groups-rings-fields

[topaz solar]

Well I mean, Z(G/(big enough to make it leq p^2) = G/…

[barren sierra]

yea

what does first order mean?

[topaz solar]

As in, can be written in a first order sentence in the language of groups

Abelian is \forall x, y (xy = yx) after all

[barren sierra]

oh

[topaz solar]

So like, class 2 would be like, $\forall x.\forall y.\exists z. xyx^{-1} = z \land z\in Z$ maybe?

[cloud walrus]

Sharp for President 2036

[topaz solar]

Or something

[barren sierra]

very last part

how do I show that if G is nilpotent, then G^k = {e} for some k?

[topaz solar]

Well, why not go a step up

[barren sierra]

?

step up?

[topaz solar]

Well, as it says, a finite central series at all <=> nilpotent

Then just show the lower central series is a central series ig

that’s probably not as easy but

[barren sierra]

so I have to show that the central series is unique?

because G nilpotent says a finite central series exists

but not necessarily that the lower central series is finite

since that's what I want to show

[topaz solar]

The way I originally remember nilpotent is iff the upper central series terminates

[barren sierra]

same but I gotta work with the definition in this problem 🙃

I did do all the above stuff so like I have that at my disposal

[topaz solar]

Just prove that form is equivalent :3

[barren sierra]

so show that central series are also lower central series?

that seems really rough

[topaz solar]

Well I think they’re different series

[barren sierra]

I'd believe that

[topaz solar]

But they always have the same length when they terminate iirc

There’s a section in DF about it

[barren sierra]

this is sounding very circular 😵‍💫

maybe I'll look at DF

[topaz solar]

Anyway, anyway, I am rambly

Can you show the lower series is central

[barren sierra]

yea I already did that

that's how I proved G^{k} = e implies G nilpotent, it's immediate from lower series is central

[topaz solar]

Ye

Hmmmmm in the same way you’d do “has a central series” iff “finite upper central,” why not just try that for lower to kinda just

Slip it on in between to terminate it :3

My class kinda just neglected the lower central stuff mostly, but like with upper it should be doable to kinda pass it in between the finite series things to extend it

[barren sierra]

Can I express the $A_i / A_{i + 1} \subseteq Z(G / A_{i + 1})$ condition in terms of commutator subgroups?

[cloud walrus]

Spamakin🎷

[barren sierra]

that seems to be the key

but idk how to do that

[topaz solar]

Well, you’re in the center iff you commute with all of G there, so like

[G, A_i] is in A_{i+1}

[barren sierra]

ahhh

[topaz solar]

Because quotients and commutators mix well

[barren sierra]

I think this gets me what I want

[barren sierra]

I need to understand this (and other identities with commutators) better lol

[topaz solar]

Note, lower central is stacking this

So

That’s what I meant by sorta slipping lower central in between

[topaz solar]

It sorta follows from normal = commutes as coset = pull it to the side in commutator expressions

[neat valley]

quite stuck on this one — does anyone know of a relatively slick solution?

[rocky cloak]

I don't know how slick it is, but with some computer help to factor it modulo 7 and 29
https://www.wolframalpha.com/input?i=factor+x^6+%2B+69x^5+-+511x+%2B+363
You can see that any irreducible factor has degree 2, 4 or 6. But also has degree 1, 5 or 6. Leaving 6 as the only option.

[marsh scaffold]

im confused with (b) the way they define wreath product and then if it is G^Y semidirect J then as G^y semidirect {e} is a normal subgroup of it so as its isomorphic to D^4 is should be cyclic right?

but apparently its not?

cause i believe any function squared in G^y will be the trivial function?

[delicate orchid]

that definition of the wreath product is a bit gross imo. Just think of $C_2 \wr C_2$ as $(C_2 \times C_2) \rtimes C_2$ where the $C_2$ acts via permuting the copies of $C_2$

[cloud walrus]

Wew Lads Tbh

[marsh scaffold]

what do we mean by permuting copies of C_2 here

[delicate orchid]

it swaps them over

it sends (a,b) to (b,a)

[marsh scaffold]

oh okay

thats quite a fancy way to say it

[delicate orchid]

not really

[topaz solar]

Why is it sigma^-1 smh smh

[delicate orchid]

yeah you have to do it backwards due to woke

[marsh scaffold]

yeah why is it that, thats one of the questions that came to me too

[topaz solar]

Is this a RIGHT ACTION wokism

[marsh scaffold]

so youre saying the author prefers right actions?

[delicate orchid]

honestly I hate this function definition so much I'm just going to straight up introduce the sane definition so I don't lose my mind

[mighty kiln]

It doesn't matter whether you're leftist or rightist as long as the math works out

[delicate orchid]

take a finite group $H$ and an embedding $\phi \colon H \rightarrow S_n$. Then given another finite group $G$ we define $G \wr_{\phi} H$ to be $G^n \rtimes_{\phi} H$ with $H$ acting by permuting the copies of $G$ in $G^n$ according to the embedding $\phi$

[cloud walrus]

Wew Lads Tbh

[marsh scaffold]

oh this is cooler

[delicate orchid]

sure your definition is slightly more general due to the fact that we can take Y to be uncountably infinite but I just don't care

[marsh scaffold]

oh so why cant we permute infinite copies

uncountably infinite copies

[delicate orchid]

yeah I realised that we can permute countably infinite lol

[delicate orchid]

I wrote up a proof that these definitions are equivalent in like 2022 one sec

complete babble

basically it boils down to a J-set X being the same as a morphism into S_|X|

[marsh scaffold]

whats the symbol with a bar one mean

the squiggle with a bar above

[delicate orchid]

just noticed I wrote the semidirect symbol backwards

[mighty kiln]

Wait isn't this what semi-direct product means

[delicate orchid]

a wreath product is just a special type of semidirect product

[marsh scaffold]

oh like what does a semidirect product do in a more intuitive sense

[mighty kiln]

How is this different from just unpacking the definition of semi-direct product

[topaz solar]

The libs when Sym(omega)

[delicate orchid]

it isn't really. But I was very stupid back in 2022

[delicate orchid]

a semi direct product takes one group acting on another, via automorphisms and form a new group using that data

[mighty kiln]

A short exact sequence
0 → A → B → C → 0
is left split iff product and right split iff semi-direct product

[delicate orchid]

that is... true....

or is it? let us investigate

[marsh scaffold]

so a wreath product is G^n doing the same stuff?

[delicate orchid]

yeah, but H acts only by swapping the Gs around. It fixes each G pointwise

that's the key difference

[marsh scaffold]

oh okay

[delicate orchid]

if you want to do a little bit of crankery - purely intuitionally a wreath product is "like permutation matrices on G^n" but a semi direct product is "like invertible matrices on G"

[marsh scaffold]

i wonder if there are like any reference towards the motivations os these constructions just like there are motivations on how groups were thought of

[delicate orchid]

this has no rigorous bearing I must emphasise

[topaz solar]

Aut(G) do be lookin like GL(R)

[delicate orchid]

yeah they're both automorphism groups

[delicate orchid]

this BECOMES rigorous if G = (Z/pZ)^k for some prime p

[topaz solar]

Abelian p-group time

[delicate orchid]

specifically because then G is the additive group of a field 😹

wait sorry I put my ^k in the wrong place

anyway redacted I kinda got distracted are u ok to tackle the exercise yourself now?

[marsh scaffold]

also on a side note for (b) is there like a geometric explanation to why we can realize D_4 like this

[marsh scaffold]

yeah i think ill try again after i got to see nnew
definitions

[delicate orchid]

yes actually - it's a square lol

hold on I'm trying to remember how B_n works

so you can realise all of its symmetries by reflecting the basis vectors I think, together with permuting the basis vectors

[acoustic igloo]

So like C_2 wr C_2 is isomorphic to <(12), (34), (13)(24)> ?

did i get that right

[delicate orchid]

yes that's exactly correct

[acoustic igloo]

thanks

[delicate orchid]

this holds in higher dimensions, the symmetires of a n-cube are isomorphic to C_2 \wr S_n

set n = 2 and we get C_2 \wr C_2

[marsh scaffold]

In 2 dimensions isn't permuting the basis vectors same as reflecting them?

[delicate orchid]

permuting them is a reflection along the line y = x

[marsh scaffold]

So different axis?

[delicate orchid]

the other reflections are along the lines x = 0 and y = 0

[marsh scaffold]

Okay 👍 thanks @delicate orchid

[rocky cloak]

D4 has normal subgroups that are not cyclic

[delicate orchid]

open problem

[rocky cloak]

Orthogonal reflections commute, and you can think of symmetries of the square (or n-cube) as permuting the axis then going some orthogonal reflections.

[marsh scaffold]

.

[cloud walrus]

[redacted]

[marsh scaffold]

And they're not the same, am i doing sth wrong?

Here's how they define wreath product #groups-rings-fields message

[rocky cloak]

I would think the definition should have been
(f, sigma)(x, y) = (f(sigma y)x, sigma y)
But maybe they're defining the semidirect product slightly differently than I'm imagining. Or maybe it's just a typo

[marsh scaffold]

This is how they define semi direct

[rocky cloak]

Yeah, that's how I would define it as well

[marsh scaffold]

So uh this is a typo in the wreath product definition?

[rocky cloak]

No, probably typo in the action

[rocky cloak]

Like it should be this

[marsh scaffold]

Oh

Okay

Wait so I also had to ask

Is this sort of definition of the action as a result is natural?

