groups-rings-fields

[chilly ocean]

hmm

why 1?

[rustic crown]

it's just a nice element lol, if it's surjective then 1 is in the image. show that this would mean that n isn't divisible by p

conversely show if n isn't divisible by p then the map is onto

another way to think would be to produce an inverse map. it's clear that multiplication by n should have inverse which is division by n.
we can't introduce factors of p in the denominators, so n shouldn't be divisible by p.

[chilly ocean]

im so confused

the way i learned it

was to say let this be an element in the codomain

and then say there must be something in the domain that maps to it

[rustic crown]

right

so if there is a/b (with b not divisible by p) which maps to 1, then we would have that na/b = 1

which means na = b

can n be divisible by p for this to happen?

[chilly ocean]

hmm

so ur saying

we get an element na/b

from the codomain right?

[rustic crown]

yep

[chilly ocean]

then

then what

lol

[rustic crown]

lol if the map is surjective, then you can take preimage of 1 which is in the codomain. Say this preimage was a/b in Z_(p)

by the definition of Z_(p) you may assume that b isn't divisible by p

[chilly ocean]

yeah

[rustic crown]

use this to show then n isn't divisible by p

[chilly ocean]

fuck man

idk why this is so hard for me

[tall jay]

it be like that sometimes

more than sometimes for me

[chilly ocean]

same

[rustic crown]

it gets better once you're used to it, but still not immune >.<

[tall jay]

I asked about this particular exercise before, and I was wondering if this argument is valid now that I finally wrote out the proof

[chilly ocean]

i still dont know how to formally say this

[rustic crown]

stare at this equation and notice what could go wrong if n was actually divisible by p...

[chilly ocean]

i cant see it bruh

fuckkkkk

[rustic crown]

.<

[chilly ocean]

whyyy

[rustic crown]

if n was divisible by p, then so would be n*a

now na = b, so they can't be both divisible by p and not divisible by p

[chilly ocean]

yeah

i get that now

[rustic crown]

take some rest maybe >.< and think about it slowly when your head clears up

[hidden haven]

Is good

[tall jay]

You think "We argue similarly for the columns" will fly?

[rustic crown]

does the previous exercise prove both left and right cancellation?

or does that also have "we argue similarly"

[tall jay]

This was said previous exercise

[hidden haven]

Then you add a line to the previous statement for right cancellativity

[rustic crown]

(btw i think you shouldn't use <=> in problem 27... like the very first step, you don't know if the backward implication is true, that's what you're trying to prove in the first place)

[kind temple]

go from bottom up, starting on the left half, then top down on the right half.

b = eb = (a^-1a)b = ... = c

[hidden haven]

[tall jay]

For exercise 28?

[rustic crown]

one line

[hidden haven]

27

They are saying you can remove implications entirely

Start with b and you have a sequence of equalities going to c

[tall jay]

Why do I need b=...=c? Isn't it just asking to prove ac=bc -> b=c?

[hidden haven]

ab = ac → b = ... = c is what c² is proving

It's not important lol

Another way to phrase the argument

[tall jay]

Ohhhhhh, oh oh, okay I see

[dusty sapphire]

How to approach?

[rustic crown]

is that true? what if G was cyclic of order 49 and H was a subgroup of order = index = 7

[dusty sapphire]

Answer is false

[rustic crown]

wait then i just gave away the answer >.<

[hidden haven]

Pranked

[dusty sapphire]

hehe

[bleak abyss]

Det got got

[dusty sapphire]

?

[hidden haven]

??

[dusty sapphire]

How to count total number of maps which satisfies this

[rustic crown]

24 >.<

[dusty sapphire]

48

[rustic crown]

lol reflections as well >.<

[lethal dune]

[rustic crown]

[dusty sapphire]

Please tell the approch

[rustic crown]

6 ways to choose the top face, 4 rotations of that, 2 ways for the orientation/reflection

this is like applying orbit stabilizer theorem from group theory, but it's intuitive enough to not say this name

[dusty sapphire]

Thank you so much.

[prisma shuttle]

hey guys

in this article: https://en.wikipedia.org/wiki/Krull's_theorem

at the end of the variants section can someone explain the application of zorn's lemma

and why the fact that the identity is not in it matters

idk i thought zorn's lemma was like related to posets but idk wut posets they are talking about here

[rustic crown]

its the set of all proper ideals under inclusion

you would need to show that every non-empty chain has an upper bound

one idea usually is to take unions of elements in the chain, so union of all proper ideals in that chain

[prisma shuttle]

so does showing that the identity is not in the intersection show it has an upper bound

by the ring R itself

[sharp sonnet]

it shows it is still proper

[prisma shuttle]

does this count as an upper bound

[rustic crown]

union?

[prisma shuttle]

oh its union

wait really

i thought it was intersection

oh wait no nvm

[sharp sonnet]

the union of a collection of sets contains any set in the union and as it is still proper it is an upper bound

[unreal portal]

quick sanity check, for an ideal $I$, $I\subset\sqrt{I}$, right?

[cloud walrus]

cgodfrey

[hidden haven]

yes

[chilly ocean]

quick sanity check, the voices are not real, right?

[delicate orchid]

they can be thought of equivalence classes on the reals but they're not real numbers themselves😌

[delicate orchid]

r^1 = r moment

[hidden haven]

Is it already make abstract algebra chill pm

[delicate orchid]

$\pm$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

and yes moldi I'm free from exams at long last we must celebrate by computing more character tables

[hidden haven]

character table of D24

GO

[chilly ocean]

Moldi when is your exam session

[delicate orchid]

do you mean D12 or do you mean D24

[hidden haven]

sem just started

[chilly ocean]

Do you mean D6 or D12

[hidden haven]

Do you mean D3 or D6

Haha I have the last laugh

D1.5

[delicate orchid]

hmmmm

[chilly ocean]

D1.5 = halfenions

[delicate orchid]

ok since D1.5 is obviously a proper subgroup of D3 it must be abelian, and as we all know all abelian group are real modules (think about it but not too much) so we can define D1.5 = <a, b : a^1.5 = b^2 = 1, bab = a^1>

it's too easy sometimes

[hidden haven]

D3 is abelian?

[chilly ocean]

Yeah

[delicate orchid]

no but it's the smallest non-abelian group

[lethal dune]

halvelian

[delicate orchid]

thus any groups smaller than it must be abelian, including proper subgroups

[chilly ocean]

So true

[lethal dune]

D1.5 cyclic

[hidden haven]

It must be true, it has 3 elements and 3 is prime

[lethal dune]

yes

[chilly ocean]

Guys I’m still kinda stuck on this I think I got the invective part right?

For part b

I still don’t know how to to surjective

[delicate orchid]

n = 1 is trivially surjective

[runic hemlock]

it's an isomorphism whenever n is not divisible by p.

