groups-rings-fields

[next obsidian]

You can just call the tensor product anything that satisfies the universal property

No basis involved

You only need a construction to show that there’s anything which has the universal property

[sinful mirage]

exactly

[next obsidian]

So?

You have a construction

Now you can throw it away

And just deal with the tensor product in terms of universal property

This has nothing to do with a basis anymore

[sinful mirage]

I mean sure,if I can prove that this construction exists

this is what i'm struggling with

[next obsidian]

But… didn’t you already say you saw it?

Via a basis

[sinful mirage]

via a basis yes

[next obsidian]

But

That’s my point

Once you have it via a basis

You have a basis free construction

By universal property

There’s no point to construct another tensor product without a basis

[sinful mirage]

how do I write it down basis-free then?

if I had a basis a priori

[next obsidian]

Literally via the universal property

You say the tensor product is an object such that it has a map from V x W -> V (x) W

Which uniquely factors bilinear maps

Universal property nonsense says anything which satisfies this is unique up to unique isomorphism

[sinful mirage]

but my professor said that what we know by choosing a basis should be a corollary of the proof of existence basis-free

[next obsidian]

I mean sure, but there’s no point to doing that

Like I don’t know what you gain by showing its existence basis-free or with a basis

[sinful mirage]

ohh

[next obsidian]

Either way you end up with a basis-free tensor product

[sinful mirage]

so i can choose a basis

and say

up to iso,I found the object i want

[next obsidian]

Yes

Now you can throw away the specific construction

And argue just in terms of the diagram

It’s like you’re defining the tensor product to be anything which has this one universal property, this is well-defined up to a unique isomorphism

You just need to know there’s at least 1 thing which actually has this property

Then you can throw away the way you made that and just go back to your definition in terms of universal property

[sinful mirage]

yeah makes sense

[next obsidian]

This isn’t even an integral domain

x^3 + x is reducible

x(x^2 + 1)

So you have zero divisors

No worries haha

[unique berry]

Texas A&M university

So I know it's really hard to see but does the bar over x and y in p(x) |--> p(y+1) mean anything?

[next obsidian]

It just means that

This isn’t a “free” variable

Because in the quotient it has relations

So this is to try and like, remind you that “hey x^2 = -1”

Because inside the quotient this is true

[sinful mirage]

I found a book,which does it nicely

except one thing,which I do not understand 😢

what's the universal mapping problem?

[next obsidian]

Factoring bilinear maps uniquely

[sinful mirage]

can someone help me understand the generating property?

f in V,that means f is a function with finite support.

why is every function with finite support of the same form?

I don't understand their proof

[thorn delta]

I’m not quite sure what you mean by generating property, but the idea is that f is non zero in only finitely many places. Let’s say f(x) = c != 0. Then f has the same value as the characteristic function (c)chi_x at x, and is 0 everywhere else. Repeat this process everywhere f is nonzero, add these characteristic functions together, and that’s it. Every function of finite support is a linear combination of characteristic functions

[thorn delta]

All functions of finite support “have the same form” only in the sense that they are all linear combinations of functions in a common basis, just like in any vector space, as far as I can see anyway

[sinful mirage]

where they say g(x)=0 if x in X/(x_1,...)

how do they get that from?

using characteristic function property?

[thorn delta]

Yea, characteristic functions are zero outside of the set that they “select to be 1”. Just plug in a value in X \ {x1,..} and check

[sinful mirage]

yes,makes sense

[sinful mirage]

my last confusion:do you agree with this proof?

I am extremely confused why they introduce g

[thorn delta]

I think they just called it “g” because we don’t know beforehand that f = g and they wanted a quick way to say “if x in X \ {x1,…} then f(x)=g(x)=0”

[sinful mirage]

but we do not need it,do we?

[sinful mirage]

isn't this legit?

[thorn delta]

Ye

[simple mulch]

May I ask a quick question?

[thorn delta]

In general when you have a long complicated function, it’s common to just write g = [that mess] to save space and extra writing

Nothing deep behind it @sinful mirage

[sinful mirage]

okay,makes sense,thanks!

[thorn delta]

Npnp

[sinful mirage]

sure,my confusion got clarified,don't worry

[simple mulch]

so basically, T is injective if and only if (if S is linearly independent then T(S) is linearly independent)?

[sinful mirage]

yes

more precisely: if S is linearly independent in V, then t(S) is linearly independent in W

[simple mulch]

thanks man!

[sinful mirage]

how to prove existence?

[thorn delta]

Think about how a matrix is defined by (1,0,0,…,0) mapping to the first column, (0,1,0,….,0) mapping to the second column, etc

[sinful mirage]

I intuitively have the idea,but I do not know how to write it down formally

it follows from every vector in V can be decomposed using v=v_i e_i

and if i know the map on e_i,so will i on any v

but idk how to formalize this into proper mat

h

like how to write down t explicitly

t(v)=sum_i v_i f(e_i)?

[thorn delta]

Ah okay. Let v be in V then v = sum ci vi, a finite linear combination of basis elements, define T(v)= sum ci f(vi)

[sinful mirage]

[thorn delta]

Then you have to show this is linear and well defined

[sinful mirage]

and now I use linearity of f to prove linearity of t

what do you mean by well-defined?

[thorn delta]

The value of T(v) depends on how you represent v as a linear combo of basis element a priori. So you have to check T(v) has one and only one unique output

[sinful mirage]

uhm,what do you mean

given a basis,the decomposition is unique

[thorn delta]

Yea, I mean that’s all there is to it

You should see why this isn’t the case when S is some linearly dependent set

Cuz when S is Lin dependent, you can write 0 in different ways so T(0) is not well defined

[sinful mirage]

for linearity, t(v1+v2)=sum_i(v1_i+v2_i) f(e_i)=sum_i v1_i f(e_i)+sum_i v2_i f(e_i)=t(v1)+t(v2)?

[thorn delta]

Yep

[sinful mirage]

[sinful mirage]

I proved that over any set,a vector space based on it exists

how do i prove uniqueness?

[thorn delta]

uniqueness of V or uniqueness of T?

[sinful mirage]

uniqueness of the pair (V,i)

the vector space V

so a vector space based on a set,is a vector space together with a map i, so that ...

I want to prove that this structure is unique

because the proof of existence of tensor products,uses as first words: Let ... be a set and ... be the vector space based on the set.

[thorn delta]

alright so usually in this setting, when we say "unique," we mean "uniquely isomorphic." So If (V, i), (V', i') are vector spaces based on X, then you want to show their is a unique isomorphism between them

[sinful mirage]

yes,exactly

but how do I do that?

sorry,I don't have practice in this

[thorn delta]

np, so lets make a map V --> V' and show its an isomorphism (i.e. has an inverse)

[sinful mirage]

so,wait,we need 2 ingredients in all proof

right?

1. isomorphism V->V', and 2) this isomorphism preserves i, in the sense map(v->v')(i)=i'

right?

and now we develop first ingredient

[thorn delta]

well, 2) is not necessary. We just want to show that V and V' are uniquely isomorphic

[sinful mirage]

why?

by preserving the structure,we mean it preserves i too

the structure here is (V,i),not just V

[cloud walrus]

ProphetX

ProphetX

[sinful mirage]

why is it different here?

[thorn delta]

im familiar with uniqueness of the tensor product, but I'm not sure exactly what you mean by "we want the \otimes to be preserved by the same linear map as V."

