groups-rings-fields

[rustic crown]

for any two a and b, we always have a^o(G) = e = b^o(G)

it's like saying that 4 as a real number of two sqaure roots +2 and -2

but in the case of groups, we don't have a way to select between all the roots

[plush quartz]

where did you get this from?

[rustic crown]

it's lagrange's theorem

take a specific example if you don't know the theorem yet

like G = Z/nZ

adding any element n times will definitely give you 0

[plush quartz]

is Z/nZ equivalent to n mod z?

[rustic crown]

yea integers mod n

anyway, how did you show surjectivity?

[plush quartz]

for surjectivity, when n is odd, we're fine. When n is even, o(G) must be odd since they're coprime. I'm confused on where to go from here since f(x)=x^2, x^4, etc is still not surjective.

[rustic crown]

why are we fine if n is odd? don't confuse this situation with reals

the main idea is this. If n and o(G) are relatively prime, then you can find another integer m such that n * m = 1 (mod o(G))

[plush quartz]

i think i made that mistake and am in too deep lol

[rustic crown]

so when n and o(G) are relatively prime, you can find an integer m which "behaves" like 1/n

so raising to the power of m should be same as taking n-th roots.

[plush quartz]

where did you get 1 from? Why can't it be 3 mod o(G) or pi mod o(G)?

[rustic crown]

in general for a and b integers with gcd(a, b) = d, you can find a solution to ax+by=d

where x, y are also integers

[plush quartz]

but since they're coprime

gcd is 1

[rustic crown]

yep

[plush quartz]

ahhh I see

[rustic crown]

so we can find m, k such that n * m + k * o(G) = 1

but yea you might need to know lagranges theorem to complete this

because g^n = h^n means that g^mn = h^mn but as g^o(G) = h^o(G) = e, that part involving k doesn't matter at all

and we get g^1 = h^1

conversely, g = g^1 = g^{nm + ko(G)} = (g^m)^n

[lethal cipher]

Okay, so I am looking at number 2.
I'm thinking about it, and I think the only automorphism of Z is the identity map.

I was initially thinking it was maps of the form f(x)=x+b for some b in Z. However, f(-x) is not equal to -(x+b).
So they are actually out

So if my assumption is correct, how do I prove that the identity is the only automorphism of Z?

[rustic crown]

there is another

[lethal cipher]

Oh, f(x)=-x

[rustic crown]

yee

[lethal cipher]

Okay. So those are great candidates. How do I show there are no others?

[rustic crown]

just start with an automorphism Z --> Z, where can 1 go to?

if f(1) = n, then the image of the map is only the multiples of n

[lethal cipher]

Oh dang! You are a genius! <1>=Z means <f(1)>=Z

Thank you!

[rustic crown]

[plush quartz]

Okay, so the idea behind this problem is if x^n=y, then taking y^1/n=x would be meaningless. Why is it more useful to interpret 1/n as m mod o(G)?

[rustic crown]

yep

[lethal cipher]

Well shoot. Closure isn't preserved here...

This function is a bust

Wait, nvm. I am silly. Z is under addition.

It does work out just fine

[past temple]

is (G/N) x N

isomorphic to G?

[rustic crown]

nope

consider G = S3 and N = A3

[past temple]

shoot

what about for modules and direct sums

[rustic crown]

still nope

G = Z/4Z and N = 2G

Z-Mod ~ Ab

true for vector spaces tho

[past temple]

shoot

ok so like

i basically want to prove that M = M1 + M2

but ur saying it doesnt suffice to show that M/M2 = M1

is there a way to still use quotients to prove the statement?

also M is a finite length module if that helps

[rustic crown]

well if M is a semi-simple module then it works

i don't think length has a lot to do with this

[rustic crown]

these are finite length mods

[past temple]

shoot

so should i try a strategy not involving quotients

[rustic crown]

seems so

[next obsidian]

You can sort of make it work with a quotient

You have to show that anything in M/M_1 can be expressed as something in M_2/M_1 where this is like the set of (m + M_1) where m in M_2

This is almost definitionally equal to showing M = M_1 + M_2 but occasionally it’s a lot more obvious in the quotient that it works

[past temple]

wait but i already showed that M/M1 = M2

how can i show then that M/M1 = M2/M1

[next obsidian]

Well

You didn’t show M/M1 = M2

You exhibited an isomorphism

But this isn’t what you want

You need the equality I stated like, as actual literal sets

More explicitly (or less?) you need the inclusion of M2/M1\cap M2 -> M/M1 to be an isomorphism where this is the map induced by M2 -> M

I think this is not going to be helpful for you

My answer to “how do I show M/M1 = M2/M1 from M/M1 ≈ M2” is that you cannot in general

If you can make it work it’ll matter a lot on the explicit isomorphism M/M1 -> M2 and you’ll have to fuck around with that a lot

[past temple]

does it help if M1 and M2 have trivial intersection

[next obsidian]

Not really, at least not for turning a random isomorphism M/M1 -> M2 into what you want

What’s the nature of that map?

Like what does it do on elements

[past temple]

its a little complicated so do u mind if i explain it to u via DM

[next obsidian]

I suppose so but I don’t think there’s an issue doing it here. If anything you can tex it here which would help if it’s complicated

[plush quartz]

Let n be in U. Then what does n look like? What is its form?

should it be n or (n_1,n_2)?

[next obsidian]

Umm

It looks like xyx^-1y^-1

Where x and y are arbitrary elements of G

[upper cape]

Wondering if someone could give me some help on this question. In the question A is a finitely generated k-algebra

So far, I've managed to show that im(phi) is isomorphic to im(psi), but i'm not sure where to go from there

[upper pivot]

ig look at one of these maps and think about what happens to each (k-alg) generator of A

Also you can reduce to the case where everything is invective so that might be useful too

[upper cape]

thanks I'll see if I can figure it out with that!

[plush quartz]

do I need to do #4 before #5?

[chilly ocean]

Not necessarily, you can assume 4) is known

But for 5) you'll have to use 4), yes

[cloud walrus]

Hausdorff

[median pawn]

This is trivial, I have proved it. I need help with:

[cloud walrus]

Hausdorff

[gritty sparrow]

The maximal ideal is the ideal of all functions that vanish at a, once you know that it shouldn't be too hard to prove that this will work (just show that any function not in this is invertible in O_a, that will prove that it is maximal)

[median pawn]

OK, thanks

Wait, I have already shown that any function not in this set is invertible in O_a

I don't see how that proves maximality though

[gritty sparrow]

All proper ideals consist of non units, now you just showed that the set of all non units is an ideal, so that means that it is the unique maximal ideal of the ring

[median pawn]

Got it, thanks!

[gritty sparrow]

Np, btw are you learning algebraic geometry rn?

[median pawn]

Nope! Is this supposed to be in algebraic geometry (complex analysis + algebra??)?

[gritty sparrow]

well it looks like they are maybe introducing sheaves, which are used mostly in either ag or complex geometry

so that is why I thought maybe ag

[median pawn]

Oh I should check it out then!

[molten silo]

How would i prove this?

[cloud walrus]

Hausdorff

Hausdorff

[median pawn]

Maybe someone could verify

[molten silo]

does associativity follow from the group law

[delicate bloom]

associativity is a group axiom yep

while you're thinking about it, try to see if you can list all the group axioms without cheating and looking it up

[molten silo]

i know them, its just what we are trying to show is not a group

i got confused

[delicate bloom]

gotcha, yeah even though the sets aren't necessarily a group, they still are elements that came from a group which have this relation

[fossil shuttle]

lots of books introduce germs or stalks without introducing sheaves

its very common

sad but true

[chilly ocean]

I'm trying to figure out the proof in this stack exchange writeup:
https://math.stackexchange.com/questions/534686/group-of-order-24-with-no-element-of-order-6-is-isomorphic-to-s-4

Prove that a group of order 24 with no element of order 6 is isomorphic to 𝑆4
I understand all parts of the proof except the statement that if there's only one Sylow 3-subgroup, this implies there exists some element of order 6. Could anyone help me understand that part?

