groups-rings-fields

[south patrol]

But I meant more like products being nonempty is choice

[tough raven]

Yes but not this one

[south patrol]

Ofc

Just aesthetically I would define it as the subgroup of Sym(X) which respects the equivalence relation

Lol

Lol wdym

[tough raven]

IDK

[tough raven]

For that matter it could be the group generated by all transpositions swapping equivalent elements.

[south patrol]

Yeah lol

https://tenor.com/view/calculation-cat-gif-26041269

[tough raven]

(Which is not the same: it is the product of S_finite (identity on all but finitely many elements) of equivalence classes.)

[tough raven]

Triangle jumpscare

[raw anvil]

why is do inversions define the odd and evenness of a permutation of $S_n$? whats the intuition behind it?

[cloud walrus]

todben

[tough raven]

I would say the intuition is which permutations are "orientation-reversing". E.g., you may have some intuition that 3-cycles permute elements "only cyclically" while transpositions "reverse parity"; the parity of a permutation captures this.

Alternatively, S_n acts on n-dimensional Euclidean space by isometries (permuting the coordinates/standard axes) and the parity literally captures whether a permutation is an orientation-reversing or orientation-preserving symmetry.

[crystal vale]

If I take a ring of real numbers R then I know R = (1), but according to this definition, it looks R is not finitely generated.

[vapid vale]

to add to the latter answer, an example: the group of orientation preserving (rotational) symmetries of the tetrahedron acts faithfully on its four vertices, so we want to think of it as a subgroup of S_4 (if you don't know group actions, try to make sense of the statement abstractly). now its clear that you can't get a permutation of the vertices where you swap two vertices just from rotating the tetrahedron. its (to me at least) interesting that every other "impossible" symmetry is odd (so this tetrahedral group is A_4). the interesting part is that the base case of swapping two vertices extends to the entire group of symmetries, but thats clear once you notice that an odd permutation is just an even permutation composed with a single transposition (just as an odd number is an even number + 1).

so you have this nice structure which is preserved by taking parity by inversions, and indeed the only nontrivial homomorphism into Z/2Z is given by the parity of inversions.

[karmic moat]

R has transcendental elements like e and pi which you can't write as a polynomial with rational integer coefficients

[crystal vale]

Yes

So which definition should I follow for a finitely generated ring?

[karmic moat]

What other definitions are there?

[crystal vale]

In terms of ideals, finitely generated ideal, so R itself an ideal

[karmic moat]

I wouldn't say those are the same definition

[crystal vale]

Yes they are not

[raw anvil]

I think I get what youre trying to say with that example. The first time I heard someone use the term "parity" was when I was solving a 4x4x4 rubik's cube when i was 12 or something in a youtube video and swapping two vertices of a tetrahedron kinda makes sense. 😃

[karmic moat]

Like that's not a definition for finitely generated ring

Anyway the identity R = (1) is true for any unital ring

[vapid vale]

the parity from rubiks cubes is the same parity here (image of the homomorphism into Z/2Z)

[raw anvil]

what is meant by Z/2Z by the way?

[crystal vale]

So as ideal it is finitely generated but as ring it is not finitely generated

[karmic moat]

Yeah

[vapid vale]

have you seen cyclic groups?

[raw anvil]

yes

group generated by spamming an element from the group to itself right

[vapid vale]

sure

[raw anvil]

$\langle a \rangle$

?

[cloud walrus]

todben

[vapid vale]

yeah, although cyclic groups don't need to be subgroups of other groups

but abstractly yes

Z/nZ is just the cyclic group C_n

the point of saying Z/nZ is that Z is the integers so by taking a quotient you are really referring to the integers 0, 1, ..., n-1

[raw anvil]

I think my prof uses the notation Z_n

the operation is addition correct?

identity 0?

[vapid vale]

yes

[raw anvil]

oh aweseme

im studying for my upcoming midterm on friday

[dense hill]

But they can be? Also why would some think they would have to be?

[twilit wraith]

I may have done this in a more complicated way than necessary

If N is the product of all the little n's then Z/NZ is isomorphic to the product of each Z/nZ

Via the mapping described in CRT

The isomorphism extends to polynomial rings and so the tuple (f1 mod n1, ..., fk mod nk) has a function f mod N in its preimage

That polynomial f can be easily lifted to a polynomial in Z and we have our result

But idk it feels really handwavy

[rocky cloak]

Argument is good, and I don't think it's overly complicated.

I guess you're brushing through some details, but it's not like those details are hard to fill in.

[twilit wraith]

Fair enough

Im not fully sure we've shown the whole ring isos extending to polynomial rings thing

So i might just do coefficientwise CRT

[rocky cloak]

I mean, it isn't even necessary for it to extend to the polynomial ring. You just pick the coefficients individually.

The ring structure of the polynomial ring isn't really relevant

[twilit wraith]

Yeah fair enough

[tribal moss]

For me, the actual intuition is that the parity counts whether the permutation is a product of an odd or an even number of transpositions. Counting inversions is just a technical trick for proving that it's impossible for the same permutation to be made both with an odd number of transpositions and an even number of transpositions.
Some textbooks elevate the inversion count to a definition of the parity just because it's the first time in the explanation where we can define it in a way that it's obvious that it at least defines something.

[mint seal]

I'm pretty stuck on the first part of this exercise 2.15.

Am I meant to deduce and use some extra information about the am + bn = 1 identity from ex. 2.13? Like that a and b can be chosen to have some parity based on m and n?

I’ve got that there exist a,b with am + b(2k+1) = 1

and I want to show there exist integers x,y with x(2m + 2k + 1) + y(4k + 2) = 1

i.e. 2mx + 2kx + x + 4ky + 2y = am + 2bk + b

and that can happen if x = a/2, y = (b - x)/2

but I’m not sure how to ensure those are integers

[tribal moss]

Hmm, n being odd seems to be crucial -- otherwise m=1, n=2 would be a counterexample.

[mint seal]

and it feels a little mean that the previous exercise has no hint about controlling a and b

but I’ve looked up that they do have the form a = a_0 + tn, b = b_0 - tm, where a_0, b_0 is any pair that works

this just feels kinda bleegggh

[tulip otter]

because you dont really have to find any explicit forms of a and b to prove that

you could've done it with the well ordering principle for example

[mint seal]

hmm I don’t see

[tulip otter]

as for this, start with the fact that (m,n)=1

[tribal moss]

I'm not sure how one would use 2.13 there either.
My first idea would be to prove separately for each odd prime p that if m and n are not both 0 modulo p, then 2m+n and 2n are also not both 0 modulo p, because the matrix (2, 0; 1, 2) over F_p is invertible.

[tulip otter]

what does that tell you? (from exercise 2.13)

[mint seal]

that there exist integers a,b with am + b(2k+1) = 1

[tulip otter]

maybe its less complicated if you keep it as n

Tho I think you will reach the desired result anyway

So your goal is to show that there exist integers r,s such that?

[tribal moss]

Are you really saying anything that MC did not already say here?

[tribal moss]

Here, in particular.

[mint seal]

😅

[tulip otter]

Well I was starting from the beginning but ok I will skip this

[mint seal]

all good. I seem to have found conditions for x and y to work, but I have no reason to believe they’re integers

but I expect with the right choice of a and b this can be assured

[tulip otter]

lol I had an error

[tribal moss]

At least, if a is not already even, then you can add n to a and subtract m from b.

[mint seal]

right.. that uses n being odd, that looks good

but then b - x needs to be even as well

and b is odd, so x must be odd

[tribal moss]

So we might need to add 2 mod 4 to a somehow.

But n is odd which means it is either 1 or 3 mod 4. So 2n is 2 mod 4.
Thus if a/2 is not already odd, you can continue by adding 2n to a and subtracting 2m from b.

[mint seal]

ooh

very nice

most of Aluffi’s exercises so far have been quite easy, but that was a lot trickier

let’s see this next part…

[tribal moss]

This does feel easier to me, still.

[mint seal]

that’s a very cool looking idea, no idea how you came up with it and I don’t really understand it. I’ve got little intuition for anything involving finite fields

[tribal moss]

Hmm, roughly what I thought was something like

I hope the matrix that takes (m,n) into (2m+n, 2n) turns out to be invertible over Z because then the two pairs will have the same common divisors, and in particular the same gcd.
Whoops, its determinant is 4, so not invertible over Z. But at least it will be invertible modulo anything where 4 is a unit, which is every odd number, so that would take care of all the odd primes. And then we just need an ad-hoc argument that 2 is not a common factor.
So it's not really about finite fields in particular, but the more general fact (from Cramer's rule) that a matrix over a ring is invertible iff its determinant is a unit in the ring.

(And I suspect there is probably some connection between the fact that the determinant is 4, and the key to your approach turned out to be to get a to have the right remainder modulo 4, but I can't quite see the details of that connection).

[mint seal]

hmm very interesting stuff

[rocky cloak]

My idea for using 2.13 would be

So there are a and b with
am + bn = 1
then
2am + 2bn = 2

as n is odd there is x and y with 2x + ny = 1, so
2axm + 2bxn + ny = 1
which you can rearrange to
ax(2m + n) + (2bx + y - ax)n = 1

But then you still need a clean argument for y-ax being even

[dull tiger]

I'm trying to prove this theorem. On this Wikipedia page: https://en.wikipedia.org/wiki/Minimal_polynomial_of_2cos(2pi%2Fn) it says that it can be derived using a product formula for n odd and a theorem using chebyshev polys. (Note that my \Psi_n(x) corresponds to \Xi_n(x) on the Wikipedia page)

Does someone maybe have a hint how it might work?

I know T_n(\cos \theta) = \cos(n\theta) so i kinda tried so approach using 2\cos(1/n*2\pi)

And then i tried T_n(x/2)=\cos(2\pi)=1, x=2\cos(2\pi/n), thus T_n(x/2)-1=0. Didnt really help though 🙂

[rose kraken]

For studying Abstract Algebra (or any math field for that matter), is it OK to read thru just the chapters of a book without doing any of the exercises and finish the book quick, then go thru the book a 2nd time—but this time—we do the exercises for sure sequentially?

Or no? Is it better to do exercises sequentially as you go thru the book the first time?

What book reading strategies do you guys personally use?

I'm starting out AA via self-study

[noble nexus]

It's fine to read through quickly first but I wouldn't go through the whole book

You can skim a section to see what the big ideas are and where things are headed

Also to get an idea of what examples you should have in mind

[quiet pelican]

My opinion on the first strategy is that (a) the first pass won't teach you much (especially as at the AA level you're probably not interacting with the textbook in a way that would allow you to go without doing explict exercises), and (b) you're more likely to lose motivation

[noble nexus]

but if you want to actually learn you need to be active

that doesn't necessarily mean doing the exercises in the book, I often don't but I'll do other things to engage my brain in what I'm learning

so taking transformative notes (reorganizing information in a way that makes sense to you, rather than just copying), working out examples, etc...

[elfin wraith]

In fairness, I typically do this and I think it works. Not the full book usually, but the chapter at least

[noble nexus]

I also like to try and work out proofs myself, maybe after reading a bit to see the main idea

but doing exercises is also just a good litmus test to see if you really understand the material (and of course helps you learn)

[elfin wraith]

I like the first pass just to let it wash over me and get the broad strokes, then I go back over it with a fine tooth comb and make sure I get it

[rose kraken]

Yeah so the reason I mentioned this is because sometimes when I do exercises, I just get completely stuck

[elfin wraith]

And this is where I actually do most of the learning

[noble nexus]

if you get stuck on exercises it's a good sign you need to review more or perhaps look at other sources

[rose kraken]

And sometimes, I insist on finishing the exercises, but then many days get.. wasted?