[sinful mirage]

st=id_v => t gamma=delta

and there is a typo in ii) it should be S delta=gamma,instead of t

i.e st=id_v => s delta=gamma

this is my point, s preserves both vector space structure and tensor product structure( otimes or delta denoted here)

https://mobt3ath.com/uplode/books/book-34422.pdf

[thorn delta]

alright, i know what you mean. Only S and T are needed to show that the vector spaces themselves are uniquely isomorphic, but you can also show that delta factors uniquely through gamma and gamma factors uniquely through delta

I kind of need to think about precisely in what sense ii) is important, but that won't stop us from proving this

[sinful mirage]

yes

agree

so we need to show first that V,V' are iso

then we have t show that i factors uniquely through this iso to i'

right?

[thorn delta]

anyway, constructing V --> V' goes like this:
we have i' : X --> V', and by the property of V being based at X, we get a unique map T : V --> V' such that Ti = i'

[thorn delta]

yea

[sinful mirage]

yes

[thorn delta]

then we can go the other way:
we have i : X --> V and by the property of V' being based at X, we get a unique map S : V' --> V such that Si' = i

take, Si' = i, and compose T on the left of both sides:
TSi' = Ti = i'

[sinful mirage]

ohh

this proof is very similar to tensor product uniqueness proof

only difference is,here the maps are not bilinear,just maps

and the unique induced map is linear

correct?

[thorn delta]

right

[thorn delta]

that's because this "uniqueness of objects with universal properties" pretty much all have the same proof. its a category theory thing

[sinful mirage]

I will try to do the proof similiarly to the tensor product proof,given your hints

[thorn delta]

aight sounds good!

[sinful mirage]

do you agree so far @thorn delta ?

and now I can go to the proof of uniqueness of (V,i), using this proposition and your hints

[sinful mirage]

basically this proposition helps me conclude ts=identity

[thorn delta]

looks good

[sinful mirage]

[lunar lintel]

just learned about quotient groups and am unsure how to find elements of G/H where G=(Z_12,+) and H=<[4]> (cyclic subgroup generated by 4)

[thorn delta]

<[4]> would be the identity for example

[lunar lintel]

wym? the entire subgroup is the identity of G/H?

[viscid pewter]

yeah

the elements of G/H are the cosets of H

so the identity of G/H is 1H

or just H

[lunar lintel]

so would the other elements just be 1+H,2+H, and 3+H?

[sinful mirage]

does this look good @thorn delta ?

[thorn delta]

good except for this

[sinful mirage]

why?

[thorn delta]

TT' = id_{V'}, not TT'i' = id_{V'}. In other words, you should have said: TT'i' = i', and by the proposition, this implies TT' = id_V'

[sinful mirage]

completely true

i messed up

i'll correct, sec

nice

now I finally have all the tools the book uses to prove existence of tensor product. it tok me 7 pages of handwriting to get there

now I can start proving existence of tensor product

[thorn delta]

hmm all this to prove existence of the tensor product?

[sinful mirage]

yes

the proof is like 4 pages in the book

https://mobt3ath.com/uplode/books/book-34422.pdf

[chilly ocean]

[thorn delta]

okay wait I guess you'll need this "based space" thing. its like their version of free vector space on a set (I shouldn't say version, its the exact same thing, different name)

[sinful mirage]

yes

the only book I found which does not use free vector space notion

and introduces every concept from 0

probably this is the free vector space introduced from beginning 😄

but I never heard that notion 😢

[chilly ocean]

so learn it

[thorn delta]

you don't technically even need this to show existence of the tensor product either

(omg im finally "active")

for vector spaces, you can define V \otimes W to be the space of bilnear maps V' x W' --> F iirc (the primes ' meaning dual space)

[sinful mirage]

I think this was a consequence of the proof

but i might be misunderstanding

[thorn delta]

hold up important edit above im dumb ^

[sinful mirage]

formally here=?if we evaluate them at a point,they become complex numbers?

their notation is so confusing

[sinful mirage]

here they set V=functions with finite support

[sinful mirage]

then here V otimes W (or Z) should be set of functions of finite support on V_1 x V_2

and then vector space by pointwise multiplication and addition

why do they not do that??

[thorn delta]

"formally adding them" just means that sums have no special interpretation like being able to add coordinate wise or something like that
For example, i could consider the vector space based on the set X = {cats, dogs, mice}. Elements of this space would be "formal" linear combinations of the symbols "cats", "dogs", and "mice"

[sinful mirage]

but how is this equivalent to what they were saying?

they were saying we construct a vector space on a set by identifying elements of set with functions of finite support

if this is true,then there is a bijection between functions of finite support and formal linear combinations generated vector space

what would be the bijection?

[sinful mirage]

they constructed the vector space on X using functions on X with finite suppor

i'm so confused

why did they even do the construction then

[viscid pewter]

ah, bugger, it's an additive group

then the identity would be 0+H = H
and the other elements are 1+H, 2+H, 3+H, yes

[sinful mirage]

more than bijection

isomorhpism,cause they're vector spaces

[thorn delta]

you can define f : X --> M_fin(X, F) by
f(x) = chi_x. This induces a map f : Z --> M_fin(X,F), and then you should be able to do this the other way to get an iinverse

[sinful mirage]

how does this induce a map f:Z->..?

what is Z here?

[thorn delta]

same Z from the proof. Z is based at X

[sinful mirage]

and how does this map f look like?

[sinful mirage]

Z based at X is by definition M_fin(X,F)

what we need to prove is that M_fin(X,F) is isomorphic to the formal linear combination set of dogs mice and cats

as vector space

[sinful mirage]

this

[thorn delta]

what do you mean? the dog, mice, cats thing was an example

[sinful mirage]

yes,but my point is,the vector space of formal linear combinations over X must be isomorphic to M_fin(X,F), if what they are writing is true

first line Z=M_{fin}(X,F), ten they randomly write formal linear combinations

how did they get that?

[thorn delta]

I’m not really sure what the point of this step is tbh. I’d have to see the whole proof. In principal, you can work directly with Z to construct the tensor product.

Anyway though X is a basis for Z, so they are really just identifying one basis with another

Make sense?

[sinful mirage]

X is a basis for Z,right,but what's the basis for formal linear combinations?

[sinful mirage]

how can I construct the tensor product with Z itself as M_{fin}(X,F)?

do you have a reference?

[sinful mirage]

actually,i'm not sure about this

a basis for Z is chi_x, not X

but I agree that X gives rise to a basis for Z

but X is not basis for Z a priori

[thorn delta]

"formal linear combinations of elements of X" are the actual elements of a space based at X

[sinful mirage]

yes,this I would like to see why

the isomorphism between formal linear combinations and M_{fin}(X,F)

I can't see why they are iso

[thorn delta]

X is a basis for Z and and Y = {chi_x : x in X} is basis for M_{fin}(X,F).
Do u see the correspondence between X and Y?

this induces an iso between Z and M_{fin}(X,F).

[languid moss]

Confused on where to start. I'm struggling a lot in my graduate course. Anything can help

[sinful mirage]

if we take a map:Z-> M_{fin}(X,F), which maps X to Y,then we are fine

but why can we do this in infinite dimensions?

[thorn delta]

there's no difference between doing this in infinite dimensions and finite dimensions.
The "space based at a set" construction does not depend on the cardinality of the set

[woven delta]

what is the star here, group of units?

