groups-rings-fields

[sonic coral]

i’ve proven burnsides p^a q^b theorem though so i could go a lot higher if i wanted to use it

[iron arrow]

If we let G ={a1,..,an} be an abelian finite group and let x = a1·...·an then show that x^2 = e.

Could I show this by using Cayley's theorem and using that every permutation can be decomposed into transpositions and because of commutativity x^2 is basically just a product of squared transpositions which are basically a bunch of fix point permutations and therefore they are the identity, and therefore x^2 = e?

[elfin wraith]

That’s way more complicated than it needs to be lol

What does x^2 mean, and what does it mean for a group to be abelian

[iron arrow]

Abelian means commutative and x^2 is just x*x

[elfin wraith]

Are you looking for a proof, or are you looking to see if your suggested proof is correct

If it’s the former there is a significantly more straightforward approach

[iron arrow]

If the idea for my proof could be correct

I know it needs to be written down better

And I imagine there is an easier way (and I am interested)

[mint seal]

I could see that idea working out

[iron arrow]

But I'm also interested to know if the way I did it is also valid

[rocky cloak]

This doesn't work because transpositions don't commute with each other in general, and you can't guarantee that the transpositions are in G, so it doesn't help that G is commutative

[mint seal]

ah

[iron arrow]

Ok that makes sense

Is the easier way just seeing that x=x^-1 or something like that?

[rocky cloak]

Pretty much yeah

[elfin wraith]

Yeah, you know that every element has an inverse, x is the product of all elements, they commute so you can pair them

[iron arrow]

Yeah I've been doing rings and fields most of the year and just entered into groups so I'm not very used to the mechanics of working with them

[elfin wraith]

There’s something to be said for idempotents but the fact that you’re considering x^2 deals with that

(Im being intentionally vague so you can fill out these details)

[iron arrow]

Thanks

[thorn jay]

rings and fields before groups is an interesting choice for sure

[rocky cloak]

Not that uncommon is it?

[iron arrow]

Other than finite fields we didn't need groups for much of anything

[elfin wraith]

It’s not unheard and I’ve mentioned here recently that I’m increasingly interested in the idea of it

Most of the objects you already know are naturally rings and fields (ok sure by extension groups but you think about them as rings)

[rocky cloak]

If you're example focused / top down kind of teaching, it seems the natural thing to do

[thorn jay]

right, that is true

[elfin wraith]

Rings (at this level) also tend to have the nicest presentations etc, and this is kinda what’s making me think this could be a nice approach

[iron arrow]

I would understand there's like a monomorphism of the type a |-> a^-1 then?

[elfin wraith]

And maybe it’s just me but I’ve never really been super happy with the whole groups are symmetries thing beyond like S_n and D_n

[iron arrow]

I would say rings and fields had absolutely no complication being explained before without group theory

[thorn jay]

commutative rings*

[iron arrow]

Nothing needed any sort of group theory introduction

[elfin wraith]

Again still too complicated, genuinely just unfold the definition of x^2 and use the fact that you can compute things

[elfin wraith]

Sure yes, but even noncom ones can have pretty nice presentations, like ones that are deformations of commutative rings (Weyl algebras, quantum plane etc etc)

Like to me it’s easier to understand what quotients do there than saying like S_4/A_4

[thorn jay]

right but im sure those satisfy a multitude of other nice equations that make them nice lol

group presentations are in general horrible to work with

[elfin wraith]

I’m not convinced this is the way to go, but it’s part of why I’m open to considering a rings first approach these days

Even if the monoid, group, ring, field build up is nice

[thorn jay]

i am enjoying the speed at which our rings and fields course is going, as it can assume some level of mathematical maturity in abstract algebra

[iron arrow]

Ah yeah I'm dumb I was trying to see why two elements couldn't have the same inverse but that's obvious

[elfin wraith]

The fact you considered that is a good sign though

The right kinda thinking

[mint seal]

yeah, not dumb, important thing to check

[elfin wraith]

Don’t share your channels please, someone will help you when they can

[karmic moat]

put it in every advanced channel damn

[thorn jay]

make a promotional poster and maybe well accept it

[iron arrow]

And the post is in spanish and about probability

For some reason

[thorn jay]

were they poofed?

[thorn jay]

such linear algebra

[iron arrow]

Well returning to the rings thing, I'm pretty grateful for having started with rings as it probably was one of the most beautiful topics I've done in the degree and one of the reasons I'll probably be trying to do more of this

[rocky cloak]

Rings -> Modules -> homological algebra
Follow the rainbow and you find gold

[iron arrow]

I don't think we'll get to modules (as a topic) but it is "scheduled" as one of the topics if we have time

Like our course is rings -> fields -> groups -> modules

[karmic moat]

rings -> modules -> algebraic geometry
follow the rainbow and you find diamonds

[iron arrow]

But modules is probably not gonna happen

[rocky cloak]

Always a next course

[thorn jay]

rings -> modules -> ??? -> UA
turn in an orthogonal direction and you'll find netherite

[karmic moat]

wtf i want netherite :(

[iron arrow]

I know I'll take abstract algebra, galois and algebraic topology but idk if I should take algebraic geometry, I enjoyed projective geometry a lot but idk if there is really any relation

[rocky cloak]

rings -> module theory -> dyslexia -> model theory -> UA

[iron arrow]

Albeit maybe this isn't the channel for thay

*that

[thorn jay]

ahhh that does make sense maybe thats what happened

[rocky cloak]

I mean there's a connection, but probably not in a "if you like this you'll like this" kinda way

[iron arrow]

Understandable

And where do you find homological algebra? There's not any courses called like that in the undergrad or the masters

[thorn jay]

better start heading in an orthogonal direction then

[karmic moat]

Usually one picks up homological algebra in a course on modules in my experience

You'll also get a taste of it in algebraic topology hopefully

[rocky cloak]

It's a central tool in algtop, alggeo and rep theory at least

[karmic moat]

On it

[iron arrow]

That's cool

[rocky cloak]

So could be baked into such courses

[iron arrow]

Mainly I want to do algtop without knowing much about it because of how much I enjoyed topology and algebra courses in general

[elfin wraith]

So far all the, admittedly limited, homalg I’ve learned has just been as I’ve went

[karmic moat]

You can always skim a book

[elfin wraith]

You just kinda need some here and there and you learn it as you go

[karmic moat]

Or there's those AMS bulletins (or whatever it is) that's the "What is..." series

[elfin wraith]

Algtop is good fun, it answers the question, “why do so many people study topology, when general topology sucks so much?”

[thorn jay]

you take that back

i will not stand for this general topology slander

[iron arrow]

General topology is my favorite course in my degree

By far I'd say

[rocky cloak]

Algtop and difftop are just different answers to "what if topological spaces where actually nice?"

[thorn jay]

Nope is just traumatized

[elfin wraith]

I’m only half joking, I really do enjoy algtop and I personally found general topology a slog, but if you enjoy it that’s great, I’m sure you’ll like algtop even more

[quiet pelican]

Every topological space is locally metric 🙃

[elfin wraith]

Metric spaces means doing analysis, I don’t get down with that

The only ordering I get down with is inclusion

[iron arrow]

Honestly I've always hated calculus classes, but real analysis and rn probability theory have been insanely cool

Still, algebra on top

[karmic moat]

> probability theory
> insanely cool

[thorn jay]

Every poset is isomorphic to some poset of sets under inclusion im pretty sure

[karmic moat]

waiter waiter one billion more nested integrals please

[thorn jay]

so we're all good here

[elfin wraith]

Nah it sounds like you’re having fun and enjoying it though, that’s good to hear

We’re all just bitter jaded and usually joking

[thorn jay]

take the set of all principal filters

[elfin wraith]

Not about analysis though, I genuinely considered doing some stochastic stuff this semester until I looked at the notes and remembered why I don’t do that stuff

[iron arrow]

What I'm doing isn't really about that, sometimes there appear like "n-variable nested integrals" but that's not the main point

[rocky cloak]

Cute probability exercise:

Take a finite non-abelian group G. Pick two elements uniformly at random. Show that the probability that they commute is <= 5/8

[iron arrow]

And I started with discrete probability theory

[iron arrow]

Lol

[karmic moat]

im salty because im not good at probability

because im not good at integrals

and my probability class was just a bunch of integrals with some convergence theorems thrown in

[elfin wraith]

Everyday im thankful my probability course was during Covid and I got to sit it online

[iron arrow]

Idk why but I've also enjoyed measure theory

[elfin wraith]

The little I know of measure theory was pretty fun, it’s not something I know well though

[karmic moat]

i dont remember exactly the grading scale after the first midterm for my probability class but it was like 70%+ is an A and some below it is A- and some below it is B+

[iron arrow]

Xd

We don't do that here

We just go from 0-10 and there's no letters

[elfin wraith]

This is just the standard grading scale in the UK lol

[iron arrow]

And it's just seeing the IMOs get perfect scores

[karmic moat]

standard in the US is like 93+ = A, 90-93 = A-, 87-89 = B+, and so on and so forth

[elfin wraith]

But yeah, that extra like hour due to it being online was absolutely clutch, I had so many thoughts in that last hour and I ended up doing crazy well, like significantly better than I really think I should have

That and analytic number theory were by far my biggest swindles of UG

[iron arrow]

Analytic number theory in undergrad?

