groups-rings-fields

[karmic moat]

Mod^R_{right}

{}^R_{left} Mod

[tardy hedge]

ok I guess it looked like i didnt need that result specifically but I did use that J^n is in I for some n which is pretty much the same sort of result as that

[upbeat dirge]

as in subgroup lattice ?

they’re useful for galois theory

yes probably

they’re just a visualization aid

[dull ginkgo]

The preimage of a subgroup under an isomorphism is a subgroup of the domain

Hijack that and you’re golden

What injectivity

That should work

What exactly are you trying to do

Isn’t that true for general maps assuming choice

Want another route:

Seems fine lol

Knowing the map pi: G -> G/H is an epimorphism, show that the preimage of any subgroup of G/H is a subgroup containing H of G, show that pi(pi^-1(K)) = K for any subgroup K of G/H.

A/N isn’t

It’s a “set of cosets”

No

They lack the identity

Surjective homomorphism in the case of groups

Let me elaborate when I get home

[rocky cloak]

Yes, G/N is all the cosets of N, and A/N are those cosets of the form aN for a in A

a is an element of A which is a subgroup of G

I interpret it as
bijection from the set of (subgroups A of G)

So the symbol A denotes a subgroup

The point is
A |-> A/N
defines a bijection between subgroups of G containing N and subgroups of G/N

It's a difficult to parse sentence indeed. Could have been worded better

But harsh maybe, but deducting stars is fair

[dull ginkgo]

Let me phrase it as:

Each subgroup of G/H corresponds to a subgroup of G containing H, and vice versa

[dull ginkgo]

In fact, let f be a surjective map from group A to group B, then every subgroup of B, K, satisfies
K ~= f^-1(K)/ker(f)

[rocky cloak]

I haven't actually read the specific book, but I think most books will have at least some wonky sentences.

Things can often be tricky to express

[coral spindle]

Respectfully I think this is actually quite clear mathematical language, but I understand that someone not familiar with it may be confused.

Since you've now understood the actual statement it's best to move along with this new knowledge

After all, we have to keep in mind that there is no such thing as a perfect book.

[dull ginkgo]

A set of sets can be a bit finicky to read at first so I prefer to use “family” of sets

You can also use lattice in this case given the structure of groups, where you can say a subgroup is smaller than another based off of if one is a subset of the other

And that you can add together subgroups and intersect them

Notation is fun (not)

Anything else you’re confused by :3

Peak

[coral spindle]

I would argue that the problem was a lack of notation!

[dull ginkgo]

Category theory jumpscare

For A being a subgroup of G?

Well multiplication of elements of A is closed within A, we are just basically “squeezing” the parts of N within A to being the identity, it doesn’t change the fact that it’s multiplicatively closed

Furthermore when the only part of A also in N is the identity, A/N is basically unchanged

The whole idea of quotienting a group is we’re essentially making a normal subgroup “redundant”, by squeezing all the elements of it to the identity

[minor gazelle]

Hello! Small question is the lemma enough for me to delete the little remark next to the second last line of showing associative property of (AB)C=A(BC)? Is there a more self-contained method proving that lemma?

[minor gazelle]

Well, i actually shuffled a bit with arguing with commutative property (confused by index) so i kinda gave up and went on for tonelli fubini

I know but still it feels nicer if its concretely justified

That was actually what I was doing but it turned out the index thing can be a bit tricky

But I will definitely be trying

It’s just index 😭

[coral spindle]

The solution is, in principle, an inductive application of associativity and commutativity. In practise this is tedious and not enlightening.

If you really want to do this, I suppose it is one of those "once, and then never again" proofs.

[minor gazelle]

I actually thought about it after trying a couple times, the manipulation isn’t that trivial.. I will try induction on it

[coral spindle]

Remember you need to do double induction, on both sets.

If you haven't done this before, well, this is a great time to learn how to do it.

You may want to identify some lemmas to prove.

[minor gazelle]

I will do a search about double induction and see how it’s performed.. this basic fact was a bit hard earlier and I was really irritated with the index… thanks so much for tips 🥰

[vagrant zinc]

What a beautiful color box, what is the name of the book or the notes?

[minor gazelle]

It’s my own note

[minor gazelle]

And this is my pointset topology note.. I used some templates on overleaf though

It’s not organized since just based on random YouTube videos and proofs are self produced so not necessarily rigorous

[vagrant zinc]

Hahahahahahahaha this demonstration is really monstrous, I literally hadn't thought about it.

I have these notes, I have this so far

[tardy hedge]

Im having trouble showing the other direction. If the torsion functors are equal, then rad(I)=rad(J)

where l-torsion(M) = {x in M st I^nx=0 for some n}

[hollow tartan]

whats the diff in thm 10 number 4 pg 176 in D&F it states that H is normal in G. But then in proposition 11. K is normal in H semi direct product with K in pg 177

[tardy hedge]

The ring is noetherian

[dull ginkgo]

GRAM MATRIX JUMPSCARE

[dull ginkgo]

this is actually interesting because these both in the background care about the norm being positive, namely the norm of the orthogonal component being positive

[vagrant zinc]

Emma, could you share with me the code of the box you have please I find it very nice.

[vagrant zinc]

[minor gazelle]

Sure I think it’s called elegant book on latex or something if it’s not the name I will find it for you tomorrow

[karmic moat]

looks like some mdframed thing

[minor gazelle]

Through I actually made a typo there but it was proven correctly in the end.. i rarely use pens anymore these days so sometimes it can be inaccurate here and there. I was in a rush extending the concept to metric space 🫣 and want it to look at least self contained so I proved those results alongside.

[vagrant zinc]

The truth is that I saw this demonstration in some book of linear algebra or differential geometry, I did not take a screenshot to see what it said about the demonstration, it seems to me a monstrous way to do the demonstration hahaha

[vagrant zinc]

Can you send me the link or the code to private to not bother the channel, I appreciate it, have a great night.

[minor gazelle]

Gram matrix is only making sense to me for being positive semi definite. Maybe it’s not that self-contained to show it that way I will switch approach..

[minor gazelle]

Btw I sent you a friend request so I can send you the link

[rocky cloak]

The difference is that in thm 10 you're only guaranteed H to be normal, and in prop 11 you're determining when K is normal

[rocky cloak]

Consider M = R/rad(I)

[frosty sigil]

Just a quick question, does the first isomorphism theorem still hold for infinite case?

[river heart]

Yes.

[frosty sigil]

Thanks!

[river heart]

You're welcome.

[muted heron]

Say you have a ring of integers O_K. Let p be a prime ideal. I want to understand why p^i/(p^(i+1) = O_K/p

[rustic crown]

one way is to wlog replace O_K with the localization A = O_{K, p}, so now A is a pid and p is principal, where it is quite easy to see.

but depending on your definitions, there might be a more direct way

[solar dock]

U(1) (sphere S^1) with angle from (0,1) modulo 2π as an example.

[tribal moss]

That's a bit too pithy for me to really get a picture of which operations you're trying to axiomatize. Where's even the ordered group there?

[hollow tartan]

Yeah I got it ty I would've been clearer though if I wrote the book. Idk maybe bc its the first time reading about it

[thorn jay]

It holds for any algebraic structure imaginable, given the right notion of "normal" substructure or "ideal"

[frosty sigil]

Thank you!

[tardy hedge]

Should we consider R/I? I think I showed it like this:

We have I-torsion(R/I) = J-torsion(R/I). Let r in rad(J), so r^n is in J. r^n is in I-torsion(R/I) (since I*r^n is in I) so r^n is in J-torsion(R/I) so J^m r^n = 0. So, r^mn r^n is in I, so r is in rad(I)

[solar dock]

I was thinking of the additive group [-1,1] but I now realize that was stupid. I do have an example for you tho: Take (S^1) with (x*y=z) where z is the unique point obtained by rotating the sphere such that x becomes (0,1) and then by rotating such that y becomes (0,1). This is U(1). Or if you want (e^{it}e^{ik}=e^{i(t+k)}) is your operation. I can act on this group with a subgroup [S^1(\mathbb{Q}):={e^it: t\in \mathbb{Q}}] for example with the following action: [\phi:S^1(\mathbb{Q})\rightarrow Aut(S^1)] [t\rightarrow e^{(it)}(.)] which is just one of the more trivial actions here. This group is ordered for example. This is arguablely a fun example.

[cloud walrus]

Bushy (Diff geo #1 Fan)

[solar dock]

A norm here would be a function

[elfin wraith]

I remember doing this as a homework problem and thinking it was much harder than it really was lol

I’m sure I spent like an hour looking at it before I realised I’m just dumb and it’s rather straightforward

[solar dock]

or some such

[tardy hedge]

Yea i guess u just need to observe how generators work under powers of ideals and then at a certain point all gens of rad(I)^n are in I

[solar dock]

This is fairly tame example I think?

[south patrol]

I thought you had a proof kiand lol

[elfin wraith]

I’m pretty sure I did some correspondence theorem bullshit but there’s likely something to that as well

[south patrol]

No need for that

This is a purely combinatorial thimg

Like it is this

[elfin wraith]

Yeah I mean I suspect there’s multiple ways to do it, I just think for that homework problem I used the correspondence theorem together with some nonsense about nilpotency

I doubt it was the cleanest approach ever

[tardy hedge]

Lol

[south patrol]

Yeah you can reduce to I = 0 quickly, nice trick often

[elfin wraith]

Yeah I’m pretty sure it was something like that

I’m also thinking back to a homework like a year ago lol

[south patrol]

"Locally nilpotent + fg => nilpotent"

[elfin wraith]

Locally nilpotent

[south patrol]

An ideal is locally nilpotent if each element is nilpotent

[elfin wraith]

I’ve not come across that but yeah I looked it up and it’s too do with localisation which I’ve never really done much with

[south patrol]

But an ideal is nilpotent if a power of the ideal id 0

[elfin wraith]

Because my comalg lecturer doesn’t like geometry and my alggeo course was just weird

[south patrol]

Dont rly need localisation for it, p sure this formulation is equivalent

[south patrol]

Lol bruh

[elfin wraith]

I know more about noncom localisation than commutative localisation and it’s just a fucking disaster in the noncom case

Hopefully Miles Reid blesses me with his knowledge next year

[elfin wraith]

Does this generalises ideal wise to noncom rings?