Like you know

Is the sth of a universal property kind of stuff to it

[rocky cloak]

I don't know if it's necessarily tied to any universal property, but it does make a lot of sense.

Like if you think of f as a tuple of element in G indexed by Y, then the action g.f is like maping x_y to x_{gy}. Which is pretty natural.

Like say f(y) = x and f is the identity for other elements of Y, then g.f(gy) = x.

[marsh scaffold]

I see, okay thanks a lot, actually I am seeing that this tuple way of defining the wreath product is much more natural

Idk why lenstra does it the way it is in these notes

[rocky cloak]

Well, the notation is a bit simpler with lenstras approach

Or more compact at least

[marsh scaffold]

Yeah ig,

Maybe I'll get more comfortable with practice

Anyways thanks again 😊

[glossy crag]

This is more #linear-algebra material, but I just wanted a quick check: I was idly recalling the construction of the determinant (for an arbitrary commutative ring R) via alternating forms M^k->R, for this you need to show the space of all forms has dimension n choose k when M is free of rank n. Clearly the coordinate functions for tuples i_1<i_2<...<i_k span it, and to show they're linearly independent you need to consider the alternating forms as a subspace of all multilinear forms, in which coordinate functions over all tuples (not only increasing ones) are independent just by plugging in values, same as for dual bases, right? Was it really as trivial as this? I remember it being more messy.

[trail cave]

https://www.youtube.com/watch?v=VdLhQs_y_E8&list=PLelIK3uylPMGzHBuR3hLMHrYfMqWWsmx5
In this intro video the prof. calls the operator on a group a multiplication operator multiple times but it doesn't seem to have the same properties as the multiplication operator of R^3 nor do I know how it would be distinct from an addition operator given the constraints. Is this just inaccurate language or is there actually a more general distinction between an addition/multiplication operator?

he defines it as just an operator which associates a third value to two others

[neat valley]

Thanks, but this is originally one part of a qualifying exam question, so presumably there is some way to do it in a reasonable amount of time without a computer or calculator.

But I can't seem to figure it out

[rocky cloak]

Well, let me know if you figure it out, because I'd be interested

https://math.stackexchange.com/questions/4723884/x6-69x5-−-511x-363-is-irreducible-over-mathbb-z

There's also this, with a similar methode using finite fields

[rotund aurora]

does this always happen for irreducible polynomials? as in, you can find two primes p and q where the factorization patterns can't match

[rocky cloak]

I think x^4 + 1 is a counter example. It's irreducible, but for every prime p, it either splits completely or splits as the product of two quadratic polynomials. So you can't really tell it's not the product of two quadratics just from the factorization pattern.

[neat valley]

This is simlar to what I did. I had a long solution which involved observing:

1. f factors irreducibly as x(x+2)(x^4+x^3+x^2+x+1) over F_3
2. f does not have any roots in F_5
3. f factors irreducibly as x(x+2)(x^4+x^3+9x^2+4x+3) over F_{11}.

Together 1 and 2 yield that if f factors over Z, it must do so as an irreducible quadratic times an irreducible quartic. 1 and 3 yield that

• the constant term a of the quadratic is divisible by both 11 and 3
• the constant term b of the quartic is equivalent to 3 mod 11
  Since 33 divides a and ab=363 = 3 * 11 * 11, it follows that a is either 33, -33, 363, or -363, so that b is one of 11, -11, 1, or -1. But none of 11, -11, 1, or -1 is equivalent to 3 mod 11. So f cannot factor over Z

The issue with this is that it involves factoring f over F_3 and F_{11}, which involves:

1. Finding the roots of f mod 3, mod 5, and mod 11
2. Performing polynomial long division mod 3 and mod 11 to get the remaining terms
3. Showing that x^4+x^3+x^2+x+1 is irreducible mod 3, and x^4+x^3+9x^2+4x+3 is irreducible mod 11.
   All of this is pretty computationally intensive, not to mention the fact that in practice you'd probably waste time checking other primes

[acoustic igloo]

is there a homomorphism from the symmetry group of a circle O(2) to D_4? i'm thinking the only homomorphism is the one that takes everything to the identity

[rocky cloak]

Well, O(2)/SO(2) = C2, and D4 has quite a few elements of order 2, so that would give you homomorphisms.

Other than that, every proper normal subgroup of O(2) is contained in SO(2), and SO(2) is divisible, so can't map to any torsion group. So there are no other maps.

[acoustic igloo]

oh so like
all the rotations that don't flip the circle could map to e in D_4, and all the rotations that do flip the circle could map to a 180-degree rotation in D_4?

that seems to work

thanks

[grizzled karma]

I’m trying to find out what part in particular you are referencing but that’s difficult since it’s such a long video. I think for the majority of the video he introduces groups using a specific kind of matrix [I think he says GLn(IR)] and its important to note that while multiplication is defined for this group, it’s matrix multiplication which does indeed have different properties than multiplication of say elements in IR. I’m not sure if this is what you’re referring to? Furthermore, it’s important to note that group operations can be more than just simple multiplication and addition, like how multiplication of two real numbers is different than matrix multiplication as mentioned above, but you can also have function composition as a group operator. Additionally, some texts use “multiplication” as a placeholder operation when defining groups (which is somewhat annoying because notationally it can lead to confusion) but it should be inferred as any binary operation between two elements of a group

[barren sierra]

0 ideas about this

Let G be a finite non-abelian simple group with all proper subgroups abelian. So let M be a maximal subgroup of G. I think the simplest thing may be show that G is not simple as my contradiction. All I've come up with in this direction is that N_G(M), the normalizer of M in G, is equal to M. I don't think this is useful though.

[sonic coral]

how would two maximal subgroups intersect

[barren sierra]

Well their intersection would be a normal subgroup of both.

but that doesn't necessarily say that the intersection would be a normal subgroup of the whole group (assuming it's nontrivial)

Ah but the two maximal subgroups are distinct and generate the whole group?

So indeed the intersection would be normal in the whole group

but G is simple so the intersection is trivial

[mighty kiln]

Why is this?

[barren sierra]

So intersection of any two maximal subgroups is trivial in this case.

Uhhh

actually good question

[mighty kiln]

The intersection of any two distinct maximal subgroups is indeed trivial though

[barren sierra]

how, if not by the way I tried to show?

[mighty kiln]

By considering centralizers of elements

But I haven't figured out how to proceed from this fact

[barren sierra]

I know that the intersection of all maximal subgroups is normal and so trivial

but not just intersection of two

[mighty kiln]

If M is maximal, and g ∈ M is nontrivial, what is the centralizer of g?

[dull ginkgo]

Nothing

Nothing in this textbook

Has been as brainmeltingly insane

As dealing with noncommutative matrices

I am still trying to fathom this shit

[mighty kiln]

The centralizer of g is something I'm sure

[barren sierra]

I mean what is it other than the centralizer of g lol

[mighty kiln]

It's M

[barren sierra]

huh?

oh yea

cause M is abelian, g in M, so everything in M commutes with everything in M

[mighty kiln]

And it isn't G since center is trivial

[barren sierra]

yea

[dull ginkgo]

conjugation action on the group?

Center is trivial

So then orbit stabilizer?

Let G act on itself by conjugation. G must have trivial center, so the action is faithful

No wait

[mighty kiln]

Oh right you can do orbit stabilizer on the set of maximal subgroups

[dull ginkgo]

Centralizers are the stabilizers

Orbits are conjugacy classes

[mighty kiln]

Idk what this gives you tho

[dull ginkgo]

oh well then normalizers ate the stabilizers

[barren sierra]

yea I don't follow this orbit-stabilizer approach

[dull ginkgo]

Okay let me state it

If there is no proper subgroups, then the stabilizer of each element (centralizer) must be either trivial or the whole group

So conjugacy classes are therefore either the whole group, or singletons

If the conjugacy classes are singletons, then gxg^-1 = x for each g. This can’t happen, since x cannot be central (center is trivial)

Wait

I’m pretty sure the conjugacy class could be the whole group (one conjugacy class) so we have no problem hhhh

[barren sierra]

yea

[dull ginkgo]

Wait that characterizes simple groups

lmao

[dull ginkgo]

Assume every subgroup of G as described is Abelian. Let K be a maximal (Abelian) subgroup of G. Then for any x in G\K, (K,x) = G. Thus x doesn’t commute with some element of K, call it y. But (x,y) must be a subgroup of K unless (x,y) = G and thus must be Abelian, so x and y commute.

[barren sierra]

that doesn't show that for arbitrary x, y that xy = yx

[dull ginkgo]

Actually made a small error

if (x,y) is abelian than it’s isomorphic to a direct product of cyclic groups, no?

[barren sierra]

should be, yea

[topaz solar]

If x in M, M < G is maximal and all maximal are abelian, then C_G(x) contains M yeah?

[barren sierra]

C_G(x) is M

[topaz solar]

Well yeah if G isn’t abelian (and x not 1)

[barren sierra]

which is the assumption

[topaz solar]

If we have two distinct maximal subgroups M, M’, then the intersection has to be trivial, since any point in the intersection has centralizer contain both M, M’

Any for any pair x, y in M, M’ both not 1, G = <x, y> :3

[barren sierra]

?

[topaz solar]

Since that’s contained in a maximal subgroup, but all maximal subgroups are abelian, and x and y don’t commute

Right?