[noble nexus]

They aren't wasted is the thing

It feels that way

But that's where the learning happens

You shouldn't expect to get through math books very quickly if you are actually absorbing the material

it's pretty easy to get a surface level understanding of math without actually knowing the details

Personally there are a lot of fields where I know a lot of the broad strokes but if you asked me to do exercises I would fail spectacularly, and it's because I haven't actually worked at problems/exercises/examples

Knowing the big picture is certainly important though, as it gives you motivation to understand the details

[rose kraken]

Ayy, I see, I see

Well, thanks so much guys! I'll probably take a hybrid approach then, yeah; I'll see a chapter thru just to quickly get an idea of where I'm headed, then make sure to perform their exercises at the end of it 👍

[noble nexus]

I'll also add that for self study I think it's really important to look at different sources, maybe not always but whenever something doesn't make sense

often I find something is confusing but then a different book or a blog post or a YouTube video explains it in a way that makes it click

Tbh this is true even when not self studying

[rose kraken]

Yeah, I totally agree with that sentiment; some time back, there was this explanation of a topic in a certain book which I was super confused by

Decided to look at another book and the topic couldn't have been any clearer 😄

It was 2 different perspectives of the same problem, so that was cool

[cyan skiff]

I wss struggling with a question and would appreciate some help, the question is "every finite field has order p^n for some prime", I can see why it must be p and can't be a composite with distinct primes bcz of cauchy's theorem and prime char of an integral domain, but can't generalise it to p^n or make sense of it.

[tough raven]

If a positve integer has only one prime factor p, it must be a power of p.

[cyan skiff]

That's all fine and good but a finite field must be abelian with respect to addition and multiplication, right so it must be isomorphic to (Z(p^n), +), doesn't that imply that it must have an element of order p^n - 1?

[tough raven]

What is Z(p^n)?

[cyan skiff]

Integers mod p^n

[tribal moss]

That doesn't make a field unless n=1.

[tough raven]

Ah. Well, it's not true that the finite field's additive group must be Z(p^n). There are other abelian groups of order p^n.

[cyan skiff]

Wait, seriously? I read in aes that it uses GF(2^8) as its underlying finite field

[tribal moss]

The additive group of the field of order p^n is the product of n copies of Z/pZ.
(The multiplicative structure is more complicated).

[tribal moss]

Yes, but its additive group is not cyclic.

[cyan skiff]

Oh makes sense but don't they have nonzero divisors?

[cyan skiff]

Ohhh

[quiet pelican]

All the F_{p^n} additive groups are elementary abelian iirc

[fast kelp]

Finite fields are generally described as having n copies of the C_p that are entwined with each other using an irreducible polynomial of degree n.

[noble nexus]

hint: think about vector space dimension

[rocky cloak]

GF(2^8) is just the name of the unique field with 2^8 elements

[cyan skiff]

That's something I don't wanna do😭, but even in that sense it considers it as a direct product of n (Zp) right

[tough raven]

It might help to just stare at the structure of the finite field with 9 elements for a while to get what is and isn't cyclic.

[noble nexus]

Yeah so any finite field is going to contain a copy of Zp, and hence is going to be a vector space over Zp

[cyan skiff]

Or maybe not direct product bcz then it would have zero divisors

[noble nexus]

so it will have p^n elements

[fast kelp]

essentially, you can imagine a root of this polynomial a, and the number (for instance) 20 is 16 + 4 so 00010100 in GF(2^8) which is a^4 + a^2.

[tough raven]

Yes (additively), which is different from Z(p^n), although both have p^n elements.

[tribal moss]

It sounds like we may need to start by seeing that a finite field must have a subfield isomorphic to one of the Z/pZ for some prime p.

[noble nexus]

it will be a direct product as a vector space, but it also has a product

between elements

[prime sundial]

I am trying to proof Cauchy theorem that say every prime dividing the order of a group G, there exist an element in G such that it's order is that prime, but what I proved it just for cyclic groups, am I getting closer to the proof?

[noble nexus]

not quite if I remember the usual proof of Cauchys theorem

The issue is you can't really reduce the general case to the cyclic case

[cyan skiff]

So additively, it is like an n-tuple coordinate vectors, but multiplication isn't pointwise?

[prime sundial]

Is it hard or long?

The proof

[cyan skiff]

With elements from Zp?

[tribal moss]

Multiplication definitely isn't pointwise.

[noble nexus]

Yeah exactly, just how the complex numbers are basically tuples of real numbers but the multiplication isn't pointwise

[cyan skiff]

Oh, yeah that's what I was struggling with. Thanks guys

[cyan skiff]

And we define multiplication in terms of irreducible polynomial or smt? Haven't gotten there yet

[noble nexus]

I don't believe so, but there's a bit of a trick that might not be obvious depending what material you've covered

[noble nexus]

unfortunately there isn't really one construction of a field with a given number of elements

but polynomials are generally the easiest

[tribal moss]

In fact the fields of size 9, 49, 121, 361, ... can be constructed in direct analogy to how you make C from R.

[noble nexus]

but essentially it's similar to the construction of the complex numbers as being polynomials where x^2+1 is considered to be equivalent to zero

[rocky cloak]

You sort of can actually. But that might not be the most common proof

[cyan skiff]

Okay, thanks guys. It will be some time before I get to constructing fields

[fast kelp]

If you read rust, I have an example implementation of this in rust.

it might help with understanding.

[rocky cloak]

At least medium difficulty I'd say. Not necessarily super long, but depends on the proof I guess

[cyan skiff]

Programming lang? If so, nahh

[rocky cloak]

Something that shouldn't be particularly hard, but may or may not be insightful is to try to construct the field with 4 elements yourself.

The elements will look something like
0, 1, x, x+1
so you just need to figure out how they can multiply to give a field

[tough raven]

Yes, rather (really exactly) like the complex numbers.

[thorn jay]

if multiplication was pointwise youd have many many zero divisors

[cyan skiff]

Okay, sounds like a neat excercise. Will work on it

[glad osprey]

Sounds interesting, can you share?

[fast kelp]

Sure thing

Let me throw it on github

https://github.com/OmnipotentEntity/gf256/blob/master/src/gf256.rs

The code was originally intended to obfuscate a secret in a reverse engineering challenge.

@glad osprey

[glad osprey]

Cool, thanks!

[fast kelp]

(The person I wrote it for never got around to implementing it, and I similarly didn't bother.)

(But the gf(256) part works)

[glad osprey]

Neat how an element fits exactly inside a u8, and addition just becomes xor

What are the const generics N and G for btw?

[fast kelp]

N is for the polynomial, G is for the generator used.

Specifically, I am implementing the obfuscation like a RAID6.

in a RAID6 you multiply through by a power of G for each disk beyond the first.

it's explained in detail starting from line 147

[glad osprey]

I see, interesting

[balmy python]

got my groups exam tmrw

lets hope i do good

ty guys for the advice along my group theory learning journey

i'll definitely be back for more groups

(and rings)

[quiet pelican]

May I interest you in a geometric group theory

[rapid cave]

Gromov's theorem is pretty cool

[sacred wharf]

geometric group theory when it meets algebraic group theory

the idea that we can distribute powers

as in

(ab)^2 = a^2b^2

only works if we assume commutativity right ?

[thorn jay]

its an iff, even

[south patrol]

Or odd

[elfin wraith]

(ab) = ab iff we assume commutiavity

You have to commute the left bracket to the right and use the fact that () =

[tough raven]

No, not an iff for any odd exponent.

[thorn jay]

i remember proving that if (ab)^k = a^kb^k for three consecutive k then G is abelian

this immediately proces the k=2 and k=-1 case cuz it trivially holds for k=0,1

[sacred wharf]

hmmmmmm
suppose (ab)^2 = a^2b^2 then G is abelian this is true because
abab = a^2b^2
ba = ab

[thorn jay]

ta

[south patrol]

Nonabelian group of order n for n odd lol and take exponent n

[sacred wharf]

ababab = a^3b^3
baba = a^2b^2
a^9b^9 = (a^3b^3)^3
hmmmmmmmmmmmm
no
baba = abab

[thorn jay]

by this logic groups of order 2 are always abelian!! surely that cannot be true

[south patrol]

Lol

[quiet pelican]

1 + 0 = 1 but 0 + 1 = 0

[thorn jay]

exactly

secret third group

second

not third

[tribal moss]

The group with half an element.

(It's the multiplicative group of F1).

[balmy python]

what are the typically used generators of Sn

and how could i prove that they generate Sn

maybe if it could be explained for a small case?

[quiet pelican]

(12), (23), (34), ..., (n-1,n) is a usual set

so is (12), (123...n)

There's also supersets of those sets (eg all the transpositions)

[balmy python]

how could i prove they generate the group?

[balmy python]

(formally)

[quiet pelican]

First I'll prove that this generates a group which contains the group generated by (12), (23), ...

[balmy python]

i guess you could say it's a renaming of elements blah blah

idk

[balmy python]

well (123...n) can be broken into two cycles right?

[quiet pelican]

so let s = (123...n)
Then s(12)s^{-1} = (23)
and also s(k,k+1)s^{-1} = (k+1,k+2)
So this is done

[balmy python]

yeah

[quiet pelican]

The proof that (12), ..., (n-1,n) work is essentially the same proof that the bubble sort algorithm works lol

[balmy python]

wait really?

[quiet pelican]

Which you can do by induction, where if the n is mapped to the kth place, we apply (k,k+1), (k+1,k+2), ..., (n-1, n) sequentially to move it to the nth place
Then apply induction

[balmy python]

thats so cool though

[quiet pelican]

What swaps does bubble sort use?
It's only adjacent elements

[balmy python]

yeah

but how does that represent any given permutation

am i being stupid

wait

no

i am

given a permutation

just apply the bubble sort

boom u get it in terms of the two cycles

i guess this is just how you can write it in two cycles

[quiet pelican]

Take the inverse of the permutation
Sort it
That gives you the original permutation
(As you're just asking "what do I need to do to the permutation to get it back to the identity")

[balmy python]

THIS IS INSANE

where did you find this out from lmao

[elfin wraith]

That’s just kinda permutations work

[quiet pelican]

I was just thinking "ok how do I prove this nicely" and somehow it came up

[elfin wraith]

It is very cool though

[balmy python]

oh yeah i guess it is how you can prove bubble sort

[tribal moss]

Could you perhaps stop pressing enter several times in the middle of sentences? It's very annoying to read.

[balmy python]

sorry 😭 idk why i do that. my bad for it being annoying

[noble belfry]

Anyone have any idea how to show that the ideal { f \in C[0,1] | f(c) = 0 } is not finitely generated?