[coral spindle]

OH I SEE HOW IT IS

[elfin wraith]

I’m sorry Boytjie

[south patrol]

I dont rly know much about noncomm rings tbh lol

Idk if the terminology changes

[elfin wraith]

I’ll guess it does, most things that occur element wise in com rings become ideal wise in the non com case

But yeah it’s hard to say if it’ll even be an interesting notion if you move away from commutative rings

[tardy hedge]

why would anyone be attracted to study noncommutative algebra

[elfin wraith]

Shits cool

[tardy hedge]

i guess i wouldnt know

it just kind of sounds like a headache

[elfin wraith]

A lot of things are significantly harder than if you just have commutativity but you also have so many more rings to work with and just a slightly different flavour to it all

[coral spindle]

because you get interesting things

[tardy hedge]

yea commutative algebra is like Z and polynomials Z and polynomials

[coral spindle]

why would anyone be attracted to studying noncommutative groups?????

Cmon you sound silly

[tardy hedge]

bro got offended

[coral spindle]

I am maybe a little bit

I'm tired of this schtick

[tardy hedge]

fair enough

[tardy hedge]

i mean like im not personally but i also wouldnt know atm why it would be interesting so

[elfin wraith]

Physics is the easy answer for applications

[coral spindle]

Have you ever seen representation theory of, say, finite groups

[tardy hedge]

I haven't learned it before

[coral spindle]

OK that does stymy me a bit

but needless to say

it is noncommutative algebra

and it is interesting

[tardy hedge]

maybe i will learn it someday

[coral spindle]

I hope so! It's wonderful

[south patrol]

But also ye i mean different flavour like

The stuff you learn in comm alg is geared towards AG stuff often ig

You will learn different things

[tardy hedge]

Yea I guess my problem is that any time i have ever seen something noncommutative in algebra, it just feels like more of a pain and i have never seen it being useful in an interesting way yet

[coral spindle]

He who has never tasted grapes says "sour"

[elfin wraith]

Intruded up a massive message about cool applications and then my phone died

If you want to study Lie algebras or their enveloping algebras, these are noncommutative, and a massive area of research, you’ve also got D-modules which are an important part of PDEs and since calculus is non commutative (see the product rule) these are also generally non commutative. The product rule specifically is modelled by the Weyl algebra which also captures the behaviour of commutator relations in quantum mechanics, so it’s helpful to understand these things

Also there’s like ya know, matrices

Phone died and now my messages won’t send, I’m being oppressed for my views on commutativity

The Weyl algebra as gives you a definition of Holonomic modules through the Bernstein inequality and these are useful in diffgeo (apparently, I know nothing about diffgeo)

[tribal moss]

Hmm, okay, you have an abelian group and (Q,+) acts on it and (Q,+) is ordered.
Perhaps I was wrong in assuming you had a concrete problem that you wanted suggestions to help solving by appropriate abstraction.

[elfin wraith]

But also as I mentioned, it just has a slightly different flavour to it, and one I think I prefer. It often is a bit more annoying but it just means you have different problems to solve

It only seems annoying because the problems you’re working on are the ones that are easy in the commutative case

[solar dock]

I mean I don't have a concrete problem right now, I'm rather trying to come up with one

[elfin wraith]

Theres also loads of interconnected stuff with like algebraic K theory and PDEs etc, there’s the Atiyah conjecture which implies stuff in noncom, Kaplanskys zero divisor conjecture

There’s a bunch of related conjectures in that area as well, with loads of dependencies and implications that says stuff about lots of different areas

[swift tundra]

Is there any reason why we sometimes introduce say, Rings, as an ordered triple? Like (R,+, x)? Some of my professors use that notation and I’m wondering if there is anything going on there besides semantics/pedagogy. Does this have anything to do with the categorical product?

Also one thing I was thinking of which might be totally wrong, but I’ve heard there is a niche study of co-algebras. Does this presentation of algebraic structures as an ordered tuple give you any advantages when doing co-algebra? If this question doesn’t make any sense feel free to ignore it I don’t really know what I’m talking about here; just wondering if the connection I had holds any significance.

[thorn jay]

Formally, a ring is an object (R; +, ×, -, 0, 1), and the ordered triple is just a shortened version of that. This comes from a formalisation of algebra known as universal algebra, where in general an algebraic structure is a tuple (A, F) where F is a collection of operations on the set A

I guess it's both for convenience (shows the operations you use) and habit

[tribal moss]

Okay; I shall cease my efforts to pry it out of you. :-)

[solar dock]

I believe some non trivial examples exist here but again, I'll have to work on it some. I can give some proper results and mayhaps problems after?

[coral spindle]

Adding to what enpeace says:

• A ring (or any other algebraic object) is not merely defined by the set. We really need to involve the operations in the definition, as the same set can have infinitely many different ring structures.
• It is often convenient to write down precisely the information that defines some object. You may see different things defined in the future with lots of parts to them, and it can be helpful for someone to say precisely what data defines them. This is just a sort of first example of this.
• Coalgebras are not different to this.

[solar dock]

If you are interested that is.

[topaz solar]

This seems like the wrong way around

[solar dock]

I mean my initial question was can I extend what I care about with norms to ordered groups, I think Lawvere metrics even take this a step further.

[elfin wraith]

As another example to the other great answers given, consider a topological space, we’re usually lazy and talk about a space X but formally it’s the pair (X,T) where T is the topology on X. It’s not enough to specific the set, you need to know what operations exist on that set

[solar dock]

Now I simply wish to see if I have some nice non trivial examples

[elfin wraith]

Are we thinking of R as an additive group? Multiplicative? A ring? A vector space?

[thorn jay]

Where this is useful in universal algebra, is that often you consider different operations on the same set (and often induced from the original algebraic structure in some way)

[elfin wraith]

Granted you still need to say more because there’s infinitely many additions on R for example, but formally you see the picture I guess

[rocky cloak]

"this course" being introduction to abstract algebra?

Depends a little bit on what you're course covers, but once you have the basics of groups, rings and modules then things like Galois theory, commutative algebra, varies flavors of representation theory, homological algebra open up.

Commutative algebra is the study of commutative rings, and deal with things like prime ideals which generalize the usual primes in Z, and completions generalizing the relationship between polynomials k[x] and the power series ring k[[x]].

It is the foundation of modern algebraic geometry, and is also very important for algebraic number theory. There is also interesting interplay with Galois theory in the study of group actions on commutative rings (see geometric invariant theory)

Homological algebra is the study of short exact sequences, and arose first in algebraic geometry, gained it's footing properly for algebraic topology and is useful more broadly in algebra for example in representation theory.

It's focused a lot on modules, so you may want to know more about those before diving into homological algebra

[long geyser]

tbh intro abstract algebra is such a foundational course that it is likely to open up everything but almost nothing at the same time

the former because it is a prereq in just about anything, and the latter because it very likely won't be the only prereq

unless you want to progress into more advanced algebra, which is discussed above me

no that stuff is more than intro

for groups it's a bit more than what you'd see in an intro course, but especially for rings there seems to be quite a bit there

I'm no algebra expert thou

[rocky cloak]

I'm not sure there is a topical path, it depends what you want / what is the main focus at your institution.

[long geyser]

I'm guessing you'd be very well setup for commalg or rep theory after those topics

[rocky cloak]

Once you have this picture course you'll have a pretty strong algebra background, and are free to jump into some more advanced algebra of your liking or perhaps some algebraic topology or geometry

[long geyser]

probably need galois theory

[rocky cloak]

Don't think Galois theory is super relevant for rep theory, but might be for commutative algebra. Not absolutely necessary I wouldn't think though

[long geyser]

yeah when I said need it was pretty loose version of need lol

it's just something that I know pops up

[rocky cloak]

Yeah, needs some basic point set topology at least, but not that much really. Depends I guess

[karmic moat]

I think the topology done in AT can be picked up as you go through but I can also see why’d it be a prereq

[long geyser]

odes are not essential for a math degree

[desert verge]

point-set topology

[muted heron]

I didn't know you could attack it with localization. I'm not there yet though

[long geyser]

the algebra you are doing, point set topology, complex analysis maybe measure theory

maybe depends on institution

[topaz solar]

maybe should be

[long geyser]

and could be different for pure/applied streams ofc, if something like that exists

[karmic moat]

I think ODEs are good for a math degree tho

[desert verge]

no

[karmic moat]

It’s good for breadth of knowledge

[long geyser]

what? if you want to do probability then you really do need measure theory

[topaz solar]

They help motivate de Rham-y things for cohomology, but that’s perhaps the wrong way around

[desert verge]

tbh where do you use complex analysis in grad courses

[topaz solar]

AG?

[desert verge]

i'm not convinced

[long geyser]

mmm, I'm not sure how wise it is to take a non-measure theory probability course over taking measure theory

it would help you obtain probabilistic intuition for sure, but there will be handwaving in that course

[topaz solar]

Sheaves come from it kinda

Also GAGA

so kinda motivates schemes

[desert verge]

yes maybe you should know what holomorphic means

but where do u use residue thm

[long geyser]

p sure complex analysis is all over some geometric topics

[topaz solar]

In evaluating number theoretic integrals, L functions, zeta functions, proving all those important theorems about analyticity

[desert verge]

maybe i'd equate it with galois theory in that it's useful in super advanced stuff later on if you specialize there

[long geyser]

yeah I stand by what I said even more, to teach that kinda stuff without measure theory requires a bunch of handwaving

I know because I was in such a course in UG and didn't enjoy it

[topaz solar]

Well, that’s not specializing in analysis, that’s AG and NT

Algebraic things even

[long geyser]

maybe it's still the course to take if it opens up a bunch of stuff for you

[desert verge]

maybe drop odes instead

[long geyser]

no you cant drop topology it's the most important topic ever trust me

but the thing is, if you want to do more advanced probability, measure theory is totally mandatory

but it could be that there's a more advanced probability course in your uni that develops measure theory as well

probably not then

right, that's the one

I'd recommend taking this over the other course if you can, you probably won't see stochastic processes to the level of the other course but you will actually be doing math

the standard probability courses in universities are a bit fake like how intro calculus courses are, as in most stuff won't be rigorous at all

[topaz solar]

Cringe!