[topaz solar]

<> here being taking as generators

[barren sierra]

no I know I just don't get how those two elements generate the whole group

[topaz solar]

since y not in C(x), they don’t commute

<x, y> is not an abelian subgroup

But all maximal subgroups are abelian so it can’t be contained in any

[barren sierra]

ah

[topaz solar]

Dunno if that helps, but it’s like a strengthening of what Miz said

So any pair x, y from M-1 x G-M generates G

:3

Which uhh similarly, conjugating points in M by points outside M does a lot, but that’s just being simple tbh

[barren sierra]

I would hate to be a student in 2021 getting this on my qual

[topaz solar]

Yeah I might take an L on this one without sitting on it for a while

If G is finite, we have Sylow subgroups right?

And, by simple, it’s not a p group so these are proper

That’s contained in some maximal M hence abelian

[barren sierra]

sure but Sylow subgroups aren't maximal and so I didn't look at the Sylow route further

[topaz solar]

Well, every Sylow subgroup is conjugate

But conjugating M which contains P splays everything everywhere, right?

Maybe

A Sylow subgroup of M > P is a Sylow subgroup of G because of properness

So something something, if you can fit em all in one M, that’s bad for being a proper subgroup

So some maximal subgroup misses some Sylow p-subgroup for some p

But uhhh conjugation throws things around like crazy, so uhh

RIP qed

Something like that work?

[topaz solar]

When in doubt, finite -> spam Sylow

[barren sierra]

I don't fully get what you said but I'll try to write out the details more and see if it comes together

[topaz solar]

Well I am known for incoherent responses

[topaz solar]

But you get what I mean by this right?

Since P < G so P \leq M < G

And since it’s full prime power in G, it is in M

[barren sierra]

yea

[topaz solar]

Not every Sylow subgroup can fit in this same M

Because otherwise cardinalities

[barren sierra]

it's the stuff after that that I don't fully follow

[topaz solar]

Conjugation of M has to throw stuff everywhere

Because again, <x, y> generates G for y not in M

So I’m thinking you have to be able to conjugate M all over to include that prime’s Sylow

[barren sierra]

what do you mean "throws stuff everywhere"

do you mean that every element in G is in some conjugate of M?

[topaz solar]

Well I’d hope something analogous happens

Or otherwise moving p-subgroups around into M

(Which is the same)

sorta

I dunno if that works though

But like, if we have maximal subgroups of wildly different sizes, that should be problematic, and all da p-subgroups are abelian

I’d be happy to not get this on a qual though

[barren sierra]

same

meaning that it will be on my qual

cause that's how life works

[topaz solar]

Do you see what I’m thinking?

I need to step up my algebra game tbh

[barren sierra]

Sort of

[topaz solar]

Uhh we have maximal subgroups of different sizes iff we can’t conjugate everything around because we can pass Sylow downward I think?

[barren sierra]

it's getting a little late so I may look at this tomorrow instead

but I think I get the vibe

[topaz solar]

And I do remember some downward passing of Sylow things but uhhh

[topaz solar]

Which, btw, this essentially depends on being finite, since Tarski monsters

[barren sierra]

Ok back to better stuff (rings!)

[cloud walrus]

Spamakin🎷

[rocky cloak]

I mean you do it in exactly the same way right.

Like take some non-zero element m, pick a prime ideal that contains Ann(m). Then m/1 is nonzero

Or is it too indirect for you that you have to use that every proper ideal is contained in a prime ideal?

[still sphinx]

https://crypto.stackexchange.com/questions/112469/sparse-packing-for-ckks Does anyone here have a clue about this one?

[dull ginkgo]

Noncommutative Matricies and Endomorphisms

[dull ginkgo]

Let’s say we have an indexing set $K$ and a ring R. Describe the ring $M_K(R)$ as an associative algebra over $R$ generated linearly by elements $e_{i,j}$ indexed by elements of $K$ such that: $e_{a, b} e_{c, d} = \delta_{b, c} e_{a, d}$ where $\delta_{b, c} = 1$ iff $b = c$ and is $0$ otherwise. \\

For a left R-module U, is $\mathrm{End}_R(U^{\oplus K}) \cong M_K(\mathrm{End}(U))$?

[cloud walrus]

The Library of Babble

[dull ginkgo]

I am not quite sure, i don’t think so?

[delicate orchid]

this is true I think

[dull ginkgo]

The proof for the finite case uses biproduct

I don’t think it generalizes because that has the implication there is a kernel for each endomorphism, no?

Because the e_{n,m} are maps that send the mth term to the nth term

[delicate orchid]

good point

hmm

yeah I don't believe it either anymore

[dull ginkgo]

Wait no lmao

M_K isn’t unital

Unless K is finite

Because the identity is 1 “along the diagonal”

[delicate orchid]

yeah that'll do it

LOL

[dull ginkgo]

lol yep

[delicate orchid]

if that's your definition of M_K anyway

[dull ginkgo]

The problem is just

[delicate orchid]

for K countably infinite would you define this as like, a limit?

[dull ginkgo]

I guess so?

But even with shit like the dual

We have one being a direct sum, the other a direct product

so there’s probably shenanigans like that here

It wouldn’t be finitely generated y’know

Eh whatever

I mean we can look at Hom(M^A, M^B)

[delicate orchid]

I think if you want only finitely many non-zero entries in each row it would have to be a colimit but w/e. That appears to be the accepted definition in the literature

[barren sierra]

Is this prime ideal not a maximal ideal? I feel this still uses maximality rather than "primeness"

[dull ginkgo]

Yeah

[rocky cloak]

It doesn't have to be maximal, just any prime ideal

You can of course choose it to be maximal if you want

[dull ginkgo]

Is M_p like M/p

Or is it like localization

Mp^-1

[barren sierra]

Localization

[dull ginkgo]

Well then can’t you look at units

units don’t lie in any proper ideal

And look at what happens to them, no?

Elements inside the ideals will be forced into the annihilator

Actually I’m a fucking idiot

[delicate orchid]

WOAH! No swearing...

[dull ginkgo]

If M is 0 as an R-module, then it’s 0 as a module for any subring of R right

If M is so for Rp^-1 (which R injects to) wouldn’t it be 0

[rocky cloak]

And the existence of prime ideals is actually weaker than the existence of maximal ideals (which is equivalent to the axiom of choice)

[dull ginkgo]

literally you can use R into Rp^-1 into End(M) (zero map)

[rocky cloak]

So you can prove this equivalence in a slightly weaker system than ZFC

[dull ginkgo]

So as long as there is a maximal (prime) ideal then

[dull ginkgo]

ultrafilter lemmon

Or Boolean ideal theorem or whatever it’s called

Boolean prime ideal theorem lmao

[delicate orchid]

it's just zorns dawg

[dull ginkgo]

The weaker one

I will keep asking this periodically

For the exterior algebra over the space of smooth sections of the cotangent bundle

We can extend the original derivative “gradient” mapping to an exterior derivative

Can we extend the original derivative gradient mapping to a “tensorial derivative” on the more general tensor algebra where when quotiented by (x (x) x) gives the exterior algebra’s one

What i mean by gradient mapping is that: V -> V^dim(M) where f(v) -> (f_x(v), f_y(v)…)

[topaz solar]

Yes

[topaz solar]

As long as you’re unital anyway

[dull ginkgo]

Actually i did it incorrectly

Actually differential forms are weirder than I thought

Nvm

[delicate orchid]

maybe ask this in #diff-geo-diff-top rather than fuckin #groups-rings-fields

[dull ginkgo]

I was originally working with a purely algebraic formulation

i think Jacobson actually covers it in later chapters

I just forgot that diff forms aren’t exterior algebra elements explicitly, moreso functions to them

I am now pondering

Assume we have a commutative ring R and a module M

then we can define the tensor algebra ring T(V)

[south patrol]

Where V = M lol

[dull ginkgo]

Yeah sorry

Meant V

[south patrol]

Dw

[dull ginkgo]

Anyway

Certain quotient rings like the exterior algebra

Are finitely graded if M is free of finite rank

I wonder when that happens for other quotients

[delicate orchid]

obvious thing to try is studying quotients by a certain quadratic form

clifford algebra type nonsense

[dull ginkgo]

Jumpscare

[dull ginkgo]

One more thing

[delicate orchid]

https://tenor.com/view/columbo-one-more-thing-personal-gif-1028468180046162187

[dull ginkgo]

So for a map between R-modules $U \rightarrow V$, it extends to a ring map $\wedge(U) \rightarrow \wedge(V)$ that also allows linear maps $\wedge^n(U) \rightarrow \wedge^m(U)$ is there an easy way to describe these maps

[cloud walrus]

The Library of Babble

[dull ginkgo]

Graded ring map

[delicate orchid]

for free modules, yes 😹

[dull ginkgo]

never said they were free

[south patrol]

Every element can be written as a sum of v_1 ... v_n and that is sent to f(v_1) ... f(v_n)

[dull ginkgo]

Can we just say the extension distributes over the exterior product

Without problem?

