groups-rings-fields

[final pasture]

moldi bad

[hidden haven]

I just have this power to affect people

Shika you are suddenly feeling like doing hw

[tough raven]

[gritty sparrow]

Sorry for the late reply, this should work

[south storm]

Oh I’m an idiot lol thanks

[vestal snow]

Thanks

[vestal snow]

I figured it out

[stone fulcrum]

@latent night
Hey we can continue here

h^n means hhhhhh... n-times

Likewise (g'hg)ⁿ = g'hgg'hgg'hg...

[latent night]

oh you're right. im dumb. the inner gs cancel.

i knew that

[stone fulcrum]

Haha yeah I know just a reminder XD

[latent night]

so its g' h^n g.

right?

[stone fulcrum]

The big deal here is that h is some element generated by an element of order n. So hⁿ = e

I feel like I need to prove that statement haha, or are you good with it?

The order of h divides n

[latent night]

i agree if h generated by a single element of order n, but couldnt you have h = ab for example where a and b have order n, but ab has some other order?

if a and b don't commute at least

[stone fulcrum]

Neat point, I didn't think of that, back to the drawing board I go!

[dusty river]

A different approach that might work would be to embed G into a symmetric group using Cayley embedding and then prove that the set of all elements of order n generates a subgroup which has the property that if some g is in it, then everything with the same cycle structure as g is in it

And then use the fact that conjugation preserves cycle structure

I am pretty sure this works for finite G, should also work for infinite G but I am not sure because there may be some annoying details with infinite cycle structures

[stone fulcrum]

Take the presentation
<g,a,b : a² = 0, b² = 0>

Then ab is generated by elements of order 2, but g'abg is not

[dusty river]

dont g'ag and g'bg have order 2

[stone fulcrum]

fk

Knew I was forgetting something

[dusty river]

But my reasoning for that subgroup having that property is that when you look at S_m itself, and take the subgroup generated by {order n elements}, if g is in it and if some other h has the same cycle structure, you use some permutation to identify things that occur in the cycle representation of h with those in g

and then write essentially the same product but with this identification

and this is now normal in S_m

and a normal subgroup intersected with a subgroup is normal in that subgroup

[stone fulcrum]

Wait that's totally how you generate the elements

[dusty river]

didn't get that

[latent night]

doesnt this give us an immediate proof? if h = a*b where a and b have order n, g'hg = g'ag * g'bg. then you can use induction on the factors of h

[dusty river]

ohh

[stone fulcrum]

Yeah that exactly lol

[dusty river]

That works

lol

[stone fulcrum]

You breaking my bad counter example gave the proof right away haha

[dusty river]

Calculated.

[stone fulcrum]

Cool question. I learned today.

[latent night]

thanks me too

[lavish sigil]

Not sure how well this fits here but I have a question about complex structures on real vector spaces. If we have a $2n$-dimensional real vector space $V$ together with a complex structure $J:V\to V$, then the complex structure extends to a $\mathbb{C}$-linear map $J:V^c\to V^c$ on the complexification $V^c=V\otimes_{\mathbb{R}}\mathbb{C}$. The map $J$ has eigenvalues $\pm i$, and we get a splitting $V^c=V^{(1,0)}\oplus V^{(0,1)}$ into the corresponding eigenspaces. Is there any intuition behind why one wants to consider this splitting, or even why we want to consider the complexification $V^c$? Since we already have a complex structure, we can already view $V$ as a complex $n$-dimensional vector space.

[cloud walrus]

gustavn64

[lavish sigil]

One can identify $V$ with $V^{(1,0)}$, for instance, but why do we want to consider $V^{(1,0)}$ in the first place?

[cloud walrus]

gustavn64

[lavish sigil]

I think it might be motivated by the study of the tangent bundles of complex manifolds, and in particular, to holomorphic and antiholomorphic functions, but is there any purely algebraic motivation?

[sturdy marsh]

the motivation behind it is to consider the holomorphic and antiholomorphic parts

it is a geometric object

I dont think there's any "algebraic" motivation

[dusty river]

Is there a characterisation of diagonal matrices of GL_n(F) using only the group structure of GL_n(F)? For example the scalar matrices are exactly the matrices that commute with everything in the group.

[humble surge]

What are scalar matrices?

[scarlet estuary]

matrices sI for s a scalar, I the identity

anyway im trying to think of multiplicative "diagonal iff ___" properties

but i cant think of anything

[dusty river]

I think something like "normalizer of the center" might work out but I am not sure

or centralizer of the center?

wait no that would be stupid

both are the whole group

[humble surge]

What's a normal subgroup of GL_n(F)?

[dusty river]

The center is one Idk if it will be easy to characterize though

The problem I am given is actually to characterize the diagonalizable matrices

and that is doable by "there exists some P st P'MP is diagonal" where ' denotes inverse

but that requires that I already have a characterization of diagonal

[humble surge]

You can't take that for granted?

[dusty river]

nope

And apparently, it is also possible to characterize "diagonal matrices all of whose diagonal entries are distinct"

Because that's also a problem

Maybe we can try to characterize the matrices which are 1 at the ij-th place and 0 elsewhere

idk if that will solve the problem though

This is the exact problem I have, in case you wanna see what I am allowed to take for granted

Ah characteristic 0 should be relevant

0-definable meaning something a bit weaker than what I said, but essentially you can't refer to entries of the matrices, only how they multiply, ie you have to only use the group structure

[dusty river]

This isn't even in the group so nvm but it might be possible to do some of the elementary matrices

[dusty river]

👀

[final pasture]

(nvm, I can't read 🐒)

[dusty river]

[final pasture]

Can you speak of the characteristic subspaces and their dimensions ? I guess this goes a bit beyond the group structure, so no ?

[dusty river]

Characteristic subspace of K^n?

[final pasture]

of a matrix

[dusty river]

What is that

[final pasture]

M's characteristic subspace associated with lambda is the kernel of M - lambda Id

(but the subspace lives in K^n, yeah)

[dusty river]

I don't think we can talk about any action of the group when we are doing definability

[final pasture]

ok sadge

[wraith obsidian]

Can you refer to elements of the field?
Like, are we in the language of groups, or do we have the language of k-vector spaces as well?

[hidden haven]

I think the problem implies we are in the language of groups

But now that I read it it is a bit vague

How would we talk about the elements of GL_n with the language of vector spaces though?

[wraith obsidian]

(i was wondering how one would pin down sI for undefinable s otherwise, but these are just the center and I'm being stupid)

Right, it's not even closed under mul by zero or even addition

Groups it is then.

[hidden haven]

Yeah

[wraith obsidian]

Every definable subset has to be characteristic, right? Because definability cannot prefer elements over images under an aut, right?

[hidden haven]

Yes that's right

[wraith obsidian]

So „truly diagonal“ matrices should be indistinguishable from diagonalizable matrices

(except if they are scalar, because they are that iff they are in the center)

[hidden haven]

Ah because there's an inner automorphism sending any diagonal matrix to a merely diagonalizable one?

[wraith obsidian]

That would be my reasoning

[hidden haven]

Seems legit, so I guess either the problem is wrong or there is some language the author forgot to mention

[wraith obsidian]

Diagonal matrices commute with each other. Could diagonalizable be pinned down as „pairs of elements commute after suitable conjugation“?

[hidden haven]

c could still magically work out somehow

We would have to say something like "largest set with this pairwise property" I think

Doesn't seem possible

any matrix commutes with its inverse

oof

[wraith obsidian]

Wait, so do you need a unary formula holding true at g iff g is in your set?

Or can you like quantify about elements in the set

[hidden haven]

Yeah that's how I interpreted it

You can quantify over the whole group

And hence over definable subsets

But not if we don't know the subgroup to be definable a priori

[wraith obsidian]

Ye sure that would be circular

[hidden haven]

Ye I think your reasoning is valid, and should disprove b, though c in not sure

[final pasture]

🤔

wdym by that

like if I pick a matrix M

how do you know if M is diagonalizable with what

okay wait I can't read

again

I think I should stop speaking, sry

[wraith obsidian]

Okay, different question. Are there other obvious definable subsets besides the center?

Like, what even is the commutant of GL?

[final pasture]

the whole set ?

[hidden haven]

We can talk about orders

So all finite order elements, subgroup generated by order n elements, etc

We should be able to take normalizer and centralizer of any definable thing

ohh commutator subgroups too

[prisma ibex]

yea all this stuff is fine in algebraic group land

I just got here by the way what is the main question

[wraith obsidian]

Hm, but commutator subgroup would be like traceless stuff, that's pretty far from being diagonal and invertible

[final pasture]

.

@prisma ibex

[wraith obsidian]

This

[prisma ibex]

ah okay

[wraith obsidian]

I mean diagonalizable matrices are inherently torsion free but that doesn't exclude like rotations around an irrational angle in R²

[hidden haven]

Yep

[wraith obsidian]

Oh, but the centralizer of such a rotation would be the one-parameter subgroup of all rots (?), and that would contain torsion elements

Or is that too naive

[prisma ibex]

I'm a little confused by the notion of \emptyset definable here

what is and is not allowed

[hidden haven]

What would one-parameter subgroup mean here?

