the heart of isomorphisms is to see some structures as essentially the same
#groups-rings-fields
tawny pine
chilly ocean
so it doesn't necessarily have to be the same "type" of algebraic structure?
you can have a homomorphism (or any kinda morphism) between any two structures?
like between vector spaces and fields?
no
tawny pine
no you're getting off track
any real vector space with dim n is isomorphic to R^n
in particular C is isomorphic to R^2
chilly ocean
the heart of isomorphisms is to see some structures as essentially the same
ah okay now I get it, by intuition of the identification of the "same" structures you mean like for example a ring of 3x3 matricies is isomorphic to a ring of 2x2 matricies
as what
winter thorn
the heart of isomorphisms is to see some structures as essentially the same
The totient function of product of two numbers is the same as the totient of those two applied individually and then multiplied.
tawny pine
which lets us see C as essentially 2d space
weak oriole
ah okay now I get it, by intuition of the identification of the "same" structure...
Nu
chilly ocean
ah okay now I get it, by intuition of the identification of the "same" structure...
no, I think your first first statement was the most accurate. The point of 'same' is different for different aglebraic structures, but for example isomorphism between just sets is any bijection, and they are 'the same' with respect to cardinality, but there isnt much structure onsets so we cant say much more. For groups for example we can say more about the structure if there exists an isomorphism between two groups
tawny pine
if V,W are findim vector spaces and we fix bases of V,W and a linear map T:V->W
then the space of linear maps V->W is isomorphic to the space of mxn matrices (use our bases to explicitly write an isomorphism)
blurring the line between linear maps & matrices
chilly ocean
the heart of isomorphisms is to see some structures as essentially the same
blurring the line and seeing that this is true
vocal wolf
If k is a field, isn't k[x]_(x) isomorphic to k[x]?
k[x] \ (x) = k \ {0}
and we can use the map f(x)/a -> f(x)/a
hidden haven
Yes
vocal wolf
Wait so why are these guys saying spec k[x]_(x) has 2 points
For example spec C[x] has more than 2 points
hidden haven
k[x] \ (x) = k \ {0}
Wait this is wrong
Polynomials with constant term non zero
Are not in (x)
vocal wolf
what
hidden haven
x+1 is not in (x)
nimble sail
Hello. What prerequisite topics do I need to have studied to learn about groups? This is the topics that I need to learn
lavish summit
it helps to know some linear algebra but generally you can just jump right in
yeah
the prerequisite is having a document to study from
also just like, knowing what a proof is
nimble sail
oki, thanks
ivory dust
I feel like I should get this but I don't
next obsidian
It’s homology
ivory dust
in what way
next obsidian
But Chmonkey
Wel this is actually just a chain complex I guess
Omegalol
Chmology is implicit
ivory dust
how is that a chain complex though
what does it go 1 monke 0 monke 2 monke 3 monke
or wait is that -1 monke in the first chain group
ivory dust
chmology
isn
isn't that chcohomology though since the arrows are going up the chain groups?
ivory dust
hm?
bleak abyss
My turn
So SL(2,C)
Actually first I'm gonna establish in general what invariant bilinear forms ought to be
cloud walrusBOT
Sloth King Daminark
chilly ocean
bleak abyss
Hmm lemme think for a sec about the right way to set this up
Yeah wait idk if \pi(I + tX) works here
Or wait yeah no lol
So general manifolds stuff
cloud walrusBOT
Sloth King Daminark
bleak abyss
Or nah this would work for functions
But I guess in the rep case it'd make sense since we map to GL(V)
So
cloud walrusBOT
Sloth King Daminark
bleak abyss
Apply it here
Okay ehh I still can't just directly compute this
Let's work simply in Lie algebras then
So
Invariant bilinear form is defined to mean
cloud walrusBOT
Sloth King Daminark
chilly ocean
bleak abyss
Lol
simple mulch
hey guys
the author defined a normal subgroup and then proceeded to define the factor group
I have to prove $G/ker\omega \cong Im\omega$
chilly ocean
remove space after first $
simple mulch
I mean the factor group of G by ker w
next obsidian
Don’t you mean G/ker w
cloud walrusBOT
mns
next obsidian
That’s probably how it’s denoted?
next obsidian
Yeah this is the first isomorphism theorem
Show that w “extends” to a map from G/ker w -> im w
simple mulch
with $\omega$ being a homomorphism
cloud walrusBOT
mns
next obsidian
In the sense that you can just define it on the equivalence classes in the “naive” way
You need to show this is well-defined
You’ll find that this is injective, and it has the same image as im w
So it’s an injective + surjective map from G/ker w -> im w
So it’s an isomorphism :D
next obsidian
well this follows by its definition and because w was a homomorphism
simple mulch
hum
but my question was
when taking $x \in G/ker\omega$ what do I have? something like $x = ng$ for some $n \in \ker\omega$?
cloud walrusBOT
mns
next obsidian
No
x is an equicalence class of objects in G
You probably denoted it either
$\overline{x},[x]$, or $x + \ker\omega$
cloud walrusBOT
Kanga Gang Boss Chmonkey
simple mulch
well I noticed the author used this set as the factor group but its the same as the quotient? right?
simple mulch
its on the image
next obsidian
Like to me this is just the quotient group
simple mulch
above
simple mulch
ye
next obsidian
No, the element Nx denoted the coset
It represents the set of elements {nx | n is in N}
It’s this entire block of elements
So in this case N = ker w
So an element is denoted by $x\ker\omega$
cloud walrusBOT
Kanga Gang Boss Chmonkey
next obsidian
It doesn’t a matter if you write x on the left or right (this is because ker w is normal)
I just write it on the left
frank lake
ok
I'm dumb
I need a bit of proofreading
so this is my proof: Assume $a$ and $b$ aren't disjoint and $a \cap b$ is nonempty, because this rejects the definition of a a partition, by contradiction P is a partition
omfg
cloud walrusBOT
ChubbyMuffins
frank lake
I feel like more stuff needs to be added
pls
I need help 😦
this is intro to abstract algebra
thorn delta
What is P?
thorn delta
it looks like you are trying to prove its a partition tho. So we don't know P is a partition
frank lake
true..
thorn delta
what's the exact statement ur trying to prove?
frank lake
I'm trying to prove that ~ is an equivalence relation on S
I don't really get it that much tho
thorn delta
ah okay. So P is a partition and ~ is defined by a ~ b if a and b are contained in the same element of the partition
frank lake
like I get that an equivalence relation is a relation that has reflexivity, transitivity, etc
frank lake
ah okay. So P is a partition and ~ is defined by a ~ b if a and b are contained ...
yeah
thorn delta
like I get that an equivalence relation is a relation that has reflexivity, tran...
yep. You'll have to demonstrate each property
frank lake
how tho?
thorn delta
Lets say P is a partition of S and let a be in S. Then a is contained in an element of the partition A \subset S.
for reflexivity you have to verify that a ~ a. Since a is in A, a ~ a.
i.e. a is contained in the same element of the partition as a
frank lake
oh yeah
how would I verify transitivity?
oh hold on
wait so if I said something like "Let P be a partition of S and let a, b, and c be in S. Then a is contained in an element of the partition A \subset S, since a and b are in A a ~ b, and since b and c are in A b ~ c, which implies a ~ c?"