[long geyser]

except discrete probability which you can develop rigorously without MT to an extent, but the course you posted was not that

[elfin wraith]

I honestly think you can black box 90% of the topology you need in basic algtop anyway

[long geyser]

but what if I'm allergic to black boxing

[elfin wraith]

You could give me any theorem from the first half of Hatcher and I think I’d have about a 50% success rate on telling you which exact connectedness conditions it requires

[elfin wraith]

I don’t think you should be tbh

[karmic moat]

You should do AT before/with hom alg

[long geyser]

I can tolerate like 1 or 2 black boxes but not more than that, it genuinely fucks with me

[karmic moat]

Hom alg never clicked for me until I did AT

[elfin wraith]

I mean obviously you should aim to understand things but black boxing is just useful and necessary, it’s generally not helpful to lose sight of the bigger picture over minute details

[karmic moat]

And it motivates why things r called cycles and boundaries and whatnot

[elfin wraith]

This is my point about ignoring most of the actual topology for basic algtop, it’s silly to get caught up worrying about exact connectedness requirements because you’re always just covering your arse for some weird pathological counter example

[sonic coral]

can you elaborate as to why? i’m registered for hom alg but havnt seen algebraic topology yet

[elfin wraith]

I think it would feel wildly unmotivated

[karmic moat]

Hom alg largely concerns itself with exact sequences and chain complexes

These are most naturally seen from the perspective of AT

[sonic coral]

i’ve seen those in com alg though

[desert verge]

i would rather do hom alg first tbh

[karmic moat]

Where these actually resemble boundary maps between topological objects

Homotopy and homology thus arise naturally as a way of classifying properties of these complexes

[elfin wraith]

A lot of the terminology makes sense when you think about homology geometrically as in algtop, but more to the point I think it would just feel like a weird amount of abstraction

I did homalg along side algtop though so I can’t say I’ve seen it without

[desert verge]

have you seen the 5-lemma?

[sonic coral]

no

[elfin wraith]

I only really know basic homalg and only because of my final project, it’s not a standard course at my UG

[karmic moat]

It also gives good reason for why the maps are called “boundary maps”, why things are called “cycles” and “boundaries”, etc

[elfin wraith]

I did combinatorial comalg stuff which relies on some algtop and homological algebra

[desert verge]

ok, you should see that (and similar stuff) in ur homalg course then, which will be useful for algtop

[thorn jay]

You can learn a lot if you're interested in stuff

[south patrol]

Imo can definitely do hom alg first lol like the amount of hom alg you do in intro alg top isnt generally that much imo, but ig keeping in mind the example of singular homology is nice even if you dont know too much about it

[elfin wraith]

Yeah I think it just provides nice motivation

[karmic moat]

Yeah I mean one doesnt need to do AT for hom alg but imo it gives motivation

[elfin wraith]

For like why you care about the 5 lemma or snake lemma, but when you’re talking about derived functors or whatever I feel like you’re just doing algebra and that’s the fun

[karmic moat]

The “motivation” is doing a lot of the heavy carrying here

[elfin wraith]

No my uni was very strict about prereqs

I took a pretty standard course selection I think (minus like 2 classes that I wouldn’t think are standard), possibly a slight amount more algebra and less analysis than some unis will require

[sonic coral]

i have been wondering this actually

[long geyser]

Is the claim here that: we have two functors, one is the identity functor, the other sends A to TA plus A/TA, and there is no natural transformation between these two functors where the components are isomorphisms?

(idk if this goes to cat theory or here tbh)

[topaz solar]

This is weird

Since like, why would you do Hom alg without AT reasons

The why is important

[desert verge]

cuz i would know a priori that it's useful for AT

[topaz solar]

Fair enough, knock yourself out ig

Thinking about chain homotopies and such sounds odd without it but

[desert verge]

if i learn homalg here and there while doing algtop i feel like i wouldn't understand the homalg very well

anyways that's what i did when i first learnt it, so doesn't rlly matter now

[elfin wraith]

The rough answer is it’s to do with holes, you can vaguely define a hole to be the failure of a boundary to bound a cycle (terms that make sense when you construct simplical complexes) and homology is capturing exactly this data

But then you realise there’s nothing special about the fact these are abelian groups given by a simplical complex, and all you need is the property of the “boundary” map (which again makes sense when you draw this stuff out for complex’s) and the ability to take quotients which is basically what gives rise to homalg

[rocky cloak]

Because you care about AG, commutative algebra, representation theory or think it's an interesting theory in its own right

[elfin wraith]

I’ve given a better explanation of that here before but I’m tired and on my phone and don’t feel like typing it out again

[topaz solar]

I suppose you could, but it feels weird to me

[elfin wraith]

I’m interested in homalg on its own, I just like doing algebra

[topaz solar]

It’s terminal…

[elfin wraith]

I don’t really feel a need to have an application, I just enjoy pissing about with modules

[topaz solar]

Consigned to noncommutative rings

[karmic moat]

I mean AT isnt required for hom alg and vice versa, I just think its nice to see the motivation of hom alg and/or basic ideas

Same reason why I think ppl should do multivar calc and lin alg together so they know where matrices come from when they do change of coordinates

[topaz solar]

Yeah it’s not required but it’s like, where’d these homotopies and stuff come from

What’s with these names etc

“Why would we expect this”

[karmic moat]

But to each their own I mean hom alg is interesting in its own right to some, I just think it’s cooler if you see it from AT as well

[elfin wraith]

I think it is interesting in its own right and I do want to do it for its own sake, but I do agree that generally it’s helpful to have seen some amount algtop before to motivate it, I know that most people aren’t motivated by “pissing about with modules” and I do think the definitions of stuff like boundary maps would feel a little plucked from think air otherwise

But equally that’s true of a lot of maths, topology definitely has that problem but we trust it’s useful and do it anyway I guess

[topaz solar]

Yeah that’s true

[karmic moat]

I think most ppl are introduced to topology in a first real analysis course tho so they get the motivation for it from that

[desert verge]

it's not like geometry is useful irl anyways

[topaz solar]

I just guess it’s a pedagogical thing

[rocky cloak]

I mean,
pissing about with modules > pissing about with CW complexes
anyway

[karmic moat]

No math we discuss here is useful in the real world

[topaz solar]

You’re 100% correct!

[elfin wraith]

If you don’t get the motivation for LA then your lecturer is fucking awful

[desert verge]

what does motiviation even mean

[topaz solar]

I just think it’s useful in teaching it it, like learning about smooth things before Lipschitz ig

[desert verge]

hardly any of this sht is useful irl

it's just useful for more math

[karmic moat]

“Where does this come from” and “why should we care” is what I consider motivation to be

[long geyser]

yeah I don't get the motivation piece either, I do math so that I get to do more math

[topaz solar]

I mean some of it should be visible in an analysis, or algebra, or like any other course

[karmic moat]

Well yeah imo the motivation comes from the fact that you see where it’s used later on in other math

If you tried teaching AT to an engineer they probably wouldnt give af bc it has no use to whatever bridge theyre engineering (idk what they do)

[elfin wraith]

Motivation doesn’t mean solving real world problems, I want to know the context behind it, what problems it solves within maths

[topaz solar]

I mean, you’d think, but then you go a gorillion layers of abstraction deep and somehow physics reappears

[desert verge]

such is the state of modern theoretical physics

[topaz solar]

“Hey is this property nearly able to be forced? Is it AVE axiomatizable?” -> quantum computing??????

Whar???

[topaz solar]

Also this, why should we expect so and so answer

[karmic moat]

If ng were here he’d talk abt cft and qft

[elfin wraith]

Linear algebra might be the most important piece of maths ever conceived, I genuinely don’t think I could name an area where LA isn’t applicable

[topaz solar]

That’s partly what I’m referencing, and group stability

[elfin wraith]

Differential equations, geometry, statistics, computing, physics, there’s genuinely just nothing that linear algebra doesn’t touch

[desert verge]

ur just generalizing R^n

[topaz solar]

The only math we understand

[elfin wraith]

Literally, maths is the process of making things a LA problem

[desert verge]

module theory kinda generalizes linalg tho

[topaz solar]

What do you think analysis is vro

[elfin wraith]

It does but not in a super helpful way

[topaz solar]

Like highkey

[desert verge]

umm ok, what ur point

[topaz solar]

Everything returns to R…..

[desert verge]

i think module theory is used in algtop

like classification of fg modules maybe

[elfin wraith]

Linear algebra is incredibly geometric too, there’s about a thousand ways to motivate the subject because it’s that useful

[desert verge]

or smith normal form

[elfin wraith]

This is the first time I’m going to recommend the 3b1b videos to someone because I think doing maths and thinking LA is unmotivated is practically criminal

[desert verge]

LA kinda boring tho

[elfin wraith]

Your proofs based course is just making sure all the other stuff actually works, and letting you apply that theory to spaces that aren’t R^n

[tribal moss]

My favorite application of linear algebra is that the algebraic reals (all numbers that are root in some integer polynomial) are closed under addition and multiplication and division, so they form a field.
This is proved by viewing various subrings of the reals as vector subspaces over Q and saying "as we recall from linear algebra" a lot.

[rotund arch]

The reason for that is because you have done calculus for many years already, of course it will be intuitive

[elfin wraith]

I don’t see how it’s different to going from calculus to analysis

[ashen heron]

you will definitely start to see that as you learn some more analysis and more math in general

[rocky cloak]

A motivation they use in the LA class at my uni is classifying conic sections, which is done by symmetric bilinear forms / orthogonal diagonalization.

The LA for engeniers course also have more examples, finding stay states for Markov chains, image compression with SVD, linear (and polynomial) ODEs

[karmic moat]

Yeah using LA in a dynamical systems course I took was cool

[desert verge]

imagine doing markov chains in a LA course

[rocky cloak]

Also just the more basic things, like solving similar systems of equations in parallel, and determining when a system has (unique) solutions.

Solving linear difference equations.

Basic 3d geometry, finding volumes with determinants

[elfin wraith]

There was a problem about the quantum plane in my noncom class that essentially just reduced to a LA problem and that was very nice

[rocky cloak]

Is that so hard to imagine?

[elfin wraith]

Like you can bully a question about semiprime ideals into a simple eigenvalue chase, LA is just too good

[desert verge]

yes, i find it a bit bizarre

[elfin wraith]

A little for me, I’ve got no clue what a markov chain is beyond “probability”

[rocky cloak]

I'm not sure when else you would learn about them.

They're a matrix you multiply by to transition from one state to another. And the eigenvectors are the steady states

[rocky cloak]

You have some set of states and for each pair of states a fixed probability to transition from one state to the other. Each transition independent of the previous.