Because then you can define an inverse: a/b-> a/(nb)

[delicate orchid]

that seems well defined tbf

that's my one concern

[chilly ocean]

hmmm

[oblique leaf]

can anyone give me an example of a Noetherian module with a submodule that isn't Noetherian?

[hidden haven]

There isn't one

[lethal dune]

a submodule S of N S ⊂ N ⊂ M is a submodule of M as well so S is f.g => N is noetherian

[chilly ocean]

can you give me a concrete example

if its possible

i think i see it

but not sure

[lethal dune]

n=1

[runic hemlock]

p=2, n=3

i guess

[oblique leaf]

@hidden haven i was just asking because my homework asks me to prove this statement, with A, B, and C being R-modules and R being a noetherian ring. I am not sure where I should use the fact that R is a Noetherian ring. It seems to me that this statement can be proven even if R is not a noetherian ring?

[hidden haven]

Yes R need not be noetherian

[chilly ocean]

wait

so

[oblique leaf]

@hidden haven Thank you!

[chilly ocean]

i would have 3a/b right ? @runic hemlock

[runic hemlock]

yeah

[chilly ocean]

where does the p take play in all of this

[runic hemlock]

a/b goes to 3a/b

[chilly ocean]

yeah

[runic hemlock]

you can define a different homomorphism, a/b goes to a/3b

the second homomorphism is well defined because if b is odd then 3b is also odd

and it's an inverse to the first homomorphism

[chilly ocean]

how tho?

yeah i get its the inverse

but how does p have anything to do with this and how is is a/b mapped to a/3b?

[runic hemlock]

It's legal to define new functions, the police wont come to your house if you do it

[chilly ocean]

lol

[runic hemlock]

no one's stopping me from defining a new function f so that f(a/b)=a/3b

I just need to check that a/3b is in the group Z_(2), which is true

[proud bear]

for example, if n=p=3, the inverse to varphi maps c/d to c/(3d) but c/(3d) is not always an element of Z_(3)

[chilly ocean]

varphi?

if n and p were 3

a/b would be mapped to 3a/b using phi

[proud bear]

Yes

and phi^-1 would have to map some element c/d to c/(3d)

[chilly ocean]

yeah

[proud bear]

but if c/d=1/1, then 1/1 would map to 1/3, which is not an element of Z_(3)

[chilly ocean]

ahh

i think i get it

thanks

[tall jay]

How does one prove that if $f \in S_n$, $f$ composed with itself $k$ times is just the identity permutation?

[cloud walrus]

beeswax

[tall jay]

I'm not sure where to start

[next obsidian]

Uhhh

You can’t?

If k is a random number

[tall jay]

k is a positive integer

[next obsidian]

I know

But again, you can’t prove anything here

[tall jay]

This was the exercise

[next obsidian]

This just follows because S_n is a finite group

Every element has to have finite order or the group is infinite

Use the pigeonhole principle to show that f^k = f^k’ for distinct k,k’ then divide

[tall jay]

Is that combinatorics?

[chilly radish]

Well the principle is combinatoric in nature but it's very simple. Consider the set
{f,f^2,f^3,....,f^n!, f^(n!+1)}
Then this set has (a priori) more elements than S_n, but S_n is closed under composition, so this is a subset of it, therefore there are some two elements f^k, f^k' that are in fact equal like chmonkey said (We can't product a set bigger than S_n using an element from S_n with the action of composition).

This is the pigeonhole principle, if you have n pigeons and n-1 holes, some hole must have at least 2 pigeons in it

[tall jay]

hmmm

Yeah, I don't think I'm seeing how we can deduce that it ends up being the identity

[chilly radish]

You have f^k=f^k' right

Suppose WLOG k>k'

They can't be equal because we assumed the k, k' are different

Now multiply both sides by f^(-k')

What do you get

Or compose ig

That's just composijg the inverse of f with itself k' times

[tall jay]

Thank you

Gonna play around with it and do some examples, so I can understand

[barren sierra]

ok so I am lost as to how free =/= faithful

(G is acting on X btw)

is the stabilizer of x = {e} for all x in X

and the kernel of an action is the intersection of all stabilizers

then how are those not the same thing

[hidden haven]

Faithful means that no element acts trivially on the entire set

Free means that no element acts trivially on any element

Free is stronger

Because you are no longer considering the permutation (12) of S_3

because it acts trivially on 3

[barren sierra]

oh so all free actions are faithful

[hidden haven]

Yes

[barren sierra]

got it

cool

that's poorly written in that part of the lecture video 💀

[hidden haven]

💀

[barren sierra]

at least to me it is

[hidden haven]

I agree

One thing that many intro books don't mention about free actions is that it is free in the sense of free group, free module etc

In the sense that if you give me a set S I can give you a free group generated by S, and similarly if you give me a group G and set S, I can give you a free action of G related to S

[next obsidian]

S_n acts faithfully on [n] but not freely

I think the best example of free actions is G acting on itself via multiplication

And IIRC all finite free actions are in some sense kind of like that?

[hidden haven]

The underlying set of the action is {(g,s) | g in G, s in S}

And G acts by just multiplying the first entry

And the given definition of free action is equivalent to saying that the action you have is isomorphic to an action of this form

[next obsidian]

Free actions are important

[lavish nexus]

how do we get = in the highlighted part

[next obsidian]

In AG for example when you act on schemes or something, if you quotient by the action and it isn’t free what you get is bad

You have to go to a stack usually

When it’s free I think you often get a scheme, and at worst it’s an algebraic space?

V(F1F2) = V(F1) U V(F2), and because F1F2 is in I(V), V < V(F1F2)

So V = V\cap V(F1F2) = V\cap (V(F1) U V(F2))

Then you just distribute

[lavish nexus]

I see I forgot the last part

ty

[compact kelp]

Is there anything special about group representations that are also lie algebra representations? or is this just something that happens with representations of matrices?

[barren sierra]

oooooo ok that makes sense but yea I didn't see that in the lecture videos or the text

[barren sierra]

does "describe" mean find some canonical group that the stabilizer is isomorphic to?

so probably some other permutation group?

for 5(iii)

this problem seems tedious as all hell

[barren sierra]

Also suppose G acts on X and there is only one orbit, is that orbit the whole set X?

it is right? since orbits are eq classes over X

[gritty sparrow]

I think they want you to find the actual elements of S3 that are in the subgroup. This is automatically a permutation group as it is a subgroup of S3, also all the subgroups of S3 have names

[gritty sparrow]

yes that is the case

[barren sierra]

cool

[barren sierra]

ah ok

still tedious

I wonder if I can have Sagemath do this for me

[gritty sparrow]

dummit and foote is full of tedious group theory exercises lol

[barren sierra]

just cause D&F has them doesn't mean my prof had to assign them

[gritty sparrow]

true, but it is worth it to get your hands dirty at least a little bit.

