groups-rings-fields

[open sluice]

you can’t know if it’s a troll until you cut its brain open to look inside

[static yew]

Idk I'm just reading the paragraph that describes what boytjie was telling me

"F[x]/p(x)F[x] is the same as the field F(a) that results from adjoining a root a of p(x) to F, that is the closure of F union{a} under +-×÷"

[coral spindle]

ayayayaya

Oh right this is just different notation

[static yew]

Wait it says a root a
What if p has multiple roots

[coral spindle]

F[x]/(p) is how I would write that

[coral spindle]

:)

That's the magic

It doesn't matter

They're all indistinguishable from the perspective of F

[static yew]

Wait is it possible to construct a poly in Q with roots of e.g. sqrt2 and sqrt3? Or maybe sqrt2 and cbrt2

[open sluice]

even if it’s like a multiple root?

[coral spindle]

Yes, but not an irreducible one.

p needs to be irreducible over F

otherwise this construction doesn't work. Typically then F[x]/(p) won't even be a field.

[open sluice]

o

[static yew]

Ah so that's where the magic breaks down

[coral spindle]

Weeeeeeeellllllllll

It's for good reasons

[static yew]

Well I don't need this magic I think anyway

[coral spindle]

Sure you do, how else are you gonna describe a finite field?

[static yew]

Ahh the exact example it gives is Q(sqrt2)

[round hull]

lol

the cayley table of course

[static yew]

Graham's number G64 is constructed entirely of threes, so that means it's a prime power and the size of a finite field

I'd like to see the table for that 🙂

[south patrol]

Cayley table of a field lol

[summer path]

[chilly radish]

anyone have a source on this, I tried googling but didn't find anything: If Aut(G)\cong Aut(H) does G\cong H? If it makes it easier consider the case of finite groups (I am almost certain infinite groups will give a lot of counterexamples, but it's hard for me to believe this is untrue in finite groups)

[static yew]

I wonder how many bits are in G64

Min size of that table is the square of that

I think xkcd gave numbers on maximum bits of storage per kg of matter based on quantum mechanics

Actually twice that cuz you need both + and ×

[coral spindle]

Well I mean there's a silly example: Aut(1) is iso to Aut(Z/2Z)

Also Z/3 and Z/4 have the same automorphism group (Z/2)

https://groupprops.subwiki.org/wiki/Automorphism_group_of_a_group
This list gives a few more other counterexamples, some of which are actually interesting

This one is cool!

@chilly radish this should be enough I hope

[chilly radish]

hm, I failed to consider this haha

yea Z/3 and Z/4 is a good nontrivial example

[coral spindle]

A really nice one is that Aut(S_3) is isomorphic to Aut(V_4)

[chilly radish]

ooh that's nice

they'd both be Z/2 right

[coral spindle]

Nah they're both S_3!

[chilly radish]

is V_4 the klein 4-group in this case?

[coral spindle]

All of the non-identity elements of V_4 are permutable

yeah

[chilly radish]

oh wait right my bad, we can permute (1,1) with (1,0) and (0,1) too

I was only thinking of permuting the factors but yea you can permute those as well

[coral spindle]

I can't help but wonder if this generalises. Would Aut(Z/2 x ... x Z/2 (n times)) be S_{2^n-1} or something

[chilly radish]

hmmmm

[coral spindle]

in which case Aut(S_n) = S_n for all n>2 and we're done

Well I say we're done when I mean there's an infinite family of counterexamples

[chilly radish]

n!=6

Can't forget the exceptional automorphism!

[coral spindle]

Ah true good point!

I completely forgot

[chilly radish]

but yea these are all good counterexamples

thanks

now i'm wondering if there's some nice subclass of groups which are classified by their automorphism group

[south patrol]

Oh wait this reminds me of another weird question lol

the one which was like

[coral spindle]

Well S_n for n>2 is such an example right?

As in, that infinite family

That's pretty nice

[south patrol]

If A,B are groups and A x Z \cong B x Z then

1. A,B are isomorphic if both are abelian
2. if they aren't both abelian, they needn't be isomorphic

[coral spindle]

Do there exists maximal families like this?

[south patrol]

I need to think about that one

[coral spindle]

Oh nice

That's cool

[chilly radish]

yea boyt, I was wondering if there's any interesting properties all such groups must share

[south patrol]

Even the abelian case is annoying idk how to do it

lol

[formal ermine]

well firstly they are all classified by their automorphism group...

[south patrol]

Wdym by classified by automorphism group

[chilly radish]

you could probably do some universal algebra memery on this

[chilly radish]

Aut(G) uniquely determines G up to iso

[south patrol]

Like cutting up groups into families based off automorphism group?

Isn't that like never the case

Can't you just uh

Hm

I was wondering if given G there's a way to produce a different group with same automorphism group

[chilly radish]

that's the requisite next question

[south patrol]

Like product by some group or smth lol but yeah

[coral spindle]

Well we were saying earlier that S_n for n>2 gives a family of groups that all have distinct automorphism groups right

So clearly such things exist

Like, families of groups that have this property

[chilly radish]

we did say that this is not true for 3 at least

or

oh you mean inside that family

yea

[coral spindle]

Sorry I'm being sloppy with my notation

but yeah I mean that class of groups

Like let's look at the full subcategory given by such a family

Is this gonna be... idk is it gonna have products even?

I think that's a hard question

I don't even know if we can answer that for the family S_n expanded to include finite products

By which I mean, do they remain distinguishable

[chilly radish]

yea, this seems like it's gonna require more heavy machinery to tackle

I might come back to this at some point but I don't think i'm currently equipped to tackle it

or at least I don't have the brain capacity to think about it too hard rn haha

[rocky cloak]

Aut ((Z/2)^n) = GL(n, 2) which has (2^n - 1)(2^n -2)(2^n - 4)...(2^n - 2^n-1) elements. Which doesn't seem like it equals 2^n-1 !

[coral spindle]

You're right! It's way too many

Idk what it would be then.

But here's a nice observation: finitely generated Z/p-modules form a nice such family (for p prime)

Yeah indeed. Since they're just F_p-vector spaces they ought to be classified by just the sizes of their automorphism group

(Fixed some mistakes there)

[chilly radish]

yea that's a nice observatin

[rocky cloak]

Another interesting question: is there some group that doesn't appear as the automorphism group of another

I'm sure there is, but how can one characterize them

[coral spindle]

According to a mathoverflow post it's yet unknown

the second part I mean

there are explicit descriptions of some

https://math.stackexchange.com/questions/1747784/groups-which-can-not-occur-as-automorphism-group-of-a-group

whoops, math stack exchange

not overflow

[delicate orchid]

reading this paper rn

[chilly radish]

this sounds like you could translate this to an inverse galois problem

[coral spindle]

Seems like an absurdly difficult question tbh

[untold turret]

"A,B nonidentical maximal ideals in R implies A+B = R". is the following a proper translation in module theory? "A,B are submodules of the R-module R. the direct sum A + B is isomorphic to the R-module R"

[coral spindle]

The sum is typically not going to be direct

The correct translation is that if A,B are maximal submodules of the R-module R then A+B = R.

[untold turret]

oh i see, perfect. thank you. what sort of sum is it?

[coral spindle]

It is just a sum

A + B = {a + b | a in A, b in B}

[untold turret]

i see, thank you!