For further reference, the brief outline given is:

To see that there cannot be just 1 3-Sylow, use that then this 3-Sylow is normal and by the N/C theorem, the centralizer must have index 1 or 2, both of which give elements of order 2 commuting with the elements of this 3-Sylow.
But I don't quite see the relevance of that

[next obsidian]

If g,h commute and have coprime ordee then gh has order equal to the product of the orders of g and h

So you produce an order 6 element from the commuting order 2 and order 3 element

[barren sierra]

Integral domain vs Division Ring

Help

What is diff

Both are non-commutative and unitary

Integral domain has cancellation

Division Ring has inverses

But can you have multiplicative inverses without cancellation???

Oh

So domains are commutative and division Rings are not

So the only difference is commutativity?

Oh right

Ok ok

Ok got it

So the differences lie in possible commutativity and existence of inverses

Cause both have no zero divisors

Dope

Also ideals are normal subgroups right?

Neat

[uncut girder]

Let $\Delta(2,5,10)$ be the Von Dyck group

[cloud walrus]

nYaminoid

[uncut girder]

It has this presentation

(im using \Delta to denote what wikipedia denotes by D)

Is it true that the commutator between x and y, which is $[x,y] = xyx^{-1}y^{-1}$ generates the full commutator subgroup of $\Delta(2,5,10)$ ?

[cloud walrus]

nYaminoid

[uncut girder]

Alternatively, is it true that $[x,y]$ and $[x,y^3]$ generate the same subgroup of $\Delta(2,5,10)$?

[cloud walrus]

nYaminoid

[uncut girder]

i dont think its true actually

[uncut girder]

Actually I think it may be true again...

wait

is the subgroup generated by [x,y] a normal subgroup?

[tropic spade]

For $\Bbb{R}$ and $\Bbb{R}(a)$ if $a$ is transcendental over $\Bbb{R}$ is $\Bbb{C}\not\cong\Bbb{R}(a)$?

[cloud walrus]

DootDooter

[sturdy marsh]

yes

what are the dimensions of C and R(a) as R-vector spaces

[tropic spade]

I'm assuming 2 for C and something infinite for R(a)?

[sturdy marsh]

yeah

so they cant be isomorphic extensions of R

[tropic spade]

That's a relief. My book has an exercise asking me to show any simple extension of R is isomorphic to C.

[sturdy marsh]

theyre probably talking about non-trivial finite extensions

[tropic spade]

Does R(a) from a sec ago count as trivial?

[sturdy marsh]

R(a) is not a finite extension

[tropic spade]

Finite as in the dimension?

[sturdy marsh]

yes

[tropic spade]

Gotcha, the next chapter talks about vector spaces so maybe they were mixing up the order of things.

Thanks for the clarification lol.

I was stuck worried about that case for a while.

[celest mantle]

Hi, i have an exercise where I have to determine the ring of functions of some plane curve (F1 = Y - X² for instance), but i don't really know what that is, could anyone help me please ?

[barren sierra]

given S_n symmetric group

is there an algorithm kind of thing to generate all the elements?

[delicate bloom]

I guess it depends on what form you want it in and/or why you want it

[barren sierra]

just curious

like is there a way to enumerate through them

[delicate bloom]

yeah, I guess what I'm trying to say is n! gets large fast

[fading wagon]

like generate all permutations?

it depends in what form you want it, but you can try to generate them in lexicographical order

[obsidian sleet]

ig u can take a generating set and just multiply them in a pattern until u get all n! ones

but yeah mero is right that gets big fast

oogissimo

[hidden haven]

oogissimo?

[obsidian sleet]

its evolving

\mapsto :oogissimo:

[hidden haven]

😵‍💫

[obsidian sleet]

😵‍💫

indeed

[robust pollen]

Stupid question I guess, but I was asked whether the following is true and I'm confused:
Let $K \subset L$ be a field extension, $A$ a $K$-algebra, and $I \subset A$ a two-sided ideal. Then is it true that $L \otimes (A/I) \cong (L \otimes A)/(L \otimes I)$?

I would think yes, because $L$ is flat, so that $L \otimes -$ is exact, i.e. preserves the short exact sequence $0 \to I \to A \to A/I \to I$. But that somehow seems to easy.
Maybe the question asks whether there is an iso of $K$-algebra, but I also don't see a problem here: surely $L \otimes I$ is a two-sided ideal of $L \otimes A$?

[cloud walrus]

expectTheUnexpected

[next obsidian]

I only ever deal with commutative stuff

But in the commutative case because of flatness tensoring commutes with quotients

I have no idea how the non commutative case works out

And yeah you prove it just using exactness

[gritty sparrow]

I think L\otimes I is a left L module, but I can't see why it would be a right L module

so I can't see why it must be an ideal

[robust pollen]

I think we're looking at those as vector spaces over K though, right? Like, L has the K-vector space structure induced by the extension K -> L
It is also clearly a K-algebra, and thus L \otimes A is a K-algebra with "factor-wise" multiplication. Under this multiplication, L \otimes I should remain an ideal.

[robust pollen]

(but it's also a right L-module with trivial L-action on the right tensor factor, i.e (l \otimes i) . x = lx \otimes i, so that I is basically only a multiplicity space)

oh sh!t, lowercase L looks like uppercase i lel

[gritty sparrow]

bruh, then it is not a right ideal

[robust pollen]

wut?

[gritty sparrow]

infact now that I'm thinking about it I can't see why L \otimesA is even an algebra

oh nvm that sorry

[robust pollen]

So the algebra structure I have in mind is basically because K-mod is a braided category lol

i.e. $ l \otimes a \cdot k \otimes b = lk \otimes ab$

[gritty sparrow]

but what I what I mean to say that something is a two sided ideal when it is a bimodule under the algebra multiplication

so I don't think L \otimes I is an ideal of L \otimes A

[robust pollen]

Why? $ (l_1 \otimes i) \cdot (l_2 \otimes a) = l_1 l_2 \otimes i a \in L \otimes I$ shows it's a right ideal, and similarly one sees it's a left ideal, no?

am I a big dumb?

[gritty sparrow]

yeah that seems right, it just felt a bit weird to me because then it seems like you could use the trivial sort of action to make L \otimes M into a right L module, which I thought was not possible

where M is some arbitrary left K module

[robust pollen]

You can, and what the trivial right action on M will do is, it will turn L \otimes M into a free right L-module with L-dimension equal to the K-dimension of M. [this should be right correct :D]

[gritty sparrow]

yeah makes sense

[robust pollen]

[Additionally, if A is an algebra over a ring R, then the category A-mod is automically an R-mod bimodule, where e.g. the left action R-mod x A-mod -> A-mod is precisely given by "tensoring with multiplicity spaces"]

[hidden haven]

"am I a big dumb?"
"yeah that seems right"

[gritty sparrow]

so all in all, it seems like your original proof should be correct, right?

[robust pollen]

But I'm always sure that my proofs are wrong.

Never trust a proof you wrote yourself.

That'll be my next tattoo, right on my forehead.

[latent night]

Can someone explain what is the difference between representation theory by quiver and "normal" represention theory? Is there a difference? My uni only has a course on the quiver thing, and the terms they use seem pretty different from whats in other texts like fulton harris

[weary terrace]

Let $\rho$ be an irreducible representation of the group $G$ and $N=ker \rho$ such that $G/N=\mathbb Z/n \mathbb Z$. What constraints do we need in order to have $Res_N \rho$ irreducible representation of $N$?

[cloud walrus]

RiesZ

[terse crystal]

Isn’t Res_N ρ=0?

[weary terrace]

Why so? e.g if N is a cyclic subgroup of G then Res_N \rho is 1-dim i.e not 0

[terse crystal]

My bad. Any irreducible F[G] module M, you want M to be a irreducible F[N] module where N=ker ρ right? But in that case M can be viewed as a F module therefore it’s dimension should be 1 I guess?

Since any g from N,x from M,gx=x

So n should be a prime number

[weary terrace]

Maybe I should've mentioned that F is algebraically closed with char F=0 (say \mathbb C)

Sorry..

