groups-rings-fields

[arctic trail]

by multiplication by integers corresponding to repeated additions / subtractions of elements of the ideal

[rain grove]

Yeah ok fkk maybe im too fresh for this still, idk about quotients, ideals, even homomrphisms...

[arctic trail]

yeah so maybe don't read this right now a.a

Wedderburn's little theorem is not a simple topic to approach

You should probably revisit it after reading up on abstract algebra

[rain grove]

Ok ty anyways, ill def try again when ill know more

[arctic trail]

Since I was already midway writing:
The ideals of Z are always generated by a single element, this is because of Bezout's theorem
(suppose p and q are in a subgroup H, by Bezout's theorem we can write their gcd as a linear combination of them, since subgroups are ideals their gcd will therefore be in H. Therefore every element in H is a multiple of the minimum natural number in H, and so is generated by that element and is of the form nZ)

[long obsidian]

Can someone confirm for me if the maximal ideal of k[x] is just <x>

[void cosmos]

wdym "the"

if k is a field, then there is no "the" maximal idea

ideal

there are many maximal ideals

in k[x]

but if ur asking if <x> is maximal in k[x] then yeah it is

[arctic trail]

?

[arctic trail]

it is maximal but not unique

[chilly ocean]

If k is a field, the maximal ideals of k[x] are those of form <p> where p is irreducible (so for example <x> and <x+1> are always two different maximal ideals)

[coral spindle]

Did you perhaps mean k[[x]]? In which case <x> is indeed the maximal ideal. But there are many other maximal ideals of k[x]

[arctic trail]

shit I've been working with k[[x]] too much

lmfao

mb

[arctic trail]

since if k is compact k[x] is not compact but k[[x]] is

[long obsidian]

What's another example maximal ideal if this isn't unique?

Ahh okay I see thank you

[chilly ocean]

<x+1>

[coral spindle]

Bequi gave an example

[chilly ocean]

And for example if k=R, then <x^2+1> is another

[arctic trail]

perhaps the most important

maximal ideal

[chilly ocean]

compact topological field?

[arctic trail]

yeah, that is, finite 😎

[chilly ocean]

What topologies do you endow k[x] and k[[x]] with?

[arctic trail]

on k[[x]] the adic topology

given from the topology on k[x]

[chilly ocean]

I've heard of it

What about on k[x]?

[coral spindle]

The x-adic topology lol

[arctic trail]

I forgot, but I think it's just the one given from treating k[x] as a vector space over k

let me check

[chilly ocean]

Vector spaces don't have topology

[arctic trail]

no wait I'm stupid

it's the x-adic topology

sorry I had a brain fart

[chilly ocean]

What is the x-adic topology on k[x]

[arctic trail]

<x> is open

[chilly ocean]

I thought adic topologies came from completions

[arctic trail]

<x^2> is open

etc

[coral spindle]

Makes sense. k[x] just gets the subspace topology’s

[arctic trail]

• the ring operations are deemed continuous

[coral spindle]

Fucking autocorrect

[chilly ocean]

This seems like an unfamiliar topology to me

[arctic trail]

It's the topology generated by the elements of the form p+ <x^n>

that is translated powers of <x>

[arctic trail]

it's the topology I like a lot cause, for the specific types of characters I like they're continuous in that topology

I should mention that instead of something like k[[x]] I usually have xk[[x]], which is a non-unital ring

here the topology is much more intuitive imo

[chilly ocean]

Sounds interesting

[arctic trail]

ahh one thing that is not entirely obvious for example is that the ideals of k[[x]] are of the form x^nk[[x]]

[chilly ocean]

Local ring

[feral timber]

I need someone to verify my solution please

Let R be a commutative ring with 1 not equal to 0. If a,b are two comaximal elements of R, then prove that a^m and b^n are comaximal for all natural m,n.

My solution:

[arctic trail]

You lost me at comaximal elements a.a

@feral timber what does comaximal mean?

[feral timber]

There exist some x and y in R such that xa+yb=1

[arctic trail]

so comaximal = coprime?

[feral timber]

Yes

[arctic trail]

ah bet

[feral timber]

Essentially I first fixed n=1 and applied induction on m, then fixed n and did the same to m

[arctic trail]

There exists a simpler proof

in the language of ideals:
aR + bR = R (meaning of coprime), hence
a^2R + abR = aR
and
b^2 + abR = bR
therefore a^2R + abR + b^2R = R
abR = ab(aR + bR) = a^2bR + b^2aR which is contained in a^2R + b^2R. so a^2R + b^2R = R

hence a,b coprime => a^2,b^2 coprime.
So a^(2n),b^(2n) coprime by induction (a^2,b^2 are in the hypothesis of the implication I wrote)
it is also easy to see that if a^k,b^k are coprime then a^t,b^t are coprime for t < k.
This ends the sketch of the proof

[feral timber]

But how does this show that a^m and b^n are coprimes for m not equal to n?

If m<n then you can take coefficient of a^m to be xa^n-m, would that work?

[arctic trail]

wlog m < n.
b^nR is contained in b^mR since we know a^nR + b^nR = R then a^nR + b^mR = R

[feral timber]

since a^n and b^n would be coprime

[feral timber]

Yes, this

[arctic trail]

that is to say

you don't need to leave ideals ever in the proof

[feral timber]

Yes, this is pretty good!

[feral timber]

Actually there was another question which said that if I and J are coprime ideals, then I^m and J^n would also be coprime for all natural m,n

I was thinking of using this to solve the ideal one

[arctic trail]

this is the same proof

there is no change

substitute aR with I and bR with J

all of the things I wrote are still valid, since ideal multiplication will still be commutative and associative

[feral timber]

I+J=R would mean a+b=1 for some a and b in I,J. Thus a^m and b^n would also be coprime, giving I^m and J^n to be coprime

[arctic trail]

errrrr

you don't have to write all that

Although

that proof you wrote essentially means "coprime ideals contain coprime principal ideals", and passing it to principal ideals

but it's not necessary

What I mean is that the general proof gives more insight as the reason why everything happens. Coprime ideals containing coprime principal ideals hides this a bit

[feral timber]

Ah, okay

[arctic trail]

@feral timber alternatively you can apply induction that (I+J)^n = I^n + J^n = R

[wooden fulcrum]

Hello. In Aluffi's Algebra: Chapter 0 it is said that Z[x_1, ..., x_n] is free over n-element sets A for functions j:A->R into commutative rings. Then in a footnote it says that it's enough for j(a) to commute with all r in R for all a in A. Would it not be enough for all j(a) to commute with each other?

[south patrol]

I'm not sure what you mean by "it's enough". Enough for what?

[knotty badger]

I think you’re right

[rain grove]

If im looking at homomorphism of groups Z_n -> Z then by definition f(0) = 0. But if im looking at homomorphism of rings Z_n -> Z then by definition f(1) = 1?

and it follows from the definition that f(0) = 0

[chilly ocean]

Group homomorphisms require f(0)=0 and ring homomorphisms require f(0)=0 and f(1)=1 yes

Although, in both cases, f(0)=0 follows from just f(a+b)=f(a)+f(b)

(And from + having inverses)

[coral spindle]

N.b. that f(1)=1 for ring homomorphisms is a requirement and does not follow from f(a+b)=f(a)+f(b) or f(ab)=f(a)f(b).

Some authors will not require this, for example if they don’t require rings to have units.

[rain grove]

Yeah even if the ring is just {0} then 0 and 1 are the same element so it still applies

[chilly ocean]

Something to note about the 0 ring is that, if you don't require f(1)=1, then there is a homomorphism from it to any other ring. But if you require f(1)=1 then there isn't any homomorphism from it into any other ring except itself

[rain grove]

huh cool... with requiring f(1) = 1, you get locked into f(0) = 1 with 0 ring

and that is impossible cause then the other homomorphism proprety don't hold

[south patrol]

0 = 1

Lol

[chilly ocean]

Yes, f(0)=0 forces f(1)=0 (because 0=1) but in a nonzero ring 0 is not equal to 1 so f(1) won't be equal to 1

[south patrol]

I guess this depends on what you meant by "with 0 ring"

[chilly ocean]

Z/1Z

[south patrol]

No I mean you didn't say if that was the target or source

[chilly ocean]

I mean 0 ring as domain

[rain grove]

{0} where 0 is both additive and multiplicative identity no

[chilly ocean]

Every ring has an unique map into {0}

[rain grove]

so you can't have f: {0} -> K if you require f(1) = 1

[chilly ocean]

But unless you don't require f(1)=1, only {0} hs a map from {0}

[rain grove]

okk

[south patrol]

Corollary: rings do not form an abelian category

[chilly ocean]

They also won't form one if you don't require f(1)=1

[south patrol]

Yeah

[chilly ocean]

Because there won't be biproducts

[chilly ocean]

They do form a semi-abelian category though, if you don't require f(1)=1

[arctic trail]

wdym?