[coral spindle]

For an example of why the sum is not generally going to be direct, take the ideals (2) and (3) of Z. Their intersection is (6)

If a ring R is a direct sum of two maximal ideals then it is the product of two fields

[untold turret]

hm... i'm not sure I understand. do we not just have (6) equals the ideals (0,2) = (3,0)?

[coral spindle]

Absolutely not no

(0,2) = (2)

(6) is strictly contained in both (2) and (3).

What do you think the notation (0, 2) means?

[untold turret]

if we're taking a direct sum of (2),(3) as modules, then (0,2) = 2*3, no? most likely i'm adding additional structure that's not there

[coral spindle]

No!

I don't understand where you're getting this from

Again I'm gonna ask, what do you mean when you write (2) or (0,2)? What does this notation mean to you?

[untold turret]

(2) to me means the principal ideal generated by 2, and (0,2) was just an ordered pair in the direct sum of modules (2), (3)

[coral spindle]

OK

Great, this is clear to me now.

[open sluice]

gotta love notational overload

[untold turret]

my bad for lack of clarity

[coral spindle]

The onus is on the author

When we talk about the direct sum of modules N and M, there are two senses:

1. the 'external' direct sum of N and M, which is in particular as a set going to consist of pairs of elements.
2. the 'internal' direct sum, which I will elaborate on now.
   These things are isomorphic.

[cloud walrus]

boytjie

[coral spindle]

So when I say "The ring R is the direct sum of ideals I and J" I mean the internal direct sum.

As an R-module, this will be isomorphic to the external direct sum I (+) J, but this is no longer equal to the ring, but merely isomorphic.

[untold turret]

i see

[coral spindle]

So unfortunately this comment still doesn't make sense even once we have this understanding in mind

(6) is an ideal of Z, whereas (0,2) and (0,3) are elements of the external direct sum of 2Z and 3Z

And they are most certainly not equal.

So I'm afraid you are going to have to continue to explain what you mean

[untold turret]

hm, i'm not sure, but maybe we can give an isomorphism between 2Z x 3Z and Z with (-1,1) = 1?

i.e., -2 + 3 = 1

so then we can produce some sort of map between the structures

[coral spindle]

What are you viewing 2Z x 3Z as? Is it a module?

[untold turret]

sorry this is not very clear

[coral spindle]

Idk what you mean by (-1, 1) = 1

That makes no sense to me

[untold turret]

adding the ideals (2) and (3) produce the whole ring because the sum contains (1)

[coral spindle]

Anyway, as a Z-module 2Z (+) 3Z and Z are not isomorphic. The second is a cyclic module whereas the first is not.

[coral spindle]

Yes

[untold turret]

can we not give some sort of representation of this fact through structures containing vectors, so that the fact that -2 + 3 = 1 is given by a map (-1,1) to 1?

where (-1,1) means -1 * 2 + 1 * 3

[coral spindle]

I'm going to write 2Z and 3Z instead of (2) and (3) for notational clarity for a bit

When we have modules N, N' \subseteq M, we have a map N (+) N' → N+N'

It's given by (n, n') → n+n'

This map is an isomorphism if and only if the intersection of N and N' is {0}.

(c.f. condition 2 in my definition of an internal direct product)

[untold turret]

we wouldn't have empty intersection in this case, since both 2Z and 3Z contain 6

[coral spindle]

So we do indeed have a map 2Z (+) 3Z → 2Z + 3Z = Z

And this map is surjective.

And yes, this isn't an isomorphism precisely because there is a nonempty intersection

I am writing $(+)$ in ascii instead of $\oplus$, which is the symbol for the direct sum.

[cloud walrus]

boytjie

[untold turret]

i am parsing what you wrote, thank you very much

right, and we don't have an isomorphism because (3,0), (0,2) both map to 6

[coral spindle]

I dislike this notation you are using intensely

3 is not an element of 2Z so it is wrong to say that (3, 0) is an element of 2Z (+) 3Z

If you compose with an isomorphism Z (+) Z → 2Z (+) 3Z then what you write makes sense.

[untold turret]

ahhhh this is what I was looking for!

thank you, very highly appreciated

[untold turret]

if A is not a finitely generated ideal, is there a surjection between the infinite product \prod Z and A?

suppose A has countably many generators

or, maybe a better question: for an element in \prod Z do we have the natural surjection sending the ith component z_i to z_i times the ith generator?

[south patrol]

What is A an ideal of?

[untold turret]

a non noetherian ring

[coral spindle]

So we have an ideal A of a non-noetherian ring R, and you're looking for a map Z^N → R (I assume countable product since you just say 'infinite')

[south patrol]

Well being finitely geenrated as an ideal of R means as an R-submodule, which can be pretty different to talking about Z

[untold turret]

yes, and yes countable product

[coral spindle]

So are you expecting this map to just be a function?

[south patrol]

And what kind of surjection do you want

[coral spindle]

Because Z^N isn't going to be an R-module typically

[south patrol]

of Z-modules?

Ye lol

I suppose the quesiton that'd make most sense to me is like

replacing Z with R

But yeah depends what you want

[coral spindle]

But A is specifically not finitely generated

Are you just trying to say that if A is not finitely generated, is it countably generated?

Because the answer is no.

[untold turret]

it would be nice if it had ring homomorphism properties, because the ultimate goal is to express the ideal as generated by an infinite dimensional vector, but i'm realizing this might only be possible if the entries of the vector are actually elements of the ring, and not just Z

[coral spindle]

Every time you say 'vector' to refer to something that is not an element of a vector space, a puppy dies.

[untold turret]

no, that's not my question but what you said makes sense

[coral spindle]

I still don't have a clue what you mean

[untold turret]

i'm trying to get at the most abstract possible representation of a non-finitely generated ideal

[coral spindle]

The most abstract possible representation is that it is an ideal of a ring

That's already peak abstraction

[untold turret]

right, well, something a bit more concrete: i think that maybe we could represent it as an infinite dimensional tuple

[coral spindle]

No typically you cannot.

[untold turret]

where each entry represents multiplication with the ith generator

[coral spindle]

You seem to be suggesting that every module is either free or a direct sum, and this isn't true

direct *product rather

[untold turret]

oh! they're not? are not the elements of an ideal a linear combination of its generators?

[coral spindle]

They are

They are not unique

[untold turret]

right

[coral spindle]

Modules are very unlike vector spaces.

[untold turret]

could we proceed in this way if we accept the representations are not unique?

[coral spindle]

Then be precise about what you want to see

[chilly ocean]

I've never been so confused in my life

[coral spindle]

Can you not maybe state your questions in a simpler way? Almost time you have asked a question here, everyone else has had to spend a lot of time figuring out what you are saying.

[untold turret]

sorry i'm not being precise but the way you give definite answers is very illuminating and clarifying about the proper way to state things and ask the question

[chilly ocean]

You could define a Z module morphism if you just take the direct sum over all elements of the ideal but I don't think that's useful for anything

[coral spindle]

You couldn't define it on the product actually, since you can't take the sum of infinitely many things

[cloud walrus]

definitely not potato

[untold turret]

yeah that's one of the problems i'm dealing with, presumably an element of a non finitely generated ideal with countably infinite many generators can be generated as a linear combination of countably infinite many generators

[coral spindle]

But this is not a problem

A linear combination of any number of elements always consists only of finitely many summands!