[terse crystal]

Wait, G/N=Z/nZ is a subgroup of multiplication group of non-zero elements of F

It’s multiplication group I thought it was addition,sorry

[weary terrace]

I actually think of using Frobenius reciprocity

$Ind_N^G (Res_N \rho) = Ind_N^G (Res_N \rho \otimes \mathbb C) = \rho \otimes Ind_N^G \mathbb C=\rho \otimes\mathbb C[G/N]$

sorry just a second

[cloud walrus]

RiesZ

[weary terrace]

Ok, so by Frobenius reciprocity I got the above

And I think I can use the idea in this post: https://math.stackexchange.com/questions/1522624/irreducible-representation-by-restriction/1522704
to prove that the restriction is irreducible

But I'm not sure how..

Wait, if N is the kernel of $\rho$, then the action of N on $\rho \otimes\mathbb C[G/N]$ is trivial

[cloud walrus]

RiesZ

[weary terrace]

So $Res_N (\rho \otimes \mathbb C[G/N])$ is the trivial representation of N, hence 1-dim, hence $Res_N (\rho) \otimes Res_N (\mathbb C[G/N])$ is 1-dim, and thus $Res_N (\rho)$ is 1-dim and thus irreducible.

Am I making **any **sense?

[cloud walrus]

RiesZ

[weary terrace]

Apologies for the tedious writing..

[terse crystal]

Does trivial representation imply irreducible representation?

I thought trivial representation can have any dimension therefore not necessarily irreducible…

[weary terrace]

The triv rep takes any element of G to the unit element, hence its action on any 1-dim space is contained in that space

But I think I made a mistake anyway..

Thanks for trying to help, it is much appreciated

[warm dirge]

question about group extensions: given Z_4, i can take Z_2 as a normal subgroup and take the factor group Z_4/Z_2 = Z_2. is there a natural construction to get back Z_4 from the normal and factor group? the direct product (Z_2 x Z_2) is obviously wrong and the semi-direct product also doesn't work (Aut(Z_2) is trivial, so homomorphisms of Z_2 on Aut(Z_2) are obviously trivial, so the only semi-direct product we can make is again Z_2 x Z_2). also is there a similar construction for Q_8 and its two possible decompositions (once N = Z_2 and Q_8/N = Z_4 and once N = Z_4 and Q_8/N = Z_2)? i think what i'm basically asking is, whether there exist products other than direct and semi-direct.

[viscid pewter]

https://en.wikipedia.org/wiki/List_of_group_theory_topics

ctrl-f 'product' only gives direct product, semidirect, wreath and free

none of them look quite right for what you want

oh, i found another product, https://en.wikipedia.org/wiki/Zappa–Szép_product

still doesn't seem quite right

[warm dirge]

i have the suspicion that constructing groups from their normal and factor groups is way harder than i thought, and i find it weird that this doesn't get taught in class

but surely there has to be some material on this question. does anyone know some reference material? i could only find some of the most random pdfs online so far

[viscid pewter]

well yes

it's like

one of the central unsolved problems of group theory

it's called the extension problem

https://en.wikipedia.org/wiki/Group_extension#Extension_problem

it's not taught in class because it's really hard

and just not well understood generally, afaict

[edgy vault]

could rectangle rabatment be having any symmetries in Group Theory?

[viscid pewter]

looks like just V4 to me

[edgy vault]

@viscid pewter do you really think such a ree group member the Tits group is not strictly a group of Lie type?

[viscid pewter]

yeah i'll be real with you idk wtf that diagram actually means

i was just looking at the literal 2d shape and it looked like it had the same symmetries as a rectangle so

[edgy vault]

@viscid pewter this diagram is about rabatment in rectangles

[viscid pewter]

yeah idk what that actually is

[edgy vault]

@viscid pewter seconds question is something else

[viscid pewter]

i don't think i know enough to help you sorry

[edgy vault]

bc tits group is sometimes regarded as a 27th sporadic group

okay

[golden pasture]

Let $S=A\left[x_0,x_1,\dots,x_n\right]$ be a graded polynomial ring, then for any ideal $I$, is there a direct proof that
[\overline I=\cap V(I)]
where $\overline I=\left{s\in S:\exists n\forall i x_is\in I\right}$ is the saturation and $V(I)=\left{\mathfrak p\nsupseteq S_+:I\subseteq\mathfrak p\right}$ is the closed set in Proj$(S)$

ik a proof of showing that $V(I)=V(J)\iff \overline I=\overline J$ via ideal sheaf which gives us the result since $V(\cap V(I))=V(I)$ but feels quite overkill

[cloud walrus]

ari 亲

[golden pasture]

im hoping to use a similar proof that radical ideal = intersection of all ideal

but im having issues with not having S_+ inside your ideals

i.e. showing that maximal elements of $\left{\mathfrak a\nsupseteq S_+:\forall n\exists ix_i^ns\notin\mathfrak a\right}$ is prime cuz if i add say $(x)$, $S_+$ may be in the ideal sum

[cloud walrus]

ari 亲

[golden pasture]

oh wait nvm got it can consider all x_i
edit: oop im still trying to figure haiz

[edgy vault]

@golden pasture um do u mean intersection of all " maximal " ideal here or ?

[golden pasture]

all *homogenous prime ideals

[golden pasture]

i think i can get it by localizing at each x_i

feels a bit

indirect

[golden pasture]

i think i got it by

for any $s\notin\overline I$, consider the maximal element of $\Sigma=\left{\mathfrak a\text{ homogenous},I\subseteq\mathfrak a,S_+\nsubseteq\mathfrak a:\exists i\forall n(x_is)^n\not\in\mathfrak a\right}$
then since $\sqrt{\overline I}$ is a homogenous ideal satisfying all the conditions, $\Sigma$ is nonempty and by zorns we have some maximal element $\mathfrak b$ and by contradiction it must be prime, hence $\mathfrak b\in V(I)$ and $s\notin\mathfrak b$

[cloud walrus]

ari 亲

[median pawn]

What is the motivation behind doing this?

[cloud walrus]

Hausdorff

[cloud walrus]

Hausdorff

[median pawn]

Could someone explain this please?

[chilly ocean]

what's the definition of D_2n you know

[cloud walrus]

Hausdorff

[hidden haven]

isnt that the exact same

[viscid pewter]

yeah, they're just omitting the = e

[median pawn]

no but why is that the quotient of the free group

[viscid pewter]

that's standard

[chilly ocean]

this is exactly what's written there. when you quotient by <<R>> the elements of R become the identity in the quotient

[median pawn]

uhhh this is the part i don't understand

First of all what is <<R>>?

I know it's the normal subgroup generated by R

but what really is it in this case

[chilly ocean]

.

[terse crystal]

The minimal normal subgroup containing R

(=Intersection of all normal subgroups containing R)

[median pawn]

Yes that is the definition

but I am not able to prove F(X)/<<R>> is the same as D_{2n}

What is <<R>> explicitly? How do you find it

[hidden haven]

That is in general hard to prove. and <<R>> is even harder to describe lmao

[median pawn]

I think we can use this

[terse crystal]

Let f be the canonical homomorphism from F(X) to D_2n then You prove that kerf contains R and kerf is contained in any normal subgroup containing R

[hidden haven]

Define a surjective homomorphism from the free group on 2 elements to D_2n such that the kernel contains <<R>>. This induces a surjective map from the quotient to D_2n. To prove that it is injective, try to find the maximum cardinality that the quotient can have by listing equiv classes

[chilly ocean]

[hidden haven]

yeah nothing similar works for normal subgroups I think

[terse crystal]

Wait kerf contains R is obvious, then you only need to prove that F(X)/<R> has order 2n

[hidden haven]

yes

2n given the convention here

and =2n is still very hard

to do directly

<= 2n is easy

[median pawn]

Okay I'm kinda slow with algebra 😦

[cloud walrus]

Hausdorff

[median pawn]

Here, phi(a), phi(b) generate D_2n

Am I right till this point

[hidden haven]

you should choose explicit generators

instead of arbitrary ones

[median pawn]

in D_2n?

[hidden haven]

yes

[median pawn]

how explicit

[hidden haven]

which generators are you choosing as phi(a) and phi(b)

[terse crystal]

Let G=F(X)/<R> You can show that [G:<σ>]=2

[median pawn]

phi(a) = sigma, phi(b) = tau is the best we can do right?