Well if we have like vectors u, v then if f(uv) = f(u)f(v) then if f(u) = f(v) then it’s 0, hm

Kernel might be interesting

[south patrol]

What do you mean

[dull ginkgo]

Nevermind

[south patrol]

Idk if extension is the right word here like the induced map ig

But sure yes

But that is just saying it is a ring map

Unless i am misinterpreting you

[dull ginkgo]

Yeah

[south patrol]

And yeah V generates the exterior algebra

[dull ginkgo]

I think any morphism of U to V can be extended to a morphism between the exterior algebras preserving the subspaces of 1-forms

[dull ginkgo]

If M is of finite rank n, and endomorphism T maps M to a submodule of strictly lesser rank, then det(T) = 0 right?

[south patrol]

Yes

[dull ginkgo]

Okay i proved it

[south patrol]

Nice

[rocky cloak]

Are you over an integral domain?

[south patrol]

Though what do you mean by rank? Are you over a domain or smth or is this free

[dull ginkgo]

image of the map M to im(M) extends to \wedge^n(M) to \wedge^n(im(M)) ~= 0

[dull ginkgo]

Free

[south patrol]

Sure

Again I'd not say extends

[dull ginkgo]

Well

[south patrol]

M doesn't embed into its nth power

[dull ginkgo]

Functorially

[south patrol]

Induces ig

Yeah like the map on top powers factors through 0

[dull ginkgo]

Yeah thanks

Yeah

Which is iso to R via End_R(R) ~= R^op = R

[rocky cloak]

But if like R=k[x]/x^2 and M=R, then the map given by multiplication by x has image with smaller rank

Or are you saying the image is free?

[dull ginkgo]

If the image is free

For PID every submodule is free i think

Well you could also use finitely generated submodule generated by less elements than the basis

[south patrol]

Yeah

[dull ginkgo]

If M is free with basis (b_i), then can we describe the “minors” of an endomorphism on M with the induced linear maps on the exterior powers

[next obsidian]

yeah

I think

[delicate orchid]

who is this guy?

[next obsidian]

let me think uhhhh

[delicate orchid]

it's gotta be the exact same mechanism as for vector spaces right

[next obsidian]

yeah

[delicate orchid]

like what's the difference libtokers

[next obsidian]

I'm just trying to recall how you do it

[rotund aurora]

there are many minors, only n exterior powers. Why should this be true? Also, I don't think minors are preserved under change of basis. Some relations between them will be preserved, but not the minors themselves

[dull ginkgo]

Well, i had to be specific with my definition of minors

It’s a very specific case

I also computed what i was looking for lol

[dull ginkgo]

If $T$ is the transformation, and you have the basis $(e_n)$ then the $(i,j)$-th minor is the scalar of $\bigwedge_{k \neq i}{T(e_k)}$ at the basis vector $\bigwedge_{k \neq j}{e_k}$ in $\wedge^{N-1}(M)$

[cloud walrus]

The Library of Babble

[dull ginkgo]

Minors are relative to a basis because they are components of a vector in the exterior algebra

The determinant is an exception because of the space it’s a vector of being 1D

The “minor’s” as described usually for matrix computations are “all but (i,j)”

[trail cave]

Thanks for the response, I should have linked the specific part I was referring to. At 36 minutes in he defines "A group G is a set with a product operation g*h (read g times h)" with some properties. Then at 39:40 he clarifies "when I say a product operation I mean that for any two elements g, h in G, there's a product gh in G". So, it sounds like it may just be that a weird way to refer to a binary operation closed in G

It confused me because it seemed like the operation on groups was more general and could include addition

[grizzled karma]

Yes. I encountered a similar issue when I took this as a class last semester. In some texts they clarify that it’s some binary operations which may be multiplication, addition, or something else. Others it may be written in a way that appears like multiplication but it should be understood that it’s some operation. A few texts I’ve seen like to use ✱ which is supposed to be a universal binary operation symbol. It’s implied to be multiplication or addition based on context.

[primal beacon]

it does! (include addition)

[hidden wind]

i forget what's that website that lists information about lots of groups

[tribal moss]

Groupprops?

[hidden wind]

yes that one thanku very much

[scenic lake]

does multiplication mod 4 mean (a x b)mod4 or (a)mod4 x (b)mod4 here

[wraith cargo]

The former

[dull ginkgo]

What do yall think is the best way to go about this

if eta is surjective and (e_i) is a basis of R^(n), then u_i = eta(e_i) is a set of generators

[wraith cargo]

What is R^(n) here

[dull ginkgo]

R to the power of n

i.e free module of rank n

[wraith cargo]

okay well then just check injectivity

and use the linearity of eta

[dull ginkgo]

That's what I'm trying to verify

But i think I have an idea

[wraith cargo]

well start by assuming you have two elements a and b of R^n that map to the same thing
So you have that eta(a-b) = 0 and now use the fact that every x in R^n can be written as a sum a_1u_1+...+a_nu_n

[dull ginkgo]

...

Not uniquely

we didn't show that yet

[wraith cargo]

okay ignore that detail then

[dull ginkgo]

I think it's relatively easy now that I think about it

Assume we chose u_i such that eta(u_i) = e_i

by linear independence of e_i

$\sum a_i e_i = 0$ if and only if $a_i = 0$

[cloud walrus]

The Library of Babble

[dull ginkgo]

$\sum a_i e_i = \eta \left( \sum a_i x_i\right) = 0$ if and only if $a_i = 0$.

[cloud walrus]

The Library of Babble

[dull ginkgo]

Now assume the $x_i$ are linearly dependent, then $\sum a_i x_i = 0$ for some $(a_i)$, and thus $\eta \left(\sum a_i x_i \right) = 0$ implying $a_i = 0$, thus proving linear independence (granting injectivity)

[cloud walrus]

The Library of Babble

[dull ginkgo]

@wraith cargo think this works?

Let me refine it really quick

[wraith cargo]

so you basically argued if you have some x = sum a_ix_i that maps to zero this implies that eta(x) = sum a_i eta(x_i) and by the linear independence of eta(x_i) we have that a_i = 0 so x = 0

[dull ginkgo]

yeah basically

Assume $\eta \in \mathrm{End}_R(R^n)$ is surjective, and fix a basis $(e_i)$. Then $\exists x_i : \eta(x_i) = e_i$. \\
Assume $0 = \sum c_i x_i \Rightarrow 0 = \eta \left( \sum c_i x_i \right) = \sum c_i \eta(x_i) = \sum c_i e_i \Rightarrow c_i = 0$ thus $x_i$ are linearly independent. The map $\zeta : e_i \mapsto x_i$ is thus an inverse.

[cloud walrus]

The Library of Babble

[dull ginkgo]

actually this doesn't depend on R being commutative

lmao

[dull ginkgo]

SOOOOOO I am confused here

If f_i form a base for a free submodule K of R^n

then f_i are linearly independent

meaning that the map A_e is injective?

I'm in general confused as all hell

Assume $f$ is an injective endomorphism of $R^n$, then let $(e_i)$ be a basis.
Assume $\sum c_i f(e_i) = 0$, then $0 = f \left( \sum c_i e_i \right)$ and by injectivity: $\sum c_i e_i = 0$ and by linear independence, $e_i = 0$ for all $i$, so $f(e_i)$ is linearly independent

[cloud walrus]

The Library of Babble

[dull ginkgo]

oh ic

[dull ginkgo]

Yeah I keep running into the issue for when the adjugate matrix is 0

fuck it, using exterior powers

[chilly radish]

Why? If det A is a nonzero divisor the adjugate can't be 0 since multiplying it by A gives det(A)I

[dull ginkgo]

I got tired and just used the functorial map on the top exterior power (iso to R) being injective (and multiplication by an element r is an injective map iff it's not a zero divisor)

[coral prawn]

If H is a subgroup of G, and G has presentation <S | R_1>, does H have presentation <S | R_1, R_2> for some relation R_2 in general?
Also, if H is normal in G, what is the presentation of the quotient group G/H?
(Free groups still causing me grief 😢)

[mighty kiln]

Adding more relations means taking a quotient

The quotient group is just <G | H>

[tribal moss]

The quotient group G/H can be presented as <S | R1, R2> where R2 is a set of (preimages of) generators of H. (But this is not necessarily a minimal presentation in any way).

Since <S | R1, R2> is always a quotient of G, you can't hope to make an arbitrary subgroup in that way.
For example the alternating group A_3 is a subgroup of the symmetric group S_3, but it cannot be a quotient because S_3 doesn't have any normal subgroup of order 2.

[cobalt heath]

Where do I ask about e.g. determinant of block matrix, and trace of discrete finitely generated subgroup of SL2C?

[tribal moss]

The usual question about "determinant of block matrix" is whether the determinant has a nice expression in terms of the block, the same way a product of block matrices can be written using the blocks. The answer to that is no: We can consider block matrices to be matrices with entries in a matrix ring. Even though that construction stacks well enough for products, the trouble is that determinants don't work well over a non-commutative ring. So there's not even any determinant concept on the upper level that we can contemplate stacking with determinants at the lower level.

"Trace of discrete finitely generated subgroup of SL2C" smells of representation theory, which would be #advanced-algebra.

[tough raven]

It ought to: there are rings such that R^m is isomorphic to R^n as R-modules for some m < n; if you map R^m bijectively to R^n, then map R^n to R^m by (x_1, ..., x_n) -> (x_1, ..., x_m), then you'll get a surjective but non-injective map from R^m to R^m.

[tough raven]

I don't think that this proves that the x_i generate R^n.
(Indeed, your proof doesn't even use the fact that the basis is finite, but this is false for an infinite-rank free R-module, for example the map (x_0, x_1, ...) |---> (x_1, x_2, ....) on \oplus_N R.)

