groups-rings-fields

[crystal vale]

subgroup of symmetric group

[brave remnant]

is this because of Sym($\Omega$)

[cloud walrus]

redoftwored

[late matrix]

its isomorphic to a subgroup of Sym(Z). consider the action of Z on itself

[sturdy spear]

not isomoprhic?

then how is it possible

{1,-1,i,-i} how this could be subgroup of symmetric group

[late matrix]

every group is isomorphic to a subgroup of its own symmetry group, because left multiplication is a bijection, and the left multiplication action is faithful

[sturdy spear]

wait, maybe symmetric group of itself

[late matrix]

so it follows from first isomorphism theorem

[sturdy spear]

idk what i am talking about

[sturdy spear]

ACTIONS, not again

probably i can argue about this result once i read ch1 of D&F. Idk about actions, faithful

[late matrix]

ngl this is quite intuitive if you think about it, we basically defined groups this way, exactly so that they represent “symmetries”

[brave remnant]

for {1, -1, i, -1} it is isomorphic to a subgroup of the symmetric group by taking like S_4 for example and picking a 4 cycle and generating a subgroup with that right? then that subgroup of S_4 is cyclic of order 4?

@sturdy spear

[sturdy spear]

oh i see.

[sturdy spear]

yeah it will be isomorphic but can it be subgroup of some symmetric group?

[brave remnant]

cyclic group generated by an element of order 4 in S_n is a subgroup of S_n which is a symmetric group

so it is a subgroup of a symmetric group

[sturdy spear]

oh

i think i should study this clearfuly

does this make sense?

i think i have made some mistake it should be

,, \langle cis (2\pi/n)\rangle

[cloud walrus]

Abstract Afzal

[sturdy spear]

since any k will work so taking k=1 is better

[crystal vale]

it is subgroup of S4

isomorphic to subgroup of S4

[sturdy spear]

Yes

[frigid epoch]

Is cis = cos + i sin

Why not just write exp

[sturdy spear]

yes

[sturdy spear]

ah yeah it would be better, Author didn't use it so didn't i

[candid patrol]

Every group G, whether finite or infinite, is isomorphic to a subgroup of the symmetric group
S(G). In the case where G is finite, it is interesting to ask what is the smallest integer m such that
G is isomorphic to a subgroup of Sm

Try it with V4 and A5 👀

[sturdy spear]

is it possible? (idk about A5)
S2 has 2!=2 elements and S3 has 3!=6 elements, so there is no integer m for which Sm has 4 elements

[candid patrol]

Yeah

[south patrol]

iirc this is some funny thing for people who don't want to mention euler's formula lol

[candid patrol]

And for Z/pZ with p prime ?

[sturdy spear]

again same issue?
n! is not prime for all n \neq 2 ( Z/2Z \iso S2)

[candid patrol]

And Z/p^nZ ?

[sturdy spear]

p^n is odd for all prime p>2 and n! is even for all n

it seems much harder to think how it is possible

[candid patrol]

Ok but an odd number can divides an even number nah ?

2^3 | 4!

And 4! < (2^3)!

[sturdy spear]

Ah yes

Let me find some examples about these groups (V4, Z/pZ, and Z/p^nZ)

I was thinking incorrectly

[late matrix]

mfw i get banned from Twitter for talking about complex numbers

[candid patrol]

So for Z/p^nZ ?

Is p^n the minimal index ?

[languid trellis]

By G^k=1, are we saying that the lower central series terminates, or are we saying that G^k = {g^k, g in G} = {1}? Or will it turn out that these things are the same/related

(G^{i-1}, G) is the commutator subgroup generated by G and G^{i-1}

[rocky cloak]

The lower central series should terminate at 1 yeah.

It has nothing to do with {g^k}

[languid trellis]

Thanks, and to show this I've shown that that the ith term of the derived series is a subset of the ith term of the lower central series by induction, so if the kth term of the lower central series is 1, then the kth term of the derived series must be 1 also (it can't be empty), then I applied the criterion of derived series terminates iff G is solvable.

[empty perch]

Hey folks, just a quick reminder that the Abstract Algebra Lecture series is meeting regularly again ^_^
We will meet to discuss the Spectral Theorem (new and improved!) in about 10 minutes over in the ⁠⁠#1055201711679082516. More information can be found in our thread: ⁠⁠#1317307081535000606

[errant portal]

Could someone tell me if I did this correctly or incorrectly? I need to prove that H is a subgroup of G where G =<F(R),+>, H ={f∈F(R) : if is periodic of period π}
Here is what I did :
(i) Suppose f,g∈H; then f(x) = f(x+πk) and g(x) = g(x+πk) for any k∈Z, so f+g = f+g. This f+g∈H
(ii) If f∈H, then f(x) = f(x+πk). Thus, -f = -f(x) = -[f(x)] = -[f(x+πk), so -f∈H.

[toxic zephyr]

okay am i just stupid or isn't that what the RCF is already?? like is this supposed to be a "hey, whoa i betcha didn't expect the answer to BE the RCF!!!!" or am i missing a bigger subtlety here

or am i fucking up the definition of the RCF and in general its just a direct sum of "acts by mult by x" matrices

the way it says "find the RCF" makes it seem like it shouldn't be a direct sum, because otherwise how would we know how it factors

ex. k=Q and f(x)=(x-2)(x-3). isnt the RCF just diag(2,3)?

[muted fog]

I'm not sure to understand your question are you able to find the matrix associated to psi in the standard basis in the first place ?

[toxic zephyr]

yeah that's ezpz

and its in the form of the RCF

my questions is more like... are we SURE that that's the actual RCF of the matrix

arent there cases where the RCF would be a direct sum of matrices of that form

[muted fog]

So you're saying for this example that matrix of psi in the standard basis is the diagonal matrix diag(2 ,3) ?

[toxic zephyr]

no def not

but the matrix for that case would be similar to diag(2,3)

so like... wouldnt that be the RCF?

[muted fog]

In that case yes, because the rational Canonical Form is just a decomposition of your space into minimal cyclic subspace. If the matrix is diagonalizable then any eigen vector is a cyclic space of dimension 1.

[toxic zephyr]

okay yeah right. so then the matrix for psi isnt necessarly the RCF? if f is irreducible, then it should be right?

or... idk im not sure

like i think the point of the exercise is for psi to be the RCF, but it seems like that's not always the case

my friend emailed the prof and asked if f needed to be irred and the prof said this

[acoustic igloo]

a field with characteristic 0 contains the rationals, right?

because you can add unity n times and take its inverse
(1+1+...+1)^-1
and it acts like 1/n

but you need characteristic 0 to be able to take the inverse

[south patrol]

Yes. Any ring R admits a unique map Z -> R, and if R is a field of characteristic 0 then all the positive integers are invertible so it extends uniquely to a map Q -> R

[south patrol]

indeed people would usually just write n and 1/n

[toxic zephyr]

wait a min my prof just said diag(2,3) is not in rational canonical form... am i an idiot

[rocky cloak]

If "rational canonical form" means decomposing into the minimal number of cyclic subspaces according to invariant factors, then it's rational canonical form would be
[0, -6; 1, 5]

[acoustic igloo]

thanks 🙏

Do indeterminates "commute" in polynomials R[x,y] even if R is not commutative?

because they want me to rewrite an element of (Q[x])[y] as an element of (Q[y])[x]

and i think i need to swap x and y

[wraith cargo]

everything here is commutative

[acoustic igloo]

oh yeah

but the book says it should work for any ring

but you're right i probably don't have to worry about it since the problem is just using Q

[rocky cloak]

This is true irregardless of whether x and y commutes.

But probably you're supposed to assume they do

[toxic zephyr]

in this case, is it because the invariant factors are just f(x)?

because it's the minimal polynomial?

[rocky cloak]

The invariant factor is (x-2)(x-3) yeah.

I wouldn't say it's "because" it's the minimal polynomial, but it also is the minimal polynomial

[south patrol]

Yes

The noncommutative version is often written R<x,y>

[muted fog]

My apologies @toxic zephyr I got the definition of rational canonical form mixed up. So the problem just follow from observing that the matrix associated to psy in the standard basis is precisely the companion matrix of f. Thus the cyclic subspace associated to 1 in k[x]/f is the whole space.

[acoustic igloo]

how should i show something is a subring?

• closed under addition and multiplication
• contains the additive identity
• contains additive inverses

like that?

thanks 🙏

[vagrant zinc]

Is this enough to prove the first part?