This is part of the definition of the infinite direct sum

[untold turret]

right, but i'm interested in the infinite product

[coral spindle]

Why so.

[untold turret]

i'd like to give a representation of non finitely generated ideals that countenances both ideals that are infinite direct sums and those that are infinite direct products, and i conjecture that if i stick to the infinite direct products, the resulting property will be easily transmissible to the infinite direct sums

[coral spindle]

Well good luck with that

[south patrol]

Direct products are less natural here because
-direct products are interesting regarding mapping into them, not out of them
-the notion of being generated is defined in terms of sums

[untold turret]

hm i see

[white oxide]

yo can i have a hint for this question pls

this is what i got so far but im stuck

my guess is that they want me to show injectivity by using the uniqueness of fbari = f somehow

[round hull]

say i(x) = i(x'). then have the map f: X -> UA send x and x' to different elements, and get a contradiction

[glossy crag]

My brain isn't working: if A and B are ideals in a Dedekind domain s.t. that A+B=1 and P is a prime dividing neither, why then AP+B=1?

[white oxide]

oh i didnt know we could define it as such but that makes sense

thanks

[void cosmos]

you guys done?

[south patrol]

Suppose $Q$ is a prime such that $Q \mid AP + B$. Then $Q \mid B$ and $Q \mid AP$. The first one implies $Q \ne P$. Since $Q \ne P$, $Q \mid A$. So $Q \mid A$ and $Q \mid B$, contradiction

[cloud walrus]

potato

[south patrol]

Or actually it's easier to use an analogy to integers lol

if a and b are coprime, then multiplying a by a prime factor which doesn't divide b won't change that

[glossy crag]

My reason was to show coprime ideals => sum =1, so I can't use this.

[round hull]

can't you do that without the thing

[south patrol]

Oh lol

Are you defining coprime by like no prime divides both then hm

Do you kow that P is contained in Q iff Q | P

[glossy crag]

Yes

[glossy crag]

Yes

[south patrol]

hm well then there's an easier way to prove the coprime thing then

which doesn't need your statement as a lemma

[glossy crag]

How so

I thought to just inductively apply this

But they're sort of equivalent anyway

[south patrol]

Well if sum isn't zero, then A + B contained in a prime P

So A,B are contained in P

[glossy crag]

and it has to be one of those primes, right

tru dat

[south patrol]

ye

Dedekind domains are hot

[glossy crag]

Forgot about the containment/dividing equivalence

[round hull]

hot potato

[void cosmos]

you guys done irght

okay so now

suppose A --> B --> C --> 0 is an exact sequence of R-modules ( R is a commutative ring with identity )

show that tensoring this with an R-module is exact

i managed to do everything except one step

showing that the kernel is contained iin the image in the right part

[coral spindle]

Are you aware that hom is contravariant and left exact?

[void cosmos]

whats contravariant

[coral spindle]

Nevermind then.

[void cosmos]

left exact yea

and i proved it

i can do it

are you the terrible flat?

[chilly ocean]

you still eat ink?

[void cosmos]

yoo hahaha

yea

on occasion

anyways

someone helped me prove "adjoint associativity"

and atiyah mcdonald proves tensor is right exact in 3 lines using it

and i dont undersatnd why

[coral spindle]

that was a joke name

[void cosmos]

i did it normally and i got stuck

at ker is a subset of img

and btw

if f:A-->B then it indcues f bar: D tensor A --> D tensor B with sending d tensor a to 1_D tensor f(a)

those are generators ik but you just need them to define the whole map

[south patrol]

yh

[void cosmos]

yea im new to this tensor stuff

anyways can someone walk me through it? the profo of right exactness

the usual way

not using adjoint associativty thingy

[south patrol]

I mean Atiyah Macdonald gives the usual way right

[void cosmos]

no

he uses adjoint associativity

and i didnt understand how

[south patrol]

Not sure what you mean by adjoint associativity

[void cosmos]

Hom(M tensor N,P) is iso to Hom(M,Hom(N,P))

[south patrol]

It's in a sense (imo) the best possible proof because it just uses the universal property of the tensor product

[south patrol]

Oh I can just explain this cause it's important

[void cosmos]

yea terrible flat helped me prove it

the normal way would be to show it manually

that the kernel of each map is the image of the preceding map

[south patrol]

Is there a reaosn you wanna do it manually lol

[void cosmos]

yes

[south patrol]

But yeah okay we can try to do it by hand but it'll be a bit of a pain I think

[void cosmos]

this secret coming out might result in the fall of nations so i will just keep it to myself

aits actually easy untill the final part

that the kernel is contained in the image

[south patrol]

Usually that is the hard part and the other is trivial

[void cosmos]

yea

i tried the same argument

with the Hom one

like using the inverse map

but it didnt wor

or okay

i can prove now adjoint associativity

so if you can help me understand right exactness using it

i will just prove both on the exam

[south patrol]

Okay so I mean $M' \otimes N \to M \otimes N \to M' \otimes N \to 0$ is exact iff $0 \to \mathrm{Hom}(M'' \otimes N, P) \to \mathrm{Hom}(M \otimes N, P) \to \mathrm{Hom}(M' \otimes N,P)$ is exact for all $P$, now replace all of those by smth equivalent

[cloud walrus]

potato

[void cosmos]

lmfao

so yea then i use adjoint associativty at each Hom

and i get Hom of something and something

which ik is exact anyways right?

[south patrol]

I wouldn't call this adjoint associativity, never heard of that term

[void cosmos]

its used in hungerfords text

[south patrol]

Just like universal property of tensor product

Okay sure

[void cosmos]

its not my wording

[south patrol]

Yes

And yes

[void cosmos]

but it doeds feel like adjointess involved

like adjointess from functional anlaysis

<Ax,y> = <x,A*y>

idk man

feels fishy to me

[south patrol]

Yes, the slogan is that "the functor $ - \otimes_R N$ is left adjoint to $ \mathrm{Hom}(N,-)$"

[ember grove]

Quick question, I saw this picture of extensions in an answer on mse, can you not conclude immediately that the fourth root of 3 is not in Q(sqrt(2)) because that would mean that the degree of the extension on top is 2 which is lower than the right extension which is a subpace of the top one when seen as vector spaces?

[void cosmos]

but @south patrol

after we simplify each Hom thing or object or whatever

why do i know the resulting sequence is exact?

[south patrol]

Because Hom is left exact

[glossy crag]

In case you don't figure out how to prove it @void cosmos, here is a very clean proof from Jacobson

[void cosmos]

yes so i need the things in the Hom functor first to be exact

like Hom("left exact sqe" ) is exact

[south patrol]

lol very clean when it's messier

[south patrol]

Sure

[glossy crag]

I consider this clean tbh, pretty straightforward.

[void cosmos]

youa re much smarter than me hahah

but yea i think knowning how to do it manually is good too

thank you ocean man'

[south patrol]

Sure yeah, I guess I view it as more elementary but less conceptual

Maybe i am too cat pilled

[glossy crag]

Yep, it's elementary, but very concise.