[noble belfry]

I’ve thought about taking |f_1| + |f_2| + …. + |f_n| and deriving a contradiction but can’t quite make it work

Let me know if anyone has any hints

[quiet pelican]

Consider the ideal J = (f_1, \dots, f_n)
Consider the subset {g \in J | g(1/n) = a_1, \dots, g(k/n) = a_k, \dots, g(1) = a_n} (where we assume wlog c =/= i/n for any i, because otherwise just take some other n element set)

[tribal moss]

Hmm, that doesn't really sound like a promising direction to me ... it feels like we'll need to consider behavior arbitrarily close to c just in order to conclude the ideal is not principal.

[quiet pelican]

I think that ||1 not being in the ideal is the necessary condition for the resulting matrix to be non-singular, and say that set has size at most 1||

[balmy python]

Can someone explain the significance of stabilisers of points in the same orbit being conjugate in the group? I know conjugation relates to normal subgroups but what normal subgroup is at play here?

[tribal moss]

If you have some g in your set, then you can add ((x-1/n)(x-2/n)···(x-n/n))·f_1 to make a different function also in your set, as long as f_1 is not identically zero.

[cobalt heath]

What are the ways Z/qZ can act on (Z/QZ)^N as a set?

[tribal moss]

One immediate consequence that feels not useless is that the stabilizers of elements in the same orbit are isomorphic subgroups.

[cobalt heath]

Sorry that I interfered..

[balmy python]

was this for me?

[cobalt heath]

No, it was my own question

[balmy python]

oh 😭

[thorn jay]

too many

[balmy python]

isomorphic with the map being conjugation?

[tribal moss]

It looks weird to write "(Z/qZ)^N as a set" rather than just specify some random set with q^N elements.

[balmy python]

I feel like i haven't grasped the idea of orbits and stabilisers

[tribal moss]

You need to do a fair amount of exercises before it starts making real intuitive sense.

[noble nexus]

Yeah examples are quite illuminating here

For instance if you take the group of rotations of say an icosahedron

The stabilizer of any point is just a cyclic group of order 5

they're different subgroups but are all sort of the same (conjugate)

[cobalt heath]

Yeah, there is details that I decided to leave out at first

[noble nexus]

and a similar thing happens in general where the stabilizers along an orbit all kinda "look the same"

tbh it's a good exercise to just think of random actions and compute or think about the stabilizers of various points

[cobalt heath]

In reality, I might be looking for Z_q-representations, but I am not sure

[south patrol]

q-adic

[south patrol]

Is q a prime? Prime power?

[cobalt heath]

Prime

[south patrol]

So like

[tribal moss]

One way might be: Consider the finitely generated ideal J = <f1,...,fn>, and set g(x) = max(|f1(x)|,|f2(x)|,...,|fn(x)|).
Then every function f in J is O(g(x)) as x->c, but sqrt(g(x)) is not, and should be in the ideal you're trying to generate.

[south patrol]

Orbit-stabiliser will tell you this

Details are left to the reader

But this is indeed a cute application thereof

[rocky cloak]

Calling the ideal I it's not too hard to see that I^2 = I. By for example Nakayama if I was finitely generated it would be generated by a single idempotent. But the only idempotent in the ideal is 0

[cobalt heath]

I can't believe I forgot orbit-stabilizer

[south patrol]

Nice argument

[cobalt heath]

But anyway, I guess there are going to be quite many

[south patrol]

Trying to think of a nice like way to write down lots of elements in there lol.

[south patrol]

Not actually too many really

Well as in the point is you can just write down all of them up to iso

[tribal moss]

Is Z_q here Z/qZ or the q-adic integers? Whoops, it said Z/qZ originally.

[cobalt heath]

Z/qZ, sorry

[south patrol]

I was unsure about this myself lol

Ah

[cobalt heath]

Cryptographers be like this, lol

[south patrol]

Yeah because it was Z/q^n and then Z_q was said, it made me assume that that Z_q actually meant the q-adics

[rocky cloak]

So what is the question? How many ways Z/q acts on a set with q^n elements? Or was there more to it?

[south patrol]

But what I have said is all for Z/q ofc

[tribal moss]

The only way Z/qZ can act on anything if is the generator maps to a permutation consisting of some q-cycles and some stationary elements.

[south patrol]

Indeed orbit stabiliser

[cobalt heath]

Yeah there is more to it but I have to find how to ask properly

Basically I want
a . (m + e) - a . m to be small, for small e

[cobalt heath]

That's why I was thinking of (Z/QZ)^N

[rocky cloak]

I think I'm more confused than before

[tribal moss]

What are a, m, e there? You need to give more context, I think.

[cobalt heath]

Yeah finding Z/qZ acting on set of Q^N elements is at least a bit standard
And there is this one I am trying to tackle, damn

[chilly ocean]

”a set of Q^N elements” is a bit of a weird phrase, “a set in bijection with R” or “a set with the cardinality of the continuum” are both much more common

[tribal moss]

They meant q, not Q.

[chilly ocean]

ohhh

[vapid vale]

why using q for a prime not a prime power 😭

[south patrol]

Real

And there is no p already

[cobalt heath]

Ah.
So suppose Z/q acts on (Z/p)^N.
For any a in Z/q and m in (Z/p)^N,
I want a (m + e) - a m to be small when e in (Z/p)^N is small.

Ah wait, the Q I meant is not the rational duh

[tribal moss]

What does "small" mean here?

[south patrol]

How many ways can Z/ℓ act on (Z/ℓ)^n

[cobalt heath]

Close to 0 when you take infinite norm of the representative.

[south patrol]

Wdym by norm

[tribal moss]

Huh.

[south patrol]

Or just like

Okay unique reps in {0,..., q-1}

Lol

[cobalt heath]

You can "put" (Z/p)^N in Z^N, and take infinite norm

[south patrol]

Ah so is this indeed crypto

[vapid vale]

lol

[south patrol]

(I would say the sup/infinity norm, rather than infinite)

[cobalt heath]

Yeah like that. Details are more involved, like it can be negative, but I don't think that changes much

[south patrol]

That sounds important tbh

[cobalt heath]

That was a typo

[south patrol]

Ah dw

[cobalt heath]

Yeah it is. LWE problem is about erroneous encryption in Z/pZ, so I am thinking of erroneous case.

Idk why they use q for primes, welp

Anyway I guess the details are all over the place, need to sort them out carefully.

[south patrol]

What does LWE sorry

Oh "learning with errors"

[cobalt heath]

Yeah "learning with errors"

[south patrol]

Cool thanks

TIL

[granite badge]

its still stuck in my head

[errant lagoon]

ah never mind, what I thought doesn't work

[spark veldt]

how do i show this?

[granite badge]

it seems hard

[tribal moss]

Yes.

We get thhat x -> x³ is a homomorphism (and then quickly that it must be an automorphism), but then everything seems to dry up.

[noble nexus]

its not too hard to see that it must be a central automorphism, so G must have some center

if its an inner automorphism, then I believe it has to have order 2, which I think leads to a contradiction since then you have x^9=1 for all x in the group which means the group has an element of order 3 or 9

actually wait, if x -> x^3 is an automorphism then the automorphism has finite order so there must be an n with x^{3^n}=1 for all x in the group, and doesn't that imply 3 divides the order of the group

nvm that has to be wrong since it would apply to the abelian case as well

[south patrol]

It shows x^(3^n) = x

[noble nexus]

oh yes of course

anyway if its inner, then you would still get x^8=1 for all x which I guess narrows the search for a potentiial counterexample

group with exponent 8 that has a nontrivial center

[south patrol]

Yeah my hope was D_8 but it doesn't work because like rs = (rs)^3 but r^3 s^3 = r^3 s

[noble nexus]

and the automorphism group needs to have a nontrivial center with an element of order 2 in it ig

[south patrol]

I wonder if you can just check the (infinite) "universal case"

[noble nexus]

(if its inner)

oh wait I think I have a potential sol

nvm I misread a paper

[south patrol]

Ngl I googled it as it is late here but will let others have fun lol

[tulip otter]

lmao I also thought about this

[vapid vale]

is there some weird commutator stuff you can do

[dull marsh]

||aabb = baba so bbaaa = ababa = aaabb. Now play around with b^2 = a^3n b^2 a^-3n (by that I mean choose something nice for n)||

[noble belfry]

why should sqrt(g(x)) be in the ideal?

i don't see how taking the max of the generating elements gives us a continuous element

i think we can argue similarly with |f_1(x)|+|f_2(x)|+... though?

I see what you're saying generally tough that
$f(x) =\sum_{i=1}^n h_i(x)f_i(x) \leq \sum_{i=1}^n M_i f_i(x) \leq Mg(x)$.
So we have that
[
\sqrt{g(x)} \leq Mg(x) \quad \implies \quad 1 \leq M\sqrt{g(x)}
]
so $\sqrt{g(x)} \geq \frac{1}{M}$ for all x, so $g(x) \geq \frac{1}{M^2}$ for all x where g(x) != 0, which is not the case since by continuity, there is some point where $|f_1|+|f_2|+...|f_n| <= 1/M^2$ (Intermediate Value Theorem)

[cloud walrus]

sudo

[noble belfry]

honestly tho i would never have thought to look at $\sqrt{g(x)$

[cloud walrus]

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

[noble belfry]

could you give some insight how u came up with it? @tribal moss

[rocky cloak]

Taking max is continuous.

[noble belfry]

Oh I see

😭

[tribal moss]

Absolute value, maximum, and square root are all continuous functions, so composing them with the (assumed continuous) f_i again gives a continuous function. And sqrt(g(c)) is obviously 0, so it is in the ideal.

You could add the absolute values instead of taking their max, that would also work.

[tribal moss]

I had a lot of false starts, but what eventually worked was thinking something like: Intuitively I suspect the obstacle to being finitely generated is that the choice of generators limit how fast the functions in the ideal they generate can move away from 0. I know from having worked with asymptotic growth rates before that there's always a function that grows faster than a given finite collection -- to be sure most of that experience is for growth as n->infty, but asymptotics at a finite limit have enough similarities that it feels promising. Now "how fast they grow" cannot simply be a matter of constant factors, since you can just multiply the generators by a large constant (which are all in the ring). However, that's the best one can do since [0,1] is compact and so every function in C([0,1]) is bounded. So intuitively we want to multiply the upper bound with a factor that gets larger the closer that bound gets to 0 -- and that is what the square root does.

Jagr's argument using Nakayama's lemma is much slicker, except I'm an ignoramus who doesn't actively remember such general tools. If you have it available in your context, definitely use it instead of the square root argument.

[tribal moss]

Ah, but then we're done, because ||that means x^-1·x³ is always in Z(G), i.e. x² commutes with everything. And then a³b³ = (ab)³ implies abbaab = ababab; cancel ab at each end and you get ba=ab.||

How do you see that it is central, though? It seems to be a not widely described property -- https://math.stackexchange.com/a/436261 states without proof that an automorphism is central iff it commutes with every inner automorphism, and the latter is clearly the case here, but the connection doesn't seem to be immediate to me.

Oh wait, it is fairly immediate anyway.

[noble nexus]

Yeah if f is an automorphism then f(x^3)=f(x)^3

[tribal moss]

If f is an automorphism that commutes with any inner automorphism, then b f(a) b^-1 = f(bab^-1) = f(b) f(a) f(b)^-1, and because a was arbitrary that also gets us bab^-1 = f(b)af(b)^-1, which rearranges to ab^-1f(b) = b^-1f(b)a. So b^-1f(b) is in the center.