[hidden haven]

yep thats what I wanted

and also no need to describe what phi does on each element of F(2), that only makes your job harder because now you need to check if it is indeed a homomorphism

[median pawn]

I have checked that is a homomorphism tho

[hidden haven]

Always use theorems, especially when they are labelled universal properties

[median pawn]

but I suck at algebra, so which theorem?

I'm slowly getting better lol it will take some time

[hidden haven]

Defining a set map on a,b is exactly the same as defining a homomorphism on F(a,b)

Its like the vector space thing

[median pawn]

Agreed

[hidden haven]

defining a set map on the basis is exactly the same as defining a linear map on all of the space

[median pawn]

Now is it trivial that ker phi = <<R>>?

[hidden haven]

since theres a unique extenison for each

no definitely not trivial

you wont be able to prove both inclusions easily

you can show one very easily though

see which one

[median pawn]

R = {a^n, b^2, abab} right?

[terse crystal]

(Any element from G has the form σ^i τσ^j τ σ^k … and τσ^iτ=(τστ)^i=σ^-i )

[hidden haven]

and the remaining arguments that cogwheels and I gave prove the other inclusion in a roundabout way

yes

[median pawn]

Is <<R>> subset ker phi supposed to be the easy inclusion

[hidden haven]

Yes

[median pawn]

Cool

One more thing, if w and w' are elements of F(X) representing the same element of G, how do we know that w' = wy for some y in <<R>>?

G = <X|R> is a group

[hidden haven]

Why do you want that?

[median pawn]

Not for what we are proving, but for showing that: two words w and w' in an alphabet X represent the same element of G iff they differ by a finite sequence of moves of Type 1: elementary contraction/expansion or Type 2: insert somewhere into the word one of the relations in R, or its inverse

→ is trivial, but for ← the author has used the w' = wy thing

I mean this probably has something to do with G with presentation <X|R> being F(X)/<<R>>?

[hidden haven]

Well what you are saying here is quite a bit stronger lol

And also not true

[median pawn]

Umm assuming w and w' represent the same element of G

Is it true then

Yeah it should be

[hidden haven]

Seems to be the implication yeah

[median pawn]

Does it follow directly from the fact that we define G = <X|R> as F(X)/<<R>>?

[hidden haven]

Yes

Lol

[median pawn]

I KNEW it, I don't understand quotients well

[hidden haven]

oof

[median pawn]

Could you maybe explain that a bit

like what exactly does a quotient group do

I feel like the idea is we are "identifying elements"

Abstractly I know it is the set of all cosets

[hidden haven]

[x] = [y] in the quotient iff they are in the same coset of N (the subgroup you are quotienting by) iff xN = yN iff x ∈ yN

It's the same as quotienting vect spaces

You've worked with those no?

[median pawn]

Yep, I have

"same element in G" means what though

Does it mean [w] = [w']?

[hidden haven]

Yes because underlying set of G is the set of equivalence classes

[median pawn]

Great, I see it now. Thanks!

[plucky flicker]

Hi guys! I have a problem here: prove that R is a local ring if and only if there exists an ideal I in R such that R\I = U(R), where U(R) denotes the invertible elements of R.
I solved the implication =>, but what about <=? Any ideas?

[hidden haven]

If such an ideal existed, it would be the unique maximal one

[plucky flicker]

i mean we don't know anything about I, just that it is an ideal

[hidden haven]

We know that it's exactly the set of all non units

[plucky flicker]

yes

[hidden haven]

That is enough

[plucky flicker]

ohh i got it

thanks

[simple valley]

are there groups that aren't a group of units of any ring (with unit)?

First I thought maybe C_3, but then it turns out Z[x]/(x^3+1)/2 is exactly that

if G is a group Z[G] is close, but it has twice as many units

so I thought maybe Z_2[G] but that can have nontrivial invertibles

e.g. in Z_2[Z_2 x Z_4] you have ((1, 0) + (0, 1) + (0, 3))^2 = (0, 0)

[hidden haven]

I think Z/5Z doesn't work

I remember discussing this with det let me see if I can find his proof

[simple valley]

I thought about it for a sec

[hidden haven]

I think this is it

[simple valley]

wait

that's exactly the table I have

[hidden haven]

Working through Riehl?

[simple valley]

more like my prof is stealing exercises from riehl

[hidden haven]

x here is a generator for the group of units

[simple valley]

I have wildly different examples for the other things though

[simple valley]

mmm partly because I'm dumb

[barren sierra]

I'm stuck on this question

[simple valley]

instead of the obvious choice of "there is only one ring hom Z -> Z" I went for some fuckery with Z[x] -> Z

[simple valley]

so is there a classification of F2 algebras or something?

I don't follow where the issue is

[hidden haven]

He did the computation with that product he found

That can never happen

Because the product = 1 says that both factors are units

[simple valley]

what product what

[hidden haven]

So one of them is x^k and the other is x^(5-k)

oof sorry

[barren sierra]

is division algorithm too overkill for this problem and is there a simpler solution?

[hidden haven]

The product of the 2 polynomials in x

The very first message in the screenshot

[simple valley]

oh

x is the generator of the group of units here?

[hidden haven]

Yes

[simple valley]

o

[hidden haven]

All computation is in characteristic 2 as he says in his second message

[simple valley]

yeah I found quickly that the only way to do without elements of even order is to have 1 = -1

[lone barn]

Are there any nontrivial ring automorphisms on the real and the complex numbers?

[simple valley]

i mapsto -i comes to mind

[lone barn]

As in conj(z)?

[simple valley]

yeah

I don't think any others exist

[hidden haven]

Many for C, none for R

C has uncountably many

Take any transcendence basis over Q, permute it any way you want, and this extends to an automorphism of C

For R, first show that Q ⊂ R must be fixed by any automorphism, then show that positive numbers must map to positive numbers, which implies that the automorphism is order preserving

[simple valley]

"transcendence basis"?

[hidden haven]

Yes?

Maximal algebraically independent set?

[simple valley]

I... have to go

[lone barn]

I don't know what a transcendence basis is, but can you give an example of such an automorphism on C? other than conj(z)

[hidden haven]

Yes, break this extension as
Q ⊂ Q(√2) ⊂ C

Q(√2) has a non trivial automorphism which does √2 → -√2

This then extends to an automorphism of C ... which is actually hard to prove

Becomes easier if you replace C with alg closure of Q

[simple valley]

(my understanding of galois theory is b a d)

[hidden haven]

Which is countable and algebraic over Q so everything follows quickly from the alg closure being alg closed

[lone barn]

ooh, this is where it gets juicy

[hidden haven]

Weirder is that this can't extend to an automorphism of R

[lone barn]

this is pretty cool actually

[hidden haven]

😵‍💫

[lone barn]

so wait

how does this extend to C again

[simple valley]

"make an uncountable choice"
"show me one"
moldi owned by constructivists

[hidden haven]

[lone barn]

Wildberger vibes

[hidden haven]

It's really hard to do without transcendence basis lol so let me just say what you need to know about them

[lone barn]

well, okay

[hidden haven]

In fact not transcendence basis but just breaking an extension into a purely transcendental one then an algebraic one

Yeah I think that was all I needed to say given a field extension E/F you can break it apart as E/K/F where K/F is purely transcendental (every new element is transcendental) and E/K is algebraic

Proof is zorn's lemma, find a maximal purely transcendental extension and then everything above must be algebraic

So do this for C/Q

It breaks as C/K/Q

√2 is in C but not in K because it is algebraic

So take the identity automorphism of K

We will extend this to an automorphism of C that does what I said

Break this again as C/K(√2)/K

K(√2)/K has an automorphism √2 ↦ -√2 because both of these have the same minimal polynomial

(exercise because I don't feel like writing this )

Now we need big lemma of field theory

If Ω is alg closed, F → Ω a field homomorphism, E/F an algebraic extension, then the homomorphism can be extended to E → Ω

Proof: do it for the case where you adjoin one element at a time, and then Zorn

So now we have a homomorphism from C → C which is identity of K and does what I said to √2

We have to show that it's an automorphism

It is injective because all field homomorphisms are

It is surjective because an element of C can only map to another root of its minimal polynomial over K

And a polynomial over K can only have finitely many roots so the homomorphism must permute the set of roots of each polynomial (injective map from a finite set to itself is surjective)

So it must be surjective

That was longer than I expected I hope no one finds a mistake, don't look too carefully

[simple valley]

your mistake is assuming that R exists

[hidden haven]

What's your problem with R now

[simple valley]

big

[hidden haven]

Isn't the construction pretty constructive

you hate big things?