[barren sierra]

yea

I guess

I'll get them dirty with sagemath, it'll be fun

[oblique leaf]

can anyone give me an example of a Z module that is neither free, nor finitely generated, nor a direct product or sum of a finitely generated Z-module with a free Z-module?

[thorn delta]

Q

[oblique leaf]

@thorn delta thank you! I understand that Q is not free and not finitely generated, but how do i show that it is also not a direct product or sum of a finitely generated Z-module with a free Z-module?

[thorn delta]

oh uhh, i kind of understood you incorrectly on that part. Im not sure

[oblique leaf]

ok no worries thanks anyway!

[gritty sparrow]

I think I have a proof: assume that R + Z^{d}=Q as a Z module, then tensoring both sider with Q we get (R tensor Q) +(Q^{d}=Q. (Q tensor Q is just Q and Z^{d} tensor Q is just Q^{d}). So split into cases: if d=0 is impossible as then we would have R=Q which can't happen as R is finitely generated. So d>0. Now d=1 as otherwise the dimension of lhs is too big. So we must have that d=1 and R tensor Q=0. Now we have that Q=Z+R hence this now says that Q is a finitely generated Z module but we now that isn't possible, hence we are done

[gritty sparrow]

edited to remove the flatness argument that was unnecessary

[thorn delta]

what is R supposed to be?

[gritty sparrow]

some finitely generated Z module

[oblique leaf]

can this be done without tensors at all?

[gritty sparrow]

probably yes, I'm just lazy

[oblique leaf]

i haven't learnt what they are yet and it seems that the problem should be possible without then

[oblique leaf]

oh ok

[thorn delta]

it looks like your initial assumption is that Q is the sum of a finitely generated Z module and a free module of finite rank?

[gritty sparrow]

d is not necessarily finite

[thorn delta]

okay okay

[gritty sparrow]

it can be any indexing set

Okay I have an argument without tensors: assume that Q=R+Z^{d}. let {e_i} denote the image of the standard basis of Z^{d} in Q. say d>1we know that no nonzero Z linear combination of say e_1 and e_2 can be zero in R+Z^{d} but in Q any two non zero rationals has a nonzero linear combination which is 0. hence d>1 is impossible, therefore d=1 or 0, but then Q is finitely generated which is again impossible

[next obsidian]

If you just want to show Q isn’t isomorphic to a free + finitely generated apply the structure theorem to the finitely generated part

Either the fg is free in which case Q is free, which we were assuming we know is impossible

Else Q has torsion, clearly false

[gritty sparrow]

nice argument

[prisma shuttle]

can someone explain why the purple part is true

i understand intuitvely why its true but how does it "follow from Proposition 2"

[next obsidian]

Lmfao wut

This is silly

Forget proposition 2, in general if I is an ideal of R you can look at the ideal I generates in R[x]

It’ll be equal to the polynomials where all the coefficients are in I

Then you can define a map R[x] -> (R/I)[x] which sends a polynomial a_0 + … + a_nx^n to (a_0 + I) + … + (a_n + I)x^n

The kernel is exactly the polynomials where every coefficient is in I, which is the same thing as IR[x] which is usually denoted as I[x]

[acoustic fossil]

Would the following be a valid start to a proof? Given two sets A and B, let X, Y be subsets of A. let f be a function A onto B. I want to prove/disprove if X is a subset of Y, then f(X) is a subset of f(Y). My thought would be a proof by contradiction. So, assume for a contradiction, X is subset of Y, then f(X) is not a subset of f(Y). This would occur when there is some element in X that maps to B, while there is no element in Y that maps to that element in B. Thus we have a contradiction?

[coral shale]

Not a clue, but to show a contradiction, you construct an explicit counterexample.

[next obsidian]

I mean you could certainly prove this by contradiction…

But I think it’s just way easier to do it directly

Also you don’t require f to be onto, this holds for any f

Also the end of your proof is really murky. You don’t explain why this is a contradiction

[next obsidian]

Also this isn’t really true. You don’t need an explicit counterexample, you need one to show something isn’t true via counterexample. You just need to derive a contradiction to do a proof by contradiction

[coral shale]

That is true, mb. However, I personally find it hard to be convinced by this proof at a glance as the quantifiers are not explicit (I would have to check carefully that we do not have a vacuous truth in there we've taken to be false or something)

[next obsidian]

Yeah I mean the end is just… not clear at all like

You would have to say something like “but this can’t be true since anything in X is in Y”

But yeah it’s just not written clearly

Also it says some things which don’t make sense, like an element in X that maps to B, but if you just adjust it it sorta makes sense

They’re just using the definition of f(X) not being a subset of f(Y)

[coral shale]

The thing is, I think it is impossible to prove this general statement via contradiction

Because there clearly IS some f for which this would be true

[next obsidian]

No you can

[coral shale]

A counterexample (even if general) seems to be necessary

?

[next obsidian]

They’re proving that X < Y implies f(X) < f(Y) via contradiction

[coral shale]

Ah ok.

[next obsidian]

If this weren’t true you have some X < Y, and f:A -> B such that f(X) isn’t a subset of f(Y)

[coral shale]

I work really badly with words

[next obsidian]

Also this shouldn’t be in this channel anyway, it should be in #proofs-and-logic or something

[coral shale]

true

[acoustic fossil]

Thanks for the advice. I think I resort to contradiction because it feels like a very consistent method of proving things. It's a contradiction because every element of P must also be an element of Q. Thus, every element in the range of f(p) must also be in range of f(q).

[coral shale]

Last thing

'So, assume for a contradiction, X is subset of Y, then f(X) is not a subset of f(Y)'

Surely this is not the opposite statement

[acoustic fossil]

noted. thank you

[next obsidian]

This is

To prove P => Q by contradiction

You start with P and not Q

Then derive a contradiction

[coral shale]

This is not P and not Q

[next obsidian]

No

[coral shale]

This sounds like P => not Q

to me

[next obsidian]

The statement is

If X < Y, then f(X) < f(Y)

Not P would be X is not a subset of Y

[coral shale]

sorry I meant:

This isn't (P and not Q)

This sounds like (P => not Q)

[next obsidian]

No

You started with P (X < Y) and not W (f(X) is not a subset of f(Y))

From here you are able to derive the contradiction that X is not a subset of Y

Or you can contradict f(X) is not a subset of f(Y)

[acoustic fossil]

WTS: If X subset of Y, then f(X) subset of f(y). Proof by Contradiction: Assume for a contradiction, if X subset of Y, then f(X) not subset of f(Y). Continue until contradiction found

[next obsidian]

You really should be doing this proof via contraposition if you didn’t want to do a direct one but whatever, you can phrase that as contradiction

It won’t be if X subset of Y

That is now showing not P => not Q

You assume you have an example of sets X,Y; X < Y, but f(X) is not a subset of f(Y)

[coral shale]

My point is the phrasing
'So, assume for a contradiction, X is subset of Y, then f(X) is not a subset of f(Y)'
as opposed to
'So, assume for a contradiction, X is subset of Y, and f(X) is not a subset of f(Y)'

[next obsidian]

Oh

[coral shale]

These 2 are surely different things. We want the 2nd and not the 1st

[next obsidian]

Okay yes

Sorry, I noted that as wel

[acoustic fossil]

sorry is this to me or @coral shale ?