[chilly ocean]

There is a nonunital ring homomorphism from {0} into any other ring

[arctic trail]

Yeah

Ohhhh

[chilly ocean]

There is no unital ring homomorphism from {0} into any nonzero ring

[arctic trail]

I see what you mean

[chilly ocean]

It is very easy to prove this. You could say it is a Trivial Lemma

[arctic trail]

Indeed

[hidden wind]

i should be learning some prereqs for these new classes i just signed up for but i'm instead stuck reading klein's lectures on the icosahedron for fun fun fun

tfw the klein 4 group first appears here it is as the 2-gonal dihedral rotation group

(explicitly embedded on a sphere)

as in burnside, it is clear the groups here are always thought as a collection of operations on some object, that is, as a what we now consider group actions, instead of just the operations in themselves, what we now consider the actual group

always interesting to see how things have differed historically mwahaha

there is something to this i suppose, though perhaps exaggerated

[hollow fjord]

Hi!
I need some help. I need to show that for every polynomial p in k[X_1, ..., X_n] the ring k[X_1, ..., X_n][1/p] is not a field (k is a field). I've tried to prove that p+1 can't have an inverse (I didn't succeed).

[chilly ocean]

p+1 can have an inverse when deg(p)=0, but in that case it is trivial that your thing is not a field

[rocky cloak]

What tools do you have at your disposal? Like if you can use that the polynomial ring is a UFD, then it's not too hard.

There's also some other more high level machinery like Zariski's lemma, that would give a short answer

[wooden fulcrum]

interesting, when we were introduced to groups I don't think actions were ever even mentioned "formally", other than the "rotations of triangle" example

[wooden fulcrum]

Enough for j to factor uniquely through Z[x_1, ..., x_n]

[dull ginkgo]

For a commutative ring R, Rx^-1 ~= R[X]/(Xx - 1). Thus your ring is iso to R[X_1… X_n, Y]/(pY - 1).

[rocky cloak]

Or Krulls principle ideal theorem, assuming you know a little about prime ideals in the polynomial ring

[dull ginkgo]

For this quotient to be a field the ideal (p(…)Y - 1) needs to be maximal

[wooden fulcrum]

I just checked the errata and there it was changed to the commuting with each other version, should've looked before asking xD

[grave sedge]

Isnt that usually a step in the proof of zariski?

But yeah using that it's a UFD (with infinitely many distinct primes!) is the natural thing to do

[hollow fjord]

thank you!

[rain grove]

So Z is a monoid and f: Z -> Z is only an endomorphism of monoids if f maps unit to unit. For example f (a) = a+1 is not an endomorphism of monoids even though Z is a monoid and f is an endomorphism

[chilly ocean]

Yes, f is a set endomorphism of Z, but not a monoid endomorphism of Z

You also need f(a+b) = f(a) + f(b) for it to be a monoid (endo)morphism

[rain grove]

is there a endomoprhism f: Z -> Z that is not f(a+b) = f(a) + f(b)?

[chilly ocean]

What do you mean by endomorphism?

[rain grove]

oh yeah sorry, then just a map

[chilly ocean]

Have you tried to find one?

[rain grove]

Oh yeah ofc you can just assign some random elements and its done

[chilly ocean]

The easiest way would be to assign f(0) to something nonzero because then f(0+0) would not be equal to f(0)+f(0)

[rain grove]

G group and a in G. Show there is a group homomorphism f: Z_n -> G that f(1) = a iff a^n = 1

So Z_n is additive group and G is multiplicative group?

I showed => and im trying to do <=, But if I just blindly go for f(a^n) = f(1) im not sure what f(a^n) is. a^n has to be element of Z_n for that to work but looking at the additive group then there is no multiplication so a^n is undefined or smth

[quiet pelican]

What do you mean by this?

[rain grove]

(Z_n,+) = {0,1,2,..n-1} and (G,*) is any group

and I have f: (Z_n,+) -> (G,*). Which means by definition f(n+m) = f(n)f(m) and also it holds that f(0) = 1, where 1 is identity of G and 0 is identity of Z_n

right?

[quiet pelican]

Yes

I'm just very confused on what f(a^n) = f(1) is supposed to mean, or how that lhs is even defined

[rain grove]

Yeah well me too, idk what that is supposed to be and idk if I can even do that

[quiet pelican]

a^n = 1 is a statement purely in the group G

we're using 1 to denote the identity of G

(which does mean we're using 1 for both the generator of Z/nZ and the identity of G)

[rain grove]

So to prove <= i just say: I define f(1) = a and there you go that is a homomorphism

because a^n = 1 it follows that f(n) = f(0) = a^n = 1

[quiet pelican]

Yes

[rain grove]

okk ty

[forest ledge]

hi

for inforamtics olimpics

[jaunty magnet]

<@&268886789983436800> is this allowed?

[tough onyx]

I've already told you not to post this. Come back when you can behave.

[jaunty magnet]

ty Hayley

[tardy hedge]

What was posted?

[brazen slate]

don't worry about it

[misty cosmos]

Can we find a group G and a subgroup H ≤ G that is not
normal, but (aH)(bH) = abH for all a, b ∈ G?

[coral spindle]

Set b=a^-1 and think about what the equation you’ve written becomes

[arctic trail]

nope

define pi: G -> G/H (set of cosets) taking a to aH.
By what you wrote G/H becomes a group, hence H is normal by H = ker pi which must be normal

[rain grove]

(1) Let $K$ be ring Show that for each $a \in K$ there exists a ring homomorphism $\phi : \mathbb{Z}[X] \rightarrow K$ such that $\phi(X) = a$.

(2) Let $A$ be an algebra over the field $F$. Show that for each $a \in A$ there exists a homomorphism of algebras $\phi : F[X] \rightarrow A$ such that $\phi(X) = a$.

[cloud walrus]

OHHELLNAH

[rain grove]

Whats the difference? Can I do this for both?

[coral spindle]

Every ring is a “Z-algebra” once you adjust the definition of an algebra to include non-fields

So yes it is exactly the same

[arctic trail]

What are examples of topological groups which are isomorphic to their Pontryagin dual?

[rotund aurora]

the additive group of a local field

[arctic trail]

thanks!

[arctic trail]

this is just a group homomorphism right?

not necessarily a homeomorphism

[rotund aurora]

isomorphic as topological groups

[arctic trail]

I'm interested in any isomorphism but that also works and is perhaps also better suited

[rotund aurora]

btw, finite abelian groups also work

but this is not as interesting imo

[arctic trail]

The Fourier transforms for locally compact Hausdorff groups will be an isomorphism though right?

Even if non-canonical / non-continuous?

@rotund aurora

Identifying \hat \hat G with G

[rotund aurora]

what is the Fourier transform?

[grave sedge]

the dual of R/Z is Z, isn't it?

[arctic trail]

let me check the dual of the torus real quick

ah.

😐

thanks

[arctic trail]

I guess it's only defined for L1 or L2 with compact support

[rocky cloak]

For a function f: G -> C, the Fourier transform is a function \hat f: \hat G -> C given by integrating against f (with respect to haar measure)

[arctic trail]

the conjugate*

[rotund aurora]

"integrating (g) against f" means int f(x)g(x) ?

[arctic trail]

but it should be f(x)conjugate(g(x))

[rotund aurora]

why

oh maybe you want it to be like an inner product, so you would want hat f(f) to be real

[delicate orchid]

yeah it is an inner product

in fact this is how we define the inner product of characters and other assorted doodads

[marsh scaffold]

Guys a little bit help please, I'm new to exact sequences

And like I've been trying to use the universal property of the direct sum

But i haven't figured out how I use h and the fact about goh to reach the conclusion

[kind temple]

consider the same situation with
0 --> G - i_G -> G + H - pi_H -> H --> 0

and i_H : H --> G + H

see how the diagrams line up and figure out how to match up the functions f, g, and h to the basic situation

@marsh scaffold

[marsh scaffold]

You want me to use five lemma?

[kind temple]

yea that works

[marsh scaffold]

No other way without five lemma?

[kind temple]

f and h are the "canonical injections" for K

[hollow fjord]

Hi!
I need to prove that in a Noetherian ring every radical ideal is the intersection of finitely many prime ideals. I have managed to prove that every radical ideal is the intersection of prime ideals. I need a hint to prove that is finitely many

[kind temple]

then show that (K, f, h) satisfies the same universal property as (G+H, i_G, i_H)

[marsh scaffold]

Aight thanks

I'll get back if I still am stuck

[kind temple]

you may need to write K as the internal direct sum of the copies of G and H living in K

[marsh scaffold]

So img(h) + img(f)?