[tough raven]

(This cannot happen for commutative rings though, as ||they have the Invariant Basis Number property that different finite bases of a module must have the same size||. Using that, you can finish the proof.)

[cobalt heath]

Yeah, that's the issue - determinants get kind of screwy on noncommutative ring.

I was hoping for something workable where elements are SL(2, C), as that is roughly what I am working on now.

Anyway, let me go to #advanced-algebra . Thanks!

[dull ginkgo]

I corrected my proof

[dull ginkgo]

Also if you want an example of this:
R[x]^2 ~= R[x] as both have a countable basis.
Thus let K = End(R[x])
K ~= Hom(R[x], R[x]^2) ~= Hom(R[x],R[x])^2 ~= K^2 as K-modules

[dull ginkgo]

eta is surjective, so for each x, x = \sum c_i e_i = eta(c_i x_i) and due to linear independence of the x_i, it’s a bijection

[tough raven]

This shows that x = eta(a linear combination of the x_i), but not that x = a linear combination of the x_i.

[dull ginkgo]

It’s basically stating that it’s injective and surjective

Injective because of linear independence of the x_i, surjective because of the problem statement

[tough raven]

I can't find this.

[dull ginkgo]

I found a simple counter example for part two: Z -> 2Z as Z modules is an injection

[tough raven]

eta is injective on span(x_1, ..., x_n), not necessarily on the entire domain R^n (unless x_1, ..., x_n generate R^n as an R-module).

[dull ginkgo]

what do you mean,

Span(…) is the image

So it’s have to be linearly independent on the entire domain

[tough raven]

No, x_1, ..., x_n are elements of the domain, and span(x_1, ..., x_n) is a submodule of the domain.

[dull ginkgo]

How could it be linearly dependent over R^n then

Since we are only considering sums in that span

[rocky cloak]

What is even the original thing you're trying to prove here?

[dull ginkgo]

Finite rank modules: surjective endo is auto

[tough raven]

This.

It seems to me like it should need IBN for commutative rings, which hasn't been used so far.

[rocky cloak]

It seems all you've proven so far is that it splits. Which you could have just proven from free modules being projective

[tough raven]

I'm not saying that x_1, ..., x_n are linearly dependent elements of R^n. I'm saying that (a priori) they might not span it.

[tough raven]

To elaborate more concretely: your inverse phi: e_i |----> x_i satisfies eta \circ phi = id_codomain, but you haven't shown that phi \circ eta = id_domain.

If you can't see the problem we're trying to point to, I suggest trying to apply your proof with "n" infinite, specifically to the map
eta: R^N -> R^N: (r_0, r_1, ...) |--> (r_1, r_2, ...)
(where R^N = {(r_0, r_1, ...) | all but finitely many r_i are 0}) and seeing why it doesn't show injectivity of eta.
(You can take e_i to be the sequence with a 1 in the i^th position and 0 elsewhere, and you can take x_i = e_{i+1}.)

[rocky cloak]

You can use the adjugate matrix to finish the proof of though. Or just apeal to IBN I guess

[dull ginkgo]

Haven’t talked about projective modules or splitting lemma yet

[delicate orchid]

ok then prove the splitting lemma for free modules and then use it

[rocky cloak]

But you know about determinants, and how unit determinant implies invertible matrix, yes?

[dull ginkgo]

[dull ginkgo]

That’s the next exercise

[rocky cloak]

Then I'm wondering what the intended methode is

[dull ginkgo]

No idea, it’s Jacobson

[tough raven]

Interesting book .

[rocky cloak]

What about localization?

[dull ginkgo]

Not sure if that’s the intended route

I’m also not following why my proof for the surjective part doesn’t work

[tough raven]

To prove that it's injective?

[dull ginkgo]

Yes

The other way around I fucked up

[tough raven]

If I give you an x such that eta(x) = 0, you haven't shown that x = 0.

You have shown that if x = \sum_i c_i x_i for some c_i, then x = 0.

[dull ginkgo]

Ohhh I see

[tough raven]

But not that any x in R^n has to be of the form \sum_i c_i x_i for some c_i.

[dull ginkgo]

I might do a Cayley-Hamilton esque proof of this

Do it over M^rank(M)

[tough raven]

Not sure how that would work myself, but great if you can.

[rocky cloak]

You have Cayley Hamilton, but don't know about determinants?

[dull ginkgo]

We don’t have it, but I did prove End(M^n) ~= M_n(End(M))

Of which you can then work with the matrix ring embedded into it :3

for finitely generated more generally

Instead of free

I am actually gonna be smart

I’m going to prove the next problem first

and use it lol

I spent a while proving functorial bullshit about the tensor algebra yesterday so I’m going to use that

[delicate orchid]

what bullshit

[dull ginkgo]

Homogeneous ideals of the tensor algebra for modules

[mighty kiln]

Functorial

[dull ginkgo]

Assume we have a commutative ring R. Let U denote the free monoid in countably many letters

Assume M is an R-module
Immediately, we have the R-algebra R[U], the monoid algebra, which also can be equipped naturally as a ring with a degree grading.

Now let’s say we have a monoid M, and we then have the tensor algebra T(M) as a graded ring.

Like the polynomial ring, we can map R[U] to T(M) through valuations of M due to the universal property of R[U] and free-ness (as a monoid) of U. These are actually graded ring / associative algebra morphisms

Assume we have a homogeneous ideal K of R[U], then we can consider the generated ideal T_K of the images of K under all valuation maps to T(M). T_K is a homogenous ideal of T(M), thus T(M)/T_K is a graded ring

There’s some interesting bullshit you can do with this

[tough raven]

M is an R-module, IG?

[dull ginkgo]

Yes sorry, accidentally wiped that

[tough raven]

Makes sense.

[dull ginkgo]

This is actually really because T(M) is a quotient of R[U] lol

[tough raven]

No I meant
I assume you're starting with an R-module M and not a monoid?

[dull ginkgo]

Yes sorry

U is the monoid

Hence the brackets

Actually now that I think about it

[tough raven]

So for example, IG you'd take the ideal of R[U], for U free on {x,y}, generated by xy - yx, and then T(M)/T_K would be T(M) but with x and y (whatever you chose to map them to in M) forced to commute?

Although honestly it seems simpler to directly consider the two-sided ideal of T(M) generated by xy - yx (which is homogeneous because it's generated by a homogeneous element)...

[dull ginkgo]

Like for the exterior algebra, K = (x^2),

[tough raven]

Oh, under all valuation maps.

Yes, sure.

[dull ginkgo]

Generated by all valuation maps yes

Or in other words, the valuation maps then the quotient map by K is always 0

:3

(Which is where we get the universal property we are looking for btw)

[tough raven]

Can you characterise Clifford algebras (x^2 = Q(x) for a given quadratic form Q) or universal enveloping algebra (xy - yx = [x, y] for a given function (x, y) |---> [x, y] making M into a Lie algebra) in this format?

[dull ginkgo]

maybe :3

[tough raven]

It seems to depend on more structure than just a module, so I'm not sure it's possible with a single homogeneous ideal of R[U].

[dull ginkgo]

The quadratic form is a homogenous element of degree 2 in R[U]

Eg: 3xy + 4x^2 + 7z^2 + 4yx

[tough raven]

Not quite the same thing

But anyway

[dull ginkgo]

The commutators [x,y] = xy - yx are degree 2

[tough raven]

That was just a tangent.

[tough raven]

Continue.

[dull ginkgo]

So they generate a homogenous ideal

K = ([x,y])

The quotient T(M)/T_K is the symmetric algebra

So there’s some rather useful bullshit here I guess

More of a fun universal property exercise though

[dull ginkgo]

Q(x) - x^2 is a homogenous polynomial of degree 2, no?

Wait Q(x) is a map from M -> R

[dull ginkgo]

I think you can if you’re smart about it actually

x^2 - Q(x)1, Hm

actually no

[coral prawn]

Oh, that explains a lot, thank you!

[dull ginkgo]

You can obviously do T(M)/(x^2 - f(x)) at your leisure but that’s different than this construction

I wonder if you can construct T(M) off of R[U]

Let M be an R-module
U = Free_grp(Forg_set(M))
Let K be the ideal of R[U] generated by elements of the form a(v + u) - av - au, auv - (av)u, auv - v(au), u(v + w) - uv - uw, (u + v)w - uw - vw

Then R[U]/K ~= T(M)

Annoying.

[tough raven]

R[FreeMonoid(S)] = T(FreeRModule(S)), since both have the same universal property.

[dull ginkgo]

Yeah that’s a subcase

[dull ginkgo]

You just need a fuckton of relations, but ironically K is a homogenous ideal

So that explains why the tensor algebra is graded from that construction

lmao

[tough raven]

If M is generated by S subject to relations R', you can take R[FreeMonoid(S)]/(R'), where each relation R' is viewed as a degree-1 element of R[Free(S)].

[dull ginkgo]

Yeah

FYI I know almost nothing about this kinda stuff and it’s me mostly pulling shit out of my ass that seems to follow relatively intuitively

[coral prawn]

I was trying to work it out on my own and stumbled across this - kind of a question about functions in general.
Given an injection f: X -> Y, does there exist a surjection g: Y -> X?
I think the answer is yes but I’d like to make sure. My thought process went something like this: If f is injective, Y must have at least as many elements as X, which implies that X must have at most as many elements as Y. This is the same as saying there exists a surjection g: Y -> X since X must have at most as many elements as Y.
Is this true in general? It would imply that there always exists a homomorphism from a group to its subgroup, which ties this into my previous question.