That's it

[elfin wraith]

Ah ok that makes sense and yes I saw them in my LA class

There was some question with that set up in a workshop, never mentioned it by name but it was definitely that

[topaz solar]

Change of variables comes in too

Since hey, volumes? sounds good for that to me

[swift tundra]

That makes sense. Thank you Boytjie, Nope, and .enpeace!

[thorn jay]

:3c

[swift tundra]

What if R is finite? I get what you are trying to say; thats just what I immediately thought when I saw this message lol

[tribal moss]

What I immediately thought now was, "but R is known to be (uncountably) infinite!"

[rocky cloak]

Well, then there finitely many, but still quite a lot (unless R has only one element)

[swift tundra]

Yeah if I don’t have enough time to count to a number then it’s basically infinite lol

[tribal moss]

A set with one element has one possible group structure.
A set with two elements has two possible group structures,
A set with three elements has three possible group structures.
...
Hey, algebra is easy!

[swift tundra]

Which is why Hilbert’s basis theorem is wrong because nobody is going to write out all those basis elements. Sure Paul Gordan, it’s “finite”

[tribal moss]

Nimber addition.

(aka, binary XOR)

[tribal moss]

A set with 4 elements has 16 possible group operations (for each choice of identity, three ways to make it into C4 and one that makes it V4).
A set with 5 elements has 60 possible group operations (it's always C5, for 5! possible choices, except divide by 2 because of the "negate everything" automorphism).
However, OEIS doesn't know a sequence starting with 1,2,3,16,60 so I suspect I've miscounted something. Is my reasoning faulty?

Oh, there are more automorphisms of C5... 🤦

https://oeis.org/A034382 https://oeis.org/A034383

[elfin wraith]

I was meaning R (the reals) not R (a ring), not a very clear choice on my part lol

[topaz solar]

The nefarious 6:

[elfin wraith]

TIL you can search the OEIS like that, I probably should have known but I did not

[elfin wraith]

Combinatorial questions start to scare me by 4 and terrify me by 6

[topaz solar]

There was this one book about function algebras and clones

I looked through it

Huge highlighted theorem in the last part: “this set has cardinality 6”

[thorn jay]

Every time I see a term related to universal algebra my brain does a neuron activation I might have a problem

[elfin wraith]

Universal algebra seems cool I should check it out at some point

Is it particularly widely studied? I’m like aware that it is a thing but I can’t say I run into it very often or hear of people studying it

[elfin wraith]

Is there any nice UA results I might like to know? Has it done anything for any of the more wildly studied areas of algebra? Any cool open problems?

[thorn jay]

It's not widely studied but there certainly are people studying it

Though a little hard to find

[long geyser]

an open problem in UA is finding reasons to study UA

[topaz solar]

I mean it’s relevant

[topaz solar]

I don’t know many in it, but that’s a general logic thing tbh

Showing some classes are varieties (in the sense of UA) is a problem I’m trying to work on

[elfin wraith]

Does UA fall more into the logic camp than the algebra camp?

[topaz solar]

Tbh

[thorn jay]

Personally I am fascinated by so-called Malcev conditions, which give equivalent conditions for particular nice properties of, say, congruence lattices with the existence of terms which satisfy some properties. An example, in particular one of the more important ones, is called congruence-permutability (which one can see in groups manifesting as the fact that NM = MN for normal subgroups M, N), being equivalent to a ternary term p(x, y, z) such that p(x, x, y) = y = p(y, x, x).

The more nice results that I've seen are in the field of commutator theory, which is fairly technical and Im still slowly working through (I took multiple breaks because I was busy with final exams lol), but it for example characterizes those algebraic structures which are essentially the same as abelian groups, and in general the study of commutators reveals a lot of stuff about the particular algebraic structure.

Also, I have been told that duality theory is a very cool area of UA.

As for applications in algebra, I am aware of UA fruitfully being used to study quandles, BCK-algebras, and in the early days even groups, but beyond that it's way more applicable in logic I'm pretty sure. It's more often used as a general framework for treating algebraic structures, but we simply don't really care about many algebraic structures besides groups rings fields modules it seems? Except the logic folk of course.

[thorn jay]

Depends on what you do. But it feels like we're so used to being devoid of logic in algebra that any entrance of logic feels like we're already in the camp of logic

Though I'd say that commutator theory is mostly on the algebra side

[topaz solar]

I mean, mal’cev conditions are fairly related to certain logic things yeah?

[thorn jay]

Yes

[topaz solar]

Saying UA is logic is maybe not it, but it’s definitely logic-pilled and attachable by logic things

[elfin wraith]

Huh that does seem quite cool but yeah it does seem to have a logic esque flavour to it

[thorn jay]

It's treading the line beautifully I'd say

[elfin wraith]

I may skim some stuff at some point though, it does seem to be cool from random things I’ve seen you mention

[thorn jay]

I love inspiring people 🔥

Careful what you do with UA though, it might make you accidentally do logic

That's what happened with me at least, was doing universal algebraic geometry and ended up writing a whole section in my paper that was just model theory

[elfin wraith]

That works, logic is something I’ve been meaning to dip my toes into for a while anyway

I’d like to learn model theory but there’s just a lot of maths out there and it’s never been a top priority (apart from when I briefly accepted a maths in pure maths and mathematical logic (sorry Boyt))

[thorn jay]

I think it's important to have at least a little experience in it

Beyond the basic logic in an intro to math class

[elfin wraith]

Yeah no I would like to know more, and I know model theory has done some cool stuff for ring theory and there’s things there but it’s just never been a high priority and doing logic at my UG was really difficult

There were logic courses but they were in the philosophy department and they limited the number of maths students who could them pretty heavily so it was kinda a nightmare to do them

I always enjoyed that they were postgrad courses in logic and the course description was like “This is an ADVANCED course in mathematical logic for PhD students in Philosophy , make sure you meet the prerequisite knowledge expected! Maths undergrads can take it if they email the professor”

[thorn jay]

This is REALLY hard like HARDER than you've ever seen and the HARDEST you'll experience
Oh yeah first year undergrads are welcome :3c of course

[elfin wraith]

It sounds like I’m exaggerating but it really was like that, but anyway yeah it meant I never got to do more than the basic propositional logic stuff in my intro to proofs class

[thorn jay]

😔 that's so sad Alexa play despacito

If you still want the algebra, you could look around in algebraic logic

[topaz solar]

Cursed to suffer model theory

[thorn jay]

Well they did say model theory = AG - fields

And UAG clearly doesn't have fields

So I guess this was all prophecised long ago

Or however the shit you spell it

[topaz solar]

Horrid

[topaz solar]

I mean, in the nice case, anyway, it’s not too far….

[thorn jay]

I have noticed lol

[topaz solar]

The bad case….

[thorn jay]

The bad case of UA is not assuming you're in a congruence modular variety

Like honestly what are those subtractive variety mfs doing

Cringeeee

Imagine not being able to do commutator theory

[topaz solar]

I mean, there’s a lotta bad cases model theoretically too

I dunno the UA versions of niceness and badness though

Or to what extent the two agree

But hey, showing this class of algebras is a variety is like 1/3 of the way to a conjecture

[thorn jay]

The problem with not assuming anything about your variety or algebra is that in many cases it's just too general; you could literally be working with sets which have zero structures

[topaz solar]

Yeahhhh

Fortunately: we have structure
Unfortunately: it’s wide open whether that’s preserved in products

[thorn jay]

Hah get REKT products are probably the only nice construction we have

And equalisers I guess

[thorn jay]

Mb forgot about reduced products those are fine

[topaz solar]

Homomorphic images

So quotients

[thorn jay]

hmm, sure

But the congruences behave less nice

[topaz solar]

Yrah

In this case, miraculously, they do

Subalgebras….also open if those are nice, but they’re “good enough”

[thorn jay]

I don't really like subalgebras

Congruences tell you much more about an algebraic structure imo

It's no wonder that 99.9% of Malcev conditions are about congruences

[topaz solar]

Yeah kinda sucks, but we’re close enough to nice here that it’s usable

Sometimes you can build upwards from them

Since like, products + build up/accessibility-ish stuff + technical logic conditions => the white whale

Unfortunately, as it turns out, harmonic analysis on products of groups kinda highkey sucks

[thorn jay]

What's that even entail?

[topaz solar]

Well, 1dim reps are always nice, but anything higher turbo sucks

Since like, you can’t just stick two reps together, because there’s a commutator

So maybe I’d benefit from the commutator theory

[thorn jay]

I was about to say lmao but I don't know honestly

[topaz solar]

(By stick together, I mean on the same space, not a direct sum thing)

[thorn jay]

So you mean sticking a rep of G and a rep of H both on V to get a rep of GxH on V?

[topaz solar]

Any matrix is a rep over Z, but not every pair of matrices is a rep over Z^2

[thorn jay]

Ah of course

[topaz solar]

See also: Voiculescu almost commuting matrices

[thorn jay]

I've got a feeling though that this does work with the free product :P

[topaz solar]

So you don’t get saved even asymptotically

Welllllll

No

Because you get a free group subset

So your harmonic analysis explodes

[thorn jay]

Aw, but it's literally the coproduct property

[topaz solar]

Even if you can glue reps

[thorn jay]

Oh right because your group isn't nice anymore

[topaz solar]

So we’re in a rather precarious position

[thorn jay]

Gosh you guys sure are picky

[topaz solar]

Well, it might not be like a free group but ykwim

[thorn jay]

It's got algebraically independent elements yes

[topaz solar]

C_p * C_q for large enough p, q obviously has issues

Maybe all of em

[thorn jay]

The coproduct is nice in exactly zero varieties which aren't commutative rings or any module variety

[topaz solar]

REAL

now, it’s nice here in gluing reps but like, we also want analysis

And analysis over the product is tractable but only if reps are

[thorn jay]

Or sets I guess BUT SETS DONT COUNT THEYRE ONLY ON THE GUEST LIST AS A FORMALITY

[topaz solar]

Morally the zero variety

[thorn jay]

I'll include them to make the category of varieties somewhat nice

[topaz solar]

Anyhow, there’s like a very precarious balance between “able to glue reps” and “algebraically tractable”

[thorn jay]

you want analysis 💔

What is tractable?

[topaz solar]

Well, it’s equivalent to rep stuff

So like, some specific group quotient stuff, commutators, centralizers etc

Harmonic analysis is really algebra and all that

[thorn jay]

Lol I'm kidding somewhat
Imagine, I got somewhat motivated to learn measure theory after seeing how you can generalised rep theory of finite groups to compact groups

[topaz solar]

And tractable as in not having too much independence of elements, free subgroups, etc

[thorn jay]

Right I see

And "not too much" is formalised in some way I presume?