@thorn delta
or more specifically
since a and c are in A, a ~ c
which verifies the transitive property
thorn delta
close. You can't really assume that a and b are in A and b and c are in A. What you have to say is this:
a ~ b and b ~ c. So a and b are in an element A of the partition and b and c are in an element B of the partition.
frank lake
then what about a ~ c
thorn delta
well, you have to show that these assumptions are sufficient to conclude a ~ c
hint: you need partition-ness for this
frank lake
close. You can't really assume that a and b are in A and b and c are in A. What ...
ohhh I kind of get it
frank lake
hint: you need partition-ness for this
I still don't get it 😅
wouldn't it just be fine to say a is equivalent to b, and b is equivalent to c, so a is equivalent to c
🤷
oh hold on
if we declare a ~ c, then that would mean a and c are in an element A and B?
so $A \cap B$
but that would be empty according to the definition of a partition hmm
or what about a ~ c is an element of A ~ B?
I got no idea
😂
thorn delta
yea u have the right idea. since $a,b \in A$ and $b,c \in B$, you can say $b \in A\cap B$. But like you said, $A$ and $B$ are elements of a partition, and distinct elements of partitions are disjoint, so what can we say about $A$ and $B$?
cloud walrusBOT
kxrider
frank lake
$A \cap B = \varnothing$?
cloud walrusBOT
ChubbyMuffins
frank lake
wait I'm kind of confused
why wouldn't $A \cap B = \varnothing$? isn't that how partitions work?
cloud walrusBOT
ChubbyMuffins
thorn delta
im just gonna place emphasis on this real quick
distinct elements of partitions are disjoint
thorn delta
A and B are elements of the partition, which comprise of subsets of S
thorn delta
distinct elements of partitions are disjoint. So this means non-disjoint elements of the partition are equal
frank lake
so ig we can say A ~ B then
thorn delta
A = B, yea
thorn delta
A and B are sets, so yea we mean equality of sets
frank lake
they almost mean the same thing
frank lake
**kxrider**
is that what you were looking for? 😂
thorn delta
yep
frank lake
this is how partitions are defined in my book
thorn delta
to sum things up, $a\sim b$ and $b\sim c$ imply $a,b \in A$ and $b,c \in B$ for some $A,B \in P$ so $b \in A \cap B$, and by def of a partition, $A = B$, so ${a,b,c} \subset A = B$
cloud walrusBOT
kxrider
frank lake
damn
thorn delta
this is how partitions are defined in my book
yea, to phrase this more symbolically/precisely, if P is a partition, then given A,B in P, if A is not equal to B, then A \cap B = \varnothing. That's what it means for P to a "pairwise disjoint" collection
frank lake
yea, to phrase this more symbolically, if P is a partition, then given A,B in P, if A is not equal to B, then $A \cap B = \varnothing$. That's what it means for P to a "pairwise disjoint" collection
cloud walrusBOT
ChubbyMuffins
frank lake
I'm gonna save this
😄
so a partition can still be possible without distinct elements then?
you can still create a partition P from the subsets A and B even if $A \subset B$
or A = B
it's alright man
thorn delta
so a partition can still be possible without distinct elements then?
not exactly. The elements of any set are distinct, but when we say something like "suppose a~b and b~c" then we get a,b is in A and b,c is in B where A, B are in P. But there is no further assumption that A is not equal to B.
so the goal was to show that the definition of a partition is strict enough that we have to have A = B in this scenario
thorn delta
nah not necessarily
thorn delta
distinct just means A is not equal to B, yea. There's another way you can think about this
if a ~ b and b ~ c then a,b in A and b,c in B for some A,B in P.
Case 1: if A = B then {a,c}\subset A = B and hence a ~ c.
Case 2: if A \neq B., then by def of a partition, A and B are disjoint. On the other hand, b is in A \cap B, a contradiction.
Therefore, we're always in the first scenario
thorn delta
npnp
frank lake
I think I get it a bit more
OHHHHHH
I GET IT EVEN MORE
if $a \sim b$ and $b \sim c$ then $a,b$ in $A$ and $b,c$ in $B$ for some $A,B$ in $P$.
Case 1: if $A = B$ then ${a,c}\subset A = B$ and hence $a \sim c$ .
Case 2: if $A \neq B$., then by def of a partition, A and B are disjoint. On the other hand, b is in $A \cap B$, a contradiction.
Therefore, we're always in the first scenario
cloud walrusBOT
ChubbyMuffins
prisma thunder
Theorem (Abel) For $p$ prime and not equal to $\mathrm{char} , F$, let $f(x) = x^p-\alpha \in F[x]$, where $\alpha$ is nonzero. Then $f$ is either irreducible or has a root in $F$. In the latter case, $f$ splits over $F$ iff $F$ contains a primitive $p$-th root of unity.
Consider $f(x) = x^2-2$ and let $F$ be a field of characteristic $p > 2$. Since $f$ is a quadratic, if one of its roots is in $F$, then it splits in $F$, implying $F$ contains a primitive $2$nd root of unity $(-1)$. If I consider $F = \mathbb{F}_5$, then $x^2-2$ is irreducible over $\mathbb{F}_5$. However, $\mathbb{F}_5$ contains a $2$nd root of unity ($-1 = 4$ in \mathbb{F}_5$), so $f$ should split over $\mathbb{F}_5$. I'm not exactly understanding what's going on here, so any clarification would be appreciated.
cloud walrusBOT
eM
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
thorn delta
the way you've stated the theorem, f has to have a root before you can talk about the splitting condition
next obsidian
^^
If you have a root beta then you get p-distinct roots beta•omega^n
For 0<= n < p
Omega the primitive p-th root of unity
But you need beta for that to start
prisma thunder
I'm not sure. That's what they said.
prisma thunder
the way you've stated the theorem, f has to have a root before you can talk abou...
