groups-rings-fields

[solar kernel]

Finite stuff is funky sometimes, but goddamn is it beautiful too

[cloud walrus]

andre

[fading acorn]

use noggin

check axioms

[indigo forge]

what is that?

[fading acorn]

your head

[tardy hedge]

thinking is hard and poo

[indigo forge]

I don't have that in my kinder

[indigo forge]

I am talking about notation

we are not seeing the same thing

I don't know what Z_n is

[fading acorn]

is Z_n a ring
I'm seeing for notations

wow wonderful

[thorn jay]

finite stuff is so nice we can understand all finite algebraic structures reasonably well

[solar kernel]

$\mathbb{Z}_n={0,1,...,n}$, technically its just a set, but we normally think of it as a ring. It could be a group under addition, a ring under addition and multiplication, or a field if $n$ is prime.

[cloud walrus]

Makon W

[thorn jay]

of course complete classification is impossible (hell, for algebraic structures on a set of three elements it's nearly impossible) but you can derive a lot of super deep and strong theorems

[indigo forge]

Okay

thank you

[tardy hedge]

hello there

[south patrol]

Given what? Being a ring is usually structure rather than a property

:trollface:

[fading acorn]

it also could have been n-adic integers and not integers modulo n

[south patrol]

Yes that's what the notation means

[thorn jay]

hoii

[south patrol]

I read this as holi and wondered what the festival had to do with this

[thorn jay]

god forbid i be joyous lol

[vapid vale]

let n be a prime

[tardy hedge]

why?

i am not letting n be a prime

[rocky cloak]

Let n be in the prime of its life

Living it up in its glory days

[karmic moat]

n is peak

[thorn jay]

let 𝔭 be any ideal

[karmic moat]

too far...

[thorn jay]

let 𝔭 be an I-primary ideal of a nonunital commutative ring G

[vapid vale]

commutative P-algebra G

[thorn jay]

oh ew

[vapid vale]

pee algebra

[thorn jay]

Let V : T → E be a linear transformation of vector spaces over the field W

[vapid vale]

Let poop be an ideal

ok i'll stop

[thorn jay]

🙏

[noble nexus]

For all n>0 there exists an m>0 such that |x-y|<m implies |f(x)-f(y)|<n

[sonic coral]

If a polynomial is reducible mod p, then does the degree of the factorization mod p tell us how the polynomial would have to factor degree wise over Q?

[formal laurel]

Not really, consider x^2-1 for p=1 mod(4)

If somehow that factorization stays consistant over every prime then you might have a case.

[sonic coral]

that factors linearly over Z/5 and it also does over Q though?

I guess i should be more careful. I am not claiming that reducible mod p implies reducible over Q, but that if it was reducible over Q, it would have to factor the same degree wise as it would mod some prime p for which it was reducible

[formal laurel]

shit, it was x^2+1.

I push - instead of +

[sonic coral]

see my second message

[formal laurel]

not even then. Just take x^4-1.

This splits into linear factors for p=5 and for Q it only splits into 2 linear polynomials and one quadratic polynomial

[elfin wraith]

Using mathfrak is the worst thing about this by far

[formal laurel]

But like, if you take a certain polynomial f irreducible over Q that splits into linear factos for every prime p. Chebotarev density theorem implies that f necessary is linear.

[thorn jay]

i take pride in this

[solar shore]

hey y'all, anyone got some ideas on how to show that aut(D_8) is isomorphic to D_8, using the fact that D_8 is isomorphic to a normal subgroup of D16?

i have so far that aut(D_8) is at most 8, but im having a bit of trouble showing the distinct permutations

actually, let me refine my confusion. i can actually show the isomorphism, but im having a bit of trouble what the actual elements are?

[tough raven]

Ah.

[tough raven]

This embedding gives an action of D16 on D8 by automorphisms (namely by conjugation). If you show the kernel of this action has order 2, then the image has order 8, so you have 8 distinct automorphisms of D8.

[crystal vale]

i got confused about ambient field, so if i have set X of some polynomials over F, then what does it mean by splitting field of X? it is given that K is extension field of F and if every polynomial f in X if it splits over K and if K = F(Y), where Y is the set of all roots of all f in X, then K is splitting field

[quiet pelican]

It is the* extension field of F such that every polynomial of X splits in it, and the field is generated over F by the roots of elements of X
*the extension field, as every such field can be proven to be isomorphic

[crystal vale]

so if K is algebraic closure of F, then how do i show K is splitting field of set of all non-constant polynomial over F?

i think K is algebraic closure of F, so every polynomial over K splits in K[x], so every polynomial over F also splits in K[x]. and every element in K is algebraic over F so F(K) = K

[crystal vale]

so when we are saying its generated by the roots of elements of X, so we already know the roots of X?

[quiet pelican]

It’s “the elements of the Extension field which are roots of X”

[crystal vale]

okay

[crystal vale]

is this argument correct? F(K) = K, and every k in K has minimal polynomial in F[x] so F(Y) = K, where Y is the set of all non constant polynomial over F

[round hull]

yo whats up with the third diagram

why are all the Fs identically oriented

and how exactly does the coloring correspond to generators

[rocky cloak]

Looks to be a mistake. The red and green lines are supposed to be mirror symmetries

[rocky cloak]

For the diagrams with a red and a green dot, the group is generated by a neighbooring red and green reflection

[round hull]

k that makes a lot more sense

thx for the sanity check

[round hull]

hm wait but in the fourth diagram if you start from horizontal red and a diagonal green how do you know to color vertical red

oh because when you reflect the red line about green it becomes vertical?

hm

[tardy hedge]

Basic question but like, for a map f: M->N and we replace M and N with isomorphic copies M’ and N’, so we now have a map f’: M’->N’, the kernels and images should be isomorphic, but how do i argue that formally

Ok nvm

Theyre just submodules so its preserved under the isomorphism

[rocky cloak]

You can use that kernel and image are functorial

[tardy hedge]

Can u pls elaborate what that means here?

[rocky cloak]

Consider any commutative square
M -f-> N
v v
M' -g-> N'

Then you get a unique map between the kernel of f and g defining a functor from the arrow category to the original category

And functors send isomorphisms to isomorphisms

[tardy hedge]

Whats the arrow category? Objects are squares like that?

I actually have to get to studying cat thoery. I have an assignment due monday

[knotty badger]

Objects are arrows, morphisms are commutative squares

You can also view it as the functor category [2, C] where

[tardy hedge]

So this is the arrow category of RMod

[knotty badger]

Yeah

[copper kestrel]

very lost on how one would do this :')

like im guessing its true

maybe not tho?

since homomorphisms always map identity to identity

but maybe it is true 😭

[rotund dragon]

some definitions don't require sending multiplicative identity to multiplicative identity for a ring homomorphism

[copper kestrel]

hm

[rotund dragon]

if your definition doesn't require that, try thinking about letting R = Z and R' = Z x Z

if your definition does require phi(1) = 1, then yeah you're right

[copper kestrel]

hmmmmmmmmm

maybe finding a homomorphism such that phi(1) = 1' x 0'

no thats dumb

[copper kestrel]

well i can prove it does for multiplication

[copper kestrel]

what should i look for in the definition to see if its that

bc my definition is that

f(a + b) = f(a) + f(b)
f(ab) = f(a)f(b)

bc we get that like 0 is mapped to 0 from additive identity, and i can show that 1 gets mapped to 1

[rotund dragon]

0 -> 0 is always true. 1 -> 1 isn't, though

[copper kestrel]

oh

[rotund dragon]

the most you can say is that f(1) is an idempotent

[copper kestrel]

ohhhh

i fear im stupid

ty

[rocky cloak]

Maybe not as dumb as you think

[rotund dragon]

ring homs not preserving the multiplicative identity lead to some funny things to consider --- i also wouldn't say you're dumb

[copper kestrel]

i just assume every idea i have is wrong at this point so i'll try the 1' x 0' idea

[rocky cloak]

Probably better to assume your ideas are correct until proven otherwise. Or at least be neutral about it's correctness

[lime badge]

I assume all my (algebra) ideas are wrong unless jagr says otherwise

[copper kestrel]

same

there are only 4 ring homomorphisms from Z x Z x Z to Z right?

[quiet pelican]

No

[copper kestrel]

okay

[quiet pelican]

Wait ring homs
Errrr

Yes

[copper kestrel]

oh

[quiet pelican]

Idempotents map to idempotents

[copper kestrel]

sick

[quiet pelican]

(This gives that there are at most 8, then you check that 4 of them don’t work)

[copper kestrel]

we unfortunately have not proved that idempotents map to idempotents

[quiet pelican]

It’s trivial

a^2 = a implies f(a^2) = f(a) so f(a)^2 = f(a)

[copper kestrel]

hm

arent there only 4 idempotents in Z x Z x Z tho?

[quiet pelican]

There are 8 idempotents

But their images are determined by the images of some subset of 3 of them

[copper kestrel]

oh wait i see

[copper kestrel]

slightly confused about this, why are we only proving the multiplication version of this?

shouldnt we do both addition and multiplication for ring homomorphisms?

or is it that its just trivial

[quiet pelican]

I’m only doing the bare minimum needed to get the number of candidates down to a workable amount

[copper kestrel]

fair

[rocky cloak]

Why use addition if multiplication do trick

[copper kestrel]

okay now im stuck again with showing that there are only 4 homomorphisms

like i showed the idempotents map to idempotents

[wraith cargo]

this is only true for integral domains!