[next obsidian]

I just adjusted this so that it actually is a proof by contradiction

It was at you

You’re right Shuri, if you took what was written literally the quantifiers are wrong

I just adjusted it as I read it so it would be a valid proof

I didn’t realize that’s what you were getting at, my bad

[coral shale]

no np. I'm kinda drowning in words here. Just bad with them lol

[next obsidian]

No it’s alright, you were technically correct

The point here is what you wrote is

If X < Y, then f(X) is not a subset of f(Y)

This is P => not Q

Because you said “if”

To prove by contradiction you assume P and not Q

So it should be written, assume there exists X < Y such that f(X) is not a subset of f(Y)

[acoustic fossil]

So the if is the mistake?

[next obsidian]

From there you derive your contradiction

Yes

You’re now trying to prove something entirely different

[acoustic fossil]

Ok great. Next, you also recommended that it would be better to prove it directly?

[next obsidian]

Yes

Take something in f(X)

It’s of the form f(x) for some x in X

But X < Y so x in Y

So f(x) in f(Y)

There’s absolutely no reason to not do a direct proof

[acoustic fossil]

Thanks, that's far more succinct. Appreciate it

[tall jay]

Speaking of proof by contradiction, I have a proof check question on an exercise I was working on earlier'

Would this be valid?

[kind temple]

no

what if it gets caught in a loop

[tall jay]

Basically, I was trying to argue that we know #S_n = n!, which is a finite number. But, if there's no positive k value which makes f^k = I, then there'd be infinite permutations, but that contradicts the finite permutations for S_n, since S_n is a group w.r.t. composition and f is in S_n. thus, there must exist some positive integer k such that f^k=I.

[kind temple]

ik, but im not convinced that there would be infinitely many permutations

by this argument

[tall jay]

Ahh, I thought I had it

How is a loop possible here?

[kind temple]

just argue that <f> is a subgroup

<f> = {f^n : n in Z}

[tall jay]

We haven't introduced subgroups yet

We've only pretty much went through the def'n of a group and permutation/diheral groups

[barren sierra]

is there a non-shit computation way to do this

or am I just stuck doing

for some x in D_8

for all g in D_8

g x g^-1

please tell me no

obviously if x and y are in the same conjugacy class then I'm done with 2 elements but like

fuuuuuuuck me

[next obsidian]

I mean these groups are small as Shit

Just compute them lol

[barren sierra]

ok but is there no like

nice symmetry or trick?

[next obsidian]

The conjugacy classes partition it

I mean maybe

[barren sierra]

it just seems like a problem that doesn't teach me anything

[next obsidian]

For A_n there is a way to do it but

It’s complicated lol

[barren sierra]

Hm

how complicated lol

[next obsidian]

You have to classify conjugacy classes of S_n that split in A_n

And that’s built on classifying them in S_n

So for n = 4 it’s just easier to compute them lol

[barren sierra]

yea

ugh ok

[hidden haven]

Start by identifying the center, that will give you some conjugacy classes and also an upper bound on the size of any conjugacy class by orbit stabilizer

[desert dome]

what are the good ways to determine if a polynomial is irreducible? I usually just use Eisenstein criterion, but it doesnt work for the polynomial I am considering

[next obsidian]

there's no way to determine it for sure using any sort of algorithm

there's a lot of techniques you can try though, it depends a lot on what ring / field you're over and the degree

[desert dome]

For example, I am thinking x^3 - 3x^2 - 3x - 1 over Q

[hidden haven]

Rational root theorem should work

[next obsidian]

It suffices to show it doesn't have any roots in Q

because if it factored it either goes as a linear and a quadratic

or as three linears

either way it would imply the existence of a root in Q

and then yeah you can use the rational root theorem as Moldi said

if it was degree 4, once you show there's no root it could only factor as two quadratics

so you could try to set up a general system of quadratics

and just bash out all the numbers and show it won't work or something like that

[desert dome]

Thank you! I'll check the rational root theorem

[cursive holly]

i understand this proof completely other than how we know for a fact that there's 2 numbers $a,b \in H$ that multiply to $x$

[cloud walrus]

ALIAS

[south patrol]

They can be the same thing

i.e. we're saying that given any two (not necessarily distinct) elements of H, their product is also in H, but these two elements could even just be e and e (for example, in the case of the cyclic group with three elements, H is just {e})

[cursive holly]

oh, so we know there's some other solutions in H that give us e, and we know e*e = e

is that right?

[south patrol]

We don't need any 'other' solutions besides e (in fact, for closure alone, I guess you don't even need any elements to be in H)

perhaps i'm misunderstanding you

But essentially this is just how you show closure - you need to show that if a and b are in H, then so is ab (in fact we don't even need to know that there are any such elements at all, for example we would similarly argue that the empty set is a subset of {0,2} because for all x in the empty set, x is in {0,2})

[next obsidian]

You could flip how you view this as well

The image of any group homomorphism is a subgroup, and if G is abelian then the map x -> x^2 is a group homomorphism

Since (xy)^2 = x^2y^2

This is useful because it lets you compute say, the number of squares in a finite field via the first isomorphism theorem

[south patrol]

Good point

[prisma shuttle]

can someone explain the purple part

why is that a homogenous ideal

cuz isn't z^5 of deg 5 but the other terms are of degree 2

[rustic crown]

right, but z^5 is still a homogeneous element

what's bad is something like <x^2 +y>

[maiden ocean]

Homogenous ideal doesnt imply that all the elements of your ideal have the same degree (this isnt possible actually, because if f in I has degree n then xf has degree n+1)

It just means that homogenous polynomials in I generate your ideal, so that we can write any f in I as the sum of f_i in I homogenous

This is the same thing as saying that the homogeneous components of elements of I are in I

[barren sierra]

"this notation is bad, I shall use this better notation" proceeds to use bad notation exclusively

That's my prof rn lol

[chilly radish]

What's the notation in question

I shall.ovjectively decide if it's good or bad

[oblique leaf]

Given a module M, is it true that every proper finite subset of the generating set of M generates a submodule of M?

[rustic crown]

any subset of M will generate a submodule of M

do you want the submodule also to be proper?

[oblique leaf]

ok thank you! aren't submodules defined to be always proper?

[rustic crown]

nope.

submodule of an R-mod M is a subgroup, which is stable under the action of R

[lethal dune]

I like the terminology 'stable'

[rustic crown]

what if your generating set has redundancies?