Thank you, that makes sense
Hmm. Alright then. I want to look at the Galois group of $f(x) = (x^2-2)(x^2+x+1)$ over $\mathrm{F}_p$ for $p > 2$, so observe the splitting field $\mathbb{F}_p(\sqrt{2}, \omega)$, where $\omega$ is a primitive third root of unity. Then $\mathbb{F}_p(\sqrt{2}, \omega)$ can be of degree 1, 2, or 4. Since any finite extensive of $\mathbb{F}_p$ are cyclic, $\mathrm{Gal}(\mathrm{F}_p(\sqrt{2}, \omega)/\mathbb{F}_p)$ is either trivial, $\mathbb{Z}/2\mathbb{Z}$, or $\mathbb{Z}/4\mathbb{Z}$.
Degree $1$ occurs when, say $p = 73$. Degree $2$ occurs when only one or the other reduces (for $p = 61$, $x^2-2$ is irreducible while $x^2+x+1$ splits, while in $p = 89$, $x^2-2$ splits while $x^2+x+1$ is irreducible) or when the roots of $f$ can be expressed in terms of the other (for $p = 5$, we have $\sqrt{2} = 1+2\omega$).
I haven't been able to find an example of a degree $4$ extension. All I know is that if $\sqrt{2} \in \mathrm{F}_p(\omega)$, then $\sqrt{2} = a+b\omega$ for some $a,b \in \mathbb{F}_p$, which we can reduce to find $a \in \mathbb{F}_p$ such that $3a^2+2 = 0$.
cloud walrusBOT
eM
prisma thunder
If no such a can be found in Fp for some p, then we have a degree 4 extension.
lilac trench
In D_8 why is it that no automorphism can map r to sr^3 even though they have the same orders
s is the reflection and r is rotation
upper pivot
In D_8 why is it that no automorphism can map r to sr^3 even though they have th...
what would r^2 get sent to
upper pivot
hmm p sure r^2 gets mapped to 1
lilac trench
huh... I get that r^2 maps to (sr^3)^2=s^2 r^6=r^2
upper pivot
It’s not commutative so you cannot distribute it like that
prisma thunder
**eM**
It looks like Z/4Z can't happen.
Quadratic reciprocity be like huehuehuehuehue
chilly ocean
polynomials in y with coefficients in C(x)
vocal wolf
What do you mean C(x)?
hidden haven
Field of fractions of C[x]
chilly ocean
quoteitns of polynomials in x each with coefficients in C
vocal wolf
Ohh i see
median pawn
How should I approach this, except brute force?
I get ac = 2bd + 3, and bc + ad = 0 for (a + bsqrt(2) i) (c + dsqrt(2) i) = 3
hidden haven
Finding the norms of the factors is usually a good first step
Of any factors that may exist
median pawn
What if I don't know what a norm is?
The book gives this exercise even before introducing what a ring is
hidden haven
You have done complex analysis 
I'm talking about complex magnitude, not norm like normed vector spaces or anything
Like the high school stuff of |a+bi| = √a² + b²
oak grove
is it appropriate to ask hamming code questions here? or should i use some other channel? they were introduced in the field context to us
prisma thunder
Like the high school stuff of |a+bi| = √a² + b²
Dag, my high school sucks lol
hidden haven
oak grove
Dag, my high school sucks lol
mine too
i didnt see complex numbers till college 
then again it was an arts school 
hidden haven
If I had to guess, Hamming codes would go in #discrete-math but idk 
hidden haven
median pawn
You have done complex analysis <:catThimc:732011802459635713> I'm talking about ...
Oh lmao that norm
hidden haven
limpid edge
yea
rustic crown
you can do induction on any well ordered set
limpid edge
if you wanna prove an assertion P(a, b)
you can fix one of the variables and induct on the other
and then do the same thing the other way around
or maybe
your inductive hypothesis is P(a, b) and you want to prove P(a + 1, b) and P(a, b + 1) for your inductive step
hidden haven
That would be one way to do it yeah, it depends on what you are trying to prove
waxen hedge
or you can also do induction over a+b
hidden haven
You can also take N x N in dictionary order and do strong induction (instead of P(n) → P(n+1), you'd be proving that if P(m) for all m<n, then P(n))
This is what det is suggesting I think
hidden haven
You can also take N x N in dictionary order and do strong induction (instead of ...
This is called transfinite induction scary name but this is all it is
Except you put any well ordered set in place of N x N (can also be a well founded partial order)
limpid edge
You can also take N x N in dictionary order and do strong induction (instead of ...
idk what transfinite induction is, but isn’t what you just described strong induction
oh
i can’t read
hidden haven
N x N with dictionary order doesn't look like N, you have to sort of go past infinity
Hence the name
It's N many copies of N strung together
median pawn
yup sounds cool
I was just trying to prove that for integers m,n and a \in (R, +, .), ma + na = (m+n)a
ik it's trivial but still
fixing one and doing induction on second seems good here
hidden haven
Ye here induction on any one should be enough
limpid edge
why doesn’t tranfinite induction work on totally ordered sets that aren’t well ordered
is the well ordering supposed to ensure the existence of a base case for the induction
hidden haven
is the well ordering supposed to ensure the existence of a base case for the ind...
Yes pretty much
You'll see it clearly if you try to prove the principle of induction
limpid edge
is the well ordering supposed to ensure the existence of a base case for the ind...
and if so why isn’t the definition of well ordered just “contains a smallest element” instead of “every non empty subset contains a smallest element”
hidden haven
It works out if you have well orders
But otherwise you wouldn't be able to prove it
limpid edge
and if so why isn’t the definition of well ordered just “contains a smallest ele...
i get that they’re equivalent , but still
hidden haven
It works out iff the order is a well order
hidden haven
and if so why isn’t the definition of well ordered just “contains a smallest ele...