That looks like the central trick of the problem is to notice how "group order not a multiple of 3" means that x->x³ must be injective so every element is a cube. The rest seems to reduce to elementary symbolic manipulations.

[rocky cloak]

So the orbits of P either has size 1 or p. If it's 1 you're done.

If it has size p and y is an element of the orbit then gy is different for the different values of g in P.

Now pick y and g such that x fixes both y and gy, and conjugate g by x

[charred mulch]

why every element of the commutator sub group is a product of commutators

[thorn jay]

that is by definition

[candid patrol]

Because the commutator subgroup is the subgroup generated by all the commutators

[charred mulch]

i move the discussion to advanced algebra

[wintry sluice]

is there anything interesting that arises from dividing a ring of matrices by a prime ideal?

[thorn jay]

every ring of matrices is a finite dimensional k-algebra and every finite-dimensional k-algebra is a ring of matrices so the best answer you're gonna get is a finite-dimensional integral domain k-algebra

which, granted, have plenty of nice properties

[rocky cloak]

What do you mean exactly?

Like if R is a commutative ring and p is a prime ideal are you asking about
Mn(R)/(p)? If so that's equal to Mn(R/p)

[south patrol]

Matrices over Z be like

[thorn jay]

rahhh assuming matrices in field coefficients

else you can see a ring as the ring of 1x1 matrices with coefficients in itself

smh

[south patrol]

Ok ig it's just that makes the original q much more boring

[thorn jay]

I think in how many ways you wanna twist it you're just gonna end up with something along the lines of "finite type R-algebra integral domains"

[tribal moss]

Matrices over Z are are ring of matrices with real coefficients.
You can't assert that a subring of the matrix ring is an algebra over the field unless you know something extra that makes it an algebra.

[thorn jay]

right that is true

lol

[south patrol]

I'm confused what you mean by the first bit

[tribal moss]

I mean M_{n×n}(Z) is a perfectly good subring of M_{n×n}(R), so saying "assuming matrices in field coefficients" won't ensure we're talking about an R-algebra.

[rocky cloak]

I mean, if someone gave Mn(Z) as an example of a ring of matrices with field coefficients I would say they have some weird choice of terminology

[south patrol]

I would say matrices with field coefficients like surely must mean M_n(F) for a field F

[alpine plank]

Trying to prove some equivalent definitions for DVRs, If R is an integrally closed 1 dimensional Noetherian local domain how do I show it's maximal ideal is principal?

[south patrol]

Let m be the unique maximal ideal. Pick any element x in m \ m^2 (exists by Nakayama) and like show this is prime iirc

Then (x) is a nonzero prime ideal contained in m, hence = m by assumption

[alpine plank]

Yeah not sure what to do after that, (x) has to be m-primary so I can get an n such that m^n \subset (x) but is this useful?

[alpine plank]

Yeah i dont think this is useful nvm, cause the n you get "should be" 1 because we took x in m \ m^2

[rocky cloak]

It should be useful for reaching a contradiction.

Like take n minimal with m^n in (x) pick y in m^n-1 not in (x). Then y/x is not in R hence not integral over R.

(y/x)m is in R. If you can show it's all of R, then x is a multiple of y and you can get a contradiction.

To finish it off ||the only way it can not be all of R is if it's contained in m.||
||But then if a1, ..., ak is a generating set for m, you can embed R[y/x] into Prod^k m by f maping to (f ai). So R[y/x] is a fg R-module||

Actually, that's overcomplicating it

||you can just embed R[y/x] into m by mapping f to f times whatever||

[alpine plank]

thanks!

[vocal pebble]

Field theory is so cool

[white oxide]

How does this show that x bar is algebraic over k

Is it because k' is a finite extension of k and thus all of its elements are algebraic over k (including xbar)

Wait nvm that's a priori

Or is it just because $\overline{x} \in k' = k[\overline{x}]$ so that $\overline{x} = b_n \overline{x}^n+ b_{n - 1} \overline{x}^{n - 1} + \dots + b_0$ for $b_i \in k$. But then $\overline{x}$ is the root of the polynomial $b_n x^{n} + b_{n - 1} x^{n - 1} + \dots + b_0 - x$

[cloud walrus]

okeyokay

[rocky cloak]

How is Sigma defined here exactly?

[rocky cloak]

It's just that k' = k[xbar] is a quotient of the polynomial ring k[x], so isomorphic to k[x]/(f(x)) I guess

And hence xbar is algebraic with minimal polynomial f

[white oxide]

Mm okay

[late marsh]

i dont know if it"s just me but my ring theory class has felt a little political lately?

we were talking about how to radicalize our ideals and like the power of people in radical ideals

[rocky cloak]

What do simple modules and conservatives have in common?

They're both annihilated by radical left ideals

[mighty kiln]

Quadruple pun

[mental lake]

Let V4 = {1, a, b, ab}, G a group, and
B = {(g, h) ∈ G × G | g^2 = 1 = h^2, hg = gh}.
Show that the function
ψ : Hom(V4, G) → B : f → (f(a), f(b)) is a bijection (first show that the function is well-defined).

can someone help w showing this is surjective

[sacred wharf]

Is the maximal ideal the ideal with the maximum number of elements or an ideal such that there is no proper ideal that contains it properly?

[elfin wraith]

The latter

[dull marsh]

Any progress?

[mental lake]

No

[sacred wharf]

Why does Max have two different meanings in a sense?

[twilit wraith]

Max with respect to containment is generally more interesting

[dull marsh]

Can you state what surjectivity of ψ means in this case?

[elfin wraith]

Because maximal with respect to set inclusion is generally the interesting concept. I don’t even really know how I’d formalise the “maximum number of elements”

At least not in a capacity that’s at all interesting

[twilit wraith]

Its clear why max with respect to cardinality would be completely uninteresting when it comes to ideals

[sacred wharf]

I meant maximal text speech isn't very good today

[twilit wraith]

In every such case the max ideal would just be the entire ring

[sacred wharf]

I mean, you could define it as the maximal proper ideal

[rocky cloak]

I mean you'd still restrict to proper ideals. If your ring was finite an ideal with maximal cardinality would also be maximal

Not that it's particularly interesting anyway

[twilit wraith]

I see

[elfin wraith]

Even still, what would this really mean? How would you compare them? Ideals are generally infinite

[sacred wharf]

You could look at the finite case, I guess

[rocky cloak]

So you're given (g, h) with g^2=1=h^2 and you want to construct f such that
f(a) = g and f(b) = h.

Do you have a guess for what f(1) and f(ab) should be? Can you verify that this is a homomorphism?

[mental lake]

i think i got it

[south patrol]

Maximal usually means this sorta ting

Eventually

[south patrol]

Ig this is relatively rare and it's not clear smth would be unique

[vocal pebble]

let a be the positive 4th root of 2. find all intermediate fields between Q(a) and Q.
here is what i tried:
so clearly any (nontrivial) intermediate field has degree 2, as [Q(a) : Q] = 4. I can only think of Q(a^2), but im having trouble proving it is the only one.
If F is another intermediate field, then F = Q[x]/(x^2+bx+c) and the roots of x^2+bx+c must be in Q(a). This happens iff (-b +- sqrt{b^2 - 4ac})/2 in Q(a) iff sqrt(b^2 - 4ac) in Q(a). Now we can multiply sqrt(b^2 - 4ac) by some integer suitably to get that there is sqrt(m) \in Q(a) for some integer m.

Now if we characterise all possible elements of the form sqrt(m) \in Q(a) (namely if we get sqrt(m) in Q(a) iff m is a multiple of 2), then we are done, as then m must be a power of 2 (otherwise we get sqrt(m')\in Q(a) with m' odd contradicting the characterisation) and thus sqrt(b^2-4ac)\in Q(sqrt(2)) giving us F = Q(sqrt(b^2 - 4ac)) = Q(sqrt(2)) as desired.
should i continue with this? i tried but it became very messy. if anyone wants to help, please join
#1438661116308885524 message if you dont mind

[quiet pelican]

Can I use the Galois correspondence?
(ie, are you expected to know it?)

[vocal pebble]

nope

i just started field theory, at max i know what separable extensions are

[quiet pelican]

At which point I'd just say "yeah I could do the messy calculation to show there are no other square roots in this field, but it's not worth my time"

[mental lake]

how do we get that conjugation from that product

u also know this

[quiet pelican]

(Essentially the proof this way would be "D_8 = Gal(Q(i, qtrt(2)/Q)", and that extension is the splitting field of x^4 - 2, and D_8 has 3 subgroups of order 4, hence Q(i, qtrt(2) has 3 subfields of degree 2 over Q, and these are Q(i sqrt(2)), Q(i), Q(sqrt 2) and only the latter is contained in Q(qtrt(2)))

[vocal pebble]

hmm that is very odd because this is given as an exercise in the field theory chapter, which is just before the galois theory chapter. im hoping there is an elementary method to do it

[quiet pelican]

I mean the elementary method is just "manually square an element of the field, multiply by squares to remove denominators, and check when the resulting diophantine equations have a simultaneous solution"

[vocal pebble]

yeah so it looks like il have to do the messy calculation i was trying to avoid

[quiet pelican]

It's not hard, but it's also not worth your time if you did similar for smaller examples

[rocky cloak]

Maybe the exercise is meant as a motivation for Galois theory.

I.e. at the end of the Galois theory chapter you'll be given the exercise again with a "see how much easier this was"

[quiet pelican]

"Remember how we tortured you? Yeah we didn't have to." /lh

[somber sleet]

Guys if I have an ideal I in a ring R, does R/I \cong R imply that I = 0?

[vocal pebble]

doesnt seem like it, in Z^{\mathbb N}, consider the ideal generated by (1,0,0,...)

[elfin wraith]

No, take Z^\infty

[somber sleet]

mmh okay, this makes me sad 🙁

[elfin wraith]

I think it depends if you’re thinking about an isomorphism of rings or modules though

[somber sleet]

of rings

I thought the statement was an if and only if

[tribal moss]

It fails in the same way for modules, doesn't it?

[vocal pebble]

yes, because R is an R-module and I is a submodule

[elfin wraith]

It may do, I feel like I remeber there being something different about the two cases but maybe I just needed a different example or something

[thorn jay]

for any nontrivial algebraic structure you have the same argument

[mental lake]

van anyone help

[tribal moss]

It would need to be an algebraic structure that has a meaning concept of quotients, so fields are out.

[thorn jay]

i was talking about take any nontrivial A, then A^∞ ≈ A^∞ / θ for some nontrivial θ (infinitely many, in fact)

[tribal moss]

If you explain what you're doing instead of just posting an image of some formulas without further explanation, there would be a higher probability that someone can find something helpful to say.

[vocal pebble]

use the definition of the semidirect product operation. you know the automorphism, so compute the product (1,x) (y,1) by definitions. that should be equal to the product (y,1)^{-1} (1,x) where <y> = C_7, <x> = C_8

[rocky cloak]

Depends what exactly the statement for modules is.

R and R/I is only isomorphic as R-modules if I=0. But in general M and M/N can be isomorphic yes.

[rocky cloak]

The natural map R -> R/I is an isomorphism if and only if though.

[white oxide]

How does x^{-1} mapping to zero show that x is not in B

Is it because if it were, then if $g: B \to k'$ is this extension, we would have $1 = g(1) = g(x x^{-1}) = g(x) g(x^{-1}) = 0$ so that $1 \in \mathfrak{m}'$, a contradiction?