[knotty badger]

isn't this part of the def of ring homomorphism

[wraith cargo]

hint: think about Z/6Z :3

[thorn jay]

they're using nonunital rings qwq

[knotty badger]

ah ok

[quiet pelican]

It is true unless you’re the sort of insane person to use non-unital rings :3
(or worse, use unital rings and not require that homomorphisms preserve the identity)

[rocky cloak]

Not even true for integral domains if you don't require it, as 0 exists

[wraith cargo]

Okay I had a bad proof in my head

Ignore I said anything

[wraith cargo]

Well ig say nontrivial

[copper kestrel]

i figured it out :3 unfortunately i am still pondering another question :')

[elfin wraith]

My adv linear (which was also intro to rings) course had unital rings which didnt require the identity, much to the dismay of litterally everyone involved in the course lol

It was taught by someone new every year for a good while, none of which had written the notes, and everyone hated it. I think every algebra class I took had a line pointing out the difference in definition from honours algebra

[copper kestrel]

i still havent figured out this stupid problem about finding the number of homomorphisms from Z x Z x Z to Z 😭😭😭😭

like i know theres 4, but i can't prove that theres 4 from the 8 idempotent elements of Z x Z x Z

i might just not solve it and submit the hw

bc i literally understand 0 parts of the solution i've been helped to create

[thorn jay]

do you know about generators

[copper kestrel]

yep, these are ring homomorphisms

[thorn jay]

oh rings lol

[copper kestrel]

i just cant describe the 4 ring homomorphisms

[thorn jay]

well, mapping each (1, 0, 0) , ..., (0, 0, 1) to an idempotent in Z gives you maps of abelian groups (sending (a, b, c) to af(1, 0, 0) + bf(0, 1, 0) + cf(0, 0, 1) ), and you gotta find out which of them are ring maps

(i.e. preserve multiplication)

[copper kestrel]

right and i dont know how to do that

i dont see how they dont preserve multiplication

especially like (1,0,1)

[thorn jay]

notice that any nonequal two of those basis elements multiplied yields zero, so at most one of them can be sent to 1

because else you'd get e.g. if both (1, 0, 0) and (0, 1, 0) are sent to 1:
0 = f(0) = f((1, 0, 0) * (0, 1, 0)) = f(1, 0, 0) * f(0, 1, 0) = 1 * 1 = 1

[copper kestrel]

i don't understand this

[thorn jay]

suppose you have a ring homomorphism f : Z x Z x Z -> Z. Ring homomorphisms map idempotent elements to idempotent elements, so the elements (1, 0, 0), (0, 1, 0), and (0, 0, 1) must be mapped to either 0 or 1

consider e.g. the case that both (1, 0, 0) and (0, 1, 0) are mapped to 1. Then, it would follow:
0 = f((0, 0, 0)) = f((1, 0, 0) * (0, 1, 0)) = f((1, 0, 0)) * f((0, 1, 0)) = 1 * 1 = 1
which is a contradiction

hence (repeating this argument), at most one of these elements can be mapped to 1, and the rest must be mapped to 0. Of course, mapping everything to 0 is also an option.

[icy ridge]

How do I like "get good" at abstract algebra? I just finished a first course in group theory and i wanna learn in my free time

[thorn jay]

you do a heckin lot of it

[icy ridge]

I'm reading dummit and foote but its getting a lil boring so im wondering what all this is for

I started reading about rings a lil more

[thorn jay]

Dummit and Foote i've heard is very terse lol

for beginners

do you know what you want to do with algebra more? I think if you want cool applications Galois theory might be something to work towards

or representation theory of finite groups

(that is less an application but it is super beautiful)

[thorn jay]

this is in general a very good idea

rings and algebras will pop up all over the place

[icy ridge]

nah im still a baby

[karmic moat]

given the phrase "Let A be a B-algebra" is there a usual name for the implied map f : A -> B

something like "structure map"

[thorn jay]

structure map is what I've seen and use myself

[karmic moat]

W

[thorn jay]

(this makes B an A-algebra by the way)

[karmic moat]

oops yea

anyway

[thorn jay]

the based option is to define an A-algebra as a ring B with a map f : A -> B

[karmic moat]

that's the definition i always use but some literature will just say "Let B be an A-algebra"

[thorn jay]

hmm, I think you should read about rings then, yes

[karmic moat]

which is fine when the underlying map doesn't really matter

[vestal trench]

try aluffi if thats your complaint with d&f, his writing is very conversational. he has an undergrad book and grad book, both are good

[karmic moat]

i heard gathmann has nice commutative algebra notes

[thorn jay]

well it usually doesn't!

[karmic moat]

at least for rings/modules

[thorn jay]

the idea is to have a ring that has an A-module structure, and whether that's given by a structure map, or wtv doesn't really matter

[karmic moat]

yea

[thorn jay]

man I love abuse of notation

[karmic moat]

sometimes i do need to reference the map tho and i've just been saying "Let f : A -> B be the map which gives B an A-algebra structure"

but i was just thinking that was too wordy

[thorn jay]

yeah you can just say "let f : A -> B be the structure map of the algebra B"

or "corresponding structure map" idk if people know of it then it should be clear

[karmic moat]

yea i mean for now ive just been using it in my solutions document lol

hasn't really seen the light of day

[thorn jay]

🔥

many such cases

[copper kestrel]

chat i did it

i finished the homework and kinda understand it

i get a ton of more exercises tmr tho 💔 and DRP exercises

[crystal vale]

if K is splitting field of f in F[x], and we know that K = F[x1,..,x_n] where these x_1,.., x_n are all roots of f in K.

if extension is different say E and say f is split over E with roots y_1,.., y_n, then F = [y1,..,y_n] so these F[x1,..,x_n], F[y1,..,y_n] are different right?

its all depends on ambient field

[novel star]

let R be a ring containing the field K as a subring. Suppose that R is a finite dimensional vector space over K under the ring multiplication. why should R be a field? (i js need a hint)

[velvet hull]

It does not seem true…. Consider R[x] / (x^2)

Are you missing some assumptions?

[novel star]

lemme check

i felt like it seemed too strong

o wait yeah

R is an integral domain

if its any help

its the third exercise of the chapter on modules in lang

[velvet hull]

Multiplication by any element in the ring is a K-vector space homomorphism

The integral domain condition guarantees that the kernel is trivial, and as it is finite dimensions it is an isomorphism

[novel star]

ah

[south patrol]

Based way is to say let A -> B be an A-algebra

[pliant forge]

why is there an arrow from 1 to 0 (in simplicial category)?

[knotty badger]

Hm I think it should be the other way around

[pliant forge]

yes because its initial/terminal

[knotty badger]

Yeah

[rocky cloak]

The 0-simplex is [0], so you have exactly one map from 1 = [0 < 1] to 0 = [0]

[south patrol]

It's not

The textbook and Jagr are correct

[pliant forge]

so are you correct or the textbook?

[south patrol]

Which category is this now

1 being terminal means you definitely are not talking about the simplicial category anyway

[knotty badger]

Category of finite ordinals and order-preserving maps presumably

[pliant forge]

i assumed it was called that from the title

[knotty badger]

Where 1 is terminal and 0 is initial

[south patrol]

Ye this is a form of the simplicial category

Oh by 0 it means empty ig lol

[pliant forge]

yes

[knotty badger]

Yes

[pliant forge]

so i dont see how theres any map to 0

[south patrol]

Then this is not the usual simplicial category

It is the augmented one if you want

[knotty badger]

There isn’t

Apart from identity

[pliant forge]

ofc

[knotty badger]

(This is sometimes called a “strict” initial object)

[south patrol]

But also this may be the notation here being annoying, as usually the objects of the simplicial category are called [n], which has n+1 elements

[knotty badger]

I’m more used to [n] having n elements

[rocky cloak]

The description of the sigma-maps only makes sense if n = [0<1<...<n] so that n is an n-simplex.

Could be that they are inconsistent with notation, I havent5 read this book

[knotty badger]

That’s convenient for the process of taking nerves for example

[south patrol]

Sure but that the convention in homotopy theory is n+1

[knotty badger]

Objects are maps from [1], morphisms are maps from [2]

I see

[south patrol]

I would say like delta^1 for maps and delta^0 for objects ye

[knotty badger]

Presumably to line up with the labelling of the topological simplices

Yeah

Like I would rather say the arrow category is [2, C]

Which has [2] being the one with 2 elements

[south patrol]

Yee 2 is fine, just [2] feels more reserved aha

[knotty badger]

Oh ok

[pliant forge]

yeah i was just unsure if outside of the defined maps there was something i was missing from its description. i just imagined it being the empty set which looks to be a sane assumption

[knotty badger]

Sometimes the empty set is the (-1)-simplex or smth

I think this is cause the dimensionality of the simplex + 1 = number of vertices of the simplex

[south patrol]

Yeah to me it seems they're just being inconsistent and would be clearer if they write [n] or smth. What they said above definitely is inconsistent with this

[knotty badger]

So you get a bunch of potential OBOEs

[south patrol]

Wdym by this

[knotty badger]

“Off By One Error”

[south patrol]

Woodwind instruments

Oh lol

Thanks

[knotty badger]

I mean a surprise gang of oboes would also be scary

[rocky cloak]

If you want F to literally be a subset, just replace whatever the image of F in K is with F and there you go

Like if i: F -> K is the embedding just replace K by
K \ i(F) union F
with the obvious multiplication and addition

[crystal vale]

Yes the professor did the same thing, but i don't get why I am doing it so? If I already have one extension in which f has root so what's the problem?