[crystal vale]

to find the all homomorphism from S3 to S3, is it fine if i just choose the image of (12) is a and (123) is b such that 2 divides | a | and 3 divides | b |, and follows the generator relations

i don't know what does it mean by follow the generator relation, yes (12)(123) = (123)^-1(12) is a relation here

so does i need to take a and b such that ab = b^-1a?

if yes then how can i gurantee that given mapping is homomorphism?

[unreal fern]

Does anyone have a problem set book for ring theory? I’m finding it hard to nail down concepts simply through reading

[thorn jay]

Do you know about group presentations?

[crystal vale]

Yes

[velvet hull]

Well, outside of the orders the generators will still be related by the internal structure of the group

[crystal vale]

If F1 and F2 is a field then there are only 4 ideals in F1 × F2, right?

[velvet hull]

So if a = (12), b = (123), then a,b will have the nontrivial relationship that baba=1, for example

And if you choose image of the generators randomly, the image may or may not actually satisfy those nontrivial relationships, even though they have to

So that’s what it’s talking about by generator relations

[crystal vale]

Yes

[crystal vale]

Yes

[thorn jay]

A presentation of S_3 is
<a, b | a^2 = b^3 = abab = 1 >

So you only need to determine every pair of elements A, B in S_3 such that A^2 = B^3 = ABAB = 1

And that uniquely determines the homomorphism, if I'm not wrong

For example, A = B = 1 works, this gives you the trivial homomorphism

[crystal vale]

Yes

[crystal vale]

Is it correct?

The idea is if S is an idea in F1 × F2 then the set of x such that (x,y) in S makes an ideal in F1, right?

[thorn jay]

I kind of doubt its that nice

Maybe if the fields have different characteristic

But for example, consider the ideal ( <1, 2> ) of QxQ

[crystal vale]

the it is Q \times Q

because if you multiply by (1,1/2) you will get (1,1)

[thorn jay]

Ah right I was thinking of the Q-module lol

Yeah you're right

If there is any element in the ideal that doesnt have a 0 in either entry then you can normalise it to (1, 1)

[crystal vale]

yes

[thorn jay]

Nice

[crystal vale]

but can i ask how it is enough to verify these condition what about to verify f(ab) = f(a)f(b) ?

[candid patrol]

But he is not looking for the automorphisms

Let f € Hom(S3,S3), then what can be ker f ?

[thorn jay]

I know?

As i said, taking A=1, B=1 gives the mapping x -> 1

Thats not an automorphism

[thorn jay]

That follows from the universal mapping property of the free group

Determining h(a) and h(b) gives a unique homomorphism
v : F(2) -> S_3
such that v(a) = h(a), v(b) = h(b)
Then S_3 ≈ F(2) / < a^2, b^3, abab >
(< > denotes the normal subgroup closure)
So we only need that
v(a)^2 = v(b)^3 = v(a)v(b)v(a)v(b) = 1
For v to factor through that quotient, in other words, give a homomorphism from S_3 to S_3

So this amounts to choosing A, B in S_3 such that A^2 = B^3 = ABAB = 1

[crystal vale]

I see

Thanks @thorn jay

[thorn jay]

You're welcome!

Group presentations are very goated 🔥

[sturdy spear]

Ah yeah, D&F mentioned that it isn't easy to conclude results from presentation of a group. Moreover, there are techniques (i think he said this) but they will be introduced in upcoming sections

[thorn jay]

Theyre great for problems like this, if the generating set and relations are known

It makes finding homomorphism an algorithmic task basically

And theyre great for high-level proofs

[thorn jay]

We're lucky that S_n is well-studied, of course

And S_3 is even dihedral

[rancid elbow]

Need help with part b here, I’m so stuck

Written out attempt so far

[sturdy spear]

I see

I haven't encountered any such thing yet. But did some presentation of dihedral group problems

[languid trellis]

How are we concluding that Aut(U) is isomorphic to Z/(n)? Also the "cyclotomic field of order n over F" is the splitting field of x^n-1.

wait i am being so silly, this is obvious

just needed to think a bit about what an automorphism of a cyclic group looks like

[kind temple]

we are not concluding that Aut(U) is isomorphic to Z/(n). it is isomorphic to the group of units of Z/(n)

[languid trellis]

yes I agree

thanks

[thorn jay]

That may just be because of the fact that youre giving an explicit quotient of a free group

So you can use the universal mapping property (for any function f : X -> G there is a unique group homomorphism v : F(X) -> G such that v(x) = f(x) )

[candid patrol]

Every sigma in An is an even product of transpositions, and the product of two transpositions is a 3-cycle by a), then An is generating by thé 3-cycles

[crystal vale]

How can I show if there is a map S_n to Z/2Z and if any transposition maps to 0 then mapping is trivial, I don't want to use here the concept of A_n and first isomorphism theorem of group?

[thorn jay]

S_n is generated by the transpositions

[crystal vale]

Yeah that was my mistake here

I mean if there is a transposition which maps to 1 then it is trivial

Not anh

[frigid epoch]

I didn't know this result

[crystal vale]

It is trivial by first isomorphism theorem

But I don't want to use that

[crystal vale]

Which result?

[thorn jay]

This

I assume

But yeah, kernel has to of size (n!)/2, which can only be A_n. But if any single transposition gets mapped to 1, then the kernel has to be at least the subgroup generated by A_n and the transposition, which is S_n

[crystal vale]

Yes but it uses the fact the first isomorphism theorem

[rocky cloak]

Consider realizing Sn as permutation matrices and taking the determinant

[crystal vale]

I don't want to use the concept here A_n

[thorn jay]

I know

[crystal vale]

How?

[rocky cloak]

Sn can permute the n basis vectors of C^n

[crystal vale]

Yes

[thorn jay]

Given that you're working with transpositions you'll probably just reinvent the concept of A_n

Lmao

[crystal vale]

Yes but my professor says that we don't want to consider the first isomorphism theorem

[frigid epoch]

But I guess it makes sense - any two transpositions are conjugate

[rocky cloak]

But all transportations are conjugate

[frigid epoch]

Let $f, g \in S_n$ be transpositions, conjugated by $h$ (i.e. $hgh^{-1}=f$). What happens when $g \mapsto 0$?

[cloud walrus]

PKThoron

[rocky cloak]

I thought you wanted a map from Sn to Z/2, but this has nothing to do with Z/2 really

[crystal vale]

Yes

Then all transposition maps to 1

[crystal vale]

Then f also maps to 0

[crystal vale]

Why?

[rocky cloak]

Because "if a map from Sn maps transpositions to 0 it maps everything to 0" has nothing to do with Z/2

[crystal vale]

Actually I misinterpreted this

I mean if f is a group homomorphism S_n -> Z/2Z such that there is one transposition such that it maps to 0 then all transposition maps to 0

[rocky cloak]

Okay, but then bringing in An or the first isomorphism theorem or anything you know about maps Sn -> Z/2 is just a big distraction

Because it has nothing to do with Z/2

[crystal vale]

Why?

[rocky cloak]

Look here for example

[crystal vale]

Yes but Z/2Z is abelian so image of conjuagte is exactly the image of that element

[thorn jay]

Doesnt have anything to do with Z/2Z being abelian

[candid patrol]

Just Check if (a b) —> 1 is a homomorphism

You have to use that Sn is generating by the transpositions in fact, Hope your professor allows it !

If it works, Then Hom(Sn,Z/2Z) ~ Z/2Z

[frigid epoch]

I mean Z^2 is generated by 2 things but that doesn't mean that if one maps to 0, so does the other

[candid patrol]

Z^2 ?

Are you talking about the same exercice ?

[frigid epoch]

No, just that I don't get the generator argument

[tough raven]

Any two transpositions are conjugate, so the image of all transpositions are the same under any homomorphism from S_n to an abelian group.

[thorn jay]

That is true

[tough raven]

OK, two people already said that 😅

[thorn jay]

But in this case you can make a much stronger statement

If you have any homomorphism h : S_n -> G sending any transposition to 1, then h must be trivial

[frigid epoch]

Wait, why for any group? 😮

[tough raven]

Conjugates map to conjugates, and a conjugate of 1 is 1.