[lethal dune]

[oblique leaf]

@rustic crown what about redundancies? isn't the generating set allowed to have redundancies

[rustic crown]

yea but in that case you won't get a proper submodule right

say M which is generated by {x, y, x+y} and the submodule generated by {x, y}

if there is at least one finite generating set then you can look at the one with minimal cardinality. that would satisfy probably what you're looking for >.<

[oblique leaf]

ooooh i see what you are saying

thank you for alerting me to that

@rustic crown also is it true that any module has a generating set that is either finite or infinite?

[rustic crown]

well any set is either finite or it's infinite

[lethal dune]

the module itself

[oblique leaf]

okok thank you!

[prisma shuttle]

hey guys for this statement

can someone explain what is the difference between (I) and (I,x)

don't they yield the same thing, namely I[x]

just the set of polynomials with coefficients in I

[rustic crown]

I[x] need not include x

[prisma shuttle]

wut

wdym

[rustic crown]

like consider Z[x] and I = 2Z

(I) = I[x] will be polynomial where all coefficients are even

[prisma shuttle]

wait so (I) = I[x] right

but (I,x) is not necessarily equal to I[x]

[rustic crown]

I haven't seen the definition I[x] before, so if you define it as what you've done above then yep

[rustic crown]

and (I, x) contains x which a polynomial having an odd coefficient

[barren sierra]

$D_{2n}$ for the transformations of a regular $n$-gon and $Gx$ for the orbit of $G$ acting on $x$

[cloud walrus]

Spamakin🎷

[barren sierra]

that's the poor notation

IMO

[hidden haven]

Gx is standard

G_x for stab x is annoying

[hidden haven]

By standard I mean across disciplines, not just when talking about groups

But G_x and Gx are easily confused when handwritten

[sharp sonnet]

the correct notation is x^G

[hidden haven]

[sharp sonnet]

g acting on x is x^g then you have (x^g)^h = x^(gh) and everything is nice

[hidden haven]

Oh for the orbit

Damn

[sharp sonnet]

yes, the stab stays G_x

[hidden haven]

Why can't we stab x

[sharp sonnet]

i can live with that

[coarse storm]

For suppose we stab x. Then we all act as the identity for x. In other words, we are x. And we would be stabbing ourselves. That would hurt!

[broken stirrup]

Hi, got a quick question. We say an element r of a ring is irreducible if r is nonzero and r=ab implies "EXACTLY" one of them is a unit.

What about prime numbers? Do we say p is a prime if p | ab implies p implies p divides "exactly" one of them or is it possibility that it can divide both of them?

[sharp sonnet]

it can divide both

[gusty halo]

an equivalent definition to the one you gave for irreducible replaces the "exactly" by the condition that r is also not a unit

and the usual definitions i know of these terms dont use "exactly one"

[sharp sonnet]

the reason for this is because units can always be rewritten by multiplying with more units but that isnt useful

[broken stirrup]

exactly, my definition is not the best in terms of the rigour but i thought it would fit my question better

[broken stirrup]

Thanks, that's perfectly sensible. Just like with usual prime numbers

[valid basin]

from the wikipedia page on the quotient group

is it true that it is unknown whether the normality of N is necessary to define the operation on G/N?

can't find any sources on this

[broken stirrup]

Yes it is, you can talk about a quotient grouo G/N if and only if N is normal subgroup

[valid basin]

this is true - but doesn't this only imply that the normality of N in G is sufficient to show that the operation is indeed well-defined in G/N?

[broken stirrup]

There are several reasons actually, my favorite intuition behind quotient groups is the one contrsuction with equivalence classes. Check out Hungerford's algebra, Theorem 1.5 for details

https://www.quora.com/Why-do-we-take-normal-subgroups-when-we-define-a-quotient-group

Also there are some good answers

[chilly radish]

'It remains to be shown' is used here in an instructive way, not to mean 'this hasn't been shown'

They just mean they haven't yet shown it to be true in the article

[valid basin]

ah right

that makes sense - thanks a lot

[grizzled elm]

for whatever reason I'm not understanding how we do the second part of this

for the first part we use eisensteins criteria, so we want a prime number that divides the non leading coefficients, namely $3$. then see that $3$ does not divide $1$ and $3^2=9$ does not divide the constant $a_0$

[cloud walrus]

Dpao

[grizzled elm]

For the second part, I think it involves using that given $p(\theta)=0$, we have $\mathbb{Q}(\theta)\cong \frac{Q[x]}{p(x)}$

[cloud walrus]

Dpao

[grizzled elm]

actually maybe we say that $p(x)$ irred. means rel. prime to $x+1$. Then use that a univariate polynomial ring over a field is a euclidean domain, so there exists polynomials $a(x),b(x)\in Q[x]$ s.t. $a(x)(x+1)+b(x)(x^3+9x+6)=1$

[cloud walrus]

Dpao

[grizzled elm]

is this on the right track?

[gusty halo]

yes, i think so

[grizzled elm]

so a(x)(x+1)=1 mod (x^3+9x+6)

[gusty halo]

so you can reformulate your equation above to $(1+\theta)(a\theta^2+b\theta+c) = 1$

[cloud walrus]

Phil

[gusty halo]

yes this

[grizzled elm]

ok great

[echo orbit]

This is from Dummit & Foote. I'm having a hard time understanding how we know this P exists and is a group.

This is the proof that there's a Sylow p-subgroup of any finite group for any p dividing its order

[cloud walrus]

grist bundle

[maiden ocean]

What in the world is Sym^n M(U)?

Or Sym^n of a module in general

google is not being particularly helpful and ive never seen this notation

[grizzled elm]

I'm having problems actually carrying this out

[gusty halo]

you can multiply it out, reduce via the given relation, and then compare coefficients to obtain a linear system of equations which you can then solve

[gusty halo]

so you using this notation, if i didnt miscalculate, you should obtain $(a+b)\theta^2 + (-9a+b+c)\theta -6a+c = 1$ which gives you the equations $a+b = 0$ and $-9a+b+c = 0$ and $-6a+c = 1$

[cloud walrus]

Phil

[gusty halo]

solving for the coefficents and pluggin them back into the generic element $a\theta^2 + b\theta +c \in \mathbb{Q}(\theta)$ then gives you the desired inverse of $1+\theta$

[cloud walrus]

Phil

[grizzled elm]

$1/4\theta^2-1/4 \theta+5/2$

[cloud walrus]

Dpao

[grizzled elm]

I think another way to do it if u do it in Q[x]/p(x) like I did is divide p(x)/(x+1) with remainder to get x^2-x+10 r -4

tyty

[obsidian sleet]

what does this notation mean

i assume it means the set of matrices in M_n(R) multiplied on the right by E_{ij}

[next obsidian]

It’s the symmetric powers, you mod out the n-th tensor power by the relations which make it so that swapping the simple tensors does nothing. So for Sym^2 for example you want x (x) y = y (x) x, it has the universal property of factoring symmetric multi linear maps. It’s almost like the exterior powers, but those are for alternating maps. A related construction is the symmetric algebra, sometimes just called Sym M which turns M into a graded ring via taking the n-th degree to be Sym^n M. when M is a free module on a basis B, this ends up being isomorphic to A[B], a polynomial ring on B so it’s sort of a coordinate free way to express a polynomial ring.

