groups-rings-fields

[latent anvil]

Like Spec is determined by the poset of containment of radical ideals, as the poset of radical ideals A and closed subsets of Spec A are dual

Err this no longer seems true...

@next obsidian is the topological data of Spec A determined the generalization relation on points?

Exercise: figure out what sham means by that

[next obsidian]

Umm... I don't think so in general. If A is Noetherian then its a Zariski space and I think the classification of opens as constructible sets closed under generization says that

In other cases

¯_(ツ)_/¯

[latent anvil]

hmmm

I'm thinking about ari's thing

[next obsidian]

I haven't really thought about it tho

[latent anvil]

I want to reduce it to like, stuff about the poset of primes

[next obsidian]

What was Ari's actual question?

[latent anvil]

Maybe the radical ideal thing is better

[next obsidian]

Like... my understanding is

[latent anvil]

Classify rings R such that Spec R = X

[next obsidian]

for any Spectral Space there is a ring realizing it

Oh classify

[latent anvil]

Determine the fibers of the spec functor

Yeah

[next obsidian]

Idfk lmao

uhhhh

[latent anvil]

So we can probably just look at noetherian R

For simplicity

[next obsidian]

yeah uh...

still don't really know lol

I have never thought about anything even close to this lol

[latent anvil]

Me neither. It's interesting though

[next obsidian]

What do you think a "residue class field" is

In the context of local rings I assume it just means k

[latent anvil]

sounds like number theory

[next obsidian]

but man that's such a jank way to say that

[latent anvil]

> class field

[next obsidian]

Okay so

pog time

not really but

if A is noetheiran local

Is A/m^n Artinian always?

or something

[latent anvil]

should de

It's a local ring where a finite power of the maximal ideal is 0

[golden pasture]

isnt residue class field like O_K/P

[next obsidian]

There’s no sheaf here

[latent anvil]

i assumed O_K was a number field

[golden pasture]

yea

O_K is ring of integers

[cyan marten]

Is it true if the ring satisfies the DCC on principal ideals?

[wind parrot]

I think you mean ACC, but yes, the ACC only on principal ideals should be sufficient to get factorisations into irreducibles

[next obsidian]

Oh lol

[frank haven]

Hey does anyone have a good way of describing inner automorphisms? Is it simply an automorphism that is of the same form as conjugation by an element?

[delicate bloom]

yeah, it's an automorphism made by things 'inside' the group itself, as opposed to an outie

[frank haven]

Oh that makes the name a little clearer actually

[carmine fossil]

Also,All automorphisms of S_n,(n!=6) are inner

For S_6, Half of the automorphisms are inner

A good way of thinking about inner automorphisms is that they are the only automorphisms which map a conjugacy class to itself(in S_n)

[cyan marten]

They do preserve classes, but there are outer class-preserving automorphisms as well.

[carmine fossil]

mb

[vestal snow]

Does f being regular on U here mean that it is regular at all points of U or that f = g/h on U and h is not 0 anywhere in U?

[latent anvil]

What does it mean to be regular at a point? Does it mean that you have a representation as a rational function in some neighborhood of that point?

Because it so it's the former

Regular functions on U may not be rational functions

[vestal snow]

Thanks

[vestal snow]

Is my proof of this correct?

Prove $O_{P,Y}$ is a local ring.

Suppose $<f,U>$ is an element of $O_{P,Y}$ such that $f(P)$ is not 0. Then there is some open subset $V$ of $U$ such that $f = g/h$ on $V$ and $h$ does not vanish at any point in $V$. It is easy to see that $1/f = h/g$ on the set $V - Z(g) ( = V \cap (Y-Z(g))$. Thus, $1/f$ is regular on the set $V-Z(g)$ which is open and contains $P$.

[cloud walrus]

Have a Banana Bitch

[vestal snow]

This proves that every element not belonging to the ideal I = {f regular| f(P) = 0} is a unit

[chilly ocean]

Very informally (which is precise for some cases like nice varieties that are irreducible) you can think of regular functions as elements on O_{P,Y} and O(U) = \cap_{p \in U} O_{P,Y} for some U \subset Y

[dapper nebula]

did you write the algebraic geometry book by gortz and wedhorn

[chilly ocean]

lol no

[vestal snow]

I'm using the definitions in the first chapter of Hartshorne

Is my proof correct then?

[latent anvil]

it's a good book so far

Ty for writing it @chilly ocean

[chilly ocean]

Can anyone help with this? Let A be noetherian then TFAE:
(1) A is artin
(2) Spec(A) is discrete
(3) Spec(A) is discrete and finite

Clearly 3 implies 2.

[latent anvil]

what have you tried?

[chilly ocean]

Also if Spec(A) is finite A must be artin

so 3 implies 1

[latent anvil]

Yup

so let's think about 2 => 3

Do you have any initial ideas?

[chilly ocean]

just to double check by discrete we do mean discrete topology

[latent anvil]

Yup

So here's a question, how are discreteness and finiteness related?

Purely topological question

[chilly ocean]

If we have compact and discrete we get finite

[latent anvil]

Right

And vice versa

Compact iff finite for discrete spaces

[chilly ocean]

So we probably want to next show that 1 impies 2

[latent anvil]

Wait so you do you see how to prove 2 => 3?

[chilly ocean]

oh sorry no

[latent anvil]

Okay, so we need to show Spec A is compact

[chilly ocean]

I'm guessing this either follows form being noetherian or i've missed that Spec(A) is always compact

I'm a little confused with how to work with the zariski topology on Spec(A)

I'm comfortable with the Zariski topology on affine space

Okay Spec(A) is always compact

So now only 1 implies 2 remains.

Its probably best to work with closed sets and show that any set is closed.

Any arbitrary set in Spec(A) is a collection of ideals.

We know that in an Artin ring every prime ideal is maximal

In general the closed sets in Spec(A) are, {V_I where V_I is all the prime ideals contain I, and I is an ideal in A}

Since our ring is also noetherian every ideal is finitely generated

I'm getting a little stuck then, maybe it is better to show that every singleton set is open

FUCK

Spec(A) is only the prime ideals

we already know the prime ideals are maximal

[next obsidian]

Spec A is always compact

Hint: write 1 as a finite sum of elements in an ideal

[latent anvil]

Ah shit sorry I got distracted lol

[chilly ocean]

its fine

you helped lots

[latent anvil]

But yeah that's correct for 2=>3

You're welcome :)

[chilly ocean]

Does this work?

1 need not always be a finite sum ?

any infinite product ring

[next obsidian]

no lol

If (f_1,f_2,...) = (1)

err

[latent anvil]

so the most important thing about the zariski topology is that you can check everything on the base of distinguished opens

[next obsidian]

Let's just say I = (1)

[latent anvil]

which chm is implicitly using

[next obsidian]

then there's a finite number f_1,...,f_n in I such that

[latent anvil]

he's arguing that you check compactness on that basis

[next obsidian]

1 = a_1f_1 + ... + a_nf_n

[latent anvil]

Ie, show that any cover by distinguished opens has a finite subcover

[next obsidian]

This let's you do it

[chilly ocean]

is there anything extra you are saysing here instead of a usual base of opens for any topology?

we can always check if something is compact on a basis sets?