[frigid epoch]

Oh duhhhh

[candid patrol]

yeah but (1,0) isn't conjugate to (0,1)

[frigid epoch]

hgh^-1 goes to j*1*j^-1 = 1

Then that's settled! (For me)

[thorn jay]

It can probably also be derived from the coxeter relations

[frigid epoch]

@crystal vale are you convinced

[thorn jay]

(If im not mistaken that every S_n is a coxeter group)

[candid patrol]

If G is abelian, then Hom(Sn,G) ~ Z/2Z

[thorn jay]

No right

Take G = (Z/2Z)^k

[frigid epoch]

Yall are smahties

[thorn jay]

Hom(S_n, G) ≈ Hom(S_n/A_n, G) ≈ Hom_Z(Z_2, G)

[candid patrol]

Fact I'm dumb

[thorn jay]

Haha its alright

[candid patrol]

Hom(Sn,G) ~ Z/2Z if 2 | |G|

else it's 0

[thorn jay]

Again, take G = (Z/2Z)^k

That means that Hom(Z/2Z, G) is at least of size k

Due to the k injections

[candid patrol]

I see

😢

[crystal vale]

Yes I got the point, thanks

[thorn jay]

Me when Hom is hard to compute in general

[candid patrol]

Wait

Hom(H,G) ~ H if H and G are abelians nah ?

[crystal vale]

I have no idea from where I have to start? Should I assume i) is not true and showing ii) is true?

[thorn jay]

No?

[frigid epoch]

Pick any two abelian groups

[thorn jay]

Take H = Z/2Z, G = Z/3Z

[frigid epoch]

Z/nZ, Z, Q, R, R^n

[thorn jay]

Hom(H, G) = 1

[frigid epoch]

Wait R is awkward

[thorn jay]

It is

Horrible group, really

[frigid epoch]

Hom(Q, Q^n)

[candid patrol]

Ok ok mb

[frigid epoch]

This oughta be Q^n cause you can send 1 anywhere

[thorn jay]

Q as group is also not very nice to work with i think

Hom(Z, G) ≈ G

Because Z is the free abelian group on one generator

[frigid epoch]

I enjoy how the injective hull of an abelian group is some product over maps to Q/Z

Just like the Stone-Čech compactification is some product over maps to I

That's just cool

[thorn jay]

Often times some closure is fairly nice

For example, the closure into a variety is always some subalgebra of a direct product of the generators of that variety

[thorn jay]

In fact, that is why every reduced finitely generated k-algebra is an algebra k[X] for some affine variety X

[frigid epoch]

"the largest [compact space containing a dense image] of it"

And just like SČC, the injective hull needs choice

As does the projective cover I think

[thorn jay]

Hehehe this does not >:3c
objectively better then

You can compare it to reducing a ring by taking the nilradical

Or abelianising a group by dividing by the commutator subgroup

Etc

(In fact, those are all the same procedure)

[rocky cloak]

Projective covers usually don't exist.

But I guess in most situation where they do it reduces to picking a basis of a vector space.

[frigid epoch]

Oh

[frigid epoch]

Whoa the fact that every module has a flat cover was only proven in 2001

[crystal vale]

https://math.stackexchange.com/questions/2792165/let-a-be-a-p-cycle-in-s-p-and-let-b-be-a-transposition-in-s-p-show

I don't get it

[quartz escarp]

Hi, how would one go about showing something like this?
$L/K$ field extension, $S\subseteq L$ and $\sigma,\tau:L\rightarrow L'$ homomorphisms.
If $\sigma|K=\tau|K$ and $\sigma(s)=\tau(s):\forall s\in S$
then it follows that $\sigma|{K(S)}=\tau|{K(S)}$

[cloud walrus]

bergi

[crystal vale]

So without loss of generality we can assume that given transposition is (1 k) and p-cycle is g = ( 1 2 3 ... p) so by using the fact that p is prime g^(k-1) is p-cycle such that ( 1 k ...) so now it is standard such that ( 1 k) , ( 1 k ...) generates S_p, right?

[chilly ocean]

refer conrad notes

[chilly ocean]

there are i think two cases i know if (|K||,|G/N|)=1 THEN (I) HOLDS

[cloud solar]

If A is a ring and for every x in A there is m,n coprime such that x^m and x^n are in Z(A) then A is commutative

Cant we just use Bezout Identity?

We can use Bezout i guess only if we know x invertible

[tawdry venture]

This is my attempt so far:

If f - g is divisible by x - y, then I should be able to write it as a(x - y) Q(x, y) where a is a constant and Q(x, y) is some arbitrary polynomial (quotient) in terms of x and y

Oh this is the hint my professor gave:

[rancid elbow]

Update. So I did what was suggested - but I don’t know how to make my proof for part b by induction. I’ve showed my non induction proof and my beginnings of an induction proof

[tardy hedge]

in a field of char p when do you have a^p = a

Idk why im asking that I just kept on thinking that was a thing but idk why

oh in the field of p elements you have a^p = a for all a in Fp

so for any field of char p the prime subfield has a^p = a

[sturdy spear]

i see, damn its interesting

[minor fulcrum]

If H is a sylow p-group of G and g ∈ G ∖ H is some other element is gHg^{-1} a sylow p-group of G also?

[crystal vale]

yes because gHg^-1 is a subgroup and it has order |H|

and it works all g in G

[minor fulcrum]

Right yeah

[drifting mauve]

Hey y'all, I've been stuck on this question for a little while. Hoping someone could give me a hand:

Let $f: G \rightarrow G'$ be a group homomorphism and $S \subseteq G$ st $<f(S)> = G' $ and $T = {a | <a> = ker(f)}$. Prove that $<S \cup T> = G$

[cloud walrus]

theaveragejoe6029

[fierce breach]

All elements not in the kernel can be obtained as a product of something in S and something in T.

[fierce breach]

I mean, from T you can get all of the kernel I believe.

And from S you can get at least one instance of every non identity value.

[velvet hull]

the problem is a slight reformulation of the first isomorphism theorem.
Hint 1: ||If f: G -> G' is a group homomorphism, then if a,b are such that f(a) = f(b) then a,b can only differ up to an element in the kernel of f||
Hint 2: ||Since f is defined on the entirety of G, then if you want to understand how any element in G looks like under f, then you only need a "representative" set that covers everything that f can become||

[drifting mauve]

I’ll be honest y’all, I’m still lost.

[crystal vale]

Is there any easy way to show the centre of M_n(R) is exactly the scalar matrices, I know they are using E_ij matrices but I don't get it much

[velvet hull]

so let's say that I have ker(f), and I have a g in G.
Suppose that I have some other g' in G, where I also know that f(g) = f(g').
Then, it follows that g' is in fact generated by ker(f) and g, because g, g' can only differ up to an element in the kernel

[drifting mauve]

How do we know that g,g’ can only differ up to an element in the kernel? Are you suggesting that for some a in ker(f), gc=g’?

[velvet hull]

suppose f(g) = f(h).
Then f(gh^-1) = f(g)f(h^-1) = f(g)f(h)^-1 = 1
i.e. gh^-1 in ker f

this is an alternative way of viewing the first isomorphism theorem, if you know about that yet

[drifting mauve]

Yeah yeah, nah I get that ofc

But how does that imply that they differ by one element in the kernel tho?

[velvet hull]

that element is just gh^-1 ...?

[drifting mauve]

Oh wait my bad.

[velvet hull]

either you understand that or you don't, it's kind of confusing for me for you understand one but not the other

[drifting mauve]

Okay, I see now. I just misunderstood what you were trying to say.

Ah, I think I’ve got it. It was just that one little bit about generating the kernel that I needed. Thanks bro.

[rocky cloak]

I mean, it's hard to imagine an easier method than
compute
Eij M and M Eij
and see what needs to be true for them to be equal

[crystal vale]

Eij M gives the ith row is jth row of M and all other entries 0 and M Eij gives jth column is ith column of A and all other entries 0

[rocky cloak]

For example
Eii M kills everything but the ith row and M Eii kills everything but the ith column.

They only overlap at the diagonal, so M must be a diagonal matrix

[crystal vale]

I have to visualise now

Yes it will be diagonal

Now I have to show every diagonal entries are equal

Now E12 M = M E12 gives M11 = M22

I got it

Thanks Jagr

If I have to find all group homomorphism from Z to Z/nZ then there are n homomorphism right? Because it is uniquely determined by image of 1

[velvet hull]

more generally, there is a (natrual) bijection between the elements of any group G and the set Hom(Z,G)

[crystal vale]

Yes, g ->f_g, where f_g is in Hom(Z,G) such that f(1) = g

[gentle crow]

https://web.williams.edu/Mathematics/lg5/ArnoldQuintic.pdf

neat!

