven]

I would prefer “proper non-trivial” for that and I thought that's at least also common

[past path]

I'm more hesitant to change something that might have mathematical significance.

It's a headache to write all the permutations of notation that might be necessary.

Like, for example, is there a point in putting an equivalence underline under the symbol for "maximal subring"? Or since it's by definition not equal, should it be excluded entirely out of redundance?

[coral shale]

afaik the conventions on ring terminology are not as well unified as other stuff

[past path]

Which is why I've taken it upon myself to fix that problem.

But

[coral shale]

eg. ring might mean commutative has 1 in some contexts but not in others

[past path]

I don't want to be careless

[coral shale]

In fact, in some many defns, subrings have to have the same identity as the original

so the null ring would not be a subring in these cases

[coral shale]

this structure-preserving definition makes more sense to me

[past path]

I'm honestly okay with removing all sorts of structure.

[coral shale]

but this defn only applies to rings with 1

[past path]

Heck, I'd even be okay with subgroups of rings.

But current mathematical notation is far too context-sensitive to handle things like that.

I'm very pleased with the way things are shaping up so far, but there are so many permutations of things to include that I'm a bit overwhelmed with the task.

[coral shale]

its hard to unify a subject in areas of active research and many ppl writing on. we're not 'done' yet and one defn in one context may have advantage over another.

As long as one is clear which defn they are using from the start of whatever they want to say, then all is good.

[past path]

Right now I have independent, distinct, and easy-to-read notation for
subring
ideal
equal/not equal
proper
maximal
principal
prime
left & right

That can all be used simultaneously or individually on the same symbol to write entire sentences in a single inequality.

One of my friends and I have been working for a few days on an overleaf package for it that will hopefully make a lot of things a lot easier.

[toxic zephyr]

so I'm doing part a, trying to prove that if f and g are ring homomorphisms R to R, then (f+g)(r)=f(r)+g(r) is a ring homomorphism and I'm struggling with the multiplication property. it just doesn't seem like multiplication should work.
like
f(xy)+g(xy)≠(f(x)+g(x))(f(y)+g(y))

am i misunderstanding something?

it's almost 2am so i feel like any second I'm going to realize i just completely messed up the definition of what f+g is

or of what a ring homomorphism is lol

[little jasper]

Multiplication of homomorphisms is defined as composition here, so (f•g)(x) = f(g(x)), not f(x)g(x)

[toxic zephyr]

wait I'm confused. then am i not showing that
(f+g)(xy)=(f+g)(x)*(f+g)(y)?

in R

i thought that was necessary for it to be a ring homomorphism over R

[little jasper]

Oh shit yeah, this does seem to be what you have to prove

[toxic zephyr]

this wouldn't be the first time my prof has sent out a question with a mistake in it so it's possible there's something missing

[little jasper]

Well (f+g)(xy) = f(xy) + g(xy)
= f(x)f(y) + g(x)g(y)
≠ [ f(x) + g(x) ] [ f(y) + g(y) ]

So doesn't seem to make sense

@toxic zephyr

[toxic zephyr]

yeah that's so weird. I'm pretty sure fg does work though...

[little jasper]

Yeah, that part works

If we were talking about module homomorphisms here then the set of module homomorphisms from some R-module M to itself does form a ring under pointwise addition and composition

[toxic zephyr]

is there some sort of alternative definition for f+g that could make it work out? it would be really surprising if he wrote this whole question and there was no way to make it work.

[little jasper]

Idts, the module thing is the best I can come up with

[toxic zephyr]

oh no... this is very concerning. 😦

oh yeah and i guess the mapping to 1 property would also fail. because (f+g)(1)=1+1=2

maybe HOM(R,R) is actually just a non-abelian multiplicative group and not a ring?

[weary bison]

Can't follow proof of cycle decomposition

[toxic zephyr]

just sent my prof an email (scheduled for 8am because i don't want him to get an email from me at 2:30am lol)
if anyone can find a way to salvage this problem that would be really great 🙏
or alternatively prove that this question could never work would be great too lmao
imma sleep now

thank you for taking a look at this and helping me @little jasper ❤️

[little jasper]

It's a non-abeliean semigroup under composition

[little jasper]

No probs :)

[thorn delta]

@toxic zephyr the mistake here is that they meant Hom(R,R) to mean endomorphisms of the underlying abelian group of R

i.e. End(A) is a ring under these operations when we interpret the elements of End(A) as homomorphisms of abelian groups

its like earlier when we were discussing F[t]-modules. I was using the fact that the "scalar multiplication" on a module (aka ring action) can be thought of as a ring homomorphism F[t] --> End(V), where End(V) is a ring under the operations laid out in this exercise you're doing

[south patrol]

you're gonna have a normal subgroup of order 7 and a subgroup of order 3

so like semidirect product of Z/7 and Z/3, so you equivalently must like classify maps Z/3 -> Aut(Z/7) up to an automorphism ig

i.e. maps Z/3 -> Z/6 of groups

Okay so the map Z/3 -> Z/6 must send 1 to either 0,2 or 4 by order reasons

the maps sending to 2 and 4 are the same up to an automorphism though

[coral shale]

whats the justification all groups of order 21 are semidirect products of C3 and C7 or C21?

[south patrol]

Well C21 is a semidirect product of C3 and C7 tbf

[coral shale]

yh is this sylow?