[elfin wraith]

I got like an 80 something in both courses, neither of which were remotely deserved. Couldn’t even tell you what an L function is

[rocky cloak]

Huge L on that

[elfin wraith]

The basics are very accessible to final year UGs tbh

[thorn jay]

get em started young

[karmic moat]

W

[iron arrow]

Understandable ig

[rocky cloak]

Not sure if number theory has a W function yet

[iron arrow]

Apostol

[elfin wraith]

We didn’t follow this book, but like, it’s in the UG series!

[karmic moat]

i think that's been taken by analysis

[elfin wraith]

I think we actually followed basically the opposite approach to apostle

[iron arrow]

My only like advanced mathematics book I possess is uh hatcher's algtop

[karmic moat]

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

though no one has ever used the same letter before in different fields in math

[thorn jay]

lambert has one i think you should ask him

[rocky cloak]

Looks like Lambert takes the W on this one

[iron arrow]

But I would really like to have a physical copy of gouvea's p-adic numbers book

Idk if there's better reads to p-adic introduction but it was beautiful to read that

[elfin wraith]

I refuse to learn about p-adics because then I’ll have to stop writing Z_n

[thorn jay]

lazy ahh

still would be annoying because R_f is standard notation for localisation by { 1, f, f^2, ... } right lol

[elfin wraith]

R_(f) usually no?

[velvet hull]

if f is a single element R_f can be accepted

[elfin wraith]

Simply don’t work with commutative rings

[thorn jay]

that would be localisation at a prime ideal if (f) happened to be prime

local rings and all that

[velvet hull]

but in the specific case of localizing by primes in Z then you would have to do Z_(p) yeah

[iron arrow]

Tbh I haven't seen anything about non com rings, is it accessible to undergrads through some book?

[elfin wraith]

Who’s out here localising away from primes

[thorn jay]

Z_p would be Z[1/p]

[velvet hull]

localizing by the monoid generated by non nilpotent elements is pretty much the only other useful thing to localize by other than prime complements

[elfin wraith]

Lam is a common book, there’s also (roughly what I learned from) Noncommutative Noetherian Rings by Goodearl and Warfield

[thorn jay]

the humble structure sheaf and associated modules

[elfin wraith]

I say roughly because I took a course that was inspired by but didn’t 100% follow said book

[iron arrow]

And do ideals in non com rings work like lateral classes in groups or is it just not defined? Just curious to know what makes them so special other than seeming ass to operate with

[velvet hull]

no the definition splits into left and right ideals separately

they are still well defined, just no longer nice

[elfin wraith]

I’m not sure what lateral classes are, but for ideals in noncom rings you have to pick a side to multiply on, left and right ideals generally don’t agree

[iron arrow]

Lateral classes as in left and right yeah

Sorry I'm just like directly translating from spanisj

*spanish

[elfin wraith]

But also modules tend to be the more interesting things to look at (which are submodules of the ring viewed as a module over itself)

I won’t lie to you and tell you it’s more useful or that most people care. There’s a reason we’re saying noncom rings rather than simply rings…. But it is fun

[iron arrow]

I don't really care if something is useful or not tbh

I just like when things work beautifully and have cool properties

[elfin wraith]

I think it comes down to what you’re looking for. A lot of people think ring theory is a necessary evil, they need to know it to do the maths they like but they think it’s a bit dry on its own, so they usually just slog through comalg. For me, I really just enjoy the actual algebra, so noncom rings feel natural, it’s just a wider class of rings where things are generally just harder but that means you get more interesting questions to ask

[iron arrow]

That seems pretty cool tbh

I mean if I get the chance to go for a non com algebra class I was gonna for sure take it

[thorn jay]

there is very good reason to care about noncomm rings and generalisations (small preadditive categories) imo

[rocky cloak]

I mean, they are sort of crucial in defining modules, since End(M) is non-commutative

[vapid vale]

i'll do you one better i took a course on automorphic representations and got an A. i retained at most 15% of the content

the class was entirely graded on the professor farming the class collectively for a solution packet to their textbook, so i did some low hanging fruit

[karmic moat]

Do you get a cut of their textbook royalties

[elfin wraith]

That’s so lucky hahah, for me it was just the prof gave, let’s say, very leading practice problems and homework’s so I just kinda knew how to answer those and I was fine. I wouldn’t even say that I don’t understand it because like there wasn’t anything in the class I didn’t understand but I certainly didn’t retain any of it

I’m sure if you put some words in front of me id remember them, or maybe not

[vapid vale]

we did reductive groups and root data in the first two weeks and the class only sped up from there lol

[ebon prairie]

is the statement possible? or that's a wrong question?

[next obsidian]

It’s normal in H not in G

[ebon prairie]

oki ty

@stable yacht pretty sure what u said was against the guidelines

[tribal moss]

Otherwise we could take N to be G itself and prove that every H is normal.

[west radish]

I think this is true and possible. Cuz - N is normal in G and H is subgroup of G then (N inters. H) = need to be present N elements then it's must be normal.

[lusty marlin]

The statement is wrong. Take N to be G, as mentioned above.

[iron arrow]

Hey guys do you know like, any sort of really cool groups? Like with cool properties, but not on the type of like the symmetric group or the dihedric group, maybe some others with other relations to algtop or other cool stuff?

[west radish]

Ouu thnx

[twilit wraith]

"briefly describe the ring structure" feels so vague, do yall have an idea on how i should describe it without being too blunt?

i mean for the first one theres just an obvious isomorphism with Z2[x]

[quiet pelican]

I have no clue what they expect for c
That’s basically the simplest it’ll get

[quiet pelican]

Otherwise, stuff like this yeah

[thorn jay]

and its weird cuz you need a pretty explicit description of the ring structure to do the next part of the exercise

[mint seal]

isn't (c) isomorphic to polynomials of degree at most 1?

[quiet pelican]

Yes but I wouldn’t really consider that a more explicit description
Because it doesn’t account for the multiplication

As an abelian group it’s Z^2, but the ring structure is different

[mint seal]

mm, fair enough

you could say something like (ax + b)(cx + d) = (ad+bc)x + bd

in that ring

[thorn jay]

at least do it for a general element then

[mint seal]

that seems like overkill since you don't have higher powers in the quotient to begin with, really.. unless you choose wacky representatives

[thorn jay]

(a + bx + cy + dxy)(e + fx + gy + hxy) = ...

[mint seal]

oh but that's for more variables

[thorn jay]

we do

its a quotient of Z[x, y]

lol

[mint seal]

not (c)

[thorn jay]

oh c

man

i should sleep more

[mint seal]

hah, me too

[twilit wraith]

i mean as a set its the constant and linear polynomials

operationally its the same as Z[x] but x^2 = 0

[quiet pelican]

See the above discussion

[somber sleet]

can someone help me how to explain that f=x^n+y^n +z^n is irreducible in C[x,y,z] using eisentsein? My idea was to look at f as an element of C[y,z], then in this case we needed to find a prime element p in C[y,z] such that p does not divide 1 and p^2 does not divide (y^n+z^n)

[rocky cloak]

I guess y + wz where w is an nth root of -1

[somber sleet]

how do you write that though?

like I got y^n+z^n=y^2n-z^2n / y^n -z^n

[rocky cloak]

What do you mean? I just wrote it
p = y + wz

[somber sleet]

yeah no fair you're right

thanks

[rocky cloak]

I meant to write y - wz, but I'm sure you'll figure it out

[delicate orchid]

"Z adjoin a square root of 0" lol

[south patrol]

Aka Z

[rocky cloak]

Thick affine line over F1 must be what they were looking for

[delicate orchid]

no no no no n o NO NO the jacobson no no no

or like. Z adjoin an infinitesimal

that's the funniest one

[south patrol]

trivial square-zero extension of Z

[iron arrow]

Sorry, about this?

[quiet pelican]

Free groups, surface groups, lamplighter group, cactus groups

(I can give definitions of any you want)

[quiet pelican]

Thompson’s groups

[elfin wraith]

This is a bit vauge so hard to answer. So I guess, not a specific group, but a way to make groups, the amalgamated free product is quite cool. Its a kinda weird way of combining two groups and this is a thing that shows up in algtop quite a lot through something called the Seifert-Van Kampen theorem, which tells you about the fundamental group (the group of loops in that space, and in fact every group is the group of loops on some space) of a space which is built out of smaller known spaces.

Elliptic curves are another cool example, it turns out that cubic projective curves* (you can be a bit more general and precise here, this isnt strictly speaking what an elliptic curve is) have a sensible notion of addition that makes them into groups which is quite interesting, like its not at all obvious that these things are groups

[quiet pelican]

the problem is that elliptic curves are boring as abstract groups
Like the interesting part is the way they relate to the AG/NT underneath

[elfin wraith]

I agree, but I still think theyre interesting groups in so far as they are not at all obviously a group

[quiet pelican]

They’re finitely generated abelian with relatively small torsion component (iirc it’s always at most size 12, and there’s a full list of which torsion subgroups can occur)

[quiet pelican]

It depends on how you define it
As the divisor class group, it’s fairly clear it’s a group

[iron arrow]

Yeah it's just that I have this little assignment about like "research a cool group" (xd) and last time it was about rings and I did p-adic rings so that's basically the idea

But there's been some pretty cool mentions here

[elfin wraith]

This is a very weird and cool result though

[quiet pelican]

The interesting part is more that that can be described in terms of the geometric construction you usually get (but even that equivalence is doable in a first course on AG)

[elfin wraith]

I forget whose, Weil and someone?