[cloud walrus]

okeyokay

[quiet pelican]

Yeah

[white oxide]

Is there any way this can be justified more precisely?

[velvet hull]

Finite fields cannot be algebraically closed

Suppose that a1, … , an are all the elements of F.
Then the polynomial (x-a1)(x-a2) … (x-an) + 1 evaluates to 1 on every element of F

[south patrol]

I would say rather that like

A non-zero polynomial has only finitely many roots

That seems to be what they are using, right?

[rocky cloak]

I guess the question is what okey was asking about.

But yes, there can't be infinitely many \xi where it's zero

[sacred wharf]

3 implies 1 can be shown using correspondence theorem right ?

[quiet pelican]

Yup

[sacred wharf]

if M has a proper submodule it must map to an ideal of R/I SAY B/I by correspondence B is an ideal in G that contains contains I

I guess i have to show proper containment

Ah yes trivial

since they are isomorphic

any non zero submodule of of M must map to a non zero ideal of R

[sacred wharf]

is this the idea ?

[quiet pelican]

Yeah

[sacred wharf]

dam correspondence theorem is actually quite nice

https://tenor.com/view/joe-biden-presidential-debate-huh-confused-gif-9508832355999336631

[elfin wraith]

It’s fantastic

You seem to have gotten surprisingly far in algebra without

[sacred wharf]

I suppose many proofs where you could use correspondence you can force another proof

the proof above didn't actually use correspondence in the lecture notes

I just thought of correspondence as a shortcut

[elfin wraith]

Yeah a lot of the proofs of that I’ve seen don’t use correspondence, but it’s easy enough to prove and useful enough that I don’t see any good reason not to use it

[twilit wraith]

which one is correspondence again

i only know the isomorphism theorems by their numbers

is that the bijection between subalgebras of A containing ker phi and A/ker phi

[south patrol]

Submodules etc

Subalgebra if by algebra you mean UA sense ig lol

[twilit wraith]

yeah i mean in the UA sense my bad

i kinda use both the UA definition and the "vector space with some spice" definition depending on context

[ebon prairie]

can anyone help me with this question "show that x^4 +1 is reducible over Zp for every prime p"

[alpine plank]

Continuing with this, If R is an integrally closed 1 dimensional Noetherian local domain with dim _k (m/m^2) = 1 how to show every non-zero ideal is a power of m? dim _k (m/m^2) = 1 implies there are no ideals between m and m^2, so if I can show dim _k (m^n/m^n+1) = 1 it should be enough (again using the fact that every* ideal is m primary and we can get maximal i such that m^i \subset (a)). It's easy to see they are isomorphic if m is principal but I don't see how to show it otherwise

Atiyah-Macdonald does it using some fact about Artin local rings, but I don't see why R/m^n is Artinian either

[alpine plank]

What have you tried?

[alpine plank]

ok wait it's dimension zero and Noetherian oops

[ebon prairie]

im not able to think what concepts i can use in this

[alpine plank]

Try checking with p = 3,5,7 and see if you get any ideas. What does being a root of x^4 + 1 mean for an element of the group of units in Z/pZ? Then think about the case when it has two quadratic factors

[alpine plank]

Also you might need to know when -1 and 2 are squares mod p

[thorn jay]

there is no sense of "containing ker ϕ" for general algebras

the correspondence theorem is about congruences

[twilit wraith]

i see

i frankly just made an assumption there that i thought would hold

[thorn jay]

because ker ϕ is a relation, not a subset of A

subalgebras just dont behave quite as nice

[twilit wraith]

how

for a group homomorphism phi from G to H, ker phi is a subgroup of G

similar for ring homomorphism, linear transformation , etc

[thorn jay]

sure but not for general algebras

in fact, there may exist nonequal congruences which share most of their equivalence classes

only when A is so-called 0-regular can you faithfully represent a congruence as its equivalence class at 0 (which is some distinguished element from your algebra)

[rocky cloak]

A good hint could be that if a and b don't have square roots mod p, then ab does

And just think about all the ways x^4 + 1 can factor

[thorn jay]

formally ker ϕ = { <a, b> ∈ A^2 | ϕ(a) = ϕ(b) }

[twilit wraith]

ah i see

[lime stump]

How can I prove this without using long division?

[quiet pelican]

Try to take a factor of x - r out of p(x) - p(r) using difference of two ith powers

[lime stump]

Ive been trying for a while. Just to make sure:
It wants me to prove $\exists v \in k : p(x) = p(r) + v(x-r)$ right?

[cloud walrus]

adapt

[rocky cloak]

No, they want you to show there is v(x) in k[x] with
p(x) = p(r) + v(x)*(x-r)

If v is in k then
p(r) + v(x-r)
is always degree 1, so can't be p(x) if it has larger degree

[lime stump]

thank you

[untold hearth]

I’m working on an exercise in ring theory, and I need to use the euclidean algorithm for k[X_1,…,X_n], with respect to the last variable. So, euclidean division of the ring k[X_1,…,X_{n-1}][X_n]. However, I thought that euclidean division only really makes sense for polynomial rings over fields. What can be salvaged in this more general case of R[X] with R an integral domain?

[rocky cloak]

In general you can do euclidean division by monic polynomials in R[X] for any commutative ring R.

What's nice about fields is that every polynomial is monic (up to unit)

[noble nexus]

yeah its all a facade. Basically the reason you can get away with this in groups/rings/modules is because you have inverses. Thus asking the question "is x equivalent to y" is the same as asking "is x-y equivalent to zero"

congruences are really the more natural objects, because the definition is the same no matter what object you are studying

A congruence is just an equivalence relation that obeys all of the structure

It just do happens that for groups, congruences can be identified with normal subgroups, for rings with ideals, etc

(For instance, given an ideal J if you define x equivalent to y if x-y is in J, that forms a congruence and given any congruence the set of elements equivalent to zero form an ideal)

[tribal moss]

So we'd define "obeys structure" as "each operation induces an operation on the equivalence classes too"?

[noble nexus]

kinda but its even easier than that I think, at least for say rings you would want x ~ y and a ~ b to imply xa ~ yb and x+a ~ y+b

and -x ~ -y and so on

for a universal algebra perspective basically just anything you can do too both sides should respect the equivalence relation

an even slicker way of saying this is that your equivalence relation should form a subring of the product R x R

(or a subgroup or sub-whatever)

[white oxide]

Can somebody explain to me why this highlighted line is true

[quiet pelican]

If x is algebraic, so is any polynomial in x
If v is algebraic, so is it’s inverse

[white oxide]

Oh wait is it because the algebraic closure is a subring

[candid patrol]

it's a field

[white oxide]

oh aight

(Continued from last ss) where do we use the definition of $u$? Since $a_0 x^m + a_1 x^{m - 1} + \dots + a_m = 0$, we have $x^m + a_0^{-1} a_1 x^{m - 1} + \dots + a_0^{-1} a_m = 0$. Since $a_0^{-1} a_i \in A$ for all $i$, $x$ is integral over $A$ and hence $A[u^{-1}]$.

[cloud walrus]

okeyokay

[tidal schooner]

Here you multiplied everything by $a_0^{-1}=a_0'u^{-1}$, and similarly for (2) you need to multiply everything by $(a_0')^{-1}=u^{-1}a_0$. The definition of $u$ ensures that both of these elements actually lie in $A[u^{-1}]$. This is necessary because they might not lie in $A$, as $A$ isn't assumed to be a field

[cloud walrus]

harmacist

[white oxide]

Ah okay, thanks!

[thorn jay]

this is where the internal equivalence relation definition of a congruence comes from

[noble nexus]

yeah

it is nice although it doesn't really translate to other mathematical structures, whereas "equivalence relation that preserves structure" can more or less be applied to anything in math

[thorn jay]

also it has been proven time and time again that consider all cosets of congruences is super useful

and it is a much less natural notion if you only know the ideal view

[alpine island]

so I know modules can be thought of as generalized abelian groups (replacing Z with another ring). What about nonabelian groups? Are there non-commutative "modules"?

[elfin wraith]

Much like how you don’t actually need to define the additive group of a ring to be abelian, it’s the same with modules. If you dropped the abelian requirement it still turns out to be abelian by the other axioms

I’m not aware of a similar notion which is truly non commutative, but it would need to look a bit different to modules, you need something more substantial than just dropping commutativity

[alpine island]

right, you need to drop some other things too

hence "modules" in quotes

[topaz solar]

Group actions?

[elfin wraith]

I guess the question is, what structure do you want to preserve?

Do you want the “vector space like” structure, or do you want something where we can view all groups as “modules” over some ring?

[alpine island]

I want to view all groups as "modules" over Z

then swap Z for an arbitrary ring

[topaz solar]

Every group has a Z action by n•g = g^n

[south patrol]

Not rly right lol

Action of what?

I mean like n.(gh) is not (n.g)(n.h)

[tribal moss]

If I remember correctly, it's having distributivity both to the left and to the right that forces addition to be commutative, so we'd need to discard at least one of them.
If we discard (a+b)m = am+bm, then there's no role for addition of scalars to play anymore, so it would just be "over" a multiplicative monoid.
If we discard a(m+n) = am+an, then multiplying by a fixed scalar is not an additive homomorphism anymore, which also sounds bad.

[south patrol]

I guess this is a nontrivial action of Z on the set underlying G lol

[topaz solar]

It’s an action but not as automorphisms

[south patrol]

Ig I mean like "acts on a group" usually means by group homs ofc

But sure

[elfin wraith]

Yeah I kinda had the feeling that anything you drop will give you a pretty uninteresting structure, but it’s late and I’m too tired to think about actual implications lol

[topaz solar]

@delicate orchid now’s your chance to speak on funny G-modules

[alpine island]

okay yeah, the second one is what fails for Z and nonabelian groups, so probably dropping that

[topaz solar]

G -> Sym(X) not G -> Aut(X)

[tribal moss]

Alternatively, I suppose we could speak about modules over a rng, in which case it might be possible for addition to be non-commutative outside the range of scalar multiplication.

That wouldn't help with considering an arbitrary group to be "something" over Z, though.

[alpine island]

that does also sound interesting

pZ modules

[tribal moss]

At least, at the trivial end of the scale, if we can't require 1m=m, then scalar multiplication could produce the additive identity of the "module" no matter what, and addition of "module" elements could be whatever. It's not clear to me whether there are really nontrivial examples, though.

[topaz solar]

Torsion groups:

[glad osprey]

Wait, is this actually a group action? (a+b)•g = g^(a+b), but a•(b•g) = g^ab

[topaz solar]

No

[alpine island]

it's a monoid action from (Z \ {0}, \times)

but not Z, the additive group

[south patrol]

Z^x = {+-1}

I guess you mean (Z, x)

[alpine island]

good point

[tribal moss]

The trouble with that is that it doesn't necessarily associate to the right: n·(g*h) might not be the same as (n·g) * (n·h).

[glad osprey]

What's (Z, x)?

[alpine island]

I mean I'm not wrong

[glad osprey]

Oh lol, monoids. I see

[tall igloo]

yeah it is the monoid of integers under multiplication

[rocky cloak]

So idk for rings, but modules over groups are also common, and groups acting on other groups come up often.

Iirc group cohomology can be generalized to this setting (though I'm not sure it's practically useful).