[south patrol]

But yes the justification is that there is gonna be a subgroup of order 7 and a subgroup of order 3 (by Cauchy or even just counting), with the subgroup of order 7 being normal by index reasons

[coral shale]

or something

[south patrol]

And then clearly intersection trivial & product is the entire thing, so we have a semidirect product

[coral shale]

ic

[south patrol]

owo

I forget what the nicest proof is that a subgroup H of index p (where p is the smallest prime dividing the order of the group G) is normal

is it just to consider action of G on cosets of H, i.e. a map G -> S_p and then show that map has kernel exactly H by order reasons or is there an even more straightforward way lol

[agile burrow]

that's the proof i always see

[south patrol]

Noice

group actions proofs are adorable sometimes idk

A semidirect product of groups G and H is equivalent to an action of one group on the other (depending on which way your like weird symbol in teh semidirect product goes lmfao)

Yeah exactly what I'm doing

Okay cambridge is cringe smh /s

But yeah uhhh I mean I could send our notes on it for example if you'd like, idk they're useful

which course is this for?

oh fair

i've not really done any like explicitly group theoretic stuff in our rep theory course

ye

oh lol

Uhhh wdym by uniqueness here

Maybe I'll just give more of an explanation

[south patrol]

Yeah so it comes down to this

So I guess are you more interested why the maps like Z/3 -> Z/6 sending 1 to 2 and -2 give the same semidirect product up to iso? (here i'm identifying Z/6 with Aut(Z/7) btw lol)

[quiet pelican]

If you want a hint, ||Consider the group of functions ax + b, where a and b are integers modulo 7||

And ||a has an inverse||

[coral shale]

brute force?

[quiet pelican]

Brute force is probably the best, use that there is a normal subgroup of order 7, and then consider the conjugation action of the order 3 element on that, and you will see there’s only 2 non isomorphic ways it can work

And one will just be C 21

[chilly radish]

Do naturally isomorphic functors have naturally isomorphic derived functors?

This seems like it should be obviously true but somehow i'm not sure how i'd prove it

wait nvm this is trivial I think

[agile burrow]

isn't it the case that if F and F' are naturally isomorphic then the induced functors on chain complexes are also naturally isomorphic

[chilly radish]

yes

and then homology is natural

or cohomology

[agile burrow]

right

[chilly radish]

so yea this is trivial

[chilly radish]

I mean this is just a consequence of the definition of natural transformations

that it's a chain map

[agile burrow]

yeah makes sense

i just drew the square lol

[chilly radish]

I think proving the defn of the natural transformation not depending on the choice of resolution might be a bit annoying

maybe not actually

wait why am I trying so hard

I think this just falls out of universality

you do the hexagon diagram chase thing

I gotta look at that proof again

[quiet pelican]

No

It’s just not on today

[south patrol]

Lol nice

[quiet pelican]

Also, you should really get Facebook

[south patrol]

Do many mathmos climb

One of my mathmo friends does it too there lol

[quiet pelican]

Many climbers are mathmos, but not many mathmos are climbers

[south patrol]

The first thing was more what I meant aha

[quiet pelican]

Then get on the group and you’ll see when it’s on

🤦

[south patrol]

In fact it has a natural ring structure

nice

i thought it would be long walks or smth given how much pacing around the room some people like me do when we get stuck on a problem

[quiet pelican]

You can do both

[south patrol]

dope

mathematicians be like "i got 99 problems and you will be one" every week

lol

responded to me and pinged someone else nice uh

give me tons of problems

jk i am busy

[quiet pelican]

Please don’t torture me like that

Unless you’re giving me the sheet 3 problems I haven’t done.
That’s fine

[south patrol]

not a well-defined question because x is in four different locations in that line

[warm urchin]

So going back to this here is my attempt. \ let G be a group of order $\kappa$ and let $F_{m}$ and $F_{n}$ be the free groups generated by M and N st $|M|=m$ and $|N|=n$ \ the number of functions from M to G and from N to G is $\kappa^{m}$ and $\kappa^{n}$ respectively \
by the universal property of the free groups on a group G for each function between M and G we have a unique homomorphism $
\varphi : F_{m} \to G $ we conclude there is $\kappa^{m}$ homomorphisms between $F_{m}$ and G and $\kappa^{n}$ homomorphisms between $F_{n}$ and G \

now this is a intuitive guess but if $F_{m} \cong F_{n}$ then the number of homomorphisms from $F_{m}$ to G is equal to number of homomorphisms from $F_{n}$ to G ? and if its true any leads on how to prove that? ( and if im approaching this the wrong way lmn)

[cloud walrus]

Susilian

[wooden ember]

Am i bein stupid right now? Im struggling to get a more common isomorphism class for a module of the form M oplus N /(1,a) given that i have a nice form for N/(a). It seems like this should be easy to do but im honestly not sure how to represent things

(In my particular case im looking for the isomorphism class of $$\frac{\frac{k[x,y]}{(x,y)}\oplus \frac{k[x,y]}{(y)}}{(1,x)}$$)

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

this is in the context of computing a cokernel for the equivalence between yoneda extensions up to yoneda equivalence and the first Ext module if that helps

[chilly ocean]

I almost never do

[south patrol]

You can take G to be a finite group for ease here to conclude and for the last bit this is because we can use the isomorphism between groups to go between maps Fm -> G and Fn -> G

[south patrol]

Fair lol

[warm urchin]

oh i see i think i got this now , thank you!

[chilly ocean]

I'm just, typically, a slow careful thinker. When I don't give out ad hoc arguments at least

I don't get emotional. It comes with some work, or the idea just arrives in my head, if I decide to think about it

I'm pretty good at catching errors, because I want to not make them

[hazy portal]

Let $G$ be a group with $k$ elements. Suppose ${\gcd(k,n)=1}$. If $g\in G$ and $g^n=\varepsilon$, then $g=\varepsilon$.