[iron arrow]

That's very cool

[quiet pelican]

Mordell-Weil for finite generation

Abelian is trivial, the fact about torsion subgroups is very non-trivial iirc

[elfin wraith]

It is indeed very non-trivial

It wasnt proven in my AG course and I rember looking up a proof just to realise quite quickly why lol

[iron arrow]

And what about the dual group of some specific group? Or is it more interesting about the duals in general

[quiet pelican]

Dual groups in general aren’t really interesting outside of specific circumstances (that usually require the hypotheses of “abelian” and some kind of finiteness condition)

[iron arrow]

Hm ok

[elfin wraith]

I mean theres the obvious options of the 26 sporadics

[iron arrow]

26 sporadics?

[elfin wraith]

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

[iron arrow]

I think there's already a group doing the monster group tho

[elfin wraith]

Yeah it is the obvious option

[iron arrow]

Seemed interesting but ig monster was already the most interesting?

[quiet pelican]

The Mattieu groups are nice if you wanna do something fairly combinatorial

[iron arrow]

Not too enthusiastic about combinatorics tbh

[twilit wraith]

oh i also had a project where i had to research a cool group

[iron arrow]

But I can check them

[twilit wraith]

i did mathieu groups

[elfin wraith]

If youre interested in algtop looking at fundamental groups could be interesting

[twilit wraith]

theres some cool results about groups that are the fundamental groups of some spaces

i mean every group is a fundamental group of some space but theres specific occurrences

[iron arrow]

I did an introduction of that in general topology and I found it interesting, maybe it is a good idea

[quiet pelican]

Surface groups!

[elfin wraith]

Like the fact that theyre a group at all is not difficult but its not clear. Similarly with the fact that every group is the fundamental group of some space

[twilit wraith]

fundamental groups of manifolds and fundamental groups of lie groups are two pretty cool results

[twilit wraith]

i mean once you do some theory with loops on top spaces then its pretty clear

[quiet pelican]

$\langle a_1, \dots, a_n, b_1, \dots, b_n \mid \prod_i [a_i, b_i] = e\rangle$

[cloud walrus]

micoi the group things (0/1/0/5)

[elfin wraith]

Yeah I just mean that like, you do need to do some work for like associativity

[twilit wraith]

ive actually never seen the proof that every group is a fundamental group of some space

you probably just construct a space or smth i imagine

[elfin wraith]

Its in the end of chapter 1 of hatcher, he basically tells you how to construct the space

[twilit wraith]

using relations on free groups

[elfin wraith]

Yeah, you make a tree based on generators and relations

[quiet pelican]

Every group has a presentation (the trivial one)
Given a presentation <S | R>, the space is the S-fold wedge of S^1, and then attach 2-cells along the words corresponding to R

[twilit wraith]

ah i see

[rocky cloak]

One group that is pretty cool is Q/Z.

Every element has finite order, but the group is infinite. Every finitely generated subgroup is cyclic.

It breaks into a direct sum of Z[1/p]/Z for all primes p. Which are called the p-Prüfer groups. Fun thing about them is that each of its nontrivial quotient groups is isomorphic to it. And all its proper subgroups are finite.

Another fun thing about all these groups is that they're divisible. Meaning for every element x and natural number n there is a y with ny = x (so y "=" x/n)

[south patrol]

Very fun

[twilit wraith]

ok i have to prove that the additive and multiplicative groups of a field are never isomorphic using 3 different observations

the finite case is obvious

but then i have to do something with whether or not -1 equals 1

i guess to just think out loud

if -1 equals 1 then phi(-1) = phi(1) = 0

so then we have that phi(-c) = phi(c) for all c in F

well i mean in general couldnt we just say that phi(-1)^2 = 0

so that phi(-1) always has to map to 0 as F is a domain

so if -1 isnt the multiplicative identity we get that phi is never injective so it cant be an isomorphism

[rocky cloak]

That covers the case where -1 does not equal 1 yes

[twilit wraith]

yeah

hmm the -1 = 1 case seems to be the hardest

im sure its just going over my head tho lol

well that means 1 is characteristic 2

[rocky cloak]

So the problem before was that the multiplicative group had elements of order 2 while the additive group did not.

Now in this case, does the additive group have elements of order 2? Does the multiplicative group?

[twilit wraith]

yes the additive group has elements of order 2

if the multiplicative group has elements of order 2, then x^2 = 1

1 satisfies this so 1 would have to map to 0 or 1

-1 also does so itd have to map to 0 or 1

but theyre the same so eh

yeah i see

ok maybe i should define specifically what my function is

phi goes from the mult group to the add group and so phi(1) = 0

[rocky cloak]

So in the multiplicative group 1 has order 1, so that is not something of order 2.

And -1 = 1, so that's no good either

Any other elements around that could have order 2?

[twilit wraith]

but those are the only elements that can possibly satisfy x^2 = 1

since F is a domain

[rocky cloak]

Boom boom

[twilit wraith]

yeah i was overcomplicating that

i appreciate it

okay this problem is just super confusing

why would i have to show that all elements of the last ring square to 0 or 1? is that not just by virtue of 0 and 1 being the only elements in that ring???

[shrewd sandal]

But they're not, eg x

[twilit wraith]

oh wow im actually losing my mind

i really read that as Z[x,y]/(x,y,2)

[shrewd sandal]

Mood mood

[twilit wraith]

ok well now the question is interesting when i read it right

who knew

now i dont know how to describe its ring structure then

i mean its all elements of the form ax + by + cxy + d

[shrewd sandal]

2d?

[twilit wraith]

wait yeah no what am i saying

uhh

d with d just being in Z2 pretty sure

a,b,c,d all in Z2

[shrewd sandal]

a,b,c,d being in Z2 yeah

[twilit wraith]

idrk what else to say about the ring

[shrewd sandal]

Well then it's finite

[twilit wraith]

i mean it just boils down to what it is as a set and then operationally how it behaves

[twilit wraith]

true

[shrewd sandal]

Well you know that all elements in the set have additive order 2 (save 0)

[twilit wraith]

sorry i should clarify in that i know how to go about solving the problem

[elfin wraith]

Quotients of Q are scary, always weird and whacky

[shrewd sandal]

Gotcha

[twilit wraith]

but the briefly describe the ring structure part is just kinda stupid to me

[shrewd sandal]

Yeah I agree with that

[twilit wraith]

idrk why its there

thats the part that i dont know what more to say on

[elfin wraith]

I guess it’s to make sure you understand what it actually means?

[twilit wraith]

the squaring to 0 or 1 thing is just checking cases

actually possibly not even casework

[elfin wraith]

Like I saw the example earlier with like Z[x]/(x^2), I guess you just have to say degree one polynomials with integral coefficients

But I do agree it’s stupid, but I’m guessing “the point” is to make sure you know what the notation means

[twilit wraith]

fair enough

but still the main issue i have is that the algebra course im in is full of people who already have some experience in it (though my ring theory is lacking), so it feels unnecessary

alas

[elfin wraith]

So it goes lol

[topaz solar]

They’re also sorta the divisible groups, once you toss in Q too

[rocky cloak]

Yeah, every divisible group is a direct sum of copies of Q and Prufer groups

[ripe harbor]

Why can we argue that (n) = Ker just because Z/Ker cong Z/(n)?

[candid patrol]

And so, C* = S^1

[rocky cloak]

Yes they are both isomorphic to Q/Z (+) Q^N

In cursed axiom of choice ways of course

[rocky cloak]

If they're isomorphic then in particular they have the same cardinality.

The kernel is equal to (m) for some m, and Z/(m) has cardinality m, so n=m

[ripe harbor]

Why is the kernel equal to (m) for some m?

[rocky cloak]

Because the kernel is an ideal of Z

[ripe harbor]

Ah

[ripe harbor]

Yeah that's sound, thanks!

[iron arrow]

Oh yeah yesterday I did an exercise myself on that, it is indeed interesting (about Q/Z) but other than the finite order and that in R/Z, Q/Z is the subgroup of the elements with finite order and they don't have a bound to how big the order can be are there that many other properties? Sorry for the late response I have been busy for like 6 hours almost

[rocky cloak]

I mean there are interesting properties. The thing about divisibility was mentioned.

Another cool fact is that Q/Z is a cogenerator in the category of abelian groups. So without fancy words: for every abelian group A and non-zero element a in A, there is a homomorphism A -> Q/Z that maps a to something non-zero.

Further Q/Z is the unique minimal cogenerator, in that it is a direct summand of any other cogenerator.

The endomorphism ring End(Q/Z) is the profinite integers. It's sort of fun to describe all the endomorphisms

And the every finitely generated subgroup is cyclic I mentioned, that's pretty cool

[tough raven]

Coxeter groups are a class of groups with very interesting structure. But they include the dihedral groups and finite symmetric groups as special cases, so maybe they're not what you want.