[latent anvil]

Yup

That's true for any topological space and any basis

Distinguished opens (sets of the form D(f) = { x in Spec A : f not in x } for f in A) are very very nice

it's usually easier to prove things about them

And they form a basis for Spec A

[chilly ocean]

ty

cool name are you irish

[latent anvil]

yes lol

But the name is unrelated

my real name is even more Irish than ☘️

well, I'm Irish-American (parents were Irish immigrants)

Your name is pretty good too

[chilly ocean]

have you been/know which part they are from

[latent anvil]

yeah, ballyhea and navan

County cork and meath

[chilly ocean]

gg

[latent anvil]

lol

anyways did you figure out 1=>2?

[chilly ocean]

yeah thanks

[latent anvil]

Nice!

[sterile garden]

Suppose I have unique factorization of ideals into prime ideals. If $\frak{p}$ is a prime ideal and $\frak{p} \supset (z)^n$, then I see that $\frak{p} \supset (z)$. But does this rely on having unique factorization of ideals into prime ideals?

[cloud walrus]

IamDerek

[sterile garden]

Or does this hold without having that property?

[oblique river]

this follows from one definition of prime ideals

if I and J are ideals with IJ contained in P, then at least one of I or J is contained in P

[sterile garden]

So it has nothing to do with this unique factorization property

[oblique river]

not really, no

[sterile garden]

Ah yes, that makes sense.

I'm not sure what I need this unique factorization property for I guess.

For reference, this is the problem I'm working on.

I've got the idea of the proof down. I think I even have a correct proof, but I'm not sure what part relies on needing that unique factorization.

[oblique river]

what's exercise 17?

[sterile garden]

It's essentially the same thing, but you assume Z[w] is a UFD

[oblique river]

oh okay

maybe the point is just you can talk about "factors" without having to handwave or anything

[sterile garden]

Can you elaborate?

[oblique river]

like, you're talking about factors of some ideal

[sterile garden]

And now that I'm thinking of it, I'm not actually sure I used the UFD property in exercise 17.

[oblique river]

I'm just saying that having unique factorization makes it easier to talk unambiguously about factors of an ideal

[sterile garden]

Can you explain what would fall apart if we didn't have unique factorization?

Maybe that's a better question

Would it help if I wrote out my proof?

[oblique river]

yes it is a better question but unfortunately I don't have much time rn

you can post your proof and maybe someone else can jump in

but I have to run

[sterile garden]

Okay, thanks!

So my "proof"...

Suppose $\frak{p} \supset (x +y \omega)$ and $\frak{p} \supset (x + y\omega^k)$ for some $k \neq 1$. Then $\frak{p} \supset (x + y\omega) + (x + y\omega^k)$. So $x + y\omega - (x + y\omega^k) = y\omega(1 - y\omega^{k-1}) \in \frak{p}$. By a previous exercise, this gives $y\omega(p) \in \frak{p}$, so $(yp) \subset \frak{p}$. Clearly $(z)^p \subset \frak{p}$. I'm told earlier on in the chapter that $yp, z$ are relatively prime, so as principal ideals they're relatively prime. So you can write it out and show that $\frak{p} = \mathbb{Z}[\omega]$ but that's a contradiction since $\frak{p}$ prime and therefore a proper ideal.

[cloud walrus]

IamDerek

[vestal snow]

How do I show that $\varphi$ being dominant is independent of the choice of the pair?

[cloud walrus]

Have a Banana Bitch

[sullen aurora]

would that follow from open sets being dense?

[vestal snow]

I don't think so

Because it is not necessary for continuous functions to take open sets to open sets

[stoic rose]

The map is dominant iff the preimage of any open set in Y is nonempty

If you restrict the map to some open set $V \subset U$ and if the map is dominant for U, that mean the preimage of any open in Y is a nonempty open of U, so it must intersect V by density of V

[cloud walrus]

radiateur-man

[stoic rose]

So the map is also dominant if you look only at V

[chilly ocean]

Can someone else me with this?

So far I have: we have an exact sequence 0 to M to N to coker(f) to 0, (f is surjective iff coker(f) is trivial)

Denote A/m by k, then we tensor this sequence with k. Since the tensor product is right exact we have an exact sequence M\otimes k to N\otimes k to coker(f)\otimes k to 0.

If we suppose that this f_m is surjective then coker(f)\otimes k must be zero

We are done if coker(f)\tomes k zero implies coker(f) is zero.

coker(f) is N/f(M)

[next obsidian]

This is an application of Nakayama’s lemma

Coker(f) is finitely generated as it is a quotient of N which is finitely generated.

As the tensor product is right exact we know that coker(f (x) A/m) = coker(f) (x) A/m = coker(f)/m(coker(f))

The latter being = 0 implies that coker(f) = m(coker(f)) so by Nakayama’s lemma coker(f) = 0

I think my proof is correct, but it doesn’t use the finite generation of M. Maybe someone else can point out if there’s an issue with it

[chilly ocean]

Why is coker(f)\otimes A/m = coker(f)/m(coker(f))?

Also I think we do need to use M being finitely generated as it’s not true otherwise

[latent anvil]

this is what tensoring with A/I does

for any module $M$ and any ideal $I$ of $A$ we have $M \otimes_A A/I \cong M/IM$

[cloud walrus]

π¯¹(Shamrock)

[latent anvil]

you have a bilinear map $M \times A/I \to M/IM$ given by choosing a representative and multiplying

[cloud walrus]

π¯¹(Shamrock)

[latent anvil]

this induces an isomorphism on tensor

and this is true regardless of M being finitely generated, I believe

[chilly ocean]

The the next part of the question asks fo find a counter example when M and N are not finitely generated

I suppose we did youse finite generation of N to use nakayamas lemma

But I still don’t see at all how to come up with a counter example when both m and n are not finitely generated

I

[next obsidian]

I mean here's the thing, if M wasn't finitely generated but N was and you still had surjectivity of M/mM -> N/mN do the following:
Since N/mN is finitely generated, take preimages of generators x_i of N/mN in M/mM, and call these y_i. Lift these y_i to elements in M, by abuse of notation call these y_i as well. Then call M' the submodule of M generated by the y_i, note that this is finitely generated as we have finitely many y_i. We can restrict M -> N to a map M' -> N, but then M'/mM' -> N/mN is still surjective because all the y_i mapping to the x_i live in M'/mM', and the x_i generate N/mN

So you can apply the finitely generated case to conclude that M' -> N was surjective, but since a submodule of M surjects onto N via M -> N then M very clearly also surjects onto N

Like, you can only ever get bigger by letting M be not finitely generated, so I'm pretty sure you can let M be whatever you want it to be

[vital quail]

@latent anvil @chilly ocean I'm quarter irish

Both my dad's parents are half irish

It's what I attribute my natural good luck to

[chilly ocean]

Do you know where from

[chilly ocean]

Do you have any idea of a counter example of N is not finitely generated

Or how to come up with one

I’d thought the way to do would be to have an infinite product ring

And have all the terms in the product above 1 be killed by tensoring with A/m

But I can’t think of a ring or module that tensor with A/m to become zero

[latent anvil]

yeah so I was actually going to give that as a hint

If there's a counterexample of any kind then there's a counterexample with M = 0 (try proving this!)

ie a ring which you can tensor with A/m to get 0

[chilly ocean]

?