[cloud walrus]

whereismydaddy

[hazy portal]

this is my proof

Assume $g\in G$ and $g^n=\epsilon$. We know ${\gcd(k,n)=1}$, then there exist $x,y\in \Z$ such that $kx+ny=1$. We get $g=g^1=g^{kx+ny}=g^{kx}g^{ny}=(g^k)^x(g^n)^y=\epsilon^x\epsilon^y=\epsilon$. Thus $g=\varepsilon$.

[cloud walrus]

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

[hazy portal]

i feel like im missing something

should i add why g^k = e?

[delicate bloom]

Lagrange's theorem - the order of every element of a finite group divides the order of the group

[ripe basalt]

is he allowed to use lagrange?

[chilly ocean]

Lagrange is easy enough to prove it

Simply consider action of G on G/H, from which the formula |G/H|*|H| = |G| readily follows

[foggy merlin]

Normally you do lagrange before looking at group actions lol

[chilly ocean]

I suppose. But you don't use the fact that it's a group action in the above argument anyways

You just observe that |xH| has the same cardinality as |H|

[foggy merlin]

Yeah. But whenever someone mentions group actions + cardinality argument I always ghink of orbit/stabilizier formulas

[ripe basalt]

yeah i mean look at how simple his proof is

you can't just assume group actions

but i guess you can show it directly pretty much anyway

[coral spindle]

This isn't relevant to #groups-rings-fields, sorry. In fact this is very off-topic. Perhaps you're looking for #computing-software.

[limber agate]

Oh my apologies, I will move it there, thanks!

[smoky cypress]

Let $f(n)$ be the number of groups of order $n$, $g(n,k)$ be the number of groups of order $n$ with $k$ conjugacy classes, what can we say about $g(n,k)/f(n)$ as $n\to\infty$

[cloud walrus]

CS person

[agile burrow]

One thing is that there are only finitely many finite groups with a given number of conjugacy classes. I think this follows from a result of Landau?

But I'm not sure what computations have been done on counting how many such groups exist

[delicate bloom]

restricting the problem to abelian groups might be more tractable, since the number of abelian groups of a given order is a multiplicative function so you'd have some dirichlet generating function for that and try to throw some ANT tricks at it. Then maybe try to use that to put bounds/asymptotics/etc to get the original case. Dunno, interesting question though

[smoky cypress]

Well for abelian groups the number of conjugacy classes is just always the order of the group

[smoky cypress]

Wait I can’t find Landau’s theorem

[delicate bloom]

oh right 😩

[agile burrow]

Let me check again, iirc it's a number theoretic result

[smoky cypress]

But on this end I am curious about the limit of number of groups of order n/n maybe

Or like the growth rate of the number of groups

[delicate bloom]

I'd play more but I have to work on my resume, keep me updated tho

[agile burrow]

Oh found it. There are finitely many solutions to $\sum_{i=1}^k \frac{1}{n_i} = 1$. Applying this to the class equation gives the result

[cloud walrus]

walter

[agile burrow]

But I agree, this sounds like a neat problem and I'll think about it more

[smoky cypress]

Oh huh I see

Maybe let k=n-i for some constant i?

Or take k in a range

[agile burrow]

Yeah, I'm sure there are variations that could give you interesting answers when you look at the limit. It just sounds like a difficult problem lol

[smoky cypress]

Yeah true

[past path]

I'm a bit stuck on an algebra problem involving ideals of internal direct products.

The symbol you don't recognize means "is an ideal of"

My approach was going to be something like this:\
If $I$ is an ideal of $R_1\times\cdots\times R_n=:R$, then $I=A_1\times\cdots\times A_n$, where $A_k=\pi_k(I)$, the canonical projection of $I$ onto $R_k\cap I$.

[cloud walrus]

Amizar

[past path]

1. Is it reasonable to simply claim I remains the direct product of each A_i, or is that something I need to justify more thoroughly?
2. How might I go about showing that each A_i is an ideal of R_i? If it weren't for part (b) I would think the problem were more straightforward, but I'm a bit confused at how we might be able to do this for rings without identities.

I think I misspoke. What I mean is, I thought my proof outline should work for any ring, and the fact that it doesn't necessarily hold for rings without identities makes me question my initial argument.

[chilly radish]

When your abelian category is concrete (e.g. R-Mod), to construct a contraction (A chain homotopy of the identity to 0, so 1=ds+sd), I really only need s to be set maps right?

I need the 'prism' maps or whatever you wanna call them to be set maps I mean

I proved this I just wanna make sure i'm not being silly

[chilly radish]

Also, here, rotman claims that P_n are free Z[G]-Modules. I'm not sure I see why it satisfies the universal property, or rather what the basis should be

[chilly radish]

I don't see how these can be free G-Modules, since the basis elements have relations between them

which can't happen in free modules

so how can he conclude this is a G-free resolution

says it here, p.529

of homological algebra

ok wait it probably is a free module just the free abelian group basis is not the Z[G] basis

so what is

[runic hemlock]

A basis is the tuples with x_0 being the identity element

It's not hard to show they have no G-relations

[wooden ember]

oh neat question

there's an asymptotic result on the number of groups of order p^n im pretty sure

so maybe your question can be answered for p groups?

yeah here it is

https://groupprops.subwiki.org/wiki/Higman-Sims_asymptotic_formula_on_number_of_groups_of_prime_power_order

[chilly radish]

Of course. That makes sense now thanks!

I was being silly here. Set maps is not enough, but them being maps of Abelian groups in the case of modules is enough

Just need them to map 0 to 0

[chilly radish]

That was in response to this

[chilly ocean]

Hi, I'm going through Sylow's theorem in Dummit and Foote right now and just wanted to make sure:
when looking at the conjugates of some Sylow p-subgroup, all conjugates will also be a Sylow p-subgroup because conjugation is an isomorphism, right?