[iron arrow]

Those are some pretty cool properties, albeit do you think there are like any sort of books on this topic or a generalization of this? (That isn't just quocient groups) mainly because the project is quite expositive so there's not place for proofs even if I'll still look at them

Albeit it does look very interesting

[south patrol]

Qp/Zp is also very important in some arithmetic geometry

[iron arrow]

I'll check

[iron arrow]

Qp as in p-adic?

[south patrol]

Namely lol the constant sheaf Qp/Zp is the prototypical p-divisible group

[iron arrow]

That would pair up really well with my last project on p-adic numbers lol

[tough raven]

Oh, another thing you could learn is BN pairs and almost simplicity, which shows that certain matrix groups G(k) are (almost) simple for any field k. You should be able to prove the steps explicitly for GL_n and black box them for a general reductive group (you can also black box what "reductive" means). The punch line is that this construction gives (I believe) most of the infinite families of finite simple groups when you take finite fields for k.

[rocky cloak]

Qp/Zp = Z[1/p]/Z right?

[iron arrow]

BN pairs

I'll need to check up some terms

Because we just started with groups

[tough raven]

Like, the definition of a group?

Sorry, my suggestion was probably way too advanced then.

[iron arrow]

Nono

I don't have a problem with advanced

It's just that I don't understand some key words you're saying so I have to check them to see if it is interesting

[tough raven]

I don't mean "advanced" as in difficult. I mean as in "needs you to know a lot first".

[iron arrow]

I mean after searching what simple means I can get it directly because I have seen normal subgroups

[tough raven]

If you're starting from what is a group now it will be very difficult and/or stressful to learn enough background to do this justice within like 2-3 months.

[iron arrow]

I know what is a group we've been doing a bunch of group theory already but we're like 2/5 of the topics on groups still

[rocky cloak]

They know rings and fields so I guess GLn(k) is within reach

[iron arrow]

Yeah

[tough raven]

OK

Actually

Does it need any theory of algebraic groups

[vapid vale]

matrix groups :)

[tough raven]

[iron arrow]

I am interested in those sort of things tbh

Like I don't mind researching from a good book

[tough raven]

Nvm

[iron arrow]

Like I did for the construction of the I-adic numbers before just centering into p-adic

[tough raven]

BN pairs is actually fine

mb

[iron arrow]

I'll take notes from all of your suggestions

And try researching a bit of my own on each

[vapid vale]

the lie group E8 is a classic

[iron arrow]

If you have any other suggestions I'll gladly take them, specially if there are some books that center on them or that there's a big section about them

Mainly because I like researching from those

[sacred wharf]

hi

Z_mn is isomorphic to Z_n x Z_m as rings if gcd(m,n) is 1 right ?

Is ring of integers isomorphic to 2Z ?

yes seems to be

[rocky cloak]

Not sure what you mean.

As groups Z and 2Z are isomorphic, but 2Z is not a unital ring

[sacred wharf]

ah hah so it doesn't have unity so it can't be isomorphic right

Because identity needs to map to identity of course

[sacred wharf]

is 2Z isomorphic to 4Z

Hmm

p(a) + p(b) seems to hold

so does p(a)p(b)

or no

nvm

[rocky cloak]

As groups yes, as (nonunital) rings no

[sacred wharf]

what are some good examples of isomorphic rings ?

[rocky cloak]

I guess you should be able to prove that mZ is never nZ unless m = ±n

[sacred wharf]

yea it's just multiplication property of ring isomorphisms right

[rocky cloak]

Pretty much

[rocky cloak]

I guess R[x]/(x^2 + 1) and C

[sacred wharf]

right the former is a way to define the latter right

[rocky cloak]

$\mathbb C \otimes_{\mathbb R} \mathbb C \cong \mathbb C^2$

[cloud walrus]

jagr2808

[sacred wharf]

is that saying

C as a right R module tensor C as a left R module

[rocky cloak]

I mean sure I guess. The module structure is the same whether you consider it on the right or left

[sacred wharf]

i thought it was defined like that ?

Like A tensor B
A must be the right module of whatever ring we're talking about

[rocky cloak]

Yeah, it's just C is a symmetric bimodule

[sacred wharf]

bimodule ?

[rocky cloak]

So it's like a right and left module at the same time

Whether I write
pi*i or i*pi it's the same thing

[sacred wharf]

That makes sense

[tribal moss]

Amazing, but checks out.

[trail cave]

can someone help me understand this notation a bit? What is the set group G is being defined on?

[tribal moss]

All elements of Fp that have multiplicative inverses.

That is, all nonzero elements, since Fp is a field.

[trail cave]

What is F_p though? Just any finite field of p elements?

[rocky cloak]

The neat thing here is that G is just isomorphic to the cyclic group Z/(p-1), but computationally it's much harder to compute discrete logarithms in G than in Z/(p-1)

[trail cave]

sorry it's been a year since I studied this I have to do a presentation on elliptical curve cryptography for a different class

[tribal moss]

All finite fields with p elements are isomorphic, so canonically F_p stands for Z/pZ.

[rocky cloak]

Yes, there is a unique field with p elements whenever p is a power of a prime.

[trail cave]

okay thx

[candid patrol]

Finite field is always unique up to isomorphism

[chilly ocean]

can somebody help me with this?

[quiet pelican]

What have you tried?

[chilly ocean]

i tried applying first isomorphism theorem by assuming a homomorphism from Z20 onto Z8

i think i got it, thankyou

[crystal vale]

Say F[[x]] is a formal series, F is field, i don't know but I am thinking about that is it ED? PID? Ufd?

[quiet pelican]

It’s Noetherian (basically the same proof as HBT)
It’s local (the maximal ideal is (x), anything with a non-zero constant term is invertible)
I think it’s a ED, with N(f) = degree in which the first non-zero coefficient of f occurs

[thorn jay]

also has a nice topology making it a topological ring

[crystal vale]

But in my book version, there is a problem given that this idea does not hold for the formal power series F[[x]], i don't see why? Because if it is ED then I think it holds the idea

[quiet pelican]

1 + p_1…p_k is invertible in F[[x]] if p_i are all primes

That’s where it fails

Because to apply the argument, you need to say “1 + p1…pk is not a unit, so has a unique factorisation”

[crystal vale]

I am dumb

Thank you

[tardy hedge]

M is fg over noetherian R , x is M-regular. Is Ann(M/xM) = (x)?

I mean i dont think so but

I am trying to show that a certain prime ideal is in Supp(M/xM)

So was trying to see if it contains the annihilator and stuff

[tough raven]

Ann(M/xM) should be something close to Ann(M) + (x), since you made x act as 0 and anything that acts as 0 on M still acts as 0.

In fact, I think Eisenbud proves that Ann(M) + (x) ⊆ Ann(M/xM) ⊆ sqrt(Ann(M) + (x)).

[tardy hedge]

Thanks, this might be helpful

[tough raven]

It was either chapter 10-12 or 17 I think

[rocky cloak]

I guess first inclusion is obvious and for the other inclusion if M is finitely generated you can use Cayley-Hamilton

I'm assuming M fg might be necessary...(?)

[worldly canyon]

Is there a constructive way to find minimal polynomial over Q of z1 +z6 (roots of unity of 7)?

[rocky cloak]

Yes, for M = Z[1/p]/Z, M/pM = 0 has annihilator Z as opposed to (p)

[rocky cloak]

Yes, from the Galois group you can determine the conjugates, then the minimal polynomial is just the product of (x-r) with r ranging over the conjugates

[worldly canyon]

thanks I will give that a try

[rocky cloak]

And I guess the fact that z1 and z6 are algebraic integers helps even more. In that the coefficients will be integers, so you can compute then numerically and round if they're hard to compute analytically

[worldly canyon]

algebraic integers?

I am referring to these two points when I say z1 and z6

[rocky cloak]

Something whose minimal polynomial has integer coefficients

[worldly canyon]

oh I see

[sacred ginkgo]

Hi. Can someone help me run through 1.13(ii)?

I know at least that alpha^k is a product of disjoint cycles each with the same length (or is the identity), but I do not know how to prove that the length l of each disjoint cycle is l = n/gcd(n, k).

[rocky cloak]

Maybe you're able to show what the order of alpha^k is?

[sacred ginkgo]

Wdym, isn't the order n?

[kind temple]

that’s the order of alpha

[rocky cloak]

I'm sure you can see that for example alpha^n can't have order n, as alpha^n is the identity

[sacred ginkgo]

Is order the value of m such that if a is a cycle, then m is the smallest value such that a^m = 1?

Well according to google, I suppose it is. So alpha^k has some number of disjoint cycles, but all of them have length l. So the order of alpha^k is l.

So alpha^(kl) is the identity, and hence kl = n?

Or at least kl = lcm(n, k) because we don't know if k divides n

[rocky cloak]

And as you might know
lcm(n, k) = nk/gcd(n, k)

[sacred ginkgo]

Ahhh I see

That makes sense, thank you!

[charred flume]

Guys help.
So from what I've searched I need to find a group with order 2, that is by using Cauchy's theorem. But this book contemporary abstract algebra restricted Cauchy's theorem to abelian groups only (I'm trying to prove this by using theorems given by the book on the chapter and previous chapters). So I'm kinda struggling on this one. I have shown already that the subgroup with order 5 is a normal subgroup.

[quiet pelican]

Do you have Lagrange’s theorem?