[round hull]

is it

[void cosmos]

and do you mean M'' tensor N --> 0? the far right part of the first sequence

[void cosmos]

for me ig

[south patrol]

Yeah sorry

[void cosmos]

anyways how do i know whats inside the Hom is exact

?

after i simplify the tensor product out

[uneven fossil]

quick question, reading dummit and foote and i'm confused what this notation means exactly. is it a group action on a set and if so, what does Z x Z mean? the additive integer lattice?

specifcially 12/13/15

[hasty sinew]

Correct

[uneven fossil]

appreciate it

[untold turret]

infinitely adding things together electric boogaloo 2 redux: is there a surjective ring homomorphism between the infinite direct sum Z^N and a non-finitely generated ideal that contains no infinite sums/products of elements?

[void cosmos]

yea you should know about it more generally as the direct product

like it should be familiar

[uneven fossil]

no actually i don't think the books talked about direct product

i'll look into it thanks

[void cosmos]

u sure?

[uneven fossil]

if so then i must've missed it

[void cosmos]

yea recheck them , you should be familiar with those

they are p easy and cool anyways

[coral spindle]

What is an ideal that contains no infinite sums/products of elements? There is no such thing as an infinite sum or product in a ring in general.

[void cosmos]

anyways im still stuck with the proof

[untold turret]

we have Z[[x]] no?

[void cosmos]

why didnt we say Hom(M tensor N , P) --> ... is exact from the first place

okay cuz that assumes what we want to prove cuz its an iff

[coral spindle]

Yes and? Z[[x]] consists of formal sums. You still cannot add together an arbitrary infinite collection of elements in Z[[x]].

[void cosmos]

okay so we use the isomorphism and now i get a sequence of like Hom(M,Hom(N,P)) ---> ...

[untold turret]

i see! thanks

[coral spindle]

There is an enormous difference between a formal sum, which you should think of as just a symbol, and an actual sum.

[untold turret]

right that makes a lot of sense

a formal sum is just a single element

[rocky cloak]

I mean, Z[[x]] is a topological ring with the x-adic topology. So I guess it makes sense to take infinite sums and long as they converge

[uneven fossil]

you were so right

both that it was there and that it was cool

[void cosmos]

yea i was like 99% sure its there

and i was just weirded out by ther fact that you had to guess what Z x Z was

i was like yea that guy missed it haha

[uneven fossil]

lmao fair enough

[void cosmos]

hf

[uneven fossil]

ofc thanks

[void cosmos]

also its weird how you know about group actions even though the section ur at rn is before group actoins

hahaha ur a time traveller

[coral spindle]

I might also point out that every element x of an ideal is a finite sum of elements of the ideal... namely x.

[uneven fossil]

nah no way group actions were section 1.7, this is section 2.3 exercises

[void cosmos]

yeahh

you caught me

[uneven fossil]

this guy

[untold turret]

right. but aren't nonformal series actual infinite sums of ring elements? in the real numbers, for instance

[coral spindle]

No they are not

Given a binary operation, we cannot define what it means to take an infinite sum

We can only do this in certain special circumstances, such as with a topological ring like the real numbers, and even then there is no way to assign meaning to an arbitrary such 'infinite sum'

[untold turret]

woah this is very cool stuff, thanks

this changes significantly how i think of non finitely generated ideals

[round hull]

awesome

[coral spindle]

Are you reading a book?

[untold turret]

yeah Steps in Commutative Algebra, but I was working on a proof for the isomorphism theorems so i've been working very hard to think about what ideals are

[coral spindle]

Well I'm going to tell you now that this approach is not helpful for understanding what ideals are.

I would suggest continuing reading.

[untold turret]

noted, thanks a lot for the help

[round hull]

when does one actually understand what ideals are tbh

[coral spindle]

A good way is through the isomorphism theorems, instead of trying to express them as a quotient of a free module(!)

[void cosmos]

when you learn about number theory i think

@south patrol you there ?

for 4a) can i choose R = Z/mZ and A = Z/mZ and B = Z/nZ?

A tensor B under R is just B while its Z/(m,n) under Z

[rocky cloak]

B is not an R-module. So tensoring over R doesn't really make sense

[void cosmos]

yea ur right

okay 1 sec

what if i take R= Z/6Z

and B to be Z/2Z

B is a Z-6 module

right?

[chilly radish]

Yes

[void cosmos]

so it works

[rocky cloak]

Yes, but does it solve the exercise?

[void cosmos]

A tensor B would be Z/2 m while under Z it would be gcd(2,6) which is 2 wtf

lmao

[rocky cloak]

You may have to be more creative with your choice of R

[void cosmos]

yea maybe Q so things just blow up

or nvm wait

[rocky cloak]

Even more creative perhaps

[south patrol]

Ye

[formal ermine]

hi potato

[south patrol]

Sorry didn't see lol

Hi

[formal ermine]

how are you doing

[south patrol]

I am okie yeah having one of those times where maff is hard but learning a lot lol

Hbu

[formal ermine]

going over some stuff for my exam in 1 and a half weeks

[south patrol]

Was gonna ask smth here

Oh okie glgl

What exam

[formal ermine]

newton okounkov theory

is mostly just toric/algebraic geometry

the only part that scares me is grassmannian stuff

toric geometry is meh

[south patrol]

Oh okay now I see who you are lol

[formal ermine]

oh yes I am illu

[south patrol]

Lol

I was wondering cause I saw German server

Bur sure yes glgl hm

[formal ermine]

like I sort of understand it

but there are so many indices

and fucking summation signs?????

when did those get added to algebra

[formal ermine]

like just LOOK AT THIS

[south patrol]

Ouch

[coral spindle]

Grody!

[rocky cloak]

Just why!

[south patrol]

Okay here is a question

[rocky cloak]

Surely there must be a more convenient way to express that

[formal ermine]

it's for showing that the grassmannian is a projective algebraic set

[south patrol]

Say I have an abelian category A. I've seen people construct the derived category D(A) by saying the naive homotopy category K(A) admits the structure of a triangulated category and working with that, whilst I've also seen people basically put a model category structure on the set of non-negative chain complexes over A and form the ("true") homotopy category of that

Am I right in thinking these give the same thing? Is there an advantage / disadvantage / major difference?

Hm

[void cosmos]

yo @rocky cloak why cant i think about like free stuff

like in a division ring everything is free and the tensor product of two free is free

like maybe A tensor B under R is the same as A tensor B under Z but one is free under R whiie not free under Z

or idk

[rocky cloak]

I'm not quite sure what you're saying...

[void cosmos]

Z/2 is a free Z/2 module but not free as a Z-module

for example

would that be a counterexample ( suppose i can get to that )

[rocky cloak]

So there are a few Rs that won't work, and Z/2 is one of them

[formal ermine]

why do we care about lie theory

[south patrol]

Uhhh lie algebras come up in topology and geometry a lot for starters

[coral spindle]

Do you mean, why do we care about Lie algebras? Or do you mean why do we care about Lie groups.

[formal ermine]

both

[coral spindle]

We care about Lie algebras because they classify Lie groups.

We care about Lie groups because we like em

[solar vessel]

real

[coral spindle]

Also physics, idk.

[south patrol]

Lie algebras also come up in topology and homotopy theory a lot in modelling certain things

I mean, I am doing a summer project where Lie algebras are a major tool used to compute things about spaces

[coral spindle]

That's cool I didn't expect that to be the case

[rocky cloak]

In general if I is an ideal in R and M and N are R/I modules. Then the tensor product of M and N is equal whether you do it over R/I or over R.