Or essentially, why can we conclude that every conjugate in S here is a Sylow p-subgroup?

[agile burrow]

Yes, conjugate subgroups are isomorphic. The inverse of conjugation by g is conjugation by g^-1. In particular, conjugate subgroups have the same order.

[chilly ocean]

Ok, thank you.

[glossy crag]

When constructing the semidirect product of H by K, the choice of homomorphism f:K->Aut(H) matters, right? I know that inside the semidirect product itself the action of K on H is always conjugation, but the "specifics" of the conjugation (what gets sent where) is determined by f and different choices of f might produce different semidirect products, right?

[agile burrow]

Yes, that's correct. For example, one could choose the homomorphism f: K -> Aut(H) to be the trivial homomorphism, in which case one just recovers the direct product

[glossy crag]

Thanks

[past path]

Does anyone have any insight into this problem?

[lavish sigil]

What is the evident way here? Is it applying f elementwise to the matrix entries?

I mean, that seems like it should work to give a homomorphism, but I thought maybe they mean some other way which I am not thinking of at the moment.

[tender bough]

is this a typo? Isn't ad_x \in gl(g)? or, ad_x: g -> g

[ruby sundial]

yeah

[tender bough]

where ad is:

[coral spindle]

yeah I reckon they probably meant g -> g

[tender bough]

cool!

[ripe basalt]

i dont think it's a typo

[ripe basalt]

https://en.m.wikipedia.org/wiki/Center_(algebra)#:~:text=The center is a commutative,of the Lie algebra L. notice what it says about the center of a lie algebra

probably because you can define g without a lie bracket but g with a lie bracket is called a lie algebra

oh g is a lie algebra

my bad

but still youre mapping the element in the lie algebra to a functional, no?

like x is going to [x, y] which is a functional on g so gl(g) no?

you map g to ad_g(x)

[chilly ocean]

It is a typo. ad maps g to gl(g), taking x to ad_x: g -> g.

[ruby sundial]

what does ad do?

is it like

XY-YX commutator?

because thats just bracket iirc

[chilly ocean]

A bracket for a Lie algebra might not explicitly be a commutator.

The expression xy - yx may not be defined.

ad_x just takes y to [x, y].

[ruby sundial]

what?

srs?

didnt know this

[ripe basalt]

ye

it just has to satisfy some other stuff

bilinearity and the jacobi identity or something

[chilly ocean]

Antisymmetry, bilinearity, and the Jacobi identity.

These are the only requirements.

[chilly ocean]

The reason I say "explicitly" is that it could be isomorphic to a Lie algebra where the bracket is explicitly a commutator. See, for example, Ado's theorem.

[ripe basalt]

isnt there a theorem for non characterstic zero

like you can decompose it into a sum of matrix algebras

or something

eh im thinking of something else

[rotund aurora]

Is there an easy example of a module over a PID that is torsion-free but not free?

(the module wouldn't be finitely generated, of course)

[agile burrow]

Q over Z

[rotund aurora]

nice

[past path]

Ok so

I have a proof for my thing

But it must be wrong

Because I don't know how the conclusion fails to hold if the rings don't have identity elements.

The symbol you don't recognize means "is an ideal of."

[agile burrow]

You've shown that I is a subset of the product of the A_k, but not equality

Try justifying why they're equal and you might see what goes wrong

[rotund aurora]

Is $\mathbb Z^\mathbb N$ as a $\mathbb Z$-module free or not? Just curious

[cloud walrus]

Croqueta

[rustic crown]

i don't think so

i've heard that it's not projective as well

but dunno an elementary proof

lemme think

there probably isn't an elementary proof as far i remember

i think i've heard that Hom_Z(Z^N, Z) = Z^(direct sum N)

so if Z^N was free, then it would have uncountable rank which means |Hom_Z(Z^N, Z)| = 2^2^|N|

so it's very far from being free as |Hom_Z(Z^N, Z)| = |N|

[rustic crown]

this is actually equivalent to non-free-ness as Z is a PID. but again it's only easy to see this under the additional hypothesis of finitely generated-ness

one way to see this in the general case is by first showing that submodule of any free module over a PID is free, this will require some sort of axiom of choice

then if you had a projective module over a PID, then it's a direct summand of a free module, and hence would be free.

[elder wave]

Alternative reasoning: Wikipedia states it as an example of a module that isn't free

[formal ermine]

is every ideal generated

[elder wave]

What do you mean by this

[formal ermine]

can we generate every ideal

what does erzeugtes ideal mean in english

[elder wave]

it's correct

but we can take the entire ideal as erzeuger if we want to

[formal ermine]

no I mean like

uh

how do I phrase this

[elder wave]

do you want to dm me the question in german

[formal ermine]

is the set of all generated ideals equal to the set of all possible ideals

like can we write every ideal as a generated ideal

[elder wave]

If all fails we can write it as the ideal generated by every element of the ideal

but i'm not quite sure what your question is still

[tender bough]

What does "common eigenvectors" mean, do the eigenvectors also have fixed eigenvalues? Let's say $\mathfrak{g}0$ is a Cartan subalgebra, $H_1, H_2 \in \mathfrak{g}0$. I reckon that then $\mathrm{adj}{H_1}$ and $\mathrm{adj}{H_2}$ have the same set of eigenvectors. Let's say $x$ is an eigenvector with $\mathrm{adj}{H_1} x = \lambda_1 x$ and $\mathrm{adj}{H_2} x = \lambda_2 x$. Are $\lambda_1$ and $\lambda_2$ equal?