[sinful mirage]

it should

[sinful mirage]

here he says cardinality

[woven delta]

(as a hint for part a, try multiplying a formal power series $\sum_{i=0}^\infty a_i X^i$ by another power series $\sum_{i=0}^\infty b_i X^i$ and try to figure out how to make $(\sum_{i=0}^\infty a_i X^i)(\sum_{i=0}^\infty b_i X^i)=1$

[cloud walrus]

'quid

[thorn delta]

well yea, the cardinality of X and B should be the same (because card(X) = dim(Z) and card(B) = dim(span(B))), but infinite vs. finite thing doesn't really matter

[woven delta]

(this gives you a sequence of equations)

[languid moss]

All the units of F[[x]]

[sinful mirage]

so even between infinite dimensional vector spaces there is always a map from basis to basis?

or only in this case,cause the cardinalities are the same

i.e. card(Z)=card(X) and card(B)=card(X) too?

[woven delta]

do you understand my hint?

[languid moss]

Didn't see your hint

let me see

[woven delta]

also I would suggest you try playing around with multiplying formal power series and see what happens, doing that is invaluable for solving this type of problem

[thorn delta]

so even between infinite dimensional vector spaces there is always a map from basis to basis?
there is an isomorphism between any two vector spaces whose dimensions have the same cardinality, so yea, there has to be a map between bases

i.e. card(Z)=card(X) and card(B)=card(X) too?
card(B) = card(X), where X = V1 x V2 x ... x Vn and B is a basis for M_fin(X,F), but card(Z) != card(X). Think about when X is finite and F is an infinite field

[languid moss]

Okay

[thorn delta]

also the replies are about to be slow for a while, prophet

[sinful mirage]

no worries

so we need to see that card(M_{fin}(X,F))=card(Z)

to have the isomorphism you proposed work

[sinful mirage]

why is this so?

[thorn delta]

More specifically it suffices to see card(X) = card({chi_x : x in X}) because X is a basis for Z and the other set is a basis for M_fin(X,F)

[thorn delta]

ok im back

to make this step more formal, Let $Y = {\chi_x : x \in X}$. Then there is a bijection $X \to Y$ given by $x \mapsto \chi_x$. The properties of based spaces imply that this induces an isomorphism $Z \to \mathcal M_{fin}(X, \mathbb F)$

[cloud walrus]

kxrider

[thorn delta]

this is the step youre stuck on, right @sinful mirage ?

[sinful mirage]

yes

thanks!

[thorn delta]

np

[woven delta]

To start you off the first equation that you have is $a_0b_0=1$, the second equation that you have is $(a_0b_1+b_0a_1)X =0X$

[cloud walrus]

'quid

[woven delta]

(the first equation should tell you why a_0 being nonzero is necessary, but it doesn't tell you why a_0 being nonzero is sufficient)

[lavish nexus]

Say I want to find a basis for an alternating bilinear form so that it takes the form diag(S, S,…)

0 1
-1 0
being S.
I think I can have the first two elements of the new basis be the first two of the standard basis (multiplying them by a unit if necessary)

But then for this example I couldn’t find any element in their orthogonal complement that is orthogonal to both except 0

nvm figured it out

[lethal dune]

this this has to do something with semidirect, like some automorphism sending h to g but can't come up with a rigorous argument

[coarse storm]

Hmm... May be construct something that gives (e, u) (g, u) = (h, e), for some u.

[sinful mirage]

can someone help me rewrite this relations,if I only have i=1,2?

i'm a bit confused

this is in the proof of exsitence of tensor product

[sinful mirage]

is it equivalent to this?

[hidden haven]

yes

[sinful mirage]

so why is this an equivalence relation?

i'm confused by their notation

when is (v_1,v_2) equivalent to (v_1',v_2')?

[hidden haven]

This is a subspace

The equivalence relation is "being in the same coset"

[sinful mirage]

then what is the equivalence relation, which generates this subspace?

[sinful mirage]

in which coset?

how is the relation explicitly defined?

[hidden haven]

(v_1,v_2) ~ (v_1', v_2') iff (v_1-v_1', v_2-v_2') is in R

Equivalently they are in the same coset of R

[hidden haven]

ie their difference is in R

[sinful mirage]

how does this lead to the relations they wrote down?

[hidden haven]

An equivalence relation can't generate a subspace

[sinful mirage]

[hidden haven]

oh this is defining the relation from already knowing those elements

when we say a = b is a relation

it is the same as taking a-b as a generator

[sinful mirage]

but then we would have 1 generator,based on this

why we have 4?

[hidden haven]

That relation combines all 4

and all their consequences

The relations that we want are

(av,w) = a(v,w) = (v,aw)

(u+v,w) = (u,w) + (v,w)

(v, w+x) = (v,w) + (v,x)

[hidden haven]

Then apply this

we take their differences as generators

that is how you get the 4 equations

[sinful mirage]

and these are consequence of what you wrote how?

[sinful mirage]

so how do these 4 arise from this?

[hidden haven]

What I wrote earlier is a consequence of all of these

The relation that these equations alone describe is not an equivalence relation

you take the smallest equivalence relation containing this

[upbeat juniper]

as a simple example, (u,2v)~2(u,v) because they differ by an element of R. namely, (u,2v) = 2(u,v) + [(u,2v) - 2(u,v)]. but the both of them are not strictly equal, just equivalent

[hidden haven]

when you do that, you get the earlier thing

[sinful mirage]

wait,so now I have 4 equivalence relations based on this

[hidden haven]

smallest equivalence relation that also respects the linear structure

no

[sinful mirage]

I will try to prove that these 4 are equiv rels

[hidden haven]

no

We say 2 things are related

if they are related by any one of those equations

[hot lake]

x ~ y <=> x-y is in U

where U is the subspace of Z generated by the elements given

[sinful mirage]

so (v1,v2) related to (v3,v4) if and only if (v1,v2)=a(v3,v4)=(v3,av4) or ...?

[hot lake]

no

[hidden haven]

yes, but then you do this

[hot lake]

(v1,v2) is realted to (v3,v4) <=> (v3-v1,v4-v2) is in the subspace generated by blablabla

[sinful mirage]

ok but how do i define subspace generated by blabla

to have a subspace generated by smt i need an equiv rel

[hidden haven]

you extend the relation to make it an equivalence relation respecting the linear structure

[upbeat juniper]

in particular this means the difference can be a sum of more than 1 of those 4 expressions

[hidden haven]

and you do that exactly by taking cosets of the subspace generated by those elements

[sinful mirage]

so,accepting this

how do i prove this is an equivalence relation?

[hot lake]

there are several ways to define the subspace generated by stuff

[sinful mirage]

which would be useful here?

[hidden haven]

subspace contains 0, is closed under addition and subtraction

[hot lake]

for any subspace U of Z, you can get an equivalence relation on Z by defining x ~ y <=> x-y is in U

it's just a matter of writing the axioms

of equivalence relations and of linear subspace

reflexivity : x ~ x <=> x-x is in U <=> 0 is in U, that's an axiom of a subspace

symmetric : if x ~ y then x-y is in U, we want y-x is in U so that we can conclude y ~ x, that's just U being stable by taking additive inverses

[sinful mirage]

being stable means?