[echo orbit]

i get how we get P now

[obsidian sleet]

Sylow strong.

[wise igloo]

Sylow strong.

[obsidian sleet]

[lethal dune]

every matrix multiplied by Eij (on right), where i varies

[obsidian sleet]

alright thanks

[lethal dune]

question doesn't make sense Ig?

"number of elements that have order 4?" what?

[next obsidian]

uhh

it totally makes sense

there's a finite number of elements with order 4

it's a combinatorial problem. Write the elements as disjoint products of cycles, then the order is the lcm of the lengths of cycles. Since you only have 4 elements, the only way you could get this is by being a 4-cycle, and the number of these are 4!/4 = 3! = 6

try to justify everything I said!

they commute

no

a 2 cycle has order 2

if the cycles go like

a_1...a_n

then raise it to the k

a_1^k...a_n^k

you need k to be a multiple of the order of each cycle which is just the length

so it's a multiple of them

and then cuz order is the least it's jsut lcm

[hybrid island]

anyone here really good with abstract linear algebra and can help me out with proofs

[dusty sapphire]

Possible smallest value of $n$ such that $S_n$ have subgroup isomorphic to $\mathbb{Z}_2* \mathbb{Z}_2* \mathbb{Z}_2* \mathbb{Z}_2

[cloud walrus]

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

[dusty sapphire]

How to find?

[lethal dune]

Z2 means non identity element has order 2

you must find distinct just elements of Sn with order 2 that you can take prod with

[rustic crown]

is *, the direct product? (free product wouldn't make sense ig?)

we can definitely do it for n = 8, just slowly go down and see if smaller is possible.

[hidden haven]

I remember proving on this channel that 6 was the best possible for (Z_2)³

It was a bit annoying I think

[rustic crown]

was there some clever idea to show S5 doesn't have it or some type of bash?

[hidden haven]

Sylow

I think

That subgroup has cardinality 8

[rustic crown]

right

[hidden haven]

8 is the largest power of 2 dividing 5!

[rustic crown]

oh lol

and it would D4

[hidden haven]

And saketh identified some other sylow 8 subgroup which didn't look like this

Well yeah because there's an order 4 element

[rustic crown]

D4 is a subgroup of S4 which itself is a subgoroup of S5

[hidden haven]

16 would be sylow 2 subgroup of S_7 or lower

[hidden haven]

Right yes

[hidden haven]

Same trick should be applicable

[lethal dune]

[hidden haven]

Show that the order 4 element belongs to a sylow 2 subgroup

[rustic crown]

yea it shouldn't be hard to show sylow 2 for S_7 is non-abelian

it will contain D4 as a subgroup

[dusty sapphire]

for 8 i think i can find but for 7 why it is not tru?

[rustic crown]

like we can extend any 2-group to a sylow 2group

[hidden haven]

Ye this is the part I wasn't sure about

How do that

[rustic crown]

remember i told a proof of sylow a month ago or something?

[hidden haven]

Ye

[rustic crown]

which precisely extended by one step

[hidden haven]

[rustic crown]

or you can also directly do with group actions i think

[hidden haven]

I see

I'll accept that

[lethal dune]

share again

[rustic crown]

H be that p-group, P be a sylow p group, act by H on G/P
so there should be a non-trivial fixed point h(gP) = gP for all h
which means hg in gP, so h in gPg' for all h
showing H is a subgruop of a conjugate of P

[lethal dune]

nice

[rustic crown]

lemme search it up

.

[lethal dune]

I've seen a similar proof on conjugates

[hidden haven]

Nice

[dusty sapphire]

I think this type of facility not available for n<8

Or can i say that this subgroup is unique, i mean only one subgroup isomorphic to (Z_2)^4 in S_8?

[rustic crown]

nah, there are plenty like that

like replace (12) and (34) with (13) and (24)

sylow 2-subgroup of S_8 would have order 2^7 or something

so we don't get uniqueness by sylow

but we do get unique (up to iso) subgroup of order 16 in S_7

[dusty sapphire]

No i mean they are isomorphic to D_4, Z_8, Z_4*Z_4 and so on

[rustic crown]

(not sure i understand >.<)

we're claiming we don't have such facility in S_7, because a subgroup of order 16 will be the sylow subgroup and that can be easily seen to contain D4 so is non-abelian

[hidden haven]

Someone should make string diagrams for group theory so that we don't have to think while applying sylow

[hybrid island]

honestly willing to pay $ for someone to help me with my abstract lin alg dm me ;-;

need to get 100%

[rustic crown]

academic dishonesty is bad

[hybrid island]

[dusty sapphire]

Btw how to show that D_4 is subgroup of 2SSG in S_6 and S_7

[rustic crown]

any 2-subgroup lives inside some sylow 2-subgroup

and D4 definitely lives inside S4, and so also inside any S_n for n >= 4

(as D4 acts on a square, making the 4 vertices permute in some ways)

you could also just say that sylow 2 group will contain an element of order 4

because there are elements of order 4 in your finite group

[dusty sapphire]

Okay but that can be Z4* Z4 how to classify that that it is D4* Z2

[rustic crown]

i'll have to think about it, idk all the groups of order 16

but it's not hard to show that it will contain D4 as a normal subgroup by above argument

maybe identify some element of order 2 outside this? (which also commutes with that copy of D4)

yea that should work

[dusty sapphire]

Okay i think about it

[rustic crown]

D4 < S4 < S6

you can pick (56) as that new element

[dusty sapphire]

Okay thank you so much

[rustic crown]

[iron vessel]

Hey guys, I need to express the sum of powers $\sum_{n} T_1^4$ in the elementary symmetric polynomials.

[cloud walrus]

Évariste Galois

[iron vessel]

However, wikipedia says that it is equal to $e_1^4 -4e_1^2e_2 + 4e_1e_3+2e_2^2-4e_2$ but for some reason I cant get the last three

[cloud walrus]

Évariste Galois

[modern hawk]

hi, does anyone have any good crash course abstract algebra course, and/or help me understand this?

[iron vessel]

I used that $\sum_n T_1^4 = s_1^4-(a\sum_n T_1^3T_2 + b\sum_n T_1^2T_2^2 + c\sum_n T_1^2T_2T_3 + d\sum_n T_1T_2T_3T_4) \ with \ a = 4, b = 6,c = 12\ and \ d = 24$.

Could someone indicate as to where I might be wrong?