[hidden haven]

Definable using first order formulas

Without parameters

[wraith obsidian]

That was meant as a visual aid, I wanted to describe „all rotations around the same plane as the initial rotation“

[hidden haven]

ahh

I see

[prisma ibex]

okay for instance is SL_n such a subgroup

is that allowed

[hidden haven]

Hmm not sure how you'd define that

[wraith obsidian]

The centralizer of a diagonalizable matrix would be all elements that are diagonalizable wrt that same basis, right?

[hidden haven]

So the claim is that diagonalizable matrices commute with exactly diagonalizable matrices with same diagonalizing basis?

That's not true for all diagonalizable matrices so you'd have to pick one carefully I think

[wraith obsidian]

Nvm, I was hoping to build off of „g is torsion-free and every element in its centralizer is torsion-free“ but being torsion-free is not even fo definable, right?

[hidden haven]

Because for the identity matrix everything is in the centralizer

[wraith obsidian]

Ah, fair

[hidden haven]

Yeah torsion free will require infinitely many formulas

[wraith obsidian]

Do we require one unary formula?

[hidden haven]

Yeah definable means one formula

Axiomatizable allows sets of formulas

[wraith obsidian]

The „distinct entries“ part could be characterized by „commutes with no element of order two“, because if two entries were equal, it would commute with the corresponding transposition matrix

[hidden haven]

Is the converse true as well?

Not commuting with any transposition → distinct entries?

[final pasture]

yes ? If two entries are the same, you can pick the transposition swapping them, no ?

wait

[hidden haven]

Are transpositions the only order 2 elements

[wraith obsidian]

That's the contrapositive tho not the converse
The converse would be „if g diagble and commutes with some order two element, do two entries have to be equal?“

[final pasture]

That's the contrapositive tho not the converse
yeah lol

I was confused too because I was like "wait, I just said what lux said"

[wraith obsidian]

No, but any order two element has minimal polynomial (t-1)(t+1) giving you a decomposition into identity and swapping subspace of K^n

So direct sums of id and -id

[final pasture]

The converse would be „if g diagble and commutes with some order two element, do two entries have to be equal?“
And the converse isn't true, because Diag(1,.., n) commutes with E_{2,2} whenever n > 2, right ? 🤔

[hidden haven]

I took converse of distinct entries → not commuting

[wraith obsidian]

What's E_{2,2}

[final pasture]

sry, wrong notation

[hidden haven]

Not invertible

If it means matrix with 1 at 2,2 and 0 elsewhere

[final pasture]

yeah, but that's because I'm dumb and I used the wrong notation

I meant matrix permuting the entry at (2,2) and the one at (1,1)

and leaving other as they are

this one is invertible, being its own inverse

[scarlet estuary]

contrapositive and converse are equivalent under the very useful "freshmans lemma"

(b implies a) implies (a implies b)

[final pasture]

Yeah but Moldilocks doesn't quite satisfy the freshman lemma hypothesis, sadly

😔

[wraith obsidian]

But left multiplication with that would transpose row 1, row 2 while the right mul would do the same to the cols
That is not the same, hence doesn't commute with diag(1,2,…)

[hidden haven]

The 2 things are the same for diagonal matrices though

[wraith obsidian]

But not for the elementary 1-2 transposition matrix

[final pasture]

yes, again I'm being dumb, sorry 😓

[wraith obsidian]

I would typeset an example but that would require me to get out of bed and it's not even 1pm

[hidden haven]

Btw the string diagrams you told me about were extremely helpful, I'm using them for a lot of stuff in mac lane and life is so much easier

[wraith obsidian]

Yw

[wraith obsidian]

Ah, the converse does not hold. Take diag(1,2) and as an order-two element diag(-1,1), which transposes (1,1) and (-1,1). These clearly commute.

[wraith obsidian]

At least we know that whatever commutes with a diagonalizable matrix g must preserve its eigenspace decomposition

In particular, if all diagonal entries of g are distinct, this ensures that all of these eigenspaces are one-dimensional, giving that an element commutes with g iff it is also diagonal wrt the same basis.

In particular, this centralizer is still abelian

A good litmus test would be in GL(2,ℝ): How could we identify diag(1,2) to be part of the set of all diagonalizable matrices with distinct entries in a way that is not satisfied by r = rotation around an irrational angle?

Also, my torsion idea was pretty stupid because diag(1,i) has order 4 in GL(2,ℂ).

[dusty river]

I didn't get how the second part followed

[dusty river]

oof

The problem mentions characteristic 0 so we should have to use that somewhere

[chilly ocean]

Sup, guys. I would like to know if you guys have any pdf of Lie Algebra, have you?

[hot lake]

fulton and harris's book on group representations has lots of lie algebra, but it's more on the combinatorics side than the geometric side

[wraith obsidian]

Because if h commutes with g, h leaves all the eigenspaces of g invariant – all of which are one-dimensional.

[orchid rover]

Lie theory here or in #point-set-topology ?

[prisma ibex]

if it's more along the lines of like the exponential map or things like that then #point-set-topology if it's more along the lines of representation theory then #groups-rings-fields

[orchid rover]

It is about Iwasawa decomposition

[orchid rover]

@bleak abyss

I was recommended to ping you by Tterra

[dusty river]

That doesn't seem right, any matrix commutes with the identity

[dusty river]

Saketh gave a solution assuming this is in the language of fields. G is a definable subset of K^n and we want to define subsets of it. But then the whole problem becomes trivial because you can talk about individual entries, and you can define things like inverses and and matrix multiplication

But this does not seem to require characteristic to be 0

This feels disappointing but it fits the rest of the problems because it is a model theory book and the algebra required for most of the things in it is pretty trivial

I am still interested in the original thing I asked though

[bleak abyss]

Lol

[orchid rover]

Hello Daminark

Do you know about iwasawa decomposition

Yes I did

Do you know anything about it

Disappointing

[vestal snow]

Do I need to know anything about the construction of fiber products of schemes to do this "by hand"?

[humble surge]

Which book?

[dusty river]

Hodges

[bleak abyss]

Sorry I'm back I was playing a lot of smash. I know vaguely about Iwasawa decomposition yeah

[orchid rover]

What is the status on computing an iwasawa decomposition?

Do efficient algorithms exist for this?

[prisma ibex]

yes

it reduces to the QR decomposition, where good algorithms exist and are well-studied

[vestal snow]

Let B be a finitely gen'd graded A-ring. Is Proj B --> Spec A a separated morphism?

[gritty sparrow]

Yes, i think it works without finitely generated as well

@vestal snow

[vestal snow]

Thanks

[wraith obsidian]

Yes, and the identity is diagonal in every basis.

[hidden haven]

Non diagonalizable matrices will also commute with it, that seems contradictory

[wraith obsidian]

The identity doesn't qualify for g in this case because g is required to have distinct entries when it's in diagonal form

[dusty river]

ahh

Now it makes sense

[orchid rover]

Thanks I guess this answers my question

[hybrid cove]

Hi guys :D. I wanned to know if the weyl group of type does have a name ?

*of type D

[balmy storm]

consider the set $S= {a,b,c}$ with the binary operation $*$ defined as follows:

$aa=a, ab= b= ba, ac= c= c*a$

$bb= a, bc= c, cb= b, cc= a$

Mentally I calculated, if associativity holds, I started with the non-commutative results to get $bc= c$ and $cb= b$ like, $b*(bc)= bc= c= ac= (bb)*c$.

But is there any general method, rather than showing that it holds or not, verifying each and every possible combinations to show the associativity holds or doesn't holds for the whole set.

Alternatively, is there any trick to easily locate one such combination for which the associativity doesn't holds so that I can conclude Associative property doesn't works so its not even a semi-group.

[cloud walrus]

S k

[fair obsidian]

I'm not too familiar with methods for testing associativity, but here's some stuff I came up with for this case:
If $$ was associative, then S would be a group, as $a$ is an identity, and each element is it's own inverse. However, $b$ and $c$ would then have order 2, which is impossible in a group order 3. So $$ must not be associative.
It looks like an example of associativity failing is
$$b * (c * b) = a \neq b = (b * c) * b$$

[cloud walrus]

The_Vman

[humble surge]

When people talk about quotienting by a kernel as an equivalence relation, they mean the equivalence relation defined by the fibers of the homomorphism, right?

[gray comet]

Ye

[humble surge]

making sure I understand this example, this is saying that the quotient group (G/K) is isomorphic to the kernel (K)?

[hidden haven]

No, the quotient group should be isomorphic to the image of phi

[humble surge]

oh shit, totally misread

[wispy scaffold]

[limpid edge]

I think that's the first isomorphism theorem

[kind temple]

it is

[chilly ocean]

:catBigEyes:

[crisp girder]

Is there any significant ideas in Category theory correspond to the 1st isomorphism theorem? From a starter in category theory🤔

[sturdy marsh]

lookup abelian categories

[sturdy mirage]

if two rings' multiplicative groups are isomorphic, and the additive groups are isomorphic, does that necessarily mean the rings are isomorphic?