[thorn jay]

I love my free object on one generator

[velvet hull]

lmao reminds me of the time someone here reminded me that Z[x] is generated by 1 element and I went "oohh"

[thorn jay]

Lmao

There's also a natural bijection between R and Hom(Z[x], R) (if R is commutative)

[velvet hull]

forgetful functors are representable yeah

[thorn jay]

And, hear me out, a natural bijection between Hom({*}, X) and X

[velvet hull]

actually what's the nec & sufficient condition needed for the forgetful functor to be representable

a good question, that I will look into another time

[thorn jay]

No clue lmao

[thorn jay]

But the bijection A^n with Hom(F_V(n), A) is really important imo, as it motivates and simply proves (at least if you define everything slightly differently) the prime spectrum and Nullstellensatz lol

[velvet hull]

oh, mind explaining a little more?

[rocky cloak]

The Hom-functor preserves all limits, so that's a necessary condition at least.

And in situations where the adjoint functor theorem applies that should be sufficient as well

[thorn jay]

I'll take the case of k-algebras for an algebraically closed field k.
Traditionally, affine varieties are defined as certain subsets of points, which we can see as k-algebra homomorphisms h : k[t1, ..., tn] -> k
It's often more easy to reason about ideals (congruences) than the actual homomorphisms, so what the idea is that we "approximate" these points using their kernels, call that obtained set of ideals S for now. Similarly, we define closed sets on S, for an affine variety X as the set of kernels of the points in X (again, points are regarded as homomorphisms).
It turns out that due to the niceness of rings, this gives you a topological space which is isomorphic to the Zariski topology as locales.

Then you can say something about the possible kernels of homomorphisms from k[t1, ..., tn] to k, given that k is an algebraically closed field, and it turns out that S is precisely the prime spectrum of k[t1, ..., tn]

For the Nullstellensatz, this is a more general phenomenon relating to similarly defined "prime" spectra, something I am researching rn as a small project, but it boils down to essentially just some fairly trivial order theory. The only thing about rings is simply that prime and radical ideals are so easily describable.

Hope this is somewhat followable haha

[velvet hull]

oh I know the statement of Nullstellensatz, I just wanna know what you meant by that

what's the bijection here specifically? whats the ambient category

what is V(n) as well

[thorn jay]

Ah, yeah

V is a variety, also called equational class, of algebraic structures

F_V(n) is the free algebra on n generators

The ambient category is V, the category of T-algebras for a finitary algebraic theory T, essentially

Hom(F_V(n), A) can fairly trivially be given a structure such that it too is a member of V

(As it then is, in fact, isomorphic to A^n)

[thorn jay]

You can think of it like the evaluation of a polynomial

Like how you have a word for groups
xyzxy^2
Which you can evaluate in a group by choosing x, y, z as elements in the group

Every homomorphism from the free algebra on n generators to an algebra A must correspond 1-1 to some choice of n elements of A, so that gives you your bijection

[velvet hull]

but then, if we're talking about the forgetful functors being representable shouldn't you say that the bijection is between A^n and Hom(A, A^n)? I can accept that it's equivalent, but i don't understand why you're presenting it in that form

[thorn jay]

Hom(A, A^n) is not equivalent to Hom(F_V(n), A)

Like for example, if V is the variety of sets, then one has cardinality |A|^{n * |A|}, and the other |A|^n, as the free set on n generators is just a set with n elements

[velvet hull]

oh, wait I think I understand what you're saying now. I think I got confused from the context of the previous conversation

F_V(n) should be isomorphic to A^n as well, but I'm assuming that this isomorphism is not natrual

[thorn jay]

Ah its alright haha, I maybe should've been more clear too

[thorn jay]

Nope

Again, take the variety of sets

[velvet hull]

oh, iso as vector spaces but not as algebras?

[thorn jay]

Nono

[thorn jay]

.

Algebra in that sense

Thats why this viewpoint is so powerful as it generalised (classical, for now) algebraic geometry to any algebraic structure

[barren sierra]

Non-zero-divisors of a ring are called regular

Where does regular come from? Like what is the origin?

[rocky cloak]

My guess would be that it comes from regular sequence which comes from regular ring and

The origin of the term regular ring lies in the fact that an affine variety is nonsingular (that is every point is regular) if and only if its ring of regular functions is regular.

[barren sierra]

but then why are those called regular points or regular functions / rings 💀

[rocky cloak]

I would guess to contrast from singularities.

Like singularities are special, an uncommon in contrast with the regular points

[minor fulcrum]

sometimes we just needed an adjective

If I'm given two group presentations and told that they're both finite, can I decide whether they're the same group

[thorn jay]

That seems very related to the word problem

[barren sierra]

if they're finite groups, this is decidable

if they're finitely presented then you can show that this problem is undecidable (in fact the word problem is undecidable)

@minor fulcrum

I think Rotman's group theory text has some stuff about this

[minor fulcrum]

I did in fact mean finite groups

[barren sierra]

yea I know what you meant, I just was adding on that a slight modification quickly turns undecidable

the brute force obvious algorithm is to write out the multiplication table for both groups

and then start trying bijections

more efficient algos surely exist but idk them off the top of my head

but it is at least decidable

[tough raven]

Well, it's only given that they're finite, you're not told the size.

[barren sierra]

Sure, but a reasonable definition of "given" is that you are given a way to enumerate all the elements and how many elements there are

[tough raven]

The question could be for an algorithm that returns the correct answer when given presentations of finite groups but has arbitrary behaviour (correct, wrong or non-halting) when one of the groups isn't finite.

That's how I usually interpret "given an assumption" for an algorithm.

[barren sierra]

yea, so "arbitrary behavior" means that I don't care what happens for non-finite groups

statement is vacuously true, not even going to think about it

[tough raven]

Yes, but in my formulation, no-one gives the algorithm the size of the group.

[barren sierra]

1000% Derek Holt's text would be a good reference as well

[tough raven]

Or more precisely, a finite list of words such that every element is equal to one of them. Once such a list is obtained for both groups, I agree that your algorithm works.

[barren sierra]

I think it's reasonable to assume you're given such a list

[tough raven]

Hmm, you actually also need to be given the multiplication table, because there might not be a way to reduce the products to something in the list.

But yeah, you can assume that. I'm unsure if the original asker did though.

[thorn jay]

How would you do it without brute force given a multiplication table)

?

[rocky cloak]

I think it's a pretty unreasonable assumption.

What would even be the point of having a presentation if you're just given everything about the group in some unrelated way

[craggy summit]

I am writing a paper for an Algebra II class right now, and am having a hard time figuring out how to show that there are multiplicative inverses for $\sum^{\infty}{i=0}{a{i}p^{i}}$ I have shown everything else is true therefore I have established this as a ring. Do any of you have ideas on how to approach this?
(also, how do you type an equation here?)

[minor fulcrum]

no, you're just given the presentation

[thorn jay]

$ around and equation

[cloud walrus]

Fearne

[thorn jay]

What are a_i and p?

[rocky cloak]

The inverse only exists if a0 is nonzero.

Anyway, what I would do is set up the equation for an inverse (write another element times it = 1) and show by induction that the equation has a solution

[craggy summit]

Shoot, they are elements of the integers where p is a prime

[rocky cloak]

(I'm assuming you're talking about the p-adic integers)

[craggy summit]

yeah

[craggy summit]

Thank you so much, that helped a ton ^w^

[untold torrent]

I read about metric space but i forget it almost

What is the best way to memorize things for long time?

[velvet hull]

it says maximal but not principal

not prime, different thing

it should be a prime ideal iff your base ring is an int domain

[elfin prairie]

can someone motivate the idea of Group Actions in an accessible manner?

[velvet hull]

groups are a way of encoding symmetry in nature.
However, often times even if an object is symmetric, it's hard to concretely describe how that symmetry interacts with that object in actuality.

so, that is the goal of group actions, to allow for things that embody a certain symmetry to interact with said symmetry

for an example, the rubik's cube is very symmetric, and the set of all permutations form a group.
On the group side of things, the law of composition is function composition.
However, physically, it doesn't make sense to "compose" two scrambles together; so instead we need a way of describing how the law of composition works for the rubiks cube:

object - rubiks cube (this is just a set of 26 smaller cubes arranged in space, there is no way of "composing" scrambles together)
group - group of permutations
action - face rotations (which tells us how the symmetry is encoded in the blocks of the cube)

[chilly ocean]

Why p+(x), p+(y) are ideals

[velvet hull]

the sum of ideals is an ideal, you can check that quickly

[chilly ocean]

Why (x)(y)=(xy)

[thorn jay]

Is your ring commutative?