[dull ginkgo]

Its universal property is basically saying K is in the kernel of every type of special morphism you want (e.g linear, symmetric, antisymmetric, etc.)

[tough raven]

The answer is yes (for sets) unless X is empty and Y is non-empty.

[dull ginkgo]

And then there’s rings and Z into Q is epi but not surjective because fuck you

[coral prawn]

Ah that checks out, thanks

So does that imply that there’s always a homomorphism from a group to its subgroup?

[tough raven]

Using notions like "as many elements as" is a bit dicey if X and Y can be infinite, since those notions are usually formalised using existence of injections and surjections (in other words, this lemma is a prerequisite for using those phrases with their intuitive meanings as you have, making the argument circular).

[tough raven]

Definitely not a surjective homomorphism from any group to any subgroup. There will exist a surjective function, but there's no reason for it to be possible to choose it to also be a homomorphism.

(On the other hand, if you don't want the homomorphism to be surjective, then certainly - "map everything to the identity" defines a valid homomorphism between any two groups.)

[tough raven]

The group S_3 and the subgroup of 3-cycles, which has order 3, is a counterexample.

[dull ginkgo]

@tough raven I have an idea

[coral prawn]

That makes sense, I totally ignored the homomorphism structure lmao

[dull ginkgo]

Assume we have ring R, and a free group F in finitely many letters.

Then if R is noetherian then so might R[F]

Because we have the multiplicative map R[F] -> R that sends f(x1… xn) to the sum of it’s leading coefficients

[tough raven]

I have to ask: is there a particular goal all of this is leading up to?

[dull ginkgo]

And if we throw a left-ideal of R[F] into that map, it’s image is an ideal I think

[dull ginkgo]

R[G] for finitely generated G is noetherian iff R is

[tough raven]

Well, if it was the free commutative monoid, then yes, R[F] is Noetherian (Hilbert's basis theorem).

[tough raven]

But why do we care about that?

[dull ginkgo]

That leading coefficient map being an ideal map is how you can prove Hilbert basis

[shell pilot]

After proving that $\phi$ is a homomorphism, how do you compare the kernels of $\phi$ and $\sigma\phi$?
My initial thoughts are comparing the orders of them because the kernel of a mapping is a subgroup of the "domain" group in the mapping. i.e. if $\phi : G\rightarrow H$ then $Ker\phi\leq G$.

[cloud walrus]

Soap_Opera

[dull ginkgo]

Let f and g be two maps, and the preimage of a set be denoted f^-1(S). Show (f \circ g)^-1(S) = g^-1( f^-1(S))

The kernel is the preimage of {e}

[shell pilot]

Here is what I understand from the definition...$ker\phi={x|\phi(x)=e}$ and $ker\sigma\phi={y|\sigma\phi(x)=e}$ so by that definition, we know that $ker\phi\in ker\sigma\phi$

[cloud walrus]

Soap_Opera

[tough raven]

Yep (although technically you should write \ker \phi \subseteq \ker \sigma \phi).

That's probably all they want for the first part.

[shell pilot]

Ah, good point

[shell pilot]

I'm still thinking about how to answer this. Was not ignoring your hint!

[dull ginkgo]

Write out using set builder notation :3

= {x : g(f(x)) in S}

[shell pilot]

$\ker\sigma\phi={g\in G | \sigma(\phi(g)) = e}$

[cloud walrus]

Soap_Opera

[dull ginkgo]

I.e u in ker(sigma) such that u = phi(g)

[shell pilot]

Oh, good idea

[dull ginkgo]

The very definition of phi^-1(ker(sigma)) :3

[dull ginkgo]

If you want some excitement in your life

1. The preimage of a subgroup of the codomain is a subgroup
2. The preimage of a normal subgroup is normal if the map is surjective

[delicate orchid]

good exercises

[dull ginkgo]

A common thing is “preimage of algebraic subobject under map is also the same type of object”

For algebra sorta objects

Like topologies are algebras in a way with union and intersection

[shell pilot]

Ok, so using your idea, $\ker\sigma\phi={g\in G | \sigma(\phi(g)) = e}={g\in G | \phi(g)\in ker\sigma}$ which would be the "preimage of the kernel of $\sigma$, but then how does that lead to comparing the kernels of each homomorphism?

[cloud walrus]

Soap_Opera

[tender wharf]

suppose g is in ker phi

can you say anything about sigma phi

[shell pilot]

If g is in ker phi then I'm not sure what can be said about sigma phi from there because it's not necessarily in the ker sigma phi

[tender wharf]

think about whats the definition of ker phi

[shell pilot]

ker phi is the set of all elements in G that map to the identity in H

[tender wharf]

right

where does sigma take the identity

[shell pilot]

Ugh, homomorphism means identity maps to identity of K

[tender wharf]

so if g is in ker phi then phi(g) is the identity yes

[tender wharf]

oh you already had this

[shell pilot]

So every element of ker phi is in ker sigma phi

[tender wharf]

yeah

[tender wharf]

yeah

[shell pilot]

From the first isomorphism theorem we have that $|\phi(G)|$ divides $|G|$ and $|H|$

[cloud walrus]

Soap_Opera

[shell pilot]

But still not sure how to "compare" the kernels of the homomorphisms

[dull ginkgo]

wdym

[shell pilot]

The question says how are Ker phi and ker sigma phi related

[tender wharf]

subseteq

[dull ginkgo]

Yes

[shell pilot]

Is ker phi a subgroup of ker sigma phi technically?

[tender wharf]

You said it yourself here

yea

[dull ginkgo]

Would you like intuition?

The kernel of the composition is bigger than the kernel of the inner map

[tender wharf]

everything phi sends to the bin will be sent to the bin by sigma, but sigma could also send more things to the bin

[dull ginkgo]

This yes

The inner morphism kernel is sent to 0, but the composition kernel is just sent to the kernel of the SECOND map (a bigger set)

[shell pilot]

Ok, but if the second map is one-to-one, wouldn't it technically be smaller or the same size as the first map?

By the way, I'm not arguing with you here because you both know algebra way more than I do, I am genuinely trying to learn haha

[tender wharf]

No worries nobody is interpreting that as an argument

[tender wharf]

yes injective maps would have trivial kernels

so if sigma is injective nothing nonidentity is sent to the bin

[shell pilot]

Thank you @tender wharf and @dull ginkgo for your help!

[lean sail]

has anyone read Lara Alcock's: how to think about abstract algebra? it looks like a good supplement to a standard abstract algebra textbook, something to help some of the concepts sink in a little further

[marsh scaffold]

hey i need some help on this

so we know that all G sets split into orbits

so they can be recognised as disjoint union of orbits

also we can split the e^sum into prod e^term

and expand each e^term into a sum

the only problem is i dont understand what the factorials do in the coefficients

like up until now i am not exactly clear on what we are trying to count

[dull ginkgo]

If we have the free monoid in finitely many characters, can we equip it with the lexicographic order, and is this order left/right multiplication invariant (depending on choice of the order direction)

[delicate orchid]

u want a < b => xa < xb right

then yeah this is fine

[mighty kiln]

I think you can just expand the right side and everything works out

[marsh scaffold]

okay maybe after the deleted reply of dary i think the factorial divisions rule out the repeated G/H types in the disjoint union

[mighty kiln]

Not only that

[marsh scaffold]

expand !!?

[mighty kiln]

Any G-set with k orbits will be contributed to by k! terms on the right

[dull ginkgo]

Then I’d have to read it from right-> left

[marsh scaffold]

wait uhh

[mighty kiln]

Unless there are repeated orbits

In which case the left side gets smaller by the same factor

[marsh scaffold]

after writing e^sum as product e^term we can write each e^term as sum stuff

[delicate orchid]

I'm defining the lexicographical ordering as "a = a_1...a_n, b = b_1...b_m, a < b iff a_i < b_i for the smallest i s.t. a_i \neq b_i". Using this you can read it left to right

[marsh scaffold]

now that stuff corresponds to the possibilities if isonorphic copies of orbits in the disjoint union right

so like we are just avoiding repition by dividing by that factorial?

[mighty kiln]

No

[marsh scaffold]

okay

[dull ginkgo]

The abject horror in the fact that you can compute Gröbner basises over R[F_n] where F_n is the free monoid in n characters

[delicate orchid]

keep ur alg-geo nonsense out of my beautiful channel

[mighty kiln]

Consider the Z/6-set Z/2 ∪ Z/3

[dull ginkgo]

Never said it was commutative :3

[marsh scaffold]

okay

[mighty kiln]

This can be Z/2 ∪ Z/3 or Z/3 ∪ Z/2

[delicate orchid]

neither did I?

[mighty kiln]

Which are two different terms on the right

So you need to divide by 2!