[deft oasis]

nope, consider R^n and R^m

as I say this I realised that I'm unsure if they are iso as multiplicative groups

[hidden haven]

I think Z^ω and Z[x] × Z^ω works
The units in both cases are exactly the tuples in which all entries are ±1 and there's an obvious isomorphism. The additive groups are isomorphic because both factors in the second ring are the same as abelian groups

I hope multiplicative group means group of units under multiplication

[unique juniper]

hmm why is this zero in a field of characteristic 2

[hidden haven]

What are a, b, c?

[unique juniper]

anything

ig

[hidden haven]

Is this the formula for some discriminant?

[unique juniper]

yes

[hidden haven]

[unique juniper]

its the general discriminant of a cubic

[hidden haven]

That seems wrong because a, b, c could be algebraically independent

In which case a²b² - c² has to be non zero

[gritty sparrow]

Also a=1,b=1,c=0 gives 1

[unique juniper]

yea- but ...

[gritty sparrow]

What made you think it would be 0

[unique juniper]

this, but i think ive realised where ive gone wrong. it would be non singular iff the cubic and its derivative have a gcd of 1, but if p = 2 then the discriminant of the derivative would be zero and then common roots and then singular points

so my bad

[snow flint]

i forgive you

[unique juniper]

thank you :)

[rain oxide]

This may be wayy, wayy to basic for abstract algebra, but... if you apply an automorphism twice is that always an identity

👉 👈

[gritty sparrow]

No, consider doubling in the group Q under addition

Applying this twice is quadrupling, not the identity

[thorn delta]

why are submodules of K cyclic?

[sturdy marsh]

I dont think theyre saying all A-submodules of K are cyclic

theyre saying that K is a colimit of that class of cyclic submodules

[thorn delta]

ohhhh wait, i remember now. A module is the colimit of its finitely generated submodules. thats why this works

and um, a finitely generated submodule of K is cyclic cuz you can just find a common denominator of all the generators

[sturdy marsh]

you probably need stronger conditions on the ring to do that

oh no nvm it's fine

ye that works

[thorn delta]

epic

[delicate orchid]

I retract my 🤨 I forgot K was a field...... :(

[humble surge]

I'm trying to understand the construction of the tensor product of vector spaces. So it's the free vector space $F(V\times W)$ on the formal products $v_i\otimes w_j$ quotiented by the bilinearity relation.

[cloud walrus]

modus ponens

[humble surge]

So is the free vector space $F(V\times W)$ just the vector space that is generated by using the formal products as a basis?

[cloud walrus]

modus ponens

[gritty sparrow]

Yes

[humble surge]

But VxW here is not the direct product, right? They just use the same symbol?

[hollow imp]

Cartesian product

It's the Cartesian product in this context

[humble surge]

But isn't VxW supposed to be one dimensional? It's the set of formal products, right? Not ordered pairs

[hollow imp]

The formal product v \otimes w can just be considered as another notation for (v, w).

And you can see that the free vector space generated by the pairs v \otimes w is naturally isomorphic to the free space generated by (v, w), it's just a matter of notation.

[humble surge]

damn structuralists

[hollow imp]

Lol

[humble surge]

They're different sets I tell you!

Thanks, btw

[hollow imp]

Np

[restive star]

Hey I found an algebra problem I'm stuck on: Show that the A_11 has no subgroups of order 2,851,200

The prime decomposition is 2^7 3^4 5^2 11 and the size of A_11 has size 2^7 3^4 5^2 7 11

anyone willing to help me out?

My thought so far was that since it's simple, I could show that if there was such a subgroup it must be normal

which would lead to a contradiction

But I'm not sure how I would do that lol

[quaint tree]

$2851200=2^{7}\cdot3^{4}\cdot5^{2}\cdot11$ and $|A_{11}|=\frac{11!}{2}=2^{7}\cdot3^{4}\cdot5^{2}\cdot7\cdot11$.

Clearly, we can see that $2851200\not\vert\frac{11!}{2}$

[restive star]

there shouldn't be an 11^2 there

[quaint tree]

I used Wolfram Alpha.

[restive star]

It's 2^7 3^4 5^2 11

[quaint tree]

Oh yes, you're right, I mistyped.

[cloud walrus]

Elliptic Curve of Rank 9001

[quaint tree]

Well I was just going to use Lagrange's theorem, but thanks to my mistyping that won't work.

[restive star]

haha RIP

I think A_11 being simple has something to do about it

[quaint tree]

I'm wondering if Sylow's Theorems might be useful here.

No, I thought I had something, but it doesn't seem to be working.

[restive star]

Yeah I looked into using sylow's theorems and wasn't able to get anything either

[vestal snow]

This doesn't seem right. When it says that pi and pi' agree on closed points, I think of that as meaning that they induce the same map of stalks at the closed points

But the way it is written, it only seems to suggest looking at pi as a purely topological map

[hidden haven]

Then isn't P_2 x P_3 x P_5 x P_11 a subgroup of that order, where P_i is the sylow-i subgroup

The product is internal product and it should be direct because everything has co prime order

[restive star]

I got this from a qualifying exam practice list so I don't think there's a typo... But I see what you mean

[north widget]

quick question

Let G be a finite group whose order is not divisible by 3. Suppose that $(ab)^3=a^3b^3$ for all $a,b\in G$ prove that G is abelian.

[cloud walrus]

Abdo Kiza

[north widget]

is it $(ab)^3=(ab)^2 * (ab) = a^3 * b^3 => (ab)^3 * (b^{-1}a^{-1})=a^3 * b^3 * (b^{-1}a^{-1}) =a^3b^2a^{-1}$

[cloud walrus]

Abdo Kiza

[scarlet estuary]

why would that prove it?

[north widget]

no

it didnt

im thinking you take (ab)^3 multiply by (ab)^-1 on both sides

[vestal snow]

Try this

(aba^-1b^-1)^3

[north widget]

oh rip

[vestal snow]

first consider aba^-1 and b^-1 as separate

[north widget]

i didnt read for all a,b

[vestal snow]

And then consider ab and a^-1b^-1 as separate

[north widget]

order not divisible by 3 means a^3n != 0?

[vestal snow]

You should get a^2 b^3 = b^3 a^2

[vestal snow]

Kinda

Unless a = e

[north widget]

ok

[vestal snow]

But in that case everything commutes with a

[north widget]

so (aba^-1b^-1)^3 = (aba^-1)^3(b^-1)^3
and (aba^-1b^-1)^3=(ab)^3(a^-1b^-1)^3

[vestal snow]

Let the order be n. Use that there exists p and q integers such that pn + 3q = 0

[north widget]

oh ok

we are saying 3q because the order isnt divisible by 3 so its coprime

i forgot name of this theorem

ive used it enough that i should know the name

[vestal snow]

It follows from the division algorithm

Once you know this, you're just an algebraic party trick away from solving the problem

I think

[north widget]

division algorithm is for rings though?

[vestal snow]

no i meant that there exist p and q such that pn + 3q = 0

[north widget]

i dont see how i could use that

what is 3q supposed to be?

[vestal snow]

That's where the party trick comes in

p and q are supposed to be plain old integers

[north widget]

oh ok

[vestal snow]

Do you want me to give you the answer?

[north widget]

this is for saying something aobut the order

nah dont give me the answer until like 10 minutes from now

[vestal snow]

Cool cool

[north widget]

wtf

ig i did the algebra wrong

after getting here

so (aba^-1b^-1)^3 = (aba^-1)^3(b^-1)^3
and (aba^-1b^-1)^3=(ab)^3(a^-1b^-1)^3

i set (aba^-1)^3(b^-1)^3 = (ab)^3(a^-1b^-1)^3

and tried to reduce it

but that sounds silly

it went like $(ab)^3(a^{-1}b^{-1})^3 = (ab)^3(a^{-1}b^{-1})^2(a^{-1}b^{-1})=(aba^{-1})^3(b^{-1})$ and then we can get that $(ab)^3(a^{-1}b^{-1})^2=(aba^{-1})^3(b^{-1})ba=(aba^{-1})^3(a)$

[cloud walrus]

Abdo Kiza

[north widget]

oh i think i got what u meant nvm

[uncut girder]

Say I have a finite galois extension characteristic p fields (can even assume the degree is p)

Is it clear that the trace function is surjective?

[north widget]

atm im struggling to get a^2 b^3 = b^3 a^2, but im guessing with ur trick id do something like (ab)^(pn+3q)=e

[vestal snow]

Try raising aba^-1b^-1 to that power

[north widget]

how did tou get that a^2b^3=b^3a^2

[vestal snow]

Ah ignore that

Just consider (aba^-1b^-1)^pn+3q

And just stare at it for a while

Your solution should fall right out

If not, I'm happy to go into more detail

[north widget]

nah idk

i just get (aba^-1b^-1)^pn * ((aba^-1b^-1)^q)^3 = e

or am i supposed to have (aba^-1b^-1)^pn * ((aba^-1b^-1)^3)^q = e

yea no clue

oh wait

ababab=aaabbb=> ababa=aaabb

(ab)^2a=a^3b^2

rip wrong way

or maybe baba=aabb

so (ba)^2=a^2b^2

yeah i give up

[vestal snow]

Okay so (aba^-1b^-1)^pn+3q = 1

Expanding the left side out gives us (aba^-1b^-1)^pn * (aba^-1b^-1)^3q which equals (aba^-1b^-1)^3q

[north widget]

why does it equal (aba^-1b^-1)^3q

oh

order is n

so it goes to e

[vestal snow]

Probably should call the identity element 1

[north widget]

where do you go after that?