A counter example would f from M to N non surjective

With the induced map o the tensor surjective

[latent anvil]

Yes, that's what I mean by a counterexample

I claim that if such a counterexample exists then one exists where M = 0, ie where N is a nonzero module but N (×) A/m = 0

[chilly ocean]

Oh okay

[chilly ocean]

Can someone help me walk through if I understand this correctly

Let A be a ring, m a maximal ideal. Then Let Am be the localization at m. This is a local ring. Its maximal ideal is m*Am. Since this a local ring it has a residue field. The residue field is isomorphic to A/m

If we replace m with just a prime ideal p, then the localization need not be a local ring

[misty socket]

@chilly ocean

[chilly ocean]

Hmm

If m is maximal is it true that the residue field of the localization is iso toA/m

I assume then in the case where we just a prime ideal p

The residue field of the localization is

The quotient by the maximal ideal containing p

And prime ideals are contained in unique maximal ideals so this is fine

[golden pasture]

$A_{\mathfrak p}$ is a local ring with maximal ideal $\mathfrak pA_{\mathfrak p}$

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[chilly ocean]

And the residue field of this local ring should be A/p

Or isomorphic

[golden pasture]

As a example we have say

$$A=\mbb Z[x],\mfk p=(x)$$
Since $(2,x)$ is a proper ideal, $\mfk p$ is not maximal.

We can understand the local ring $\mbb Z[x]_{(x)}$ as the ring of rational function which is defined at $0$, i.e. functions $\frac{f(x)}{g(x)}$ where $f,g\in\mbb Z[x]$ and $g(0)\neq 0$.

It is a local ring as we have the ideal $\mfk m=x\mbb Z[x]_{(x)}$ of rational functions $q(x)$ where $q(0)=0$, clearly all other elements are invertible. Hence we can construct the quotient ring

$$\frac{\mbb Z[x]{(x)}}{x\mbb Z[x]{(x)}}$$

which consists of rational functions $\frac{f(x)}{g(x)}$ with $f(0)\neq0$ and $g(0)\neq0$. This is clearly a field.

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[golden pasture]

$\frac A{\mfk p}$ is in general not a field

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[golden pasture]

in the example above, it is jus the integers

[chilly ocean]

I know A/p need not be a field only domain

[golden pasture]

yup

[golden pasture]

so this isnt true

[chilly ocean]

Sorry

I said above I forgot to say it again

A a ring take the localization at p a prime ideal, which we denote by Ap.

p is an ideal in A and hEnce contained in a maximal ideal

I feel like it should be contained in a unique maximal ideal?

[golden pasture]

ig isnt

it

[chilly ocean]

Shit

[golden pasture]

let's go back to this example, you have the maximal ideals (p,x) where p is a prime

[chilly ocean]

Okay so it’s only tru that every element is contained in a maximal ideal

There are ideals that need not be contained in a maximal ideal

[golden pasture]

also false

[opal osprey]

Ok so, in this first number theory lecture the professor presented to me a way to define C that I have never seen before

[golden pasture]

you can prove by zorn's lemma

[opal osprey]

That being C := |R[i]/<i²+1>

[golden pasture]

i mean if you assume not AC

then pls go #foundations

[chilly ocean]

Okay so every ideal is contained in a maximal ideal

[golden pasture]

ye you use that in algebra

[opal osprey]

So you take all the one variable polynomial of real coeficiente and quotient it by i²+1

[golden pasture]

yup

[chilly ocean]

But need not be uNique

[golden pasture]

yea

[chilly ocean]

And the uniqueness is what’s going wrong here

Lack of

[opal osprey]

But this is a new thing to me, because I've seen C before as being R² equiped with a certain product that makes it a field, and also as a field of 2×2 certain special matrices.

How do I prove all these constructions are field-isomorphic?

[golden pasture]

write an explicit isomorphism

[chilly ocean]

As in is it the case, if we have p a prime ideal contained in a unique maximal then the residue field of Ap is A/m where m is the unique maximal ideal containing p?

[opal osprey]

Yeah, so I guess that for the case of |R² with a product and the field of 2×2 matrices it's easy for me to see the isomorphism.

But what about |R[i]/<i²+1>?

[golden pasture]

probably not true either

[chilly ocean]

Damn

[golden pasture]

A_p/pA_p is generally "larger" than A/m

[chilly ocean]

Is there any nice result about what A_p/pA_p looks like

Or does it actually turn out to be not too hard to just calculate

[golden pasture]

here "larger" im using it in a intuitive sense lol

you probably need a really funny ring to get like unique maximal ideal containing p

[dapper nebula]

local ring

😃

[golden pasture]

oh nvm

[dapper nebula]

lol

[golden pasture]

i mean this kinda lolz

but yea

can like

$k[x,y]_{(x,y)}$

[cloud walrus]

a cute cat ٩(˃̶͈̀௰˂̶͈́)و

[golden pasture]

only one maximal ideal

but we have the primes (x) and (y)

[golden pasture]

you can think of this ring in a similar manner as this one

or can do k[[x,y]] for algebraically closed k

[golden pasture]

basically this, but bigger

[chilly ocean]

Ty king

[dreamy snow]

Hello! I'm doing a problem on the sylow theorems where I'm trying to show that if a group's sylow-p-subgroups intersect trivially, then n_p (the number of p-sylows) is congruent to 1 mod |S| (S being a p-sylow).

By the sylow theorems n_p is congruent to 1 mod p. Does this mean one has to show that |S|=p?

I tried to look at the number of elements in the union of the p-Sylows and did some calculations with n_p but it's not really getting anywhere.

I can send what I've written so far. If anyone has any ideas on how to approach the problem I'd be super grateful.

[carmine fossil]

I think you might have to do some sort of a group action

Like conjugation on the set {G_1,G_2...G_(np)} where G_i are the sylow p groups

[chilly ocean]

a + bx ↦ a + bi

[opal osprey]

But it's weird because all sorts of polynomials with a higher degree than 1 can still be complex numbers in this definition, right?

[carmine fossil]

R[i]/(i^2+1) is "made up" of polynomials of degree atmost 1

[opal osprey]

Ooh, ok I see how that can be done

Because if you have a given polynomial ax_n•x^(n)+...a_(0)

Damn

[carmine fossil]

Do you understand what R[i]/(i^2+1) means?

[opal osprey]

Yup

You are taking all the polynomials of real value coefficients

[carmine fossil]

The equivalence class of that would have a representative of form ax+b for some a,b in R

[opal osprey]

And whenever you have i² in it

You just substitute by -1

That's basically the idea

I was just being dumb when I asked how is it that all the polynomials have degree at most 1

You can just "cancel" a lot of stuff with this equivalence relation.

like, x³+x²+1 = x•x²+x²+1 ~ -x+(-1)+1 ~ -x

so x³+x²+1 is in the equivalnce class of -x

And etc

[nova plank]

I thought this was expert trolling for a second, and i was the imaginary unit so drake was trying to take the quotient by the zero ideal

But I overestimated him, obviously

[opal osprey]

C := |R[x]/<x²+1>, better now lol?

i is just a formal symbol

Anyway, yeah thanks

Idk how I didn't see that

[dreamy snow]

Thanks! Yeah I was kind of avoiding group actions because they're not very intuitive for me at this point. All the more reason to use them though

[robust breach]

Hi i have a small question, is a field (like the real numbers) a principal ideal domain ? The ideals of (R, +, *) are {0} and itself, but is R generated only by a single element ?