In that setting I think one considers actions as G -> Out(H) such that extensions correspond to H^2(G; H)

When H is abelian Out(H) = Aut(H) and you recover ZG-modules

[alpine island]

Sadly I don't know any group cohomology, but this does sound neat

[rocky cloak]

Some references for further reading exists here
https://mathoverflow.net/a/129731/157483

[alpine island]

Alright, I'll give that a look. Thanks.

[chilly ocean]

Modules are almost of actions of monoids on abelian groups. Except they are Ab-enriched monoids. So something that comes close to it is monoid actions on groups, also called groups with operators

[alpine island]

Those are pretty interesting, I'd never heard of them before

I think the difficulty with trying to construct what I want to construct is that the action of R isn't by endomorphisms, in the same sense g |-> g^2 isn't always an endomorphism for nonabelian groups

[delicate orchid]

This comes up in representation theory over cyclotomic fields. In particular the Galois group (which is a cyclic group) of your extension acts on the conjugacy classes in precisely this manner and the number of irreducible representations is in 1:1 correspondence with these orbits AND the characters of these representations are constant on these orbits. Serre covers this

[south patrol]

Sure ye I just mean it isn't a group action

But this is cool heh

[vocal pebble]

advertising my thread here:
https://discord.com/channels/268882317391429632/1438661116308885524

if anyone knows any field theory and has something to say / wants to answer some questions feel free to drop by!

[prime sundial]

;et G a group, if p a prime divides the order of G, then does |{x in G : x^p=e}| divisble by p?

[quiet pelican]

Yes, by a slightly more careful look at the standard proof of Cauchy’s theorem

(The way you prove it’s not 1 is by proving it’s at least 1, and it’s divisible by p)

[iron arrow]

Hey sorry guys, is it true that if we have two isomorphic rings R1 and R2 by a mapping f, then let A1 be an ideal on R1, then R1/A1 is isomorphic to R2/f(A1)? If not, if we make R1 and R2 be fields, do we have the implication?

[quiet pelican]

The first statement is true

[iron arrow]

Ok it was that I read something in my class notes that if I have a field isomorphism i between K and K', then we have that a polynomial f(X) in K[X] is irreducible iff i(f(X)) irreducible in K'[X], and I was wondering if what I stated was a valid way to justify it easily

(Due to K being a field, therefore being principal, therefore any irreducible polynomial being maximal, therefore K/(f) being a field, and isomorphic to K'/(i(f)), therefore K'/(i(f)) a field, therefore (i(f)) maximal, therefore irreducible)

[quiet pelican]

It's easy to see that if f(X) is reducible, then if(X) is, as if f(x) = g(x)h(x), then if(x) = ig(x) ih(x)
The converse follows by symmetry

[iron arrow]

Is this valid too?

[quiet pelican]

Yeah it works

(It's just a little overkill)

[iron arrow]

Yeah I get it

Thanks

[iron arrow]

I guess but I like these type of arguments a lot, but I get it

[sacred wharf]

good examples of non commutative rings that don't involve matrices ?

[quiet pelican]

Non-commutative polynomial rings, like k<x, y>

[sacred wharf]

https://tenor.com/view/poul3-f-r3veil-p-gif-10988460833755738340

what r those ?

[quiet pelican]

Endomorphisms of anything where adding endomorphisms makes sense (this generalises matrices though so like if you don’t count it)

[sacred wharf]

ohh yeah that is a good example

the ring of endomorphisms

[quiet pelican]

Pretty much exactly what you think they are
Ie k-linear sums of strings of xs and ys

You can also get deformations of rings like k<x, y>/(xy - yx = 1)

You also have universal enveloping algebras of Lie algebras

(Essentially answering the question “ok but what if I really want my Lie bracket to literally be xy - yx” in the “best” way possible)

Tbh at this point I feel legally obligated to throw this at you /hj
https://arxiv.org/pdf/1212.0914
(it’s lecture notes from a series aimed at grad students)

Just read (the first bit of) section 1 🙃

[sacred wharf]

bru i'm dying in algebra 2

how am i gon do grad algebra

[quiet pelican]

It’s just examples so it’s relatively simple

But you’re unlikely to see much non-comm algebra if you’re not in an explicit course on it tbh

[elfin wraith]

Or you go into like representation theory…

[elfin wraith]

It’s just like the polynomial ring except that you can’t commute x and y. Think of a free group

[quiet pelican]

Oh yeah group rings are another example

^allegedly a group theorist who would’ve done Rep theory in another world

[rocky cloak]

Path algebra of quiver or tensor algebra of species

Group algabra or skew group algebra is good

[elfin wraith]

The quantum plane is a good example, but I guess that’s just a quotient of the free algebra

[rocky cloak]

I mean everything is

[quiet pelican]

Yeah I didn’t say because it’s essentially a more complicated Weyl algebra

[elfin wraith]

Yeah I mean fair lol

[quiet pelican]

(This is the Weyl algebra, mq)

[rocky cloak]

Less complicated weyl algabra*

[elfin wraith]

(The first Weyl algebra if we’re getting pedantic)

[quiet pelican]

misremembered which example was the quantum plane

(But I still think Weyl algebra is more “natural”)

[sacred wharf]

bru why people just saying words i never heard of

[elfin wraith]

Yeah quantum plane you just pick up some multiplicative factor

[quiet pelican]

It’s kinda the sort of thing where most examples will come further into algebra

[elfin wraith]

Weyl algebra you’ve got relations coming out your ass and they’re not even just multiplicative

[quiet pelican]

Hence this

[sacred wharf]

So basically the ring containing the polynomial ring of X and the polynomial ring of Y, x,y can't commute

[quiet pelican]

You’re not quantum brained enough 🙃

[elfin wraith]

Ye

[elfin wraith]

I mean it’s just calculus right

[sacred wharf]

damn idt i'm doing more algebra any time soon

[elfin wraith]

Just a ring of differential operators, it’s the product rule

[quiet pelican]

Up to a multiplicative factor, the first Weyl algebra over C is literally the algebra of the position and momentum operators

Like the 1 should be something like iħ

[elfin wraith]

Yeah but subtraction is still less nice than just multiplication

Like picking up a factor of q may as well be commutative

[sacred wharf]

why would x,y not commute ? hmm i guess you could let x,y be r,s from D_4 as an example

[quiet pelican]

Because we say so

[elfin wraith]

Because you don’t introduce any relations that say they do

[quiet pelican]

You’re not quantum brained enough 🙃

[elfin wraith]

This is what a free object is

[quiet pelican]

Both of them are fairly natural though

[alpine plank]

Has someone mentioned quaternions?

[elfin wraith]

You get fun things like twisted polynomial rings with derivations (also called Ore extensions irrc?) which are similar

[quiet pelican]

Yeah iirc

[knotty badger]

operator algebras from QM

[quiet pelican]

I already gave the basic example 🙃
But yeah

[sacred wharf]

quarternions when they meet halfnions

[alpine plank]

Typo 😭

[quiet pelican]

(And also sometimes these are just universal enveloping algebras, eg the algebra of angular momentum operators)

[elfin wraith]

Ore extensions are cool because they let you define Gröbner bases for non com rings (apparently, someone was telling me about this in the ZX calculus server, idk how that possibly works but I’m assured it does )

[white oxide]

How hard would this be to prove without f.g. assumptions (i.e. checking closure under addition and subtraction the old fashioned way)?

[quiet pelican]

Iirc there’s a computational way to do it, but it essentially comes from a minor variant of how this proof works

[white oxide]

Why the hell do we care about chain conditions

Where are they used and where are they necessary

[quiet pelican]

They’re essentially “good” finiteness conditions for rings
Like they kinda feel like a form of compactness for rings

[white oxide]

Are they often used in hypotheses in algebraic geometry

[elfin wraith]

They’re used all over the place, not just AG. Being Noetherian or Artinian is massively powerful in terms of what you can say about a ring

As mico says they’re kinda like a finiteness condition, which makes life so much easier. They’re so nice in fact I took a course last year called non commutative Noetherian rings theres just kinda that much to say about them

[white oxide]

Ah okay I guess I just have to wait and see

Is this what they mean by their proof? Let $N_0' \supset N_1' \supset \dots \supset N_{j}'$ be a composition series of $M'$ and $N_0'' \supset N_1'' \supset \dots \supset N_k''$ be a composition series of $M''$. We claim that
[\beta^{-1}(N_0'') \supset \beta^{-1}(N_1'') \supset \dots \supset \beta^{-1}(N_k'') \supset \alpha(N_0') \supset \alpha(N_1') \supset \dots \supset \alpha(N_j')] is a composition series of $M$. Now, $\alpha(N_0') \subset \beta^{-1}(N_k'')$, for $\beta \circ \alpha = 0$ (and submodules contain $0$). Suppose that we could an insert a submodule $S$. Then, depending on the placement of $S$, we could take the image of $\beta$ to obtain a larger chain of $M''$ or the preimage of $\alpha$ to obtain a larger chain of $M'$, contradicting maximality. Thus, this chain is a composition series of $M$.

[cloud walrus]

okeyokay

[formal laurel]

Basically not having the noetherian hyphothesis makes everything a lot harder to prove. And in the case of algebraic geometry most rings of interest are quotients or localizations of polynomial rings over fields which are noetherian Hilbert's theorem.

[south patrol]

Artinian rings are (perhaps tautologically) very important for some things like deformation theory – a local artinian ring with residue field k is an "infinitesimal thickening" of k

[shut gull]

which book is best among herstein topics in algebra, abstract algebra dummit & foote & artin alegbra?

[marble hinge]

I like them all! 🙂 I only read parts about group theory though. D&F is a larger book, covering more material.

[next obsidian]

Ermmm it also needs to be a k-algebra. I think

It’s kind of unclear, this is often a thing but then randomly sometimes is excluded?

I think when it’s excluded that’s an error tho… but idk

[shut gull]

ty

[marble hinge]

so far I've got the following impressions: Herstein is an older book, so it has this a little bit old-fashioned, quaint style. But it is still quite readable, Herstein is enthusiastic and sometimes provides illuminating insights along the way (not only a stream of definition/theorem/proof). Also I really like the exercises, they are mostly proof-based, relatively difficult but often reveal something important. Dummit and Foote feels quite encyclopaedic, I have a feeling that it covers everything that one might need for undergrad algebra. Some people consider it verbose, but I really like the explanations and the flow. It also doesn't feel dumbed down (despite being verbose), and I like the style. Also has tons of exrcises! Artin -- I have read less of that, one difference to other books is that he starts with linear algebra in the first chapter and often uses matrix examples for groups and in problems. Has less exercises for each section (maybe 5-10). There is a MIT course by Artin himself that uses this book as a reading text (probably the book was developed from his lecture notes, and he just started using the book as his lecture notes)

Other people noted that Herstein doesn't use group actions (D&F introduce them quite early and base some explanations on them, like looking at Cayley's theorem or conjugacy classes from groups actions' point of view). Artin uses them too but names them differently.