Base case alone isn't enough
Try to come up with a proof
limpid edge
oh is it like
if some subset doesn’t have a smallest element
then we will never reach it during the induction
is that the handwavy idea
hidden haven
Yes
Proof proceeds by assuming there's a P that satisfies that strong induction condition, and then proving that P holds for all elements of the well ordered set by ||taking the set S on which P doesn't hold, if S is non empty then take the least element of S and show a contradiction||
limpid edge
hm I read the proof online and kinda got it
i should learn what ordinals and the peano axioms are first
hidden haven
Strictly speaking you don't need ordinals, they are used to do transfinite induction proofs because instead of having to prove the strong induction condition, you can get away with proving a slightly stronger version of the weak induction condition
By dividing into cases
The well ordering thing is the entire idea really
Just so happens that every well ordered set is order isomorphic to an ordinal
So we can phrase things in terms of those
limpid edge
Just so happens that every well ordered set is order isomorphic to an ordinal
oh that’s how they connect lol
hidden haven
Yep, and a unique ordinal at that
So they give a classification of well ordered sets
cloud walrusBOT
Hausdorff
median pawn
**Hausdorff**
For me, an integral domain is a commutative ring with unity, with at least two elements, and no zero divisors
hidden haven
Show that total degrees add when you multiply polynomials
hidden haven
Any 2 non zero polynomials have non negative degree, so does their product. Consider the cases when both have degree 0, and where at least one has positive degree
chilly ocean
take two guy with non zero big term. product of guy has product of big term. nonzero!
median pawn
Any 2 non zero polynomials have non negative degree, so does their product. Cons...
ahh wait, suppose pq = 0 where deg p = m, deg q = n. deg(pq) = m+n, m + n = 0. so m = n = 0?
hidden haven
Yes
hidden haven
So they are both constant
hidden haven
so in fact p = 0 or q = 0
No
median pawn
either one yeah
smoky hedge
Just so happens that every well ordered set is order isomorphic to an ordinal
So, in ZFC every set would have an order type?
hidden haven
Be careful here because deg 0 is either undefined or -infinity
hidden haven
So, in ZFC every set would have an order type?
Every well ordered set
Every ordered set also has an order type, but then it's not represented by an ordinal
median pawn
Be careful here because deg 0 is either undefined or -infinity
right so what do we do
smoky hedge
But the well-ordering theorem says every set can be well-ordered
Or maybe im not understanding this
median pawn
right so what do we do
the goal is to prove that p = 0 or q = 0
but then either m or n is not defined
so that's weird
hidden haven
the goal is to prove that p = 0 or q = 0
We are assuming that R is a domain
hidden haven
But the well-ordering theorem says every set can be well-ordered
Yes but not a unique one
So you can't say that a set alone has an order type
It could have different order types under different well orders
median pawn
We are assuming that R is a domain
yes agreed. if p = c_1, q = c_2, and pq = 0 gives c_1c_2 = 0, which means c_1 = 0 or c_2 = 0 since R has no zero divisors
but.. suppose c_1 = 0. then p = 0. and deg p = ???
smoky hedge
Ok, i see. So for a specific well-ordering of a set we can assign an order type to some set
median pawn
our degree argument doesn't make sense if p = 0 or q = 0
hidden haven
our degree argument doesn't make sense if p = 0 or q = 0
It does, R domain → product of deg 0 polynomials is deg 0 hence non zero
frank fiber
is a spectral secuence determinated by the first page?
Because the second page is the (co)homology of the first and so on
next obsidian
Let (A,m,k) be a local ring, and F,G two infinite rank free modules
Suppose that F -> G is a map such that the induced map F (x) k -> G (x) k is an isomorphism
Is F -> G then also an isomorphism?
If somehow it helps, you can assume A is Artinian
gleaming quarry
Does lR[X] support all real numbers as coefficients for polynomials?
I want to normalise the polynomial 3X^2+9X-1 such that the leading coefficient is 1
Real Numbers
ahh thanks
that makes this task quickly done
next obsidian
If somehow it helps, you can assume A is Artinian
I can prove this when A is Artinian, but the proof is kinda silly, you only use flatness (although that’s equivalent over an Artinian local ring, lol). I bet it’s false in general
arctic dirge
hello, i need some help. ive got this exercise for unique factorization domains, but i think that there may be a typo.
considreing 21 in Z a UFD. the divisors as defined in the picture would be 3,7,-3,-7 and the required product 3^1 * 7^1 * (-3)^1 * (-7)^1 is 21^2. But the only units in Z are 1 and -1, so 21^2 and 21 are not associates. What could be th eproblem?
latent sand
Just wanted to ask a quick question to make sure I understand how to generate ideals. If we're in Z[x] (Z being the integers), then the ideal (x) is polynomials with integer coefficients, but the constant term of each polynomial is zero right ?
next obsidian
Yeh buddy
median pawn
Why is it true that the elements of GL(n, Z) are exactly those nxn matrices with entries from Z with determinant = +1 or -1?
For n = 2 I remember brute forcing this
but how do we prove it for general n?
lethal dune
GL(n, Z) makes sense even if Z is not a field?
next obsidian
It’s because a matrix is invertible iff its determinant is invertible
And in Z the only invertible elements are +-1
@median pawn
median pawn
GL(n, Z) makes sense even if Z is not a field? <:stare:826920038623346729>
Indeed
median pawn
It’s because a matrix is invertible iff its determinant is invertible
Hmm okay let's see why that is true
Suppose A is invertible. Then there exists B with AB = BA = I. det(AB) = det(A)det(B) = 1. So, det(A) is invertible
Now suppose det(A) is invertible. So there exists c such that c det(A) = 1. Hmm? Now?
We can also find a non-zero matrix C such that det(C) = c, that's fine. Then, det(AC) = 1. But that does not mean AC = I necessarily
@next obsidian
next obsidian
You use cofactors matrices
You aren’t gonna come up with this yourself
You like take a matrix with each entry being the minor of a certain sub matrix
Then divide it by the determinant of the matrix
By some sort of black magic this is an inverse
If you’ve seen cramer’s rule, this is basically what Cramer’s rule is lmfao
median pawn
Do you mean A adj(A) = det(A) I?
median pawn
By some sort of black magic this is an inverse
All of algebra is black magic
median pawn
No wait how does it help
median pawn
c det(A) = 1, this is where I am
next obsidian
It doesn’t matter
Like
Use c
Put c into every entry of AdjA
Then your result becomes A(c Adj A) = c det A I
= I
that’s what 1/det A is, it’s just c lol
chilly ocean
chilly ocean
hi
median pawn
how to you define the determinant
using that permutation business
next obsidian
I think so
kind temple
is it different for rings
chilly ocean
$$T^*(e^1\wedge\cdots\wedge e^n) = (\det T)e^1 \wedge \cdots \wedge e^n$$
cloud walrusBOT
TTerra
next obsidian
HURBBBBB
median pawn
like, is it the same as the unique alternating n form i know and love
for commutative rings with identity it's the same right?
next obsidian
Why must you be this way
chilly ocean
geometer moment
next obsidian
You just do it recursively
median pawn
**TTerra**
what is this magic
chilly ocean
also, c squared needs to learn this definition anyways
kind temple
next obsidian
When you first took linear algebra
median pawn
All of algebra is black magic
right?