[marsh scaffold]

understandable

but what im doing is

splitting stuff up into isomorphic classes

so z/2 would be in a e^term series

or maybe im doing sth wrong

[delicate orchid]

just as an aside this question fills me with fear

[mighty kiln]

It's just unpacking

[marsh scaffold]

why it looks pretty cool

[delicate orchid]

oh nvm it's just transitive G-sets

[marsh scaffold]

yeah

[mighty kiln]

The coefficient of t^n on the right is the sum of ∏_k 1/[Aut(Xk) m!] over ordered m-tuples of transitive G-sets with total n elements

(then summed over m)

[marsh scaffold]

ah i see

gah damn

thanks darnymuckhi

[dull ginkgo]

This website has a proof of hilbert basis theorem

Is this correct, I am a bit hesitant

[marsh scaffold]

is that aops

[dull ginkgo]

Yes

Because the m_i need not be increasing

You can slightly modify it though for it to be fine

But this proof seems incorrect

[dull ginkgo]

Particularly the last part about being for any j

[delicate warren]

Wish me luck, link or give any important starter info I should know before diving in

[coral spindle]

Looks like a good intro book, judging by the contents

[chilly radish]

A lot of people don't like Gallian

[wraith cargo]

I read Gallian a long time ago when I was starting out and didn't like it but I don't remember why anymore lmao

[chilly radish]

I never read it

[wraith cargo]

I'm forever a Dummit and Foote apologist 🫡

[languid saddle]

what about aluffi

[coral spindle]

Gallian looks like Fraleigh but slicker

[hollow imp]

artin seems pretty good

[lone niche]

iirc group actions or something else were done weirdly in Gallian (maybe it was fraleigh idk). I had to TA from that book once.

[dull ginkgo]

jacobson mfers

[barren sierra]

How do I show that all elements of the automorphism group of S5 has only inner automorphisms?

[rotund aurora]

This is the outline Rotman follows:

• If sigma in Aut(S_n) preserves transpositions then sigma is an inner automorphism

[barren sierra]

no idea how I'd think of that but I'll try to show that and then ig show all automorphisms preserve transpositions

[wraith cargo]

We had this as HW problem and I think it came down to showing Out(G) is trivial via an Out(G)-action on the conjugacy classes of G

Oh wait no I had those mixed up

I think what croqueta said is the way to go

this is for another problem lol

https://mathweb.ucsd.edu/~asalehig/math200a-22-f-hw5.pdf
This HW set gives a way to show if for n>6

[rotund aurora]

If you can derive that all automorphisms are inner (n!=2, n!=6) from what I said it is clear how one could have thought about it

well yeah you can just prove that automorphisms preserve transpositions (n!=2, 6)

[wraith cargo]

I think Aut(S_2) = Inn(S_2)
It isn't true that Aut(S_n) = S_n for n=2,6 tho

[lean sail]

im going through this book right now

[delicate warren]

im going through the Chapter 0 and havent even touched groups yet

[lean sail]

same here, I am on problem 19

[delicate warren]

I tried my hand at a proof though

[topaz solar]

Well, brute forcing by listing all automorphisms works

[tardy hedge]

have fun abstract algebra is awesome

[vagrant zinc]

I have it too, I am progressing very slowly in this area of mathematics.

[grizzled karma]

My first semester of abstract only went up to chapter 11. Hopefully that gives some context on a good pace

[delicate orchid]

No, Aut(S_2) is trivial and Aut(S_6) = S_6 x C_2

Oh nvm u said isnt. I shouldn’t post while intoxicated

[tropic spade]

Trying to understand some wording. In the quote p is the permutation representation of an action . of a group G on a set A: "In particular an action of G on A may also be viewed as a faithful action of G/ker(p) on A." They're saying the map sending (g ker(p), a) to (g . a) ker(p) is a faithful action?

Or do they mean something else?

[coral spindle]

The action rho : G -> Sym(A) has a corresponding action rho’ : G/ker(rho) -> Sym(A) defined by rho’(g ker rho) = rho(g). It is faithful.

This is also known as the first isomorphism theorem

Nerd emoji

[tropic spade]

Lmao

That makes sense.

Thank you 🫡

[acoustic igloo]

would it be correct to say that if G has no normal subgroups, an action of G on A must be either faithful or trivial?

[rotund aurora]

An action of G on A is given by a homomorphism G-->Sym(A)

[acoustic igloo]

yes

[rotund aurora]

the kernel is a normal subgroup

[acoustic igloo]

so the kernel must be G or the identity of G

i think that means my statement was correct

right?

thanks

[sly crescent]

No, Aut(S_6) = S_6 : C_2

[smoky grove]

A field is it should be commutative on both operations?

[rain grove]

Let $H$ and $K$ be finite subgroups of $G$. Show that $|HK| = \frac{|H||K|}{|H \cap K|}$. $Instruction$: Let $m$ be the number of all different cosets of the form $hK$, where$ h \in H$. Show $|HK|=m|K|$. Then show that $hK = h'K$ holds for $h,h'\in H$ exactly when $h(H\cap K)=h'(H\cap K)$. From here conclude that $m = [H : H\cap K]$ and finally use Lagrange's theorem

[cloud walrus]

OHHELLNAH

[rain grove]

How do I show |HK| = m|K|

Usually I need to show equality of sets, where i can find a bijection or just prove subset + supset but these are not sets

[rocky cloak]

What you can do is just show that HK is the union of those cosets. They all have size |K| and there are m of them.

(and cosets are disjoint)

[delicate orchid]

Put a lil twist on that thang…

[rain grove]

Ok i got that, what about the step "From here conclude that $m = [H : H\cap K]$ "

[cloud walrus]

OHHELLNAH

[rain grove]

I managed to prove that hK = h'K <=> h(H \cap K) = h'(H \cap K)

[rocky cloak]

Right, so then the number of cosets hK is the same as the number of cosets h(H cap K)

[rain grove]

Yeah how that follows from there

[rocky cloak]

hK |-> h(H cap K) is a bijection, that's what you just showed

[rain grove]

aa yes ty

[languid trellis]

I solved this problem by definined an R-bilinear map I x I -> R mapping (a,b) -> ab, then applying the universal property and showing that the resulting map is injective. One question, however, why do we need the ideal to be principal? I see why we need I to be an ideal, so that we get closure under multi. and addition but I have no idea why principality is necessary, I'd appreciate any idea (:

[rocky cloak]

Let R = k[x, y] and I = (x, y)

Then consider the element
x(x)y - y(x)x
multiplying by xy gives
xy(x)xy - xy(x)xy = 0

[languid trellis]

is k[x,y] a PID?

surely not

[rocky cloak]

No, I is not a principle ideal

[languid trellis]

right I see

Can we explicitly link "not being principal" with having nonzero torsion elements in I \otimes I? This feels like a counterexample which "just works because it does" atm

[tough raven]

(Of course, to show that this is a counterexample, you also need to show that x (x) y - y (x) x is non-zero in I (x) I. )

[languid trellis]

OH wait. D&F literally gives the reason in the next few exercises

[rocky cloak]

Well, I think the better perspective is that if I is principle, then I is just isomorphic to R. So it's just R(x)R = R.

In general there's no reason to expect the tensor product of torsion free modules is torsion free

[languid trellis]

oh yeah of course

thank you

[tough raven]

It's not true that I (x) I has torsion for any non-principal ideal of any integral domain, so perhaps not.
(Counterexamples include any non-zero ideal I of a Dedekind domain R. It can be shown (if I'm remembering correctly) with some effort that I (x) J is isomorphic to IJ (product of ideals) as R-modules, and the latter embeds in R and is hence torsion-free.)

[rocky cloak]

Not terribly hard:

R - (x, -y)^t -> R^2 -(y,x)-> I -> 0

is a projective presentation of I. Tensoring with I gives
I -(x,-y)-> I^2 -> I(x)I -> 0
(x, -y) |-> x(x)y - y(x)x

And (x, -y) is not in I*(x, -y)

[languid trellis]

do we have any general strategies for showing that an element of a tensor product is not a simple tensor?

I suspect that this depends too much on the underlying modules we are working with, but I figured i'd ask anyway

[coral spindle]

You can try exhibiting a bilinear function with a domain that is more restricted than what its linearisation via the tensor product achieves

[dull ginkgo]

Admittedly this is like asking to determine if a single element alone is a basis element or not. It depends highly on the structure of the object, and especially upon how it’s given

[languid trellis]

I'm not quite sure how to go bilinear-map wise in this question, and I can't work "naively" because 2(x) 2 can't be factored as 4(1 (x) 1) or whatever. do you have any advice?

[languid trellis]

i foreshadowed this lol

[dull ginkgo]

2 (x) 2 can be

It’s like multiplication

[languid trellis]

2 (x) 2 = ?

[dull ginkgo]

wait the tensor is over the poly ring

[languid trellis]

(1 (x) 1) isn't in I \otimes I because 1 isn't in I

[dull ginkgo]

Oh in I

Hmmm

[rocky cloak]

I guess you can think about the multiplication map
I(x)I -> I
a(x)b |-> ab

[dull ginkgo]

Tensor algebra my beloved

[languid trellis]

oh yeah

[coral spindle]

There is no more natural bilinear map on an ideal than multiplication :)

[languid trellis]

4+x^2 is irred in Z[x]

and 4+x^2 * 1 doesn't exist

because 1 isn't in I

i literally wrote 4+x^2 is irred in Z[x] o nmy page but forgot that multi. identity doesn't exist in the ideal

oh my god

thank you all

[dull ginkgo]

huh

[coral spindle]

Miz discovers multiplication is bilinear

[dull ginkgo]

It’s the “no more natural” part

[rocky cloak]

Name a more natural map, I'll wait

[dull ginkgo]

Oh yeah the ideals are viewed as R-modules

Tensor product of R-algebras since they are closed under multiplication

and then you map the tensor algebra onto the ring

[rain grove]

\textbf{Let $G$ be a finite group and $H \leq G$. Show that there exist elements $a, b \in G$ such that $a \not\in H$, $b \not \in H$, and $ab \not\in H$ if and only if $2|H| < |G|$}\

For => side im tryin to prove: $2|H| >= |G| \Rightarrow \forall a,b \in G$ $ a,b \in H$ or $ab \in H$

So from lagrange theorem and $2|H| \geq |G|$ follows that there are only 2 cosets (if $H \neq G$), so then i got 3 options: $a,b \in H$, $a,b \not\in H$ and $a\in H, b\not\in H$... How to prove that $a\in H, b\not\in H \Rightarrow ab\in H$?