When you say it is generated by 1, do you mean like the ideal is 1R ?

[cloud walrus]

slimvesus

[robust breach]

No it's alright i understood what you meant thanks

It's still a bit weird to me because i'm used to write Z as <1> and just applying addition and substraction an infinite number of times

So i thought well you can't exactly express R the same way

Yes

I'm not sure if i get it

i mean yeah that's true

Nah it was as you said, i only thought of Z as a group and not as a ring, I'm not sure if i can write it clearly but when we say generated by a single element, i think of <1>. 1Z or 1R doesn't look like anything is generated, only that we take the same ring and give the same ring back

It's like well i understand it now thanks to you but I still don't exactly have the intuition

Intuition will come with a bit of time. i have the answer to my question so thank you again

[unique berry]

Is there a preferred notation for the order of an element in a group? My book simply denotes it o(a).

[carmine fossil]

|a|?

[unique berry]

a^(o(a)) = e

Well o(a) is the smallest positive integer such that a^(o(a)) = e

[carmine fossil]

d and f uses |a| for order of element a

[unique berry]

I mean that makes more sense

I was just wondering if the o(a) was standard or not

[carmine fossil]

Don't think it's standard

[latent anvil]

Your book is not unique in using this notation

But I think the bars are more common

[prisma ibex]

|a| and ord(a) seem to be the most common in my experience

[median pawn]

Can someone help me check if my proof is right?

[cloud walrus]

the-last-knight

[median pawn]

The first part seems good, It'd be nice if someone could confirm my proof of the double implication

[chilly ocean]

i don't see any issues with it

i think you can cut it down in size a bit, maybe by doing the double implication like a series of iff's

but that's just personal taste

[vestal snow]

Does this mean order?

[nova plank]

I think it's a twitter hashtag

[viscid pewter]

i think it's a sharp

[vestal snow]

I'm in an exam so I can't post the entire problem for context. Do some authors use this for order?

[nova plank]

Bruh

[vestal snow]

Does anyone know what this notation means?

[viscid pewter]

i've literally never seen anything like it

[nova plank]

And you are in an exam

[vestal snow]

It's a qual that I'm doing for fun

The guy who made the exam is not my instructor

[nova plank]

It's still an exam

If you are allowed to ask questions, ask the invigilator

[vestal snow]

There is none lol

[burnt pond]

could be the number of elements in the double coset

[golden pasture]

usually i see #S to mean number of elements in S

[golden pasture]

so this

[next obsidian]

I think this is asking for the size of that double coset

[chilly ocean]

Let $R$ be a ring, $I$ an ideal of $R$. We then can consider the category $R/I$-mod of $R/I$-modules.

We call $C$ the full sub category of $R$-mod whose objects are $R$-modules $M$ with $Ann_R(M)\subset I$.

Give a a functor from $C$ to $R/I$-mod which is an equivalence of categories

[cloud walrus]

lime_soup

lime_soup

I assume the idea would be to send $M$ to $M/Ann_R(M)$

[sturdy marsh]

M/Ann(M) would not be an R/I module in general

I think you want the inclusion to be the other way

if I \subset Ann(M), then M is an R/I mod

[chilly ocean]

Opps yes! that inclusion is the wrong way around

How do I see this?

M is an R-module
M/Ann(M) is an R-module

then because every thing in I anniliates M, and we have already quiotented M by this already, if we are free to make all of these zero in R/I

[sturdy marsh]

you don't need to quotient M by anything to make it an R/I module

if 'I' annihilates M, then M can be given the structure of an R/I module

[chilly ocean]

Okay so M is already an R/I-module

Now why can we see that M/Ann(M) is a still an R/I-module?

Ann_R(M) is certainly an R module

but also an R/I module

[sturdy marsh]

a quotient of a module by any submodule is still a module over the same ring

[chilly ocean]

If we quoitent and A module by an A-submodule we always get an A submodule?

[sturdy marsh]

yes

[chilly ocean]

functoriality of this is fine since compositions work nicely with quoitents

not super obvious that thisis an equivalence of categories to me though

[sturdy marsh]

do you need to quotient at all tho

just send M to M

[chilly ocean]

oh

[sturdy marsh]

in fact I don't think the functor you defined is actually a functor

[chilly ocean]

yeah

i don't even think it makes sense

Ann_R(M) is in R not M

thanks very much

[sturdy marsh]

ig it makes sense and actually works if you interpret M/J as M/JM

but Ann(M) M = 0

so accidentally, you got the correct functor

but yeah send M to M

[carmine fossil]

x^2+2 is a irreducible in R,which is not of that form

[gritty latch]

(x) is not a max ideal of Z[x] right

[carmine fossil]

Yes

[gritty latch]

hm

[thorn delta]

x^2+2=2(1/2x^2+1)
or am I dumb?

[carmine fossil]

nvm

I was being dumb

[gritty latch]

whats the easiest way to find the number of elements in Z2[x}/p(x)

just list them out?

[carmine fossil]

Yes

number of elements will be 2^deg(p)

[gritty latch]

is that true for any p?

ill try proving it ig

[gritty latch]

how do i show that minimal prime ideals in a UFD are principal?

[golden pasture]

i assume minimal primes excluding (0)?
show that every prime ideal contains an irreducible element

[opal osprey]

I've started to learn Ring Theory this week

And I tried to prove this fact here:

Without using any constructions with quotient rings.

The professor mentioned that a proof that goes in this way and I've already proved this fact using this tip.

But my first idea was to do this.

Is this proof right?

Should I be more specific in some steps?

[scarlet estuary]

how do you know A has a subfield containing I?

[opal osprey]

It should be the preimage

The image of an ideal is not always an ideal

Lemme fix it

Thanks for noticing my mistake tho

[opal osprey]

Hmmm

I don't really know

That example could make sense

But it actually doesn't

I should just get rid of that in the proof

And just write "there's a homomorphism from A to a certain field |K"

Because it's clear that "|K" is a non-trivial ring

And well

We have examples of fields that exist

So it should be no problem to just get rid of that passage.

[scarlet estuary]

i dont understand how your argument works then?

oh i get what youre trying to do

yeah clean this up a bit and this idea should work

but your specific constructions are necessary

[opal osprey]

Damn

There's another little mistake I didn't notice

This should be right now.

[scarlet estuary]

how do you know that there exists a homomorphism into a field K?

[opal osprey]

Because every field is a non-trivial ring.

You take |R

Idk

And you can construct |R

[scarlet estuary]

okay... but not all two rings have homomorphisms between them

what if there's a ring that has no homomorphisms into any fields?

[opal osprey]

But the hypothesis states that this is for all non-trivial rings B.