[celest brook]

P(R)

[simple valley]

everything is metatheoretically countable, so powerset axiom is sus

[lone barn]

Okay, so here's what I understood, please check if it's correct. You send every z to itself, except those that are an algebraic number times √2, you send them to -z. Correct?

[hidden haven]

Your mom is a counterexample

[simple valley]

idts

[hidden haven]

We didn't actually do this, the extension from K(√2) to C was very non constructive

[simple valley]

you would have to send sqrt(sqrt(2)) to isqrt(sqrt(2)) as well

[hidden haven]

Or -isqrtsqrt2

[lone barn]

what, why?

[simple valley]

so that it's a field homomorphism

[celest brook]

uncountable number of uncountable sets

[hidden haven]

Their 4th powers must be equal

[simple valley]

for x = sqrt(sqrt(2)) you want f(x^2) = f(x)^2

[hidden haven]

Wait what I said is pointless nvm

[lone barn]

yeah i guess you would want that indeed

[hidden haven]

What did you mean by this btw Are you referring to skolem's paradox?

[simple valley]

something like that

[hidden haven]

I see

[simple valley]

you have term models where the size hierarchy just ends

[hidden haven]

Imagine throwing away the power set axiom because of that though, if anything our theory isn't rich enough

I say we throw away set extensionality

[simple valley]

cantor's diagonal argument talks abut being unable to construct a surjection onto a powerset, and we interpret this as saying that the powerset is "larger"

maybe we're just bad at constructing surjections

[hidden haven]

Lol I wouldn't really interpret skolem's paradox as that

[simple valley]

well, the solution to a paradox is naturally that internally the system has limited tools for introspection

and you may not be able to accurately "feel" the size of a set with surjections

[hidden haven]

Like we have one meta theoretically countable model but we still consider all models of ZFC models of ZFC, and also applying a theorem of ZFC in the metatheory is somewhat uncomfortable

Ye true

[simple valley]

I'm not even really talking about ZFC

this comes up in a variety of logics

[hidden haven]

See I just feel like you're wrong but I don't know enough about this stuff to prove it so wait until my bros ultra and ng come online and sully you

Ye I see

[simple valley]

we can move into #foundations I feel

[shell brook]

So say I have some subset of permutations for S_n (not a subgroup) and I want to show that I can (or can't) generate S_n with them. Is being able to show that the composition of some number of these permutations gives a 2-cycle/transposition enough to show that they generate Sn?

It is pretty easy for me to show that no single element of the subset is a transposition, but I don't think thats enough to claim it doesn't generate Sn

[next obsidian]

I think any n-cycle and a transposition generate S_n

The set of transpositions generate it

Etc

[shell brook]

what

[next obsidian]

Wut

[shell brook]

do you mean any single transposition?

[next obsidian]

You can decompose any cycle into a product of transpositions

And with an n-cycle and a transposition you can construct every possible transposition

I think

[shell brook]

I am not disagreeing with you I just do not understand

[shell brook]

I understand this

[shell brook]

Not sure about this 1

[next obsidian]

You have to be clever

But you can make any arbitrary transposition

[shell brook]

So if the composition of two of my cycles can make any arbitrary transposition, then I can generate Sn?

[next obsidian]

Wel it might not just be of two

[shell brook]

sry to give more context I'm trying to show that riffle shuffles can/cannot generate S_52

[next obsidian]

I think I’m misremembering slightly

(12) and (123…n)

Definitely generate S_n

I think when n is prime you can do some shenanigans to WLOG it to something of this form

Or just like

Actually just renumber things

And you’re fine

I think

[shell brook]

[next obsidian]

No I think you do need prime for this

Anyway

Idk wtf a riffle is

[shell brook]

like

when u split a deck into two

and let the cards fall down

https://www.youtube.com/watch?v=O1i1wihL4c8&t=138s

[next obsidian]

Are you like

Assuming you perfectly split it

[shell brook]

No, arbitrary split

[next obsidian]

And the shuffle is perfect?

[shell brook]

So you split it at p

no so you could have like one element from p, and then all of 52 - p cards, and then the rest of p

[next obsidian]

Okay but if you did do 26-26

Do they have to go like one at a time

[shell brook]

No

[next obsidian]

Perfectly zigzagging

This totally generates S_52

[shell brook]

No so you could just stack them

[next obsidian]

Just show any transposition exists

Just like

[shell brook]

Right but any transposition does not exist as the consequence of a single riffle

as far as I can tell

[next obsidian]

Yeah

[shell brook]

like you cant flip the first and last card

[next obsidian]

So?

You can take as many riffles as you want and chain them

[shell brook]

okay and if that gives a transposition I'm good?

that was my original question

[next obsidian]

Yeah

[shell brook]

okay

awesome

thx chmonkey

[final oasis]

what are exact sequences useful for?

I have seen it used a few times but in general what does it add intern of insight

[sudden jolt]

quick likely very stupid question. Let f \in k[x_1, ..., x_n] and f \notin <x_1, ..., x_n>. I am now supposed to show <x_1, ..., x_n, f> = k[x_1, ..., x_n]. Why is this not a counterexample. k[x] where (x+3) \notin <x> but <x, x+3> != k[x].

[golden pasture]

tbh more of they appear very often and have many nice things you can do with them

[golden pasture]

<> is the ideal generated by the elements id assume? (x+3)-x=3

[sudden jolt]

<> is the ideal generated by the elements yeah

[golden pasture]

3 is a unit so the ideal contains units and there you go

[sudden jolt]

Oh wow I see jesus

You're right

[delicate bloom]

tell him i said hi

[sudden jolt]

Will do

[final oasis]

could you show something cool to do with them

[past temple]

when is a finitely generated module over a pid is noetherian

[golden pasture]

homology/cohomology (derived functors) stuff are pretty cute and useful

[final oasis]

what information do they cary, Is it just to illustrate structure

[golden pasture]

welll pid are noetherian sooo

[golden pasture]

somewhat yea

it's kinda a thing where you get more and more comfortable using it over time

and soon you're using it on a daily basis

[final oasis]

I have been watching some of Richard Borsherds yt videos , he uses them a lot , so I am somewhat comfortable seeing it
I guess it's not part of dna as yet

[lethal cipher]

I'm really struggling on #2

I tried just making a matrix a and b and see if that'll get me there. But it gets really messy, really quick

[final oasis]

what is an inner automorphism

[paper flint]

Inner automorphism induced by an element a is defined by f(x)=axa^(-1) for all x in the group

[lethal cipher]

For some $b\in G$ an inner automorphism induced by b is $\phi_b(x)=bxb^{-1}$.

[cloud walrus]

dackid

[lethal cipher]

Yea, what Manan said.

[paper flint]

2, hmmm

[lethal cipher]

So either, we need to find such a b, or show that such a b does not exist for the function $\phi(A)=(A^{-1})^T$

[cloud walrus]

dackid

[lethal cipher]

So the claim is $\exists B\in SL_2(\R)$ so that $(A^{-1})^T=BAB^{-1}$ for all $A\in SL_2(\R)$.

[cloud walrus]

dackid

[lethal cipher]

wait, I see it

Just let A be B

[kind temple]

yea i was just doing this

[lethal cipher]

Then the claim is $(B^{-1})^T=B$

[cloud walrus]

dackid

[lethal cipher]

So we need the inverse of B to be it's transpose

[paper flint]

Looks like B=I works?

[kind temple]

nah

[lethal cipher]

Only for B though. It falls apart as soon as we choose any other element

[paper flint]

Right

[kind temple]

if you let A = B^-1 you get some other conditions

[lethal cipher]

oh, that B^-1 is B^-1's transpose

[paper flint]

But isn't that a sufficient proof that this can't possibly be inner aut induced by a single element?