[cloud walrus]

Évariste Galois

[iron vessel]

Then I used the fact that $\sum_n T_1^3T_2 = \sum_n T_1^2\times\sum_n T_1T_2 - 2\sum_nT_1T_2T_3$

[cloud walrus]

Évariste Galois

[loud cypress]

I don't really understand what it means here by "unambiguously". Does the order matter with this notation, or is it talking about something entirely different when talking about ambiguity?

[dusty sapphire]

different representation for the same cycle.

[loud cypress]

so like order doesn't matter right?

[turbid pond]

Perhaps a vague question: why do we define quotient groups/rings using normal subgroups/ideals?

I can see that in both cases we have this sort of GN=NG (resp. RI = IR) 'commutativity' going on.
and we can get equivalence classes because the normal subgroup (resp. ideal) respects the group (resp. ring) operations.
but im not sure if i fully get why they're good choices for quotienting

[hidden haven]

Because they are exactly the kernels, and the first isomorphism theorem holds

Or a more elementary and maybe less satisfying answer would be that the equivalence classes of those have a natural inherited group (ring) structure

[hidden haven]

The first isomorphism theorem is what we really want in the first place. It (or a slightly more general version than the one in the sticker) is the defining property of a quotient

[rigid cave]

would you even have a group if you quotientet by a non-normal subgroup`?

ye operation no well defined

[hidden haven]

The not well defined one is good but I find it slightly less satisfying because then it feels like maybe there is some other operation that makes it work

[rigid cave]

ye right

[turbid pond]

very insightful, thanks guys

[turbid pond]

also love the fact that i didnt realise its a sticker until you mentioned it xd

[delicate orchid]

It’s the best sticker

[hidden haven]

Don't tempt me to send induction

Lmao

Nice

[delicate orchid]

That’s the 2nd best

All groups are abelian so I can just quotient by any subgroup I want

[turbid pond]

so just extending the same idea:

for a homomorphism between two vector spaces V -> W over a field:

• the kernel is a subspace of V
• theres a canonical way to realise any subspace as a kernel of some homomorphism

[hidden haven]

✓

[chilly ocean]

It should not be legal to ask this kind of question when you've read most of hatcher

[hidden haven]

Toki minmaxing

[south patrol]

all groups are abelian anyway

[hidden haven]

Is this some high iq joke that that I don't understand or a dumb one?

I ask because this was said here 5 mins ago

[rigid cave]

it's not a joke, it's a fact

[hidden haven]

Next you'll be telling me how all rings are commutative

[lethal dune]

aren't all rings commutative anyway

[coral shale]

all commutative rings are commutative

[lethal dune]

you can never make me say non-commutative rings are nice

[gusty halo]

matrix rings

[lethal dune]

no

[delicate orchid]

Matrix rings are swag AF

[hidden haven]

To what extent have you even studied them

[delicate orchid]

Representation over a non-commutative ring

[lethal dune]

(modules over non-commutative rings are not nice)

[hidden haven]

In what sense

They form an Abelian category

You have Freyd Mitchell only if you allow modules over non comm rings

And you can't do so much of homie alg if you don't assume Freyd Mitchell

You can but it is bad

Because you can't take elements

You have to do all that members stupidity

You can't do localisations as cleanly I guess

And like anything in commutative algebra lmao

But is abelian category not nice enough

[lethal dune]

[delicate orchid]

Nope

[lethal dune]

it has 'abelian' in it's name

so it's nice

[hidden haven]

bruh

[delicate orchid]

I’ve changed my mind

[hidden haven]

At least you are consistent

[lethal dune]

even diagrams commute

you don't have 'nice non- commuting diagrams'

[hidden haven]

Have you never heard of chain homotopies

[lethal dune]

no

[hidden haven]

They are the basis for homological algebra and algebraic topology

And form non commutative diagrams

[lethal dune]

[hidden haven]

Damn if you don't like non commutative then wait till you hear about non associative

It's all a Lie

[lethal dune]

even hearing the name makes me shiver

[hidden haven]

in ecstasy?

[gusty halo]

Lie theory is actually pretty cool

not that i know a lot tho

[lethal dune]

no 'no associative'

we'll need Lie in this sem

F

idk I might change my mind later

[hidden haven]

Commutative category theory when

fg = gf 😌

[delicate orchid]

Non-closed algebras when

[hidden haven]

Closed algebra is when you have addition, multiplication, and exponentiation

This is a category theory joke

[lethal dune]

btw I like category theory now

[delicate orchid]

Category theory.... more like set theory...

Anyway chat, say you’ve got a non-commutative ring, do the commutative elements form a sub ring?

Been thinking about this one for about 20 seconds

[hidden haven]

What do you mean by commutative elements?

Center?

[delicate orchid]

I guess you could describe it as the centre of whatever monoid the elements form under multiplication

[hidden haven]

Ye that is called the center of the ring

And it is a subring

[delicate orchid]

Niceee

[lethal dune]

btw is there a adjoint functor of the forgetful functor of R-mod to abelian groups?

[gusty halo]

should be tensoring with R

[lethal dune]

i.e. given an abelian group, what is the most general construction of R module over that group?

[lethal dune]

okk

oh yeah

right, nice

[gusty halo]

its even a monadic adjunction

[clever rampart]

Let G be a primitive permutation p-group. How does one show that G must be Z_p acting on p elements?

I have read a little about how one can consider the stabilizer H of a point and if there is a subgroup K (not equal to G) strictly containing H, then G must be imprimitive. This implies that if |G| = p^n, |H| = p^{n-1}. From here I can deduce that the action is on p elements and I can intuitively see in this case that now the stabilizer H must in fact fix all elements, but I don't see it formally.

[cloud walrus]

Ultramothematics

[maiden ocean]

I just realized that deleting that msg means chmonkey probably cant tell that he wasp inged by me specifically now

lol

@next obsidian Ok ill do it directly in case u go looking thru ur inbox for mentions

[next obsidian]

Eh

But yeah

You have to sheafify this tho

[maiden ocean]

Yeah

I know

I have to show that the if the original module of sheaves is locally free than so is the symmetric power and its so bad because there slike two layers of sheafification going on

Fucked up

[next obsidian]

No no no it’s fine

The question is local so I’m pretty sure it should follow from the fact that if M is free so is Sym^n M

At the very least, over a Noetherian base you can test for freeness on stalks so it suffices to look at the presheaf

But I don’t think you even need that

[maiden ocean]

Yeah thats what i was doing

I know its easy if everything is Noetherian because then you're taking Sym^n of free objects

so then thats free

[next obsidian]

That one is simple tho you just grab bases

[cloud walrus]

Ultramothematics

[maiden ocean]

Its like

ugh

[next obsidian]

It’s finite free

I assume?