(aba^-1b^-1)^3q=(ab)^3q(a^-1b^-1)3q=e

[vestal snow]

One sec

Ahh I made a mistake

Give me a minute

Yes so aba^-1b^-1 has order dividing 3q

Wait nvm

[north widget]

im thinking maybe its a brute forcing thing

im gonna look up solution on mse

[vestal snow]

It should have been (aba^-1b^-1)^pn+3q=(aba^-1b^-1)

not e

[north widget]

nah

because pn+3q=0

[vestal snow]

That was my bad

[north widget]

so (aba^-1b^-1)^0 = e?

oh wiat not

[vestal snow]

I meant to say pn+3q=1

[north widget]

yea

thats the theorem

[vestal snow]

I'm sorry about the confusion

[north widget]

i think the name starts with a b

i think we can bruit force though

(ab)^3=ababab=aaabbb=a^3b^3

(ba)^2=a^2b^2

using that (aba^-1b^-1)^3q=(aba^-1b^-1)

we have ((aba^-1b^-1)^3)^q=(aba^-1b^-1)

(a(ba)^2b)^q =(aba^-1b^-1)

[vestal snow]

I'm sure there's some slick way to do this

[north widget]

yea imma look it up

https://math.stackexchange.com/questions/1505189/to-show-that-group-g-is-abelian-if-ab3-a3-b3-and-the-order-of-g-is-no

ok

they use fact that x->x^3 is injective homomorphism

they get ab^4=a(ab)^3b=a^4b^4=ab^3a^3b

so they brute force it and get a^3b^3=b^3a^3, since f is injective they have f(ab)=f(ba) =>(ab)^3=(ba)^3=>ab=ba

x->x^3 is injective homomorphism because the kernel of the function is trivial since x^3=>x=0

[crystal lion]

hey, guys!

do you guys know about a book detailing the conjugacy classes of SL(V) and PSL(V) for arbitrary finite vector space V?

it looks like in GL(V) the conjugacy class g^(GL(V)) is completely determined by its Jordan-Chevalley decomposition, but it is not clear to me when would a conjugacy class of GL(V) split into several conjugacy classes in SL(V)

[echo plinth]

can someone help with this one?

[paper flint]

@echo plinth Wrong channel. Also, is this a quiz?

[chilly ocean]

the exact same question was posted in another channel...

[paper flint]

[prisma ibex]

multiple accounts posting the same thing

[north widget]

can you ping me when someone answers you

[potent patrol]

in macdonald's symmetric functions and hall polynomials, i see this in section 2. is the index on the coefficients a supposed to be "lambda COMMA mu" i.e., indexed by lambda and mu, or is it the pointwise product of the partitions?

lmk if this should be moved to the combinatorics channel

it makes sense to me that it would be "lambda COMMA mu" but macdonald introduced the product notation earlier in the book and so it has caused me some confusion

[maiden ocean]

Whats going on with the stuff about factors of the tensor product being the compositum

[limpid edge]

aren't ideals and normal subgroups supposed to be analogous structures for rings and groups?

because in that case why is it for ideals the condition is "closed under left/right multiplication with elements from the ring" while for normal subgroups it's "closed under conjugation with elements from the group"

[hollow imp]

The conditions are to make the quotient ring/group well-defined

If you look at it, ideals are required to also be closed under conjugation under the additive group, as the additive group is abelian

[limpid edge]

how does an ideal have to do with the addition operation?

[hollow imp]

Ideals being closed under addition automatically implies that addition on the quotient ring is well defined because the addition group of a ring is abelian, which means an ideal is a normal subgroup of the addition group.

It being closed under multiplication is an additional condition made so that multiplication in the quotient ring is well defined.

[limpid edge]

"Ideals being closed under addition automatically implies" yea but who said ideals had to be closed under addition?

assuming "closed" here means r + k \in I for r \in R and k \in I where R is a ring and I is an ideal

[hidden haven]

Closed under addition means adding things in the ideal gives something in the ideal

What you are describing would be called absorbing addition from R

[limpid edge]

oh I think my confusion was from the fact that all ideals are subrings

so the "closed under addition" part isn't specific to ideals but all subrings

I get it now

[light tusk]

Larry C. Grove "Classical Groups and Geometric Algebra", chapter 1.

The author shows that transvections generate SL(V) (Theorem 1.4), as well as some conjugacy results.

[thorn delta]

Just making sure: the zariski topology on P^1_k is basically just the cofinite topology, just like for one-dimensional affine space?

[crystal lion]

it didn't answer my question unfortunately, but thanks for the suggestion!

[crystal lion]

for anyone interested the case is this:

n>1, q a prime power.

1. When F_q has no n-th roots of unity, in other words (n, q-1)!=1 mod n, then SL(n,q) is a direct factor of GL(n,q), so in particular none of the conjugacy classes split

2. when F_q has an nth root of unity then the conjugacy classes that split are exactly the ones that have a single Jordan block of size n

[steady axle]

I tried to write out solution to this problem in detail. can someone take a look.

suppose $f$ be element of some $S_n$ such that $f^3=(1, 2, 3)$. If $f(1)\neq 2$ or $3$ then $f^3(f(1))=f(1)$ and so $f^3(1)=1$ a contradiction. If $f(1)=2$ then $f^4(1)=f(2)=3$ and similarly $f(3)=f^4(2)=1$ so we cannot have $f^3=(1, 2, 3)$. Similary we get contradiction if $f(1)=3$ as we get $f(2)=1$ and $f(3)=2$.

[cloud walrus]

bert

[steady axle]

there should be a sleeker solution

[hot lake]

I don't get how you deduce that f⁴(1) = f(1)

ah I see

if f(1) is not 1,2,3 it is fixed by (1,2,3) = f³

also not very clear how you get f(2)=3 when f(1)=2

[steady axle]

$f^4(1)=ff^3(1)=f(2)$ and $f^4(1)=f^3f(1)=f^3(2)=3$

[cloud walrus]

bert

[hot lake]

yeah

[wooden ember]

so after the exercise i constructed $r$ as the product $hs$ where if $s$ is a flip about the line going through the vertex with index 1 of a rigid polygon and its center then $h$ is a flip about the line going through the midpoint of 1 and 2 and the center of the rigid polygon. I got this just from messing around with a square, pentagon, and hexagon. Does this construction hold up geometrically for all polygons and is there some intuitive reason why it works?

[cloud walrus]

Little Narwhal

[wooden ember]

okay so i checked the general case with some matrices it does check out

i think it's kinda cool that you can get a rotation from mirroring twice

[uncut girder]

Look back in ch 1 to see how they computed the tensor product of two etale Algebras

Its contained in some proof

[untold sapphire]

This is part of a more general phenomenon of reflection groups

[wooden ember]

Interesting

[chilly radish]

Can someone confirm this proof outline for me? The question is:
Let $G$ be a group with $|G|=105$, suppose that the 3-Sylow subgroup of $G$ is normal, then $G$ is abelian.

Proof is as follows:
We have that $105=3\cdot5\cdot7$, therefore the unique 3-Sylow subgroup $P$ of $G$ (It is unique as a consequence of the Sylow Theorem which states all p-Sylow groups form a conjugacy class, therefore if it is only conugate to itself it must be unique) is cyclic of order 3, and therefore its automorphism group is isomorphic to $C_3$, which is of order 2.

Consider $\iota:G \rightarrow AutP$ which takes an element $g$ to the corresponding conjugation automorphism (This is well-defined as $P$ is normal and therefore conjugation invariant), its image is either of order 1 or 2. It cannot be of order 2 as by the 1st isomorphism thm. And Lagrange this would mean that
$$\frac{|G|}{|Im \iota|}=\frac{|G|}{2}=|ker \iota|$, but 2 does not divide 105, therefore the image of $iota$ is trivial, but this means that $P<Z(G)$.

Note that $\sfrac{G}{P}$ is of order $35=7\cdot 5$, I've shown previously that groups of this form (of order $pq$ with $p<q$ primes where $q\nmid p-1$) is cyclic, and so in particular $\sfrac{Z(G)}{P}\leq \sfrac{G}{P}$ is cyclic as a subgroup of a cyclic group.

Finally, the center is always normal, therefore from the 3rd isomorphism theorem we have that
$$\sfrac{G}{Z(G)}=\sfrac{\sfrac G P}{\sfrac{Z(G)}{P}}$$

Is cyclic as the quotient of a cyclic group, but it is known that if this quotient is cyclic, then $G$ is abelian, as required.

[cloud walrus]

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

[chilly radish]

I don't.think I actually used the uniqueness of the 3-Sylow.group. In hindsight the question might have stated uniqueness and you had to deduce normality

Oh and of course i've shown before that $\iota$ is a homomorphism

[cloud walrus]

ShiN

[terse crystal]

I am looking at your answer, and I just read the first 8 lines, I am not sure how you eliminate the situation when G has 7 3-Sylow groups

[hidden haven]

Uniqueness comes from normality

Though that first isomorphism theorem looks wrong

Swap image and kernel

[terse crystal]

Never mind I am a dumb 3-Sylow group being unique is said in the question ...