It is from Lang's algebra

[rocky cloak]

If your definition of extension has F being a literal subset, then F needs to be a literal subset.

If your definition is different you can do it differently

[crystal vale]

Okay

[crystal vale]

So the author defined that a field extension K \subset L is separable when the extension is algebraic and the roots of the minimal polynomial of every l in L are all simple, what is meant by simple?

Does it mean that multiplicity should be one ?

I think no

Got it

[rocky cloak]

Yes

[crystal vale]

can anyone tell me what does it mean of 2 in simple language?

if choose l in L, and f is minimal polynomial of l over F, so all roots of f is simple, so all roots or the roots which are in L?

because how do i know other roots of f if it is not in L?

[barren sierra]

(ii) means that the minimal polynomial of l in L has no repeated roots in any field extension of K

so in particular, in the algebraic closure of K

so there may be roots not in K, that's fine

So for example, take L = complex numbers and K = reals

then the minimal polynomial over the reals of i in the complex numbers is x^2 + 1

[crystal vale]

yes

[barren sierra]

both of its roots are in the complex numbers, not the reals

that's fine

the point is that all the roots of x^2 + 1 are distinct

[crystal vale]

i see, thanks

why it is suffice to prove for l? I think minimal polynomial p, is minimal polynomial for all each roots so it is same argument

are they two equivalent? I can say splitting field is algebraic extension

[barren sierra]

I think minimal polynomial p, is minimal polynomial for all each roots so it is same argument
yes this is correct

[barren sierra]

what is the definition of a splitting field, that would probably answer your question

[crystal vale]

so it is clear splitting field is algebraic over F

[barren sierra]

yea so the question is what is your set S in definition 1.3.1

[crystal vale]

let me think

i think take the set of all min(l, K) for all l in K, i.e, set of all minimal polynomial of all l in L

[barren sierra]

Yea, so S just has to contain min(l, K)

I think

S could be bigger and I think that's fine

[crystal vale]

yes

but if K is normal by definition 3.23 how do i show it will satisfy the ii) condition of 1.3.1?

[rocky cloak]

It won't necessarily. Being normal and being seperable are two different things

[crystal vale]

No

I am asking different thing

[rocky cloak]

What are you asking?

[crystal vale]

I am asking if it is normal by definition 3.23 then is it satisfy the condition of 1.3.1 ?

[rocky cloak]

A splitting field will be algebraic yes

And have polynomials split

L/K is normal iff for every l the minimal polynomial splits iff L is the splitting field for some family of polynomials.

[prime sundial]

Does anyone have links to good examples of how abstract algebra is used in physics? Especially group theory

[crystal vale]

How do I show it, if L is the splitting field for some family of polynomials then for every l the minimal polynomial splits ?

[karmic moat]

these wikipedia articles

[prime sundial]

Thanks

[rocky cloak]

I think the easiest way is to go through the equivalence that any embedding of L/K into an algebraic closure of K has the same image

[lime badge]

Let k be a field and G a finite group.

Does the k-dimension of the center of the group algebra k[G] have any straightforward relationship with the order of the center of G, i.e. |Z(G)|?

I know that the center of G is the union of all the trivial conjugacy classes and I think the k dimension of the center of k[G] is the number of conjugacy classes of G, but I don't immediately see whether this implies anything about |Z(G)| vs dim(Z(k[G])

[rocky cloak]

I don't think there's any immediate relationship other than
dim Z(kG) >= |ZG|

[lime badge]

Well ok, |Z(G)| <= dim(Z(k[G]))

haha, indeed

Thanks!

[copper kestrel]

do the dotted lines in this structure mean "element of"?

also are the nonarrowed lines meaning subset?

oh my god it literally says its element of

i cant read

[granite badge]

ig it should be the right channel even though the level is pretty low

[cloud walrus]

ζnozzi

[granite badge]

we're asked to find the "iterates" values of

[cloud walrus]

ζnozzi

[granite badge]

but i don't know what that is and i can't find it😿

though the next question says prove that for all n in N

[cloud walrus]

ζnozzi

[granite badge]

could anyone help?

[quiet pelican]

$5^{*n} = 5 * 5 * \dots * 5$ where there are $n$ 5s

[cloud walrus]

micoi the group things

[granite badge]

thanks!

[lapis saddle]

Hi, I've been pointed to here, and I have a question. How do I prove this:
1)let H,H',K,K' Be subgroups of G where H' is mormal in H and so is K' in K.
Show that (H \cap K)•K', (H'\cap K)•K' and (K'\cap K)•K' are subgroup of G.

2. Also show that (H \cap K)•K'/(H'\cap K)•K' is isomorphic to (H \cap K)•H'/(H\cap K')•H' and show those quotients make sense
   I've done 1) easily but I'm having problems with 2 because I can't show normality

[quiet pelican]

Note that you want them to be normal in (H \cap K) \cdot K’ (for the first one, for example) and not G

[lapis saddle]

Yes

That I know, but how, when I do like gHg^-1 with g being in (H'\capK)•K'

I don't get anything that looks like is in H

Should I try to do the gH=Hg thingy, show that the equivalence classes are equal?

[rocky cloak]

One thing that can be useful is that if both N and Q are normal subgroups then so is N*Q

[lapis saddle]

N and Q?

You mean H' and K'?

[rocky cloak]

Well I mean arbitrary subgroups, but the things you want to show are normal are products of two things

[lapis saddle]

Yes

[rocky cloak]

In particular
(H\cap K') * H'
is the product of
H\cap K' and H'

[lapis saddle]

Yeah H' is normal but H\capK' is too?

[rocky cloak]

Well normal in H\cap K at least

[lapis saddle]

I mean yeah all the elements in H\cap K' are in K'

[rocky cloak]

Anyway, I guess the easiest way is just comparing gN and Ng

Like just write down an element of the top group and the bottom group and try to "commute" them

[lapis saddle]

Hm let me see

[solar shore]

man all the primes lol

[lapis saddle]

Yeah I hate it too but that's how they write it

[solar shore]

nah i get ya

the cdot’s the subgroup product right?

[lapis saddle]

Yeah it's the {hk s.t h is in H and k is in K}

[solar shore]

ie $(H \cap K) \cdot K’ = {xy \mid x \in H \cap K, y\in K’}?

o

man i type slow asf otp

[lapis saddle]

Yeah

[solar shore]

anyways yeah gN = Ng i guess?, g in (H cap K) * K’ and N = (H’ cap K) * K’

[lapis saddle]

I'm trying that rn

[solar shore]

any progress?

[lapis saddle]

I've taken an arbitrary element of the left coset and inverted it and basically gotten an element of the right coset

I think that's right

Correct me if I'm wrong

[solar shore]

show the process? im not sure that taking the inverse out of nowhere is correct

or do you mean like swapping the places of the elements

[lapis saddle]

Well like the intersection is a subgroup so the inverse must also exist in the subgroup

[solar shore]

right, but did you just say let xy be in the left coset, then consider (xy)^-1?

[lapis saddle]

Yeah, is that like wrong?

[solar shore]

yes, namely because cosets arent generally subgroups

so you dont know that if xy is in the coset gN, then (xy)^-1 is also in there

[lapis saddle]

Then Idk

[solar shore]

okay

lemme see if i can help

for notation’s sake im gonna let (H cap K)K’ be A and (H’ cap K)K’ be B

since we’re tryna show A/B is well defined, then B must be normal in a, ie, for every a in A, we must have aBa^-1 = B

and it is sufficient to show aBa^-1 is a subset of B

so proceed w that

take a in A, then a = xy for x in H cap K and y in K’

similarly, let b in B, so b = st for s in H’ cap K and t in K’

and consider aba^-1

can you show that that’s contained in A? try writing out all the elements and use normality to move elements around

[lapis saddle]

xysty^{-1}x^{-1} and we must show that that is in B, right?

Which is (H' \cap K)K'

[solar shore]

yes

[lapis saddle]

Wouldn't y^-1x^-1 be outside of K'

I don't think there's a way to write things in a 1 to 1 way

[solar shore]

it is at the moment

you need to use normality of K’ and H’ in K and H respectively to move the elements around

[lapis saddle]

How could I move elements around?

I vaguely remember that but I don't think you can do that here

[solar shore]

okay so we got xysty^-1x^-1

so lets try to shuffle all the K’ elements to the right

can you tell me where ty^-1 is?

[lapis saddle]

Both in K'

So K'

Because the operation is closed in K' as it is a subgrup of G

[solar shore]

perfect

now x^-1, where is that?

[lapis saddle]

H \cap K

[solar shore]

in particular its in K, correct?

[lapis saddle]

Yeah it has to

[solar shore]

now recall that K’ is normal in K right?