[kind temple]

yea

[marsh scaffold]

And then as g is surjective

uhh i should be able to write for some k in K

Uhhh

k-g(k) is in img(..)

[marsh scaffold]

What's the thought process behind this

[kind temple]

wdym

[marsh scaffold]

Like how did we come up with this exact sequence and compared them

I actually don't have much intuition for exact sequence

[kind temple]

idk if i am the best person to offer intutive advice on this, but in the ideal situation, K would be exactly equal to G + H. the SES's look really similar

and you can just pair them up

and use the 5 lemma

there is still some work to do tho

with constructing the middle map

[marsh scaffold]

Hmm okay

It's just that I haven't proved 5 lemma yet

And like if there's not a way to do it without that then why is it necessary

[kind temple]

you kind of need to write K as an internal direct sum anyways. so you can just say that K = im(f) + im(h) = G + H

[marsh scaffold]

Right

[kind temple]

where = is isomorphism

this only relies on the fact that the internal direct sum and external direct sum are isomorphic

using universal properties here doesn't seem like the best way to go

[marsh scaffold]

Okay so the only reason why we need h:H->K such that goh=Id is cause it gives an embedding into K?

[kind temple]

yea

[marsh scaffold]

Would it work if we had any embedding?

[kind temple]

it needs to be a right inverse to g

[marsh scaffold]

Oh okay

And that's sufficient?

Yeah

Okay thanks a lot!

[kind temple]

yea. if it were any embedding, i think showing that im(h) and im(f) intersect trivially would fail in general

[marsh scaffold]

Just to double check this is it right?

And then five lemma?

Oh

,rotate

[kind temple]

(f + g)(x,y) = f(x)g(y)?

[marsh scaffold]

Yeah

f(x)+g(y)

[kind temple]

that should work, yea

[kind temple]

but if you can't use the 5 lemma, then doing it this way is fine as well

[marsh scaffold]

Aight

Thanks again

[hollow fjord]

Hi!
I need to prove that in a Noetherian ring every radical ideal is the intersection of finitely many prime ideals. I have managed to prove that every radical ideal is the intersection of prime ideals. I need a hint to prove that is finitely many

(I send it again so it won't be fotgotten)

[marsh scaffold]

uh does prove it for the nilradical and then consider the map A->A/p not work

idk if its equally difficult for the nilradical or not

yeah i guess these two statements are equivalent in this case

[chilly ocean]

In showing that the ring of complex numbers is isomorphic to the matrix ring M with 2×2 matrices with real entries of the form: m(a,b) = [(a,b),(-b,a)], are following steps sufficient:

1. Has multiplicative identity I, and m(a,0) = aI (contains Reals)
2. m(0,b) = bm(0,-1) where (m(0,-1))² = (-1)I (contains Imaginaries)
3. showing m(a,b) + m(c,d) = m(a+b,c+d) = m(b+a,d+c) = m(b,d) + m(a,c), and m(a,b)m(c,d) = m(ab-cd, ad+bc)
4. showing that m(a,b) = √(a²+b²)m(cosA,sinA) that causes anti-clockwise rotation upon matrix multiplication
5. a+bi |-> m(a,b) is a bijection

Hence a ring isomorphism between C and M exists where a+bi |-> m(a,b)?

[kind temple]

all you need to show is that m is a ring isomorphism if you want to show that C and M are isomorphic. you can easily produce the inverse, so all you have to do is show that m is a homomorphism

just show that m(1) = I, m(z + w) = m(z) + m(w) and m(zw) = m(z)m(w)

[rain grove]

Let $V$ and $V'$ be vector spaces over the field $F$. Show that for an arbitrary basis ${b_i \mid i \in I}$ of the space $V$ and an arbitrary set of vectors ${a_i \mid i \in I}$ from $V'$ there exists such a homomorphism of vector spaces $\phi : V \rightarrow V'$ such that $\phi(b_i) = a_i$ for all $i \in I$.

[cloud walrus]

OHHELLNAH

[rain grove]

How is this true? If $a_1$ and $a_2$ are linearly dependent than It could happen that $\phi(0) \neq 0$

[cloud walrus]

OHHELLNAH

[rain grove]

Oh wait no... I just found a v != 0 so that \phi(v) = 0

So then thats ok?

[chilly ocean]

What makes you think so?

[rain grove]

I think I see, thats just an element that is mapped to 0, the 0 vector is still mapped to 0.. i think so

[chilly ocean]

Yes, there can be other elements that are mapped to 0

[rain grove]

Thats pretty cool, vector spaces are simple, they have basis and they are the isomorphic if dimensions are the same and now there is also always a homomorphism between any vector spaces

[old hollow]

If we have J ⊆ I as ideals in R, is there a natural surjection R/J —> R/I?

[chilly ocean]

What do you mean by simple?

[chilly ocean]

Yes, have you heard of the correspondence theorem?

[old hollow]

Yes I have

Hm lemme see

For correspondence theorem, I’m thinking, consider R —> R/J. Then ideals in R containing J correspond to ideals in R/J. This would mean, in my example, that I is an ideal in R/J

[rain grove]

there is not always a homomorphism from any group to any group, but for vector spaces it is (if they are over the same field at least)

[old hollow]

But then how do I get a surjective map R/J —> R/I?

[chilly ocean]

This is not true. If G and H are any groups, f:G->H given by f(x)=0 for all x is a homomorphism

[rain grove]

Ok right, what about nontrivial homomorphisms?

[chilly ocean]

There also isn't any nontrivial homomorphism from any V into the zero vector space. But alright, we can have nontrivial G and H with only trivial homomorphism from G to H and that can't happen for nontrivial vector spaces (asuming axiom of choice)

[chilly ocean]

For the ideal in R/J corresponding to I in R, it will turn out that the codomain of the quotient will be isomorphic to R/I

[grave sedge]

I dont know if you're allowed to do it, but you could take a primary decomposition of (0)

[grave sedge]

This for the nilradical, for the general case you can look at the primary decomposition of that ideal

[old hollow]

The codomain of what quotient?

(R/J)/I?

[chilly ocean]

Yes if by I you mean the ideal of R/J corresponding to I in R

[old hollow]

Oh, right yes

Oh hmm

I suppose that makes sense

R/I is the same as flattening J and then R’/I’

or something like that

I feel like a nice diagram would make this clear

[mighty kiln]

@knotty badger is this universal property? :P

[knotty badger]

Uh is what

Also am I becoming the universal property agent lol

[mighty kiln]

^

Mathcord's most trusted yoneda lemmist and universal propertist

[knotty badger]

Oh this

I mean you can use the universal property of quotient for this

Ring homs R/J -> R/I naturally correspond to ring homs R -> R/I which send J to 0

The quotient map R -> R/I is one such ring hom, since J is contained in I

So you can convert the quotient map to a ring hom R/J -> R/I

Moreover since the quotient map R -> R/I is surjective, and you can write your new map as a composition R -> R/J -> R/I, the map R/J -> R/I must be surjective too

[lean sail]

this morning i did this proof. @glad osprey helped me through it a few days ago. someone mentioned that there is also an inductive way to do the proof that is more mathematically rigorous than what i did here. i was wondering if someone wouldn't mind helping me understand how to do that. here is the proof i did, for reference

[knotty badger]

It’s a consequence of the fact that if f o g is surjective, then f is surjective

[south patrol]

In other words the natural map R -> R/I factors through R/J

[knotty badger]

Yes, cause of the universal property of R/J

[rain grove]

Let $V$ and $V'$ be vector spaces over the field $\mathbb{Q}$. Show that every homomorphism of additive groups $\phi : V \rightarrow V'$ is also a homomorphism of vector spaces. In other words, show that from $\phi(x + y) = \phi(x) + \phi(y)$ for all $x, y \in V$ it follows that $\phi(qx) = q\phi (x)$ for each $q \in \mathbb{Q}$ and $x \in V$.

[cloud walrus]

OHHELLNAH

[knotty badger]

By a^-n, you mean (a^-1)^n?

[rain grove]

I have seen this proof many times but now that I have to think about it... why can I do this:

$\phi(x) = n\phi(\frac{1}{n}x)$

[cloud walrus]

OHHELLNAH

[rain grove]

1/n x is element of the group so then what is that element? I get nx = x+x+...x n times but 1/n x? Or is that element of the vector space and 1/n is just rational scalar

[chilly ocean]

1/n x does not always exist in a group or vector space

You can only do it if you know 1/n x exists

For vector spaces over Q (or any characteristic 0 field) it does exist and coincides with multiplying by the scalar 1/n

[rain grove]

hmm ok and that group homomorphism proprety is extended to the vector space map and i have to prove that map is a v.s homomorphism

[lean sail]

no i didnt mean that

[knotty badger]

Then what does a^-n mean

[knotty frigate]

(a^n)^-1? but arent those equal?