[cloud walrus]

Mattuwu

[formal ermine]

can we write every ideal in the form of $(a_1, \ldots, a_s)$

[cloud walrus]

illuminator3

[elder wave]

is that supposed to imply that there's finitely many of the a_i

if so then no

[rustic crown]

🙈

[rotund aurora]

Where do you deduce that from?

(the uncoutnability of the hypohtetical rank I mean)

[next obsidian]

Free modules are classified by their rank

[rustic crown]

if the rank was countable, then Z^(direct sum N) is also countable

[next obsidian]

So Z^N would have to be an uncountable direct sum of Z

[rotund aurora]

(the uncoutnability of the hypohtetical rank I mean)

[rustic crown]

but |Z^N| = 2^|N| = uncountable

[rotund aurora]

oh yeah, I was totally missing that

[formal ermine]

wdym

[rotund aurora]

idk why i was thinking Z^N was countable lmao

[next obsidian]

You’re searching for the word “finitely generated ideal”

[elder wave]

all ideals in your ring are finitely generated iff. your ring is noetherian

[formal ermine]

ahhhhhhh

that's what I was looking for

thanks

[elder wave]

(that's even a common definition for it)

[formal ermine]

oh lol

[next obsidian]

Anyway I’m pretty sure you can show Z^N isn’t projective

I’m trying to come up with a proof of a map you can’t factor

Okay

Consider the map Z^N -> direct sum Z/nZ

Given by sending (a1,a2,…) to (a1,2a2,3a3,…)

Wait umm

[rotund aurora]

Idk why you would say that instead of sharing the link of the Wikipedia article, but ok

[next obsidian]

Gah

[rustic crown]

[next obsidian]

Nvm this won’t work I gotta think more

[elder wave]

The article i was referring to is the german one for "Free Module"

oh the english one uses it as well, didn't see that

[past path]

I see. If I is a subset of this product, how would we get an element of the product not contained in I?

Could we not use the canonical injection to map elements of A_k into I?

Since they're explicitly defined as the canonical epimorphism of elements in I?

I'm not sure how we would have anything but equality.

[next obsidian]

Yeah I’m just gonna give a size argument

Consider a map f:Z^N -> Z

Then f necessarily is the sum of the maps f•i_n

What I mean from this is, let e_i be the like i-th “basis” element

It is 1 in the i-th term and 0 elsewhere

[rustic crown]

but this sum isn't finite >.<

[next obsidian]

Then f(e_i) has to be zero for almost all i

Uh… is that even true

[rustic crown]

nope

[past path]

Is there a notational way I can refer to A_k as subrings of R instead of factor rings?

[rustic crown]

at least false for vector spaces

[next obsidian]

Yeah this is also false

[rustic crown]

like Q^N --> Q

[next obsidian]

Smh

[past path]

It's a bit cumbersome to define (0,...,i_k,...,0) every time I want to refer to an element of A_k living inside R.

[agile burrow]

Let me phrase it like this. We have a tuple of elements a_k in A_k. By definition, for each k there exists an element b_k in A such that pi_k(b_k) = a_k. The issue is that there may not be one specific element such that the projection applied to each factor is a_k

[past path]

How could that be?

[agile burrow]

Well if you want to prove such an element exists then you'd want an identity element

And if you want a counterexample then you look to nonunital rings

[next obsidian]

Yeah idk this is trash

[past path]

I imagine that would be the case based on the way the questions are phrased

[agile burrow]

Yes

[past path]

Is there a convenient notation I could use to refer to elements of the rings interchangeably?

[agile burrow]

Idk I use subscript and superscript sometime but indexing is hard

[past path]

The whole "canonical projection" thing makes sense but it's a bit awkward.

[next obsidian]

https://math.stackexchange.com/questions/1727159/direct-proof-that-infinite-product-of-copies-of-mathbbz-is-not-projective

[past path]

I've been restructuring ring notation from scratch lately, but I don't know if I necessarily want to go that far just yet.

[next obsidian]

Here’s a proof

It doesn’t appear to use choice only that you can split maps to a projective and then some minor set theory

[agile burrow]

This is interesting

I'm forgetting, is the direct sum at least free/projective

Or wait is that the definition of free

Idk how infinite works

[next obsidian]

Lmao

It is

Yeah

Well I guess it depends

You could say “has a basis” but then it’s iso a direct sum on the cardinality of the basis

[agile burrow]

Right, finite sums and all that

[next obsidian]

Yeah

Generation says the map from the direct sum is surjective

Indepencence says it’s injective

[south patrol]

Yhh

[past path]

How would we have a tuple of elements in A_k that isn't contained in I?

[rustic crown]

if you know about tensor products, then just write it as i_k ⊗ 1 in R_k ⊗_R R

what was the actual question tho?

[past path]

I don't know enough about tensor products and they've not come into this class.

Like I said earlier, this doesn't make use of what seems to be a necessary fact, so I'm not sure I've done this properly.

[rustic crown]

the key is the statement you claim without much justification

[past path]

I'm sure it is.

[rustic crown]

why are i, j in the ideal I?

[past path]

It's taken me an uncharacteristically long time to solve this problem.

Wait, I thought that one was actually justified.

Oh

one sec

Why are they in I?

[rustic crown]

if i_k is in A_k then you only know that there is some tuple (i1, ..., i_k, ..., i_n) in I

[past path]

Wouldn't that direct product necessarily be included?

[rustic crown]

not necessarily (0, ..., i_k, ... 0)

[past path]

Could you help me properly understand what is involved with a direct product?

[rustic crown]

okie so lemme say this first

[past path]

Ah, I see.

[rustic crown]

you know about groups, right?

[past path]

Of course.

I think I see where you're going with this.