[charred flume]

yes I can use lagrange theorem

[quiet pelican]

If so then let N be the subgroup of order 5
This must be normal, as gNg^-1 is also a subgroup of order 5
So by correspondence theorem (also known as 3rd isomorphism theorem) we need to find a subgroup of order 2 in G/N
Can you manually prove cauchy’s theorem for elements of order 2 in groups of order 20?

[charred flume]

Okay I'll try

I've prove that there exist bN which is an element og G/N with order 2

since G/N is cyclic so I can also form that subgroup

[quiet pelican]

G/N is not necessarily cyclic

[charred flume]

N,bN

[quiet pelican]

Those don’t generate G/N
they generate a subgroup of G/N, as long as b^2 \in N (ie b has order 2 in G/N)

[charred flume]

i prove that a factor group of a cyclic group is cyclic and since 5=|N| has order 5 which is prime then it is cyclic

[quiet pelican]

N is cyclic, but for the theorem you’re quoting, you’d need G to be cyclic

[charred flume]

oh yeah you're right i got confuse on that

wait let me try again

[quiet pelican]

The argument I have for why a group of order 20 should have an element of order 2 is mostly combinatorics

[charred flume]

Ive found a theorem in the book which uses totient function. So I kinda use that and found that there must exist one element with order 2 if the group is order 20

[quiet pelican]

Yeah
The way I’d do it is errr
Every element has order either 1, 2, 4, 5, 10, 20
If an element has order that isn’t 1 or 5, some power of it has order 2
But there are 1 + 4n elements of order 1 or 5, where n is the number of subgroups of order 5 in G/N
But 20 =/= 1+4n contradiction

[knotty badger]

Another way you can do it is

You consider pairing each element g up with its inverse g^-1

The only times this fails are when g = g^-1, i.e. g^2 = e

Meaning either g is the identity (which happens once), or g is an element of order 2

[noble nexus]

Yeah that's the cleanest way to do it

[twilit wraith]

oh i see

[knotty badger]

Thus you obtain the formula $|G| = \underbrace{1}_\text{identity} + # \text{elements of order 2} + 2k$ for some integer $k$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[twilit wraith]

since theres an odd amount of nontrivial elements in a group of order 20

[knotty badger]

In particular, taking residues modulo 2, the number of elements of order 2 in any finite group is congruent to $|G| - 1 \pmod 2$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

So if G has even order, then there’s an odd number of elements of order 2, so in particular there must be at least one

More generally, the proof of Cauchy’s theorem demonstrates that, for $p$ prime, the number of elements of order $p$ in a finite group $G$ with $p \mid |G|$ is congruent to $p - 1$ modulo $p^2 - p$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[white oxide]

I am a little confused about this Theorem in Atiyah Macdonald. Here, $K$ is a given field, and $\Sigma$ is defined as the set of all pairs $(A, f)$ where $A$ is a subring of $K$ and $f$ a homomorphism of $A$ into $\Omega$, where $\Omega$ is an algebraically closed field. $(B, g)$ is a maximal element of $\Sigma$. In the proof, they show that $B$ is a valuation ring of $K$, but they don't show that $K = \text{Frac}(B)$. Why is this?

[cloud walrus]

okeyokay

[rapid cave]

by this definition of sigma, isn't the maximal element just K?

[rocky cloak]

Is K = Frac(B) not part of the definition of being a valuation ring?

Or what is it they're showing?

[rocky cloak]

Presumably there is a fixed subring and map everything is supposed to extend

[white oxide]

I mean I'm just confused because they have to show that it's a valuation ring, which includes showing that K = Frac(B). Also on the previous page there wasn't anything mentioned about all these subrings having a field of fractions equal to K (unless I'm confused):

[rocky cloak]

I mean if for every x in K either x or 1/x is in B, then obviously Frac(B) = K.

Since everything in K is either x/1 or 1/x

[white oxide]

Ah you right

Thanks

[rocky cloak]

If g is not injective it cannot extend to a map from K. So you can have other maximal elements

[white oxide]

So in defining a valuation ring, we could've just said a subring B of a field K is a valuation ring of K if for each x in K, either x or x^{-1} in B

[white oxide]

Why is $\mathfrak{m}' \cap B$ a proper ideal of $B$? If not, then $\mathfrak{m}' = B$ so that $\mathfrak{m}[x] \subset \mathfrak{m}' = B$, but I don't see the problem here?

[cloud walrus]

okeyokay

[white oxide]

And that doesn't imply that x is in B, since 1 is not in m

[rapid cave]

if m' = B then 1 in m'

contradiction to m' being maximal in B[x]

[white oxide]

Oh lmfao

Ofc thanks

Also why does k' = k[\overline{x}] imply that \overline{x} is algebraic over k

Ah I guess this follows from a result in field theory which I forgot: if E/F is an extension field and a in E is such that F[a] = F(a), then a is algebraic over E

Maybe I should probably try to reprove this

[wooden rain]

I want to show that a permutation has order 2 iff its cycle decomposition is a product of commuting 2-cycles.
However, I don't know what it means that 2-cycles commute. Should I treat a cycle as the permutation where each element outside the cycle is mapped to itself?

[knotty badger]

Yes

[wooden rain]

Thanks!

[south patrol]

When you say cycle decomposition, presumably the cycles should commute like by definition essentially

but yeah also 2-cycles are like by definition permutations of this form - they are not confined to being a permutaiton of 2 elements

[knotty badger]

The idea is to consider the action of the subgroup generated by the permutation on your set

This gives the cycle decomposition

[wooden rain]

Yeah. It's an introductory exercise haha I started going through a textbook today

[white oxide]

Does anybody know what they mean by this highlighted line?

[mint seal]

a complete system of residues mod n means all the distinct equivalence classes from 0 through n-1

mod n

residue is another name for a remainder

[white oxide]

Oh sorry I meant what do they mean "as k runs over"

I guess I should have also highlighted "so does -2k" because I'm confused as to what that means as well

Oh is it just that -k is congruent to -2k mod n

Wait nvm what

Oh it's just that k |-> -2k is surjective mod n

Right?

[tulip otter]

So k goes from 0 to n-1, and {0,...,n-1} form a complete system of residues which means that it forms a set of representatives of all distinct equivalence classes of Z/nZ

Now n is odd, so (2,n)=1

Hence {0,2,...,2(n-1)} forms a complete system of residues and so does {0,-2,...,-2(n-1)}

[tulip otter]

Correct, tho you should also add that k runs over all of Z/nZ

[copper kestrel]

i'm a bit confused on how we even start proving anything in H n N, i don't think there are any theorems about proving anything with H n N besides knowing that if H and N are subgroups then H n N are a subgroup as well

[quiet pelican]

I don’t think this needs any theorems, just the definitions of every term in the question

[copper kestrel]

ah

still a bit confusing unfortunately, like i understand that we have the equivalences of normal subgroups

i just don't understand like. how to conceptualize H n N besides the obvious definition of it

[candid patrol]

Like, take x in H n N, for any h in H, is hxh^-1 still in H and N ?

[copper kestrel]

i guess that makes sense

i didnt realize i just had to do that

[mental lake]

Let R ⊆ S be commutative rings with unity, such that 1_R = 1_S and J ⊂ S[X] is an ideal. Then I = J∩R[X] ⊂ R[X] is also an ideal (you do not need to prove this).
Show that if J is prime, then so is I. Give a counterexample to illustrate that the converse statement does not hold.

can someone give me a counterexample

cause idk

ive done the first bit already

[thorn jay]

"commutative rings with unity" carries beams to Norway

[tardy hedge]

what do u mean by this

[thorn jay]

@tardy hedge

[mental lake]

blame deepl

[thorn jay]

💔

[tardy hedge]

S = R[y] maybe?

J = (xy) and I = (0)? maybe i did smth wrong but

[rocky cloak]

One think to notice is that if J = (x), then I = (x) and R[x]/I = R and S[x]/J = S

And an ideal is prime if the quotient is a domain

[tardy hedge]

oh lol

[rocky cloak]

(0) might be prime in R[x] though

[tardy hedge]

Yeah I wanted I prime but J not prime. unless i misunderstand what converse means

[rocky cloak]

Oh, derp yeah

[mental lake]

yeah this is what u have to get

[tardy hedge]

i think my example is ok then right?

[mental lake]

ill work it out later if im stuck ill ping uif thats ok

[tardy hedge]

Wait (0) might not be prime if R not a domain

oh nevermind we can choose the counterexample

lol

[next obsidian]

You can generalize your example in a way to see that this phenomenon is very common.

Let k be a field and R a ring such that k < R. Let I be any non-prime ideal of R then I\cap k = (0) which is prime (even maximal)

[tardy hedge]

How do you know (0) maximal?

[next obsidian]

Bro it’s a field

The only ideals are (0) and (1)

And the intersection definitely ain’t (1)

[tardy hedge]

I didnt know (0) ideal in k

was thinking in R

[next obsidian]

I wrote I\cap k lol

My point was just that if you make it so there’s no choice but for the intersection to be prime then you can make a bunch of examples super easily

[tardy hedge]

Ok I see

[tardy hedge]

Did you need to say I proper?

[next obsidian]

I mean sure

Lol

[tardy hedge]

just checking

i get what you mean now

[noble belfry]

guys

what the hell is a tensor product lol

[thorn jay]

its the left adjoint of the hom functor

[tardy hedge]

im sure thats so helpful

[quiet pelican]

Tensor product of what?