[chilly ocean]

@thorn jay yes

[thorn jay]

Then (x) = { r • x | r in R }

And
(x)(y) = { a • b | a in (x), b in (y) }

So you can see that a = r • x and b = s • y for some r, s in R

And then, if you multiply them, your ring being commutative:
ab = rs • xy is in (xy)

So (x)(y) is contained in (xy)

The converse is just choosing the right value of a and b

[barren sierra]

A good exercise would be to generalize this

[cloud walrus]

Spamakin🎷

Spamakin🎷

[barren sierra]

so you can find a counterexample for the other direction for a noncommutative ring

[thorn jay]

"Finitely generated"
Life is a prison

[wraith cargo]

Why even live if your rings aren't noetheiran

[thorn jay]

So true

Imagine not caring about the fact that your rings are not finitely generated as rings though

[delicate orchid]

imagine caring about rings

[thorn jay]

Imagine caring

[hidden wind]

https://tenor.com/view/moti-hearts-gif-8240660592853947517

[delicate orchid]

imagine

[hidden wind]

hiii i’m here to care

[barren sierra]

is this even true for rings with infinitely many generators?

probably is

but whatever, all rings worth thinking about are Noetherian

[thorn jay]

No

[delicate orchid]

_ _

[barren sierra]

oh wait yea it should be true for infinitely many generators, same arguments go through cause all sums in question are finite

[thorn jay]

Yeah

[barren sierra]

my point still stands

[thorn jay]

I suppose that the class of commutative rings is generated by Noetherian rings

[hidden wind]

i have faith that the ring-flowers will bloom

[thorn jay]

Every Z[x_1, ..., x_n] is Noetherian, by induction, so take the direct limit of the ascending filtration
Z[x1] -> Z[x1, x2] -> Z[x1, x2, x3] -> ...
To get the ring Z[X] where X is countably infinite, which does indeed generate the class of commutative rings
Yay

[thorn jay]

I cant wait until the flowers start to bloom ngl

[tardy hedge]

Real life picture in the math discord 😧

[hidden wind]

i touch grass

[tardy hedge]

Bloom by Radiohead. Good song 😊

[tardy hedge]

😵‍💫🤯

[thorn jay]

Radiohead is great

Dont listen to them often though

[tardy hedge]

Im a huge fan lol

I love everything thom yorke has put out

[thorn jay]

Personally more into shoegaze and all it's subgenres not gonna lie

[tardy hedge]

Do you know Deafheaven? They’re black metal and shoegaze mix

If you dont like heavier stuff you wont like it but I love their album Sunbather

[thorn jay]

I know them but not the biggest fan

They have a couple cool stuffs but idk, doesnt really do it for me

[tardy hedge]

Thats fair

[thorn jay]

Cant be the heaviness; I'm an avid enjoyer of Electric Wizard

[tardy hedge]

Funeralopolissss

[thorn jay]

The wizaarddd 🗣️

[tardy hedge]

Actually i should check out more of their stuff

[thorn jay]

Dooo

Great for doing math to in my experience

[barren sierra]

I should listen to more shoegaze

[thorn jay]

Yesssss

The album "Margaritas Podridas" be Margaritas Podridas is great, has a lot of punk and noise rock influences too. If you want similar stuff to that but more dream pop, check out Señior Kino, it's the same singer.

Also Whirr is great for more heavy stuff, and if you wanna go more into the grunge direction, I'd recommend Superheaven.

I'll stop here because if I wrote out all my recommendations we'd be here for a while

Ah, maybe for dreampop, Film School is great

These are just starting off points, of course, most streaming services will give you fairly good recommendations similar to them

[tardy hedge]

Interesting, I haven’t heard of any of them before

Recently I’ve been getting into aphex twin

[prisma ibex]

Groups rings and fields

[thorn jay]

We're talking about math rock i swear

[chilly ocean]

[tardy hedge]

Tbh the problem is all the server ppl i know hangout here

When i go to discussion page im like who tf are these ppl

[thorn jay]

New people

Scawy

qwq

[coral spindle]

Hell yeah

[thorn jay]

I love breakcore

Both actual and sewerslvt type

[tardy hedge]

Im hyped to move to a bigger city where more cool shows happen

I randomly stumbled into a drum n bass show once and realized i fw it

[thorn jay]

Goooodddd

Join us

[tardy hedge]

Ye i love music and going out dancing, seeing live bands etc

So fun

[thorn jay]

Are you into jump up dnb, or more dark/atmospheric stuff

[tardy hedge]

Im not sure, ive only recently started to get into edm genres

[thorn jay]

Cuz there's a big diff between the stuff i usually listen to and what's played on festivals

Haha

[tardy hedge]

I know i fw aphex twin for sure

[thorn jay]

You'll like VISION stuff

They're a record label for like experimental dnb

Founded by Noisia

Groningen represent 🔥 🔥 🔥

[tardy hedge]

I feel like the discord police will get us soon

[thorn jay]

Uhm well you see

The logo of Noisia and VISION is three circles

Circles is math

[tardy hedge]

Oh shiiiii

[distant summit]

Isn't his proof of (c) flawed?

He says "if g1 H = g2 H, then psi(g1 H)=psi(g2 H)", but this is not injectivity is it? This is trivially true of all functions

[rocky cloak]

The proof is correct.

That is indeed not injectivity, but well definedness. Just like they say

[distant summit]

Yes I just realised that

My mistake

Oopsie

[distant summit]

Please could you check my own proof of injectivity?

To show that $\psi$ is injective, we can show that $\ker{\psi}={e_K}$, where $e_K = e_1 H$, the identity element of $G_1 / H = G_1 / \ker\phi$.

Clearly $\psi(e_K)=\phi(e_1)=e_2$, so $e_K \in \ker \psi$ as claimed.

Now consider $gH$ where $g\not\in \ker\phi$. Then $\psi(gH)=\phi(g)\neq e_2$. So $g\not\in \ker\phi \iff g\not\in e_1 H=e_K$ implies $gH \not\in \ker\psi$

Thus $e_K$ is the only element of $\ker\psi$

[cloud walrus]

Douglas

[rocky cloak]

Looks good, assuming you have established this have that ker = {e} iff injective

[tardy hedge]

Im stuck on this, i know that since b is purely inseparable we have b^(p^n) is in F for some n

Im trying to show that a or b is in F(a+b)

char p fields

[rocky cloak]

Notice that (a+b)^p^n = a^p^n + b^p^n

Now ask yourself what is the minimal polynomial of a over F(a^p^n)

[charred river]

Guys I am learning maths for an exam in my country ....i came here in search of resources but couldn't find any ....I am not a math major ....I am practically a math noob but I need math to ace the exam ....so since it's about groups rings and fields ...I think this is about abstract algebra ....can some of you help me with lecture videos you have or notes something a beginner can learn to solve in abstract algebra ...I have searched youtube and everywhere can't find things that I can understand.....like I need to know the fundamental of the fundamentals so if someone knows to do it kindly guide me in the right direction it's a big thing to ask being a complete stranger but no harm in asking right .
Thank you 🙏

[rocky cloak]

Does your course follow a book or have a more specified curriculum?

[charred river]

It has a defined syllabus of the topics what I have to study books aren't mentioned

[rocky cloak]

I guess picking up an abstract algebra book would make sense then.

I think Dummit & Foote is one. Fraleigh is the one used at my uni. Otherwise I guess just look at the contents of some and see if there's one that covers your syllabus.

Or ask your professor if there is a book they would recommend.

Kinda weird to have a course with no recommended resources

[charred river]

Thats the thing it's not a course ...it's an standalone exam for job ....so it has only syllabus and no recommended books

Will start with your book suggestions as of now

[rocky cloak]

Exam for a job covering abstract algebra, colour me intrigued

[charred river]

And it's not only about abstract algebra I have to cover real complex analysis linear and modern algebra calculus differential fuilds and dynamics ...it's like I have to study undergrad in maths .....

[charred river]

It's kinda big deal of an exam in my country

[charred river]

I have mentioned only few here ....I should have a mastery over various topics a math undergrad would have

So only running clueless and seeking help where ever I can

And you where the 1st one to respond thank you for that

[tardy hedge]

ive still been stuck on this

lost in the weeds of everything rn

lol

[rocky cloak]

Well the minimal polynomial must divide
x^p^n - a^p^n
since a is a root of that.

But it must also be seperable

[tardy hedge]

the degree of the min poly has to be a multiple of p

[tardy hedge]

This poly is separable

[rocky cloak]

It is not, it's purely inseperable

[tardy hedge]

oh shoot yeah

so its x-a^(p^n)

so a^(p^n) = a so a^p^n + b^p^n = a+c = d where c in F and d in F(a+b) so a = d-c is in F(a+b)

this is weird. im going to have to think about all these concepts more

[rocky cloak]

a^p^n = a is a bit strong

But F(a) = F(a^p^n) would be right

[thorn jay]

Fake, no one cares about algebra besides algebraists

[tardy hedge]

But a solves x-a^p^n?