If choice of A affects what B happens to be, that is

[kind temple]

and if you let A = B^t

i think just playing around with operations you can do to B and then plugging it back into the function you get more conditions on what B can be

[kind temple]

i dont think so

its like, if you're given one, it has to satisfy certain properties

hopefully they lead to a contradiction

[lethal cipher]

Actually B^T does not give any new information

[paper flint]

Hmmmm

[lethal cipher]

We already have that B is equal to its transpose and inverse by the first two parts

Oh, this is actually huge

So let's say B is the matrix a,b//c,d

Then with all this information, we get that a=d and both c and b must be equal to 0, since b=-b and c=-c

So B must be a scaled identity...but it can't be

Most importantly, if it is not the identity, then it is not in SL_2(R)

[paper flint]

Right

[kind temple]

but the identity doesnt work

[lethal cipher]

And we know for a fact the identity does not work, so this is not an inner automorphism

[kind temple]

sick

[lethal cipher]

Precisely!

[kind temple]

i wonder if you can do this more generally

[paper flint]

Are you using Gallian, Dackid?

[lethal cipher]

Sure am

[paper flint]

I started the Isomorphisms chapter yesterday

Ah I see

[lethal cipher]

Honestly, isomorphisms are very clean to work with

[paper flint]

Yes 😌

I liked the chapter

The previous one was "le count"

[lethal cipher]

"Poof, you've proven the thing"

[paper flint]

😌

[lethal cipher]

Permutation groups were not too bad, but they definitely had some tricky moments

[paper flint]

I just find them very tedious at times

[lethal cipher]

The book overkills it a bit.
Since every permutation can be written as a product of disjoint cycles or a product of transpositions, you usually just need to prove a certain property for one of those.

[paper flint]

Right

[dire shuttle]

As someone who just spent the past 5 hours studying for an exam over permutation groups, I can say they are incredibly tedious.

[paper flint]

Indeed

Also, goodluck for your exam!

[lethal cipher]

I also just realized that we didn't need to show B=B^T at all. It does not enforce the requirements on B

B=B^-1 is actually all we needed

[dire shuttle]

Thanks! Luckily this professor lets us see the set of proofs we will have to potentially do beforehand, so its not too bad.

[lethal cipher]

Not gonna lie, it feels pretty good to finish hw the day I got it

This almost never happens in proof classes xp

I must admit that I am a little hung up on 4 though.
I am pretty sure that it is only the identity and -x, but I am not sure if there are any others.

[kind temple]

can you show the same result for SL_n? i cant seem to get that B = c I

[lethal cipher]

I am not so sure. It also sounds like a hell of a mess

[kind temple]

yea... wouldnt resort to going into entries of B, but something about all the conditions should enforce that B = c I

[lethal cipher]

I'm not sure if I have enough knowledge of LA to say

[lethal cipher]

I'm working on #4 now.

So I know f(x)=x and f(x)=-x works. But I am not sure if those are the only ones

Unlike Z, 1 and -1 do not generate Q

[lethal cipher]

This was a key component for finding the automorphism group of Z

[delicate bloom]

what about f(x)=kx for k in Q^*

[lethal cipher]

No this is not. Since f(xy)=kxy, but f(x)f(y)=k^2xy

So it is not a homomorphism (so it is not an isomorphism)

[median pawn]

I understand the first part of the proof, need help with the second

[cloud walrus]

Hausdorff

[proud bear]

but Q is an additive group right?

[lethal cipher]

Whoops

Yea, those will work too in Q

[rustic crown]

That's just the universal property of quotients! you have a map from F(X) --> H and the normal subgroup <<R>> dies under this map. So this map factors through F(X)/<<R>>

[median pawn]

Hmm

[rustic crown]

okie

[cloud walrus]

Hausdorff

[median pawn]

q is the canonical quotient from F(X) to F(X)/<<R>>

[rustic crown]

yep

[cloud walrus]

Hausdorff

[median pawn]

How does this show that varphi is a homomorphism

[rustic crown]

okie... what the universal property says is that there is a unique such varphi... you can easily check that its a group hom

[median pawn]

I don't think that I have studied that universal property before lol

Should I directly just check that varphi is a group homomorphism?

[rustic crown]

$\varphi(q(w)) = \varphi(w + \langle\langle R\rangle\rangle) = \phi(w)$

[cloud walrus]

det

[rustic crown]

you know the image of every coset of w, if such a map existed

[median pawn]

that's correct

[lethal cipher]

Hmm, so here is the thing. We do have automorphisms. But I need to find a way to verify this is all the automorphisms

[rustic crown]

yep, but universal property of quotients is like one of the first things you do... it sometimes goes by the name of (generalized) first isomorphism theorem

prove it separately maybe... it's a nice fact to know

[lethal cipher]

I am having a hard time figuring out some invariance to show we have exhausted all of the automorphisms

[cloud walrus]

Hausdorff

[median pawn]

Does this show that it's a group homomorphism @rustic crown

[rustic crown]

oh oops my bad i shouldn't have denoted the operation by +

• feels too commutative lol

[rustic crown]

but yea that's the argument

[median pawn]

Okay replace + everywhere by \cdot

Then?

[rustic crown]

yep it's good

[median pawn]

lol i still don't see what the author was up to

[cloud walrus]

det

[median pawn]

he said proving phi(w) = e for all w in <<R>> suffices? but how?

[rustic crown]

yep, cause all the other work you do is just verifying the universal property again

[median pawn]

noo i don't see it hmm

[rustic crown]

maybe try proving this, and notice that what all work you did after checking that <<R>> dies is pretty general

[median pawn]

ok cool

one more thing

[median pawn]

what is the operation on cosets? + or \cdot

like do i write

w1 <<R>> w_2 <<R>> = w_1 w_2 <<R>>?

[rustic crown]

for abelian groups people use + and for nonabelian or when they really want to think multiplicatively, they use \cdot

[median pawn]

cool, so it really doesn't matter what notation i use

the group operation is understood from the context

[rustic crown]

yep

no one says let (G, m, e, i) be a group where m : G x G --> G, e : 1 --> G and i : G --> G

[median pawn]

yes lol

thank you det!

oh wait there is a small prob

[rustic crown]

also one more thing... what you did here is like showing that <<R>> dies is a necesseity

[median pawn]

in this proof where did we use the assumption that phi(r) = e for all r in R?

[rustic crown]

that's required to show the map is well defined

you can't directly define the map varphi(gN) = phi(g) without checking what happens to a different representative

say aN = bN then we know a = b * n meaning phi(a) = phi(b) * e = phi(b)

[median pawn]

agreed, makes sense

[lethal cipher]

Let me know when y'all finish.
No rush, I just don't want to interrupt

[rustic crown]

universal properties are nice, they let you construct maps to and from objects without a lot of work... you just avoid all the repeated work that we need to do otherwise... another nice thing is that as the structure gets more and more complicated checking it's properties more than once is just a pain lol

[proud bear]

for q_1,q_2\in Q, we have that phi(q_1)=k_1q_1 and phi(q_2)=k_2q_2 for some k_1,k_2. try to show that k_1=k_2

[lethal cipher]

Well for starters: f(q_1+q_2)=k(q_1+q_2) for some k.
Since we have an automorphism, k(q_1+kq_2)=k_1q_1+k_2q_2.
So k=k_1 and k=k_2

[proud bear]

yeah

[lethal cipher]

And that does it actually

[proud bear]

it should i think

[lethal cipher]

That helped a lot. Thank you Stain

[lethal cipher]

So I made a pretty fatal mistake here. Phi(B) does NOT imply B=B^-1 it implies B^-1=B^T

So this isn't quite the finishing argument

I may need some help trying to fix this argument. Unfortunately, this breaks everything I did

[thorn delta]

do u have a question

[lethal cipher]

Unless Q is something other than rationals, Q is an Abelian group. It is not an example of this

[thorn delta]

Q means quaternion group usually

[lethal cipher]

Oh. I have absolutely no idea then :p

[proud bear]

dackid, have you made progress on your problem

[lethal cipher]

No, not at all

I don't quite see how to fix this dilemma

[proud bear]

so you have $B\inv=B^T$. let $B=\begin{bmatrix}a&b\c&d\end{bmatrix}$. see what relations you get for $a,b,c,d$

[lethal cipher]

Yea. You get a=d and b=-c

Hold on though it seems I am on to something

So if you plug B^TB into phi, you get that B^-1B^-1^T=I_2

So we get B^-1=B^-1^T, and this does actually get right back where we wanted

[proud bear]

?