[thorn jay]

or right adjoint i forgor

[tardy hedge]

i actually do need to review this

i never learned the proof tbh

[thorn jay]

yes left adjoint

[white oxide]

Do you know why this equality holds?

[noble belfry]

a left module and a right module and a ring R

[thorn jay]

its basically the idea of currying: a map f : X → Hom(Y, Z) is the same as a map f* : X x Y → Z

[thorn jay]

similarly, a homomorphism f : Z → Hom(M, N) is the same as a map f* : Z ⊗ M → N

this for modules over a commutative ring cuz then youve got an internal hom (Hom(M, N) can be given a module structure)

[tardy hedge]

i have thought about tensor products a lot cause they were really confusing to me for a while. Practically speaking, the tensor product of M and N over R is the "minimal" R-module that has the property that bilinear maps out of M x N (to an R-mod L) always can uniquely factor through an R-mod map out of M tensor N

[tribal moss]

... assuming we're talking about tensor products of modules, which the OP has not clarified yet.

[noble belfry]

yeah modules

[tardy hedge]

they did clarify

[thorn jay]

^

[tribal moss]

Oops, sorry, missed that.

[noble belfry]

what other stuff have tensor products

[thorn jay]

balanced tensor product over noncomm ring

[noble nexus]

ya unfortunately with modules over noncommutative rings tensor products are kinda scuffed because u get an abelian group

[tardy hedge]

In practice however, tensor products can do all sorts of interesting things

[thorn jay]

its also a general category thing

[tribal moss]

Algebras, vector spaces, abelian groups.

[noble nexus]

unless you have two bimodules

[thorn jay]

two of those are modules

[noble nexus]

basically most linear objects have a kind of tensor product

[rapid cave]

All of these are modules

[noble belfry]

abelian groups r Z modules

[thorn jay]

R-algebras have term reducts which are modules, so you lose information :>

[tribal moss]

But since they can be taught without calling them "modules", sometimes bringing modules into it is not the best way to explain tensor products.

[thorn jay]

true

[tardy hedge]

Maybe, but I first learned tensor products through the most general module theory approach

like ur prob right tho cause i only NOW feel like i barely understand them

[noble nexus]

its best to learn them through vector spaces first

[tardy hedge]

i will always say dummit and footes exposition is so all over the place

[tribal moss]

There's a reason why vector spaces are generally taught before general modules, rather than the other way around.

[tardy hedge]

like they motivate it nicely in the start with extension of scalars idea, but i think that made me personally lose sight of what is really going on

[noble nexus]

its best to learn the universal property, show it exists with quotients, then iimmediately forget about the quotient definition and do a bunch of examples

imo

[tardy hedge]

yea dummit and foote also do not emphasize universal properties too much

[thorn jay]

universal property is the only thing that should be necessary for proofs anyways

[noble nexus]

its also good to know what elements of a tensor product look like and how to manipulate them, but really that's also just the universal property

[knotty badger]

Hm that’s unfortunate

[tribal moss]

Especially since the quotient construction is well nigh unusable for showing that such-and-such tensors are not equal.

[tardy hedge]

yeah. I do like dummit and foote but as i learn more algebra i can see how they do things kinda weird sometimes

[noble nexus]

I do think universal properties in general are pretty confusing for people early in their math career, but tensor products are the one place where its really hard to have them click without it

its a funny thing in math where when you start learning, universal properties are really confusing, but then when you have more experience universal properteis are actually very clarifying 😆

[knotty badger]

Why do you think universal properties are confusing

[tulip otter]

note that k=0 in the product formula of f(nz) gives the term f(z), so f(nz)/f(z) has the same form as that of f(nz) except that the index starts from k=1 instead of k=0, also note that (n-1)/2 and (n+1)/2 are consecutive integers, the author decided to split the product like this because of the next steps

[noble nexus]

I don't, but people just learning math certainly do

[tardy hedge]

its abstract

[noble nexus]

I don't think its good to introduce them until they are needed

[tardy hedge]

more abstraction than ppl are used to at that point

[noble nexus]

which I would say is either tensor products in a second course on linear algebra, or free groups

[knotty badger]

Is that all

[tardy hedge]

i think so

[noble nexus]

yes as someone who has tried to teach universal properties to ppl before its something that just takes a while to get

[tulip otter]

this is for the first equality that you highlighted

[tribal moss]

There's a difference between introducing the universal property of tensor products, and introducing a general concept of "universal properties". The latter should ideally only come after many examples of particular universal properties have been digested.

[tardy hedge]

Actually, for me its one of those things where ive found it hard to tell if i really "get it" yet

Like, I think I understand them pretty alright now, but i cant totally tell for some reason

[noble belfry]

what should i look at to get universal properties

[thorn jay]

the nlab labe of course, that stuff is very beginner friendly

[noble nexus]

I would probably start by looking at the universal property for the direct product as its a pretty gentle introduction

if you've done groups its also worth maybe looking at the universal property for free groups

then look at the one for tensor products

[elfin wraith]

I still dont think I fully appriciate universal properties tbh

[noble nexus]

the really nice thing about universal properties is that they tell you what a construction "ought" to be

[rapid cave]

Universal property for associated sheaf

[noble nexus]

and also once you really internalize them, you don't worry as much about which particular construction is correct

[elfin wraith]

I understand them well enough, like I could tell you what they are and what they mean but, I dont know exactly what it is, I just always feel like im missing something around there

[knotty badger]

is-does duality

[noble nexus]

it also tells you what structure is important in a particular object

I think for tensor products a lot of people think the actual tensor product object itself is the structure, but really its the tensor product together with the bilinear map from the two spaces you started with

and once you internalize that, it makes it a lot easier to parse why all the constructions are the same

[elfin wraith]

I think people are often just introduced to them pretty poorly as well tbh, I feel like tensor product is particularly bad for that

[knotty badger]

I don’t especially like the “unique morphism that makes this diagram commute” approach

[elfin wraith]

I had far too many people just draw a trinagle and say, its that and then move on. Which like is fine when youre comfortable with it, but utterly unhelpful at first

I didnt understand the tensor product until I finally saw someone just define it by generators and relations for bimodules

Like I knew what it was, but I didnt understand it

[tribal moss]

Tensor product is probably not a good first introduction to the words "universal property".
On the other hand I think the universal property that tensor products have is essential for understanding what tensor products are good for -- but one doesn't need to emphasize the particular words "universal property" about it for that purpose.

[knotty badger]

As the quotient of a free vector space you mean?

[thorn jay]

free vector space, lol

[knotty badger]

That’s more to do with the construction than the universal property, I suppose

[thorn jay]

well, showing existence is formally important, and the construction may give an idea of how to think of the object

[knotty badger]

Yes

[elfin wraith]

I think they have to come together to be really appriciated tbh

At least as an introductory level

[noble belfry]

this has been my understanding so far

[elfin wraith]

Like I get that the whole point is "it doesnt really matter how you construct it" but if you never construct it ever then ???

[noble nexus]

idk I've never seen someone just state the universal property and not prove existence

[thorn jay]

at some point the universal property gives an idea for how to construct stuff which i think is a good goal to measure comfortability with universal properties

[tribal moss]

I don't think anyone is arguing that an existence proof (via quotients) should not be presented.

[elfin wraith]

Im not saying they are, im saying I had this happen multiple times in my UG

[tribal moss]

Oof.

[elfin wraith]

Yeah I agree, which is why I said this is fine when you understand it but like, for an intro, just not a good idea

[noble belfry]

my prof usually just says the proof is trivial and it has to be the case🥀

[thorn jay]

lmao

i think they just dont feel like writing down three boards of technically unpacking of definitions

[elfin wraith]

This was my experience, draws a triangle "it is what it has to be" sure, but this is my first time seeing it lmao, write something down

[noble belfry]

ok i reviewed the universal property of free groups

it makes sense and seems kind of cool

[knotty badger]

Yeah it’s nice

[noble belfry]

universal property of modules:
Given M an A-module, F free on S, and map $\phi: S\to F$, for any map $f: S \to M$ there is a unique $\tilde{f}$ module homomorphism such that $f = \phi(\tilde{f})

[knotty badger]

Hm

I don’t quite understand this statement

[noble belfry]

neither do i

its in my notes tho

[thorn jay]

this looks like nonsense lol

[knotty badger]

Is S a set here?