(a^n)^-1 and (a^-1)^n i mean

[knotty badger]

Isn’t that what you’re trying to prove

[lean sail]

yes

[knotty frigate]

oh

mb

[knotty badger]

Then what’s your definition of a^-n

[lean sail]

i define this in my proof as $(a^{-1})$ getting multipled by itself $n$ times

[cloud walrus]

proofman

[chilly ocean]

Do you only want to prove it for positive n?

[lean sail]

yes, $n$ is a positive integer in my case

[cloud walrus]

proofman

[chilly ocean]

And you want a rigorous proof

Do you know how induction works?

[lean sail]

yeah i wouldn't say im an expert, but i have an idea

[knotty badger]

so that’s (a^-1)^n

[chilly ocean]

I will tell you the steps, and you will do the proof

You want to prove that, for every natural number n (no reason to not include 0), that (a^n)^-1 = a^-n

[chilly ocean]

Step 1: prove it for n=0. This is called the base step.

[lean sail]

im not sure what you're getting at, are you saying that there is something wrong with the proof?

[knotty badger]

no im just confused why you think a^-n and (a^-1)^n are different definitions

(a^-1)^n exactly means “a^-1 getting multiplied by itself n times”

[lean sail]

i never explicitly said i think they are different, i just said "no" when you asked if i meant to type the other one

[knotty badger]

I was under the impression you thought they were different

[lean sail]

.

[knotty badger]

yeah, that’s why

[lean sail]

yes, i acknowledge, understand why you thought that

[knotty badger]

anyway yeah you should start with this

[lean sail]

i actually tried this the other day and got stuck, i will share my work

.

[lean sail]

i got stuck at the inductive step

[glad osprey]

Try again, but try to show that a^(-n) a^n = e

[lean sail]

Hey I was thinking more about what you said, why exactly are they the same? I mean, I never thought they were different but I also never asked why they are the same… in the book it just defines the exponents to work this way…

[lean sail]

Oh interesting. Ok, I’ll give that a try.

[knotty badger]

As in for me, the way I’ve seen the notation “a^-n” defined is as (a^-1)^n

So they’re equal by definition

[lean sail]

Yeah, they didn’t explicitly state that in the book I am using, but they define, $a^{-n}$ as the product of inverses… maybe it’s a matter of taste by the author, I have no idea

[cloud walrus]

proofman

[south patrol]

that is what Pseudo said though

[lean sail]

yeah you’re right, so in a different style, they do say that explicitly

[cloud lynx]

if we have a cyclic group, does a unit of this group generate the group?

[lusty marlin]

What is a unit of a group?

[chilly ocean]

It never does unless it is the trivial group

I'm assuming by unit you mean the identity element

You might be confusing it with the element denoted by 1 in the usual way C_n is written

[cloud lynx]

for example the units of Z are 1 and -1

[chilly ocean]

That is not a group theoretic property

[cloud lynx]

and these two elements are generating Z

[chilly ocean]

You need ring structure to talk about that kind of unit

[lusty marlin]

Unit isn't a term used for groups, at least not in any resource I have seen

[cloud lynx]

a element which is a divisior of all elements of the group

[cloud lynx]

lol

[chilly ocean]

Groups don't have the notion of divisibility you are thinking about

[lusty marlin]

What is a divisor here?

[chilly ocean]

Well..

You could define divisibility in Z-module terms I suppose

But I think that's not what you are thinking about

[cloud lynx]

ah alright yeah not groups rings

[chilly ocean]

So maybe you are thinking about the rings that are generated by 0 elements

That is closest thing to cyclic groups in the ring world

[cloud lynx]

hö

ah

[chilly ocean]

Those rings are of the form Z/nZ

And have underlying group the cyclic group C_n

And yes, 1 will generate their underlying group

[rocky cloak]

But the generators of Z/nZ as a group are exactly the units of Z/nZ as a ring, so there's that

[chilly ocean]

That is true

[rocky cloak]

Mostly just because ideals and subgroups are the same for Z/nZ

[cloud lynx]

ahh okk

ty

[arctic trail]

I mean that makes sense

lmao reminds me of number theory class

[lean sail]

okay, after days of work, i think i finally have something that resembles a proof by induction @knotty badger @chilly ocean @glad osprey @south patrol (@ anyone else who helped me):

[south patrol]

Yes this works well.

Though the n = 0 case is easier, indeed I would just say it is clear

as it is the statement that e^-1 = e

[hidden wind]

i had a logic prof who told me with a straight face there’s never any need to verify the base case and i’m still confounded as to what he meant

because from context i don’t think he meant that the base case is obvious or clear

[delicate orchid]

the base case is usually so completely o- oh

then no he's just being a nerd

[rocky cloak]

If you just have an assertive voice, theres really no need to prove anything

[hidden wind]

i have tried this at exams to varying success

at my intro to complex analysis exam i got full marks by skipping half the details; and my intro to proofs course i wasn’t allowed to take the exam because i didn’t bother answer the boring routine parts of a midway assignment 😭

i got held back a year for that

still haven’t finished my bachelors because of it

math students are supposed to take it first year but i did physics my first year and really don’t feel a need to take that course, but it’s a formal requirement, which is stupid, as if you can pass any of the later courses if you don’t know the material of that one :S

[lean sail]

great, thank you for reading through this.

[lean sail]

where did i say that the inverse of e is e? im just not seeing that

[south patrol]

a^0 = e

[lean sail]

ok, i see, it was in that initial "consider"

[lean sail]

how do you think i could improve that?

[rain grove]

Wait isn't $x^{-n} := (x^n)^{-1}=(x^{-1})^n$

[cloud walrus]

OHHELLNAH

[chilly ocean]

That's what they are proving

[rain grove]

as far as i know x^{-n} is a way of writing the element (x^n)^{-1}... I mean there is no proof its a notation for that element

[grave sedge]

You need to prove powers commute with inverses

The element you are referring to is the inverse of x^n

You want to show its also equal to (inverse of x)^n

[nocturne shard]

I think the proof reads very well, nice job

[rain grove]

that follows from (xy)^{-1} = y^{-1}x^{-1}

[grave sedge]

Sure, but "that follows from" surely means it's a proof

You can call it trivial but it's still something you technically need to prove

And you just gave a proof

[chilly ocean]

I'm talking about the second equality

[knotty badger]

the only thing I can think of which relates to this is well-founded induction

[rain grove]

Yeah that you need to prove I agree

[knotty badger]

which is a formulation of induction that doesn’t explicitly mention base cases

and you can occasionally use it to prove something inductively without even needing a base case

[lean sail]

Thank you!

It seemed trivial to me, and then @chilly ocean asked about if I can prove it and I wasn’t able to 🤷🏻‍♂️

[arctic trail]

In transfinite induction this is indeed true

[hidden wind]

whaaat

where can i find an introduction to this

[arctic trail]

if forall b we have that (P(a) forall a < b) => P(b) then P(b) forall ordinals b

[knotty badger]

yes that’s an instance of well-founded induction

i think it’s like

[arctic trail]

(I didn't read the above)

[knotty badger]

if < is a well-founded relation, then you can do induction in this form

[chilly ocean]

You usually/sometimes consider 0 and limit ordinals as base cases though

[knotty badger]

so if you can show (P(a) for all a < b) => P(b), then you’re done

so, both < on the naturals and < on the ordinals are well-founded relations

[arctic trail]

Yeah, it's the easiest way to setup that we have that property

[knotty badger]

i think the axiom of foundation makes elementhood a well-founded relation on sets

so you can do $\in$-duction

[cloud walrus]

Pseudonium

[arctic trail]

man

[knotty badger]

but there are other well-founded relations you can consider than just these

[arctic trail]

is that supposed to be epsilon induction?