[lapis saddle]

Yes

[solar shore]

so take (ty^-1) x^-1 is in K’ x^-1

by normality of K’ in K, then ty^-1 is also in x^-1K

hence ty^-1 x^-1 = x_0ty^-1 for some x_0 in K

thats the way to move elements around

try to see if you can finish up

[lapis saddle]

Okay okay

Hmmmmm ys maybe is the next shuffle, sorry if I'm slow

[solar shore]

no worries

so rn we have xysx_0 ty^-1

ys is a good instinct

what can you tell me about that one?

[lapis saddle]

s is in H'\cap K and I'm trying to get (H'\cap K) to be on the right

But if I do that then I get s_0 y with s_0 being in K so maybe it's dumb

Oh never mind

[solar shore]

😋

[lapis saddle]

Nvm again because now x is in H \cap K

So I dont have anything of H'

Like I get
xs_0x_1yty^-1

[solar shore]

oh wait one sec

i parroted my own notes wrong lol gimme a moment

[lapis saddle]

No worries xd

[solar shore]

ah shoot sorry dude i just realized i got a meeting for work lmao

ill be back later to help if still needed

[cloud walrus]

whateverowo | 64598

[lapis saddle]

What do you think?

[solar shore]

hey sorry had a busy night

looks great though

[white oxide]

Can I have a hint for 19 please?

[rocky cloak]

Can you use that (ab/D) = (a/D)(b/D)?

[ebon prairie]

if the order of group is n
and there is exactly one subgroup of order d where d is any arbitrary divisor of n
does it imply G is cyclic?

[kind temple]

for each divisor d, right?

[ebon prairie]

yes

[twilit wraith]

No, consider S3

we have <(1)>, <(12)>, <(123)> and S3 itself

[candid patrol]

No for an arbitrary d
Yes for every d

[twilit wraith]

i just gave an example where even every d doesnt work

1,2,3,6 are all the divisors of 6

and a subgroup of each order exists in S3

but its not cyclic

you can make this work by taking a nontrivial semidirect product of two cyclic subgroups of coprime order im pretty sure

given such a semidirect product exists

[candid patrol]

I know, but it works if the subgroups are uniques

[twilit wraith]

ah i see what you mean

since (23) is another subgroup of order 2

[candid patrol]

S3 has 3 2-Sylow yes

[rocky cloak]

Notice that n is a divisor of n

[rocky cloak]

Oops thought you said element, not subgroup.

Still true, but less obvious

It can be done by counting the number of elements of a given order

[rapid cave]

I am trying to find the number of quadratic subfields of the cyclotomic field $K = \bQ(\zeta_n)$, first assuming $n$ is odd.
Now, by Galois, quadratic subfields of $K$ correspond to index $2$ subgroups of $(\bZ/n\bZ)^\times$ so I just need to count those.
If $n=p^r$ is a prime power, then $(\bZ/p^r\bZ)^\times$ is cyclic and hence has a unique subgroup of index.
For $n=p_{1}^{r_1} \cdot ... \cdot p_{k}^{r_k}$, by CRT $$(\bZ/n\bZ)^\times \cong (\bZ/p_{1}^{r_1}\bZ)^\times \times ... \times (\bZ/p_{k}^{r_k}\bZ)^\times$$
which means that every index $2$ subgroup of $(\bZ/n\bZ)^\times$ must come from an index $2$ subgroup of $(\bZ/p_{i}^{r_i}\bZ)^\times$. \
This means that the number of quadratic subfield of $K$ is exactly the number of prime divisors of $n$. But I think I am missing some quadratic fields as for example $\bQ(\zeta_{15})$ has $\bQ(\sqrt{-3}), \bQ(\sqrt{5}), \bQ(\sqrt{-15})$ as subfields.

[cloud walrus]

ExpertEsquieESQUIE

[quiet pelican]

You have index 2 subgroups that look like diagonal subgroups

Ie, for the Z/15Z case, we have Z/4Z x Z/2Z and the index 2 subgroup generated by (1, 1)

No, that’s the one generated by (2, 0) and (0, 1)

[rapid cave]

it comes from the index 2 subgroup of (Z/3Z)^\times

[quiet pelican]

This is the last one that hasn’t been mentioned and is the one generated by (1, 0)

[rapid cave]

oh I see

[rocky cloak]

One way to think about it is that an index 2 subgroup is the same as a homomorphism onto C2, so in a product of cyclic groups you just have to choose which generator maps to 1 and -1.

So there will be 2^(number of prime factors) homomorphisms and then subtract 1 for the one that is not surjective

[rapid cave]

thanks

[iron arrow]

Well guys thanks for the help during this semester, I got a 93 on my algebraic structures final

[supple ice]

93/100 ?

congrats 🎉

[iron arrow]

Yup

Well in my country it's 9.3/10 but tried to adapt to the usual US grading system

[tulip otter]

whats the natural isomorphism here? I was thinking about $\sum a_{(v)}X_1^{v_1}\dots X_n^{v_n}\mapsto\sum a_{(v)}$ where $(v)=(v_1,\dots,v_n)$ but this map isnt injective right?

[cloud walrus]

ali yassine

[quiet pelican]

The natural map is “evaluation at (0, …, 0)”

[quiet pelican]

This is evaluation at (1, …, 1) and isn’t even well defined (what if that sum has infinitely many terms?)

[tulip otter]

but how is this injective or even surjective? I mean in this case any element of k[[X_1,...,X_n]] will be mapped to 0

[tulip otter]

ohhh right i see

[quiet pelican]

Constant term

And it’s injective, from R/m

[tulip otter]

ah wait yes for some reason I unintentionally/implicitly thought that there is no constant term

tysm

have a great day/night

[dire wren]

Any nice proof of part (vi) ? I have a proof, but it is long and I think there should be a nicer proof.

The proof given in the book is approximately "we will show something later that makes this trivial, but it's too strong, and it's a good exercise to think of a simpler proof"

My proof involved considering maximal cyclic subgroups of order being a prime power.

[quiet pelican]

If G is not cyclic, then there is a divisor d of its order such that there are (at least) two different subgroups of order d (the contrapositive of this isn’t too hard to prove)
If one of these is not cyclic, we are done by induction, as trivially d is less than the order of G
If both of these are cyclic, then take generators x, y
If x^k = y^l for some minimal k, l, then (by changing generator if needed), k = l =/= 1, and so xy^{-1} has order k, and we never have x^i = (xy^{-1})^j unless both sides are the identity
So then <x, xy^{-1}> is of type C_{ord x} x C_{ord xy^{-1}}
Which has a subgroup of the desired type, as ord(xy^{-1}) divides k which divides ord(x), and is not 1

[dire wren]

How did you get k = l =/= 1?

Also what do you mean minimal when you have two powers?

Why can't I have something like
{a,b|a^4=b^8=e,a^2=b^4,ab=ba}

[quiet pelican]

x^k = y^l so they have the same order
Then mess around with cyclic groups for a while

[quiet pelican]

The size of the group generated by a and b need to be the same

By assumption

(Because that number is d)

[dire wren]

Oh, right

[quiet pelican]

Minimise k, and then fixing that k, minimise l

(I’m being slightly sloppy because I then immediately proved they’re equal, and that doesn’t need minimality)

[dire wren]

Ok, makes sense.

[cerulean swan]

Question: why are class groups useful. The most "useful" (i.e. what it's been used for so far in my courses) case which I know is when we have trivial class group and then we have a PID. However, I've read that we don't really care about the group structure of the class group, i.e. If a number field has class number 4, we don't really care whether the class group is the Klein four group or cyclic. However, I have recently read that given a number field, we have $Cl_K\cong Gal(H/K)$ where $H$ is the Hilbert class field. Admittedly, I don't know much about the Hilbert class field, but I'm wondering what else class groups are useful for, and if their particular group structures give us any insight (which would make sense if we care at all about $Gal(H/K)$). Another thing that I read is that the class group is just more of an indicator object, in the same way that say homology groups are nice indicator objects we can attach to spaces: i.e. when we have trivial homology groups for $n>0$, it is much easier to get info about a space than when we have stranger homology groups (not saying that nontrivial higher homology is uninteresting, just that trivial homology is low-hanging fruit, though perhaps I'm completely off base here).

[cloud walrus]

Sleepybear

[rapid cave]

first, this fits more #advanced-number-theory

[rapid cave]

About what you mentioned about the class group being isomorphic to the Galois group of the Hilbert class field. The isomorphism is given by the Artin symbol, which sands a prime ideal to the corresponding Frobenius element, which contains a lot of information about how the prime decomposes in the extension K.

This is a specific case of one of the main theorems of Class Field Theory

[severe otter]

what is galois group

[rapid cave]

Given an extension of fields L/K (which means that K is a subset of L and both are fields), you can consider automorphisms of L that are identity on K. These automorphisms form a group with composition. When the field extension is seperable and normal, it is called Galois and this automorphism group is called the galois group of L/K.

[severe otter]

Wait so Z/Z2 means integers mod 2 because its subset of Z which is divided by 2 or is just by convention

[rapid cave]

this is unrelated

Z/2Z is integers modulo 2 yes

[severe otter]

True but you mentioned fields L/K but does that also apply to this

[cursive spindle]

If you care about geometry, the ideal class group can be though of as an example of the Picard group

More specifically the picard group of the spectrum of some Dedekind domain

This touches the last part of what you said. You can view it in terms of sheaf cohomology

[rapid cave]

Z is not a field and this is just similar notation. what you describe is quotients

[severe otter]

What does seperable and normal mean in this context

[cursive spindle]

One other place where the ideal class group comes up is the class number formula which is super useful

[rapid cave]

wdym class number formula? I forgot what it is

[lilac magnet]

I’ve read that we don’t really care about the group structure of the class group, i.e. if a number field has class number 4, we don’t really care if the class group is the Klein four group or cyclic.
Where’d you read this? because I very strongly disagree

[cursive spindle]

Fancy residue formula for dedekind zeta function

[rapid cave]

this?