[hot lake]

transitivity is also given by the axiom that U is stable by addition

it means if x is in U then -x is in U

[sinful mirage]

so if x~y and y~z, then x-y is in U and y-z is in U. then x-y-y+z is in U, because subspace is closed under -

right?

hence x~z

[hot lake]

no

you do (x-y)+(y-z) to get x-z

not (x-y)-(y-z)

[sinful mirage]

ah right,lol,my bad

I use subspace is closed under +

[hot lake]

yeah

[sinful mirage]

ok yeah,now I see

so I arrived finally after 1 and a half day to defining the free vector space quotient this equiv rel

[hot lake]

you can define the subspace generated by a set S by saying it's the set of all finite linear combinations of elements in S

[sinful mirage]

now I need to see why this does the job for existence of tensor prod

[hot lake]

where you do all the operations in Z

so you take all scalar multiples of the generators

and then you can do all the finite sums of those

and that gives you the elements of U

you can also say that U is the intersection of all subspaces of Z containing S but it's a little less concrete

[hot lake]

with the universal property for tensor products ?

[sinful mirage]

yes

I proved uniqueness,and now I am proving existence

took me 9 pages so far to even understand what free vector space is and why it always exists

now getting through defining the equiv rel and proving that it satisfies universal prop

3pages of proof to go through

[hot lake]

that's a lot of pages

[sinful mirage]

but at least I will have a eslf contained handwritten very explicit document of existence and uniqueness of tensor product

[upbeat juniper]

idk if it's just me but this exposition looks very... offputting

[sinful mirage]

it's the only exposition, which I found that builds it up from 0

i.e. 0 prereqs

they define everything and all theorems and proofs used in the construction

[upbeat juniper]

oh

[sinful mirage]

other books used notion of F[X] immediately from first line

and I had no idea what F[X] is and why it even exists

[hot lake]

at least it's a textbook in mathematics and not a textbook for physics

[sinful mirage]

but this book proved it

it's a pretty self-contained book in my opinion

i like it

homological algebra books be like:take the free module, 0 def

[hot lake]

yeah

did it define quotient spaces somewhere

[sinful mirage]

yes

[hot lake]

nice

[sinful mirage]

what does adding U mean??

(v_1,..,v_m)+U?

I add a set to a point?

[hot lake]

remember the definition of a quotient set

elements of Z/U are cosets of U

or equivalence classes for the relation x~y <=> y-x is in U

[sinful mirage]

yes,but a coset is gU

why v+U?

[hot lake]

in vector spaces a coset is v+U

[sinful mirage]

a left coset of U,is gU

ohhhhh

so gU=g+U in vector space

where g is an element of V

[hot lake]

gU is for when you are doing groups and you write the operation with a multiplication

[sinful mirage]

yes,this is what i was used to

[hot lake]

well then it's the same but the group operation is +

and + is commutative so left coset = right coset

[sinful mirage]

could you help me to see why this map is linear in the first argument?

i'd try to do linearity in second argument

in book it's extremely confusing

i can show how I tried

[hot lake]

10.1 or 10.2 ?

[sinful mirage]

I don't understand either of tem

this is what I understand

[hot lake]

first you have to know the definition of the map and the definitions of scalar multiplication

[sinful mirage]

and this should be equal to tensor(v_1,v_2) + tensor(v_1',v_2) somehow

but idk why

[hot lake]

you want f(v1,w) + f(v2,w) = f(v1+v2,w)

[sinful mirage]

yes,where f=otimes

I started from your RHS

and try to get LHS

[hot lake]

by definition of f, that's ((v1,w) + U) + ((v2,w) + U) = (v1+v2,w) + U

by definition of + in Z/U, that's ((v1,w) + (v2,w)) + U = (v1+v2,w) + U

(the leftmost + is now + in Z)

[sinful mirage]

why?

we have 2 U on LHS

how did it turn to U

[hot lake]

I mean I rewrote f(v1,w) into (v1,w)+U

[sinful mirage]

yes

[hot lake]

and f(v2,w) with (v2,w)+U

[sinful mirage]

(v_1,w)+U + (v_2,w)+U

[hot lake]

and you get ((v1,w)+U) + ((v2,w)+U)

[sinful mirage]

but why is this equal to (v_1+v_2,w)+U?

[sinful mirage]

yes

[hot lake]

where the innermost + is addition in Z/U

[sinful mirage]

I don't see how you got this

from which def?

[sinful mirage]

so far good

[hot lake]

then you use theorem 1.11

[u]W is notation for u+W I think ?

yes

so you use theorem 1.11

[sinful mirage]

ok,yes

[hot lake]

and that tells you that ((v1,w)+U) + (in Z/U) ((v2,w)+U) = ((v1,w) + (in Z) (v2,w)) + U

next we say that (v1,w) + (v2,w) - (v1+v2,w) is in U

by definition of U as a subspace that contains this very element along many others

and so ((v1,w) + (v2,w)) + U = (v1+v2,w) + U

and this shows that f(v1,w) + f(v2,w) = f(v1+v2,w)

[sinful mirage]

yes

could you please help on the scalar multiplication in the first entry too?

I'll do this for the second entry

so that really is bilinear

[hot lake]

it's the same but instead of using the definition of + in Z/U and the fact that U contains (v1,w) + (v2,w) - (v1+v2,w)

you use the definition of scalar multiplication in Z/U

and the fact that U contains c * (v,w) - (c*v, w)

[sinful mirage]

so you mean here theorem 1.11

[hot lake]

yes

[sinful mirage]

okay,great

i'll prove first theorem 1.11 and then get back to my proof

thanks

[hot lake]

you didn't prove theorem 1.11 yet

[sinful mirage]

no,I just saw it for the first time you showed me

[hot lake]

chapter 10 did tell you quotient spaces were crucial to understand

[sinful mirage]

oh lol,lemma 1.1 is exactly the theorem you proved

[sinful mirage]

here

[hot lake]

yeah about how subspaces make great equivalence relations

[sinful mirage]

wow,reading this makes everything have much more sense now

I just realized that I did not understand properly the definition of quotient space

and just to be completely sure and precise

here W=vec subspace

not just set subspace

[hot lake]

yeah subspace means vector subspace

[ancient rivet]

"The multiplicative group of millennials modulus the color green is homeomorphic to which baseball team?"

[serene radish]

Can someone help me on this? If I have a k-cycle, and a k>m-cycle whose elements are subset of elements of a k-cycle, is their product a single k >= t-cycle?

[hot lake]

sounds unlikely

[serene radish]

That's sad

[sinful mirage]

the nicest thing in math is that I can always ask why and most probably find an answer

def. x, or thm y. or lemma z.

It's just I need to have patience to search the answer

in physics we just handwave a lot

[hot lake]

also, closure under additive inverse (that I mentioned earlier) is obtained by using c= -1 in the closure under scalar multiplication property

[sinful mirage]

so here,just to make sure, [u]w+{coset}[v]_w=[u+v]w means (u+w)+{coset}(v+w)=(u+v)+w

right?

where +coset=+ on coset space

[hot lake]

= (u+v)+W

[sinful mirage]

ah

sorry

yes

my last equality is wrong

[hot lake]

well it's still true

[sinful mirage]

it's true,but not needed

[hot lake]

anyway it talks about equivalence classes

[sinful mirage]

hmm so it's basically just the definition of coset vector space,lol

or well,the proof that it is a vector space

this is what I was missing

[hot lake]

and lemma 1.2 says that the equivalence class of u is u+W

[sinful mirage]

yes,that one I proved

[hot lake]

where u+W = {u+w | w in W}

[sinful mirage]

ah,i should be way more careful with w W

cause it might be misunderstandable if I write a coset as u+w

it is u+w, for all w in W, not just u+w

[hot lake]

no it is {u+w | w in W}

[sinful mirage]

what's the difference,between this nad what I said?