BB^-1B^-1^T=B -> B^-1^T=B -> B^-1=B^T

[proud bear]

anyway, compute BSB\inv for some shear matrix S

[lethal cipher]

Shear matrix?

[proud bear]

$S=\begin{bmatrix}1&s\0&1\end{bmatrix}$

[proud bear]

and compare what you get to S^-1^T

[lethal cipher]

Damn no it doesnt. I got so excited too

[delicate bloom]

maybe use phi(phi(A))=A if you haven't yet

[lethal cipher]

Oh, well that one is a dead giveaway 🤦‍♂️

Pretty sure that would mean B has to be the identity

[delicate bloom]

-I would work too

[lethal cipher]

True. Are there others?

[delicate bloom]

but it's not the automorphism (A^-1)^T so that's no good

well we know it's an orthogonal matrix so I think that basically forces it to be I or -I

[lethal cipher]

Remind me what orthogonal means

[delicate bloom]

B^-1 = B^T

[barren sierra]

idk what Q is so idk about that part

but I cannot shw that there is some a

[delicate bloom]

like a rotation or reflection matrix

permutations as well

[lethal cipher]

Ah, okay.

[delicate bloom]

actually there are more I think

[lethal cipher]

That have determinant 1?

[delicate bloom]

yeah

1/sqrt(2)[1,1; 1,-1]

or actually

that's a rotation nevermind

just 45 degrees

I guess we could try to prove that the only 2x2 orthogonal matrices with det 1 are rotation matrices

[lethal cipher]

Definitely not within the scope of this class/ what they'd expect from us

We are missing something

[delicate bloom]

seems like a pretty fundamental fact, the kind of thing that's not hard to prove

[terse crystal]

B=

(0 1

[delicate bloom]

rotation matrix in R^2 just leaves lengths unchanged, so we can take an arbitrary vector and show that |Rv|=|v| and that would prove it

[terse crystal]

-1 0)

[delicate bloom]

$$|Rv|^2 = (Rv)^T (Rv)= v^TR^TRv = v^T R^{-1}Rv = v^Tv = |v|^2$$

[cloud walrus]

Meroseous

[delicate bloom]

ok done

Maybe you could take the route of just working out the algebra with A^2=I for a 2x2 matrix

[lethal cipher]

No I tried. It is not kind

[delicate bloom]

pretty sure I've worked that out before by hand in the past

I don't think it's that bad

like you can use tricks to simplify it

[terse crystal]

Are you still discussing that inner automorphism?

[delicate bloom]

a^2+cb=1, ab+db=0, etc they're not bad at all

[lethal cipher]

Yes

[delicate bloom]

just factor a bit

[terse crystal]

B=(0 1 ; -1 0) doesn’t work?

[proud bear]

i think it does

[delicate bloom]

doesn't square to the identity

doesn't work

phi(A)=(A^-1)^T so if you apply it twice you get the identity

maybe I'm wrong

as matrices you get two things squaring to -I so together they make I

yeah good point

[proud bear]

,w {{0,1},{-1,0}}{{a,b},{c,d}}{{0,-1},{1,0}}

,w ({{a,b},{c,d}}^-1)^T

[cloud walrus]

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[terse crystal]

Who told you it’s square must be identity, why not -I

[delicate bloom]

yeah already convinced myself

[delicate bloom]

from here down I explain why I'm wrong and you're right

[lethal cipher]

Well. Ngl, I am not sure I would have found that example myself

[terse crystal]

Oh I see

[delicate bloom]

well we're looking at it as $B^{-1}(B^{-1}AB)B = B^{-2} A B^2$ so if $B^2=-I$ we have $-IA(-I) = A$

[cloud walrus]

Meroseous

[delicate bloom]

that's cool, that shows it's an inner automorphism and you're done lol

proof by example, my favorite kind of proof

[lethal cipher]

Yea, that is certainly one way to do it :p

But still, never woulda found that on my own

[delicate bloom]

I didn't find it cause I wasn't careful about that last part there, oh well, can't be perfect lol

[lethal cipher]

Yep. Thanks for the help guys. I appreciate it

[thorn delta]

anyone have a hint on showing that A = B^G? my field theory is rusty

[waxen hedge]

Notice that B^G is included in L^G !

[thorn delta]

ah okay so $B^G \subset L^G = K$, so if $x \in B^G$, then $x \in K$. But $x$ is integral over $A$, which is integrally closed so $x \in A$

[cloud walrus]

kxrider

[chilly ocean]

what makes a field "algebraically closed"?

[thorn delta]

every poly in it splits completely

[chilly ocean]

splits?

[kind temple]

every polynomial has a root

[chilly ocean]

a root within that field right?

[kind temple]

ye

[chilly ocean]

ah okay

so C is algebraically closed?

[kind temple]

thats what the fundamental theorem of algebra says, yea

[robust pollen]

Me again, with the dumb question. But I'm always really dang confused when things seem obvious, because in my mind, I'm always wrong.
Anyway, enough therapy. I'm asked to show $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \times \mathbb{C}$ as $\mathbb{R}$-algebras.
However, in my mind there is absolutely nothing to show, therefore there must be a catch, right? I mean, the map $w \otimes z \mapsto (w, z)$ does the trick, no?

[cloud walrus]

expectTheUnexpected

[waxen hedge]

Yes exactly !

[thorn delta]

nice, thanks for the hint C:

[median pawn]

I've a basic question

Suppose we have a group $G = \langle X|R\rangle$. By definition, $G \cong F(X)/\langle\langle R \rangle\rangle$. Suppose $H$ is another group. If $\phi: G\to H$ is a map, why does it suffice to check $\phi(r) = e$ for every $r\in R$, for the map $\phi$ to be well-defined?

[cloud walrus]

Hausdorff

[waxen hedge]

Is your map well defined ?

[robust pollen]

wtf wait. It sends e.g. x \otimes 1 to (x, 1) for x real, but since x \otimes 1 = 1 \otimes x, it also sends it to (1, x) \neq (x, 1). So I guess it's not well-defined. Huh. Now I'm even more confused

[thorn delta]

You just want the map to be defined independently of the representatives you choose from G. If g = h are elements of G (with g, h being representatives in F(X)), then gh^-1 is in <R>, and therefore g = rh for some r in <R>, and the idea is that if phi behaves like a homomorphism, then phi(g) = phi(rh) = phi(r)phi(h) = phi(h)

but your question is worded kind of imprecisely, because R is not contained in G. Rather <R> is the identity in G

[waxen hedge]

What's a map from a tensor product ? What is ir equivalent to ?

[robust pollen]

a linear map from the tensor product is the same as a multilinear map from the product of the tensorands (we are in Vect)

[rustic crown]

right... and what's a map to a product of spaces? can we describe it in a different way?

[median pawn]

Makes sense! So phi(r) = e for all r in R is indeed sufficient

I guess it's necessary and sufficient in fact

for phi to be well defined

[median pawn]

The stuff I'm looking at (pushouts of groups) is using the canonical presentation so far

G = <G| R(G)> where R(G) = kernel of the canonical homomorphism F(G) → G

So the set of relations, R(G) is a normal subgroup of F(G)

[thorn delta]

i don't think so. Its definitely necessary, but phi needs to "behave like a homomorphism," but since phi might not well defined, you cant just say phi is a homomorphism.

[waxen hedge]

Yes exactly
So in your map above, is the map
$\mathbb{C}\times\mathbb{C}\rightarrow\mathbb{C}\times\mathbb{C}, (z,w)\mapsto(z,w)$, $\mathbb{R}}$-bilinear ?

[cloud walrus]

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

[robust pollen]

Hom(A, BxC) = Hom(A,B) x Hom(A,C), so we'd get something like Hom(C \otimes C, C x C) = Bilin(CxC, C)^2, right?

[rustic crown]

Yep... we need to find a nice pair of billinear maps from CxC --> C

[robust pollen]

So, (w,z) \mapsto wz is R-bilinear, as is (w,z) \mapsto (\bar{w}, z), etc..