[coral spindle]

Equal as an R-module, you mean, right? I think this is part of the confusion

Well yes equal literally as a set

but isomorphic as an R-module

[south patrol]

For example (I need to learn more about it!) Quillen shkwed there is a correspondence between rational homotopy types and differential graded lie algebras in some sense

[hasty sinew]

Interesting

[rocky cloak]

Well assuming R is commutative I guess. But first and foremost they are isomorphic as abelian groups.

But yeah any R/I module is naturally an R module

[delicate orchid]

this..... so much this...

[coral spindle]

Yeah I'm just pointing this out because of the comment about free modules. I don't mean to intrude

[sweet echo]

How should i be interpreting $\Ext^1(u_i,A)$ (near bottom)? The $u_i$ are inclusions from $T$ into $T^d$

[cloud walrus]

ΣΑC

[sweet echo]

oh as a morphism from Ext(T^d,A) to Ext(T,A)?

[void cosmos]

yo anything tensor itself is just itself right?

[uneven fossil]

this actually just turned out to be the cartesian product 😭😭

[void cosmos]

cuz u have (a,b) --> ab as a bilinear map anyways?

[void cosmos]

yea it is 😄

[uneven fossil]

how do i make sense of Z_2 x Z then

one is cyclic in a multiplicitive sense while the other is cyclic in an additive sense

[void cosmos]

well

its just the cartesian product with the structure in the textbook

and the operations are pointwise

[delicate orchid]

V \otimes V is definitely not always isomorphic to V

[void cosmos]

why

[delicate orchid]

I wish it was cause then the tensor algebra would be nice and simple

[void cosmos]

(a,b) -->ab doesnt work?

[delicate orchid]

a dimension argument will suffice

[uneven fossil]

a supposed generator for the group would be <a, b> where a is in Z_2 and b is an integer i guess

[formal ermine]

ong

[formal ermine]

V* tensor V* is Hom(V, V*)

[void cosmos]

okay

now hwy

is not sending (a,b) --> ab

not work

is it cuz thats just induces a homomoprhism? and may not always

have an inversE?

[formal ermine]

yeah

beat my wife to it

what's the inverse supposed to be?

[round hull]

what the fuck

[void cosmos]

what

[delicate orchid]

my wife is mean to me!

[void cosmos]

okay

i was about to lose my mind over this

cool

[formal ermine]

you guys aren't on reddit enough to understand my level of humor

[round hull]

so i should never go on reddit then

[formal ermine]

like

if someone comments r/beatmetoit

[delicate orchid]

holy shit shut up hahahaha

cringe

[formal ermine]

then the next person has to comment r/beatmymeattoit

[summer path]

i don't know why, but i read "beat my wife into it" followed by "what's the injection supposed to be?"

[formal ermine]

and the person after that r/beatmywifetoit

[void cosmos]

i cant do it

[summer path]

illumi get your head out of the gutter

[void cosmos]

i cant find a counterexample

[summer path]

[void cosmos]

im so frustrated honestly

[delicate orchid]

R^2

[void cosmos]

everything literally works

[delicate orchid]

no

R

no, R^2

[void cosmos]

i meant for the problem i posted

[delicate orchid]

that's the one

[void cosmos]

.

wait

[agile burrow]

Think about what tensoring with Q over Z does

Actually don't think about that, idk if it's helpful here

[delicate orchid]

^

wait no it isn't haha

[agile burrow]

But think about it anyway because it's kind of cool

[delicate orchid]

next question boss

anyway idk just take like

C[G] \otimes \C[G] for some finite group G that'll probably work

ohhh or can we be funny

[void cosmos]

yo how do u tensor over Z[x]

i think this works

[delicate orchid]

Z/2Z \otimes Z/4Z \otimes Z/2Z \cong Z/2Z? remains to be seen

[agile burrow]

Yeah Moamen I think you're onto something

Tensoring Z[x] with itself over Z gives Z[x, y]

[void cosmos]

how do i know that

[rocky cloak]

Yes

[agile burrow]

You can show Z[x, y] satisfies the universal property, or just construct the homomorphism explicitly

[void cosmos]

no i meant how would i even come up with Z[x,y]

[agile burrow]

But tensoring Z[x] with itself over Z[x] just yields Z[x]

[void cosmos]

like okay maybe sending f(x) --> f(x,0) or something

yea

it works

but i wouldnt be able to see Z[x,y]

[delicate orchid]

it makes sense if you consider what the elements actually look like, p \otimes q for some polynomials p,q

[void cosmos]

yea sending Z[x] x Z[x] --> Z[x,y] like (f,g) --> f(x)g(y) or something

[delicate orchid]

then you can wrench out all of the Z coefficients, and then use the nice additivity of \otimes to seperate it into things that just look like x^n \otimes x^m

[void cosmos]

idk

[maiden ocean]

Universal property it

[delicate orchid]

lame

[void cosmos]

are u talking to me

[maiden ocean]

Yeah

[void cosmos]

thats what im trying to do 😄

what is it then

if im not doing this

im trying to find a bilinear map

from the product to Z[x.y]

is that NOT the universal property or what

[maiden ocean]

It's easiest to just show Z[x, y] has the universal property

[void cosmos]

im sorry what does that mean

[agile burrow]

You want to show that if you have a bilinear map from Z[x] \times Z[y] to some abelian group M, then you get a unique homomorphism Z[x, y] -> M such that the composition Z[x] \times Z[y] -> Z[x, y] -> M agrees with the original bilinear map

[maiden ocean]

Of the tensor product of Z[x] and Z[y]

What walter said

[void cosmos]

im so confused rn

im trying to show that Z[x] tensor Z[x] is Z[x,y]

[maiden ocean]

Do you know what the universal property of the tensor product is

[maiden ocean]

Best to denote the second one Z[y] but yes

[void cosmos]

so what i am trying to do is find a bilinear map from Z[x] x Z[x] to Z[x,y] so that it induces a map from the tensor product to Z[x,y] so that the diagra m commutes

[summer path]

i think it might help to look up the commutative diagram for universal property

[void cosmos]

thats the universal property for me

[maiden ocean]

Yes

[void cosmos]

isnt that what i have been trying to do?

or is that something else

[maiden ocean]

The universal property says that bilinear maps Z[x] x Z[y] to A correspond to homomorphisms Z[x] (x) Z[y] to A

In a nice way

It suffices to show that Z[x,y] has this property

[void cosmos]

?

[maiden ocean]

No

[delicate orchid]

moaman is asking what the map at the top of the triangle is I think

[maiden ocean]

You are trying to construct a map Z[x] x Z[y] -> Z[x, y] descending to a map out of the tensor product which happens to be a homomorphism

[void cosmos]

moamen is mega frustrated :d i think the map at the top is (x,y) -->< x tensor y

[void cosmos]

yea

and what your saaying is you want me to find a 1-1 correspdoncone ? between bilinear maps from the product and linear maps from the tensor?

or what

[maiden ocean]

Okay I see what you mean

[void cosmos]

yea sorry im just very confused

can u do an example

of "universal property it"

cuz i thought thats what im doing

[maiden ocean]

No I misunderstood you you are trying to do the right thing

[void cosmos]

so i was universal propertying it?

[maiden ocean]

If your goal is to show that given any bilinear map Z[x] x Z[y] -> A there is a map Z[x, y] -> A commuting with the map you suggested then yea

[void cosmos]

fuck .. no

[tribal moss]

The proper wording would be: show that Z[x,y] together with a certain map Z[x] × Z[x] -> Z[x,y] has the universal property that defines the phrase "tensor product".