[cursive spindle]

yes

[rapid cave]

it doesn't really use the strucuture of the class group

[cursive spindle]

Yeah but it's still there

[cursive spindle]

I can see why would someone say this if they haven't seen class field theory yet

[rapid cave]

CFT is a big quest xD

[cursive spindle]

It is yeah

but that's the fun part about it

[rapid cave]

yeah

I still need to get past cohomology

[dusty raft]

Why is taking \bar{L_s} necessary here we know Lbar/Kbar is normal and finite extension and by the same process in the red part we can show every element lands in one of its conjugated coming from D's image

Oh shoot

This is not number theory

[rapid cave]

xD

it is now

[dusty raft]

Hmmm but anyways approximation lemma is crt in a sense

I just don't get why serre takes the maximal sepravoe subextension

[lilac magnet]

Ig that makes sense

[dusty raft]

From what I understood everything that is done with the generator of this separable extension can be shown for any element in \bar{L} in general, then as Lbar/Kbar is finite and normal you can just show it on the generators

[rapid cave]

I think this is just done to make L/K seperable

[dusty raft]

Lbar/Kbar you mean?

L/K is already separable

It's galois i mean

[rapid cave]

yeah just writing bar is hard

[dusty raft]

Okay so then why is showing the the automorphim of barL_s lifts to one in the decomposition

Group enough

That's actually the part bothering me

[rapid cave]

if s(a)bar is nonzero, then s(a) is in the prime ideal defining Lbar

[dusty raft]

Huh wdym by prime ideal defining Lbar. The prime ideal is already fixed no?

[rapid cave]

if s wasn't in the decomposition group, then s(a) being both in s(P) and in P gives S(P) = P

[dusty raft]

Uhh but why does any automorphim of Lbar lift to L/K first of all in that regard

[rapid cave]

every such automorphism comes from sending a to some root of P(X)

since p(X) is a polynomial over Kbar with a being a root

[dusty raft]

Yeah but we need to talk about automorphims of Lbar not Lbar_s

Right?

[rapid cave]

but you can restrict to Lbar

[dusty raft]

Hmmmmmm why is Lbar_s normal over Kbar

You can restrict as long as those automorphims fix Lbar_s no?

Ohkay maybe because P(X)bar for abar completely splits into linear factors in L_sbar?

So it's the splitting field of the minpoly of abar?

[rapid cave]

I guess

[dusty raft]

Hmm i don't see it still. Why is it that

Why are the bars(a) in barL_s

Oh cause of maximality right

abar is separable so it's minpoly is separable if it didn't have any such bars(a) just adjoin it

A larger subseparable extension

[rapid cave]

yes

[dusty raft]

Cool

Thanks

[rapid cave]

if you want this is a slightly different proof

from Neukirch's ANT

[dusty raft]

Yeah I'll say it's the same he just supplied the details to why picking that sigmabar of the theta bar works for everything

Serre didn't

[rapid cave]

what book is that

[dusty raft]

Serre's local fields

[rapid cave]

ohh nice

I have it on my list

just haven't opened it yet xD

[dusty raft]

Right, I have been told that it's the fastest way to get along with local cft

And other stuff

Which makes sense as the proofs are really terse, but so far I have been trying to make sense of them only to spend hours on a single line

[rapid cave]

good luck

[dusty raft]

Thanks gng ❤️‍🩹

[rancid timber]

I’m having a hard time understanding how they prove that sigma=pi

I don’t think I get how the contradiction supports the idea that there isn’t a P outside of sigma and in pi

For a P outside of sigma, it should be in its own orbit, and so we can re apply all the previous arguments to it to get that the second orbit has size congruent to 1 mod p

[rotund dragon]

the argument that's being made there is that if there was a P-orbit {P'} of cardinality 1 in Sigma, then P is a subset of N(P') and so P = P'. so P is in Sigma which is a contradiction

[rancid timber]

Couldn’t P be any sylow p subgroup?

So what’s stopping the fact that there could be a second sigma with another P as its representative

Wait hold on

Wait nvm I get it now

Thank you 🙏

[crystal vale]

If F is a field and F(x) is a rational field of F[x], what does the basis of F(x) as vector space over F look like?

[quiet pelican]

Do you have any reason to think there’s a nice one?

[fiery sentinel]

Hi

[crystal vale]

No

So my guess is [F(x) : F(x^2) ] is 2

[quiet pelican]

[crystal vale]

But the problem is I am not sure about it, I think the basis is {1, x} in this case, but I have to show it

[rocky cloak]

Something you can show in general is that if a has minimal polynomial of degree n then
K(a)/K has basis
1, a, a^2, ..., a^n-1

[crystal vale]

yes, i think then Y^2 - x^2 is a minimal polynomial of x over F(x^2)

thank you

[hasty lake]

Question about homomorphisms: the book I’m using defined homomorphisms as functions that preserve the product, but can a function also be a homomorphism if it preserves the sum?

[delicate orchid]

I'm going to presume you mean a homomorphism of groups, which is just any map f between two groups G, H with operations # and * respectively such that f(x#y) = f(x)*f(y)

I don't know why you'd think the symbol used for the operation would matter

[hasty lake]

Yes, I did mean group homomorphisms; it’s not so much about the symbol, but rather the fact that addition and multiplication are two different things and it didn’t feel right to just force this operation into the definition

Thx for the answer though

[thorn jay]

? only if you work in a ring

groups dont have a notion of addition. only in abelian groups the multiplication is conventionally called addition

[cerulean swan]

can someone pls give me a slight hint with this? I know basically that I just want to show that the ideals lying over $p\mathbb Z$ are principal, generated by some $a+b\sqrt{-1}$. With this I have $N(p\mathbb Z[\sqrt{-1}])=|N_{\mathbb Q(\sqrt{-1})/\mathbb Q}(a+b\sqrt{-1})|=p=a^2+b^2$. I have found that we have $p\mathbb Z[\sqrt{-1}]=(p,1+\sqrt{-1})(p,1-\sqrt{-1})$, but not quite sure where to go from here

[cloud walrus]

Sleepybear

[cerulean swan]

and especially unsure of how to use the fact that the unit group of the field is cyclic. I know that when we find the primes lying over $(p)$ we have $\mathbb F_p[x]/(x^2-1)$, so I'm thinking the cyclic unit group may arise here, but not sure. Also potentially has to do with Frobenius map $x\mapsto x^p$ but even more unsure of how to apply that here

[cloud walrus]

Sleepybear

[rocky cloak]

So one thing to notice is that (Z/p)^* is a cyclic group of order divisible by 4. What does this mean (try to come up with something related to -1).

If p is prime in Z[i] then Z[i]/p = (Z/p)[i] would need to be an integral domain. Is it?

[cursive spindle]

You can do this via considering the quadratic character of (Z/4Z)^x

[sweet salmon]

Hello, I'm just startingto learn abstract algebra. Is there anyone could why is associativity required for group? Why not commutative?

[vapid vale]

groups are generally motivated by their actions on things, i.e. groups list symmetries of things (i.e. funtions that preserve something). think dihedral group acting on n-gon, symmetric group acting on a set of n elements, etc. then the group operation is just composition of these symmetries. and composition of any functions is going to be associative

in that if you do a sequence of things to an object, all that matters is the order you do the sequence, and not some notion of grouping

commutativity is not required because many symmetries of things do not commute with each other (such as reflection and rotation), so its a stronger condition than wht we want for the things we're trying to formalize

[sweet salmon]

Can I say, when we list all of the symmetries, then we naturally get the properties of group? Which commutative is not one of them?

[vapid vale]

i think thats a fair statement

[sweet salmon]

Okay, thank you @vapid vale

[thorn jay]

to paint a picture of why we want associativity: there is a nonassociative analogue of groups that is symmetric, called a commutative loop. Even for noncommutative groups we have a pretty good idea of their structure (and they pop up everywhere), and we have numerous classification theorems about them. However, almost nothing of the like is known about these commutative loops! Dropping associativity simply makes basically everything fall apart

[kind temple]

it would suck if the order in which you wrote down the characters in this sentence changed the meaning of this sentence.

e.g., (ab)c != a(bc)

[foggy axle]

Yes we lose a lot of the nice things we can do with groups and they turn into much less nice objects of study 🙁

[sweet salmon]

okay, I think it makes sense for me. thanks a lot

[thorn jay]

between dropping associativity and dropping inverses ALWAYS drop inverses

monoids still have Green's relations

[vapid vale]

all good points but i should add that before the notion of an abstract group was invented, the main ways people considered groups were symmetries of algebraic equations (galois groups) and differential equations (lie groups), so function composition is the answer to "why" as a question of causation

[thorn jay]

yes i was just giving additional information
its nice to know what happens if you do decide to ignore common sense

[vapid vale]

ya
not trying to do an "actually my answer is better" lol

[thorn jay]

erm ackshyually my answer is 1000000x better

[foggy axle]

Ah yes

And on the topic of function

[supple oracle]

axler emphasizes an analogy between the space of linear operators in V (End(V)) and the complex numbers. would an intro abstract alg book, say artin, explain this analogy further

i assume not..?