next obsidian
No
kind temple
but i dont like that oneeee
chilly ocean
you're learning differential topology c^2 you need this
next obsidian
Ignore him
kind temple
which side do i pick
chilly ocean
both
kind temple
😎
kind temple
i love multilinear alg. we covered it in my linear alg class
and then
everybody started calling them "tensors"
and im like "whats a tensor"
kind temple
and then they said, "oh, its a quotient of a blah blah blah stoopid stuff"
its literally just an alternatining k form
chilly ocean
lol
median pawn
I've a question
cloud walrusBOT
Hausdorff
median pawn
I know that it's an integral domain
chilly ocean
because algebraists are insane
next obsidian
It’s just…
Number theory shit
Lol
They’re a simple thing where you can compute explicitly like if that’s the ring of integers of some quadratic extension of Q
Or if you have to add something else I think
median pawn
Ooh alright!
chilly ocean
hey chmonkey
next obsidian
Hi
waxen hedge
I've a question
it's the ring of integers of $\mathbb{Q}(\sqrt{d})$
It plays a role similar to $\mathbb{Z}$ in $\mathbb{Q}$
cloud walrusBOT
Adrien
waxen hedge
this ring has a lot of good properties : it's a Dedekind ring (in particular it is noetherian and an integral domain), it's a free $\mathbb{Z}$-module of rank 2, …
cloud walrusBOT
Adrien
median pawn
Thanks!
Hey! Would anyone like to work through (Mostly Commutative) Ring Theory in Shahriar Shahriari's Algebra in Action with me? It's 6 chapters, I plan to go through most or all of the exercises. Ping me if you're interested!
plucky flicker
Why is the sum of two injective submodules the homomorphic image of the direct sum of them?
hidden haven
You don't need injectivity for this
The sum is just the direct sum with extra relations
hidden haven
The surjection from N_1 dirsum N_2 to N_1 + N_2 is given by (n, n') → n+n'
hidden haven
the intersection must be trivial
no
plucky flicker
ok, i'm lost lul
hidden haven
If it is, then it becomes an isomorphism
if they do intersect then it is just a surjection
hidden haven
The surjection from N_1 dirsum N_2 to N_1 + N_2 is given by (n, n') → n+n'
you can see it from this
it is clearly a surjection
plucky flicker
so M + N = {m + n : m \ in M and n \in N}
hidden haven
yes
plucky flicker
isn't the direct sum the same + the extra condition that every vector has a unique m + n representation?
hidden haven
yes
plucky flicker
like for vector spaces
hidden haven
but we don't have that here
M + N is inside the large module
and may not be a direct sum
M dirsum N is being done abstractly
Where we formally treat them as different modules, 0 intersection
like taking product of 2 groups
and it is defined by (m,n) for m in M and n in N
with pointwise operations
hidden haven
The surjection from N_1 dirsum N_2 to N_1 + N_2 is given by (n, n') → n+n'
Then you have this
hidden haven
plucky flicker
Here is a definition: a ring is called (left) hereditary if the homomorphic image of any injective module is injective.
So, it means that it should work for all module homomorphisms or there exists a module homomorphism such that the image of every injective module is injective?
maiden ocean
The former, I'm not really sure what you mean by the latter
Homomorphisms are maps from one module into another
@chilly ocean module such that any map into it lifts along monomorphisms
plucky flicker
Homomorphisms are maps from one module into another
Yes, i know that
maiden ocean
Then what do you mean by "there exists a module homomorphism such that the image of every injective module is injective"
plucky flicker
So it should be true for all module homomorphisms? Like if we have M as an injective module then f(M) is injective for all f?
maiden ocean
From the definition as stated thats what id assume
(this isnt the definition of a hereditary ring ive seen but idk it could be equivalent i am not super knowledgeable on hom alg)
maiden ocean
actually yes it is equivalent
Yea
This might seem less weird if u think of it as quotients of injective modules are injective lol
chilly ocean
what kinda algebraic structure is the extended real line?
I'm pretty sure it can't be a ring or a field
what is it???
limpid edge
from wiki:
R not even a semigroup, let alone a group, a ring or a field as in the case of R
I remember in axler chapter 1 there was an exercise asking you to show it’s not a vector space over R
chilly ocean
> R not even a semigroup, let alone a group, a ring or a field as in the case of...
does the same thing happen with the riemann sphere?
maiden ocean
Technically they are like
wheels
But i have never seen anyone actually cite this fact directly lol
limpid edge
so it's its own kinda algebraic structure?
sure? but i don’t think the same structure appears in other objects (that i’ve seen), so it’s probably not appropriate to label it as “it’s own algebraic structure”
chilly ocean
wheel theory 
chilly ocean
sure? but i don’t think the same structure appears in other objects (that i’ve s...
i see
that's true I guess
maiden ocean
the algebraic properties of P^1(C) as a wheel are just not that interesting or important to most fields
chilly ocean
I've heard that wheel algebra is non-distributive, is that true?
maiden ocean
idk what u mean by that
Wheels are given by a unary operation / that does distribute over multiplication and addition
Maybe "wheel algebra" here is referring to something other than that just the wheel itself
chilly ocean
oh I guess I must've misread something
random pendant
Let $\pi$ bethe Permutation representation associated to the action of G(finite) on G/H where H is a subgroup of G. What is the character of representation induced on G by G/H?
cloud walrusBOT
Averisera
random pendant
Frobenius reciprocity gives the induced representation when we are given the subgroup but how to calculate induced representation when quotient group is given?
chilly ocean
is there another example?
thorn delta
an invertible matrix Rn --> Rn
primal pier
What is a nil ideal? I assumed by definition it's an ideal that's nilpotent (and this is the definition given by Wiki) but I have it here as a theorem, which tells me there's another definition I'm not familiar with.
I'm just shopping around for various definitions, basically
Ah, never mind, I think this is an error in statement which mixed badly with my poor recollection of algebra. It's distinguishing nil ideals and nilpotent ideals.