[rocky cloak]

No....?

[tardy hedge]

Tbh ive probably really lost it at this point

One sec

Is x-a^p^n not the min poly of a over F(a^p^n)?

[rocky cloak]

Only if a = a^p^n

a is a root of
x^p^n - a^p^n = (x - a)^p^n

So the minimal polynomial is a seperable polynomial that divides this

The only such polynomial is
x - a

[tardy hedge]

Oh, so we’re saying [F(a) : F(a^p^n)] is 1?

[rocky cloak]

Yup

[tardy hedge]

That was like . Weird

[low cliff]

hello, i need help w this question, n>=2, show that every ideal of Z/nZ is principal

[south patrol]

What can you say about subgroups of Z/nZ?

[low cliff]

their forms: dZ/nZ where d is a divisor of n

[low cliff]

i really dont know

[south patrol]

exactly

but now you are done

lol

:)

[low cliff]

howw

it's generated by d?

[south patrol]

Exactly

And every ideal is in particular a subgroup

Here in fact this means the subgroups are precisely the ideals

(This is also clear ish anyway as if x is in a subgroup H and k an element of Z/nZ, then kx is just x + ... + x (k times) and hence also in H. So H is N ideal)

[low cliff]

oh ok thank you

[wooden rain]

What does N stand for here (third bullet)? Did the author accidentally make a mistake?

[sonic coral]

natural numbers

[thorn jay]

"Number field" :(

[wooden rain]

I see, so it was a mistake. Thank you

[thorn jay]

The notation is not standard anyways

Mathcal for number sets 🥺

[thorn jay]

The first paragraph has horribly written english

Lmao

Well "horribly" might be a hyperbole

"A well-defined ODE must the following features"

[thorn jay]

Also, "mathematical expression on the right hand side"? ?? Thats just 0

[wooden rain]

Fair enough. Surprisingly, it comes from the website of a university in Bath

I'm looking for an introduction to ODEs for a mathematician, not a physicist 🥲

[thorn jay]

Personal nitpick is the term "mathematical expression" but ig im just nitpicky

[thorn jay]

Lmaoo

It does seem that way

[thorn jay]

Also, what is "Eq"

I just, aeugh, find something else

[wooden rain]

ok, it seems the pdf was actually written for engineers

[thorn jay]

That whole first bullet is a mess

[wooden rain]

I'll look for something else

[thorn jay]

Oh thats why it makes me want to vomit

Explains

[thorn jay]

(Joking)

[wooden rain]

anyway, I won't flood this channel with unrelated stuff anymore. I was just wondering if the author might've meant some exotic field

Thank you for the help 🙏

[thorn jay]

Sorry for my small rant abt it

[tardy hedge]

yea i mean what is going on there

so weird

[thorn jay]

Idk theyre all so small

But they add up, and just overall produce a very low quality work thats clearly not polished or proofread

Making it frankly not worth my time at all

[tardy hedge]

If a is separable over F, why is every element of F(a) separable over F?

F char p

Because otherwise any finite extension of char 0 field is separable extension right … because irreducible polynomials must be coprime to their derivative

[south patrol]

Unless I am mistaken, this is a bit nontrivial. The way I would do this is to phrase separability in terms of embeddings into e.g. an algebraic closure: let [K:F]_s = Hom_F(K, F bar) and you can show K/F is separable iff [K:F] = [K:F]_s, and moreover the latter satisfies a form of tower law. Then in your case, if b is an element of F(a), you can cut up as F(a, b)/F(b)/F and applying the tower laws you see that F(b)/F is separable.

But yeah iirc this does need more field theory than one might expect

Yeah googling again, this may also be one of the quickest ways lol

[tardy hedge]

Oh dang

[south patrol]

This is essentially a more Galois-theoretic-ish approach though lol

And I guess it is unsurprising as like this is a useful lemma when doing Galois theory

[tardy hedge]

Im already unsure on what that first line of math means

I was trying to do q5. It seems like something of what ur saying is related to q7 there

[south patrol]

Yeah lol I am saying the last thing

But lol I mean like

[tardy hedge]

Bruhhh

[south patrol]

Isn't q5 assuming that the separable elements form a field

Like is that smth you're expected to already kmow

Lol

[tardy hedge]

No idea

My prof puts random things on homework

[glad osprey]

I think this corollary answers your question?

[wooden rain]

It's all cool, dw

[south patrol]

Yeah this is more or less my proof lol

There is another funny approach. You can define Galois extensions as splitting fields of separable polynomials and then show that those are all separable rather explicitly lol

And then you just embed F(a) in a Galois extension

But somewhere in there you will have basically done what I said

[tardy hedge]

14.28 using separability degrees are multiplicative ?

I havent thought about that yet

That is q6 in my hw

[south patrol]

Ye my point was like separability degree is nice in terms of maps to alg closure ig but that is ur q7

Anyway I think since q5 just assumes that separable elements form a field you can just use that fact aha

But like still a good question

[toxic sapphire]

I'm really struggling to show that given a cyclic group G of order n, there is a unique subgroup of order d for each divisor d of n

I've shown existence, I'm just not sure how to prove uniqueness

[south patrol]

It is easier psychologically if you think about Z/nZ

What are the elements of order dividing d?

[toxic sapphire]

I can't figure it out

I don't know how to find which elements have order that divide d in general, I wouldn't even know where to start, what can we say about the order of elements in Z/nZ in general?

[south patrol]

So you want elements that are zero after multiplying by d, and try to phrase in terms of Z

[toxic sapphire]

yes, I have done Lagrange, let me recall it and I'll think for a moment

[south patrol]

Actually this isn't even lagrange, ignore lol

[toxic sapphire]

hmm I'm struggling to see how Lagrange could be helpful here. it tells us that the order of a subgroup divides the order of the group

[south patrol]

But yes ths

So like if b is an integer and db is 0 in Z/nZ, what can you say about b

[toxic sapphire]

wait are we focusing on additive Z/nZ for now?

[south patrol]

Yes (otherwise not a group ahyway)

But it is easier psychologically, and equivalent to any other cyclic group of order n

[toxic sapphire]

ok right I'm following now. it could only be a multiplicative group if n were prime?

[south patrol]

Well it never is - you have to chuck away 0 regardless of n

(Unless n = 1 I guess lol)

[toxic sapphire]

oh right. we would call this like the group of units modulo n

of course 0 is never coprime with n

ok sorry for the aside

[south patrol]

Np it's like probably good to check lol

[toxic sapphire]

if db = 0 then the order of b must divide d?

[south patrol]

Sure, but I was basically going from the 2nd to the first

If db = 0 mod n, what can you say about b (as an integer)

[toxic sapphire]

it must be a divisor of n also?

I'm not sure what you're getting at at all I'm sorry. maybe I don't understand something really fundamental here

[south patrol]

Well it's that if db = 0 mod n, you can write db = an right for some a

[toxic sapphire]

yes

db is a multiple of n

[south patrol]

so b = a (n/d)

So b is a multiple of n/d

That's the important thing

Now can you guess what the unique subgroup is?

In fact you presumably already know one example and this argument shows it is the only one lol

[toxic sapphire]

I don't think I do. I've actually already gotten effectively this far and didn't know what to do from here

I mean I have an idea, it's like every n/dth element or something?

ok so the group must be generated by a(n/d) for some a, right? now we should show that it's actually the same regardless of the choice of a

[south patrol]

Well for some choices of a you won't get a group of order d, e.g. if a = d. But we have shown every element of order dividing d is an element of the subgroup generated by n/d

Now check how big that subgroup is

[toxic sapphire]

the subgroup generated by n/d has order n/d right?

[acoustic igloo]

the polynomial x^2 is reducible right?

over Z or whatever

[tardy hedge]

Yea

[chilly ocean]

Why R belongs to m

[tardy hedge]

R is the intersection of all maximal ideals

So every element in R is in every maximal ideal, so its in m

[south patrol]

Not quite

[toxic sapphire]

oh wait no no, it has order d

[south patrol]

Exactly

So how do we conclude

[toxic sapphire]

have we shown that H has order d if and only if H is generated by n/d

[south patrol]

Yes, so like

Say H is a subgroup of order d. Then every element of H has order divisible by d (Lagrange - this is where it is useful lol) and hence is in the subgroup K generated by n/d

So H is contained in K

But H and K have the same cardinality

So H = K

[toxic sapphire]

ah okay that makes sense. would it be equivalent to suppose that H and H' (subgroups of G) both have order d, which implies they're both generated by n/d, so they're the same?