[foggy axle]

I guess if the analogy is in that they both form a ring and a have a nice way to express conjugation then that's the whole focus of abstract algebra is developing that abstraction

I wonder what the analogy is

[supple oracle]

conjugation <=> adjoint, self-adjoint operator <=> real number, positive operator <=> nonnegative real, unitary operator <=> element of unit circle

[noble nexus]

you explore this analogy more when you learn about operator algebras basically

specifically C*-algebras

but it's a lot more of an advanced topic

[supple oracle]

where would that be then? in like operator theory or smth/

[noble nexus]

usually you see this stuff near the end of a functional analysis book, which in turn requires topology/measure theory

b/c it involves a lot of analysis

essentially the reason why is that operators behave a lot more like complex valued functions rather than just the complex numbers

and spaces of functions are typically infinite dimensional and require analysis

[supple oracle]

hmm Okay thanks

[noble nexus]

but for example, the spectrum of an operator (for finite dimensions this is just the set of eigenvalues) is analogous to the image of a function

the trace of a matrix is analogous to the integral of a function

[supple oracle]

after I finish reading my linalg book, i plan on going on roughly this sequence: abstract alg -> real analysis -> pointset topology -> complex analysis -> measure theory -> functional analysis -> differential geometry?? -> ????

[noble nexus]

Yeah that sounds like a good path

[supple oracle]

oh that's nice

[supple oracle]

idek what comes after this Lol

[noble nexus]

along with this, projection matrices correspond to characteristic functions of sets (the characteristic function of a set A is just the function that's equal to 1 on A and 0 otherwise)

[supple oracle]

characteristic function as in the indicator function?

[noble nexus]

the similarity comes from the fact that in both cases, if you square or conjugate them you get themselves

yeah

[supple oracle]

idempotence?

[noble nexus]

ya

[supple oracle]

cool

[noble nexus]

oh I should say orthogonal projections rather than projections

[supple oracle]

yeah ladr has a few exercises with projection operators or operators in general that satisfy P^2 = P

[noble nexus]

but yeah there are a lot of cool similarities

[supple oracle]

great hopefully i make those connections in the future loll

[noble nexus]

there's a whole theory of "noncommutative integration"

[supple oracle]

what do people do after they go throught hese fields btw

like Idrk what's after 😭 math seems too diverse

[karmic moat]

usually by then they have a specific field that's caught their eye

[noble nexus]

lots of things really, functional analysis and differential geometry lead to a lot of places

and by the time you get there you might be interested in a whole other field

[supple oracle]

from the things ive heard in functional analysis it seems extremely interesting

I guess I'll see when I get there thx

[noble nexus]

yeah I find it quite fun, basically puts together linear algebra and analysis

[supple oracle]

yeah that's what i like

i really liked ladr

[vapid vale]

this looks like a good path but if you're still taking linalg i would not be hard set on any one sequence

the next classes i took after algebra and analysis were completely decided by what i enjoyed and found interesting in abstract algebra

[supple oracle]

of course, i've changed the order of this list multiple times

i have a lot of flexibility cuz im just reading books so i can read a certain book whenever i want lel

[vapid vale]

yeah for sure

[rapid cave]

Let M be a f.g. module over a noetherian ring A. Is it right that any A-submodule N of M is also f.g.? Is there a restriction on the minimal number of generators of N?

[thorn jay]

being Noetherian is closed under extensions iirc

[rapid cave]

It is yeah

[thorn jay]

so as A is noetherian, every A^n is a Noetherian A-module, hence every fg module is Noetherian

[rapid cave]

Yeah

What about my second question?

About the minimal number of generators.

[noble nexus]

No there isn't

[thorn jay]

thought so

[noble nexus]

Even C[x,y] has ideals with arbitrarily large number of generators

[rocky cloak]

Hmm, if there is a bound on the number of generators for ideals of A, does it then follow that the bound for A^n is n times this bound?

[thorn jay]

if we have a chain x1 < x2 < ... and some element y such that both [0, y] and [y, 1] are Noetherian, then { xi } stabilizes at the maximum of the indices where {xi \cap y} and {xi + y} stabilize

so yes

probably??

wait no doesn't follow

Consider a submodule N < M, such that N and M/N are Noetherian. Let A < M be another submodule, such that A \cap N has generators { a1, ..., an } and (A + N)/N has generators represented by e.g. { b1, ..., bm } which we may choose to all lay in A. By the correspondence theorem, we have then that <b1, ..., bm> + N = A + N.

Now, let A' be the submodule generated by < a1, ..., an, b1, ..., bm >. Then clearly A' < A, and so A' \cap N < A \cap N. But all the generators of A \cap N lay in A' and in N, hence A \cap N < A' \cap N. Indeed, A' \cap N = A \cap N. Furthermore, we have:
A' + N = < a1, ..., an, b1, ..., bm > + N = < a1, ..., an > + <b1, ..., bm > + N = <b1, ..., nm > + N = A + N
as <a1, ..., an> already lays in N. Finally, we conclude that A' = A as the lattice of submodules is modular

@rocky cloak

in particular, if the minimal generating sets of N and M/N are bound by n and m, then the minimal generating sets of M are bound by n + m

[crystal vale]

if i take r in R, then r is alg over C, so it has minimal polynomial but since irreducible polynomial in C are linear so r in C, right?

[quiet pelican]

Essentially yeah

[crystal vale]

essentially yeah

[delicate orchid]

essentially yeahjective on objects

[quiet pelican]

“If you were giving that proof to me in a course I’d consider that a fair outline, but I’d want more detail”
In particular I want you to be specific with where finite dimensionality comes in

[crystal vale]

to show r is alg over C

[cloud walrus]

Luka.s

[barren sierra]

If you know what monomial orders are, then you can do this via Grobner bases

so my "better way" is that you can ask a computer to compute this

[mystic ether]

yeah I've heard of this. I feel this is way worse though

we started our course with Gröbner bases and it's like 10 million lines long to find one

and even longer to prove that something is a Gröbner basis

[barren sierra]

yes but if you know how to use computer algebra systems

[mystic ether]

Explicitly turning something into a Gröbner basis was actually the worst homework assignment in the course lol

yeah fair

[barren sierra]

then you don't have to do any of that

the computer can do it

[mystic ether]

Yeah makes sense. Unfortunately im studying for an exam

[barren sierra]

and so for small examples like this it's not much better to do via computer

[mystic ether]

so no matlab there

[barren sierra]

matlab wouldn't be what I use

I'd use Sagemath or Macauley2

[mystic ether]

If I get further into AG I'll prob learn one of these eventually. For now just good to sanity check there's not really an easier "by hand" way than just polynomial long division

which is super skull but it is what it is

[barren sierra]

but like I think there are ways to come up with polynomial maps f such that for a polynomial p(x), f is injective if and only if p(x) is irreducible

and checking irreducibility of polynomials is hard

so finding these kernels would be hard also

[barren sierra]

so like you'll only ever get examples like this

which are checkable by hand

[mystic ether]

Alright I'll just pray my prof is chill like that

lol

[barren sierra]

cool that you're learning computational AG stuff

not enough people know it and IMO it's the best way for people to get their hands on alot of examples in algebraic geometry

which is useful

[mystic ether]

Yeah the Gröbner basis stuff was cool in theory although I had a somewhat negative experience because I think he didn't check the "find the Gröbner basis" problem he gave us because I was running the algorithm (by hand, on paper) for like 10 pages (different than this problem).

and it wasn't terminating

so I crashed out and gave up

lol

[twilit wraith]

For the latter proof, i realized i forgot to show that all elements of Hom(Z/nZ, A) are of the form phi_a

but i honestly am not fully sure how i would do so

well i guess if f(x) = a for some a in A is a Z-module homomorphism, then of course na = 0 has to be true since 0 = f(0) = f(nx) = nf(x) = na

ok so we have that f(kx) = ka

well wait nvm

since f has to be a group homomorphism i can just consider where f(1) goes

but then if f(1) = a for some a in A then we just have that f = phi_a

[vocal pebble]

Well 1 generates Z/nZ

[twilit wraith]

since f(k) = ka

alright cool i just had to say it out loud ig

or via text

either way it worked

[vocal pebble]

The first part of this exercise can be viewed as a special case of:
An R module M is naturally an R/I module iff I is contained in Ann(M).

R= Z and the R module would be the submodule generated by a, nZ the ideal I, and phi_{a} would be ring action which we are trying to check is well defined.

[crystal vale]

any hint?

[quiet pelican]

Consider the extension K(x)/F where x is a root of f, and compute its degree in two different ways

[crystal vale]

yes i thought in this direction but i don't know how to find the degree of K(x)/F?

[stark cloak]

Im reading through the wikipedia article on relation of SU(2) and SO(3) using mobius transform. I'd like to understand this particular approach, not the one using quaterions.