Yep, I forgot the nil ideal -> nilpotent ideal step is nontrivial.
nvmd lol
lethal cipher
By the looks of it, a nil ideal is an ideal s.t. all of its elements are nilpotent.
lilac trench
groups of prime order are cyclic?
nvm I see it
but are Sylow subgroups always distinct
as in the only element they share is the identity
rotund shoal
Hey I have a quick question: So I'm reading a proof that any finite subgroup of the multiplicative group of a field is cycle, but I don't quite understand one of their initial steps:
They prove it by citing the fact that if |{x in G s.t. x^d = 1}| <= d for every d | n, then G is cyclic, and say that this applies "because the polynomial x^d - 1 has at most d roots". I don't get that logic tbh
holy shit I'm such a fucking idiot lmao
I legit didn't make the connection between x^d = 1 and x^d - 1 = 0.
lethal dune
how to show x⁴+1 is irreducible in lF_5[x]?
woven delta
You can brute force it by plugging in elements of F_5 to show that it doesn't have a decomposition with any degree 1 polynomials, and write (ax^2+bx+c)(ex^2+fx+h) = x^4+1 and show that there aren't values that satisfy that.
That's one method that will probably work, maybe there are better ones
lethal dune
saw it somewhere, but is there really no better way?
woven delta
You can have a computer just multiply all the degree 2 polynomials in F_5[x] if you want a way that's guaranteed to work and isn't hard to code lmao
The brute force method isnt that hard anyway so I would just do that
lethal dune
not that hard with sage math, also not the approach I was looking for but thanks
rustic crown
it's reducible
you can do it very easily without bruteforce
in fact you can say that x^4+1 is reducible modulo every prime
wanna see?
lethal dune
wait it's reducible?
next obsidian
What about mod 1
lethal dune
lmao sure
kind of feels like something from number theory, like x⁴+1 is reducible if p² cong 1 mod 8
rustic crown
you actually just need p^2 = 1 mod 8
which is true for every odd prime
so you want to see if x^4+1 is irreducible or not
if it was, then Fp[x]/(x^4+1) is finite field of size p^4
but you can show that x^4 + 1 already factors completely over a finite field of size p^2
so that can't be irreducible!
cloud walrusBOT
det
rustic crown
x^p^2 - x factors completely over F_{p^2}
this forces x^4+1 to factor completely!!
?
(p = 2 is trivial, so this shows what we want)
lethal dune
cute
and what about x⁴+10x+1 in ℤ [x]. I had used the fact that x⁴+1 is irred iver F5, since it doesn't work anymore...
1*
rustic crown
okie this is probably a very bad way
nvm i think it was circular
i was going to show that galois group is S4
just bash lol
it's more that enough for this problem
by rational root theorem, check 1 and -1, they clearly not roots
so the factorization is
(x^2 + ax +1)(x^2 -ax + 1)
or
(x^2 + ax -1)(x^2 -ax - 1)
but these clearly doesn't work because x term in both is 0 and not 10
lethal dune
shell brook
whats the rule for classifying finite abelian groups, like I don't really see where all the different decompositions into Zn's come from? I know you can do like the maximum one which is just from the prime factorization, but for example, if like Z3 x Z3 x Z2 x Z5 is a valid decomp, is Z3 x Z6 x Z5 valid?
I think if someone has this theorem that is written in a comprehensible way that will be a sufficient answer to my question lmao
the one in my notes make no sense
thorn delta
part of the classification theorem says that you can write a finite abelian group as a product of cyclic groups of prime power order. But ye, if you have Zn x Zm for n,m relatively prime, you could always combine them together to get Zmn.
thorn delta
Znm = Zn x Zm for gcd(n,m) = 1?
well it is, but gcd(n,m) = 1 is the simple case
thorn delta
err, i take that back, what do you mean by gcd shit?
shell brook
okay like where do these come
lethal dune
these cyclic groups always confuse me, like why tf we use ℤn for multiplicative group of order ϕ (n), but Zn for cyclic of order n
shell brook
err, i take that back, what do you mean by gcd shit?
Sorry I just meant that they had to be coprime lol
It's been a logn day
thorn delta
oh, but that's why I said "relatively prime"
shell brook
Like is that actually the algorithm to get all these decomps?
you just start with the one based on the prime factorization
and combine relatively prime pairs
lethal dune
so the factorization is
(x^2 + ax +1)(x^2 -ax + 1)
or
(x^2 + ax -1)(x^2 -ax - 1)
specific question, why should the factorization be in the form?
thorn delta
nah, so going back to Z3 x Z3 x Z2 x Z5 and Z3 x Z6 x Z5. These are the same group. So to be more precise, the classification theorem says that a finite abelian group can be uniquely written as a product of cyclic groups of prime power order.
so lets look at 144 = (4^2)(3^2). What are all the different ways to write product of cyclic groups of prime power order such that the order of the whole product is 144?
shell brook
10 
thorn delta
well okay i guess you have the answer in front of you lmao, but do you see why each one of those groups have to be distinct, and why that makes up all of them?
rustic crown
specific question, why should the factorization be in the form?
leading coeffcients can be assumed to be 1. the constant terms need to multiply to 1, so they are units. both 1 or both -1. since cubic term is 0, the we need to something like +a and -a
shell brook
Like it looks like they just combine any two Zn, Zm that with gcd(n,m) = 1?
I have tried that on the actual assignment question I have in front of me and it seems right
thorn delta
Like it *looks* like they just combine any two Zn, Zm that with gcd(n,m) = 1?
nope, they don't do that anywhere, all of the cyclic groups here have prime power order.
If one of the factors was Znm for gcd(n,m) = 1, then nm is not a prime power
lethal dune
to cancel out cubic/ x terms.. ok
shell brook
Oh
oh I get it
prime power means something
Right cause like Z9 x Z16 is 3^2 and 4^2 lmfao fuck
yeah yeah okay that makes sense
thorn delta
lets compare Z9 x Z16 and Z9 x Z8 x Z2.
The classification theorem says that every abelian group is uniquely the product of cyclic groups of prime power order. (9)(16) = 3^2 * 2^4 = 144 and (3^2)(2^3)(2) = 144, so these groups have the same order, but they have to be different because we've decomposed the factors into different prime powers
shell brook
Yess i get it
thorn delta
epic
shell brook
okay thanks so much
thorn delta
npnp
thorn delta
so that can't be irreducible!
how do you get that x^8 - 1 divides x^{p^2 - 1} - 1?
thorn delta
ah yea okay
next obsidian
whats the rule for classifying finite abelian groups, like I don't really see wh...