So hmmmm

[scarlet estuary]

for example, there is no homomorphism Z/nZ -> Z

[opal osprey]

I don't really know what did I do wrong

[scarlet estuary]

in fact, a homomorphism A -> B can only exist if char(B) divides char(A)

[opal osprey]

Maybe it's the way it's stated?

[scarlet estuary]

the problem is that you havent established that such a K actually exists

how do we know there's a homomorphism from A into a field?

isnt it possible that A only has homomorphisms into rings that arent fields?

[if not, justify why not]

[opal osprey]

Yeah, it's just that I thought "all non-trivial rings" would include fields as well.

[sturdy marsh]

this problem is a bit unfair if they don't know zorn

[scarlet estuary]

eh fair

@sand gull

oops

wrong ping

sorry bro

[sturdy marsh]

I didn't mean to ping, my bad

[opal osprey]

Zorn's Lemma has to do with maximal elements in a partially ordered sets I guess.

[scarlet estuary]

in that case though i dont think this argument can be salvaged

[opal osprey]

I've read about it in Munkres

[scarlet estuary]

if you dont want to go that far

unless im missing something

[opal osprey]

But I've never really used this theorem any time lmao

[scarlet estuary]

uh the point is that

[scarlet estuary]

we're only considering Bs where there actually exists a homomorphism phi: A -> B

but you dont necessarily know whether one of those Bs is a field

(the answer is that yes, one of those Bs is in fact a field, but thats not particularly easy to prove)

unfortunately im not sure theres a way to fix this specific argument without that fact :/

[opal osprey]

Aaah

So I really misinterpreted the problem

It's just that the way it was stated I thought that's what it meant.

[opal osprey]

Does it really uses Zorn's Lemma?

Maybe I could try to use that for the first time in my life lmao

[sturdy marsh]

yes

[opal osprey]

It would be a fun challenge

[sturdy marsh]

it is equivalent to the existence of maximal ideals

which is equivalent to Zorn

[opal osprey]

Ok

This looks really non-trivial

I'll think more about it

[sturdy marsh]

the proof is not hard

if you know the lemma

[opal osprey]

Oook

But lemme think by myself

I guess it's gonna be fun

Also

Yeah, I really misinterpreted the problem

The way I did it didn't really make any sense

[scarlet estuary]

note that this "maximal ideal" approach only works if A is commutative

unless im missing something

[sturdy marsh]

unital is what you need

[scarlet estuary]

oh yeah

[sturdy marsh]

not commutative

[scarlet estuary]

unital

ops

wait dont you need unital and commutative?

[sturdy marsh]

yeah you state the theorem as left/right maximal ideals

[scarlet estuary]

ah

duh

okay thats fair

even that eventually uses the idea of a quotient though

which they said they wanted to avoid

in the proof

[opal osprey]

So the problem would actually be stated as "If A is a ring where, for all non-trivial rings B such that there exists φ : A ---> B an injective ring homomorphism, then A is a field"

It was a translation problem

[scarlet estuary]

uh

[opal osprey]

I kind of translated it the wrong way and got the wrong idea

[scarlet estuary]

youre missing the injectivity of all phi

[opal osprey]

Yup

[scarlet estuary]

from your statement

but yes

[opal osprey]

Thanks

[scarlet estuary]

well thats still a weird way to phrase it

[opal osprey]

Gotta think more about that thanks

Thanks for the suggestions

[scarlet estuary]

i'd probably phrase it as

[sturdy marsh]

no field satisfies this

oh wait I misread

[scarlet estuary]

If A is a ring where, for all non-trivial rings B such that there exists φ : A ---> B, we have that φ is injective, then A is a field.

perhaps more succinctly:

If A is a ring where all homomorphisms φ from A into nontrivial rings are injective, then A is a field.

[worn grove]

can anyone help me with ii

[sturdy marsh]

what have you tried?

[worn grove]

i completed i

[sturdy marsh]

no I mean for ii

[worn grove]

oh i proved it was injective by statting different x and y values would give different outputs

however i ahve no idea about surjective

[sturdy marsh]

look at your argument carefully again

[opal osprey]

Ok so, the poset I'm thinking about working with is exactly the set of all non-trivial rings B where there exists that injective homomorphism φ from A to B.

This set would be partially ordered

Now I don't really see how every chain would have an upper bound.

[sturdy marsh]

what are you trying to prove?

[opal osprey]

This

[worn grove]

that there is a preimage or x,y for every image f(xy)

[sturdy marsh]

zatza look at your argument for the injective thing again, you did something wrong

[sturdy marsh]

try proving that every ring has a maximal ideal

[worn grove]

really i cant ssem to see the probem other than the extra y

any hint

[sturdy marsh]

extra y?

hint: addition is commutative

[worn grove]

it wouldnt make a difference would it

oh igot it

x=3 y=2 x=2 y=3

[sturdy marsh]

yup

[worn grove]

give same value

[sturdy marsh]

yup

[worn grove]

damn didnt see that

how abouy the surjective

how would you do that

[sturdy marsh]

try it

it isnt hard

[worn grove]

like i cant find inverse

because of y

[sturdy marsh]

you're trying to solve x + y + 2 = n

for each n

[worn grove]

x = n-y-2

[sturdy marsh]

sure...

[worn grove]

y = n-x-2

like that

[sturdy marsh]

just find an x and y that work

[worn grove]

oh would the above stated inverse be valid as a surjection proof?

[sturdy marsh]

you just need to show that there is some x and y such that f(x,y) = n

for each n

so say I give you an 'n'

can you give me an x and y such that f(x,y) = n

[worn grove]

y = n-x-2

x = n-y-2

dont theese two do that?

am i missing anything?

[sturdy marsh]

you gave me x in terms of y and y in terms of x

but you need to find both of them

okay let's make this simpler

find x and y such that f(x,y) = 3

[worn grove]

hmmm

y +x = n-2

[sturdy marsh]

like give me numbers x and y

actual integers

like 1 2 or 3

[worn grove]

y +x = n-2

2 3

x = 2

y=3

[sturdy marsh]

oh wait am I looking at the wrong function

I meant g

g(x,y) = x + y + 2

[worn grove]

yes

[dim escarp]

Why is this in Abstract Algebra?

[sturdy marsh]

so you want x + y + 2 =2

idk

[worn grove]

i didnt know where to post it

[dim escarp]

Groups, Rings, Fields, Vector Spaces, Modules. What is this one relevant with

[worn grove]

its part of my basic computer science course

[dim escarp]

At least I guess you are not supposed to be here

[worn grove]

bro i dunno tbh

[dim escarp]

Not advanced

Try #prealg-and-algebra #proofs-and-logic

[worn grove]

will do thanks m8

[opal osprey]

There another question I had in mind

For example

Suppose you had a ring R and you equip it with an equivalnce relation ~.

If this equivalence relation is not defined in the usual way as the quotient of R by a 2-sided ideal.

Then what can we say about R/~

In the sense that

If we define the operations on R/~ in the usual way as [x]+[y]=[x+y] and [x]•[y]=[x•y]

Can we still guarantee R/~ is a ring?