[noble belfry]

this is the diagram

[noble belfry]

S is a basis

oh

i didnt specify that

M is free on S

[thorn jay]

oh the universal property of free modules

[knotty badger]

Ah ok now I see what you mean

[knotty badger]

F is free on S

I don’t think M has to be

[noble belfry]

wait what

oh i see

[cloud walrus]

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

[knotty badger]

Yep, that’s the one

[thorn jay]

if M is a module, and f: S → M then there must be a unique homomorphism f* : F → M such that f* ∘ φ = f

yeah

[knotty badger]

In other words, you can convert between functions $S \to M$ and homomorphisms $F \to M$ in a reversible way

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

Given one you can obtain the other

That’s what the free module on S “does”

It lets you upgrade functions to homomorphisms in a reversible way

[tardy hedge]

thats exactly what i was saying too earlier in the conversation

[tardy hedge]

I think you said something some time ago that made things a little clearer for me, something like you need to specify both F and its associated S->F map

[thorn jay]

yeah, you need specify exactly what subset is your basis

thats what the function φ is doing here

[noble nexus]

in fact for vector spaces thiis is the universal property that people are most familiar with

that is, if you specify where the basis vectors go you get a unique linear map

[thorn jay]

thats why linear maps have faithful representations as matrices for some choice of basis :>

[knotty badger]

Yeah

[tardy hedge]

then if F satisfies that universal property wrt S->F then it must be the free module on S

[thorn jay]

famously, vector spaces have many bases so this shows that modules can be free wrt many maps φ : S → F

[tardy hedge]

this fact is using what exactly, something about uniqueness but I can't verbalize it properly

[noble belfry]

oh

wait

i get why he didnt put any proof

it rlly is trivial

[tardy hedge]

so thats why we say something like free module on S is unique up to isomorphism ..?

that mightve been dumb. Idk im actually tryna understand this better ngl

[knotty badger]

Unique up to unique isomorphism in fact

(But only if you include the map S -> F as part of the data)

[thorn jay]

yes

there are many different choices (M, φ : X → M) which are free over X

[tardy hedge]

to me that looks like the basis is fixed but ur embedding it differently...?

[thorn jay]

think of it as the cardinality of the basis being fixed

[knotty badger]

There’s a difference between knowing that a module is free over S, and knowing how a module is free over S

For the latter you need to supply a specific map S -> F that makes it free

[thorn jay]

yes

[knotty badger]

Whereas for the former you just need to know that one such map exists

If you have two modules “that” are free over S, then they’re isomorphic, but potentially in more than one way

[thorn jay]

(for two choices of basis in a vector space this isomorphism is exactly the change of basis isomorphism)

[noble belfry]

wait im confused :dead:

why does it have to commute

[knotty badger]

However, if you also know “how” they’re free over S, then there’s only one isomorphism that works nicely with the way they’re free over S

[noble belfry]

since F is free on S, each element x in F has a unique expression
$ x = \sum a_i s_i $

Let
[ \tilde{f}(x) = \sum a_i f(s_i)
]
Then we have a module homomorphism from F to M

[cloud walrus]

sudo

[tardy hedge]

Idk I guess in my head rn i was thinking basic example like R^2 over R having different bases, I was trying to reconcile what you guys are talking about with an example like that

[knotty badger]

Yes so

R^2 is free over the 2-element set

But specifying how it is free is equivalent to choosing a specific basis

[noble belfry]

but why do we have to have it that $\phi: S \to F$ has $\phi(s) = s$ (which would make it commute)

[cloud walrus]

sudo

[noble belfry]

or do we not need that

[knotty badger]

There are many isomorphisms of R^2 with itself

But if you have two bases for R^2, there’s only one way to convert between them

[tardy hedge]

so S here in the R^2 example really is just thought of as some two element set

[knotty badger]

Yes

The map S -> R^2 picks out two basis vectors

[tardy hedge]

Ok, I think this makes more sense. I appreciate it

Are vector spaces distinguished because there are always many ways V can be free over S?

[thorn jay]

wdym distinguished?

oh, well

[tardy hedge]

Yeah not sure the right word to use there

[thorn jay]

no, any automorphism will give a new basis

[knotty badger]

Vector spaces are one of the few algebraic theories where every example is free

Or more precisely, free over some set

[thorn jay]

many structures with many automorphisms

[thorn jay]

i cant think of any besides Set and K-Vec

[tardy hedge]

But maybe vector spaces are unique because there always exists automorphisms?

[knotty badger]

There are some degenerate examples

[thorn jay]

im pretty sure basically any free structure has nontrivial automorphism group

[thorn jay]

pointed sets? lol

[knotty badger]

Yeah

[tardy hedge]

This is difficult for me but i have learned 0.01% more at least

Lol

[tall igloo]

surprised there's no axiom of choice emoji in this server

[thorn jay]

https://mathoverflow.net/questions/157974/varieties-where-every-algebra-is-free

hah

very nice post

uses some surprisingly deep universal algebra

i suppose its not that surprising this question is hard lol

[tardy hedge]

Ts shit so abstract

[noble belfry]

wait i understand tensor products better now i think

wait i need to go over the universal property of quotients for this proof to make sense 🥀

[knotty badger]

Yeah I think one of my friends was studying it for a project in her phd

Classifying all such theories where every algebra is free

[elfin wraith]

Where do you even start on a problem like that

Like sure I guess you can start with properties that force tgere to be non-free algebras, but that still just seems wildly general

People are crazy smart

[tardy hedge]

For reaaaaal

[cosmic lion]

why are noetherian rings studied, what are they used for? i just learnt their definition and want to know why would they be important

[tardy hedge]

Ppl like to ask that question^

[tardy hedge]

Dont you think ideals being finitely generated is nice

[worldly canyon]

if alpha is algebraic over K, and [K(alpha):K]=n, then can we say that [K(alpha,t):K(t)]=n? Why?

t is transcendental element

[tardy hedge]

I think ultimately u study noetherian rings cause its isolating a good property that are common to many examples ppl study

[elfin wraith]

They show up all the time in areas like number theory and algebraic geometry, which Im not qualified to speak about but they show up there lol. From a purely ring theoretic perspective, being Noethetrian is just a pretty powerful property to have when it comes to proofs, but its still pretty general which gives you a lot of examples. I like to think about it as a finiteness property in a sense. In a very handwavey way it means your ring isnt "too big"

[rapid cave]

The minimal polynomial of alpha in K doesn't split in K(t)

[worldly canyon]

yeah I am trying to prove this

[elfin wraith]

.

[noble belfry]

wait so broadly the universal property of tensor products is just that given a bilinear map $f: E \times F \to G$, we have a R-homomorphism $\tilde{f}: E \otimes F \to G$ which extends $f$?

[cloud walrus]

sudo

[noble belfry]

or am i imisinterpreting

[tardy hedge]

Like it doesnt adjoin any new factors right

[tardy hedge]

Yeah also the R-mod hom is unique, and there is a bijection between R-mod homs into G and bilinear maps into G

[worldly canyon]

I guess the specific case I am trying to show is this.. alpha is root of x^2 + 1. I want to show that [f_3(t)(alpha):f_3(t)] = 2, I'm not sure how to prove it

[cosmic lion]

what do you mean by finiteness property

[noble belfry]

the bijection is because for bilinear maps into G we get a unique R-homomorphism by the above and R-mod homs as defined are bilinear right

[worldly canyon]

if the minmal polynomial of alpha in K doesnt split in K(t), this would get it done, but why is it true that minimal polynomial of alpha in K doesnt split in K(t)? Not sure how to show this formally

[tardy hedge]

I guess the R-mod hom being unique shows the association is injective and then given the R-mod map, precomposing with MxN -> M(x)N inclusion shows its a surjective association

[noble belfry]

yeah exactly

wait now that i know what a tensor product is and the universal property of it

[worldly canyon]

intuitively I understand why it would be the case but I am struggling to formalize an argument

[elfin wraith]

Nothing specific, I mean that in a handwavey way. Your module isnt finite, but all of its submodules are finitely generated, which tells you that your module is "small" enough that a lot of proofs work out nicely. Small here is of course not precise because these things are generally uncountably large, but still

[noble belfry]

can someone explain why tf these matter for physics and machine learning and stuff

actually machine learning tensors are not math tensors i guess

they are just multidimensional arrays

[noble nexus]

In physics you see tensor products in a few different contexts

one is for differential geometry where tensors are used because of the prevalence of multilinear maps

[karmic moat]

Extension of scalars

[noble nexus]

then in qm the tensor product of Hilbert spaces describes composite systems

[thorn jay]

well the case where the signature has no constants is pretty smart. You reason that the initial algebra has to be empty, so as the trivial algebra is free, F(1) must be trivial and so the variety is idempotent (t(x,...,x) = x for all terms t). This is lucky as idempotent varieties are extremely well understood (relatively) because they were of interest in solving some conjecture relating to constraint satisfaction problems

[tardy hedge]

For your last point, its more like based on how tensor product is constructed we have that the inclusion MxN -> M(x)N is bilinear

[thorn jay]

so then you kinda go by a case by case basis, ruling out unwanted cases

[tardy hedge]

It doesnt really make sense to say R-mod homs are defined to be bilinear

[thorn jay]

of course a result by McKenzie sneaks in somewhere, that man is a beast of a mathematician

[noble belfry]

in this diagram

isn't f tilde bilinear

wait ohh

[tardy hedge]

Its an R-mod hom

[noble belfry]

for every bilinear map into G, there's a corresponding R-mod hom f~ so thats our injection

but then for every R-mod hom

[noble nexus]

If you're talking about noncommutative rings then in general it'd be just an abelian group hom

but probably best to just start with commutative rings

[thorn jay]

or fields for that matter

[noble belfry]

what u said is true yeah

that its the $\varphi$

[noble nexus]

the motto is that "the tensor product linearizes bilinear maps" meaning every bilinear map into G extends to a unique linear (meaning module homomorphism) map from the tensor product

[noble belfry]

that makes sense i get it

[noble nexus]

One of the main motivations behind the tensor product imo is that it lets you basically do multilinear algebra for free if you know linear algebra

[cloud walrus]

sudo

[noble nexus]

a priori you might assume that you'd have to invent a whole field of multilinear algebra to study multilinear maps

[tardy hedge]

Where is multilinear algebra used? I still havent found a need to learn much of it for anything

I tried learning tensor algebra for something but i ended up didnt needing it

[noble nexus]

Differential geometry, also a lot of multilinear maps are just very natural

[tardy hedge]

Oh ok yea

[alpine plank]

Also a lot of nice things are constructed as quotients of the tensor algebra

[tardy hedge]

Whats an example?