Now i get the stereographic projection part and how both $g_{\theta}$, $g_{\phi}$ rotate the complex plane, but im absolutely unable to understand what are these variables in the article supposed to mean.

the author first uses $\zeta$ with $e^{i \phi}$ rotation. Then describes analogous rotation with $\omega$, and then reuses $\zeta, \zeta'$ variables in the sentence with $\omega$, which doesn't really make any sense to me.
Any ideas what the author meant?

article link:
https://en.wikipedia.org/wiki/3D_rotation_group#Using_Möbius_transformations
The confusing part comes after: "which, after a little algebra, becomes".

[cloud walrus]

MINECRAFT_LOVERBOY

[stark cloak]

I guess its supposed to mean, another transformation of complex plane.

take complex plane, invert stereographic projection S, project onto x=-1/2 plane, rotate with g_theta, and then invert the prior operations except for g_theta.

[rocky cloak]

If there exists an intermediate extension
F < E < K(x)
do you know a relationship between
[E:F] and [K(x):F]?

[crystal vale]

I just know [K(x) : F ] = [ E: F] [K(x) : E]

[delicate orchid]

yeah, so in particular [E : F] is a divisor of [K(x) : F]

[tardy hedge]

Using Zorn’s lemma to prove that every vector space has a basis, i did not notice where the field structure is needed

[south patrol]

I guess point is that in a vector space V, if you have a maximal linearly independent subset S then anything not in the span of S can just be chucked into S as a contradiction

[tardy hedge]

Maybe if the vector not in the span its linearly independent from them is using it or smth?

[south patrol]

But over a general ring it will not be linearly independent from the others in general

[thorn jay]

exchange lemma or smt

[south patrol]

Ye

It does work for division rings tbf lol

But yeah

[delicate orchid]

or as I like to call them, fields

[south patrol]

Probably good to think about things like Z (+) Z/2 as a Z-module

Or even Z/2 lol

Milne's algebraic number theory book

[delicate orchid]

:lloran2:

[tardy hedge]

Think about Z/2 in what sense? 1 is not in the span of 0 but {1,0} is not independent over Z?

[south patrol]

I guess Z/2 has no nonempty linearly independent subset

Lol

Z (+) Z/2 is a little more satisfying but works similarly

[delicate orchid]

what you planning on doing the cheeky (1,1)+(0,1)+(-1,0) or something like idk how it would be more satisfying

[south patrol]

Just like it has nontrivial linearly independent subsets

Lol

[delicate orchid]

oh ok sure lol

[rocky cloak]

Your essentially proving that any semisimple module is a direct sum of simple modules. And thats true for any ring its just that vector spaces are semisimple

[crystal vale]

ah, i don't see how it helps me here

[quiet pelican]

Now find ||a spanning set of size mn||

[crystal vale]

i think from that, if f is reducible over K, then for root x of f, [k(x) : K ] < n, it implies [K:F] <mn, but it contradicts that mn divides [K:F]

and [k(x) : K] < n, because f is reducible over K

[quiet pelican]

[crystal vale]

thank you

btw anyone has idea how Lang introduced separability? I don't get it

i don't get it

[rocky cloak]

So if E/F is an algebraic extension, then the number of embeddings of E into the algebraic closure of F is called its seperable degree.

A finite extension is seperable iff the seperable degree equals the degree.

The motivation is that embeddings of
F(a) into the algebraic closure are exactly given by sending a to a root of its minimal polynomial, so this is detecting if the minimal polynomial of a has repeated roots

[twilit wraith]

I like intuitively am kinda seeing whats going on here but im struggling to formalize it

[crystal vale]

which fixed F ?

[twilit wraith]

its just that image of the most obvious map into Z/mZ that this has is expressed kinda weirdly

[crystal vale]

so you have to consider the image of 1 only, right?

[twilit wraith]

yeah

i mean i know 1 maps to m/(n,m) but the way i justify why this map is indeed a function is not super clear to me

i guess i just show that all the elements of the given Hom module have to map elements of Z/nZ to multiples of m/(n,m) in Z/mZ because somehow those are the only elements x in Z/mZ that satisfy nx = 0

[rocky cloak]

What do you mean?

F should be fixed yes. So embeddings of extensions

[crystal vale]

my idea is 1 maps to x, where order of x should divide the order of n. And show that this Hom(Z/mZ, Z/nZ) is a abelian group and show that its order exactly gcd(m,n) and it has an element of order gcd(m,n)

[twilit wraith]

the abelianity of the Hom(Z/mZ, Z/nZ) is clear as the morphisms can just be given as left multiplications

from elements from a commutative ring

showing its order exactly gcd(m,n) is whats tripping me up

[rocky cloak]

First step should just be describing what the elements are

I.e. what is the set Hom(Z/m, Z/n)

[twilit wraith]

well its Z-module homomorphisms phi_a(k) = ka for all a in Z/mZ such that na = 0

i guess itd be helpful to show that there are gcd(m,n) elements in Z/mZ that satisfy nx = 0

[rocky cloak]

So for example m maps to km mod n

[twilit wraith]

wait isnt it n to m

eh either way i get what u mean

[rocky cloak]

Yeah, whichever was which

[twilit wraith]

so this gives us that Hom(Zm, Zn) is abelian

i guess i should approach this from element orders

[rocky cloak]

No, I mean it is, but not relevant to what I was saying

[twilit wraith]

oh

[rocky cloak]

What is m in Z/m ?

[twilit wraith]

0

[rocky cloak]

So then km mod n must be...

[twilit wraith]

0 as well

i see

well i did already know that km = 0 from a previous exercise

[rocky cloak]

So k can't be just anything

[twilit wraith]

true

but what exactly it looks like im having a hard time seeing

[rocky cloak]

Well what would it mean for km to be 0 modulo n

[twilit wraith]

km is a multiple of n

so k divides n

[rocky cloak]

k dividing n is maybe a bit strong

Maybe consider m=2, n=4

What k has km a multiple of 4?

[twilit wraith]

0 and 2

[rocky cloak]

Oh yeah, k divides n. I was thinking n divide k, that wouldn't be right.

Sorry

[twilit wraith]

thats alright

[rocky cloak]

Anyway, we can say more. What do we need to multiply m be to get a multiple of n?

[twilit wraith]

i mean n is obvious

im trying to think of examples tho

i guess n/gcd(m,n) is correct here

[rocky cloak]

That's right k would need to be a multiple of n/gcd(m, n).

Then we might ask how many such there are modulo n

[knotty badger]

You have the identity gcd(m, n) lcm(m, n) = mn

[twilit wraith]

well clearly gcd(m,n)

[rocky cloak]

Boom boom

[twilit wraith]

i see

man i cant do ENT for the life of me

[rocky cloak]

So at least if we verify that all of those work, then we have exactly that many

[twilit wraith]

i see

i guess i can just show that arbitrary multiples work

[twilit wraith]

btw is there a clean way to prove that n/gcd(m,n) is minimal here

so i dont risk that theres a proper subgroup that is the image of the isomorphism im building

[rocky cloak]

Well km is a multiple of n and m, so a multiple of lcm(m, n).

The smallest choice would then just be
km = lcm(m, n)
which means
k = lcm(m, n)/m

Then use that lcm(m, n) = mn/gcd

[twilit wraith]

i see

but lcm(m,n)/m = n/gcd(n,m)

[twilit wraith]

alright, here is the proof i ended up making

pretty sure it works but might as well check

[alpine island]

In a commutative ring, does "a+u is a unit for all units u" imply a is nilpotent?

[thorn jay]

where u is a unit?

[velvet hull]

I think you need to restrict on u some more, otherwise what about u = -a or u = -a + [any nonunit]

[alpine island]

oh, yeah I realized I missed a word there. whoops.

The converse is true, unit + nilpotent is a unit

[rocky cloak]

Consider for example a=x in k[[x]].

More generally this will be true for any a in the Jacobson radical

[alpine island]

I see

well jacobson radical and nilradical get confused sometimes anyways, I wasn't far off

[rocky cloak]

I don't think it's equivalent to being in the Jacobson radical though, gonna need a think on that

[alpine island]

I know if 1-xy is a unit for all x then y is in jacobson radical

this is weaker since it only lets x be a unit

[rocky cloak]

Alright it's not equivalent.

Take R = k[[x]][y]
the units are power series in x with non-zero constant coefficient, but the Jacobson radical is just 0.

So x satisfies unit + x = unit

[mystic ether]

This is an iff, it’s proven in Atiyah Macdonald ch1

So either Atiyah Macdonald is wrong which I suppose is entirely possible or there’s something wrong with jagr example

But I think the characterization is right. It’s used for example in proof of Nakayama from Cayley Hamilton

[alpine island]

no this is true. The thing I was looking at is weaker

[mystic ether]

Ah I only caught the end of the conversation I guess

Ignore me then

[vapid vale]

what sort of questions motivated the inception of invariant theory? afaik it starts with boole and cayley in the 1840s, but most of the fundamental ideas seem to come from klein and hilbert and whoever in the late 1800s.

but what form did it take in the first 20-30 odd years

i've seen people point to things like the discriminant being, up to scaling by a square, an invariant of linear transformations

[supple ice]

early invariant theory was basically: I changed coordinates and my polynomial got uglier so instead of admitting I chose a terrible basis I’m going to invent numbers that don’t notice

[vapid vale]

Right, is that like the hyperdeterminant or whatever

Did it really take 30 years for it to become interesting lol

[karmic moat]

there's an 1864 quote by Sylvester

As all roads lead to Rome, so I find, in my own case at least, that all algebraical inquiries sooner or later end at the Capitol of Modern Algebra over whose shining portal is inscribed "Theory of Invariants"

according to wikipedia, cayley established invariant theory in an 1845 paper, based off an 1841 paper of boole

https://www.sciencedirect.com/science/article/pii/0315086086900911

okay according to this paper sylvester and cayley were close lifelong friends so maybe the quote above is biased

[vapid vale]

I’ve skimmed the paper by Cayley

[karmic moat]

oh sorry yeah i meant this is a survey paper on it

idk why it didnt embed

"The rise of Cayley's invariant theory (1841–1862)"

[vapid vale]

oh this is precisely the time period i was interested in, thank you!