It's weird how we usually think about quotient rings in a rather specific sense where x~y iff x-y is in a certain ideal I.

Where we could equip R with any equivalence relation we wanted.

[thorn delta]

yea i was wondering about this a while back. We give sufficient conditions an equivalence relation can have to form a quotient ring/group (relation defined by ideal/normal subgroup), but I've never seen the necessary conditions

[opal osprey]

This seems to be a rather common theme when one is working with algebraic structures

You usually quotient it by some interesting subset of your original one

With respect to some operation

For example

The quotient space of a given vector space X by a subspace Y is just X|Y where x~y iff x-y is in Y

Why is defining quotients like this a more natural thing to do?

Yeah

I was conjecturing that

But I didn't know how to prove it

I'll try to read that out later

[sturdy marsh]

if the operations are well defined then you have a ring

[opal osprey]

But before I'll try to give a proof by myself.

[sturdy marsh]

and a --> [a] is a ring hom

use first iso thm

[dim escarp]

Why did my prof's module lecture have to be looking so verbose and hard

[patent harbor]

@opal osprey
Let us take the case of groups, for instance.

Claim 1.
Let (X,•) be a group. Let R be an equivalence relation on X. One would like to define a group structure on the quotient X/R, so that this particular natural map, f:X—> X/R st. f(x)=[x] is a group homeomorphism.
However this is not always possible for any equivalence Relation. The product may not generally be well defined. The conditions on R for when this is possible is exactly this: [e] is a normal subgroup (e denoting identity) and xRy iff x•y^{-1} is in [e] for each x, y in X.

Claim 2.
One might call this the theorem of divisibility for groups (it generalizes our particular fixation on normal subgroups to something we might call “normal subsets”).
(To be added)

[south storm]

I have a question in regards to Boolean algebras, a book I own defines a Boolean algebra as a set B together with two binary operations +,• such that both are commutative, distributive over each other, B contains identity elements 1 for • and 0 for + and for each element a there exist an element a‘ such that a+a‘=1 and a•a‘=0. Then it states that each formulae in a Boolean algebra can be checked by looking at all possible assignments of 0 and 1 to the variables. Why is that the case? Thanks for anyone who answers.

[delicate bloom]

because you're brute force checking it

[south storm]

But you aren’t checking all cases, what if it contains other elements than 1 and 0?

[patent harbor]

maybe then it wouldnt be called a boolean algebra, but a heyting algebra

[south storm]

I‘m not sure what you mean

[delicate bloom]

yeah doesn't sound like a boolean algebra if it has other elements than 1 and 0

[south storm]

That‘s how it’s defined in my book tho, and that‘s also the definition I found on Wikipedia when looking this up.

[south storm]

It even gives an example of a Boolean algebra consisting of four elements in the exercises

You looked up the branch Boolean algebra

Not a Boolean algebra

https://en.wikipedia.org/wiki/Boolean_algebra_(structure)?wprov=sfti1

[delicate bloom]

from the page you linked

[south storm]

And two elements

Not of two elements

[delicate bloom]

you sound like you're conflating number of variables with the number of values the variables can take

but I'm not sure what you're thinking

[south storm]

,rotate

[south storm]

This is undoubtedly the definition stated in the book, and what I’m asking is regarding this definition why you can check all cases.

[chilly ocean]

Surely you can admit

Even if you only have seen Boolean algebra refer to the 1’s and 0’s

Surely you can admit there could be generalizations

Especially

[chilly ocean]

When shown examples

[south storm]

Why did you guys get into a fight with me and leave why can you check for only 0 and 1s?

Nvm you back

[delicate bloom]

sure, the example was shown after I said that though lol

[south storm]

You could have just read the wiki article and I mentioned the example before you said it, but nvm that why can you check only for 0 and 1s?

[chilly ocean]

[south storm]

Wdym

Anyways if someone answers pls ping, the question is #groups-rings-fields message

[golden pasture]

wdym all possible assignments of 0 and 1

[south storm]

Like if you have x+y•z‘=1 then you can see that‘s true by plugging in all possible values for x,y,z being 0 or 1

This example isn’t true tho

[golden pasture]

evidently not

boolean algebras are lattices containing 0 and 1 + some other properties

[golden pasture]

could send a screenshot of the book?

cuz the statement is obviously false

[south storm]

Well maybe a typo or I’m misinterpreting one sec let me send

[south storm]

In the line with see Problem 18.15

[golden pasture]

hm

i feel like it implies that if you come up with some statement

[south storm]

It then gives an example of doing that lol

[golden pasture]

you can verify it by plugging in 0 or 1

instead of trying all possibilities

[south storm]

Maybe who were writing this didn’t know what he was talking about that‘s in a small section in the end of the book

[golden pasture]

this does makes sense tho

[south storm]

Wait how

[golden pasture]

like jus check if it works for the boolean algebra over {0,1}

[south storm]

Yeah my question is why you can check it for only the specific B algebra and it‘ll hold for any

[golden pasture]

hm i cant quite think of a proof rn but intuitively feels true lol

[south storm]

I mean kinda the intuition behind this I feel like is just set theory and I don’t see how it would

At least how the books presents it, it seems like a generalization of the algebraic structure in set theory with unions and intersections

[golden pasture]

you can realize every boolean algebra in terms of sets

[south storm]

And like checking for 0 and 1 would only be checking for the empty set and the „universal set“ if you‘d be working in some weird set theory with one

[golden pasture]

(or even better in terms of clopen sets of some compact hausdorff topological space)

[south storm]

Hmm I still don’t feel convinced do you know maybe where I could find a proof?

[golden pasture]

hm not too sure either :p the only place i saw boolean lattices is in AM exercises lol

[south storm]

What does AM stand for?

Algebraic model theory lol can’t think of anything better but I would have no idea what that’d would be about

[delicate bloom]

atiyah mcdonald

commutative algebra book

[south storm]

Ah alright

Apparently algebraic model theory is actually a thing https://www.springer.com/gp/book/9780792346661

Looks fun

[tiny pagoda]

Model theory is already somewhat of a cross between algebra and logic

[gritty latch]

if i have the subring S of Q of elements m/n, m,n integers and n is odd, and an ideal I in S of the form m/n, m,n integers m is even, how many distinct elements does the quotient ring have

the answer is 2 but no idea how to get there

[gritty latch]

<@&286206848099549185> any help?

[opal osprey]

If your conjecture is that there are only two elements in the quotient ring.

So I think the way to go would be "Suppose, without loss of generality, that you have x \in S/I; then, if x≠0, we will show that x=1"

So you suppose that you have an x that is not in the equivalence class of 0

You should prove that it's in the equivalence class of 1.

Or vice versa.

[gritty latch]

why 1 tho?

[opal osprey]

Because every ring with two distinct elements necessarily only has a zero element and an identity.

So if your conjecture is that this quotient ring S/I only has two elements.

Therefore when you pick an element in S, either it's related to 0 or to the identity.

[inner acorn]

Let Z2(G) = { x : [g, x] is in the centre of G}
I want to show this is a normal subgroup of G, but I'm somewhat struggling.
(I've shown it's a subgroup)

Here [g, x] = g'x'gx is the commutator

[golden pasture]

g is some arbitrary element?