[hidden haven]

I am a dumb 3-Sylow group

[terse crystal]

😂😂

[terse crystal]

It’s correct,except a small error . I think it should be “p doesn’t divide q-1”

[ivory trail]

is there an obvious way to show Z/60Z localized at 2Z is Z/4Z? (i'm doing ravi vakil's alg geo notes and this is on page 144.) a better question would be is it easy to tell what Z/nZ localized at a prime looks like. (I was doing Z/12Z at 2Z by hand, but the zero divisors make it extremely annoying. i gave up halfway through checking the equivalence relation on half the fractions)

[cloud walrus]

Brofibration

[sturdy marsh]

localization commutes with direct sum

[chilly ocean]

you commute with direct sum

[sturdy marsh]

oh lol they wrote down the iso in the screenshot

what do you need help with

proving Z_(2) (x) Z/3Z = 0?

hint: every element in Z_(2) is divisible by 3

@ivory trail

[ivory trail]

the direct sum decomp is pretty much the chinese remainder theorem

trying to figure out if there's way to do it without that

so if there's an easy way to compute (Z/60Z)_(2)

[sturdy marsh]

there is

you do chinese remainder

chinese remainder theorem isnt really doing anything

it just makes it easier to analyse

[ivory trail]

then i'm not sure what the point of the underlined statement is

like how do you "just check" the stalks

[sturdy marsh]

he's asking you to compute the localization

prove that the localization at 2 is Z/4Z

[ivory trail]

yeah but the conclusion of this whole aside is the isomorphism given by the chinese remainder theorem

so i am trying to figure out how to prove that that is the localization without using it

[sturdy marsh]

Umm there's a map Z/60Z ---> Z/4Z

this map sends everything outside (2) to a unit

so it factors through the localization

it is surjective

assume a/b is in the kernel

this implies a is 0 in Z/4Z

[sturdy marsh]

so is this the nontrivial part

assume something is outside 2Z/60Z. Then it cant be 0 or 2 mod 4

so it does get mapped to a unit

so it does work

what

@ivory trail do you see any holes in the argument

im a bit sleepy so it could definitely be BS

[ivory trail]

let me think

[sturdy marsh]

we got injectivity for free for some reason

which is fishy

[gritty sparrow]

a maps to zero in Z/4Z but we still have to argue that a=0

[sturdy marsh]

aah

lmao

AHAHAHAHAAHAH

[ivory trail]

maybe vakil just didn't do it

without the chinese remainder theorem

because that's definitely the most obvious way

[sturdy marsh]

a/1 = 0/1 in the localization iff sa = 0 for some s outside (2)

right?

ye

15 is outside (2)

a is 0 mod 4

@ivory trail

[chilly radish]

I just didn't wanna write the first implication from the isomorphism thm but it's just dividing by the order of the image and multiplying by the order of the kernel

[ancient night]

What is the name of this property on S?
$$
(\forall m \in S)(\forall n \in S)(\forall p \in S) (m \leq n) \Rightarrow (m \cdot p \leq n \cdot p)
$$

[cloud walrus]

JohnDark

[dusty river]

Ah I see, mb

[chilly radish]

I think translation-invariant

[ancient night]

@chilly radish thank you!

[keen oasis]

I need some help with this. What i was thinking so far and got stuck:
Suppose K is the subgroup of index 56. S8, K are finite so the order of K must be 720 (8!/56). Initially i was thinking about Sylow theorem to prove the existence of this subgroup but 720 cannot be written as p^n where p is a prime(same for 20). Any suggestion? Thanks!

And in general how can i find the number of cyclic subgroups of a given group and their order?

[steady axle]

can someone take a look at my solution to this problem?

(a)I know this result from Kummer theory but I try to prove it explicitly here. Let $K \cong F[x] /(x^2+ax+b)$. By completing the square $x^2+ax+b=(x+ \frac{a}{2})^2-(\frac{a}{2})^2+b$. Define homomorphism $F[y]\rightarrow F[x] /(x^2+ax+b)$ by $y\mapsto x+\frac{a}{2} $. We have isomrophism $F[y]/(y^2-((\frac{a}{2})^2-b))\cong F[x] /(x^2+ax+b)$. We also have $F[y]/(y^2-((\frac{a}{2})^2-b))\cong F[\sqrt{(\frac{a}{2})^2-b}]$. \
(b) Consider $F_2 [x] /(x^2+x+1)$. It is a degree $2$ extension of $F_2$ but cannot be obtained by attaching square root of any element of $F_2$ because all square roots are contained in $F_2$.

[cloud walrus]

bert

[frank fiber]

a direct summand of free modules is free?

[gray comet]

yes

wait

You said direct summand, not direct sum

Then no

[hidden haven]

Ye this works. For the first one you can also say that adjoining a root of a quadratic is the same as adjoining its discriminant (will be the same proof probably since the quadratic formula is derived from completing the square)

[fair obsidian]

I don't know of a general method, but one strategy might be to think of a group of order 720 and then try to find it inside S8. I don't think the cyclic group of order 720, or a product like $\bZ/60\bZ \times \bZ/12\bZ$ is in S8, but can you think of any others?

[cloud walrus]

The_Vman

[fair obsidian]

@keen oasis

[crystal lion]

I think you need to guess the subgroup of index 56. What is special about 720 that you can use is that 720=120*6, both of which numbers are factorials of natural numbers. I'm sure you can guess the solution from here.

As for the question about cyclic subgroups of certain sizes, the question can be asked differently this way: can you find an element of order 20? How should it look like?

@keen oasis

[final pasture]

(yeah the number is way too big for the group to be abelian, as a subgroup of S_8, I guess )

[crystal lion]

Btw, this is closely related to Landau's function, which is an integer sequence $g(n)=$the size of the largest cyclic subgroup of $S_n$

[cloud walrus]

kadmate

[fair obsidian]

(720 itself is also the factorial of a natural number)

[crystal lion]

And g(n) is asymptotically known by Landau, very intersting imo

[crystal lion]

I never thought about 720 being both a factorial of a natural number and the product of two factorials

Now that I think about it, that shouldnt be very rare though

[fair obsidian]

Yeah, I didn't realize it was the product of two factorials until you mentioned it

[delicate bloom]

it isn't that rare?

I would think it'd be pretty rare but idk

[crystal lion]

Well, it is, but there are infinitely many examples i think

[nova plank]

👀 really?

[delicate bloom]

when do we have nontrivial solutions to a!b!=c!

c=a+b is sort of the first thing I would think about

then in that case it's looking at binomial coefficients equal to 1 which is a trivial case

[crystal lion]

$n!!= (n! - 1)! n! $

[delicate bloom]

don't put a space before the last dollar sign

bot doesn't like that

[crystal lion]

$n!!= (n! - 1)! n!$

[cloud walrus]

kadmate

[delicate bloom]

what is n!! here, is that (n!)!

[crystal lion]

Yeah, I'm not very axtuce in that channel sorry

[crystal lion]

Yeah, im a bit lazy :D

[delicate bloom]

nice I see cool example

are these the only cases?

sorry if I'm derailing whatever was going on continue if this isn't that interesting lol

[crystal lion]

My intuition is that they should be the only ones

But I cant think about a Quick explanation

[delicate bloom]

noticing we can also generalize this to have infinitely many examples where n! is the product of m factorials

n!!! = n!(n!-1)!(n!!-1)! for instance lol

[ivory trail]

can you elaborate more? i can see how Z/(60) is mapped into (Z/(60))_(2), so that a/1 = (a mod 4)/1 for all a in Z/(60), but what about other denominators, like 3/7 or something? what equivalence class do they fall into, i.e. what a/1 are they equal to?

appreciate the help btw

[sturdy marsh]

@ivory trail in general everything will not be of the form a/1 in the localization.

im not sure why you need that tho

[keen oasis]

I think you need to guess the subgroup of index 56. What is special about 720 that you can use is that 720=120*6, both of which numbers are factorials of natural numbers. I'm sure you can guess the solution from here.

As for the question about cyclic subgroups of certain sizes, the question can be asked differently this way: can you find an element of order 20? How should it look like?
@crystal lion
So S5×S3?
And about the order, yes 20 is to bigger, if i thake a permutation from S8 the its order will be the lcm of nr of elemnts in its disjoint cylcles. So it can be 20, i think the greatest order is 15

[crystal lion]

Yes, $S_5\times S_3$ can be embedded into $S_8$ for instance. Fun fact: it is a maximal subgroup of $S_8$, so any subgroup strictly containing it is the whole group itself. Or as observed $S_6$ is of the same order (which however is not maximal). The only thing left to do is finding these groups in $S_8$,but that is easy.

I think you've pretty much figured out the second, just prove it more rigorously, 15 is the order of the largest cyclic group in $S_8$.

[cloud walrus]

kadmate

[crystal lion]

@keen oasis

[keen oasis]

Thanks a lot!

[steady axle]

are $\mathbb Z[1/2,1/2^2,\dots]$ and $\mathbb Z [1/3,1/3^2 , \dots ]$ isomorphic as modules over $\mathbb Z$?

they are not isomorphic as rings

[cloud walrus]

bert

[waxen hedge]

Suppose there exists an isomorphism of abelian groups, and consider the orders of elements

[steady axle]

isn't order of any element infinite? (because 0 is the identity and not 1 )

[chilly ocean]

1 has to map to 1.