You can write it in two ways, either as products of primes powers, with possible repetition so like Z/2^3Z x Z/2^3Z x Z/2^2Z x Z/5^2Z x Z/7Z or some shit
you can also do it in the form of Z/a1Z x Z/a2Z x ... x Z/anZ where each ai divides ai+1
next obsidian
ah okay
shell brook
you can also do it in the form of Z/a1Z x Z/a2Z x ... x Z/anZ where each ai divi...
this
next obsidian
Yeah so
shell brook
this seems like? u need to order them or something first lol
next obsidian
you can break apart these things if they're coprime
so write it in terms of prime powers
then basically you just greedily do it
let me think of the best way to explain this
okay so like
Let's try to do it by induction actually
based on the number of factors
So like isolate it as say Z/p^nZ x (other shit)
and say, pick p to be the smallest prime
and n to be the largest power of p that appears
okay so (other shit) can be written inductively as
Z/a1Z x ... x Z/anZ
So we are dealing with Z/p^nZ x (Z/a1Z x ... x Z/anZ)
We good?
next obsidian
😢
shell brook
ill ask another time cuz i prob should figure this out lmao
and write me a blog post
next obsidian
then you can write the rest like that
then you just need to work that single factor in somewhere
basically now all you do is
you take the factors of p out of the various ai
and then kick them back one
then you're done
then put the Z/p^nZ into the front
lethal dune
what is the point of discrete valuation and discrete valuation ring, like asking for an overview
willow mason
i think they are important in algebraic number theory
next obsidian
DVRs are 1-dimensional regular local rings
The fact that you can describe every ideal so amazingly and that they’re noetherian local rings makes them very useful
In algebraic geometry they show up a lot as well, for a curve over a field or something of that sort, the curve is smooth iff all its local rings are DVRs
wooden ember
i mean u can just do a <:diligentClerk:913954051052404756> and write me a blog p...
It’s easier to first determine elementary divisors. By elementary divisors I mean look at the prime factorization of the order of your group and look at partitions of the highest prime power of each prime divisor
So like if your group has order 2^3 * 3^2 you could have elementary divisor lists 2,2,2,3,3 ; 2,4,3,3 ; 8,3,3 ; 2,2,2,9; 2,4,9; 8,9 right?
Then to get the invariant factors (the orders of the cyclic groups Z_a1, Z_a2,…,Z_ar with a(i+1) dividing ai) you can form a table for each elementary divisors decomposition: in this table make the columns a particular prime divisor and in those columns write down elementary divisors divisible by that prime in decreasing order. You’ll get columns of different size so just append 1s to the shorter columns to get a nice rectangular table
Then you take the products of the rows one by one to get your invariant factors
The reason we order our columns is to make sure the divisibility condition is fulfilled
And we can take products of rows because each row elements are pair wise comprime so that we can use the fact that Z_mn = Z_m x Z_n iff (m,n) = 1
Hope this helps
simple mulch
Suppose I have V,W vector spaces and T : V -> W linear
I have T injective, which means ker(T) = {0}
how does the author claim nullity(T) (which is the dim(kerT)) = 0 ?
kind temple
you said it yourself, ker(T) = {0}
the dimension of ker(T) is the dimension of the trivial vector space, which is 0
simple mulch
why 0 doesn't counts as an element of ker(T)?
why isnt dim(ker(T)) = 1?
oh yeah lol
kind temple
dimension is cardinality of a basis
smoky ivy
is this channel free now?
Let $R$ be a ring and $p=\sum_{i=0}^{k} a_{i} x^{i}$ and $q=\sum_{j=0}^{\ell} b_{j} x^{j}$ polynomials such that $a_{i}, b_{j} \in R$ for all $0 \leq i \leq k$ and $0 \leq j \leq \ell$. Define the product of $p$ and $q$ as the polynomial
$$p q:=\sum_{r=0}^{k+\ell}\left(\sum_{i+j=r} a_{i} b_{j}\right) x^{r}
$$
Let $R=\mathbb{Z}{6}$ (quotient ring), $p=a x^{2}+b x+c$ and also $q=d x^{2}+e x+f$ such that $a, b, c, d, e, f \in \mathbb{Z}{6} \backslash{0}$. Determine all the possible polynomials $p$ and $q$ such that the product $pq$ is the zero polynomial $\sum_{i=0}^{4} 0 x^{i}$. Is there any way to solve this elegantly instead of breaking it down into separate cases?
cloud walrusBOT
lewis
smoky ivy
can you expand on this
rustic crown
yea wait
cloud walrusBOT
det
rustic crown
this is basically reducing mod 3
if p * q = 0, then the same should hold when you go down mod 3
but Z/3Z is an integral domain, and so is (Z/3Z)[x]
so either p = 0 or q = 0 mod 3
smoky ivy
wait, i've never seen the notation of Z/6Z
rustic crown
ah, ignore that then, it's the standard notation... because of quotient rings
smoky ivy
oh, okay
rustic crown
you'll see this when you get to that
rustic crown
so let's assume that p = 0 mod 3
we can do the same mod 2
so we'll get p = 0 or q = 0 mod 2
now both p = 0 mod 3 and mod 2 isn't possible
because it forces all coefficients to be 0 mod 6
which contradicts that a, b, c are non-zero
smoky ivy
so we'll get p = 0 or q = 0 mod 2
so wait, are you reducing the problem to mod 2 here?
rustic crown
so we must have p = 0 mod 3 and q = 0 mod 2
smoky ivy
saying if it's 0 mod 6, it has to be 0 mod 2?
rustic crown
yep
smoky ivy
okay, got it
rustic crown
specifically we're using the ring homomorphisms,
(Z/6Z)[x] --> (Z/3Z)[x]
and
(Z/6Z)[x] --> (Z/2Z)[x]
which is a more precise way of saying "just go mod 3 or 2"
so p looks like 3 * something and q = 2 * something else
smoky ivy
okay, yeah, i get the way of solving it a bit i think, but how would i determine all the possible polynomials now?
rustic crown
um so they say that a, b, c are non-zero mod 6
and if they are 0 mod 3, then they are forced to be all 3
so p = 3(1+x+x^2)
for q, each coefficient can be 2 or 4
smoky ivy
for q, each coefficient can be 2 or 4
why can't?
rustic crown
oh my bad
smoky ivy
aren't they forced to be 2 or 4
rustic crown
yea 😓
smoky ivy
okay, i got a bit confused there
wait, i didnt fully follow
so technically, we have that pq should be 0 mod 6
rustic crown
yep
smoky ivy
so it should also be 0 mod 3 due to ring homomorphism
so it's either p = 0 mod 3 or q = 0 mod 3
and we just divide it into those cases
im just not 100 percent sure on how to reason it tbh
it is pretty intuitive, yeah
this channel is occupied
rustic crown
if image of p under the map (Z/6Z)[x] --> (Z/3Z)[x] is 0, then it lies in the kernel
is that what you're looking for?