[wispy scaffold]

@inner acorn Write down [g,hxh'] and try to find [h'gh,x]

[inner acorn]

I figured it out

xD but thank you

[wispy scaffold]

lol yw

[inner acorn]

I was being derpy, and was determined to try and use gH = Hg
but after playing with a'Ha, the result naturally fell out

[next obsidian]

I feel like this subgroup should probably be characteristic by virtue of how its defined

[cloud walrus]

Have a Banana Bitch

[vestal snow]

Never mind, I figured it out

Does anyone have suggestions on where to learn about Witt vectors from?

[prisma ibex]

I found Rabinoff's "The Theory of Witt Vectors" (https://arxiv.org/pdf/1409.7445.pdf) useful when I was learning about them

[inner acorn]

another fun question, given a group presentation <x, y | y'xy = x^2>
how would I find a free sub-semigroup of rank 2?

[stoic rose]

The semigroup generated by any element?

[vestal snow]

@prisma ibex Thanks, I'll check it out

[inner acorn]

@stoic rose oops, corrected xD

actually, it's been more than 15 minutes

<@&286206848099549185>

[stoic rose]

What is y'?

[inner acorn]

y' is y^-1

[chilly ocean]

If a < b and c < d, a + c < b + d is a theorem or axiom in the reals?

[oblique river]

it's a theorem

[chilly ocean]

oh ok, what is it called? I am assuming it's a family of theorems

[oblique river]

I don't think it has a name

I would call it "how inequalities work"

[scarlet estuary]

wait

is it a theorem?

i mean every axiom is immediately a theorem but

you know what i mean

if youre defining ordered fields the canonical way, typically you have the axioms:
a < b implies a + c < b + c
0 < a and 0 < b implies 0 < ab

so i guess it is a theorem under these axioms

but a very easy one to prove

if a < b and c < d, then a + c < b + c and b + c < b + d, so a + c < b + c < b + d

[latent anvil]

i always assumed you were able to prove this but im glad to have confirmation nami

[scarlet estuary]

[chilly ocean]

[latent anvil]

A friend posed a question to me and I haven't thought about it that much

For a finite group $G$ define $\mu(G) = |G| - \sum_{H < G} \mu(H)$

[cloud walrus]

shamrocK^n(X)

[latent anvil]

so it's the unique operator satisfying $|G| = \sum_{H\leq G} \mu(H)$

[cloud walrus]

shamrocK^n(X)

[latent anvil]

Compute $\mu(S_{2021})$

[cloud walrus]

shamrocK^n(X)

[latent anvil]

So far I've done some small examples. $\mu(C_p) = p-1, \mu(C_6) = 2, \mu(S_3) = 0, \mu(Q_8) = 0$

[cloud walrus]

shamrocK^n(X)

[latent anvil]

I thought maybe it vanished on noncyclic examples but unless I screwed up my computation $\mu(A_4) = 2$

[cloud walrus]

shamrocK^n(X)

[latent anvil]

Anyone want to think about this problem with me?

[stoic rose]

Hum, it's interesting

[latent anvil]

It seems tricky

He sort of suggested you should be able to come up with a clean definition of μ

And then prove it satisfies to recurrence, so deduce μ has that definition

[stoic rose]

I could try to write some sage code to compute it for small groups

[latent anvil]

Good idea

I was doing it by hand but that's error prone

I think you should be able to compute μ on abelian groups also

The subgroup structure of abelian p groups is easy

And if G, H are of coprime order subgroups of G×H are products of subgroups of each

[stoic rose]

Hum I think for cyclic groups it's phi(n)

[latent anvil]

that's plausible for me, by mobius inversion

Not sure if what I said is exactly right

[stoic rose]

That just follows from formula for sum of phi(d) over divisors d

[latent anvil]

oh yeah lol

okay yeah what am I saying, mobius inversion isn't good here

[next obsidian]

joe pap

[latent anvil]

hi chm

[next obsidian]

joe papa

Also I think like

[latent anvil]

How are reu apps going?

[next obsidian]

If you sum mu(H) over all the subgroups

you get this telescoping sum

[latent anvil]

yeah, I had that thought earlier

[next obsidian]

no that's for prime power

nvm

lol

[latent anvil]

But it depends on the subgroup structure of G

Oh I meant in general

[next obsidian]

oh sorry this was for cyclic

[latent anvil]

you can like, expand it all out

[next obsidian]

yeah for general I don't think so

Also I'm trying to show that the functor separated presheaves to sheaves actually gives a sheaf

so it's not started yet

it's...

blech

most of the work so far was really easy and intuitive if you just pretend the open covers are like open covers you know

but this one is a bit weirder

I think maybe if I pondered the equivalence relation longer it seems like trying to do some like stalky locally equivalent thing or something but

¯_(ツ)_/¯

[delicate bloom]

I know this isn't the formula but I'm imagining the formula is something like this $$\mu(G) = \sum_{H \le G} (-1)^{f([G:H])} |H|$$

[cloud walrus]

Merosity

[latent anvil]

Yeah, me too

Eer

Kind of

[delicate bloom]

for some f, like a mobius function or something not sure

[latent anvil]

hmm

[delicate bloom]

actually how are we counting subgroups here

like if we have two of the same subgroups

do we count them as one or two

[stoic rose]

Wait I think I got the answer

It should be 0 for all non cyclic groups if I'm not mistaken

[latent anvil]

I thought so too but then I thought I found a counterexample....

[latent anvil]

We're not counting up to isomorphism, if that's what you're asking

[stoic rose]

Phi(n) is the number of generators of the cyclic group and each element of G generates a unique cyclic subgroup

[latent anvil]

Can I sanity check this first?

[stoic rose]

So this is satisfied

[latent anvil]

I think A4 has subgroups e, C2, C2, C2, Klein four, C3, C3, C3

[carmine fossil]

A4 has no C3

[latent anvil]

Wait what?

3 divides 12?

[carmine fossil]

Mb

I read as S3

[delicate bloom]

yeah that's what I was asking

we might as well though, just put an extra factor to count the subgroup frequency

[stoic rose]

There are 4 C3s

[latent anvil]

oh!

of course

and that accounts for my 2

nice okay it's more believable now

okay so

prove it by induction

all terms in the sum corresponding to noncyclic subgroups vanish

now sum over cyclic subgroups

you can now sum over elements

with a multiplicity term corresponding to phi(|x|)

does that kind of make sense?

[stoic rose]

Each element generate a unique cyclic subgroup

So the sum of number of generators of all cyclic subgroups is the number of elements

That's as simple as that

[latent anvil]

yeah, makes sense to me!

I was trying to go the other way

but you can just prove the recurrence

so... yeah

[mellow forge]

Is the langlands program still chugging along?

[vital quail]

yup

[river nebula]

Hi, we know that in commutative rings when two ideals are coprime (comaximal), then their intersection equals their product. I wanted to know under which rings does the converse holds.

[vital quail]

Well look at the 0 ideal

[river nebula]

I know that the converse holds in ring of integers Z, and not in Z[x]. But does it holds in every PID? can we say that the ring is a PID or a field if it holds?