[hidden haven]

You can look at how many times 2 divides an element

@steady axle

[hidden haven]

This isn't true if they're modules over Z

The 0 endomorphism is also allowed

[steady axle]

Excellent hint. We look at the image of 1/2 if it is x then image of 1/2^2 is x/2 and so on. But we cannot go on dividing x by 2!

[hidden haven]

Yep

[steady axle]

Did u have something else in mind

[hidden haven]

Nope, that's exactly it

[opal osprey]

Let $R \subset \mathbb{C}[x]$ be a subring that properly contains $\mathbb{C}$, I want to prove that $\mathbb{C}[X]$ is a finitely generated $R$ module.

[cloud walrus]

MisterSystem

[opal osprey]

I think that I should pick a p(x) \in R a non constant polynomial

And prove that {1,i, p(x)} generates C[x], but I am not sure

[gritty sparrow]

That won’t work, adding i p(x) and 1 will not generate anything new as an R module as they are already in R. Instead you want to add elements in C[x] that are not in R. Here is a hint: given g in C[x] what does division with remainder by p give you?

[opal osprey]

yeah, I can always write g \in C[x] as g(x) = p(x) q(x)+r(x) with q(x),r(x) \in C[X] and 0 ≤ deg r(x) < deg p(x)

I was thinking about using the fact that C[x] is an euclidean domain

[gritty sparrow]

Well, here is one more hint to get you a bit further: ||can you think of a finite set of polynomials that will generate the polynomials of degree<deg(p)||

[opal osprey]

{1,...,x^{deg p(x)-1} }, but I would have to prove that such a set is in R

[gritty sparrow]

Nope

[opal osprey]

Ah

You are right

[gritty sparrow]

Also we aren’t done just yet, can you fill in the last part of the proof

[opal osprey]

Ok so let deg p(x) = n

We want to prove that there exists u_1(x),...,u_n(x) such that for all g in C[x] we have that g(x) = a_1 u_1(x)+ ... + a_k u_k(x) for a_1, ... , a_k in R.

Let p(x) in R be such that deg p (x) > 0.

So, for all g(x) in C[x], there exists q(x), r(x) such that g(x) = p(x) q(x) + r(x) with 0 ≤ deg r(x) < n.

We then consider the set {1, ... , x^{n-1}}.

Now, I don't really know what should be the last step

[gritty sparrow]

Well the r(x) part of g(x) is easily generated, the only problem is the q(x). So how do we generate q(x)? We do the same thing again

We may have to do this division by p(x) many times, but the degree of the thing we are dividing by keeps eventually it will be lower than n, so then generate it by the set

Does this make sense to you?

[opal osprey]

Yup

I just need to formalize this

Thanks

[gritty sparrow]

Np

[opal osprey]

The fact that this process of iterative euclidean division eventually stops has to do with the fact that C[X] is noetherian ?

[gritty sparrow]

I think of it more like the reverse. It has to do with the fact that it is a euclidean domain and then euclidean domains are all noetherian and the proof will hinge on the fact that division gives a remainder of smaller size

[opal cedar]

why are a=(1 +√5)/2 and b=(1−√5)/2 distinct in the group Zp where p is a prime not equal to 2 and 5? I made some observations that a+b=1 and a-b=sqrt5=0, but not sure how to go from there

[latent anvil]

since 2 and p are coprime, it suffices to show 2a ≠ 2b

similarly we just need to show 2a-1 ≠ 2b-1

This is to say, 2sqrt(5) ≠ 0

But if 2sqrt(5) = 0 then 2 = 0 or sqrt(5) = 0, neither of which is the case

@opal cedar

[opal cedar]

👍 Thanks! I didn't realize that p!=2 can be used toward the argument! Thank you so much.

[vestal snow]

Does k(A) here mean the fraction field of A?

[gritty sparrow]

Yes

[latent anvil]

This is a great theorem

10/10

[fossil shuttle]

Eisenbud speaks about this very poetically in his commutative algebra book

He points out that it wasn't too long before this result was published that Peano had challenged people's intuition about dimension with his space filling curve.

Peano's curve snaking through R^2 was like a "snake in the Garden of Eden", forcing analysts and geometers to leave paradise and deal with point set topology But the results of Krull and Noether show that for algebraic geometry "there is no snake in the garden"

[chilly ocean]

what field studies the snakes in the garden?

[kind temple]

herpetology

[kind temple]

i mean, its not wrong...

[latent anvil]

feel like rougher versions of DG

[chilly ocean]

DG doesn't study the snake DG is the snake

[latent anvil]

o no

[fossil shuttle]

is that why they call studying differential forms "riding the dragon"

[latent anvil]

they do!??

[chilly ocean]

do they call it that

[fossil shuttle]

no but we could start

[latent anvil]

we should

[chilly ocean]

im riding the moving frame 😎

[latent anvil]

I was riding the dragon all night

[chilly ocean]

[kind temple]

maybe we shouldn't lmao

[latent anvil]

no I think we should

> try to study differential geometry

[chilly ocean]

petition to rename "moving frame" to "dragon" and to change cartan's profession to "dragon tamer"

[latent anvil]

> find differential forms super confusing

> "aw man I've been riding this bad dragon for a while"

[chilly ocean]

die

[latent anvil]

[chilly ocean]

shamrock warned arc

[latent anvil]

hahaha

I haven't been very active on here recently

might as well

[chilly ocean]

im still laughing

but

[fossil shuttle]

Shamrock did you end up giving that presentation to your homological algebra class on Dold-Kan or spectral sequences or whatever

[latent anvil]

😌

On dold kan yeah

Months ago

It was fun!

Didn't end up talking about the monad pov

[fossil shuttle]

been thinking a lot about simplicial methods in homological algebra the past couple months. I keep coming back to it. it's really fascinating

[latent anvil]

Ooh nice

[fossil shuttle]

sheafified_sarah on twitter has posted a lot about the philosophy of chain complexes and lately i dmed her this long ass rant about dold kan hoping she'd find it interesting

i don't think that's actually her name, i can't remember her actual @ but i'm sure you know who i mean

[sly storm]

Hello. Is the semidirect product of $Z_n$ and $Z_2$ with respect to the automorphism of inversion on $Z_n$ the group $D_{2n}$? If so, why?

[cloud walrus]

MathPhysics

[latent anvil]

It's easiest to think of Dn in terms of an internal semidirect product

Dn is presented by two generators r, f with the relations r^n = 1, f^2 = 1, frf = r^-1

Here r is a minimal rotation and f is a reflection

[chilly ocean]

||deez nuts||

[latent anvil]

The subgroup N generated by r is cyclic of order n and it's normal due to the relation frf = r^-1

The subgroup K generated by f intersects this subgroup H trivially and HK = G

So G is isomorphic to the semidirect product of H and K, with K acting on H via conjugation

But we know frf = frf^-1 = r^-1, so the action of K on H is just by conjugation

[sly storm]

Thanks.

[steady axle]

Can I get a hint?

[rustic crown]

use the geometric series to get an idea,

then simplify the expression, and then verify that thing works all the time (so that we don't need to worry about the convergence of that series)

[steady axle]

wow that works so well

[rustic crown]

[tawny bridge]

Hi, ive been struggling with finding an example of a torsion module over a PID which has annihilator equal to the null ideal, is there one?

[hidden haven]

Direct sum over natural numbers n > 1, of Z/nZ, as modules over Z. Any element has finitely many non zero entries so has torsion (multiply by lcm of corresponding n) but no integer annihilates everything

[tawny bridge]

oh it makes sense, thanks! 😄

[sullen island]

question about notation: if we are viewing $A \otimes B$ as an $R$-module, why is $S$ the subscript instead? im not really understanding the notation here, couldnt we have written $A \otimes_R B$ ?

[cloud walrus]

xy

[fossil shuttle]

because the tensoring is over the ring S.

when we talk about tensoring two modules $M,N$, we are always more explicitly talking about tensoring a right $S$-module $M$ with a left $S$-module $N$ for some ring $S$, and if the ring is ambiguous, we must specify it. The structure of the resulting gadget is different depending on what ring we tensor over, as in in the tensor product all elements of the form $ms\otimes n - m\otimes sn$ must be zero for all $m \in M, n\in N, s\in S$. Thus the choice of ring $S$ matters.

[cloud walrus]

diligentClerk

[fossil shuttle]

It's not possible to define $A\otimes_RB$ in the example you've given, because $B$ isn't a left $R$-module, and $A$ isn't a right $R$-module.

[cloud walrus]

diligentClerk

[sullen island]

ahhhh right, so just because "R acts on the resulting structure" doesnt mean that $R$ is in the subscript, the subscript denotes whatever ring that gives you the "linearity relations"

[cloud walrus]

xy

[sullen island]

got it, thanks!

[fossil shuttle]

\yep

[minor badger]

anyone here familiar with sl_2-modules and weight modules? I would need some clarification about a part of a proof

[void knot]

,tex Suppose that $G$ is an abelian $p$-group such that [G\cong C_{p^{e_1}}\times...\times C_{p^{e_r}}] with $e_1\ge ...\ge e_r \ge 1$. Why does the subgroup [G^p=\left{g\in G:: g^p=1\right}] has order $p^r$?