[hot lake]

u+w is some vector

u+W is a set of vectors

[sinful mirage]

ah yes

theorem 2.16 is the one saying that cosets are equivalent to equivalence classes

do you guys agree with my statement,that (u+W)+(v+W)=(u+v)+W is a consequence of teorem 2.16?

[sinful mirage]

basically I think it's this reformulated more explicitly

[hot lake]

did you prove that + on V/W is well defined ?

[sinful mirage]

no

what does that mean?

[hot lake]

it means proving that if you pick other representatives for [u] and [v], it doesn't change the result of +, aka [u+v]

basically when we define a function we want to define a relation that for each input has only one output

[sinful mirage]

i only heard the term well defined for linear maps

[hot lake]

the way we defined + on V/W was by saying that for any u,v in V, (u+W) + (v+W) should be (u+v)+W

[sinful mirage]

that they give same result independent of basis

[sinful mirage]

yes

[hot lake]

so the actual relation is defined by

x+y = z if there exists vectors v,w in V such that x = v+W and y=w+W and z=u+v+W

when you want to prove that this is the relation of a function

you have to prove that u+v+W doesn't depend on the choice of v and w

v can be any element of x

w can be any element of y

how come v+w+W is independent of those choices ?

well you say that any two choices v and v' differ by some ... uuuuh ... k in W

and any two choices w and w' differ by some k' in W

my choices of letters are terrible

[sinful mirage]

no worries

[hot lake]

and then we want to show that (v+w) ~ (v'+w')

[sinful mirage]

they differ by k and k' by def of them being in the same equiv class

right?

[hot lake]

that is, v+w-(v'+w') has to be in W

[sinful mirage]

this is why i am allowed to chooes k and k'

[hot lake]

but that's (v-v')+(w-w')

= k+k'

and that's in W

because W is closed under addition

everytime you want to define a function on a quotient space V/W

you actually first define a function on V and then prove that the result doesn't change if you change the input by something in W

[sinful mirage]

i'll finish the proof of + satisfying the axioms of vector space,then write down well definedness

and then try the same for scalar multiplicaton

i'll write how I think scalar multiplication well-definedness should be checked

hope I got it right

[hot lake]

if you try to define the quotient group G/H for a subgroup H that is not normal in G, you won't be able to prove that the group law on cosets is well-defined

so you should never skip proving that some operation is well-defined

[limpid edge]

why do we speak of “modding out” and “modulo” our normal subgroup when discussing quotient groups

I know that in Z/nZ you do “modulo arithmetic” so there’s that

but still

[hot lake]

because we already do that when talk about quotient groups of (Z,+)

[limpid edge]

oh is that literally it

[sinful mirage]

is this proof ok?

the well definedness

[median pawn]

Have you seen this set and these operations elsewhere?

Not sure if they mean anything. By the way I've checked that it's a commutative ring with identity (1,0), it is also an integral domain. Haven't checked that it's a field yet

[cursive temple]

those are basically ||complex addition and multiplication||

[median pawn]

oh lol

that was a good one

then it is in fact a field also

[chilly ocean]

[limpid edge]

lol that reminds me of this math comp. problem

[weak oriole]

Big bren

[sinful mirage]

@hot lake I just had a big revelation

the 4 relations were exactly chosen so that otimes is bilinear

bilinearity is a direct consequence of the relations chosen

so mindblown

a priori I had no idea why we choose them as such, and now after the proof I see they are the reason otimes is bilinear

[chilly ocean]

youre talking about the tensor?

[thorn delta]

Yea basically tensor just turns bilinear things into linear things

[sinful mirage]

can someone help me to see why from the univ property it follows what they claim?

I don't see why S restricted to X is f

S is a map from the vector space based on Z to W

f is a map from the set(which the vector space was based on) to W

they have different domains

they only map to the same thing

[thorn delta]

think of the map i : X --> V as an "inclusion." it sends elements of X to themselves, but inside of V

[sinful mirage]

why is it inclusion?

thats so abstract

how can it be inclusion if V is a vector space X is not

ohhhhhhh wait

is it because,the elements of V are formal linear combinations

so the trivial combinations with coefficients 1=X?

so X is a subspace of V?

does this make sense?

[thorn delta]

yea, its an inclusion of sets, so for example, if X = {cats, dogs, mice}, we could have i : {cats, dogs, mice} --> span{cats, dogs, mice}
defined by i(cats) = cats, i(dogs) = dogs, i(mice) = mice

[thorn delta]

the based space universal property is really just a high-brow way of saying that V is a space which has X as a basis, and therefore, any map V --> W is defined by what it does to a basis, i.e. where it maps X

[sinful mirage]

but then what I said makes sense,right?

so since X is the trivial linear combination of elements of X, X is contained in F[X]

[thorn delta]

X is not a subspace of V, but you safely think of X as a subset of V. In particular, a basis for V

[sinful mirage]

but if it is not a subspace

then why can i apply universal prop?

[thorn delta]

the universal property says if you have a set map i.e. function f : X --> W then you get a linear map T : V --> W which restricts to f on X

[sinful mirage]

where does it says it restricts to f on X?

[thorn delta]

j = T o i = T restricted to X

remember, we think of i as including X into V

[sinful mirage]

ok,so pose it differently

s restricted to x is f := s\circ i=f?

if yes,then i see

[thorn delta]

yep!

[sinful mirage]

so the idea now is

I have this thing

and I want to prove that the generators of the subspace,which I quotiented by (before), lie in the kernel of this map S

and I have to do this by direct computation?

so let's say my generators of the subspace are $<cv_1,w_1>-c<v_1,w_1>$, first pair. there are 3 more. then i need to prove that $S(<cv_1,w_1>-c<v_1,w_1>)=0$

[thorn delta]

yea, but its not too bad. If we are talking about the tensor product of just two vector spaces for example, V1 x V2, you just need to verify that elements of the form
(v,w1 + w2) - (v,w1) - (v,w2),
(v1 + v2, w) - (v1, w) - (v2, w), and
(cv,w) - (v,cw)
lie in the kernel of S

[cloud walrus]

ProphetX

[sinful mirage]

and similarly for other 3 generators

yes

can you help me proving one of them?

and I try the others

like one of 4,any

I try other 3

[thorn delta]

Sure, let $(v, w_1 + w_2) \in V \times W$. Then by bilinearity of $f$,
\begin{align*}S(v, w_1 + w_2) &= f(v, w_1 + w_2) = f(v, w_1) + f(v, w_2) \ &= S(v, w_1) + S(v, w_2) \end{align*}
and therefore
$$S(v, w_1 + w_2) - S(v, w_1) - S(v, w_2) = S((v, w_1+w_2) - (v,w_1) - (v, w_2)) = 0$$

[cloud walrus]

kxrider

[sinful mirage]

how did you get S=f?

f=s circ i

first line i'm lost

[thorn delta]

f = s on VxW

[sinful mirage]

how does this follow from f=s circi ?

[thorn delta]

$$S((v, w_1 + w_2)) = S(i(v, w_1 + w_2)) = (S\circ i)(v, w_1 + w_2) = f(v, w_1 + w_2) $$

[cloud walrus]

kxrider

[thorn delta]

is that a bit more clear?

[sinful mirage]

only the first equal sign not

except,yes

why is S(v,w_1+w_2)=S(i(v,w_1+w_2))?