Yes, what about the 0 ideal?

[vital quail]

Intersection equals product

Moreover in Z, (a)(b) is what, and (a) cap (b) is what? Then if they're equal what do you get?

[stoic rose]

Right, for the zero ideal its intersection with itself equals its product with itself but it's not coprime to itself

So you need at least to ask for nonzero ideals

[golden pasture]

the qn is what type of rings does the converse hold

in UFDs it should hold i believe

but does that imply a UFD idts

[stoic rose]

The converse never holds unless you specify nonzero ideals

[golden pasture]

wait (0) works i tot?

oh mbmb misread

[urban acorn]

I've been thinking about this problem for a while, I didn't get very far in terms of any kind of result, but I do enjoy this problem, I do feel like I got some progress in that things feel a bit clearer now.

Computing this for small groups is indeed a good idea.

[urban acorn]

I computed it with sage up to S5, and it's 0 from S3 and on, it seems.

[chilly ocean]

Did not expect that

[urban acorn]

I'm tired right now, so I may have done something wrong in my code, but it yields 0 for A4.

It might just vanish for non-cyclic groups.

[prisma ibex]

Yes, the Langlands program is still EXTREMELY active and there's some really big things that I expect will be coming in the next decade or so

[latent anvil]

This is not true

μ(Cp) = p - 1

[urban acorn]

I meant I computed the symmetric groups up to S5

not all of them

I actually used the fact that mu(Cn) = phi(n) to simplify the computation

@latent anvil

[latent anvil]

Ah gotcha

[sturdy marsh]

mu vanishes on all nonabelian groups of order pq

[urban acorn]

given the subgroup lattice structure, mu can be expressed as a Z-linear combination of the orders of the subgroups

it's something like you count the amount of upward paths from H to G and then that's the coefficient of |H|

I mean, counted as +1 or -1 depending on whether there's an even or odd amount of edges respectively

something like that, not exactly

[latent anvil]

(jfyi, someone solved it earlier, so don't scroll up if you don't want spoilers)

I was also wrong in my computation if you saw that, the correct version is μ(A4) = 0

[urban acorn]

I skimmed the chat history earlier

but for whatever reason I missed that

damn

I don't know how I didn't think to just check the recurrence relation with the hypothesis that mu vanishes on non-cyclics

I gotta say that's kinda really fucking awesome though

[wind cypress]

My abstract algebra course/module starts in 9 hours 🙂

Rings and fields are the first topics

[nova plank]

Have fun

[wind cypress]

I already do

I have the lecture material already

And I know the teacher, he is good

[past topaz]

Abstract algebra, is that the same thing as modern algebra? I'm at UC Davis, and we don't have Abstract Algebra, but we have Modern Algebra

[oblique river]

yes

[wind cypress]

The course/module I am taking is just called Algebra actually

Algebra 1, meaning rings and fields to be more specific.

[shy thunder]

If we have G/H iso K and G/K iso H, then is it true that G iso H x K?

[next obsidian]

No

Take G = Z/4Z and H = K = Z/2Z

[sly crescent]

Is this the right place for symmetry group discussion?

[latent anvil]

Yup

[sly crescent]

What are orders of the symmetry groups of birectified 7-orthoplexes with the following polytopes inscribed in them?

1. Rectified 7-simplex
2. 3_21 polytope

Wait, there might be multiple ways to do #1.

[carmine fossil]

Yes?

[odd abyss]

Describe all ring morphisms

How to do that

?

[carmine fossil]

A ring morphism maps 1 bar Z/70Z to some a bar in Z/35Z

And the ring morphism is defined by the choice of a bar

Do you understand that?

[odd abyss]

yes

and ?

[carmine fossil]

Now,just check if this map is well defined

[odd abyss]

I do not know how to check that

[carmine fossil]

For example 0 bar and 70 bar are the same in Z/70Z

70 bar =(1 bar)+(1 bar)+(1 bar)..=70(1 bar)

So a bar + a bar...=70(a bar)(in Z/35Z) should be (0 bar)(in Z/35Z)

[odd abyss]

But didn't you show that it was well defined here?

@carmine fossil

[carmine fossil]

You mean this?

[odd abyss]

you did not show that it was well defined here : think

[carmine fossil]

Let's say a bar= b bar,then b=a+70k for some k

[odd abyss]

what 70 bar mean ?

i don't understand

what bar mean

I'm freench and I don't understand english

very well

[carmine fossil]

Equivalence class of 70 under the equivalence relation aRb iff 70 divides a-b

[odd abyss]

Ok you lost me

No matter

thank you

[carmine fossil]

Do you understand what Z/70Z means?

[odd abyss]

70 bar is that

[carmine fossil]

I mean,like do you understand the elements of the group Z/70Z

[odd abyss]

yes

what I don't understand is "BAR"

[carmine fossil]

The elements are equivalence classes,right?

[odd abyss]

You write something in literal writing which can be written in mathematical writing

[carmine fossil]

(sorry,was being lazy)

[odd abyss]

English is not my mother tongue, I do not know the equivalent mathematical words

[carmine fossil]

anyway,I will restate it properly

Let $\bar{x}=\bar{y}$(implying x=y+70k,for some integer k) and let the ring hom be $\bar{1} \mapsto \bar{a}$ , then $\bar{x} \mapsto x \bar{a}$ and $\bar{y} \mapsto y \bar{a}$
Now,for the map to be well defined,
$ x \bar{a}= {x+70k} \bar{a} \implies 70k \bar{a}=0 \bar{a}=\bar{0}$

[cloud walrus]

MoonBears-D-

[odd abyss]

ok thank you

I understand

[carmine fossil]

This case is simple, because that's true for any choice of a

Try finding the ring homorphisms between Z/20Z and Z/3Z

[odd abyss]

that is a bar ?

[carmine fossil]

Yes

[odd abyss]

OK

I will try for 7/20Z and 7/3Z

[lone violet]

If i have $A_4$ the set with the identity, 3-cycles and the product of two transpositions and i want to prove $A_4$ is a semi group of $S_4$ to prove $A_4$ is closed i have to verify element by element or there is a way to generalize?

[cloud walrus]

alef0

[lone violet]

subgroup* sorry

[carmine fossil]

Just use the fact A_4 is the set of even permutations

[prisma ibex]

Right, use the fact that A_4 is the subgroup of even permutations (you're implicitly using the fact that the composition of even permutation is even)

Maybe a slicker way to say this, consider the group homomorphism S_4->Z/2Z that sends a permutation to its sign: it sends even permutations to 0 and odd permutations to 1

The A_4 is the kernel of this group homomorphism, so it's automatically a subgroup

(in fact this automatically tells you that A_4 is a normal subgroup of S_4)

[lone violet]

Oh right, thank you guys!

[molten silo]

help

[carmine fossil]

What have you tried?

[molten silo]

i dont know it means

[prisma ibex]

7-cycles are elements like (1 2 3 4 5 6 7), or (1 3 5 7 2 4 6) and so on

[molten silo]

so permutations

[prisma ibex]

Not all permutations

(1 2)(3 4 5 6 7) isn't a 7-cycle

(it's a composition of a 2-cycle and a 5-cycle)