[cloud walrus]

RaD0N

RaD0N

[thorn delta]

the order of each element divides p, so it can't be anything else

oh wait, i guess you mean why p^r where r is given as above hmm

[rustic crown]

just write down the element (g1, g2, ..., gr). this is in that subgroup if and only if each g_i^p = 1 in C_{p^{e_i}}.
you should have enough experience with these to conclude that g_i would be one of h^0, h^1,..., h^{p-1} where h = g^{p^{e_i - 1}} for a generator g.

so each g_i has p options which means that tuple has p^r options.

(also the notation you used is kinda misleading G^p usually means {g^p for g in G}, this does form a subgroup when G is abelian)

[void knot]

Unfortunately I don't have that much experience with that. Neither I understand why it is true(

[misty socket]

What exactly don't you understand from what they said?

Do you understand why we can represent an element of G as (g1, g2, ..., gr)?

[void knot]

Yes

If $g_i$ is a generator of $C_{p^{e_i}}$ then $g_i^{p^{e_i}}=1$.

[cloud walrus]

RaD0N

[misty socket]

The fact that g_i is a generator here doesn't really matter

This is usually called Lagrange's theorem

[void knot]

Then $h_i=g^{p^{e_i-1}}$ is a possible element.

[cloud walrus]

RaD0N

[misty socket]

Right, and note that (h_i)^p = 1

[void knot]

$h_i\in C_{p^{e_i}}$

[cloud walrus]

RaD0N

[void knot]

Yes.. I noted that.

So there is $p$ powers of $h_i$ in $G^p$. Actually that group was denoted by $G_p$

[cloud walrus]

RaD0N

[void knot]

but i ranges from 1 to r

Ohh my gosh.... now I understand. I'm so tired(

I'm grateful to u

[thorn delta]

maybe im misunderstanding something, but sometimes I see the unique gluability condition for a sheaf on a space X stated as the sequence
$$ 0 \to \mathcal O(X) \to \prod_\alpha \mathcal O(U_\alpha) \to \prod_{\alpha,\beta} \mathcal O(U_\alpha \cap U_\beta)$$ being exact for an open cover ${U_\alpha}$ of $X$ interchangeably with the condition that for each open set $U$ of $X$ and open cover ${U_\alpha}$ of $U$, the sequence
$$ 0 \to \mathcal O(U) \to \prod_\alpha \mathcal O(U_\alpha) \to \prod_{\alpha,\beta} \mathcal O(U_\alpha \cap U_\beta)$$ is exact.

which one is correct, or is there even a difference in general?

[cloud walrus]

kxrider

[fossil shuttle]

the second one is correct.

we need to impose the condition for every set U, not just globally

I'd be surprised if they were equivalent, I don't think so

[gritty sparrow]

I think that if you take X={1,2,3} with the open sets being {1}, {2}, {1,2}, and X and define F(U)={constant functions U->R} if U is a subset of {1,2} and {the zero function X->R} otherwise we will get an example of the first condition but not the second.

[thorn delta]

thank yall for earlier!

So.. on 5.1.8., I am having trouble getting anywhere. Probably missing something simple. If $(f\circ F)(p) = (g\circ F)(p)$ then of course $f_{F(p)} = g_{F(p)}$. Since $F^#$ is a local ring map and $(f-g)_{F(p)} = 0$, we have $$F^#(g)_p = F^#p(g{F(p)}) = F^#p(f{F(p)}) = F^#(f)_p.$$ So in particular, $F^#(f)(p) = F^#(g)(p)$. Not sure how this might get me any closer to showing $F^#$ is the pullback of $F$ though.

[cloud walrus]

kxrider

[fossil shuttle]

Do you get the hint?

[chilly ocean]

What is this book btw

[fossil shuttle]

yeah this book looks interesting

[thorn delta]

well, yea i think so. is computing F_p#(g_F(p)) not what they mean by passing to stalks?

[fossil shuttle]

if you like this you might also like Manifolds, Sheaves and Cohomology by Wedhorn

[thorn delta]

These are lecture notes from a mini course im taking (on varieties and schemes)

[fossil shuttle]

i don't remember exactly how this works, i need a minute to work this out

Ah. Right

You obviously need to use the fact that these are locally ringed spaces, the "locally ringed" being crucial

and the fact that it's a morphism of local rings

so in particular that means that if you mod out both local rings by their maximal ideals you get an induced map between the quotient fields

now do you know what that quotient field is in general?

[thorn delta]

Its isomorphic to R

[fossil shuttle]

right. yeah

so follow up

if you have some function g in O_N(V) and you take its germ at F(p)

and you pass to the quotiend field

then the residue class of g in the field will be, essentially, a real number

what real number is that?

[thorn delta]

it would be g(F(p)). i think i see where youre going with this

[fossil shuttle]

right. yeah so if you think you get the idea i won't tell you the whole punchline, but right, the germ is exactly the value of the function, and you're trying to prove that these two functions have the same value. so using this equivalence you can try and use the induced ring homomorphism to prove that these two functions you're trying to prove have the same value determine the same residue class in the respective quotient fields

[fossil shuttle]

Did you get it?

[thorn delta]

no, not quite. I have something like
$$(F^# f)(p) \equiv (F^# f)_p \equiv \overline{F^#p}(f{F(p)}) = \overline{F^#_p}(f(F(p))) = (f(F(p)))_p = f(F(p))$$

[cloud walrus]

kxrider

[fossil shuttle]

what's the over bar notation. the residue class?

[thorn delta]

haven't convinced myself the last two equalities are valid. The bar is the induced map and i use \equiv whenever i switch out representatives mod m_p

[fossil shuttle]

gotcha

[thorn delta]

(the latex cut it off too)

yea, and even if the last two equalities are correct, this gives F^#f(p) = f(F(p)) mod m_p which i don't think is enough.

i thought i understood the general sentiment of your approach, but I'm a little confused about how we're supposed to recover the values of F^# after passing to the quotient field

[fossil shuttle]

ok. let me think for a minute

[thorn delta]

i see an error in what i wrote above

[fossil shuttle]

you know what, I think you might need an extra assumption. I mean I'm still thinking through this but

I think it would be useful if you could assume that all these maps were $\mathbb{R}$-linear. I wonder if instead of locally ringed they meant to write locally $\mathbb{R}$-ringed with $\mathbb{R}$-algebra maps in the natural transformations

[cloud walrus]

diligentClerk

[thorn delta]

that the F^#_p are R-linear?

[fossil shuttle]

Yeah. We might not need that tho i'm just thinking out loud

Ok. Look. What are the ring homomorphisms $\mathbb{R}\to \mathbb{R}$?

[cloud walrus]

diligentClerk

[fossil shuttle]

is it only the identity lol or could there be some more pathological shit going on?

[thorn delta]

ok i thought the F#_p were always R-linear since we think of the stalks as R algebras but idk

[fossil shuttle]

Here's what I mean. I think I have a proof if we assume that by locally ringed spaces they actually meant the category of sheaves of $\mathbb{R}$-algebras and natural transformations of $\mathbb{R}$-algebra maps but maybe there's a way to do it without that.

[cloud walrus]

diligentClerk

[fossil shuttle]

Yeah, I agree intuitively we think of them as $\mathbb{R}$-algebras they just don't mention that

[cloud walrus]

diligentClerk

[fossil shuttle]

but yeah if we assume $F^\sharp :\mathcal{O}{N,F(p)} \to \mathcal{O}{M,p}$ is $\mathbb{R}$-linear

[cloud walrus]

diligentClerk

[fossil shuttle]

then modding out by the maximal ideals, the induced map is

$F^\sharp : \mathbb{R}\to \mathbb{R}$

[cloud walrus]

diligentClerk

[fossil shuttle]

which should just be the identity map

sorry i mean your overbar notation, the induced map between local rings

So, $F^\sharp(f)(p)$ is some real number which we can also think of as the residue class of $(F^\sharp(f))p = F^\sharp(f{Fp})$ in the quotient field

[cloud walrus]

diligentClerk

[fossil shuttle]

But the map $F^\sharp$ induced on quotient fields is just the identity, so as a real number this is the same as the residue class of $f_{Fp}$ in $\mathcal{O}_{N,Fp}$

[cloud walrus]

diligentClerk

[chilly ocean]

[fossil shuttle]

and as we discussed, this is the same as the value of $f$ at $Fp$

[cloud walrus]

diligentClerk

[fossil shuttle]

which proves $f(Fp) = F^\sharp(f)(p)$, as desired.

[cloud walrus]

diligentClerk

[gritty sparrow]

All ring maps from R to R are identity (it is a little tricky to prove)

[fossil shuttle]

that's fucked

[gritty sparrow]

Basically all ring maps will preserve the order as well

[fossil shuttle]

ohh ok

[gritty sparrow]

Then you can do some stuff to get that it will be the identity map

[chilly ocean]

wait nvm R as in real ok

[fossil shuttle]

yeah ok. i see that if it preserves the order it should be easier to prove

[gritty sparrow]

Preserving order is because if x>0 the x=y^2 for some y hence f(x)>0

[fossil shuttle]

Uggh, ok. cool. yeah, real closed fields

[gritty sparrow]

And f will be identity on Q. So we can easily see that if f(x)>x then taking x<q<f(x) we get f(x)<q<ff(x), a contradiction

[fossil shuttle]

right. nice

[thorn delta]

so is the idea something like $$(F^#f)(p) \equiv (F^# f)p = \overline{F_p^#}(f{F(p)}) = \overline{F_p^#}(f(F(p))) = f(F(p))$$

[cloud walrus]

kxrider

[gritty sparrow]

What are the bar’s over these things exactly?