[rustic crown]

in general it's false that subgroup of G x G' looks like H x H'

where H is a subgroup of G and H' is a subgroup of G'

the simplest example is the "diagonal" of G x G which is {(g, g) : g in G}

[past path]

We've proven that it's true for specific cases, like finitely generated abelian groups and p-groups

I see.

I see what you mean by that.

[rustic crown]

it's false for any non-trivial group tho :p

[past path]

Hmm.

[rustic crown]

so let's first prove it for a smaller product

say R and S are rings, show that an ideal of R x S looks like I x J

the proof remains the same, but easier to write down and think concretely without using "..."

[past path]

Yes, I suppose it would.

And since the direct product is associative it should generalize to any finite case immediately, right?

[rustic crown]

yep

[past path]

So I can entirely restructure my proof into a case of two with that caveat

That would make the notation so much nicer that I think I could tackle it pretty well.

[rustic crown]

(but i think once you understand the case for n = 2, you can easily write for any arbitrary n)

[past path]

Yeah.

But you could just say

[rustic crown]

maybe try to think in the chat, but loudly :p

(so i know what you thinking)

[next obsidian]

This is insanely nit picky, and in practice it does not matter but it’s interesting to think about if the direct product is actually associative

Because it actually isn’t, if you ask for literal equality, it’s associative up to the most obvious isomorphism in the world

((a,b),c) -> (a,(b,c))

[rustic crown]

(i said to myself i shouldn't say anything categorical here :p)

[past path]

Well, yeah. I suppose I could say "associative and commutative, up to isomorphism."

[next obsidian]

Yeah I just bring it up because this starts driving home some ideas that become important depending on what you end up doing

And also is good to think about, it kinda drives home the idea that “=“ really isn’t the right question to be asking a lot of the time

[past path]

I suppose that's fair.

[rustic crown]

natural isomorphisms are equality

[past path]

To some, lol

[strong valve]

elaborate

[rustic crown]

[strong valve]

what does it mean when you put text inside parenthesis?

[rustic crown]

i say that softly >.<

or as if you're reading my thought

[past path]

like this, really

[strong valve]

oh so like thought bubble in a manga?

[past path]

Ahhhh. I think I'm beginning to understand why this doesn't necessarily hold in some cases.

[rustic crown]

[past path]

Does the canonical projection send things to the multiplicative identity or to 0?

If it's 0, then we can just multiply (i_1,i_2) by (1,0) to get (i_1,0) in the ideal, so its projection is in the ideal.

[rustic crown]

Not sure what this means >.<

It sends (a,b) to a

[rustic crown]

Yep this is exactly how being an ideal and having 1 in your ring helps

[past path]

Er, the canonical injection, I mean.

The one that maps an element of A to an element of A x B.

Does it map a to (a,0) or (a,1)?

If it's the latter then I think we have what we need.

Er

I mean

[rustic crown]

Similarly you could use the element (0,0,...1,...0) to kill of everything but the kth component

[past path]

if it's the former

Right. That's what I'm trying to do.

But the canonical injection maps to a,0 not a,1. Right?

[rustic crown]

There isn't any canonical injection tho

R --> R x S can't sending r to (r,0) is not a map of rings as it doesn't preserve the identity

[past path]

Ahh, so that only works for groups.

OH

[rustic crown]

And you can't send r to (r, 1) as in this case the map isn't even additive

[past path]

So if you have (r,0) in your ideal

Then you can subtract (r,0) from (r,s)

Er

Wait

that doesn't really help

[rustic crown]

Just read this again without the "If it's 0, then"

[past path]

If we have i_1,0 in the ideal

Ohhhh

Because i_1,0 is isomorphic to R_1

And 0,i_2 is isomorphic to R_2

Er, with poorly named variables.

[rustic crown]

Not sure I understand what you said

I'm kinda sleepy, so I'll just write the argument

Say you have the ring R x S and some ideal K in it

Since R x S --> R the canonical projection is surjective it sends the ideal K to an ideal I in R. Similarly you get an ideal J in S.

Now it's clear that K is contained in I x J

We need to show the other inclusion

so given i in I and j in J, we want to show (i, j) lies in K.

By definition, i is in the image of K under the projection, so there is an element (i, j') in K and similar and element (i', j) in K for some i' and j' in R and S respectively

Since K is an ideal, and R is a unital ring, we can multiply (i, j') by (1,0) and still land inside the ideal K, as ideals absorb multiplication by elements from your ring

So you get that (i, 0) is in K and similarly as S is unital, you get that (0, j) is in K

Finally their sum (i, j) is in K which shows K = I x J

Me sleep now

[old hollow]

if G is a group of size 2n, and H is a subgroup of size n, then is G/H isomorphic to Z/2Z?

[south patrol]

Provided G/H actually is a group (i.e. provided H is a normal subgroup), sure

G/H is a group of order 2 and hence isomorphic to Z/2Z

[wind steeple]

Actually a subgroup of index 2 is always normal

[south patrol]

Good point aha

[old hollow]

interesting

[south patrol]

(if you've not seen that a subgroup of index 2 is always normal, it might be worth pointing out that in general, if G is a group and H is a subgroup of index p, where p is the smallest prime dividing |G|, then H is normal; this special case comes from the fact that 2 is the smallest prime overall lol)

[ruby sundial]

I mever thought of this

I sorta see it though?

H is of index 2 if G/H has order 2

[chilly ocean]

Isn't this one of those classic intro group theory exercises?

It's like the "a space is Hausdorff if and only if the diagonal is closed" of algebra.

[ruby sundial]

i never learned it

[chilly ocean]

(Algebraic geometers don't @ me.)