[tardy hedge]

Ong

[topaz solar]

Well, somewhat

[thorn jay]

Mathematicians try not to handwave "too much of X" challenge

[topaz solar]

Like, the condition we want tells us “no subgroup has bad finite dim reps”

You can have no bad finite dim reps even though subgroups do though, so it’s a bit stronger

[topaz solar]

See F_2 \wr Z

[elfin wraith]

“Sufficiently nice”

[thorn jay]

Jumping off a cliff

Hopefully my landing is

Sufficiently nice

[topaz solar]

F_2 has too much independence, but F_2 \wr Z doesn’t to mess up finite dim reps

But the F_2 subgroup means the left regular rep sucks so infinite dim can

[topaz solar]

Something something Z action vastly affects equations in the group

The exact formal version of this condition is stated in terms of the reps, though

The translation to properties of G? Idk, cry

[thorn jay]

Pff

Make a Malcev condition

[topaz solar]

The “white whale” is this

[topaz solar]

As stated earlier, we dunno if it’s preserved in products or is mal’cev and such

[thorn jay]

Hmm

[topaz solar]

Since like, that’s based on how the class of groups looks, but the reps kinda get trolled under what you do to verify it’s mal’cev

So it’s very nontrivial

[thorn jay]

Hmm I figured

Would be an interesting overlap of rep theory and UA though

[topaz solar]

I’ve tried some model theory attacks around it, some analysis, so the finite dim rep thing is “almost mal’cev”

But that doesn’t control subgroups being nice

That doesn’t control products

[thorn jay]

Have you read W. Taylor's paper on Malcev conditions?

[topaz solar]

No

What about it

Also note, “almost mal’cev” isn’t enough without a lot stronger work

[thorn jay]

It characterizes classes of varieties which "form" Malcev conditions

[topaz solar]

Right right

But, as stated, we don’t know it’s a variety

Products…..

[thorn jay]

Right of course

[topaz solar]

It’s just shy of every possible attack

But if you tried to make a counterexample you’d run into a similar wall of “we can’t just glue this like so”

[thorn jay]

Hmhm

[topaz solar]

Like, a possible counterexample route? Yeah maybe you can turn it into XYZ, but sorry some analysis stuff blows up

So you lose equivalence to actually reverse the thing to make the counterexample

[thorn jay]

Algebraic sets can be extended to universal algebras in the way you'd imagine (with or without coefficients in the equations). There is a "pseudo malcev-condition" that characterizes the property that the union of two algebraic sets is again algebraic. Namely:
"There exists an infinite set T of equations with variables in x0, .., x3 such that A ⊨ T <-> x0=x1 ∨ x2=x3"

I think there is a much higher chance of the group needing to satisfy something like this rather than a usual Malcev property

But I'm very sure that there is a lack of research concerning these types of conditions, sadly

[topaz solar]

Indeed

[thorn jay]

Huh

well Ferb I know what we're gonna do today

[cursive spindle]

algebraic nonsense

[thorn jay]

I think that a meaningful extension of mal'cev conditions would be something actually cared about

[topaz solar]

It should actually be a full on variety and mal’cev btw

Since, as stated, it is mal’cev in a reduced form + technical conditions

And those technical conditions can be arbitrarily strong, it just needs uniform bounds

[thorn jay]

The old "needs uniform bounds"

[noble bronze]

maybe if it was 1976

[thorn jay]

Don't mock me qwq there are still universal algebraists

[noble bronze]

i know, my phd had a universal algebraic flavour

just studying malcev conditions is something you'd do in a nursing home

[thorn jay]

Yes yes

But what if you did it

But general

Eh

Eh

[topaz solar]

Well, long standing open problem is if a specific bound is uniform

[copper kestrel]

grad school project

[topaz solar]

Well, it was a master’s project

[copper kestrel]

[topaz solar]

(Where I bothered to write out some conditions where things work)

[copper kestrel]

thats good!

[thorn jay]

Seems that as of 2015 there are still open problems regarding them lol

[austere prairie]

waht sup chat

https://cdn.discordapp.com/attachments/903430333888884736/1394168459725639820/image.png?ex=6875d40f&is=6874828f&hm=19385a4030853ad8b7d94f5b9cfc8a263cef71dbdc6c0c80549d00704302f2ea&

anyone wanna give this a try?

[austere prairie]

the problem was just an applcaiton of QR decomposition, sadge

[brazen ravine]

Is it possible to prove $y = z \implies y + x = z + x$ assuming the group $(G, +)$, or it is only possible in ordered groups?

[cloud walrus]

Hurricane

[rocky cloak]

This doesn't have anything to do with ordering, and it's not even about being a group. It's essentially just what it means for two things to be equal.

Perhaps you meant the converse implication, which is true for groups. Still not related to ordering though

[brazen ravine]

I don't mean the cancellation law, I am just wondering whether it is possible to derive this from the definition of group, or does this have to do with the definition of equality in FOL?

[rocky cloak]

It would come down to the definition of equality yeah.

If two things are equal and you do the same thing to them they're still equal

Alternatively you can see it as a consequence of the definition of a function

[vivid kestrel]

so let $k$ be a field, $D$ a central simple division algebra over $k$ and $L$ a maximal subfield of $D$. in lecture notes, we proved that $D \otimes_k L$ is isomorphic to $L^{n \times n}$ as $L$-algebra, where $n^2 = dim_K(D)$. how does one conclude from the above, that $dim_L(D) = n$ ?

[coral spindle]

It's standard that if k <= L is a field extension and D is an n-dimensional k-space then D (x)_k L is an L-space of the same dimension

This really has nothing to do with the algebra structure here

Just choose a basis of D and observe that it lifts to D (x) L

Maybe it's possible to see this more abstractly...

There is a s.e.s 0 -> k^n -> D -> 0 and after tensoring, since the tensor product is right exact we get an s.e.s. 0 -> L^n -> D (x) L -> 0 since we can prove that k^n (x) L = L^n without much difficulty.

[cloud walrus]

eggman

[vivid kestrel]

sorry I know that, I forgot a square over the n in the original message

[coral spindle]

OK I see

Yes

OK

Have you shown that n = [L : k] yet?

I.e., that n = dim_k L?

I don't remember the proof of this. It's not very enlightening I don't think

This is then just a corollary of that and the tower law.

[vivid kestrel]

Hm no, that is precisely what is being proven

[coral spindle]

Well yes lol

I can see it's equivalent

[rocky cloak]

So you've proven dimD = n^2 without saying anything about L?

That's interesting

[vivid kestrel]

but it is just stated that one sees dim_L(D) = n from the above isomorphism

[coral spindle]

OK

[vivid kestrel]

That was proven previously by tensoring with the closure of k

[coral spindle]

L^n (x)_k L

this is the kind of tensor you need to worry about

If you can argue that L (x)_k L is of a certain dimension over L, then you're done.

Indeed L is a [L:k]-dimension k-vector space

so L (x)_k L is a [L:k]dimensional L-vector space

so L^m (x)_k L is a m[L:k]-dim L-vector space

Setting m = dim_L D

Ach nevermind this is just the tower law 🙄

OK. Let me just say that we need more information about L. There is going to be some result that I can't remember that has to do with it being a maximal subfield

Darn, I can't find this result anywhere I want in the stacks project bit about central simple algebras

OK, it's proved here https://stacks.math.columbia.edu/tag/074S

but it's in opaque language

I'll think on how to find a nice argument, but my bet is still that you've missed a nice lemma somewhere that your lecturer expects you to be able to refer to

[vivid kestrel]

Hmm, I'll look if something sticks out. We did have double centralizer before, and the fact that maximal subfields are their own centralizer if that's relevant

[rocky cloak]

D(x)L acts on D by d(x)l * d' = dd'l, and this is a simple modules since it is generated by any nonzero element

So since D(x)L is Mn(L), D=L^n

[vivid kestrel]

Ah, so since L^n is the unique simple module of L^nxn D = L^n as modules over that ring and therefore as vector spaces since that ring contains L?

thanks!!

[naive lance]

Hello, i have a question regading the proof of the coleman mandula theorem, at some point we use that the the liealgebra decomposes as direct sum of U(1)' and semi simple lie algebras. This is stated as a resulte from maths under conditions which are fufilled in the proof. but i cant find any reference to the statment so if somebody knows which theorem about lie algebras they use or has a ref to to proof it would be cool.

[rocky cloak]

There's something called Levi decomposition that says a reductive lie algebra decomposes as an abelian + a semisimple lie algebra like you described

[minor gazelle]

Can I show associative property of such monoid in this way

I actually don’t know how to show associative property between maps however I saw a method like this

[rocky cloak]

No, there's quite a few problems with your image.

Some minor things: it says o is "a binary operation", but presumably it's specifically composition, because it's not true in general that any binary operation gives a monoid.

You're writing
Hom(A, A) -f-> Hom(A, A)
when I'm assuming you mean
A -f-> A
since f is a map from A to A.

The larger problem is that your diagrams aren't actually related to associativity.

Associativity is the fact that
(fg)h = f(gh)
and these compositions don't appear in your diagrams

[minor gazelle]

I actually don’t know what this diagram is so I just tried

But how to show associative property in this case and define composition properly?

So I have to say o is the operation

[rocky cloak]

So I'm assuming A is a set or something like that.

Then it's useful to know that two functions are equal if they agree on every input. So you can try to prove that

[(fg)h] (a) = [f(gh)] (a)

for all a, by unpacking the definition of composition

[minor gazelle]

Just a set

[rocky cloak]

Well, maybe I don't understand what you're asking, but you should say what o is.

If you say "Let o be a binary operation" it seems to imply it could be any binary operation when it is in fact a very specific one

[minor gazelle]

So Becasue g(h) gives some input for f and it’s exactly in the domain of f so that’s why?