[void cosmos]

my understanding of the universal property is is that IT INDUCES the map

like it does it for me

[delicate orchid]

Let f : Z[x] x Z[y] -> M be a bilinear map
Can you find a linear map g, of the form g(p \otimes q), such that g(p \otimes q) = f(p, q)

[void cosmos]

like once i give a bilinear map

the universal property gives me the homomoprhism for free

is that wrong

[tribal moss]

There are two maps invoved.

[maiden ocean]

It does

[delicate orchid]

we don't know if this module has the universal property yet

[void cosmos]

yea i think thats the problem./.

[maiden ocean]

^

[tribal moss]

The "tensor product" comes with a map from A×B to T.
Then if you give it another map A×B -> C, it produces a map T -> C for you.

[south patrol]

Or you can be cool and say that Z[x] (x) Z[y] and Z[x,y] both have the property that a map into A is equivalent to a choice of two points of A hehe

[delicate orchid]

so by doing this you show it has the universal property, and is thus the tensor product

[void cosmos]

this map is (a,b) to a tensor b right?

[maiden ocean]

I mean that basically is the proof potato

[glossy crag]

a) Why should there be a sigma fixing beta and moving alpha?
b) What do we even need separability for?

[tribal moss]

Yes, that's what it's called.

[south patrol]

Okay sure lol

[void cosmos]

okay so

[south patrol]

I guess it is ultimately ges

[void cosmos]

now this is for any problem lilke this: to show that it is actually the tensor product i show it actually achieves the universal property

which says that given a bilinear map from the product it factors through the map tropo said basically

or that it induces a map from the tensor to the abelian group such that the diagram commutes

[tribal moss]

I think you may be confusing that concrete construction of a tensor product we talked through the other day, with the abstract chraracterization of a tensor product via a universal property.

[void cosmos]

yea i think im confusing alot of things haha but i think i get it now

so let f: Z[x] x Z[y] --> A be bilinear , why cant i just say define f bar as f bar(f tensor g) as f(f,g)

[untold turret]

i've stopped thinking of generators and started thinking only in terms of containment relations between ideals and it clicked. i was thinking of PIDs and rings with ideals of nontrivial height and these led me astray heheh

[void cosmos]

? excuse the redudant naming

[delicate orchid]

or you can be actually cool and just $p \otimes q = (\sum a_nx^n) \otimes (\sum b_ny^n) = \sum\sum a_nb_m (x^n\otimes y^m) \mapsto \sum\sum a_nb_m x^ny^m$ I won :iwon:

[cloud walrus]

wew ladz

[delicate orchid]

something probably breaks here

[void cosmos]

this by definition forces the diagram to commute right?>

[delicate orchid]

is that well defined

[void cosmos]

its bilinear to its well defined ig

[tribal moss]

Before you start doing this, you need to define what the map Z[x] × Z[x] -> Z[x,y] that you will be composing with is.

[void cosmos]

yes

yea yea

so for example (p,h) --> p(x)g(y)? would that work

[delicate orchid]

just thinking for something like a Q-module you could replace f tensor g with xf tensor g/x, could run into well definedness issues

[tribal moss]

Yes, if h=g :-)

[void cosmos]

wtf yea

sorry

yea thats what i meant

[tribal moss]

Okay, right.

[void cosmos]

okay so now the map this corresponds to "from the tensor" is sending f tensor g to fbar(f,g)

again excuse the redundant naming

call it f bar

[tribal moss]

Instead of apologizing for the reduncant naming, why not use different letters for different things?

[void cosmos]

yea mb

so this works ?

but now this is only half right? cuz this is just a homomorphism not an isomorphism

yet.

[tribal moss]

I think you'll just confuse yourself if you keep calling it "f tensor g" instead of f(x)g(y).

[glossy crag]

Anyone?

[tribal moss]

The larger problem is that if you have a random element of Z[x,y] that might not actually be a product of a polynomial in x with a polynomial in y.

[void cosmos]

yea so instead i must find an inverse

[tribal moss]

So you need to define your fbar that makes it clear that (a) it has a single well-defined value on every element of Z[x,y] and (b) it is a homomorphism, and (c) finally, that its composition with the canonical map is the bilinear map you started with.

[rocky cloak]

Yes, these should give the same thing. In the former you already have a triangulated structure and then just do a verdier localization. So this seems like the most straight forward way to do it if you want to study the derived category (this is the approach I would go for).

I would guess the latter approach is more useful if you're already working in the language of homotopy theory, i.e. the derived category is just one of many homotopy categories you work with and want to compare.

But I'm no homotopy theorist so, I don't really know why people take the second approach.

[tribal moss]

Fortunately Z[x,y] happens to be a free module over Z, so you can define fbar by specifying where it maps each element of a basis for Z[x,y] and asserting that it's linear.

[void cosmos]

are you talking about the map f bar thats from the tensor to Z[x,y]?

im confused again lol

[tribal moss]

Unless I've lost the thread, you're trying to show that Z[x,y] is the tensor product of Z[x] by itself.

[void cosmos]

what ur asking is show this is well defined

[south patrol]

Sure, fair enough thank you!

[void cosmos]

right?

[south patrol]

I guess yes this is just because I am coming from the perspective of homotopy theory

[tribal moss]

(I hope we're talking tensor products over Z, but I can't quickly find a place to confirm that in the scrollback).

[south patrol]

But yeah afaik the triangulated structure is still p useful because it keeps the suspension functor as part of the structure right

[void cosmos]

yea we are

why would that matter now?

cuz it might not be bilinear in another ring right?

cuz the scalars change?

[south patrol]

Which we want in analogous situations like the stable homotopy category

[south patrol]

Yes

You'd be hard pressed to find a non trivial Q-linear map out of Z[x]

[tribal moss]

Just making sure I'm not talking about a different situation than the one you are.

[void cosmos]

yea

okay so again

[void cosmos]

is that what you meant?

[untold turret]

if a ring has krull dimension n, is there a sense in which there are n classes of minimal prime ideals?

[tribal moss]

Well, I would prefer to say I'm asking you to define it at all.

[void cosmos]

wait what

f is just mutiplying the polynomials out

[tribal moss]

What is "f" now?

You keep calling everything f.

[rocky cloak]

If a is not in K(b) then the splitting field of a over k(b) is nontrivial. So there is an automorphism moving a bit fixing k(b).

Without seperability we can't guarantee, that the automorphism moves a.

[void cosmos]

the map from Z[x] x Z[x] to Z[x,y]

[void cosmos]

i laughed lol

mb

[void cosmos]

okay let me rename things

the original bilinear map

from Z[x] x Z[x] to Z[x,y]

sending (a,b) to a(x)b(y)

[tribal moss]

Okay.

There are two other maps we need to give names.

[void cosmos]

and now i want to show that this induces a map g from the tensor product (that i dont know of ) to Z[x,y]

such that

the diagram commutes

[tribal moss]

No no no.

[void cosmos]

okay

[tribal moss]

Read my lips.

[void cosmos]

i cant we are on discord

[tribal moss]

You are trying to show that Z[x,y] is the tensor product.

[void cosmos]

okay

[glossy crag]

Yeah, I worked both out, thanks. Separability still feels like overkill though, it's enough for the minimal polynomial to have more than one root, no?