[ruby sundial]

a space is seperated if diagonal is closed immersion

[chilly ocean]

I know.

[ruby sundial]

im restating for myself lol

and to be corrected

im still a 👩‍🎓

i went through proof of this a week ago and barely followed. i have to back track it sometime

its mostly because i am lost in the sauce of definitions

we went over valuation criterion for seperatedness at the same time

[chilly ocean]

This was my definition of "separated variety", although, as the quotes might suggest, I only did varieties. The word immersion didn't come up once, we just said subspace.

I know about immersions from differential geometry, though, so the statement makes some sense to me and sounds right.

I am a little rusty.

[chilly ocean]

Welcome to AG!

[ruby sundial]

you know more than me im pretty unconfident

yeah i dont feel welcomed lol

immersion in dg is when differential is injective right?

[chilly ocean]

Yup.

[ruby sundial]

also

[chilly ocean]

Same as being locally an embedding.

[ruby sundial]

there is a term in back of my mind

oh what?

[chilly ocean]

Immersions and local embeddings are the same.

The normal form for immersions gives you this.

Every immersion locally looks like x -> (x, 0).

[ruby sundial]

i had no clue

oh yeah

constant rank is a term

i think it means for any point on manifold the rank doesnt change

rank of the differential

[chilly ocean]

Correct. For immersions, the rank is always equal to the dimension of the domain.

[ruby sundial]

but idk how to verify this, i dont know many explicit examples of maps between manifolds

yes my knowledge is very gapped

[chilly ocean]

I just wrote one down.

[ruby sundial]

what you are saying the x,0 map

like yes i know i can talk about maps from Rn to Rm

[next obsidian]

Separated is a property of a morphism

A map f:X -> Y is separated if the diagonal map X -> Y x_X Y is a closed immersion

[ruby sundial]

yeah

[next obsidian]

Hausdorff is then the special case when you’re looking at the map f:X -> pt over the final object since the fibered product is just the product in this case

[ruby sundial]

ive been told you say a scheme is seperated if its seperated over k

[next obsidian]

Well it depends

[ruby sundial]

like X->spec k

[next obsidian]

If you say separated without any other reference I think it is likely over Spec Z

[ruby sundial]

oh

[next obsidian]

But if you’re working in a context you’re fixing a different base then it should be over that base

So if you’re dealing with varieties it’ll be over spec k

And if you’re “working over a Noetherian base S” then it should be the map to S

Anyway, sorry, I just wanted to re-hash that separated isn’t an absolute notion!

However it can have really interesting absolute properties

[ruby sundial]

wdym pt over final object?

[next obsidian]

Here’s an example:
If X is a scheme which is separated over any affine base then the intersevtion of affine opens in X is again affine

[next obsidian]

In topological spaces the space which is just a point is the final object

So I just mentioned that because the fibered product over a final object is just the normal product

Because the condition about the maps having to agree after going down to the base you took the fiber product over is trivial, since ofc they have to agree!

[ruby sundial]

If someone asked me if intersection of affine open sets is also affine I would have said yes

[next obsidian]

Nope

[ruby sundial]

I didnt know you needed seperated

affine opens are just the covers of the scheme right?

[next obsidian]

Not really?

A scheme is a locally ringed space covered by affine schemes

[ruby sundial]

yes

[next obsidian]

An affine open is just an open set with the induced scheme structure being isomorphic to Spec A for some A

[next obsidian]

But like this is not true for a couple reasons. 1: an affine open is just a single open, not a cover. 2: not every open cover is by affine opens, eg {X} covers X so take X non-affine

Okay so I have an example of a non-separated scheme

[ruby sundial]

yeah i didnt mean the cover but the components of the cover sorry

[next obsidian]

Take a DVR A, maybe even k[x]_(x)

[ruby sundial]

okay

[next obsidian]

Spec A is two points, m and 0

0 is open, and m is closed

Does that seem good?

[ruby sundial]

ok

[next obsidian]

Oky so we can glue two copies of Spec A at the open point

So our resulting space has 3 points

{m,m’,0}

Where 0 is the now-identified open points

And m,m’ are the two different copies of the closed point

[ruby sundial]

ok

[next obsidian]

Okay so we have an open cover

{m,0} and {m’,0}

[ruby sundial]

yes

[next obsidian]

Each of these is just Spec A

Let me think for a second

Fuck aaaa I think this is separated

It isn’t affine though

It’s hard to come up with a non-separated scheme

The standard example is a line with two origins

But the issue is that the intersection then is non-affine

OH

Okay okay okay I think I have an example sorry

[next obsidian]

So btw this construction is the smallest (cardinality) non-affine scheme. You can compute that the global sections is A, so if it was affine it would have to be Spec A, but that only has two points!

Sorry sorry so

Take A^2

And then remove the origin

[ruby sundial]

yes

ive done this example

[next obsidian]

You have A^2\0 which is open

Take two copies of A^2 and glue them along A^2\0

You get a plane with a doubled origin yeah?

[ruby sundial]

so k[x]/x times 2 are open covers

oh wait

[next obsidian]

I want A^2!

Because A^1\0 is affine

But my example here

You have a cover of this plane with a double origin via the two planes right?

This is an affine cover

Their intersection is A^2\0

And that’s non-affine!