[minor gazelle]

🫣🫣now i realized how misleading my wording is

You’re absolutely right

I am always trying to improve I am fixing it now! 🥰

[white oxide]

Silly question, but if $A$ is a ring and $\mathfrak{a}$ an ideal, is it possible to have $\text{im } \iota \otimes 1 \neq \mathfrak{a} \otimes M$ where $\iota \otimes 1: \mathfrak{a} \otimes M \to A \otimes M$

[cloud walrus]

okeyokay

[rocky cloak]

That is usually the case, so yes

[white oxide]

Damn okay

I guess I shouldn't trust my intuition when it comes to tensor products lol

[rocky cloak]

Easy example would be
A = Z, M = Z/2 and a=2Z

[white oxide]

Ah yeah makes sense, thanks

[solemn garden]

In this question, how do you show that 1 is a T_0-linear combination of s_1’, …, s_n’?

by assumption, it is a T-linear combination of them

[minor gazelle]

Is this considered correct now? 🫣 i tried to see h(x) as a independent entity so to justify the associative property

[solemn garden]

the $\circ$ is an operation on functions, so instead of $(f \circ g) \circ (h(x))$, it should just be $(f \circ g)(h(x))$

[cloud walrus]

soup_norm

[solemn garden]

since you’re applying the function $f \circ g$ to the argument $h(x)$

[cloud walrus]

soup_norm

[solemn garden]

and same for $f \circ (g \circ h(x))$, which should actually be $f((g \circ h)(x))$

[solemn garden]

other than that, it’s right

[cloud walrus]

soup_norm

[thorn jay]

No

That is right

Just apply g^{-1} to both sides

[tardy hedge]

when talking about "actions" in algebra being faithful it means the action is injective, so distinct g gives distinct permutation of A

no its just jargon u eventually get used to once u learn more math

u not an idiot

G/ker(phi) is removing that redundant structure of G that acts the same way on A

yup

[glad osprey]

What book is that btw?

[tardy hedge]

dummit foote my guess

i recognize that font

I never read their group theory sections

[glad osprey]

Looks pretty decent, the group actions chapter in Fraleigh wasn't that great imo

[tardy hedge]

The book i used when first learning group theory was some super obscure one

not sure why any class would choose to use it

beachy and blair abstract algebra

yea

take something in G/ker(phi), like gker(phi), that acts on A like gA

well whats happening here is that elements of G/ker(phi) can be represented by elements of G, and so the action of G/ker(phi) is defined thru that

an element of G/ker(phi) is a coset gker(phi) where g is in G

gker(phi) * a = ga

[glad osprey]

You can also think of a group action of G on A as a homomorphism phi : G -> Aut(A), in which case it's clear that the map G/ker(phi) -> Aut(A) is injective, ie. a faithful action on A

Automorphism group

I'm sure D&F mentions this equivalent definition somewhere

You can skip ahead, you don't always have to read a textbook linearly

[tardy hedge]

a section on group actions surely talks about automorphism group

and what shedow said

[glad osprey]

I haven't read D&F, but I feel like if you can understand something easier by taking a sneak peek at a later chapter, then you should do so. The definition of a group action as a map G -> Aut(A) is pretty useful to keep in mind

Maybe it's called something other than Aut(A), maybe Sym(A)?

[tardy hedge]

you could look at different textbooks too

that is helpful sometimes

[thorn jay]

This is just the first isomorphism theorem

[thorn jay]

Sym(A) is the usual notation if A is a set, ye, Aut(A) usually assumes some structure on A

[glad osprey]

Yeah, it sucks, but it's hard to put everything you need to know in a linear order, so sometimes you just have to skip a bit back and forth

[tardy hedge]

so basically Aut(A) means the set of permutations (or bijections) A->A, and this can be made into a group through function composition. If you have a group action of G on A then each g in G can be associated with an element of Aut(A) (i.e a permutation), by g -> (a -> ga)

and that mapping is a group homomorphism

first isomorphism theorem in the sense of the map G-> Aut(A)

applying first iso theorem there

i find it unlikely that there is jargon that hadnt been defined previously?

[thorn jay]

A group action of G on X can be thought of as a homomorphism ρ : G → Sym(X), to the set of bijections from X to itself. This in particular is given by the mapping
g ↦(x ↦g ⋅ x)
Applying the first isomorphism theorem this gives an injective homomorphism ι: G/ker ρ → Sym(X), i.e. a faithful group action of G/ker ρ on X

[tardy hedge]

page 112

first page abt group actions in dummit foote

it talks about exactly what we're saying but ur right it doesnt use Aut(A) or Sym(A) its "S_A"

they write a map G->S_A and that discussion there is what we are talking about here

[thorn jay]

You can't think of a way to turn a homomorphism from G to S_A into a group action of G on A?

[tardy hedge]

Because the induced map G/ker -> SA is the "same" map but now its injective

so its the same group action but distinct g gives distinct permuations

You should read section 1. 7 dummit and foote

it talks about this

[thorn jay]

Nvm I misread what you were confused about

[tardy hedge]

yea ik what u mean

I guess its the same map in the sense of the image is the same

but its injective now

i guess in algebra its more abt thinking how the structures interact versus thinking abt individual elements

[thorn jay]

No

[tardy hedge]

I understand how you are confused but its hard to explain what im trying to say i guess

[thorn jay]

This is not a rigorous definition of "equivalence" because an equivalence of group actions is something else

[glad osprey]

I looked at D&F and tbh I feel it could be written a bit clearer. It talks about "permutation representation" for some reason, and it doesn't clearly spell out that the axioms of a group action are literally the axioms for a group homomorphism G -> Aut(A)

I think Aluffi explains it much better in Algebra: Notes from the underground

[minor gazelle]

Another question, maybe I am slow if this hold then immediately (M_n(F), •) is monoid

The definition is that matrix multiplication is the composition of linear transformation

[dull marsh]

Show that S_n, n >= 2 is generated by any triplet (a, b, c) of cycles of lengths n, n-1, 2 respectively

Don’t know where to start. I thought of conjugating the transposition by the other cycles but looks like doing just that doesn’t give enough transpositions

[velvet hull]

S_n is generated by any single n cycle and any single transposition - it suffices to show that it generates every transposition

not sure why a cycle of length n-1 is included there

[south patrol]

Though be careful that the choice matters here

Unless n is prime

[rocky cloak]

Assume wlog that a = (1, 2, 3, ..., n)

What you can do is conjugate c by a until c touches an element not touched by b.

Conjugate c by b until c is of the form (i, i+1).

Show that Sn is generated by a and this new c

[dull marsh]

Are you sure? What about n = 4 and (1 2), (1, 3, 2, 4)

[velvet hull]

oh, right, there was a coprime condition

[south patrol]

Sniped

[velvet hull]

that's why n-1 was there

I was thinking of the (1,2) and (1, 2, ... n) case

which doesn't generalise I don't think

[rocky cloak]

The same problem is still there, i.e.
(1, 3) and (1, 2, 3, 4) doesn't generate S4

(1, 2, 3, ..., n) and (i, i+k) generates Sn if k is relatively prime to n

[dull marsh]

I see, so just exploit the fixed point of b, thanks

[vast verge]

Can someone explain why I got it wrong?

[velvet hull]

why does Z4 have 4 elements of order 1 or 2?

[vast verge]

You're right

1 and 3 have order 4 right?

I thought they had order 2..

[south patrol]

yes order 4

[vast verge]

0 - order 1
1 - order 4
2 - order 2
3 - order 4

I believe this is correct?

[south patrol]

though writing (+) here feels kinda dodgy

usually only used for abelian groups

[vast verge]

That's how I'm being taught for direct products

[south patrol]

ok ye sure just letting you know

[steep mesa]

If I have (n+1) polynomials p_i in n variables x_j over a field F, must there be some nontrivial polynomial in the p_is with coeffs in F which is equal to 0?

If the above is true (and I'm pretty sure it is), then if we add the additional requirement that every subset of n of the p_is is algebraically independent, must there be a unique (up to a scalar in F) solution of minimal degree?

[minor gazelle]

First time I felt algebra is so interesting I literally saw something so similar to generating sigma algebra and it’s called generating submonoid🫣

[vivid kestrel]

In a quaternion algebra for elements a, b non zero over some field k, is the splitting field of x^2 - a over K always a splitting field for the quaternion algebra?

[vivid kestrel]

Why I think believe it to be true: Let $F$ be the aforementioned splitting field, then there is a $k-algebra monomorphism f : A \xrightarrow{} F^{2 \times 2}$ which by tensor product magic (i hope) induces an $F-algebra monomorphism$ $g : T \otimes_k A \xrightarrow{} F^{2 \times 2}$ which must then be an isomorphism for dimension reasons

[cloud walrus]

eggman

[vast verge]

How do I find the right classification of the elements of this group?

[vivid kestrel]

hm, think you can disprove it by noticing that the former group has no element of order 27 if these are indeed the cyclic groups denoted by Zn here

[vast verge]

I know, but I'm trying to find the number of elements of order 3 and 9 in the group just to test myself

[vivid kestrel]

oh my bad

[vivid kestrel]

$A$ here is supposed to be the aforementioned quaternion algebra

[cloud walrus]

eggman

[sonic coral]

the order of elements in this group are 1,3,9 so if you’ve counted all the ones of order 1 and 3 then you can find how many of order 9

[vast verge]

I know that there is:

• 1 element of order 1
• 8 elements of order 3
• 18 elements of order 9 (by process of elimination)
  However, I'm still not sure whether 18 is correct

[sonic coral]

well you need the lcm of the orders to be 9 like you have written

you could have (1,b) where b is order 9 in Z_9

or you could have (a,b) where a is order 3 in Z_3 and b is order 9 is Z_9

if you count the number of possibilities you should get 18 to verify that it is what you should expect after counting all the order 1 and 3 elements

[vivid kestrel]

ping

[vast verge]

Could someone's explain what they're doing in the highlighted parts?

[sonic coral]

in the cyclic group Z_n, the number of elements of order k where k | n is given by phi(k)

[vast verge]

Thanks, it's been a while since I've seen that

[sonic coral]

so they’re just counting the number of elements of the given orders that would be required to have the correct lcm

[vast verge]

So I've forgotten

[vast verge]

Can someone explain how I got this wrong please?