[void cosmos]

so the top right of the diagram is actually Z[x,y]

[void cosmos]

okay but whats C here

is C also Z[x,y]

[rocky cloak]

I don't know if that's the main reason they're useful, but it is true

[tribal moss]

You have defined a map f: Z[x] × Z[x] -> Z[x,y].
Your adversary now comes with a module C and a bilinear map g: Z[x] × Z[x] -> C.
Your task is to produce h: Z[x,y] -> C such that g = h o f.

[void cosmos]

i thought this C was Z[x,y]

and i show that the tensor product is homomoprhi to that C

and has inverse so its iso

[tribal moss]

No, C can be anything the adversay fancies.

[void cosmos]

thats what i had in my mind all along

[rocky cloak]

Minimal polynomial over k(b) should have more than one root. But I guess it would be akward to frame the hypothesis that way

[glossy crag]

Maybe not though, what we need is for the minimal polynomial of alpha over K(beta) to have more than one root, so there is probably no simple way of guaranteeing it besides assuming separability.

[glossy crag]

Yep, came to that conclusion myself.

[delicate orchid]

yes, and in this diagram the Z is completely arbitary

[void cosmos]

lmfaoo

[tribal moss]

What I called C is called Z in that diagram.

[void cosmos]

damn

i was literally in koko land

[tribal moss]

Not to be confused by the Z that means "the integers".

[void cosmos]

i thought this C was the thing i was trying to prove the tensor product is

yea that makes sense

ie literally any bilinear map that involvles the product induces a linear map that involves the tensor product

whatever it goes

ig

okay

[void cosmos]

but isnt phi always (v,w) --> v tensor w?

[delicate orchid]

interesting what happens if you do put Z[x, y] as Z though, you get an upwards map via the other universal property

oh wait not quite

shame

[tribal moss]

"tensor" here is just another name for the map that the diagram calls phi (and which I called f before).

[void cosmos]

yeaa thats right

they call it tensor AFTER they obtain the image

like tensor is just the image of phi..

yea okay

okay so now i want to show that Z[x] tensor Z[x] is Z[x,y]

[tribal moss]

It's literally just a typographical variant of the name phi, notated with an infix symbol instead of parentheses.

[void cosmos]

so to do that i have to show that for every bilinear map from Z[x] x Z[x] i can find a linear map from Z[x,y] to the same modfule or group o rwhatever

[tribal moss]

Yes.

Apologies for trying to define new names for everything. I was annoyed by you calling them all f. Shall we use the names in the diagram instead of the ones I proposed?

[void cosmos]

yea sure

didnt mean to annoy you

you can say im new in math

so forget it

problem is tho..

since Z is arbitrary

how tf do i define the map h

[tribal moss]

The map h is given to you by an opponent.
The map you should be trying to define is h-tilde.

[void cosmos]

yea

yea yea i figured but

h tilde is going to be composed with phi

such that its = h

so i need to know what h does XD

[tribal moss]

You can't.

[void cosmos]

yea cuz its arbitrary ..

cool

[tribal moss]

h is something arbitrary that your opponent chooses!

Now the domain of htilde is Z[x,y], and the way to define a Z-linear map to the opponent's module (which I'm going to keep calling C because Z is in use already) is to define what htilde(v) is for each v in a basis for Z[x,y]. Do you know a basis for Z[x,y]?

[void cosmos]

x and y

ig

like powrs

powers*

[tribal moss]

That would mean that you can write any element of Z[x,y] as a linear combination of x and y.
How would you do that for, for exampe 3x²y + 5?

[void cosmos]

yea ayea mb i didnt mean like just x and y

more like {x^1,x^2,....}

and y aswell

[delicate orchid]

what about tropo's example

[void cosmos]

yea lmao mb boys

{x_i * y_i}

or maybe diff indices

[delicate orchid]

x^i * y^j but yes

[void cosmos]

yea the indices are meant to like

index the basis for Z[x] and Z[y]

like x_1 is first element in basis of Z[x] and etc

mb

[tribal moss]

Oh, I thought we were talking about powers of the variables?

[void cosmos]

ugh its just my stupid notation

{x^1,x^2,x^&3...} is basis for Z[x]

and similary for Z[y]

so x_i would be like ith element in Z[x] basis whcih would be x^i or x^(i-1)

u get me

[tribal moss]

I'm not sure, it sounds like you're going in an unhelpful direction.

[void cosmos]

okay forget it

the basis is {x^i*y^j for i,j in N}

cool?

[tribal moss]

Yes, if we agree on what that means.

[void cosmos]

yea

okay so the hting is

i dont know hwo to define this h tilda

at all

cuz

i dont know where its supposed to go

[delicate orchid]

you can just define where it sends these basis elements

[south patrol]

Tbh it's probably best to show that Z[x,y] is isomorphic to Z[x] (x) Z[y] as Z algebras [that is, rings lol]

[void cosmos]

yea but where

[south patrol]

And that's easier in fact imo lol

[delicate orchid]

my mind is cast back to many a moon ago

[void cosmos]

to some Z-module C ig

[tribal moss]

We've just spent an hour on setting up a different approach ...

[void cosmos]

😦

[south patrol]

Oh lol oop

[delicate orchid]

well, now we can ask a similar question for Z[x] x Z[y], what would the basis of this one be?

interrupt me if this isn't where you were heading tropo

[void cosmos]

yo just send x^i*y^j to h(x^i,y^j) lol

[south patrol]

As Z-modules I'm pretty sure Z[x] and Z[x,y] are isomorphic anyway right

so lots of structure lost

[tribal moss]

But just to make sure, am I still right that we're talking about tensor products of Z-modules rather than tensor products of Z-algebras, @void cosmos?

[void cosmos]

yea im with you man

algebras are the next 2 sections

i knew this was going to be rough af

[void cosmos]

wouldnt it be the cartesian product of the two baseses ?

[tribal moss]

I hope what you mean is send x^i·y^j to h(x^i,x^j)?

Because h doesn't know anything about y's.

[void cosmos]

wait what did i saay

oh you mean ys

yea yea mb

but isnt Z[x] and Z[y] the same?

but yea yea ur right

its better this wya

yea

so now i need to show that this is LINEAR right?

given h is bilinear

right?

[tribal moss]

Okay, so with that specification you've actually managed to define a htilde.

This htilde is implicitly linear, because we're tacitly using "htilde is linear" to explain what it does to things in Z[x,y] that are not basis elements.

[void cosmos]

yea

so we should be done right? we just have to prove that this commutes with the diagram

and is a homomorphism? ( which it is cuz its linear? )

[tribal moss]

Yes, for modules, "linear" and "homomorphism" is the same.

[balmy belfry]

HI

YOU GUYS ARE TALKING ABOUT HOMOMORPHISMS?

[tribal moss]

What we still need to see is that (a) htilde o phi = h, and (b) there's only one htilde with this property.
Of these (a) means we need to show htilde(phi(p,q)) = h(p,q) for every p and q in Z[x].

[delicate orchid]

alright captain capslock simmer down

[balmy belfry]

xd

[tribal moss]

Indoor voice, please. But yes, most of abstract algebra is about homomorphisms.

[balmy belfry]

is there one overall definition of "homomorphism" or are they each different for each structure

[delicate orchid]

a structure preserving map

[balmy belfry]

like people tell me group homomorph and linear maps are not equal

[delicate orchid]

or an arrow in the corresponding category

[balmy belfry]

yet linear maps are homomorphisms?