[ruby sundial]

ok im still analyzing sorry

so we went over the glueing construction in class one day

[next obsidian]

Sure

[ruby sundial]

and all i remember is that you have two open subsets and an isomorphism between them

[next obsidian]

Right

[ruby sundial]

with 3 properties that need to be satisfied

[next obsidian]

Yeah

So I wouldn’t worry about those details rn

Altho that’s very important

[ruby sundial]

i dont entirely understand the construction besides that it mirrors how its done for manifolds sorta

[next obsidian]

What matters is just that you can find two planes

The only way they differ is their origin

[ruby sundial]

yes

[next obsidian]

Cuz we identified every other point of them

These are both affine

Cuz they’re A^2

But their intersection is just the set you glued them along

Which was A^2\0 which isn’t affine

[ruby sundial]

yes

[next obsidian]

It doesn’t mirror manifolds really because you can make non-separated stuff which is like being non-hausdorff

[ruby sundial]

i meant construction of glueing schemes

[next obsidian]

But maybe if you want to visualize it, it’s just like gluing any two spaces

Like you can probably imagine say

Taking a box

[ruby sundial]

like how isomorphism exists between open subsets that intersect

[next obsidian]

Right

That exists basically like

You know how transition functions exist?

For manifolds

[ruby sundial]

yeah

thats the intution right?

[next obsidian]

It kind of says like when you change charts

It won’t change

[past path]

Is an element of a ring always in the ideal it generates?

Particularly a nonunital ring.

[next obsidian]

Those conditions with the crap about like

Intersections

And the cocycle condition “the thing about the compositions”

Is to basically guarantee that when you go and change charts or whatever it all works

[ruby sundial]

ugh

[next obsidian]

Like you need that in order to glue your sheaves

[ruby sundial]

cocycle is a word that ive been hearing often

[next obsidian]

Like here’s the idea even better

[past path]

I read the definition of principal ideal, and it's either only meant to be used for rings with identity, or it excludes some elements from their own ideals.

Which wouldn't make sense but the definition technically doesn't include it.

[ruby sundial]

but i think about cohomology and idk how there is a connection

[next obsidian]

It’s about Čech stuff

You’ll eventually get it

For now just take the definition to be like

The stuff are compatible in triples

[ruby sundial]

yeah we went over how to compute cech cohomologies

but i need more practice

[next obsidian]

We have a sheaf on X_1, X_2, X_3

[ruby sundial]

we should move to ag

[next obsidian]

The isomorphisms on the opens say U_1, U_2, U_3

[ruby sundial]

#algebraic-geometry

[past path]

Ok I think it's just poorly worded when it comes to rngs.

For some reason the textbook we're learning from uses almost exclusively rings without identity.

[thorn delta]

Hungerford?

[past path]

Yupppppppp

It's painful that that's enough to give it away.

[thorn delta]

Yeah the generality is a bit excessive in the ring section

[past path]

Perhaps

I'm not a fan of the way it's written.

I don't mind solving the exercises (most of the time)

But I rarely read the definitions and theorems because they make my eyes glaze over.

I've literally had an easier time deriving algebraic theorems from scratch than trying to figure out wtf the textbook says.

That said, currently I'm trying to find the best way to show that (2,0) is not in the ideal generated by (2,2) in 2Z x 2Z.

[thorn delta]

Relatable. My professor was epic enough to assume all rings have identity for the modules chapter

[past path]

Oh that's lovely.

My professor is pretty awesome as well, but it's rather a pain to assume everything is identityless.

Fine, I'll prove the ring squares to itself. Happy?

[thorn delta]

Lol yea I remember the statement of CRT started like “let R be a ring with R^2=R”

[past path]

As for the ideal thing, my first thought is a rather exhaustive approach, but I'm not sure if there's a more straightforward way to show the exclusion.

[thorn delta]

[past path]

Yeah I've just chosen not to read those statements.

So, ((2,2)) is the set of all integer and ring multiples of (2,2), right?

And sums thereof?

What's the most convenient way to claim that (2,0) is not in ((2,2))?

It's just a piece of the puzzle I'm trying to put together, but it seems to be a rather straightforward piece.

I'm just not sure what tools I have at my disposal other than common sense.

[thorn delta]

Hmmm. Using the relations of the ideal you should get some constraints that can’t occur if (2,0) is in ((2,2)). Like a linear algebra problem. I’m on mobile so I can’t really check myself atm

And I can’t think of an easier way hm

[past path]

I've reasoned out that the ideal generated by (2,2) is the set (a,b) of even integers whose difference is divisible by 4.

Which would prove the claim, but I'm not sure if there's a nicer way of writing that down that takes less space.

Without knowing the precise definition of an ideal of a ring without identity, I'm not sure how to proceed.

Is it all ring multiples? All integer multiples?

The set containing sums of both?

It's rather a pain.

[thorn delta]

It’s just an abelian subgroup which is closed under multiplication by the ring. Isn’t this the def in hungerford?

[past path]

Honestly not even sure.

So if I instead defined it to be "the set (a,b) where 4 divides b-a" it might be easier to prove, then?

[thorn delta]

Defined what to be that?

[past path]

Ah, the set of elements (a,b) whose difference is a multiple of 4

This is a really gross proof of both sides of the problem

Is there a way I could make this less awful looking

[chilly ocean]

You really don't need to keep restating definitions.

The entire first four lines of your proof for the first part can just be reduced to "We prove the n = 2 case, from which the case for arbitrary n follows by the obvious induction."

Do you really have to make a new line for every sentence?

You also don't need to restate things given to you in the problem. You are only wasting space by writing "where pi_1, pi_2 are the canonical epimorphisms" and such.

[next obsidian]

I can kinda understand what you’re trying to do, but you’re trying to write a proof for a human

This looks like you’re trying to write formal math like lean code for this

[tough raven]

But were they monic?

[thorn delta]

i see what you did there

[fading ibex]

outside of classifying finite groups, what are some significant uses of Sylow theorems?

[thorn delta]

passing qualifying exams

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Here Sylow means p-Sylow where q = p^r

[cloud walrus]

Sloth King Daminark