[thorn delta]

the inclusion i : X --> V is by definition a map i(x) = x. So if x = (v, w1 + w2) then i(v, w1 + w2) = (v, w1 + w2)

[sinful mirage]

oh correct

[sinful mirage]

so we have this, first 2 lines I agree. then with last line too

everything good so far

but why does this tell me this is in the kernel?

[thorn delta]

oh the = 0 got cut off on the last line

[sinful mirage]

ah,because S(...)=S(0)?

[thorn delta]

basically i took the first equation, and subtracted the far right hand side from both sides to get the bottom equation

then applied linearity of S

[sinful mirage]

ahh

and then S(...)=0

[thorn delta]

yup

[sinful mirage]

hence (...) in Ker S

this makes sense

now I will write down the other 3

thank you so much for the patience

2days of constructing the tensor product. slowly,but surely!

one quick question tho,still

i(x)=\chi_x=1 or 0

i(x) not equal x

[thorn delta]

yea, so i doesn't always have to be an inclusion, but it has to act like an inclusion. Let me think about how to make this make more sense

like the fact that $X$ is not a subset is not really important. Once we have a map like $i : X \to V$, we can essentially choose $i$ to be an inclusion like so: define $i' : i(X) \to V$ defined by $i(\chi_x) = \chi_x$. If $V$ is based at $X$, we can show that $V$ is based at $i(X)$, and $i(X)$ is a basis for $V$

[cloud walrus]

kxrider

[thorn delta]

and this is really just because we're identifying elements of X with elements of V: x <--> chi_x

[sinful mirage]

ohh wait

I tink

there's the confusion

the book denoted the elements of i(x) with same as x

i.e.

i(v,w)=(v,w)

strictly speaking,this is bad notation

since i(v,w):=<v,w>

different ( )

right?

so what you are saying is basically that i(v,w)=<v,w>

where (v,w) in X, <v,w> in F[X]

i.e. inclusion

[thorn delta]

honestly, this is probably a better way to think about it for now

In practice though, literally thinking of $X \subset F[X]$ doesn't hurt

[cloud walrus]

kxrider

[thorn delta]

the reason why is because there is a bijection between (v,w) elements of X and <v,w> elements of F[X] (just given by a change in notation), but its not worth trying to overthink this

[sinful mirage]

is the composition of a bilinear map with any map bilienar?

[next obsidian]

This doesn’t make sense

Write it down and you’ll see where it doesn’t make sense

Bilinear composed with linear is bilinear

I mean, I guess you can try to make sense of two bilinear things but it doesn’t work like you want

You like end up squaring the scalar

[sinful mirage]

yes,found my mistake

@thorn delta do you think this is fine

?

using universal property

[thorn delta]

ye

[sinful mirage]

i'll prove other 3 too,then use extension theorem

[long obsidian]

In a module M over a unitary ring R, is this so $1_R * 0_M=0_M$?

[cloud walrus]

fajitas

[chilly ocean]

maybe

what does the definition of a module say

[long obsidian]

R a general unitary ring and M possible a non-unitary R-module. I wanna show that there exists unique $M_{triv}$ and $M_{unit}$ submodules such that the following hold

1. $M_{unit}$ is unitary

2. $M_{triv}$ has trivial multiplication: $r*m=0,\forall r\in R, m\in M$

3. $M\cong M_{triv}\bigoplus M_{unit}$

But I'm at the beginning so right now im just trying to show that $M_{unit}$ is unique

[cloud walrus]

fajitas

[sinful mirage]

if the generators of a subspace are in the kernel of S, why is the whole subspace in the kernel of S @thorn delta ?

I proved now that all generators are in kernel of S

and almost ready to apply the unique extension theorem,but idk why it's enough to see for generators

[long obsidian]

I thought I would try to do a trick like have the identity of R multiply a sum of elements from two different unitary submodules M1 and M2 and express one of the elements in M1 as a combination of elements from M2

[hot lake]

a kernel is a subspace

[sinful mirage]

I mean,i have a subspace generated by those 4 relations

I proved all of thoes 4 relations are in the kernel

[hot lake]

and the subspace generated by X is the smallest subspace containing S

[sinful mirage]

why is the subspace generated by them in the kernel?

[long obsidian]

I think that comes with the def of generating

[sinful mirage]

the fact that the kernel is a subspace,does not tell me that the subspace generated by the 4 relations is the kernel

all I know is the 4 relations are in the kernel

why is then the subspace generated by them in the kernel?

[thorn delta]

because the generators are contained in space which is closed under the vector space axioms

so the space those generators generate are inside the kernel

[sinful mirage]

so,basically because U is a subspace?

[thorn delta]

yea

[sinful mirage]

I see

[hot lake]

we would have to see what your definition of "subspace generated by _" is to do a proof

[sinful mirage]

subspace generated by relations=linear combinations of generators

i have 4 relations,which assure me bilinearity

that's a set, collection of 4 vectors

[hot lake]

okay so the kernel is a subspace so is closed under finite sums and scalar multiplication

so if it contains those 4 vectors

[sinful mirage]

and then I take linear combinations of those

[hot lake]

then it contains their scalar multiples

and the finite sums of that

[sinful mirage]

ohhh

yes

[hot lake]

so it contains the subspace generated by the 4 vectors

[sinful mirage]

yep

so basically I have to prove lemma

kernel is subspace

and then apply lemma

[hot lake]

yeah that's probably in chapter 2

[sinful mirage]

yep,i'll go back to that

[thorn delta]

if x1, x2, ..., xn are in span({v1, v2, ..., vn}) then span({x1,...,xn}) \subset span({v1,v2,...,vn})

this is the same concept

[thorn delta]

because x1, x2, .., xn generate span({x1,x2,..., xn}). that's the intuition via more elementary linear algebra

[sinful mirage]

yes,i'll prove this both

I am sorry for not knowing these,but I never had a formal linear algebra course and I can't read all the book cursively,since it would take a lot of time

I try to catch up on things that I strictly need for proofs,even if tis might sound horrible to a mathematician

[hot lake]

chapters 1 and 2 are really fundamental to the rest of the book

[sinful mirage]

so you recommend no matter what happens for me to read chap 1-2

before checking anything in the book?

[hot lake]

yeah if you have time for it

[sinful mirage]

my only issue is

the prof did this proof

he said this is undergrad linear algebra

this is why i've been struggling with the proof for days,to decipher what he means here

he just draws commuting diagrams without justification

[sinful mirage]

I will probably understand this after finishing my proof

for instance,now at least i see 'exists iff' condition

I managed to do the proof!

thanks a lot for the help @hot lake @thorn delta @hidden haven !

❤️

[chilly ocean]

algebra

[sinful mirage]

@chilly ocean you mad

1 proof

took me 14 pages

wtf

14 pagessss

only about existence and uniqueness of tensor prod from universal prop

crazy proof

[chilly ocean]

[sinful mirage]

so after doing all this proof as an outlook

tensors are elements of the tensor product space $V \otimes W$, i constructed,right?

[cloud walrus]

ProphetX

[sinful mirage]

more precisely (2,0) tensors are elements of this space

[thorn delta]

yes, and you would write $v \otimes w$ to denote the equivalence class of $(v,w)$ in $F[V \times W]/U$ for example.