[cloud walrus]

OHHELLNAH

[dull ginkgo]

Biblical misread

what that does that mean?

[dull ginkgo]

2|H| < |G| is equivalent to there is at least 3 cosets.

And we have a, b, ab not in H… how many cosets does that give us

[tough raven]

Along the lines of this, we have the presentation for I as an R-module:
I =〈x, 2 | 2 x = x 2〉.
Hence I ⊗ I has the presentation
I ⊗ I =〈x ⊗ x, x ⊗ 2, 2 ⊗ x, 2 ⊗ 2 | (2 x - x 2) ⊗ x = (2 x - x 2) ⊗ 2 = 0, x ⊗ (2 x - x 2) = 2 ⊗ (2 x - x 2) = 0〉,
or
I ⊗ I =〈x ⊗ x, x ⊗ 2, 2 ⊗ x, 2 ⊗ 2 | x (x ⊗ 2) = 2 (x ⊗ x) = x (2 ⊗ x), 2 (x ⊗ 2) = x 2 ⊗ 2 = 2 (2 ⊗ x)〉.

So I ⊗ I is generated by those four simple tensors. Moreover, we can use these relations to move coefficients from 2 ⊗ 2 to x ⊗ 2 or 2 ⊗ x to x ⊗ x, with the result that every element of I ⊗ I has a unique representation of the form
p x ⊗ x + l x ⊗ 2 + m 2 ⊗ x + n 2 ⊗ 2,
with p in Z[X] and l, m, n in Z.

Now, in particular, for a = xp + 2m, b = xq + 2n, p,q in Z[X], m, n in Z, we have
a ⊗ b = pq x ⊗ x + pn x ⊗ 2 + mq 2 ⊗ x + mn 2 ⊗ 2.
This can be equal to x ⊗ x + 2 ⊗ 2 only if pn, qm are divisible by x (i.e., p,q are divisible by x), and then its standard form is
a ⊗ b = (pq + 2(p/x)n + 2(q/x)m) x ⊗ x + mn 2 ⊗ 2.
So pq + 2(p/x)n + 2(q/x)m = 1. But the latter is impossible for degree reasons unless pq = 0. But if p = 0 or q = 0, then the LHS is even and cannot be 1.

In summary, this was a big mistake, but at least I got some idea of how to compute explicitly in tensor products of ideals.

[tough raven]

If a in H, b notin H, then you already have a in H, so you don't need to get ab in H.

[dull ginkgo]

This is not true.

Consider the cyclic group Z and the subgroup 3Z.

[3Z : Z] = 3

But (1 + 3Z) + (2 + 3Z) = 3Z

Unless you are allowing a = b

Oh wait finite group

Just swap with a different cyclic group for the same result

[tough raven]

That should be allowed.

[rocky cloak]

That's not what you're asked to prove though. If a is in H, then you're already done

What you should be proving is that if neither a nor b are in H, then ab is in H. (Use that a and b most be in the same coset)

[rain grove]

To prove:

$\exists a,b \in G$ $a,b \not\in H$ and $ab \not\in H \Leftrightarrow 2|H| < |G|$

So I thought for the right side I will prove the contrapositive and contrapositive is:

$2|H| >= |G| \Rightarrow \forall a,b \in G$ $ a,b \in H$ or $ab \in H$

Is that not correct?

[cloud walrus]

OHHELLNAH

[rocky cloak]

if by "a,b in H" you mean a or b is in H, then that's correct

[rain grove]

hmm no I mean a and b in H

[rocky cloak]

Remember the negation of
"P and Q" is "not P or not Q", as opposed to "not P and not Q"

So the opposite of "a and b not in H" is "a or b in H"

[rain grove]

yeah i see what i did wrong, i just coupled a,b together and forgot the and between

[tardy hedge]

Oh HELLLLLLL Nahhh!!!'

[hidden cairn]

!rotate

Is there a command to rotate these images?

My question is: how does one see HK = S4

Please read only the last paragraph in pg 94 and its rest in pg95

[dull ginkgo]

,rotate

[hidden cairn]

I’m confused at how |HK| = 24 implies HK= S4

Because it doesn’t check if HK is a subgroup before concluding that

,rotate

[hidden cairn]

,rotate

[hidden cairn]

Please read only the last paragraph in pg 94 and its rest in pg95

I’m confused at how |HK| = 24 implies HK= S4

Because it doesn’t check if HK is a subgroup before concluding that

[wraith cargo]

HK is a subset of S_4 of the same size as S_4 so it must be S_4

[hidden cairn]

Ah I see

Thank you

[marsh scaffold]

,rotate

[marsh scaffold]

What do they mean here by a "group homomorphism" that dosent agree on G

Like if it's a group homomorphism then it has to agree on G right 💀?

[barren sierra]

agree means equal

[marsh scaffold]

Oh so the two homomorphism are equal over H

Ahhh

[barren sierra]

So you want two homomorphisms $\phi_1, \phi_2\colon G \to K$ such that they agree on $H$ and don't agree on all of $G$. So you want for all $h \in H$ we have $\phi_1(h) = \phi_2(h)$ (agree on $H$) but there exists $g \in G$ such that $\phi_1(g) \neq \phi_2(g)$ (don't agree on all of $G$)

[cloud walrus]

Spamakin🎷

[marsh scaffold]

Okay so I somehow I need to look at cosets of H or sth?

[dull ginkgo]

Not necessarily

[marsh scaffold]

Hmm so what's the way

[dull ginkgo]

I’ll give you a hint. If we have a group G^2 (direct product), we can find an easy one :3

What are two important morphisms from G^2 into G?

[marsh scaffold]

Uh projection?

[dull ginkgo]

yes

What subgroup would be equal in both?

[marsh scaffold]

Diagonal ?

[dull ginkgo]

Yep

[marsh scaffold]

Uhhh

[dull ginkgo]

There you go

Is that not what you’re looking for

Or is it looking for an explicit example for each K and H

[marsh scaffold]

G,H is fixed

We have to find K

[dull ginkgo]

I see

What if we make it so that both have H in the kernel

[marsh scaffold]

I mean if it helps this is from the group actions section

[dull ginkgo]

Okay in that case

[dull ginkgo]

Here’s a silly solution:

Consider the two group actions of G on G/H, one is left multiplication and the other is conjugation

Both are therefore maps from G to Sym(G/H)

but H is closed by multiplication

So any element of H acting on H in both ways is an element of H

[marsh scaffold]

Wait left multiplication in H isn't a group action? On elements of H?

Ah

Okay

[dull ginkgo]

Wait let me refine that

[marsh scaffold]

Sym(G/H)?

[dull ginkgo]

Symmetry group of the set of left-cosets

Actually I think they are equal

or not well defined

Yeah it’s not well defined, shit

g(xH)g^-1 is not necessarily a left coset

But you can consider two sided cosets xHy

[marsh scaffold]

Oh O_o

[dull ginkgo]

actually there has to be a better way to do this

Let me try to think of a better example

OH! (G/H)^2

send G to G/H, and then we have two injections into (G/H)^2, a left one to (x, 0) and a right one to (0,x)

[marsh scaffold]

[dull ginkgo]

The kernel of both is H, so they agree on H (just sent to 0) but differ everywhere else

[marsh scaffold]

But we need a group

G/H is not necessarily a group

[dull ginkgo]

Oh fuck H isn’t normal

I am not doing too great, usually I’m rather okay with this stuff lmao

[marsh scaffold]

np 🙂

Would it help if I posted what question this question was a "part of"

[dull ginkgo]

Probably

[marsh scaffold]

Like there are 2 bits to this

Ok

2.23

I was able to do a via realising that if an element of one copy if maps to another copy then the entire one copy maps to another

So in constructing an H automorphism map x to its copy and vice versa

But keep all the other elements in their respective copies

But it would never be a g automorphism

[dull ginkgo]

Hmm

The problem is the “not all” so you need to consider more than just H

[marsh scaffold]

For part (b)

[dull ginkgo]

Yeah

[marsh scaffold]

I see

Oh wait i think it somehow includes conjugating with an element of the centralizer of H ?

Or sth

[dull ginkgo]

But what if there isn’t any

[marsh scaffold]

Yeah

That

So H is normal in N(H) right

[dull ginkgo]

Yes,

[marsh scaffold]

Okay......I thought I was onto sth 💀

[dull ginkgo]

The normalizer is the stabilizer of H under the conjugation action on sets of elements of G

[marsh scaffold]

Right

Also like is part (b) even related to part(a)

Cause they look similar kind of statementa

But I sps they're not