However, I personally don't see a problem with reading Herstein and reading about group actions separately

[shut gull]

true

thank you

[south patrol]

I mean definitely not

What if you want to lift from k to W(k)

I don't really get your point

[next obsidian]

Ok sure yeah

But I mean like uh

Wait I’m getting things backwards

Sometimes the category of Artinian local k-algebras is defined to have k map isomorphically to the residue field and sometimes not

Idk

Ghost ping

[south patrol]

Ah ye OK

No I misread what u said sorry lol

I think though like

If k has char 0 then every artinian local ring w residue field k admits a unique map from k such that composite is identity

If k has char p and is perfect the analogue holds for W(k) as the base

[next obsidian]

Ok but this is just like

Okay “just” is a hard word to say here

[south patrol]

I mean it is fairly easy yes lol

[next obsidian]

But the key is that you’re automatically equicharacteristic

And then apply theory

To lift up

[south patrol]

I think a nice way to do this is work with square-zero extensions and stare at the cotangent complex

To avoid using Cohen structure theorem or smth ott lol

[next obsidian]

How?

Lmao

I was definitely appealing to Cohen theory stuff

But also I love that theorem so

[south patrol]

Ok my embarrassing confession is that my proof goes via like simplicial commutative rings but I am sure there is an easier proof lol

I guess like use infinitesimal lifting property repeatedly

[worn trout]

Does anybody have sources discussing the algebraic properties of higher order primes?

[formal laurel]

I've used the contangent complex a lot throughtout my life and I'm too ashamed to admit that I still have Ilusie theorem's as a black box (the one that says there is some obstruction in ext^2(something something) that is zero only if you can lift some algebra).
That's like a whole gap in my knowledge that I've been too lazy to fix.

[south patrol]

That could refer to many theorems in obstruction theory aha

But yeah I think it is kinda just tricky. Though I do really like the approach to this via formal moduli problems if you know about those

[formal laurel]

I understand the main idea in the book by Hartshorne, but I really don't understand Ilusie argument. I know that somehow the naive and the simplicial contangent complex are related by some abstract stuff

[formal laurel]

can you elaborate?

[south patrol]

I guess it depends a lot on your taste, but the idea is that often these deformation problems make sense much more generally, say for simplicial commutative rings instead of commutative rings, and have nice categorical properties as such. Then often these Ext group things have conceptual interpretations

[formal laurel]

Then, it is not that explicit to construct this obstruction, right?

[south patrol]

Perhaps not lol but often makes it more straightforward to like compare obstruction classes between problems etc

I guess like, at least in my experience, obstruction classes are like often quite hard to get a hold on besides mapping obstruction classes to one another or smth

[chilly ocean]

It’s interesting to me that lots of properties of group actions can be determined purely by their action groupoid(e.g. freeness, transitivity, having a fixed point), but some others such as faithfulness seem to require more information than just the action groupoid.

In general, what properties of group actions can and can’t be determined by their action groupoids?

[delicate orchid]

is that really true that you can't spot faithfulness from the action groupoid? Like each morphism is labelled by an element of the group acting so can you not just check to see if every non-trivial element appears as a label on a morphism? Although I suppose this is slightly more structure than the standard action groupoid so I might've just answered my own question there

[chilly ocean]

To be clear, I do mean just from the action groupoid yeah, so we don’t know if say two morphisms are the same group element or not in general. Idk any counterexamples, say isomorphic action groupoids where one is faithful and one isn’t, but I’d expect you’d at least need to know whether two morphisms are the same element from the group or not

[delicate orchid]

yeah like you can tell if an action is free from the groupoid (all automorphism groups are 1) but faithful seems stranger

[chilly ocean]

Yeah I’m not sure how to go about that

I’ll try to construct a counter example(e.g. two actions, one faithful, s.t. their action groupoids are isomorphic)

[tribal moss]

How about something like S_3×S_3 acting on {1,2,3} cup {a,b,c}, such that in one case each S3 factor acts on one of the sets, and in the other there's one factor acting simultaneously on both sets, and the other never does anything?

[delicate orchid]

I think that works

[chilly ocean]

Yeah both groupoids have 2 connected components with 3 objects each and every vertex group is the product of the cyclic group of order 2 and S^3 right? Cause for example take 1, the automorphisms of 1 are (123, _) and (132, _) for _ in S^3 acting on {a, b, c}. This should be true in either case right, e.g. the action ignoring the second S^3 or not?

[tribal moss]

Yeah -- the intuition I'm building on is that each orbit creates a separate component of the action groupoid, and that component can't "see" the other orbit and there's nothing to encode how the actions on the two orbits relate to each other.

So modifying the action on just one orbit by applying a group automorphism (here, interchanging the two factors) will produce an isomorphic action groupoid.

[chilly ocean]

Yeah, that makes sense

The motivation I had for this question was mainly that I was wondering what would happen if, instead of caring about, say, group actions(on sets, or spaces, etc.) we just cared about arbitrary groupoids(or say topological groupoids for spaces) and treated them as if they were action groupoids of some action(even if they need not actually correspond to one); e.g. we can still make sense of a groupoid having a fixed point or being free or being transitive even if the groupoid doesn’t actually correspond to any group action, but in particular we can’t make sense of a groupoid being faithful. Do y’all know any other properties that are or aren’t reflected in the action groupoid that might be of interest?

[tough raven]

This is not a formal principle, but I think in general you can at best determine the orbits and the stabiliser group and size of each orbit (and the stabiliser group is only "well-defined up to conjugacy"). For example, given a connected groupoid with object set S and stabiliser subgroup H of some point s0 in S, take any group structure on S with s0 the identity (use a cyclic group for S finite and a free group for S infinite); then H \trianglelefteq H⨯S and the action groupoid of H⨯S/H is your groupoid. So you can find a transitive group action with that groupoid where the stabiliser subgroup of every point is the same.

I suppose any semidirect product S \ltimes H would work equally well...

I do feel like an action groupoid might be a good way to take a quotient of a group by a non-normal subgroup.

[chilly ocean]

interesting, thanks

[worldly charm]

im a bit lost about O(3) (real orthogonal group, so for the euclidean space)
im getting sources that O(3) is a (nontrivial) semidirect product of SO(3) and C_2
but also sources claiming that its a direct product out of those groups
chatbots seem to be on the side of semidirect product, some math exchange (like https://math.stackexchange.com/questions/4357162/direct-product-representation-of-o-3-contradicting-the-fact-tht-it-is-not-abel) but in my head i had it as direct product, as well as some lecture scripts at uni

so my question is which is correct? and how come it seems like there are contradicting answers to be found

my guess is that maybe in general its a semidirect product but for the special case of O(3) being the real orthogonal group it becomes a direct product

i was writing up nicely the proof of goursats direct product lemma (about it forming a group isomorphism between specific quotientgroups (and thus the sumdirect product being a fiberproduct) and its application on finite subgroups of O(3) classification how it then inspires schönflies notation
and well i cant use that if O(3) is not a direct product :/

[quiet pelican]

It should be both I think
Because {I, -I} commutes with SO(3)
But also I think you can get it as a non-trivial semidirect product via eg {I, diag(1, 1, -1)}

[knotty badger]

Mhm

For odd n, O(n) is iso to SO(n) x C_2 as a direct product

[quiet pelican]

(And yeah my proof generalises to any odd n)
(For evens the first case runs into the problem that -I \in SO(2n))

[knotty badger]

For n = 2, one way to see this is impossible is that O(2) is nonabelian, but SO(2) x C_2 is abelian

[quiet pelican]

In general I think looking at the centres of both groups works, but computing the centre of SO(2n) and O(2n) isn’t fun

[worldly charm]

i assume there isnt one unique group isomorphism from SO(3) x C_2 -> O(3) ?
using C_2 isom ZZ_2
till now i assumed i can just use (R,0) -> R , (R,1) -> I dot R
and for I using -idm_3
but i assume i can also use R dot I
or use a different I right? aslong as I^2 = idm_3 and I not in SO(3) ?

[rocky cloak]

An isomorphism
SO(3) x C2 -> O(3)
would need to take C2 to the center, and there -I is the unique order 2 element.

The SO(3) component can map in different ways though, so therefore uniqueness fails

[knotty badger]

Yes exactly

The centre of O(n) always has two elements

Whereas the centre of SO(n) x C_2 has either 2 or 4 elements

[knotty badger]

Uh unless n = 2 in which case you have infinitely many

[dire turret]

hi! this might be the wrong channel for me to ask this question so if so lmk

i have this very specific problem: suppose i have a map f : Z_12^12 -> Z_12^14 represented by a matrix that is a monomorphism. then 0 -> Z_12^12 -f-> Z_12^14 -?-> Z_12^2 -> 0 is a short exact sequence i believe. There can be many ? maps which i guess probably differ by a Z_12^2 automorphism, what i am looking for is an algorithm of some kind that can give a matrix representation of ? (any). does anyone know of anything or know where i should look?

[delicate orchid]

I'm 99% sure this sequence splits (unless you're working with a short exact sequence of rings in which case you're on your own) so I think your map f is the canonical projection followed by any automorphism of Z_12^2

so matrix wise it should look like a 12 by 12 matrix of 0s direct sum something in GL_2(12)

[rocky cloak]

Pretty sure you can do this in Macaulay2, and I guess similar computer algebra systems as well.

[elfin wraith]

I know magma is pretty good for those kinda things

[dire turret]

oo thank you i'll have a look

[rocky cloak]

It would be the canonical projection up to automorphism of (Z/12)^14, but which automorphism would depend on f

[delicate orchid]

ah good point

I forgot matrices can fix subspaces 🤪

[south patrol]

No they can't

[tribal moss]

Once a subspace is broken, it stays broken.

[harsh gale]

Is there some topic or other source on useful examples of groups or rings?

[formal laurel]

It depends on what you are looking for.

[harsh gale]

I want some practical application (in mathematics)
Something like Heisenberg group or semidirect products of groups, groups with complex structure but with a easy way of doing computations

[humble palm]

Quaternions

[proud vigil]

idk if this counts but permutation groups were super relevant in breaking enigma, especially conjugation group actions

[copper kestrel]

the invariant factors are the same as the prime factors in this instance correct? maybe we dont use Z_16?

oh this is the pic before i added Z_4 x Z_4

[wraith cargo]

they're all combinations of divisors

but a priori due to chinese residue theorem they'll be powers of primes

[copper kestrel]

whats the chinese residue theorem and whats a priori

[rocky cloak]

In (a) the invariant factors are all 2, in (b) they're 2, 2 and 4. Etc

In general there are two ways to make the product unique.

One is to use invariant factors, the other is to write a group as a product of cyclic groups with order powers of primes.

Since 16 is itself a power of a prime these give the same decomposition. But for example
Z/6 = Z/2 x Z/3
the invariant factor decomposition is simply Z/6, while the factorization into prime powers would be Z/2 x Z/3

[mint seal]

it's perhaps more commonly known as the Chinese remainder theorem, but in modular arithmetic remainders are also known as residues, so the alternative name is reasonable.

"A priori" is a Latin expression meaning roughly "from what is earlier (before you know other things, using only starting information)"

[copper kestrel]

so which would you say is the "correct" decomposition? i know it needs to be unique so my best guess would be the direct product with all Z_2's

[wraith cargo]

wdym "correct"?

[rocky cloak]

I don't understand the question. In this case the two decompositions are identical, so both are correct.

In general it depends what you need them for

[copper kestrel]

i see'

*!