[waxen hedge]

If I remember well you have to use that : $\mathbb{C}$ is isomorphic as $\mathbb{R}$-algebra to $\mathbb{R}[X]/X^2+1$

[cloud walrus]

Adrien

[waxen hedge]

At least that's one way to do it

But you need to know how the tensor product interact with quotient, algebra of polynomials, etc

[rustic crown]

right. so we have those two maps... and we only need to check now is that this forms in isomorphism.

[ocean raven]

tried all possible ways but could not find how to prove

[rustic crown]

one way i can think is by directly checking that images of the 4 things are linearly independent

[robust pollen]

I don't understand this right now 😦

[south patrol]

I'd try a 'bezout's lemma'-style argument, if you know that, @ lost_boy

[ocean raven]

never heard of it

[robust pollen]

So, but the map w \otimes z \mapsto (wz, \bar{wz}) seems now to be a morphism of R-algebras. And since it's injective (this is clear because C has no zero divisors), we must have that it is onto as well (since both spaces have the same dimension)

[rustic crown]

So under the map we get, we'll have
\begin{align*}
1\otimes 1 &\mapsto (1, 1)\
1\otimes i &\mapsto (i, i)\
i\otimes 1 &\mapsto (i, -i)\
i\otimes i &\mapsto (-1, 1)\
\end{align*}

[south patrol]

Oh fair, well essentially try to get 1 as a sum of multiples of 5 and 12

Then you'll be able to write a in terms of a^12 and a^5

[cloud walrus]

det

[ocean raven]

oh

[rustic crown]

i don't think if you take bar{wz} then it will work... because the elementary tensors will get mapped linearly dependent things

[robust pollen]

I see, but I don't 🙈 . Oh, I'm big dumb. The map I wrote down is not injective, since of course wz=0 does not imply w=z=0 lmao

[rustic crown]

wait no that's fine

it implies that one of w or z is 0

so the elementary tensor w otimes z = 0

[ocean raven]

thanks for the help

[robust pollen]

wait yeah but it can't be injective. wut.

[rustic crown]

it sends a basis to a basis. so it's definitely an isomorphism

only thing left is to see why it's a map of algebras and not just vector spaces

i'm a little rusty... but i think that also follows because tensor product is a coproduct in R-alg

[waxen hedge]

Yeah I'm sorry, I don't know what will be covered in your course but you'll probably be able to understand it later

[robust pollen]

Actually, I'm a bit more advanced, but my head does not work at all today. I have no idea why. I mean, after thinking for more than a second, by "interact with quotients" you probably mean "exactness of tensor product (over fields)" and by "interact with polynomial algebra" something like "L \ot_R A/I \cong (L \ot_R A)/(L \ot_R I)", for any pair of R-algebra A, L, and two-sided ideal I, right?

[waxen hedge]

Yes this one

[robust pollen]

which in our case would give C \ot_R C = R[x]/(x^2+1) \ot_R C = (C \ot_R R[X])/(C \ot (x^2_+1)) = C[x]/(x^2+1) = C[x]/ (x+i)(x-i), and then?

[robust pollen]

For any commutative ring? Huh

[rustic crown]

yep

[rustic crown]

from here you can use chinese remainder theorem

[robust pollen]

I forgot how that goes 😄

[robust pollen]

Also true for the category of algebra objects in a monoidal category?

[rustic crown]

not sure... i've heard those words but can't say

[robust pollen]

But how does the computation above actually give that it's an iso of algebras?

[rustic crown]

about chinese remainder theorem... say you have 2 ideals I and J such that I + J = (1) then we'll get a map R --> R/I x R/J. It's kernel is I intersect J and in the case when I and J are coprime, this equals the product ideal IJ. This gives us an isomorphism R/IJ --> R/I x R/J

[robust pollen]

it's also wrong, because the category needs to be braided in order for the tensor product of algebras to be defined, I guess.

[robust pollen]

Oh, cool. Thanks

[rustic crown]

so C[x]/((x+i)(x-i)) = C[x]/(x+i) * C[x]/(x-i) = C * C

[waxen hedge]

That's the proof, sorry I'm in the subway I can't answer really fast haha

And latex on mobile is a pain

[robust pollen]

No worries 😄 Thanks guys 🙂 Now I just have to see why the iso is automatically of algebras

[vestal snow]

Is the identity point on an elliptic curve always of degree one?

[rustic crown]

for this map, i sent w tensor z to (wz, bar{w}z)

[waxen hedge]

I went screenshotting some of my course's notes in order in answer x) (in French unfortunately)

[robust pollen]

c'est ne pas un probleme, j'ai essaye de lire SGA

many thanks again, oof

[median pawn]

hey, i have a question about pushouts once again

[cloud walrus]

Hausdorff

[median pawn]

This is what Lemma V.11 says

Also for reference, the definition

[cloud walrus]

Hausdorff

Hausdorff

Hausdorff

[chilly ocean]

Hello. I have a quick question. Im currently learning a little bit algebraic geometry, especially algebraic groups, but there are a few things i dont quite understand. I thought that GL2 would act on K² by the normal matrix-vector multiplication, so $A \cdot x = Ax$, so its the automorphism group of a 2d vector space $V$. THen, PGL2 would act on P² by $[A] \cdot [x] = [Ax]$, but i dont understand how the center of GL2 wouldt act on Sym²V? In particular, i dont really understand how Sym²V looks like. I read something online about tensorproducts, and that Sym²V is spanned by elements ${ x^2, xy, y^2}$, but i dont know how to connect that to GL2. How exactly does this action look like?

[cloud walrus]

Gewisser Fler

[chilly ocean]

This is from the book "Algebraic Geometry" by J. Harris, page 117

[delicate bloom]

lol quick question --wall of text--

[hot lake]

Then, PGL2 would act on P² by [A] \cdot [x] = [Ax] that would be an action on P1

first, GL2 acts on K² by A.x = Ax

then it acts on the tensor product K² * K² by A.(x * y) = Ax * Ay

the subspace of symmetric products is preserved by that so it induces an action on Sym²(K²)

oh yeah you can just see Sym²(K²) as Vect(x²,xy,y²)

say K² = Vect(x,y)

and say a matrix A = (a b // c d) acts on it by A.x = ax+cy and A.y = bx+dy

then in Sym²(K²), A.x² = (A.x)² = (ax+cy)² = a²x²+2acxy+c²y²

for example

and A.xy = (A.x) (A.y) = (ax+cy)(bx+dy) = abx²+(ad+bc)xy+cdy²

and similarly for A.y²

and that describes how GL2 acts on Sym²(K²)

[chilly ocean]

Thank you very much

[hidden haven]

@median pawn The map G_1 to H is defined by looking at the presentations <X_1' | R_1'> and the presentation of H. You just map each x in X_1' to itself in H

[barren sierra]

woke up and got coffee, still can't solve this

Q = {+-1, +-i, +-j, +-k} in the quaternions

[delicate bloom]

just the second part of confirming it?

[barren sierra]

no

the first part

finding such an a

given a group described as such

is this not just definition?

[delicate bloom]

nah

you have xy=0 but how do you know yx also is 0?

[barren sierra]

oh

ok I'll work that out in a bit

but like that first question

how does non-abelian and (xy)^2 = (yx)^2 => there is a =/= 1 such that a^2 = 1

very lost

[delicate bloom]

what have you tried

[barren sierra]

no amount of symbolic manipulation has helped

[robust pollen]

Show first that for all x, y you must have x^2 y = y x^2. From this you can conclude that x^-1 y x = x y x^-1 holds.
Use all of this to show that (x y x^-1 y^-1)^2 = 1

[barren sierra]

x^2y = yx^2

hm

ok

[robust pollen]

I didn't come up with this myself lmao. Here: http://www.qbyte.org/puzzles/p129s.html

[barren sierra]

thanks

[delicate bloom]

lame

cheating takes the fun out of it

[chilly ocean]

all rings are commutative so this is immediate from the defn of zerodivisor qed

[warm holly]

if we have equivalence relation S included in equivalence relation R, does the fact that partition X from which S is induced is included in partition from which R is induced comes from definition or I need to prove it? It seems really trivial but I don’t know if I can use it as the fact