[thorn delta]

i feel a bit hung up on a ring hom R to R being the identity and having two fields which are each isomorphic to R somehow becoming the identity

the stalks are local rings, F# is a local ring map, and the bar map is the induced map on the residue fields.

[fossil shuttle]

yeah sorry i was abusing notation but like, all the isomorphisms are unique. We can still call the map $\overline{F}^\sharp : \mathcal{O}{N,F(p)}/\mathfrak{m}{Fp} \to\mathcal{O}{M,p}/\mathfrak{m}{p}$

[cloud walrus]

diligentClerk

[fossil shuttle]

but each element of both of those rings can be identified with a real number in a unique way, and with respect to that identification $\overline{F}^\sharp (r)=r$

[cloud walrus]

diligentClerk

[fossil shuttle]

and yeah your equation is right kxrider

[gritty sparrow]

Also the solution is to look at the map of stalks from F(p) to p. Then modulo the maximal ideals. We get a ring hom from R to R and the image of f in the first R is f(F(p)) and in the second it is F^#f(p). Now as the ring map is identity between the two R’s f(F(p))=F#f(p)

[thorn delta]

okay, so how does the first \equiv become equality though?

oh wait i guess the identification is unique

[gritty sparrow]

Oh sorry, @fossil shuttle gave an answer

[fossil shuttle]

you don't need the germs themselves to agree, the question is about the value of the functions

it's alright, another way of explaining it is probably helpful

[thorn delta]

i was about to start a rabbit hole with an affine scheme question, but I think I'm satisfied for tonight. thank yall

[wooden ember]

im having a bit of trouble verifying that set of generators and relations really does generate the group you want: the relations in particular confuse me a bit im never sure how to detect if your set ends up "collapsing" through some hidden relation or not. For example how can I tell whether or not the generators that I have come up with do generate $S_3$? $<\tau_1,\tau_2 | \tau_1^2=\tau_2^2=1, \tau_1\tau_2\tau_1=\tau_2\tau_1\tau_2>$

[cloud walrus]

Little Narwhal

[wooden ember]

for example the trivial set would also satisfy these relations so how do i make sure it is truly s3 im generating

[chilly ocean]

What is abstract algebra?

[void knot]

In S_4, how to see that V is isomorphic to C_2xC_2?

[carmine fossil]

Show they have the same presentation

[chilly ocean]

you can also define a mapping, its not hard to guess

and just check its an iso

[void knot]

sorry, but I feel dumb

I can't figure it out

[chilly ocean]

can you guess what (0,0) in C2xC2 should map to in V?

[void knot]

to 1

[chilly ocean]

yes

now you have to find what (1,0) and (0,1) should map to

not many choices

so either you can just try every possibility or think about it a bit to skip possible calculations

[void knot]

ok. I'll try to find the isomorphism between Z_2xZ_2 and V

[chilly ocean]

if you can figure out which two permuations after composing them give the third one then you're done

kinda

[void knot]

done. thx)

[fast tree]

are integers under addition cyclic?

pretty sure they are cuz they're generated by 1

[hidden haven]

Yes

[fast tree]

:nyoom

[paper flint]

@carmine fossil Can you confirm if (Z,+) is cyclic?

[chilly ocean]

[hidden haven]

His sidekick already confirmed it

[carmine fossil]

I cannot

According to my sir's notes

[chilly ocean]

it's not cyclic

it's

loopy

new word for cyclic grp

[hidden haven]

I don't see any loops

[chilly ocean]

try drugs

[carmine fossil]

I can confirm (N,+) is a group

Oh wait,Exclude 0 out of N

So (N-{0},+) is a group

[nova plank]

👀

[dawn gorge]

👀

[nova plank]

Ik you are trolling, but it really is

[carmine fossil]

I am not trolling

I am just following my teacher

Who is infallible

[nova plank]

Let $f: \bN \setminus{0} \to \bZ$ be a bijection. Then $(\bN\setminus{0}, +)$ is a group where
$$a+b = f^{-1}(f(a)+{\bZ} f(b))$$
and $+{\bZ}$ is regular addition.

[cloud walrus]

Lunasong the Supergay

[robust pollen]

If $H$ is a (finite-dimensional) Hopf algebra, then it's well-known I guess that its coadjoint representation $H^_{\mathrm{adj}}$ is a $H$-module algebra, i.e. an algebra object in $H$-mod.
I know that the (internal) category of modules over $H^{\mathrm{adj}}$ is the Drinfeld center of $H$-mod, so the following question seems obvious but I don't know if it's true:
Is the Drinfeld double of $H$ isomorphic to the smash product $H # H^*{\mathrm{adj}}$??

[cloud walrus]

expectTheUnexpected

[robust pollen]

Is this just a computation I'm too lazy doing? Or am I a big dumb?

[winter vigil]

Question : Let $G=<x>$ be an infinite cyclic group. Then $<e>,<x><x^{2}>,...$ are distinct subgroups of G. Heres my approach : Since $G=<x>$, then $x^{j} \neq x^{k}$ for $j \neq k$ where $j,k \in \mathbb{Z}$ So it implies that $<x^{j}> \neq <x^{k}>$ so no two generators of subgroups of G are the same, thus all must be distinct.

[cloud walrus]

Otoro

[hidden haven]

a ≠ b doesn't imply < a > ≠ < b > in general

For example < n > = < -n >

[winter vigil]

hmmmm are there any hints ?

[gray comet]

I mean, what are you trying to prove?

That x^k \neq x^j for distinct k and j?

[winter vigil]

well maybe perhaps <x^j> \neq <x^k>

since if no two subgroups are equal, then all are distinct

[hidden haven]

For distinct non negative j and k

[hidden haven]

Look at the smallest positive power of x contained in both the subgroups

[winter vigil]

wouldnt that be x^j and x^k respectively ?

oh identity e ?

[gray comet]

Positive power

So not e

[hidden haven]

Yep. Try proving that, and see what you can conclude.

[winter vigil]

ohhh okay, I will try it now thanks

[unique juniper]

anyone know how to do (b)

[oblique river]

use the fact that $\mathbb{F}_p^\times$ is cyclic of order $p-1$

[cloud walrus]

Buncho Bananas

[oblique river]

under the isomorphism $\mathbb{F}_p^\times \to \mathbb{Z}/(p-1)\mathbb{Z}$, $-1$ gets sent to $(p-1)/2$ and multiplication turns into addition

[unique juniper]

ok

[cloud walrus]

Buncho Bananas

[unique juniper]

but then what?

[void knot]

,tex Let $G$ be a finite group and $P$ be a $p$-Sylow subgroup of $G$. If $H$ is a subgroup of $G$ then how do we prove that $P\cap H$ is $p$-Sylow subgroup of $H$?

[cloud walrus]

RaD0N

[void knot]

I'm feelin dumbo.. shall I move to infantary(

[limpid edge]

for the forward direction, what can you say about the order of the element x where x ^2 = -1 ?

[unique juniper]

ive managed!

but

x^2 + 1=0

(x+1)(x-1)=0

and then we can show that it must be an element of order 4 iguess

[limpid edge]

yea!

and then to finish you can use ||lagrange||

[unique juniper]

yeah! thank you

[limpid edge]

also this is related to quadratic reciprocity, if you’re interested you should look into it

it’s more number theoretic though

[unique juniper]

number theory is kinda scary

maybe in the future

[limpid edge]

no but what you’re doing now is like a generalized version lol

[unique juniper]

oh

[maiden ocean]

@oblique river sorry if you're busy by my wifi keeps going out so i would like to ask while i can

what is going on with the underlined portion

why can we write f as that

[oblique river]

np let me take a look

I think the idea is that f is a 2-variable polynomial

in x and y

[maiden ocean]

Yeah

[oblique river]

but since f(P) = f(0,0) = 0, it has no constant term

and we're splitting it up like "take all the terms which only have x in them and factor out an x" and that's phi(x)x

and then everything else is a multiple of y because the only terms which aren't a multiple of y are taken care of with phi

[maiden ocean]

I see

[oblique river]

like if you write f as a sum of c_{i,j} x^i y^j

then c_{0,0} = 0

and all of the x^i y^0 terms are in phi

and the rest of the terms are all divisible by y and so that's the bit which goes into psi

[maiden ocean]

This makes sense I think

Why is the constant term of phi delta_y(P) then?

[oblique river]

one sec brb

is that a typo? should it be delta_y f(P)?

I'm assuming it is based on context

[maiden ocean]

Uh probably yeah

[oblique river]

yeah

to see this you should use what i like to call the Nike method: you just do it

take the y-partial derivative of f

the x terms vanish

and you're left with (del_y psi(x,y))y + psi(x,y)