[copper kestrel]

sorry i didnt quite know what i was asking as i thought the invariant factorization was unique so i thought only 1 decomposition of invariant factors was the "correct" one

[rocky cloak]

But your picture only talks about invariant factors, so I guess that would default to "correct"

I only brought up the other decomposition because you talked about prime factors

[wraith cargo]

so this is a good question

but it's important to understand that none of these represent the same group

an invariant factor decomposition is unique up to isomorphism

[copper kestrel]

wait youre actually so right

omg

writing this down rn

[wraith cargo]

like the point is that these are different (isomorphism classes) of groups of order 16

[copper kestrel]

yeyeyeye

chat im learning

[wraith cargo]

:)

[copper kestrel]

ty yall :] i really appreciate yalls help

[glad spoke]

is there a slick way to prove Hom_Ring(ZG, End(A)) = Hom_Grp(G, Aut(A)) or equivalently something with Hom_ZG(ZG (x)_Z A, A)?

without being explicit at some point?

like using e.g. tensor hom and yoneda or smth

[rocky cloak]

So the functor taking G to ZG is adjoint to taking unit group. So from there it follows just from Aut(A) being the unit group of End(A)

[noble nexus]

well the group ring functor is left adjoint to the unit group functor although that's basically a restating

[glad spoke]

thats something i want to avoid i guess

[noble nexus]

To prove that I believe it's not too hard

[glad spoke]

yeah

[noble nexus]

Idk why you would want to prove it another way though

[glad spoke]

idk, i was just wondering

[noble nexus]

any map from ZG to A restricts to a homomorphism on G so there's at most one extension, then showing there is at least one follows from the fact that ZG is free

[rocky cloak]

I guess the question then is how you're defining ZG?

Because if not as an adjoint I don't know how you can avoid an explicit definition

[glad spoke]

yeah, ur right, i cant avoid it

[karmic moat]

I'm reading this thing and it says that a chain complex (C*, d) is filtered if there is a filtration {F_p C_i} such that the differential preserves filtration, ie d(F_pC_i) ⊂ F_pC_{i-1}. Does this mean every filtration of C_i has to have the same indexing set?

Or can I have something like this (if each has a finite filtration)

[south patrol]

They should have the same set. A better way to phrase the definition imo is that it should just be an increasing sequence {F_p C}_p of sub-chain complexes

[karmic moat]

Okay ty

[south patrol]

Often tbh it is convenient not to even force the maps to be injective but less standard lol

[karmic moat]

Hmmm i see

[tough raven]

That's a bit difficult but how about this: HomRing = {f : f is linear and f respects multiplication} = {f: G -> A (by free module) and f respects (1 and) multiplication on the basis G (⨯) G of ZG ⨯ ZG} = {f : G -> A and f(1) = 1 and f(mul(g (⨯) h) = mul(f(g) (⨯) f(h)) <=> f(gh) = f(g) f(h)}

OK let's be more explicit

There is an adjunction between Set and Z-Mod given by free and forgetful. These are both monoidal categories.

The free functor carries (unit and) products in Set to tensors in Z-Mod (I think you can say it is monoidal) and the forgetful has a map (1 -> Forget(1), i.e., {*} -> Z and) Forget(A) ⨯ Forget(B) -> Forget(A (⨯) B) (it's probably lax or op-lax monoidal).

It follows that both functors map monoid objects to monoid objects. You can now show that the maps of monoid objects are adjoint (this is what my original argument did).

Finally the inclusion Grp -> Monoid has a right adjoint X -> Units(X), and if you compose both adjunctions you get the group algebra-group of units adjunction.

[tardy hedge]

Gawd Damn

[tough raven]

Hom_{Z-Mod-Mon}(Free(M), A) = {f in Hom_{Z-Mod}(Free(M), A) : f ∘ Free(mul_M) = mul_A ∘ (f (⨯) f) : Free(M) (⨯) Free(M) = Free(M ⨯ M) -> A} = {f in Hom_{Set}(M, Forget(A)) : mul_M ∘ f = Forget(mul_A) ∘ (Forget(A) ⨯ Forget(A) -> Forget(A (⨯) A)) ∘ (f ⨯ f)} = Hom_{Mon}(M, Forget(A))}

[tardy hedge]

How’d you do all that

[tough raven]

The notation is terrible but it's too annoying and time-consuming to make good notation

so 🤷

[tough raven]

Possibly the left adjoint can be just lax or oplax monoidal as well, but I couldn't tell.

[tardy hedge]

Its just unfamiliar math to me

[tulip otter]

given a group G such that Aut(G) is cyclic. Prove that G is abelian

any hint for this?

I am thinking about the homomorphism f:G->Inn(G)
x\mapsto i_x where i_x is the inner automorphism given by i_x(y)=xyx^{-1}. f is clearly surjective whose kernel is Z(G), the center of G. So G/Z(G)\cong Inn(G)

but Inn(G) is cyclic (since Aut(G) is cylic), hence G/Z(G) is cyclic

so now i have to show that G/Z(G) is cylic implies G is abelian ig

let me see if i can cook from here

but first is everything correct until now?

btw by Inn(G) I mean the group of inner automorphisms (conjugation maps) of G if thats not a standard notation

[formal laurel]

yes

[tulip otter]

I see tysm, now I will try to continue from there and then I will write what I come up with here

(or I ask for a hint if I am stuck)

well there exists x in G such that G/Z(G)=<xZ(G)> and
G=\sqcup x_i Z(G)=\sqcup x^k Z(G) where \sqcup denotes the disjoint union. So any element of y of G can be written in the form of y=x^k z for some natural number k and some z in Z(G)

so given 2 elements g,y in G such that g=x^k z and y=x^a z_1,
gy=x^k zx^a z_1=x^a z_1 x^k z=yg

the manipulation done is valid since the powers of x commute with each other and z,z_1 are in Z(G) so they commute with any element of G

hence G is abelian

is this correct?

[formal laurel]

Yes

[tulip otter]

ohhh nice

tysm for taking the time to check my work

is there some other way to do this that you have in my mind?

[glad osprey]

ali is cooking 🔥 (except when letting g and y be elements in G, there's a letter literally right next to g you haven't used!)

[formal laurel]

Not anything simpler. though the proof can be written a bit more clearly.

[tulip otter]

hahaha thanks. Yea h is probably sad rn 🥀

for some reason whenever i use x my mind tells me to use y and then z

[tulip otter]

I see. If you have another way and dont mind sharing it then that would be interesting (even if its more complicated).
Also can you please tell me more about how I couldve written the proof more clearly in order to improve that for the future exercises etc..

[formal laurel]

I don't think there's another way of proving it, at least if you relaxed the conditions by saying something like "aut(G) is abelian" then it is no longer true that G is abelian. Idk.
But let me give it some thought.

and also, the proof is good, i'd give it a 10/10, just be a tiny be careful with the choice of variables.

[tulip otter]

ohhh i see. I will keep that in mind

I just made a quick search on google and I didnt find an alternative proof

so maybe its a waste of your time to try and think about an alternative proof

anyways tysm for your feedback

have a great day/night!

[crystal vale]

What is meant by (a: x) ?

I got it

Any hint? How do I find such finitely generated ideal a_0 \subset a ?

I want a finitely generated ideal a_0 such that a \subset a_0 + (x)

I see

a \subset a + (x), a +(x) strictly contains a, therefore a + (x) is finitely generated, implies a + (x) can be generated by { s_i + r_ix | 1 ≤ i ≤ k}

Hence, take a_0 be the ideal generated by s_i's, s_i \in a

[alpine plank]

Similar exercise: If R is an integral domain such that every prime ideal of R is principal, show R is a PID

[twilit wraith]

not sure what to do for the infinite case

the finite case is handled by induction by the first condition

i just dont see why i cant have an ideal thats minimally generated by an infinite list of elements that dont divide each other

i guess if i have some ideal I i can always find a sequence of elements that do divide each other by the gcd condition

so pick some a1 in I then b1 in I \ (a1)

then a2 is a gcd of a1 and b1

keep doing that recursively

then (a1) is properly contained in (a2) et cetera up until some (aN)

wait i might have it

let bN be in I \ (aN)

then a(N+1) is a gcd of aN and bN so that (bN) is a subset of (a(N+1))

but since (a(N+1)) = (aN) that implies bN is in (aN), a contradiction

so I = (aN)

shit no that just implies that I is a subset of (aN)

but aN is a subset of I by construction so equality

[tough raven]

I will remark that the key point in the proof is that we can choose a set of representatives for the cosets of Z(G) (i.e., a section of the quotient map from G to G/Z(G) if you're familiar with that language) which commute with each other.

Hmm can this be analysed in general using group cohomolgy ideas 👀

Extensions are classified by H^2(G/Z(G), Z(G)) where Z(G) is a trivial G/Z(G)-module. A section is a group hom iff the cocycle is 0 and one exists iff the cohomology class is 0. A section commutes iff the cocyle is symmetric, and this doesn't depend on the cocycle (because coboundaries are symmetric, because the quotient group is abelian and the action is trivial). So central extensions of an abelian group are given by the subgroup of symmetric classes in H^2.

So this result can be formulated as: for Q cyclic, the symmetric subgroup of H^2(Q, N) for any trivial Q-module N is everything. (In other words every cocyle is symmetric.)

It would be interesting to see what happens for other groups.

[tulip otter]

right. Also from what I know a section of a morphism f:A->B is a morphism g:B->A such that gof=id where id is the identity map, ie a section of f is a left inverse of f. I assume this is what you meant by section here too since your statement makes sense if section is taken to be like this

[tulip otter]

from here onwards you lose me

(since i am not familiar with most of the words)

tho i will check the definitions of these and try to understand what i can from this

[knotty badger]

wait i thought a section was a right inverse

[tulip otter]

hmmm idk, at least i saw it defined it like this in lang.

[tulip otter]

ok from the example right below the definition and the definition of wikipedia they are adopting different definitions

since wikipedia calls the dual notion of a section to be a retraction while lang says that topologists use retraction to mean section ?

[knotty badger]

"section" for me comes from "cross-section"

like how functions X -> Y are equivalently sections of the projection X x Y -> X

[tough raven]

yeah that was for me 😅

[tough raven]

f∘g but yes

[tulip otter]

just so that i understand this, here you mean a sequence X x Y->X->Y->X x Y where the first arrow is the projection, the second is the function X->Y and the last is inclusion?

[tulip otter]

ah well i thought it was gof because of the definition by lang which i sent above

[tulip otter]

yea ik but well i want to understand this interesting stuff too lmaooo

[knotty badger]

how do you include Y into X x Y?

what i mean is that if you unravel the definition of a section of the projection X x Y -> X, you obtain exactly the notion of a morphism X -> Y

[tulip otter]

I was thinking about something like fix x_0 in X, and then include Y into X x Y by y\mapsto (x_0,y). Something like this

[rocky cloak]

Does Lang perhaps happen to have the convention that gof means "first g then f"?

[knotty badger]

that does work, but doesn't let you get a section

[tulip otter]

no, for example here foλ_j means first λ_j then f

(here λ_j is the canonical injection λ_j: A_j->A)

[tulip otter]

the thing is that i was trying to identify Y in X x Y since the section of the projection X x Y->X is a morphism X->X x Y right?

well it doesnt matter if the image of the section is Y ig

[knotty badger]

it's a morphism X -> X x Y that's a right inverse of the projection

[tulip otter]

right

[knotty badger]

it'll help if you unfold what that means