[knotty badger]

mhm yeah

[arctic trail]

I dislike that writing 💀

it's not even an epsilon

[knotty badger]

hehe

[arctic trail]

OH SHIT IT'S A PUN

[knotty badger]

i think it’s just usually the case that N or the ordinals suffices

[arctic trail]

ON HOW IT WORKS

[knotty badger]

yeah lol

[arctic trail]

AND IT READS INDUCTION

[knotty badger]

pfft

[arctic trail]

I only did epsilon induction once

and I forgot how it worked (as in how to setup an epsilon induction*)

[knotty badger]

iirc it’s the axiom of foundation that lets you do it

[arctic trail]

no I mean

[knotty badger]

oh right sure

[arctic trail]

how do you know that a proof will be using epsilon induction

it's tricky

[knotty badger]

i think it’s a similar setup to this

[arctic trail]

it's weird

because sometimes the setup isn't immediately obvious

[knotty badger]

but instead of < it’s $\in$

[cloud walrus]

Pseudonium

[arctic trail]

and the proposition you use is a modified version of the thing you want to prove

[knotty badger]

yeah it definitely is weird

[arctic trail]

I think I remember seeing one on transitive sets

Ah no it was the von Neumann hierarchy

[glad osprey]

We've had this discussion before, and the difficulty is that it's somewhat subjective what constitutes a proof, and therefore whether an induction proof contains a base case or not. But it's my opinion that even though you can do well-founded induction without an explicit base case, you atleast have to consider the base case as an edge case, since the inductive hypothesis must hold even when the premise is "empty"

[knotty badger]

yes that’s what happens in practice - when you do well-founded induction, you end up having to prove base cases for “empty” premises

i do still think it’s cool that it, like, automatically detects what base cases you need

and there are genuine situations where you can skip any base case

the one im thinking of is existence of prime factorisations

[glad osprey]

yeah, that's a good example

but I guess 2 being prime is a base case there

You could prove the proposition P(x) = True over any set without any base case but otherwise, I'm envisioning an induction proof programmatically as:

if premises is non-empty:
    choose some hypotheses in premises;
    return proof using hypotheses;
else:
    return base case

[rain grove]

Let p be a prime number. Show that the finite-dimensional spaces V and V' over the field Z_p are isomorphic exactly when |V | = |V'|

[chilly ocean]

What have you tried?

[rain grove]

Just as I posted I got an idea

The other side is clear, if they are isomorphic then |V|= |V'|?

[chilly ocean]

Yeah any isomorphism is in particular a bijective function

[rain grove]

So I think I get why Z_p works, there only p possible elements of the field so there p^n possible ways you can write an element

But why is there isomorphism now?

[grave sedge]

Two vector spaces over the same field are isomorphic if and only if they have the same dimension

[rain grove]

Why it's not true for non-primes?

[chilly ocean]

Z_n is not a field when n is not prime

[coral spindle]

It’s true for all fields with a finite number of elements. There are fields with a non-prime number of elements, but you have probably only seen the ones with a prime number of elements.

A similar fact is actually true for fields with infinitely many elements, but that is a story for another day.

[rain grove]

hmm cool

[serene dune]

p^n

F_p^n

vector space over F_p

[dull ginkgo]

Are you referring to the field finite subgroup cyclic theorem thing

[grave sedge]

No, the fact that vector spaces with the same (large enough) cardinality have the same dimension

If i had to guess

[coral spindle]

Indeed, aka that the theory of vector spaces is categorical in appropriate cardinalities

(Nothing to do with category theory, before you ask. It’s a model theory term.)

[serene dune]

can anyone help me to intuit PSL groups

[dull ginkgo]

Oh so just invariant basis number

But for the infinites

[grave sedge]

No not really

All vector spaces have a well-defined dimension

That's the invariant basis thingy

The thing above is the fact that this dimension is uniquely determined by the cardinality of the vector space as a set

(assuming some conditions, because say k and k² are not isomorphic as k-vector spaces but have the same cardinality for infinite k)

[hidden wind]

mhm but there are finite fields of non-prime order

[grave sedge]

They are not Z/n

[hidden wind]

what do you want to intuit about them

[serene dune]

give me anything related to it

imma follow the path

i know nothing
saw the name too much, interested in general

[rain grove]

If I have 2 n-dimensional vector spaces over field Z_2, then there are 2^{n}-1! different isomorphisms between them?

[grave sedge]

No

[rain grove]

What if they are 2 dimensional?

[grave sedge]

In that case yes

[rain grove]

oh sry like this (2^n)! - 1

[grave sedge]

You mean (2^n-1)!?

[rain grove]

fk yes

even then no?

[grave sedge]

Still too many

Remember you need f(x+y) to be equal to f(x)+f(y)

[hidden wind]

are you familiar with the elementary geometry of the hyperbolic plane?

[grave sedge]

So you can't just take all bijections which send 0 to 0

[serene dune]

no

[hidden wind]

ok that intuition is perhaps a bit too specific anyways

[rain grove]

Ok I did it for 2 dimensions with straight up defining all isomorphisms and trying f(x+y) = f(x)+f(y), is there a better approach?

[grave sedge]

Yes

[serene dune]

alr

[grave sedge]

Having an isomorphism is equivalent to sending some fixed base of the first space to any base of the second

So you just need to count how many bases there are

ordered bases

And an ordered basis is equivalent to picking a nonzero vector, then any one which is not in the span of the first, then one which is not in the span of the first two, and so on

[hidden wind]

if S is some bounded region in the euclidean plane with area s and if A is any two by two matrix with real entries, then the area of the region AS (obtained by multiplying each point of S by A) is |s det(A)|

so if det(A) = 1 the transformation is area-preserving

now of course areas are positive, but det(A) may be negative, and it turns out it is so precisely when the transformation is orientation-reversing, i.e. if you draw some circular arrow in your region (or anywhere on the plane), after transforming by A you’ll find the arrow is going around in the opposite direction

[rain grove]

Yeah thats way less work, nice

[hidden wind]

so quotienting away the kernel +-I amounts to ignoring orientation

[hidden wind]

that is one way to think about it :S but there are many other

personally i prefer to think about PSL(2,R) in terms of its action by Möbius transformations on the hyperbolic plane (in poincarés disk model)

the group (P)SL(2, Z) also shows up all over the place

and certain quotients of this corresponds to the rotations of platonic solids

[languid trellis]

Hi all, slightly lost on how to compute the quotient. I have tried computing the inverse of (1+θ+θ^2), I tried doing the division in Q[x] and then passing to the quotient Q(θ) = Q[x]/(x^3-2x-2), I have tried computing θ^2 in other terms but none of these have worked so far. Does anyone have an idea for an approach? Pls ping (: thx ❤️

[coral spindle]

Hint: f(x) = x^3 - 2x - 2 is coprime to the denominator g(x), so by the Euclidean algorithm you will find polynomials p, q such that 1 = pg+qf. Substituting in theta, 1 = p(theta)g(theta). Can you see how this might help you?

[languid trellis]

I already wrote this down, hoping to find (1+θ+θ^2)^-1, p(θ). Is the strategy here just to try stuff and see if it works?

oh I should find p and q in Q[x] because it's a euclidean domain

and then apply eval. map

[coral spindle]

Exactly!

This is a “once and then never again” kind of exercise

If you know the algorithm and can apply it, you’re good

[languid trellis]

I'm not looking forward to calculating it but I feel like I should so here we go

I suspect that it works exactly as in Z

thanks boyt (:

[rain grove]

Can someone help with this:

Let $\phi$ be an endomorphism of a finite group $G$. Suppose that $|\ker \phi| = m > $1. Let us take $b \in G$. Show that the equation $\phi(x) = b$ either has no solutions or has $m$ solutions. Also explain why there are elements of $b$ such that the equation has no solutions

[cloud walrus]

OHHELLNAH

[rain grove]

Im sure I have to use propreties of finite groups but I can just remmember the Lagrange theorem and that for every element a there exists n so that a^n = 1

[quiet pelican]

Suppose f(x) = b has a solution x = a0 and let k \in Ker(f)
Then what is f(kx)

Further suppose f(x) = f(y0)
What is f(x^-1 y0)?

[dull ginkgo]

Describe the cosets of the kernel :3

[rain grove]

Ok From the first one I get that there are m solutions but I still don't see why no solutions

[quiet pelican]

Well then the image of f has size |G|/m
(As each element of the domain has an image, and can be put into groups of m based on what they map to)

[rain grove]

i don't undertsand

Why image has size G/m?

[dull ginkgo]

If there are no solutions it’s outside of the image

(Which is a subgroup)

[dull ginkgo]

The function splits the image into cosets (elements who get mapped to the same image)

All the cosets are the same image, so it splits G into m-sized sets, so there must be G/m elements in the image (G/m many cosets)

[rain grove]

What is outside of the image?

[dull ginkgo]

Any element that isn’t the output of a function

Therefore there is no element being sent to it

I.e no solutions

[rain grove]

Can the argument be like: If it has a solution a_0 then it has m solutions and if it has no solutions then well it has no solutions. end

[dull ginkgo]

If it has a solution then the set of solutions is of the form xKer(\phi) for \phi(x) = b

[rain grove]

Ok ill try to look at what the cosets are

[rain grove]

But is this argument ok?

[rain grove]

Ok I get it

I mean I get why the solutions look like that

[dull ginkgo]

Yeah

[rain grove]

Ohh so the image has G/m elements and because there are elements outside the image the map \phi(x) = b can have 0 solutions if b is outside the image

[dull ginkgo]

Yeppers

So quick uh question here

For a vector space M (which is always free), we have the Symmetric Algebra from the relators (uv - vu), the Alternating algebra from the relators (x^2), and the Exterior Algebra from the relators (xy + yx)

Now for vector spaces the exterior algebra is canonically isomorphic to the first two mentioned dependent upon the field’s characteristic (char 2 implies it’s the symmetric, else it’s the alternating)

Is this true for general rings or can there be a ring where the three are basically distinct

[chilly ocean]

Would the ring Z + x*(F_2[x]) give a counterexample?