[maiden ocean]

It doesnt say that

[next obsidian]

Just assume it

If you ever need it for infinite rank just figure that out some other time

[maiden ocean]

Ok then yes thats easy

[next obsidian]

Infinite rank free stuff are way less important

[maiden ocean]

I need this for like exterior power things i think or some shit

like sheaves of differentials

[next obsidian]

And Sym on them are kind of weirdchamp

[maiden ocean]

at least in this book

[next obsidian]

Yeah those are all finite rank tho

[maiden ocean]

Oh ok

Then yea

ill just assume finite free

[next obsidian]

Like thinking about it

I’m pretty sure M ≈ Sym^n M

If M is infinite rank

They should both be free of the same cardinality

[maiden ocean]

Thats horrific

[next obsidian]

I mean it comes to down bases things

Like if {x_i} was a basis if you fix an ordering

You get a basis of the form of all of the like n-fold products of them

But because of symmetry you can order it

So you just assert that the indices increase

If the x_i draw from an infinite set I

You have at least I many

By just putting x_i n times

And then the number is a subset of I^n

So it should still be the same size as I

Does that make sense?

[maiden ocean]

I guess

[next obsidian]

Idk if you’ve ever seen the symmetric powers in a diff geo context

But if you’ve seen the exterior power

You saw it for vector spaces and from a basis you combinatorially create a basis for the exterior powers

By like ordering it in increasing, but also that each index is strictly increasing

The story is the same for free modules

[maiden ocean]

I think i see

Wait

Uh

Sym^n should be identifiable with like degree n homogeneous polynomials in A(U) over I

right

[next obsidian]

Exactly yeah

[maiden ocean]

or is it degree n-1

[next obsidian]

Degree n

[maiden ocean]

I mtrying to think bc Sym^1 is just A(U) again right

[next obsidian]

This is why the symmetric algebra is actually a polynomial ring

Nah

That’s Sym^0

[maiden ocean]

oh ok

wait how is Sym^1 defined

[next obsidian]

Sym^1 is just M

It would be the 1st tensor power of M

[maiden ocean]

Oh duh

Yeah

[next obsidian]

Mod out by the relation that lets you swap stuff

[maiden ocean]

Ok sorry

I think i was confusing by indexing with I over M and over A(U) because like free stuff

blah

[next obsidian]

Lol

But yeah, this should tell you Sym^n M ≈ M

For infinite rank M

I don’t think it’s natural tho lol

[maiden ocean]

Good because there is nothing natural about this :catscream:

Omg no nitro

ok whatever

[next obsidian]

Oh noes

I run out in like 6 days

D:

[dapper nebula]

peter

[next obsidian]

Peter rabbit

[maiden ocean]

anyway the basis is like {x^n | x in I} right

[next obsidian]

Not quite

[maiden ocean]

and also 1

[next obsidian]

Because you can swap stuff

You fix an ordering of I

And enforce that the indices have to go up

So like if your basis was

x_1,x_2,x_3

A basis would be

x_1 (x) x_2, x_1 (x) x_3, x_2 (x) x_3

For Sym^2

[dapper nebula]

what is this

pbw theorem

[next obsidian]

Symmetric powers

Of finite free modules

[maiden ocean]

is x1, x2, x3 supposed to be I

[next obsidian]

Those are the basis elements of M

I = {1,2,3}

[maiden ocean]

Oh ok i was thinking about this wrong ugh

[next obsidian]

Each of these correspond to monomials

If you just multiplies them

If you think of x1,x2,x3

as indeterminate

These are the 3 degree 2 monomials

[maiden ocean]

So its like degree n homogeneous polynomials over A(U) in |I| variables?

[next obsidian]

Yeah

[maiden ocean]

Ok

[next obsidian]

But that’s why you say the indices increase

If you’re writing it as like tensors

Since x_1 (x) x_2 = x_2 (x) x_1

In the polynomial way of thinking

It’s just saying that x_1x_2 = x_2x_1

Since your indeterminates commute

[maiden ocean]

mhm

[next obsidian]

I guess thinking of it this way

T^n M

The n-th tensor power

Is like the degree n homogeneous polynomials in the non-commuting polynomial ring thing

I forgot, there’s a cooler name for it but

Where you do the same shit as usual except the indeterminates don’t commute

[maiden ocean]

ya

[next obsidian]

I think they usually call it R<x1,…,xn>

Cool stuff

[maiden ocean]

and quotienting out by S_n is more or less forcing it to commute

[next obsidian]

That’s exactly what it is yeah

And I guess a neat thing here is like

Transpositions generate S_n

So we could also just like assert that they commute pairwise

And after stringing those everything commutes

Just an idea that popped into my head

[maiden ocean]

ok so the basis is like

simple n-tensors of indeterminants, and theres one indeterminant for each I

[next obsidian]

Yeah

[maiden ocean]

Got it!

[next obsidian]

I wouldn’t say indeterminates tho

They’re just basis elements of M

But also you need to remember that you fix an ordering of I

And say that they have to weakly increase

Otherwise you end up with a basis for the tensor powers

Here’s it symbolically

If {x_i}_i in I is a basis, fix an ordering of I

[maiden ocean]

what does weakly increase mean

Oh like the basis has to be ordered?

[next obsidian]

Then a basis is
x_i1 (x) … (x) x_in where i1<= … <= in

Yeah

[maiden ocean]

Right

because symmetric

[next obsidian]

Exactly

Fixing the ordering makes each thing appear uniquely

[maiden ocean]

Ya

[next obsidian]

IIRC the exterior powers

Has the same basis

Except you need strongly increasing

Since if there’s a repeat the thing is 0

[maiden ocean]

Ah

Ok

this makes sense

[next obsidian]

So you can combinatorially write down the dimension

Just in terms of dim M

[maiden ocean]

Yeah

It should be like n choose |I| or something idk

combinatorics isnt real

[next obsidian]

For Sym

I think it’s like

If r is the rank, n is well, n

I think it’s like

r+n-1C n-1

Or something like that

[maiden ocean]

Right

Or something

Wait

Hm

[next obsidian]

Just count

Degree n monomials

In d variables lol

[maiden ocean]

ok anyway the important part is that the Sym^n M(U) are all free over A(U) of the same rank

yeah

[next obsidian]

Yup

Plus like

The bases all line up

Since once you assume M is free

As a sheaf

Any basis of M over U pushes down to a basis over all the opens even lower

Like if B is a basis over U, then B_V is a basis over V

[maiden ocean]

[next obsidian]

Plus Sym^n is a functor

So given a map of sheaves

You can define a map on Sym^n of the sheaves

It suffices to define it on presheaves

And the functorial nature of everything just means everything lines up

[maiden ocean]

Ok thats pretty neat

[next obsidian]

You just need a bunch of square to commute

This is a useful thing

Often times functors on modules lift to sheaves of modules

By doing everything locally

[maiden ocean]

I see

hm now i have to define the exterior power and show the same thing for it