[cloud walrus]

kxrider

[sinful mirage]

fine,but then how does this fit into the picture a tensor is a bilinear(multilinear map)?

i.e. a tensor is a map VxW->R

here a tensor is an element of $V \otimes W:=F\left[V \times W \right]/ \sim$

[cloud walrus]

ProphetX

[thorn delta]

are you familiar with dual spaces?

[sinful mirage]

yes

[cloud walrus]

ProphetX

[sinful mirage]

dual space: space of linear functionals on V

i.e. linear maps V->R

[thorn delta]

So, I think it goes kind of like this. Let $\mathcal L (V \times W, \mathbb F)$ be the vector space of bilinear maps $V\times W \to \mathbb F$. Let $(v,w) \in V\times W$. Then you can define $T_{v^, w^}$ by
$T_{v*,w^}(x,y) = v^(x)v^(y)$. You should get $T_{v^, w^} \in \mathcal L(V \times W, \mathbb F)$. So you can define a map $f : V^ \times W^* \to \mathcal L (V \times W, \mathbb F)$ by $f(v^, w^) = T_{v^, w^}$. This should be bilinear so it induces a unique map $S : V^* \otimes W^* \to \mathcal L (V\times W, \mathbb F)$.

[cloud walrus]

kxrider

[thorn delta]

okay and now you need to find an inverse for S

i don't remember if finite dimensionality is needed here somewhere

If V and W are finite dimensional then from here, you just prove that ker(S) = 0 so S is injective. Then since dim(V*\otimes W*) = dim(L(V x W, F)), S is surjective (finite dimensionality used here), and therefore an isomorphism

[sinful mirage]

why would this fail in inf dim?

[thorn delta]

but the takeaway is that, at least in finite dimensions, tensors in $V^* \otimes W^$ are bilinear maps such that $(v^ \otimes w^)(x,y) = v^(x)w^(y)$ for all $v^ \otimes w^* \in V^* \otimes W^*$ and $(x,y) \in V \times W$

[cloud walrus]

kxrider

[thorn delta]

its probably reduces to the fact that infinite dimensional vector spaces are not isomorphic to their dual

[sinful mirage]

but the tensor prod construction I did works for infinite dimensions to

right?

[thorn delta]

yep

just not the particular identification I wrote above

[next obsidian]

You just show manually that bilinear maps are the same as maps into the Hom

This is just currying

This works in all dimensions

Then use the fact maps from a tensor prouduct are the same as bilinear maps

Then you get it

[sinful mirage]

what is t_{v*,w*} here?

[thorn delta]

hm

[sinful mirage]

where does it map from where to

[thorn delta]

T_{v*,w*}(x,y) = v*(x)w*(y). You can show that this is a bilinear map V x W --> F

[sinful mirage]

it induces a unique linear map between those two

last line

right?

[sinful mirage]

ah yes,this one i don't see yet,sec lemme think

[thorn delta]

yea, and it satisfies that S(v*\otimes w*) = T_{v*,w*}

[sinful mirage]

also,here's a typo,right? t should be $:=v^(x) w^(y)$?

[cloud walrus]

ProphetX

[thorn delta]

where? I thought i defined T_{v*,w*}

[sinful mirage]

yes,but the def of t does not involve w

this is why i'm confused

I think it has to be a typo

[thorn delta]

ohh yea oops

that's a typo, yea

[sinful mirage]

thanks for help! it's getting pretty late here,so i can't check bilinearity of V explicitly and invertability

i'll check that tomorrow and get back if I didn't manage

thanks a lot for the help

[thorn delta]

npnp

[thorn delta]

You're thinking of something like $(V\otimes W)^* \cong Hom(V \otimes W, \mathbb F) \cong Hom(V \times W, \mathbb F)$ right? I guess I was suspicious of $(V\otimes W)^* \cong V^* \otimes W^*$ in infinite dimensions but I think i can see how this works

[cloud walrus]

kxrider

[next obsidian]

No

Hom_bilinear(V x W, F) = Hom( V (x) W, F)

Just like, by definition

Uhh… F here is a vector space

[thorn delta]

yea i know

but thats not what im talking about

[next obsidian]

And Hom_bilinear(V x W,F) = Hom(V,Hom(W,F))

You just send a map

f(v,w)

To the map v -> (w -> f(v,w))

This is currying

[barren sierra]

Suppose $G$ and $H$ are groups. If $G \simeq H$ and $g$ generates $G$ we must have that $\phi(g)$ generates $H$ right?

[cloud walrus]

Spamakin🎷

[barren sierra]

oh shit sorry for interrupting 💀

[next obsidian]

I wasn’t trying to commute the dual with tensor product

[next obsidian]

No, only if phi is the isomorphism

But if it is, then yes

[barren sierra]

Yea it is

Cool thanks

[thorn delta]

okay, so you're talking about something different than the isomorphism $V^* \otimes W^* \cong Hom(V \times W, \mathbb F)$ i was walking through with prophet?

[cloud walrus]

kxrider

[next obsidian]

Yeah

You just use that tensor is the same as like bilinear maps

This is more natural I think as well, since it’s only using the univ property of tensor

And in a sense using the univ property of bilinear maps

[thorn delta]

ah okay sure i agree

[sinful mirage]

ah btw before I go to sleep

maybe it will be useful for some people who try to learn this by themselves and struggle with itlike I did

tried to make it as self-contained as possible

[chilly ocean]

[lunar lintel]

How can I find the minimal subring of R which contains all elements of S={a+b\sqrt(2)+c\sqrt(3)| a,b,c \in Z}? Think it should be just linear and multiplicative combinations of these elements but dont know how to prove

[lethal cipher]

I dont know the full answer, but I do recommend you multiplying two elements out and see what happens.
You are going to be getting some unwanted square roots, so perhaps that can help narrow things down.

[lunar lintel]

Seems like sqrt(6) needs to be in as well

[lethal cipher]

Well, if sqrt(6) is in there, then it is not in S

[lunar lintel]

right, im trying to find minimal subring that contains S

[lethal cipher]

So whatever coefficient is being multiplied by sqrt(6) needs to be 0

[lunar lintel]

wouldnt sqrt(6) need to be in the subring? otherwise multiplication wouldnt be closed right?

[lethal cipher]

Well, that can never be in S, so no

But if the term being multiplied by sqrt(6) is 0, then it is in S

[lunar lintel]

do i need sqrt(6) to be in S? i just need a new subring that contains all of S, but I can add sqrt(6) to the new subring, no?

maybe im misunderstanding something but i was under impression that sqrt(6) could be in my new minimal subring that includes the set S

[lethal cipher]

No no, you are right. I just didnt quite realize what minimal subring was. I also didnt pay attention to the containing bit

I thought it was the opposite, you needed to find a subring in S

[lunar lintel]

so it would be ok to have sqrt(6) in the new subring containing S right

or neccessary

[lethal cipher]

It is completely necessary. In order for two arbitrary elements of S to be in your subring, they must multiply to a number in the subring. So there must be an integer coefficient for sqrt(6)

[lunar lintel]

that makes sense. to articulate the subring would it be enough to just have {a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} | a,b,c,d \in Z}? any multiplicative combination just yields one of the terms with a coefficent i think

[lethal cipher]

You also can't exclude any integer multiples of sqrt(6) because sqrt{2} multiplied with c sqrt(3) for any integer gives c sqrt(6)

So this is indeed the smallest subring possible.