The constant elements behave like char 0 but the deg>=1 terms behave like char 2

But I'm not very familiar with these algebras

[arctic trail]

what do you mean?

I don't understand what the question is?

You mean substituting vector spaces to modules?

[dull ginkgo]

Nevermind I figured out what I meant

Just generalizations of different functorial algebras

[arctic trail]

If so if M is an irreducible module over C[G] the group algebra of a finite group, then the character of the exterior algebra on M should be distinct to that of the symmetric algebra (hence not isomorphic)

ah

[dull ginkgo]

I wonder if it’s easy to show the symmetric powers of a free module is free

[arctic trail]

seems simple

[dull ginkgo]

Over a general ring.

[arctic trail]

a free module has a basis E = {ek}:k.
The nth symmetric power will have a basis {B <= E: |B| = n}

all that's left should be to prove that

[dull ginkgo]

Well you’d have to show they span it, and that they are linearly independent

That is my problem

[arctic trail]

I mean, proving that they span it should be simple

since they're all finite sums

so you can prove that their finite sums of simple 'tensors' (in quotes cause of the action)

[dull ginkgo]

The exterior alg one uses maps to the tensor powers

[arctic trail]

I think they have a name but I forgot

for what I'm thinking when saying simple tensors

like those which can be written under the form u x v x ... x w

Given that they're free each one of those vectors can be written as a linear combination of the basis vectors of your free module

[dull ginkgo]

Simple tensors, that’s the name

[arctic trail]

ah nice, I love when math names are the obvious ones

[dull ginkgo]

Yes by definition but you have to PROVE that

Which is the tricky part

Going from the def of the algebra

[arctic trail]

I meant if q = u x v x ... x w

each of the u,v,...,w can be written as a linear combination of the basis vectors of the base space

I wasn't talking about q itself

[arctic trail]

unwrapping everything will give you linear combinations OF tensor products of the basis vectors of the base space

Writing this all neatly finishes the spanning part

To prove the linear dependence part

Given a tensor written as the linear combination of tensor products of basis vectors of the base space, in such a way that there exists no repetition in the tensor products of the basis vectors. It being equal to 0 implies that the homomorphisms given through the subsets B I described before, to the ring that the modules are over. You get that all coefficients are 0. Hence these are linearly independent and thus free

[dull ginkgo]

It doesn’t feel that simple

[arctic trail]

I think it might just be easier to prove that

Sym(n,R^k) =iso R^(number of multisets of size n of {1,...,k}) =(according to google cause combinatorics is solved for these easy questions) R^(k + n -1 choose n)

[serene dune]

so it preserve the shape, regardless of its parity ?

[arctic trail]

Man I love the Mobius group

It's the group of continued fractions 🫡

and everything makes so much sense when you think of continued fractions in terms of mobius transformations and compositions

I would love to learn more about PSL cause I also don't have any intuition on them

@hidden wind is there any intuition on PSL through its group action on some object?

[hidden wind]

i’ve been meaning to get to this looks like great fun

[hidden wind]

well yeah for the real entries and n=2 case, how about the entire hyperbolic plane

[arctic trail]

I don't have intuition on those geometric objects

😓

is there any other, non-geometric interpretation?

[grave sedge]

Projective spaces?

[arctic trail]

Projective spaces are a-okay for me

[grave sedge]

Idk if you want some specific PSL(something) or in general

But PSL is a subgroup of PGL

And PGL is the group of automorphisms of the projective space

[arctic trail]

I'm more comfortable with PGL(2)

since I just think of them as mobius transformations

Do these groups show up outside of geometry?

[grave sedge]

For finite fields they are quite relevant in group theory i guess

Since they are simple (the PSL)

[hidden wind]

well you mentioned comtinued fractions yourswlf

[arctic trail]

I meant that's just for PGL(2)

[hidden wind]

sorry i can’t type

[arctic trail]

Sure I've read that

But is there like a canonical group action that these groups have?

what does an isomorphism of projective space look like?

[grave sedge]

For PGL(2) it's a cross-ratio preserving map

Since PGL(2) is the automorphism group of the projective line

[arctic trail]

And what properties does the action of the subgroup, PSL, have on the projective space that PGL doesn't?

[grave sedge]

In some cases it's the same

Like if the field is algebraically closed

[arctic trail]

Also, is there a way to represent automorphisms of projective space in n dimensions like the mobius transform which is the ratio of two affine polynomials (I guess) in n > 2?

[grave sedge]

Well

Mobius transformations use the fact that the projective line is more or less a field

Where by more or less a field i mean precisely the field with an extra point at infinity

Actually you can always do that

Since projective spaces have coordinates

[arctic trail]

yeah

[grave sedge]

So for projective planes (i.e. PGL(3)) i guess you can represent the elements as pairs of rational functions in the two coordinates where numerators and denominators have degree ≤1 and the denominators are the same

And consider a bunch of cases for behaviour at infinity

[arctic trail]

rational functions in two coordinates?

so like

[grave sedge]

Not really nice now since infinity is a whole line and not just a point

[arctic trail]

[p1(z):p2(z)]?

[grave sedge]

Well, no

[arctic trail]

ohhhh

[grave sedge]

(p1(z1,z2),p2(z1,z2))

[arctic trail]

I see

[grave sedge]

If you want homogeneous coordinates then it's just a triple of linear functions

[arctic trail]

wait, non-homogeneous coordinates?

hmm

[grave sedge]

The mobius thingy with rational functions is with one non-homeogeneous coordinate

[arctic trail]

so instead of (az+b)/(cz+d) I can think of [az+b:cz+d]?

[grave sedge]

Yeah

[hidden wind]

caroline series has an article titled the modular surface and continued fractions:

[grave sedge]

That's for PGL(2)

[hidden wind]

The aim of this note is to clarify the somewhat elusive connection between geodesies on the modular surface M (the quotient of the hyperbolic plane H by the modular group G = SL (2, Z)) and continued fractions. This connection was, for example, noted by Artin [3] who, by an ingenious use of continued fractions, deduced the existence of a dense geodesic on M. Our results may be regarded as a rationale for Artin's method.

[grave sedge]

It's more or less its definition

[hidden wind]

the actions of SL and PSL on the hyperbolic plane are the same

[arctic trail]

So transformations of P(V^n) take a point (z1:...:zn) to some (p1(z1,...,zn):...:pn(z1,...,zn))?

[grave sedge]

Yeah

[arctic trail]

Wait but isn't this dependent on the representative of each coordinate?

because of the constant term?

[grave sedge]

you have no constant term

[grave sedge]

This should be [az1+bz2:...

[arctic trail]

OHHHH

So when I write (az+b)/(cz+d) I am essentially considering P(V^2) as V with a point at infinity and the embedding z -> (z:1)???

[grave sedge]

Yeah

[arctic trail]

This is starting to make a lot of sense to me!

Wait

These automorphisms are also automorphisms of projective varieties then? I haven't read much algebraic geometry yet

[grave sedge]

They are, i'm not sure they are the only ones

[hidden wind]

isn’t this somehow analogous to PGL -> PSL

[grave sedge]

One case where the "projective line" automorphisms are also the only "something variety" automorphisms is the riemann sphere (i.e. complex projective line) when seen as a riemann surface

[grave sedge]

I guess i should have said "something manifold" instead of "something variety"

[grave sedge]

PSL is a subgroup of PGL

[arctic trail]

P(F^n) has coordinate ring F[x1,...,xn] no?

[grave sedge]

No

[arctic trail]

shiet

[grave sedge]

It's not an affine variety so it has no "coordinate ring" i think

[arctic trail]

you can define a coordinate ring for a projective variety I think?

[grave sedge]

F[x1,...,xn] is the coordinate ring of the affine space

[arctic trail]

I remember homogeneous ideals I forgot

[hidden wind]

i don’t really see any clear appearance of PGL in the geometric stuff i’ve been looking at, but if it’s the projective automorphism group then these geometric objects should all inherit a group action from that… as they may be modelled in projective geometry

[grave sedge]

Yeah i'm not an expert in this

[hidden wind]

god i should really get to projective geometry how have i not found the time to properly look at that yet

[arctic trail]

let's goooo

I remember seeing this in my commutative algebra book

[arctic trail]

I think I started to really understand continued fractions when I discovered the mobius group by myself through them

[hidden wind]

so swag

[arctic trail]

Then Lagarange's and Galois theorem's make so much more sense

as to why like quadratic even show up

[hidden wind]

i don’t think i know these

[arctic trail]

Periodic continued fractions correspond to quadratic irrationals

and similarly quadratic irrationals have periodic continued fractions

[hidden wind]

right

[arctic trail]

So like @hidden wind you can think of the group of invertible 2x2 matrices as acting on the continued fractions themselves

[hidden wind]

mhm

what is the significance of a scalar multiple of such a matrix for the action?