[south patrol]

Yes

[toxic sapphire]

😮‍💨 thank you so much for your patience. I don't think I'm usually this bad at understanding algebra but it's just not happening today I guess

[tardy hedge]

Algebra is hard

[thorn jay]

Shit happens, algebra is a little silly goober sometimes

[acoustic igloo]

and hence is in the subgroup K generated by n/d
why?

[south patrol]

Well we did that above

[acoustic igloo]

oh

[south patrol]

if dx is a multiple of n then x is a multiple of n/d

[thorn jay]

It gets easier when you make everything universal, trust me bro

[tardy hedge]

Man i thought u were srs for a sec

[south patrol]

/srs or /j

[thorn jay]

Uh, well depends on how you look at it i guess

[tardy hedge]

Idunno what that is

[thorn jay]

Once you overcome the abstraction of UA, it becomes easier to think about the other algebras

[tardy hedge]

I was talking to potato too

[tardy hedge]

Yeah that is interesting though

Cause category theory kind of is like that too isnt it

[thorn jay]

But then again, with stuff like ring theory, the questions you ask are fundamentally different

Funnily enough, classical algebraic geometry is actually at heart really close to UA, so naturally you've got universal algebraic geometry :P

Criminally underdeveloped and undergeneralised

[south patrol]

/srs is like serious

/j for joking

Lol

But here I said it as a joke

[south patrol]

Interesting, I've not heard of this

[thorn jay]

Its so cool

[tardy hedge]

Is there any research being done in UA

I wonder if any school actually has a UA research group

[thorn jay]

I made a small generalisation akin to prime spectra and you recover the construction of free algebras

[south patrol]

Nice

[tardy hedge]

Yea why do we always automatically assume UA is poop

It probably is interesting

[thorn jay]

And I want to look into something like localisation (so you can look down the lattice, rather than up using quotients) and perhaps use sheaves to get a duality between HSP(K) and affine K-schemes or whatever

[tardy hedge]

Bro said he only knows ua^

[south patrol]

I think like a lot of the original stuff for algebra seems better with things like monads and category theory etc but I'm sure UA has links to e.g. logic /model theory that are important

[thorn jay]

Category theory is more general yes

[tardy hedge]

What is a monad

[thorn jay]

UA explicitly focuses more on the logic side of algebra

[tardy hedge]

Hmm

What is the logic side?

[thorn jay]

But mixing it with category theory, like the category of equational classes to define Mal'cev conditions, direct limit constructions etc yields great results

So imo a balance is necessary

[thorn jay]

The structure of equations basically

Also all the induced lattice structures everywhere

[south patrol]

Monads can sort of be thought of giving one way to express notion of "Ds are Cs with extra structure". E.g. the groups are algebras for a monad in Set

[tardy hedge]

Bruh

[thorn jay]

Every finitary monad is an equational class

As in, universal algebra is the study of finitary monads over Set

Just like how group theory is the study of group objects in Set

[tardy hedge]

Oh cool

Wow this is pretty abstract

[south patrol]

The interest for me in monads is that basically if you can recognise a category D as algebras for a monad in C then that can tell you a lot about D lol

And C may be way simpler

[thorn jay]

Did you know the category of compact Haussdorf spaces is monadic over Set?

[tardy hedge]

Alright hold on

[south patrol]

Yeah it's great

Lol

[tardy hedge]

😂

[thorn jay]

ITS NOT FINITARY SO I WILL NOT ACCEPT IT AS ONE OF US

[south patrol]

I like the funny corollary that this means that a continuous bijection between compact Hausdorff spaces is a homeo

Very overkill and not general enough but funny

[tardy hedge]

You guys know a lot of math

I wish i had more time to think about it and not rush through everything

It was more fun as a hobby last year

[thorn jay]

This also holds just for Haussdorf spaces right?

[south patrol]

The optimal I believe is: a continuous bijection from a compact space to a Hausdorff space is a homeomorphism

[thorn jay]

I like thinking abt it

[tardy hedge]

Me too but i feel like i could think about it better when it was a hobby

Right now its all too fast for me

[south patrol]

Basically you want to show that the map is a closed map (sends closed to closed) and you do this by 1) closed subspaces of a compact space are compact 2) continuous maps send compact to compact 3) compact subsets of Hausdorff spaces are closed

[thorn jay]

Thats fair

[south patrol]

Which is a really cute proof aha

And then for compact Hausdorff spaces like compact = closed so it is more symmetric etc

[tardy hedge]

There are literally like 500 things i want to review

[thorn jay]

Compact lattice element in P(X) under inclusion?

[south patrol]

Same

And learn lol

[tardy hedge]

Yea i imagine that feeling never ends

I wonder if its all worth it type shiii

[thorn jay]

Or compact in op(X) under inclusion

op(X) being the open sets

Wait no

Ugh i cant read

I should try to sleep again

[tardy hedge]

Nooo

[thorn jay]

Its 01:42 qwq

[thorn jay]

Felt

[tardy hedge]

U know a lot

To me ppl like u jagr and boytjie are like bruh

Potato, are your peers also that knowledgeable?

[thorn jay]

I had the bright idea of wanting to learn algebraic groups before algebraic geometry
So im learning algebraic geometry
Then i started learning universal algebraic geometry
Then i needed topology
Then i wanted suddenly algebraic topology
??????
Is my brain broken?

[tardy hedge]

Hahahaha

How much algebraic topology do u know?

[thorn jay]

Vague concepts and the fundamental group

[tardy hedge]

Honestly it does seem really cool

[thorn jay]

I know the definitions but thats kinda where it ends yk

[tardy hedge]

Getting the cool pictures in topology is fun

Or visuals yk

[south patrol]

Idk lol

[thorn jay]

Imagine having to draw them in latex tho

[tardy hedge]

Algebra is like satisfying but rigid

[tardy hedge]

Lol bruh

[thorn jay]

Idk, i somehow see them as pretty pictures

[tardy hedge]

Yeah i ask cause the phd students at my school are knowledgeable for sure but i dont think like u

[south patrol]

I think I have quite broad interests but like obviously will less deep knowledge in other bits so idk

[thorn jay]

So true

[south patrol]

But thanks

Uwu

[thorn jay]

I also think youre cool and knowledgable

🔥

[tardy hedge]

Yeah talking to real algebra experts is crazy

[south patrol]

Thank youu

[tardy hedge]

My galois theory prof is like that

My algebra prof in first term not really in the same way

[south patrol]

In my mind I am not even like an algebraist rly lol idk like jagr and boytjie are more so

And wew

[tardy hedge]

What are you?

Yea jagr is so good bruh

[south patrol]

I am nothing

Idk like homotopy theory and algebraic geometry maybe

[thorn jay]

Ive got a friend who stubbornly insists that i am a logician

[tardy hedge]

Jagr will always be the 🐐 to me

[thorn jay]

Thats so awesome!!

[tardy hedge]

He got aura bro

I stg

[next obsidian]

What am I

[tardy hedge]

The cheeky bastard

[next obsidian]

Wow

Wow

Wow

Wow

https://tenor.com/view/you-wo-joker-joaqu-gif-23924191

[tardy hedge]

Was it you that had beef with that mathnetworks guy 😂

Or ngroupoid

[next obsidian]

What

I don’t think it was me….

[tardy hedge]

It was probably ngroupoid then

[next obsidian]

Unless I forgot about my beef

Did I win?

[tardy hedge]

Yeah

[thorn jay]

UA is also a more natural context to study finiteness conditions in

[next obsidian]

Knew it

[thorn jay]

In my opinion

[tardy hedge]

I mean tbf he had a point but he got too upset about it

[next obsidian]

Chmommon Chmonkey W

[thorn jay]

(Finitely generated things being compact elements of the lattice)

[tardy hedge]

Ong i been living on this server for like >1 year

[thorn jay]

Been here since october

[tardy hedge]

Thats crazy

I actually joined in sept 2023 but i got kicked once 😂

Wew was so much more active back then

[thorn jay]

What happened 😔

[tardy hedge]

To which part

Anytime i see wew here now its like rare pokemon or smth

[thorn jay]

Uh

Ironically to the second

Actual to the first

[tardy hedge]

I think i was particularly bored one night and was trolling

I dont remember how bad it was tbh

I apologized and asked wew to let me back in

@delicate orchid do u remember that? 😂

I dont remember i think i just needed an invite link actually

Like i wasnt banned just kicked?