[lethal cipher]

Yep, you do need to show it, but that is closed under multiplication

[lethal cipher]

And do go through this logic to show that this is indeed the smallest subring containing S

[lunar lintel]

is your explanation sufficient?something along the lines of "the new subring must contain a+b\sqrt{2}+c\sqrt{3} and must be closed under multiplication hence d\sqrt{6} needs to be as well. If we neglected any term, S is not contained or multiplication is not closed"?

[lethal cipher]

That is not quite what I said.

For any integer d, d sqrt(3) * sqrt(2) needs to be in our subring.

[lunar lintel]

whoops i typoed i meant d\sqrt(6)

[next obsidian]

Just take Z-linear polynomials in the things you’re putting in

[lethal cipher]

Chmonkey knows what they are talking about. Listen to them

[next obsidian]

So like here you’ve adjoined in two things

Take any polynomial in Z[x,y] and plug in (sqrt(2),sqrt(3))

This will be the minimal subring

This is basically what you said

All sums, all products, then you just do that as much as possible

What you end up getting is polynomial in sqrt(2) and sqrt(3)

Me thinkies that this does just end up as Z-linear combinations of sqrt(2),sqrt(3),sqrt(6) tho

But that’s my chmonkey instinct speaking

So like

So your polynomials basically only have an x,y, or xy term

[lunar lintel]

what do u mean by this

[next obsidian]

Take the polynomial ring Z[x,y]

Think of each element as a function

f(x,y)

Just plug in that vector

So it’s like f(sqrt(2),sqrt(3))

[lunar lintel]

ok that makes sense, but how did u conclude this line from that?

[next obsidian]

So like

Think about what happens when you plug in sqrt(2) for x

If you had an x^2

This becomes just a 2

So like, when you look Z-linearly if you had like say

ax^2

You could instead replace this with 2a

So each time you have an x^2 you can replace it with 2 and then you’ll get the same element when you plug in (sqrt(2),sqrt(3))

Similarly for y^2 and 3

So my point is, you only need to look at polynomials with x, y, xy, and a constant

And like after plugging in (sqrt(2),sqrt(3)) these terms correspond to sqrt(2),sqrt(3),sqrt(6)

Maybe think about what generic terms in this ring looks like

It’ll be like shit like

3sqrt(2)^3 + 7 sqrt(3)sqrt(2)

But this is the same as 6sqrt(2) + 7sqrt(6)

[lunar lintel]

that part makes sense, its just proving the minimality that is not good

[next obsidian]

Well not really

[lethal cipher]

Sorry, it took me a minute, but yes.
For any integer d, d sqrt(6) is in our subring. Since the additive group is closed, [a+bsqrt(2)+csqrt(3)]+dsqrt(6) must also be in our subring, for any integers a,b, and c.
So this is indeed the smallest possible subring.

[next obsidian]

You can show that the ring in terms of the polynomial stuff can be described as Z-linear combinations of sqrt(2)?sqrt(3),sqrt(6)

But you’ll see that the description in terms of polynomials is minimal

Cuz any ring containing that stuff will have that a as a subring

They have to exist just from sums and products closure

[lunar lintel]

ok ill try writing this all up thank u both for help

[odd dirge]

How do I start proving this? I've got no idea where to start
S,T ⊂ V(arbitrary vector space)
span(S ∪ T) = span(span(S) ∪ span(T))

[thorn delta]

are you aware of the fact that S \subset span(S)?

[odd dirge]

yes

[thorn delta]

alright, so taking this one step at a time: you can show span(S ∪ T) ⊂ span(span(S) ∪ span(T)):
You know S ∪ T ⊂ span(S) ∪ span(T) ⊂ span(span(S) ∪ span(T)), so...

[viscid pewter]

How do I start proving this? I've got no idea where to start

let s1, ..., sn be the elements of S, t1, ..., tm the elements of T

then what form must the elements of span(S U T) have? what form must the elements of span(span(S) U span(T)) have?

this will help you show RHS < LHS, the other way round is easier

[thorn delta]

I think you can prove this using only the property that if S is a set, and V is a vector space, and S \subset V, then span(S) \subset V

[viscid pewter]

my instinct says you can only prove one direction with that

[thorn delta]

i would say "watch me," but I'm not going to spoil it for blazer

[viscid pewter]

ah

[odd dirge]

after thinking about it, I am still clueless how to get span(S ∪ T) in between those deductions

[thorn delta]

are you talking about mine or kaisheng's?

[odd dirge]

yours

[thorn delta]

are u aware of this fact (the S \subset span(S) property was a mistake earlier, I meant this property that I am replying to)

[odd dirge]

but isn't S subset of span(S)?

[thorn delta]

yea, that's true, but more generally, if S \subset V for a vector space V, then span(S) \subset V

S\subset span(S) is a special case

[wind locust]

Hey, can someone check if my reasoning is correct? It's a pretty basic question but its 5am so i cannot be sure. We have shown in class that if $N_1,\cdots,N_k$ are Noetherian R-modules then so is $\sum_i N_i$. Right after this, we use (without proof) that if $R$ is Noetherian then so is $R^n$. Is the following reasoning valid?
Suppose that $R$ is Noetherian. Then $R^k\cong\underbrace{R \oplus R \oplus \cdots \oplus R}_{k\text{ times}}$, and we can treat this external direct sum as an internal one (i.e. pass to an isomorphic module with submodules $S_1,\cdots,S_k$, each isomorphic to $R$ and whose internal sum is the whole module) and this internal direct sum is actually just the sum of the submodules, and by the result proven before, this must be Noetherian.

[odd dirge]

I don't think I am aware of span(S) subset of V

[cloud walrus]

iruneachteam

[thorn delta]

alright, so we know V is closed under taking linear combinations of its elements, so if we take linear combinations of elements of S, which are contained in V, then we stay inside V. Thus span(S) \subset V

[odd dirge]

alright, this makes sense

[thorn delta]

so using that, it follows from this chain of inclusions: S ∪ T ⊂ span(S) ∪ span(T) ⊂ span(span(S) ∪ span(T))
that span(S ∪ T) ⊂ span(span(S) ∪ span(T))

make sense?

[odd dirge]

I am not sure why

span(S ∪ T) ⊂ span(span(S) ∪ span(T))
is derived from the chain of inclusions

[thorn delta]

what i mean is that each inclusion in that chain is not too hard to see, and the result is that:
S ∪ T ⊂ span(span(S) ∪ span(T))
which would imply
span(S ∪ T) ⊂ span(span(S) ∪ span(T))

[viscid pewter]

S ∪ T ⊂ span(S) ∪ span(T)
and span(S) ∪ span(T) ⊂ span(span(S) ∪ span(T))

ie. A ⊂ B and B ⊂ C
so A ⊂ C

ie. S ∪ T ⊂ span(span(S) ∪ span(T)) which is what we want

[odd dirge]

okay so first,
span(S ∪ T) ⊂ S ∪ T if understood correctly,
thus
span(S ∪ T) ⊂ span(span(S) ∪ span(T))

[thorn delta]

nono span(S ∪ T) ⊂ S ∪ T is not true.

S ⊂ span(S)
T ⊂ span(T)
This implies
S ∪ T ⊂ span(S) ∪ span(T)

[odd dirge]

S ∪ T ⊂ span(S) ∪ span(T) ⊂ span(span(S) ∪ span(T)) this part I understood

I just don't know why it implies span(S ∪ T) ⊂ span(span(S) ∪ span(T))

[viscid pewter]

if A is a subset of B and B is a subset of C, A is a subset of C