[arctic trail]

Like 0

What you're more interested in are matrices of the form
(a 1)
(1 0)

these correspond to: a + 1/z

A continued fraction is essentially an infinite composition of these functions

they have determinant -1 which explains some behavior on the convergents of a continued fraction of a number

And a lot of the proofs I saw during number theory class were essentially bringing back this group without explicitly saying they were in sneaky ways

[rotund aurora]

The basic ideas are not too deep imo, you can learn them in a day or two

[delicate orchid]

squish onto the sphere

[rotund aurora]

No. This is like saying that automorphisms of a group are also automorphisms of a subgroup

[hidden wind]

lovely, but the additional structure it seems able to support is quite something

for klein, all geometry is projective geometry

[arctic trail]

wdym?

I didn't quite understand

[arctic trail]

isn't that Cayley?

[rotund aurora]

Maybe I missed context. But you asked if automorphisms of projective space are also automorphisms of projective varieties, no?

[hidden wind]

both

[arctic trail]

I meant that they're also the automorphisms in the "projective space is a projective variety" sense

[rotund aurora]

Ah, sorry

[arctic trail]

And I guess I wanted to ask if they are exactly the same automorphisms

[rotund aurora]

But yeah it's true, it's also true (complex) analytically

[arctic trail]

cause I have no idea what is meant by a projective space automorphism otherwise 💀 (as in what makes categorizes an automorphism)

[rotund aurora]

You are in a category, it's automorphisms in that category

[arctic trail]

I meant categorize like describe

automorphisms in IR^n are distance preserving maps

[rotund aurora]

These are not the automorphisms as a variety tho

[arctic trail]

yeah, all I wanted to know is what do people mean when they say automorphisms of projective space if they're not referring to them as projective varieties. Or all they always referring to them as projective variety

[hidden wind]

my guess is that the projective automorphisms are incidence-preserving

[arctic trail]

yeah apparently it comes from defining morphisms through linear maps

So they're not a priori defined as being the automorphisms when considering them as a projective variety

[rotund aurora]

There are different ways to define "projective space". You dont need to start from a vector space

[hidden wind]

i am definitely contenting myself too easily with guesses

[arctic trail]

they're probably equivalent why would I even write this if they weren't they wouldn't be the same 🤦‍♂️.
Ah I guess I meant the notion of morphism / automorphism

I choose to think of P(V) since it's easier on my brain to work with

[rotund aurora]

Ye. But the scheme theoretic definition is subtler. But all I'm saying is that starting from a vector spaces assumes unnecessary structure if you want to deal with varieties in general, unless all you want to do is "linear"

[arctic trail]

I just wanted to understand what was meant by automorphism, as in with regards to which type of structure but I guess there's that theorem which states that the automorphisms are the same no matter the structure we consider so it ultimately doesn't matter

Anyway

I feel like I have a much better understanding of the mobius group now which is lovely

[lean sail]

can someone tell if this question is asking for a proof? or just if we know what the inverse is?

[surreal dagger]

You construct a candidate and show that it satisfies the definition of being an inverse

[lean sail]

here is what i have so far:

i know i havent proven anything, but i can kind of see where it goes... hence, my last line

[surreal dagger]

Maybe youre overthinking it, but in my opinion you just take the product of (a_n)^-1......(a_1)^-1, multiply it with (a_1)....(a_n) and check if it satisfies the definition of being an inverse

[lean sail]

but unless you knew how socks-shoes work, it wouldn't be super obvious.

[surreal dagger]

What does that mean? Is this a saying?

[lean sail]

yes, its a saying. shoes-socks = $(ab)^{-1} = b^{-1} a^{-1}$

[lean sail]

do you think you can expand on this a little bit? although the question isn't asking for this, i am curious. it is not obvious to me how to multiply these two things together.

[lean sail]

wait nevermind! i think i got this... let me try.

@surreal dagger i realize this is not a proof, but i think this is your idea:

[surreal dagger]

Yeah, but this is exactly what the exercise wants.
Youre asked to come up with a concrete inverse, then you constructed a candidate and showed that it satisfies the definition of being an inverse.

[lean sail]

thank you, i think you answered my question.

[lean sail]

i am curious though, what would a proof look like for this? i am not even sure how to start. i think it looks kind of like induction might help here? (maybe) but then, i have no idea how to deal with the repeated multiplication of different elements.

[surreal dagger]

Why would you say this is not a "proper" proof?

[lean sail]

because it doesn't seem mathematically rigorous to me. especially because of all the dot notation, it looks like i am leaving out steps.

i think maybe this is a short proof: "G is a group. let a1, a2, ..., an in G. since G is a group, consider the repeated application of socks-shoes. we are done"

we can repeat socks-shoes for $n - 1$ times

i already proved socks-shoes on a previous problem

[delicate orchid]

Calling it “sock shoes” makes it 10 times less rigorous to me lmfao

You’re cancelling letters in a word

[surreal dagger]

What is sock shoes anyway?

[lean sail]

@delicate orchid @surreal dagger

[delicate orchid]

I know I just think that’s incredibly dumb

[lean sail]

i mean you can laugh at it, but multiple authors call it that

[delicate orchid]

I’ve yet to encounter one

[cobalt heath]

Wh

[delicate orchid]

I’d just call it “obvious” personally

[knotty badger]

My course did this

I think it’s a nice name

[lean sail]

i first heard it called that when in linear algebra

i like it too, the analogy makes sense

[rain grove]

Show that the ring Z_n contains a nonzero nilpotent exactly when n is divisible by the square of some prime number

[surreal dagger]

Anyway I think its fine.
Also you have to strike a balance on how "rigorous" you want to be for (obvious) details.
I think its good to be very rigorous in the beginning, but In this case I dont think there is much to gain really expanding this proof further and any grader/reader should be convinced by your proof

[lean sail]

good points!

[rain grove]

I did:

(=>) $a^k = 0$ (in $Z_n$) then $a^k = r*n$ (in $Z$) $n$ can be factorised with primes: $n = p_1^{l_1}p_2^{l_2}\dots p_m^{l_m}$

[cloud walrus]

OHHELLNAH

[cobalt heath]

You would like to use "nonzero" part as well

[rain grove]

I guess | is a transitive relation so all p_i must divide a^k?

and then because p_i is prime and p_i | a^k it either divides a^{k-1} or a so then eventually it must divide a

[surreal dagger]

Maybe its a good heuristic to strive for a level of detail that would convince a strict tutor that grades your imaginary course.
So in the beginning the tutor doesnt trust you and you need to be more thorough, but down the line you can just assume more things (if youre confident that you can make it "rigorous" if you had to).
As in next time, you just assume what the inverse of (a_1,...a_n) is, instead of proving it again

[rain grove]

How?

[lean sail]

this makes sense

[cobalt heath]

Well, you should figure it out

[rain grove]

ooh I assume $n = p_1^1\dots p_m^1$ and then show it can't be

[cloud walrus]

OHHELLNAH

[rain grove]

unless a = n

[delicate orchid]

or note that 0 is always a nilpotent element in any ring and thus must be excluded for the statement to be true

[hidden wind]

do you know how to prove this fact?

[lean sail]

yeah i think i got it

[hidden wind]

yeah that works

[hidden wind]

surely then it is also clear how you can verify that these indeed are the inverses, even without induction, for each particular given n

[lean sail]

yeah i came up one idea, @surreal dagger gave me another idea

[hidden wind]

lovely

[lean sail]

here i typed them up if you wanted to double check them for me

[hidden wind]

mhm the second one looks good to me

[lean sail]

the first one i detailed out way above ^

im not sure if you saw that conversation

[hidden wind]

strictly speaking induction is indeed unnecesary here as long as n is given, it is clear that your sequence of expressions will always have exactly n lines, and so a verification will always terminate

[lean sail]

i think both ways are valid, you can say, repeat socks-shoes $n - 1$ times.

or $n$ times?

[hidden wind]

doesn’t really matter

[hidden wind]

yeah any lenient reader used to this terminology (i had never heard it before yesterday) would give that a pass as it’s clear you’re refereing to a proof that would work, but you aren’t actually giving such a proof there

[lean sail]

do you have some idea how you might go about using socks-shoes to do a proof? i know the steps to do the so-called "extended application", but it is unclear to me how to make it clear to the reader.