Anyway

This server is like extremely valuable to me now

I take it for granted tbh

[thorn jay]

Same and totally not because of the postgraduate role i somehow got i feel like an impostor

[thorn jay]

Its human nature

[tardy hedge]

True

Legit like what even is this

[thorn jay]

Smt i wanna learn abt

[tardy hedge]

Potato could inform us

[thorn jay]

You*

Im gonna try and get my 5 hrs of sleep

Again

Im failing

[glad osprey]

"... sufficiently many applications of the suspension of disbelief functor"

[south patrol]

Lol this is an area I am more familiar with ig

[tardy hedge]

I would ask what is it but there is prolly so much background knowledge im still not that aware of

[south patrol]

I should say I do derived algebraic geometry

😭

[tardy hedge]

Really? I was just reading about that

I was like wtf

Is this

[south patrol]

Yeah lol

[tardy hedge]

It looks ridiculous

in a good way though

It got me interested

[tardy hedge]

Could you explain what it is to someone at my level

I stg weed makes me go to the math rabbit holes

Ong if anyone feels like explaining math to me im down

I mean imma head to bed soon now but

[south patrol]

Sorry I was washing up lol

[south patrol]

Uh so tbh you can think of this from many different angles lol but basically you can think of algebraic geometry as being about commutative rings and gluing them, whilst derived algebraic geometry generalises these to more "homotopical" things. One easy example of these more homotopical things would be (over a field of char 0) using (connective) cdgas, which are basically chain complexes with a multiplication and a very common objects in other areas of maths

But anyway there are many advantages to considering such things. One is that you can deal with non-smooth things better

Another is that deformation theory works a lot better in some respects - this is roughly studying ways to "infinitesimally thicken" a space (or deform some other object) and when you have more flexible rings you can deform stuff in more ways lol

Another is that there is spectral algebraic geometry (which may be viewed as a type of DAG but often there is a distinction) which is particularly useful for studying stuff arising in homotopy theory, basically because homotopy theory already studies generalised rings and it's cool to study them via some modification of algebraic geometry

[charred river]

I care for my job man

[tardy hedge]

That sounds really cool

@south patrol

[prisma ibex]

homotopy groups of spheres are too complicated so let's just study what they stabilize to after higher enough suspension, then they are still too complicated

[boreal inlet]

Task failed successfully

[thorn jay]

Whats with the messages here which get deleted

Lol

[candid patrol]

It was me sorry

Wrong server

[thorn jay]

Ah alright

[wraith cargo]

Lol have you seen the Adams SS calculations

People who do motivic homotopy stuff are mentally ill lmao

[prisma ibex]

Unfortunately yes

[wraith cargo]

I tried to read some papers of Zhouli Xu and that stuff is impenetrable

[prisma ibex]

I don’t think I’ve seen a single use for stable homotopy groups of spheres above like idk \pi_7

Other than like exotic spheres or whatever

[south patrol]

But yeah I guess a rough idea is that instead of just studying spaces you can also study these guys called spectra (forming a category Sp) which are super related but also more "algebraic". The nice thing is there is a good notion of "algebra" within Sp - for example you can study commutative rings inside Sp and their categories of modules. This truly generalises usual algebra stuff because you can view "ordinary" commutative rings as things in Sp

but yeah because Sp is more algebraic it is much more pleasant to work there for many things lol

[frigid epoch]

wait, comm rings are topological spectra??

[crystal vale]

Option a is incorrect, right? Because it is finite Boolean ring so every non-identity element has order 2 so its additive structure has order of 2^n, and it is the same as the cardinality of the ring so its cardinality cannot be 100 because 100 ≠ 2^n for every n in N.

[sly crescent]

But you have seen a use for π₇?

[glad osprey]

Yep, that sounds correct 👍 can you post your answers to b, c and d too? I wanna see if I got them right

[crystal vale]

I am trying

[prisma ibex]

yeah absolutely

[frigid epoch]

Tell us unc!

[rocky cloak]

I'd say
(B) ||False, if positive characteristic the additive group has too many torsion elements while in characteristic 0 there are too few.||
(C) ||2x is irreducible over Q, but reducible over Z. True for monic polynomials though||
(D) ||True if and only if R is an integral domain||

[crystal vale]

How to think for option b? I know F cannot be finite and I am thinking about some property in additive structure F which is not in multiplicative structure

[prisma ibex]

\pi_7(S)=Z/240Z is explained by the octonionic Hopf fibration and is the source of the 8-fold Bott periodicity for topological K-theory KO

[rocky cloak]

Probably helpful to split into the case where F has characteristic 0 and characteristic p

[frigid epoch]

ah, I remember that periodicity because my prof wrote an infinity sign instead of 8, and then corrected it saying "well, as a mathematician you don't have to write the number 8 often anyway"

[prisma ibex]

it is also related to 7-dimensional Chern-Simons theory and dual heteroic string theory among other things in physics

[frigid epoch]

in some talk, he did excitedly talk about how pi_90(S^136) has been figured out

[glad osprey]

Nice, B was the only one I wasn't sure about

[prisma ibex]

I guess \pi_11(S)=Z/504Z shows up for a similar reason in physics

but that's kind of where things top out

[crystal vale]

c is correct and d is incorrect, right?

[velvet hull]

c is almost correct. think about the leading term

[crystal vale]

Oh I need monic polynomial for that other wise it is not true

2x + 1 is irreducible in Q[x], right? And also it is irreducible in Z[x] so not all non-monic polynomial irreducible polynomial in Q[x] implies it is reducible in Z[x], I have to find one counter example

[velvet hull]

what about ||2x+2||

[crystal vale]

Oh yes

[rocky cloak]

A polynomial is irreducible over Z iff it's irreducible over Q and the coefficients are relatively prime

[crystal vale]

Oh yes primitive polynomial

I was forgot it

And we can generalize this to a ring of fractions

[crystal vale]

If I take the field of fraction Z_2(x), can I say 2 is a characteristic of it ?

Z/2Z(x)

[rocky cloak]

It is yes.

1 + 1 is still 0

[crystal vale]

Okay I am just verifying that there is an infinite field which has characteristics p

[rocky cloak]

Yes, Fp(x) is a good example of that.

Also the algebraic closure of Fp is also infinite.

[crystal vale]

I see

[hidden wind]

because there are infinitely many irreducible polynomials in Fp[x]

[rocky cloak]

That's one way to see it.

You can also quite explicitly for any finite field construct a polynomial without roots. So an algebraically closed field is never finite

[distant summit]

What exactly is the thinking here?

That if ab is order 6, then <ab> is a subgroup of order 6 and hence G=<ab>=Z_6?

[thorn jay]

Yeah

Counting argument

[distant summit]

okie

[tardy hedge]

Sup Groupies

[tardy hedge]

Without roots in the field?

Create separable polynomial of degree higher than # elements in field?

[rocky cloak]

That would also be a strategy I suppose, though seems more difficult

[velvet hull]

Or just consider (x-a1)(x-a2)…(x-an) + 1 where you range an over every single element in the finite field

[south patrol]

Lol so you can view abelian groups as spectra. The recipe sends an abelian group A to the spectrum HA (often just called A still lol), which represents singular/ordinary cohomology with coefficients in A

and then you get a similar thing with commutative rings

[tardy hedge]

Whats that?

Gauss lemma?

And the coefficients are rel prime? WhyV

[coral spindle]

If they have some common factor

then you can just pull it out and reduce it in Z[x]

[south patrol]

Basically if you can factor a poly nontrivially over Q then you get a nonttivial factorisation over Z. The reverse isn't generally true because Q[x] has more units than Z[x] (namely Q^x instead of just {+-1}) but once your polys are primitive that disappears

[tardy hedge]

Do u guys get annoyed if i start asking a lot of questions like this

Ive kind of had a realization that i should just endlessly ask questions if i dont know things

I mean like^^ obv work thru things myself but i mean like, not being scared to ask basic questions anymore

Ive gotten better at not caring l

[tardy hedge]

Oh cause a constant times polynomial over Z is a reducible poly over Z? Lol

[tough raven]

Nice, character growth arc.

[tardy hedge]

Yeah! Im happy its happening

In lectures i just ask away bruh

In first term i was so hesitant lol

Silly undergrad mentality

[next obsidian]

I think it depends. If someone mentions a concept or something and you just ask what that is I think nobody should be irritated.

I only get a little irritated if someone asks for help on a hw problem or something and it’s like, something that you can solve by just trying the obvious things. It looks like you aren’t even trying, and that’s not a good habit to be in lol

[tardy hedge]

Yeah for sure, I agree

[tardy hedge]

For first point - because u can pull out the lcm of the denominators?

Oh wait uh

[south patrol]

Gcd