[velvet hull]

you are undercounting, as long as at least one of a,b has order 15 the other can have any order you want from {1,3,5,15}

[south patrol]

Ye like order of (a,b) is lcm of orders of a and b

[vast verge]

|b| must divide 20, so |b| can only be 1 or 5 in this situation

So, to my knowledge, these are the only "valid" pairings that give a lowest common multiple of 15

[south patrol]

There are 4 elements of order 5 in Z/20, not 3

4, 8, 12, 16

That givds you anothee 8

Indeed note the number of elements of order p is always -1 mod p by Lagrange's theorem

[vast verge]

Oh thank you, I think that was what I was getting wrong

Doing all those order counting problems in my head makes it easy to get it wrong

[south patrol]

yeah nws

[soft tiger]

I'm trying to prove this about the symmetric difference on the power set. I know that it forms a Boolean group.

\begin{proposition}
Let $\triangle$ denote the symmetric difference on $\mathscr{P}(X)$. Then $\triangle$ is the unique group operation $$ on $\mathscr{P}(X)$ such that
[
(\forall A,B \in \mathscr{P}(X)),, AB\subseteq A \cup B
\end{enumerate}
\end{proposition}

[cloud walrus]

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

[soft tiger]

I don't see how to prove this from the assumptions given

[thorn jay]

Are there any restrictions on X size wise? Is it finite?

[soft tiger]

X is arbitrary

[thorn jay]

Damn

[karmic moat]

identity

[kind temple]

i don't think one exists. there are a lot of exercises in that book

it would be quite a project

[cedar vault]

Well, many of them could be passed off as one liners in a manual

[cursive spindle]

You can probably find some bits and maybe combine

[cursive spindle]

Every time I trust my intuition on tensor products I just get everything wrong

[vast verge]

I don't understand how I got the highlighted part wrong

[kind temple]

15 has order 20 in Z_100, not order 4.

so <15> is isomorphic to Z_20, not Z_4

[vast verge]

Does that mean the typed solution is incorrect then?

[kind temple]

yea

[vast verge]

Okay thank you

[kind temple]

also, i hate that they chose to swap the order of Z_2 and Z_12

[vast verge]

I thought it would be okay for me to do it since it would remain isomorphic (my working is on the third image)

[delicate orchid]

it's mathematically fine, but morally ur going to the 7th circle

[next obsidian]

Of fun

[minor gazelle]

Is my proof correct? A disclaimer I haven’t proven the intersection preserves the submonoid since I am learning language of group

Though I’m essentially saying take the elements from smallest submonoid (since defined as intersection) and then such that the product x•y in every submonoid containing A

Since Generator is arbitrary so immediately it holds for the smallest submonoid containing A

Or I should just do the intersection preserves the submonoid first

I should make it separated

This look weird

[rocky cloak]

It's a little unclear what's happening in the proof. I'm assuming <A> is defined as the intersection of submonoids containing A.

And that you use "generator" to mean one of those submonoids.

So the structure is correct I think, but it would be nice to write it out a little clearer.

And I guess you should say e is <A> because e is in all T

[minor gazelle]

I kinda want my note to develop naturally so I want somewhat to separate the proof for that generated submonoid is a submonoid and monoid is closed under intersection so it’s more natural for my brain to accept…

And should I call T a submonoid? Or generating submonoid

I only know generator becasue of sigma algebra

[tribal moss]

I think we need to see definitions both of <A> and "generated submonoid" in order to understand that proof.

[thorn jay]

Is there even a distinction between generated submonoid and submonoid?

Using the term "generated" usually includes that it's "by the set X"

[minor gazelle]

I defined it as intersection though I will be sending in a couple min I’m doing chore 🫣

[tribal moss]

Yeah, but then again every substructure is generally generated by itself, so that's not much of a property.

[thorn jay]

Yeah that's what I mean

[minor gazelle]

Here

My wording can be weird, I rephrased it after the vids on YouTube..

A study techniques that rephrase anything after you learned it 🫣

I want to separate the proposition such that the <A> is a submonoid and the smallest submonoid containing A and the proposition that the monoid is closed under intersection

Please definitely point out if there’s any awkward wording🥰

[rocky cloak]

You're missing the condition A \subset T in your definition.

It would also be good to include what "generator" and "generated submonoid" means somewhere.

[minor gazelle]

Fixed! Is it proper to call T a generator?

[rocky cloak]

I mean, there's nothing wrong with it I guess, but it's not a terminology I've seen used before

I guess it's kind of clashing since the elements of A are usually cashed the generators of <A>

[minor gazelle]

How’s now

And is the study method for algebra fine? So you see some video so you prove what you watch

[coral spindle]

I would suggest reading a book instead but you do you

[rocky cloak]

Trying to prove a lot for yourself is generally good I think. But I guess it comes down to which videos you watch, and how you're organizing it otherwise

[minor gazelle]

But I am not sure if I am dedicated enough

[coral spindle]

and how are we supposed to judge that lol

[minor gazelle]

Like it was coincidental i like analysis and book is expensive so if i like it ill buy one. A person I befriended here suggested me a book

If I would be able to solve something like a midterm of any school’s exam for respective subjects then I’ll buy it? Is that a good plan?

[twilit wraith]

I suppose so

I think just reading about a topic online and enjoying it constitutes buying a book

[coral spindle]

I cannot encourage piracy in any way. Have you ever googled the name of a book by the way?

Some books that Google might know:

• Fraleigh's "A First Course in Abstract Algebra"
• Dummit & Foote's book
• Artin's book.

[minor gazelle]

Yes I can just get it there

Certainly I’m not gonna do it but thanks for the information 🥰🥰🥰

Note is really I haven’t had any idea beyond this I can have a relative more mathematical environment 🫣

[dire wren]

Is there something like a feild except we don't require 1 to have an additive inverse ?

I feel like that would have interesting behaviour

(so 1 would not have an additive inverse and 0 would not have a multiplicative inverse)

[coral spindle]

Yes, some people call these semifields (although a few different things are called 'semifields'). I saw a talk a while ago that connected these to some other object in an interesting way, but you'll have to forgive me because I don't remember the details. They are quite niche.

[rocky cloak]

https://en.m.wikipedia.org/wiki/Semifield

[dire wren]

This doesn't seem to contain the thing I was talking about

[rocky cloak]

In ring theory, combinatorics, functional analysis, and theoretical computer science (MSC 16Y60), a semifield is a semiring (S,+,·) in which all nonzero elements have a multiplicative inverse

[dire wren]

A semiring does not have additive inverses at all

[rocky cloak]

Then I don't understand your question.

You want the field to have additive inverse, but for every element except 1 somehow...

[vast verge]

Can someone explain the highlighted part please?

[rocky cloak]

Like the positive rational numbers is not an example of what you're interested in?

[dire wren]

It is not an example of what I am interested in

[rocky cloak]

Then I think you need to explain further what you want

[rocky cloak]

This is not true (unless there are some unstated assumptions on m and n)

For example Z/3 (+) Z/6 is not cyclic and has only one element of order 2

[south patrol]

This is false

Or for another example, Z/3 (+) Z/7 has no elements of order 2 but is cyclic

[vast verge]

Here's the associated question

[dire wren]

there exists binary + and × functions on a set S
a + b = b + a
(a+b)+c = a+(b+c)
a*(b+c) = (a*b) + (a*c)
a*b = b*a
(a*b)*c = a*(b*c)
for all a,b,c in S
There exists 0 and 1 in S such that for all a in S
a+0=a
a*1=a
for all a not equal to 1 in S there is b in S such that
a+b = 0
for all a not equal to 0 in S there is b in S such that
a*b = 1

[dire wren]

@rocky cloak

[velvet hull]

the latter part is a confusing statement, but a proof for this result would look something like this -

• in C*, the only elements with finite multiplicative order must lie on the unit circle
• you can only fit cyclic groups into the unit circle, and it's easy to see that C* has a subgroup isomorphic to any cyclic group

[rocky cloak]

Then
1 + a^-1 * (-a) = 0

So a^-1 * (-a) is the additive inverse for 1

[dire wren]

Proof?

[rocky cloak]

Multiply both sides by a

[dire wren]

How do you show 0*a = 0

?

[rocky cloak]

More detailed:

say a*0 is different from 1 (otherwise 0 would have a multiplicative inverse)

Then a*0 = a*(0+0) = a*0 + a*0
so subtracting a\0 on both sides gives
a*0 = 0.

[dire wren]

Oh, I didn't think of first step

[rocky cloak]

Then
a + (-a) = 0
multiply by a^-1 gives
1 + a^-1 * (-a) = 0

[dire wren]

Hm

But why is it impossible for 0 to have multiplicative inverse?

I just never postulated it to exist

I never stated it doesn't exist

Oh got it

Cause 0 into anything would be 1

So the thing would consist of almost 3 elements

And that is not fun

[rocky cloak]

Another thing to note is that 1+1 cannot have an additive inverse either, so 1+1 = 1.

Which means a+a = a after multiplying by a. Which means
a = 0 for any a not equal to 1

[dire wren]

Yes

[rocky cloak]

But you can have {0, 1} with OR and AND I guess

[tribal moss]

I will stick my neck out and say, no that is not proper. It invites confusion with the existing convention of calling the elements of A "generators".

[tribal moss]

"then <A> is a generated submonoid" is still a very weird wording that I don't know the meaning of. Every submonoid has some set that generates it (if nothing else, then the set of all its elements), so if "generated submonoid" is intended to mean something different from merely "submonoid", then you need a better definition for that.

But your proof only purports to prove that <A> is a submonoid (which is fair, I suppose), and doesn't contain anything that appears to claim it is a "generated" submonoid, whatever that means.

[vast verge]

I literally have no idea how this solution was acquired, can someone help?

[velvet hull]

since G is abelian, when you multiply every element in G every element will cancel with its inverse, given that its inverse is distinct

try and figure it out from there

[rocky cloak]

The case where all ni are odd and the case where you have a unique element of order 2 should be the easy ones.

Then there is some work required for when there are several even ni's

[vast verge]

Oh right I remember doing that in an exercise from a few weeks ago but I forgot

But it only works if there are no element of order 2

But every group of even order must have an element of order 2

[tribal moss]

That's the "none of the n_i are even" case,

[vast verge]

And in this case we get an even group if one of the nis is even

[tribal moss]

A hint might be to consider each component of x separately.

E.g. how does $x_i$ relate to $\sum_{g\in\bZ_{n_i}} g$?

[cloud walrus]

Troposphere

[vast verge]

Well the answer would be x is the identity if and only if ni is odd. Otherwise, x would be the unique element of order 2

By how inverses cancel out after you sum everything together

[velvet hull]

you only need to think about the sum of each individual component

and when you sum over the entire group, the result is that each component Gi is exactly the sum over Gi multiplied by |G|/|Gi|

[tribal moss]

That sounds like you're forgetting that you're getting a copy of each g in Z_n_i for each choice of what all the other components can be.