[hidden wind]

you have it in the n=3 case here

though no need to use both left- and right- associativity

one side will suffice

i am always fond of pointing out when one can get away without induction but here the proof is much simpler with

inductive step looks like…

[ (a_1\cdots a_na_{n+1})^{-1} = ((a_1\cdots a_n)a_{n+1})^{-1} = a_{n+1}^{-1} (a_1\cdots a_n)^{-1} = a_{n+1}^{-1} a_n^{-1} \cdots a_1^{-1} ]

[cloud walrus]

rødbet

[lean sail]

very nice, i like it!

[narrow zinc]

Could one construct a ringmorfism as follows
/phi : Q[X] -> Z_p[X]
with p prime

[dull ginkgo]

Actually wait

Misread, X can be sent to anything. Let me think for a bit

[narrow zinc]

I'm pretty sure there's a morfism from Q to Z_p because the former is the field of fractions of Z and the latter is te division ring by a ideal pZ of Z

[dull ginkgo]

False

Maps between fields must be either zero or injections (fields have no nontrivial/proper ideals thus no kernels)

And Q and Z_p have differing characteristics

[narrow zinc]

Ahh yea I made a mistake

[dull ginkgo]

Q[X] is a PID, consider the kernel, which would be generated by a polynomial (q(X))

[narrow zinc]

I was using something that's obv. only applicale for injective morfisms, sorry

[dull ginkgo]

However, Q[X] has char 0 and F_p[X] has char p. Thus p is IN that kernel

So q(X) | p therefore q(X) has degree 0 so is a constant, and that constant must divide a prime, so is either 1 or p

If it’s 1, then (1) = Q[X] so it’s a zero morphism

HOWEVER, p is a UNIT

so (r/p)(p) = r would be in that ideal

Thus in both cases, the only possible map is a 0 morphism

@narrow zinc does this work for you?

[narrow zinc]

I must admit I could only follow about half of it 😆

It sounds right tho

[dull ginkgo]

Every ideal in Q[X] is generated by a single element, a polynomial

But the element p is mapped to 0 under the morphism right

[narrow zinc]

Ah. Yes

[dull ginkgo]

So the polynomial must divide a constant, therefore the polynomial itself must be a constant dividing p.

[narrow zinc]

And that is simply 1/p?

[dull ginkgo]

But (x/p) is always an element of Q, so

f(x) = f((x/p)p) = f(x/p)f(p) = 0

[narrow zinc]

So all polynomials would be mapped to 0?

[dull ginkgo]

Yes

[narrow zinc]

I see

Yes the morfism for Q to Z_p I was constructing would map

a/b -> a . b^(-1)

But obv. this doesn't work for b=p , which I overlooked

Thinking 'sure, inverses are perfectly well-defined, as Z_p is a field....

Ignoring the very obvious... my eternal enemy

The intent was to use it to prove a criterion for irreducibility of elements in Q[X]

It's one not proven within my course material, and not left as an excercise, so probably beyond the course perview, but it's bugging me

[dull ginkgo]

Eisenstein’s?

[narrow zinc]

Eisenstein's is the next criterion, so I think it may be related

[dull ginkgo]

What’s the criterion

[narrow zinc]

I can send a picture, but it's in dutch

Mostly symbols though?

[dull ginkgo]

I suppose

[narrow zinc]

Wait huh it twisted

[dull ginkgo]

OOOOH

Hmmm

[narrow zinc]

That mapping could be very useful, but of course it doesn't hold for x/p

Can you read it properly?

[dull ginkgo]

Any polynomial in Q[X] is a polynomial in Z[X] divided by a constant

[narrow zinc]

It actually doesn't pertain to polynomials with non-Z indices
Edit: I see now that this isn't even a limiting constraint because of your comment right above this one...

Right.

Hm so is that mapping not used?

I'd argue you could avoid the x/p mapping by indeed transforming the polynomials to ones in Z[X] and keeping the dividing constant at the side

[dull ginkgo]

It is, because we’re mapping to fields

[narrow zinc]

Then you could always inject the factor-polynomials to Z[X] and then to Z_p[X] ?

[dull ginkgo]

One is an injection straight into Q (it’s localization), so it’s not of much use but we can also map to Z_p via Quotienting mod p

[dull ginkgo]

Here’s a simple explanation:
It’s a ring morphism, so if a polynomial is factorable in Z, then it has to be in Z_p IF neither of the polynomials are sent to 0

Since the map would factor over the product

But if one of the polynomials was sent to 0, then obviously the whole thing would be 0 mod p

[narrow zinc]

Thus f will always be reducible in Z_p[X] if it is reducible in Q[X]

Q.E.D.

Nice thx

[dull ginkgo]

You have to make sure though

Like here’s the thing

let’s take

px + 1

then mod p it’s a constant

[narrow zinc]

Right but that's in the given

[dull ginkgo]

Yep, the degree is preserved

[narrow zinc]

Yes

[dull ginkgo]

So that’s it

That is to say, if f(x) factors into polynomials who have degree >= 1 mod p, then it’s factorable mod p

Since those factors won’t be constants (units)

[narrow zinc]

Right. Of course, the use of the criterion is in the opposite direction, to prove irreducibility...

[dull ginkgo]

So it’s the contrapositive

It would be a contradiction if Q(x) was reducible but not irreducible mod p with the criterion met above

[narrow zinc]

Yes. I had a feeling that working with contrapositive would grant a simpler proof

[dull ginkgo]

:3

[dull ginkgo]

Hopefully that makes sense

TLDR if f(x) in Z[x] is reducible into factors that are not constants mod p, then it’s reducible in Z_p[x] since the quotient morphism is multiplicative (the degree being preserved is when all of the factors keep their degree too, so can’t be constants)

[crystal vale]

Can an empty set be a group? Because an empty set follows every Property E(x), now some authors defined a group as non-empty set, I understand but I am confused here

[chilly ocean]

No, it would not have an identity

[dull ginkgo]

Most say no

Vacuously for algebraic shit usually it’s a no

Kinda like how the “trivial ring” of just 0 is not a division ring by most accounts because it doesn’t have a multiplicative identity along with it

[chilly ocean]

There is something called a heap that allows the empty structure (and any nonempty heap is equivalent to a group)

[chilly ocean]

That's not true, it does have a multiplicative identity

[dull ginkgo]

FIELD OF ONE RLEMENT JUMPSCA

[chilly ocean]

The reason it is not a division ring is because you want products of nonzero elements to be nonzero

But in the trivial ring the empty product is zero despite there being no zero factor in that product

[dull ginkgo]

But in the same vain it doesn’t have any nontrivial or proper ideals so

[dull ginkgo]

Frankly just say no so you don’t have to say “assume group G is nonempty…”

Every god forsaken time

[chilly ocean]

It doesn't have a maximal ideal and fields should have a maximal ideal

Everything seems to work much nicely if you don't let 0 be a field

Quotients are fields iff they are by a maximal ideal

This works iff you don't let 0 be a field

Else quotient 0 by (0) which is not maximal and you'd get a field contradicting that

The zero ring just has too small of a dimension to be a field, fields should have dimension 0

[lean sail]

@hidden wind i just threw this together with inspiration from you

@delicate orchid SOCKS AND SHOES!!! 🧦 👟

[delicate orchid]

https://tenor.com/view/cat-eating-camera-gif-26348428

[narrow zinc]

Do you by any chance have a proof of Eisenstein's criterion at hand?

[dull ginkgo]

Not a reference but I think I have an intuitive one

Assume f(x) satisfies the criterion’s requirement (p not dividing the leading coefficient, p dividing all other coefficients, NOT dividing the last coefficient twicefold)

But is also reducible, thus equalsh(x)g(x) for neither being constant polynomials

the leading coefficient of f(x) is the product of the leading coefficients of g and h

So is the last coefficient

[narrow zinc]

Of both last coefficients

[dull ginkgo]

By induction and using the “convolution def” of the coefficients for f(x) leads to the last two coefficients of g(x) and h(x) BOTH being divisible by p

But the product of them is the last coefficient of f(x)

So it’s a contradiction

[narrow zinc]

Ahh that sounds familiar. I may have already asked this like a year ago, maybe even in this server

Yeah that's real easy to follow

[dull ginkgo]

Eisenstein’s criterion can’t usually be used immediately if your polynomial has 1’s for any of the coefficients after the first

But you can set x = t + n or smth to try to get rid of em

And then you can use it hopefully

[narrow zinc]

I forgot that that's even allowed, gotta make sure to use that. Eisenstein pretty frequently shows up as something to use on exams, I believe

[dull ginkgo]

Usually it’s used to show x^p + 1 is irreducible by the and that’s about what I know it’s used for

[crystal vale]

Yes but I don't know when someone says it is vacuously true