smoky ivy
i mean yeah, technically what we're looking for is the image of p that maps to 0
so the elements of the kernel
of the map
same for q i guess
i think the main problem here is to concretly turn it into a formal proof
for me at least
rustic crown
i don't see what's not formal in this proof lol
smoky ivy
nono, i didn't say your way of proving it is not formal
im just not sure on how to formulate it correctly
i guess
my way of solving it was way less elegant
rustic crown
not sure what you're looking for... but maybe the isomorphism
(Z/6Z)[x] = (Z/2Z)[x] * (Z/3Z)[x]
helps you a little.
say p is sent to (p1, p2) and q is sent to (q1, q2)
we want p1q1 = 0, p2q2 = 0 in the corresponding rings
then use that these are integral domains and stuff...
it's pretty much the same thing, but said via the isomorphism
rustic crown
yep that's right... just one tiny thing maybe... you probably wanted to write Z[x,y] as (Z[y])[x] and not (Z[y])(x)
i'm not sure if latter means something outside of field theory.
prisma thunder
yep that's right... just one tiny thing maybe... you probably wanted to write Z[...
Rational functions in terms of x with coefficients as polynomials in y over Z
I think that’s what that is
vocal wolf
is there a list of theorems/techniques one can use to simplify quotients of polynomial rings?
im struggling with even the most basic Q[x,y]/(x-y) = Q[x]
vocal wolf
yeah thats my difficulty in this case
ker subset of (x-y)
so do you basically just spam the isomorphism theorems? and work out the details if you need to?
hidden haven
what is the map you have defined?
vocal wolf
p(x,y) -> p(x,x)
hidden haven
Right, so x-y is clearly in the kernel
to show the other inclusion
try division
with remainder
vocal wolf
ahh good one
vocal wolf
is there a list of theorems/techniques one can use to simplify quotients of poly...
but yeah can anyone give a short list?
Isomorphism theorems, division, grobner bases
oak grove
Can someone help me track the errors here? my teacher is letting me do corrections
I felt alright about this but I'm at a 30% on this question
im sorry my handwriting is so bad 
rustic crown
so the splitting field is the smallest field containing all the roots of a polynomial
you want the polynomial to factor completely
oak grove
so the problem would be not mentioning roots?
like if the root is not present in F, then F(a) would be the smallest field which allows us to fully factor f
rustic crown
yea, but a priori it only tells you that you have one factor
consider the polynomial x^3-2
rustic crown
just adjoining one root isn't enough
it will factor as (x - alpha)(x^2 + alpha *x + alpha^2)
and the second factor would be irreducible
so the problem says that, this shenanigans doesn't happen if f is deg 2
oak grove
okay
I see what you mean
Ive missed the purpose of the problem
for a degree two polynomial, even if neither root is in F, the splitting field will be F adjoin a single element
rustic crown
yep
oak grove
okay 😄 thank you det 🙇
oak grove
just adjoining one root isn't enough
wait, im still confused 🤔
so if you are missing both roots, how can you be sure that adjoining a single one will capture the other?
does like
multiplicity two or one root is in F capture all cases?
rustic crown
say f = (x - alpha)(x - beta)
we know the spilitting field is F(alpha, beta)
we need to show this is same as F(alpha)
so can you show why beta can be written in terms of alpha (and stuff from F)?
oak grove
so can you show why beta can be written in terms of alpha (and stuff from F)?
I do not, but Im gonna play with it
oak grove
is it because the root of the discriminant is added to the field through the extension?
rustic crown
F(alpha) = F(beta) = F(sqrt(D))
rustic crown
then you can reach either through operations
yep! that's the entire reason... alpha + beta lies in F because f in F[x]
beta = c - alpha for some c in F
rustic crown
issokie uwu
sinful mirage
here the authors jumped a step, right? In defining S
since Z,delta is a tensor product, there exists a unique linear map S:Z->Y, so that the diagram commutes
my point is for any vector space Y it exists
hence,in particular for V
hidden haven
They aren't defining S, you are supposed to define it
and make it satisfy those conditions
unique berry
I refuse to believe that this is the simplest way to show that finite fields aren't algebraically closed
thorn delta
its just the first part that explains why finite fields aren't algebraically closed. Its essentially the same kind of reasoning that there are infinitely many primes
small bison
Get ducked idiot
unique berry
Lmao that’s a pretty sick proof
I mean I think so too but at the same time I'm just tryna get my homework done today
unique berry
oh I see
unique berry
wtf is T is it just anything
unique berry
When I see capital letters I think sets
Why do wikipedia editors have to be so weird with notation
unique berry
but x is even more standard
next obsidian
¯_(ツ)_/¯
unique berry
oh well whatever yeah, the proof is actually really nice
Still would like to learn more about the other stuff?
I know the frobenious automorphism is used in the proof of hasse-weil bound
no idea what the other stuff is
sinful mirage
does anyone know a book,where I could find proof for existence of tensor product of vector spaces?
chilly ocean
a&m
sinful mirage
A&M=?
chilly ocean
atiyah macdonald
introduction to commutative algebra
they do it for modules in the second chapter
take base ring a field and you have vector space
sinful mirage
any proof which does it explicitly for vector spaces? 
this uses words like 'free module' and other undefined notions for me
chilly ocean
i think "free" is going to come up no matter
next obsidian
Just replace that with the word vector space on a basis of size “the set”
next obsidian
does anyone know a book,where I could find proof for existence of tensor product...
Anyway
For vector spaces the tensor product you can just make in some stupid way
Lol
sinful mirage
wdym?
next obsidian
If you want to make V (x) W
sinful mirage
I am trying to prove existence and uniqueness from universal property
next obsidian
And V has a basis {xi} and W has basis {yj}
sinful mirage
if i choos basis i just take linear combinations of tensor prod of basis vectors
sinful mirage
yeah,i'd like basis free
next obsidian
If you want a basis free thing you’re just gonna do what you do for modules
sinful mirage
cause if I choose a basis,then the whole power of commuting diagram loses sense
next obsidian
?
sinful mirage
and I am trying to understand commutative diagram proof
sinful mirage
I mean aren't commutative diagrams useful,cause they give natural identifications(basis free)?
next obsidian
No they’re useful because they’re useful
Like
Once you know you have some construction
You can throw it away and work only via universal property
If you want a basis free interpretation, you know there’s one you made via a basis
sinful mirage
why would I do that if I need to choose a basis anyway later on?
next obsidian
This seems like it’s going in the other direction
It’s like you’re saying you need a basis so why make it basis free