[vast verge]

You're basically asking me about this simpler case, right?

[tribal moss]

For example let's take Z_3 (+) Z_3.
The sum of all the elements is x=(0,0)+(0,1)+(0,2)+(1,0)+(1,1)+(1,2)+(2,0)+(2,1)+(2,2).
In the first position of x we have 0+0+0+1+1+1+2+2+2, not just 0+1+2.

[tribal moss]

I was asking how that case relates to the full case.

[vast verge]

Oh my bad my bad

Well what we care about are the even Zni's

Since the odd ones will just be the identity again and again

I think?

[tribal moss]

Yes, so the question is, what will the positions with even n_i be?

[vast verge]

They will either be the identity or their unique element of order 2

Depending on how many times the entire groups gets repeated

[tribal moss]

Yes. And the number of times it gets repeated is ...

[vast verge]

Well if you're looking at Zni, it gets repeated n2n3...*nk times

[tribal moss]

Yes.

(At least when i=1).

So do we agree that the sticking point is whether the product of all the other n_j is odd or even?

[vast verge]

What do you mean by sticking point? Sorry, I don't understand

[tribal moss]

The question that determines whether the i'th component of x is the identify or has order 2 is whether the product of all the other n_j is even or odd.

[vast verge]

Yeah I understand that

[tribal moss]

So x_i has order 2 iff n_i is even and all other n_j are odd.

[vast verge]

And we require all other $n_j$ to be odd because if they weren't then $x_i$ would be repeated an even number of times and it would also become the identity

[cloud walrus]

Tropical Greens

[tribal moss]

Yes.

[white oxide]

Can I please have a hint for 6a? This is what I have so far:

Given ((b_1, \dots, b_m, 0, \dots, 0) \in B_m) and (1 \leq i \leq m), ((0, \dots, b_i, \dots, 0) \in B_i), and so (b_i = n_i \pi_i(\textbf{x}i) = n_i x{i,i})

Thus, ((b_1, \dots, b_m, 0, \dots, 0) = (n_1 x_{1,1}, n_2 x_{2,2}, \dots, n_m x_{m,m}, 0, \dots, 0)).

However, I don't think this can be written as a linear combination of the (\textbf{x}_i)'s.

[cloud walrus]

okeyokay

[white oxide]

Wait hold up

Nvm this is chill since since $\pi_j(\mathbf{x_j}) = \pi_j(0, \dots, x_j, \dots, 0)$

[cloud walrus]

okeyokay

[cedar vault]

I can prove it for finite X.
If X is finite and * is a group operation satisfying the property given, we have the following:

ø must be the identity. || ø*ø ⊆ ø, so we get ø*ø=ø||

singletons have order 2. ||Indeed, if S is a singleton, S*S ⊆ S and S*S ≠ S, so S*S=ø||

The product of any two singletons is indeed their symmetric difference.
||If S and T are two singletons with S = T, then S*T=ø i.e their symmetric difference.
If S≠T then ø≠ S*T ⊆ S U T but since neither S nor T = e, S*T ≠S or T. So S*T=S U T i.e their symmetric difference.||

Also note S*T = S Δ T = T Δ S = T*S i.e singleton products commute

Using the above fact and induction, show that any element of (P(X),*) is a product of its constituent singletons.

Using the above fact and that singletons have order 2, it directly follows that A*B for any arbitrary subsets A,B is their symmetric difference.

[cedar vault]

If you can prove ø is the identity without using finiteness you are done for the general case since the proof will carry over

[thorn jay]

I'm not sure I see how the step "using the above result and induction" can be generalised to infinite sets

The empty set must be the identity by the fact that ø * ø ⊂ ø, which means that ø * ø = ø

[cedar vault]

I must have hallucinated that, i dont think its true either

[cedar vault]

Thanks for pointing this out, i knew i was missing something simple

[thorn jay]

Happens to the best of us lol

[rocky cloak]

Hmm, so if S is a singleton and A is any set not containing S, then
S*A < SuA and also S*S*A = A, so A < S*A. Since S*A isn't A it must be that S*A = SuA.

So if one of the sets is finite, the operation is symmetric difference.

For any A, there must be a B with A*B = X. Then B must contain A^c. Take a finite set S in the intersection of A and B (assuming the intersection is nonempty)

Then X = A*B = S*S*A*B = (A\S)*(B\S), so the union of still everything when you remove S -> S is empty

Now take any set C in X and any A in C. Then there again is a B such that A*B = C.

A and B can't intersect by the same argument as above. And if S is a finite subset of B outside C then
A*B*S = C*S
A*(B\S) = CuS
Which contradicts CuS < Au(B\S)

So anytime C is a disjoint union of A and B, A*B = C

Then for any A and B if there intersection is N we have
A*B = (A\N)*N * N * (B\N)

So we just need N*N = Ø...

Take a finite subsets S of N*N, then N*N = N*S*S*N = (N\S)*(N\S) which doesn't contain S, so S is empty

So it's kinda like induction I guess. It's reducing to singletons/finite sets

[tribal moss]

It feels like you can probably get away with doing just Ø and singletons as the first two steps, and then moving on to larger sets, finite or infinite, in one go using your S*S trick.

[rocky cloak]

Yeah, you just need singletons

[tribal moss]

So, putting things together:

1. Ø is the identity.
2. {x}*{x} = Ø
3. {x}*A is the symmetric difference, by considering {x}*{x}*A as above. Similarly from the right.
4. For arbitrary A and B, an element that's in neither is also not in A*B, by the original specification.
5. If x is in both A and B, then A*B = A*{x}*{x}*B = (A*{x})*({x}*B), reducing to case 4.
6. If x is in A but not B, then A*B = {x}*{x}*A*B = {x} * ({x}*A)*B, and using 4 on ({x}*A)*B and then 3, we see x is in A*B.

And this works not only for P(X), but for any subset of P(X) that contains all the singletons -- say, the Borel sets in P(R).

[thorn jay]

I.e. for any Boolean algebra B there cannot exist a group structure on B such that ab < a ∨ b that is not the symmetric difference, or the addition of the corresponding Boolean ring.

[tribal moss]

Hmm, the argument we have so far depends on the Boolean algebra having enough atoms.

This is not always the case -- for example if we take the Boolean algebra of periodic infinite binary sequences, it has nothing to play the role of singletons in this argument.

[thorn jay]

I didn't see the "that includes all singletons" clause lmao

[tribal moss]

If we didn't have singletons (and Ø) we coud do something like f: R -> P(R)

f(a) = (-infty,0] cup [a,infty) for a >= 0
f(a) = (-infty,a] cup [0,infty) for a < 0
and then define A*B = f(f^-1(A) + f^-1(B)) on the range of f.

[amber burrow]

where is the (n-1)! coming from?

[tribal moss]

Assuming you mean the (n-1)! in the last line of your image:
If f is an n-cycle, then there are n-1 different elements that can be f(1); then n-2 elements that can be f(f(1)), then n-3 different elements that can be f(f(f(1))), ...

[thorn jay]

Nice!

[amber burrow]

no, in the paragraph

why are there (n-1)! ways to make this identification

[tribal moss]

Because there are n-1 cosets other than H, and you can assign them to the n-1 numbers from {2,3,...,n} any way you want.

[amber burrow]

oh

but then why does phi need to be transitive?

[rocky cloak]

Gamma acts transitively on Gamma/H. g takes H to gH

For a subgroup of Sn the stabilizer of 1 will have index equal to the size of the orbit of 1. So it's only n if the action is transitive.

[amber burrow]

but then isn't that only n different possibilities - 1 for each representative taken from a given coset

oh wait

is it because you might identify other cosets differently?

like lets say your cosets are H, g_1H, g_2H, g_3H,g_4H

[rocky cloak]

Yeah. It's just a bijection between {1, ..., n} and G/H that send H to 1

[amber burrow]

ok yes that makes sense

[white oxide]

Can somebody please give me a hint for this question? I really have no clue how to proceed, especially since the universal property is given for modules over the same ring

[south patrol]

I assume you are fine with how you can form these tensor products M (x)_A N, but not sure how to put the B-module structure on M (x)_A N (and similarly other way round)?

[white oxide]

I'm assuming the B-module structure on M (x)_A N is b(m (x) n) := m (x) nb right

I'm just confused because if we write M (x)_A N = C and N (x)_B P = D then we're asking that C (x)_B P is isomorphic to M (x)_A D. But the only techniques that I've seen for showing tensor products are isomorphic only apply when they're all over the same ring

Or maybe I can establish some bilinear map out of (M x N) x P into M (x)_A (N (x)_B P)? but then I'd have to consider (M x N) x P an A-module and I'm not sure if I'm allowed to bestow my own structure upon it

Am I supposed to use the fact that A (x) N \simeq N \simeq B (x) N?

[south patrol]

Yes, exactly

Small remark: because this text uses commutative rings, the order doesn't really matter, but you need to be more careful in the non-commutative case

[south patrol]

Yeah this is a good point

It's also slightly unclear because each of these objects is an (A,B)-bimodule

[rocky cloak]

You can try to show that both satisfies a universal property of multilinear maps out of MxNxP

[south patrol]

You can also use tensor-hom adjunction to show that maps out of the two are the same, which is exactly the same thing repackaged

[white oxide]

I don't know what that means but I'll try jagr's recommendation

Thanks both

[rocky cloak]

Does "click" mean something different from making perfect sense to you?

Like isn't clicking when you see that it makes sense

Have you tried just computing an example?

Like some small group and subgroup and just see what all the cosets are

[thorn jay]

Just take any group and subgroup?

[south patrol]

Computing examples scares me

[thorn jay]

Yeah...

What do you mean I have to work with concrete objects

[south patrol]

I guess what I have in mind is like idk I feel like I used to be like eh examples are boring and often can just be done by computer etc

but now for me examples can be extremely hard to come by or hard to compute lol

maths is hard

[rocky cloak]

Well, the doing it yourself is kinda part of the process

[south patrol]

I have a proposition in smth i'm writing where I have 0 examples where the conditions aren't held lol

but i suspect it isn't even usually true

[thorn jay]

At least they didn't say "let me ask gpt"

[thorn jay]

Could it be that
Perhaps
The conditions always hold?

[south patrol]

I would love to know this but suspect they don't even usually hold

[thorn jay]

For sufficiently loose definition of 'always'

[south patrol]

hmm

probably many things like this!

[glad osprey]

Goldbach's conjecture (I have a nasty counter example)