[void cosmos]

okay but we havent defined phi right?

[tribal moss]

Different for each structure, but with common themes that are mostly informal at this level.

[balmy belfry]

hm alr

[delicate orchid]

you're ignoring the big word "group" before "homomorphism" for unknown reasons

[tribal moss]

Yes, but we were calling it f at the time.

phi(p,q) = p(x)q(y)

[void cosmos]

okayy

okay now

bare with me here please

[balmy belfry]

so homomorphism isnt just some type of map that linear maps happen to fall into, its a thing that changes from thing to thing and homomorphism and linear map are actually equal in its field

idk that just sounds complicated

[void cosmos]

now this is was to show that the tensor product IS Z[x,y]

[rocky cloak]

A "homomorphism of Xs" is a map that preserves X-structure.

So a homomorphism of groups preserves group structure, a homomorphism of vector spaces preserves vector space structure, etc

[void cosmos]

this ISNT showing that the tensor product is isomorphic to something else right?

to do that iw ould then put that something else as my C?

[balmy belfry]

ah

what does it mean preserve though?

[void cosmos]

like what 5.2 is saying here ?

[balmy belfry]

i mean the sets it connects are already their structure

[delicate orchid]

it commutes with any operator on the structure

[balmy belfry]

what does the map between them have to do with anything

[delicate orchid]

\phi(x)\phi(y) = \phi(xy)

\phi(nx) = n\phi(x)

so on

[balmy belfry]

oh

[tribal moss]

As far as I'm concerned, the only tensor product we're working with here is Z[x,y], and there is no isomorphism between that and any other tensor product involved in the discussion here.

[void cosmos]

yea yea

i meant now in general

cuz

in general

what we were doing rn is showing that tensor product IS none other than Z[x,y]

but if i were want to show that for example

Z/nZ tensor Z/mZ is Z/(n,m)Z

isomorhic

would i take my C to be Z/(n,m)Z?

and find an inverse map?

[delicate orchid]

no, again

oh wait that could work

[balmy belfry]

i mean i get these are the conditions of the homomorphism, but why "preserves" operations, what operations? Vector spaces have associatity and commutitivity on addition, how does the map "preserve" that, it has no operations of its own

i just dont get it

[void cosmos]

look at this guy

[delicate orchid]

I think this is what I was getting at earlier by somehow being able to leverage the universal property of Z[x, y] to get a natural inverse to Z[x] (x) Z[y]

[void cosmos]

this guy took his C to be B

[tribal moss]

You would then let Z/(n,m)Z be the module in the top right of the diagram you showed.

[void cosmos]

and then just said oh and cuz the universal property says so we find a homomrphism instantly

[void cosmos]

okay can u check out the last thing i showed

[void cosmos]

.

[delicate orchid]

those are conditions on the operator, not the operator itself

[void cosmos]

this proof

[delicate orchid]

by commuting with the operator all of those conditions naturally hold - assuming you're mapping into something with the same structure

[void cosmos]

didnt he take his B to be the C of the diagram?

5.2 is above

[delicate orchid]

this is the part where I really just want to pull the category card lol
but that's circular and unhelpful even if it's easier

[open sluice]

the linear transform of a sum is the sum of their linear transforms

[void cosmos]

like i think he took what he wanted to prove the tensor product is to be his C

so that he gets the homomorphism fro mthe property for free

and then constructs an inverse

[open sluice]

so adding in vector space 1 works similarly in vector space 2

[balmy belfry]

😭 i dont understand, T(a + b) = T(a) + T(b)
and T(ax) = aT(x) are just arbitrary conditions we tacked on, how do these axioms force the map to preserve anything

[tribal moss]

Oh, the rings are not necessarily commutative?

[delicate orchid]

a vector space is a set with the operators of addition, and scalar multiplication

[void cosmos]

yea we are using bimodules here

[rocky cloak]

You can think of the operations as giving the thing its structure. For example if x, y and z are vectors such that x+y = z, that says something about how x, y, and z relate and says something about the structure of the space.

Then a structure preserving map should also satisfy
f(x) + f(y) = f(z). I.e. the image of f has "the same structure".

[delicate orchid]

a linear transformation respects those operators, by which I mean exactly what you wrote

it doesn't matter what order you do them

[void cosmos]

thats what i was trying to do

in the first place

so what did he do

[delicate orchid]

learning about a comm alg operator in a non-comm alg setting

b-0b-b-b-b-bb-ased

[balmy belfry]

oh

[delicate orchid]

and wdym what did he do can u not go read it

[void cosmos]

u read

what i said

and asked

[tribal moss]

That looks like he's phrasing what he's proving as "we already have an R otimes B from god knows where, and now we will show it happens to be isomorphic to B".

[balmy belfry]

is this for linear maps or all homomorphisms

[delicate orchid]

the same principle applies for all homomorphisms

[tribal moss]

"linear map" is another word for "vector space homomorphism".

[void cosmos]

yea tahts what i meant as my questioni think

how is that different from what i tried to do?

[balmy belfry]

thanks

oke bye

[delicate orchid]

he's going from R \otimes B to a concrete module (B in this case)
you were doing the other way around

[rocky cloak]

So like I said before a "homomorphism of Xs" should preserve the operations that define the structure of X. The structure of a vector space is defined by addition and scalar multiplication, so a homomorphism of vector spaces (aka a linear map) should preserve those.

A group of defined by its group operation, so a homomorphism of groups should preserve that

[void cosmos]

mhmmmmmmmmm

yea that makes sense ig

[tribal moss]

There are two different ways to achieve the same end result:

1. suppose we already have something that satisfies the universal property, and show that such-and-such is isomorphic to it.
2. show directly that such-and-such itself satisfies the universal property.

These are really the same thing (because the way universal properties work, anything that is isomorphic to something with an universal property has that universal property itself), but the details of the proof will be worded differently according to which of them we say we're doing.

[void cosmos]

okay now last question im sorry

but why

isnt what i wastrying to do

finding a map from Z[x] x Z[x] to Z[x,y]

the same

[void cosmos]

like it was the same way as this proof

ig

[delicate orchid]

a mfin commutative triangle in that proof would make it so much easier to read it's unreal

[tribal moss]

I am not quite sure what you "were trying to do" -- I only came in after Moth said "universal-property it", and that is the starting point for everything I've been saying.

[void cosmos]

yea

[void cosmos]

.

[void cosmos]

yea i think now i get what u wre trying to say

[delicate orchid]

we were showing that Z[x,y] transforms like Z[x] (x) Z[y], satisfying the definition that a tensor is something that transforms like a tensor

[void cosmos]

yea

[tribal moss]

Aaargh

Get out!

That's not what "transforms like" means.

[delicate orchid]

I know I'm just doing a little trolling

that "defintion" is more for the physical applications of tensors

[void cosmos]

tropo if i were to find an inverse to the map induced (from the tensor product to Z[x,y] ) from finding a bilinear map Z[x] x Z[x] to Z[x,y]

[summer path]

wew doing his best to make timo right

[void cosmos]

would that be the same as showing its isomorphic to the tensor product?

ie approach number 2?

[delicate orchid]

oh timo was completely correct

[void cosmos]

and is that what he did in the proof i sent?

[delicate orchid]

well, constructing an isomorphism between two things tends to show they're isomorphic