[crystal vale]

to show if K is alg over F and there is a ring R such that F \subset R \subset K, then R is subfield.

let a be non-zero in R, and since K is alg over F, R is alg over F, hence R[a] \subset R is a finite dimensional vector space over F.

Now mapping x -> ax (R[a] -> R[a]) is injective because R[a] is integral domain, and since it has finite dimension so this linear mapping is sujrective hence ax = 1 for some x in R[a] \subset R, hence R is subfield

is it correct?

[karmic moat]

if a is an element of R then R[a] is just R

[crystal vale]

how?

[karmic moat]

what do you mean by R[a]?

[crystal vale]

oh it is not R[a], it is F[a]

[karmic moat]

looks fine to me then

[crystal vale]

okay thank you

[lament flame]

Does the existence of an nontrivial idenpotent (a^2=a) in a ring R always imply that R has a nontrivial decomposition S(+)T, or does R need another property?

[south patrol]

Maybe should be central but otherwise should be good, as then multiplication by a is a projection

[rocky cloak]

An idempotent always breaks R into
Ra (+) R(1-a)
as a module.

If a is in the center then these ideals are also subrings, making R a product of rings

[crystal vale]

my professor taught separable polynomial such that each its irreducible polynomial is separable, and for irreducible polynomial separability defined as it has no multiple roots in its splitting field. But in dummit they defined any arbitrary polynomial is separable if it has no multiple roots in its splitting field, so which one is good ?

[elfin wraith]

I’m not sure I follow what you mean, how did your class define separable polynomials?

[crystal vale]

he first defined separability for irreducible polynomial such that it has no multiple roots in its splitting field, then for then for arbitrary polynomial if its all irreducible polynomial is separable, (x-1)^2(x-3) is separable here

[rocky cloak]

I mean both are good.

For the first definition you get that a polynomial is seperable iff it's splitting field is. Which is nice.

For the other you get that a polynomial being seperable doesn't depend on what the ground field is. So I guess if you're using the first you should probably say "f is seperable over K".

The latter is also good in that
K[x]/(f) is a seperable K-algebra iff f is seperable

[crystal vale]

how in first, ground field matter?

[tardy hedge]

I'm trying to compute colon ideals (I : v) where v is a monomial and I is a monomial ideal plus extra generators that are linear forms (eg. x1+x2+x3)

I know if I is a monomial ideal there are formulas for this

anyone have an idea of where I could look?

[tardy hedge]

because a polynomial being irreducible depends on what field its over

[crystal vale]

ah nice point

[copper kestrel]

(sorry for interrupting, feel free to ignore!)

So i plugged my values in for part a and got y = 999^{239} mod 1457

then saw that 128 + 64 + 43 + 8 + 4 + 2 + 1 = 239

but computing 999^2 mod 1457 = 1413 mod 1457 was difficult, and then i have to compute 1413^2 mod 1457, etc. is there any easier way to do this or is it just going to be computational slop?

i didnt see this channel was active while i was typing, sorry about that

[tardy hedge]

its fine you can post questions whenever

[tardy hedge]

it will probably help if ppl have context to this question

[copper kestrel]

youre so right, this is basically just encryption and decryption. it's called "RSA Encryption"

we know y=m^s mod n

just doing part a rn, idk how to even do part b

[rocky cloak]

Binary exponentiation is generally the fastest way yeah, seems like they might be expecting you to use a CAS.

I guess in this case you might be able to do it faster by using that 1457 = 31x47 and use CRT

[rustic trench]

"standard RSA" would be sth like

N = p*q
e = 0x10001 (can be picked somewhat arbitrarily, usually 0x10001 or 3)
d = e^-1 (mod ord(K))
c = m^e

[rustic trench]

what is r?

[rocky cloak]

Presumably the decryption key

[rustic trench]

luckily you don't need part b for part c

[rustic trench]

so

r = 239^-1 (mod (30 * 46))

ig?

[lament flame]

As a module of the ring generated by a?

[rocky cloak]

No, just R-module

[lament flame]

...I'm gonna reread this when I'm more awake lmao

Thanks tho

[rocky cloak]

You mean 239^(-1)?

[rustic trench]

yes

if you don't have CAS (which I mean, they tell you to use it kinda?) do stuff mod p and mod q and then CRT

[copper kestrel]

i figured it out!

[copper kestrel]

yah i ended up using eulers theorem to get 239^{\phi(1380)} = 1 mod 1380, then just need to compute that and then we'll find r!

[rustic trench]

sage oneliner

r = pow(239, -1, Zmod(1457).unit_group().order())

[tulip otter]

I will prove (a) $\implies$ (b) $\implies$ (c) $\implies$ (a)\\I am not sure how to prove (a) $\implies$ (b). So if $(f(x))$ is prime then $k[x]/(f(x))$ is an integral domain and i have to find a multiplicative inverse for any non-zero element of $k[x]/(f(x))$. let $g(x)\notin (f(x))$, then i was thinking about the inverse $\sum_{k=0}^{\infty}(1-g(x))^k$ but the problem is that this sum doesnt seem to have finitely many terms (i didnt actually check, i guess it will turn out that it has finitely many terms and i have to prove this but i am not sure).\\Now assume $(f(x))$ is maximal, then $(f(x))$ is prime and it follows that $f(x)$ is irreducible since $k[x]$ is a UFD ($k$ is a field).\\Finally, assume $f(x)$ is irreducible. Then $f(x)$ is prime since $k[x]$ is a UFD and hence $(f(x))$ is prime.

[cloud walrus]

ali yassine

[tulip otter]

Is everything correct apart from the (a)\implies (b) part?

I also thought about using the equivalent definition of a maximal ideal which states that an ideal m of a ring R is maximal if there doesnt exist any ideal I in R with I neq R such that m\subsetneq I.
I tried contradiction by assuming that there exists such an ideal and then i would probably have to use the fact that f(x) is irreducible. But i dont see how (f(x))\subsetneq I would imply f(x)=g(x)h(x) for non-constant polynomials g(x) and h(x) of k[x].

because i dont really know anything about say the generators of such an I

[valid fox]

Btw this holds in general in a ufd

[tulip otter]

yea because so far nothing really relies on the properties of polynomials

just the properties of a ufd in general

[valid fox]

Sorry this is wrong, this holds in a PID (and K[x] is a pid)

[rocky cloak]

I guess the canonical way to construct the inverse would be to use euclidean division / Euclids algorithm.

I guess there's also a trick to do it without constructing the inverse: ||notice that k[x]/(f(x)) is a finite dimensional||

[valid fox]

It might be easier to do (b) => (a) =>(c)=>(b)

And use that K[x] is a pid, if you have proven that

[rocky cloak]

You should not expect this sum to be finite

[copper kestrel]

i believe i made a mistake doing the phi function bc now i have to compute 239^{351} mod 1380

which does not sound fun

it says we can use a computer but i'll ask my professor if he actually wants us to use a computer

[twilit wraith]

U can use Chinese remainder theorem here right

[copper kestrel]

whats that

[twilit wraith]

The theorem that gives how to solve systems of modular congruences

The idea is that I can solve the left mod each of the prime power factors of 1380

Which is 4 * 3 * 5 * 23

So do 239^351 mod each of those factors

[copper kestrel]

ohhhhhh

[twilit wraith]

Hopefully this made the process quicker lol

Kinda heinous to calculate that congruence without FlT and CRT

[copper kestrel]

i havent done it yet but when i do do it i'll report back 🫡

[twilit wraith]

Sounds good

[rocky cloak]

You're supposed to compute 239^-1 mod 1380

Which you would do with Euclids algorithm, not binary exponentiation

Don't even need to compute phi(1380), that's something that would be infesible if you were working with actually big numbers

[copper kestrel]

hm, im not even sure how i would do that

[rocky cloak]

You're not familiar with Euclids algorithm?

[copper kestrel]

i am but like we havent ever calculated the inverse of a value with it

especially not in mod n

[rocky cloak]

So it produces x and y such that
239x + 1380y = 1 (or more generally the gcd)

So then x will be the inverse of 239 modulo 1380

[copper kestrel]

ohhhhh

[copper kestrel]

im trying to think about how someone would set that up for the division algorithm, bc like i know we want x,y in 239x + 1380y = 1, but i dont know which thing is which in the algorithm

[rocky cloak]

Maybe hard to explain in a comment, but here's someone computing an example
https://youtu.be/fz1vxq5ts5I?si=VqjTQ9y_KO0AyJqH

[copper kestrel]

i'll take a look and report back!

ty!