[alpine plank]

Exterior algebra, symmetric algebra

Universal enveloping algebras

[worldly canyon]

I guess my question is, I can see why this is true, but not splitting in K(t) isnt necessarily being same thing as being irreducbile in K(t) right? So how do we know the minimal polynomial are the same, because thats what would be desired to show the property right

[noble belfry]

just to recap a module is an abelian group thats p much acted on by a ring

tensor products r smthn u get from defining a free group and forcing stuff to obey bilinear relation on two modules

[rapid cave]

but not splitting in K(t) isnt necessarily being same thing as being irreducbile in K(t) right
it is the same

[worldly canyon]

Why? consider x^4-1 in Q. It doesnt split in Q, but its reducible in Q

[rapid cave]

What I meant what reducible

sorry for the confusion

[vocal pebble]

1, alph, alph^2,...,alph^{n-1} spans [K(alpha,t):K(t)] so it suffices to show that this set is independent.

If it is not independent, then you have a polynomial over K(alph) satisfied by t contradicting the fact that t is transcendental over K(alph).

[tardy hedge]

Pretty much

[twilit citrus]

is the set of subgroups of an infinite group always uncountably large?

nevermind

[quiet pelican]

No, Z

[thorn jay]

the set of subgroups of Z is very countable

lol

[twilit citrus]

I don't think so

yea

[thorn jay]

it's very kindly and obviously indexed by N even!

[twilit citrus]

yes

[tardy hedge]

1Z 2Z 3Z 4Z 5Z 6Z 7Z 8Z...

i am counting them

[twilit citrus]

how do we show that the number of subgroups of an absolute galois group of an infinite extension is always strictly larger than the number of intermediate fields

[tardy hedge]

thats better

[twilit citrus]

this was in continuation of the 'nevermind...'

lol

I wasn't disagreeing

[tardy hedge]

im not gonna know the answer to this but what is an "absolute" galois group? And infinite extension meaning for F > K F is infinite dimensional over K right

[thorn jay]

golf clap

[tardy hedge]

yea actually i didnt even end up learning galois theory that much cause my prof completely rushed it at the end

so sad

[tardy hedge]

thank u so much

[twilit citrus]

absolute galois group is the galois group of the extension to the separable closure of a field I think

I am trying to understand some intuition the theory has be bogged down in symbols so much so that even subgroups of Z have started to appear uncountable...

[formal laurel]

This is very specific to the behaviour of profinite groups.

[twilit citrus]

it doesn't happen otherwise?

then what's the need to specifically look at the closed subgroups in the infinite galois correspondence?

ah wait

[formal laurel]

Well, if not the there's no Galois correspondance

[twilit citrus]

is the absolute galois group always an inverse limit

?

[formal laurel]

yep, an inverse limit of finite groups therefore it is profinite

[twilit citrus]

ahhh I see I see

that makes sense

okay

[formal laurel]

You can't have Z as a Galois Group of something for instance

For any group G you can take its profinite completion which is the inverse limit of all its finite quotients

[twilit citrus]

for any group G?

[formal laurel]

In Z this is the a product of all the p-adic intergers.

[formal laurel]

yep

[thorn jay]

it always has at least one finite quotient which is G/G

and clearly it defines an inverse system

[quiet pelican]

Yes, but it may not reflect much of G if G is badly behaved

[twilit citrus]

I see that makes sense

[quiet pelican]

Like the profinite completion of Q is 0

[formal laurel]

Indeed

[twilit citrus]

oh wow

[thorn jay]

Q is a terrible group

[formal laurel]

And like, not every profinite group is the galois group of some field

[quiet pelican]

Corollary: all divisible groups are bad

[thorn jay]

I see no problem with this statement

I do not like injective objects

[quiet pelican]

what about surjective objects /j

[thorn jay]

projective objects 🤝 enpeace

[twilit citrus]

you mean to say that not every profinite group corresponds to an inverse system of groups corresponding to finite field extensions associated with an infinite extension?

[quiet pelican]

Ok but what if I want a projective object in C^op

[formal laurel]

Yep, exactly that.

[thorn jay]

no

those aren't true projective objects

fakers

[twilit citrus]

oh great

[formal laurel]

C^op isn't real

[twilit citrus]

do you have an example

that illustrates that?

[formal laurel]

I mean, there's no easy example. But like if you only care about absolutely galois groups, by that I mean taking K/L where L is the algebraic closure of K, then every finite group is out the table since only the trivial group and the ciclic group of 2 elements are the only possibility.

Profinite groups are really wild objects. For instance we know almost nothing about the absolute galois group of Q.

[twilit citrus]

do you know a place where I can read about some of these difficult examples?

[twilit citrus]

wow insane

[formal laurel]

There's the book "Galois and fundamental groups" by Tamás Szamuely and well a general tool for studying these groups is the Galois cohomology which is the group cohomology but taking the profinite cohomology into account, these aren't exactly accesible so I wouldn't recommend read it them if you don't know much about abstract algebra but there's the book by Neukirch "cohomology of number fields" and "Galois cohomology" by Serre

I would like, download this books and wait like 2 years of learning homological algebra, number theory, some algebraic topology and so on after starting reading them btw.

[tough raven]

Does one of these books specifically have an example of a profinite group that is not the Galois group of any Galois field extension (not necessarily absolute, just algebraic)?

[twilit citrus]

thanks, I'll see what works

[formal laurel]

I can't remember but there's the book "Field arithmetic" and I'm kind of sure there's an example in there. Let me search for it.

[tough raven]

Thanks! That sounds really interesting.

[twilit citrus]

wow

I guess I could download at the very least and check these out yea

what do they mean by 'of the form of Gal(K/L)'? Do they mean that the automorphism group doesn't always fix the base field here?

[formal laurel]

I got mixed up. I got things wrong lol. It appears every profinite groups appears as a Galois group
https://www.ams.org/journals/proc/1974-042-02/S0002-9939-1974-0325587-3/S0002-9939-1974-0325587-3.pdf

This is no true for absolute Galois groups though.

[twilit citrus]

as in it corresponds to some infinite extension...but doesn't always appear as an absolute galois group necessarily? As in not the galois group of the extension to the separable closure but maybe some other kind of infinite extension?

[formal laurel]

Take for the algebraic clousure of the finite fields the fixed field of the subgruop generated by the frobennius automorphism.

[thorn jay]

and if its finite and nontrivial, then K is elementary equivalent to R like wtf

[tough raven]

Ah. This closely mirrors the theorem for finite groups: every finite group is a Galois group but you can't fix the base field.

[rocky cloak]

Kind of an open question whether you might be able to fix the base field

[formal laurel]

Yep, this is what the result is saying.

[tough raven]

Yeah that's what I meant

Too lazy to be precise lol

[formal laurel]

No but you can take C(t)? and every finite group is the Galois group of some finite extension of C(t)

Or what do you mean by fixing the base field?

[rocky cloak]

Oo, is there an easy construction?

[formal laurel]

It is kind of similar as the one constructed in the article i put down. For C(t) you can take a "ramified" cover of the riemann sphere and its galois group is the profinite completion of the fundamental group of this ramified cover, since all of this are basically the complex plane minus some points then the absolute Galois Groups is the free profinite group in the set C. Then you can recover every finite group by just taking a representation by a free group.

C(t) is the field of rational functions with complex coefficients btw.

[rapid cave]

by just taking a representation by a free group.
wdym by this?

[rocky cloak]

Hmm, so from what I'm reading you take the Riemann sphere (whose rational functions is C(t)), remove some points (for some reason), then you get a space with fundamental group a free group. So you can create a covering with whatever finitely generated Galois group you want.

Now I just need to understand why removing some points and taking a covering space should correspond to a Galois extension

[rocky cloak]

Writing your group as a quotient of a free group

[formal laurel]

The book Fundamental groups and Galois groups explain this more clearly, but the idea is that the category of compact riemann surfaces and fields with transcendental degree 1 over C are equivalent.

Of course, for a finite extension ramified at only the prime t in C[t], the correspondance are taking place in the complex plane minus the origin remove. You are taking covers of t->t^n and these corresponde with C(t^1/n). and then you take profinite completion

[rocky cloak]

Maybe I'll try understanding that theorem for my next week project

[formal laurel]

These are also technically ramified at the infinite point, so basically you are trying to understand covers of the riemann sphere that ramifie at two points.

[formal laurel]

It is an excelent result. It is basically the only tool we have to completely describe the absolute Galois group in non-trivial case.

[tall igloo]

what is transcendental degree? like how many transcendantal elements you have to add to generate it?

[formal laurel]

Just the cardinality of some transcendental base.

C(t)/C has transcendental degree 1 as also the field C(x,y) over C with the relation y^2=x^3+1

[tall igloo]

by C(t)/C you mean the field extension right?