groups-rings-fields

[copper kestrel]

i just didnt know what the entire problem meant i'm gonna be totally honest

[thorn jay]

oh lol, that's valid

[knotty badger]

Yeah that’s the standard construction of tensor products

And free objects generally

[wraith cargo]

yeah this is how it's originally given

it's called like the chord construction or something?

proving that it's a group is very cool

Very elegant application of algebraic geometry

[alpine plank]

tangent-chord construction

[alpine plank]

mfw associativity is the hard part

[wraith cargo]

there's a purely analytical proof in a book of Knapp and it's genuinely hell

tho if you use Riemann-Roch it's immediate

[alpine plank]

like messing around with Weierstrass elliptic functions?

That sounds scary

[wraith cargo]

no I don't fully remember how it goes but you write out the formulas for the coordinates of both sides or something like that

it's like a 5 page calcualtion

very like elementary proof but it's annoying as fuck

[alpine plank]

Ohh yeah baby Silverman makes you do that too I think

[noble nexus]

using elliptic functions is actually quite nice

[wraith cargo]

mfw the true redpill is using the fact that elliptic curves are isomorphic to their own jacobian

[noble nexus]

its the analogue of sine and cosine for the circle really

though I suppose it only works for C

[alpine plank]

You can also use Pappus theorem

Oh wait that's not what it's called

I forgot what the actual thing is called

[alpine plank]

Cayley-Bacharach

[alpine plank]

Showing this is the same as the geometric group law is still a bit of work no?

[noble nexus]

yea I believe so

[chilly ocean]

Given a group action(say, on a set), say G on X, we can define the action groupoid to be the groupoid whose vertices are elements of X, s.t. a morphism from x to y is a pair (g, gx) where gx=y. The orbits of a group action are essentially the connected components of this groupoid.

Equivalently, it is the set of all gx for some specified x in X, for all g in G.

[thorn jay]

why would someone who doesnt know what an orbit is know what a groupoid is

[chilly ocean]

I learned about orbits from Topology & Groupoids

Also I think it's a very neat perspective, so at least if they google that too they'll learn more about it. It's quite natural to think of orbits as like, vaguely connected components imo, and this is literally true when thinking of the orbit as a subgroupoid of the action groupoid

[noble nexus]

I feel like that is vastly overcomplicated

[chilly ocean]

Idk maybe, that's why I included the "or equivalently [...]" in the next message

[rocky cloak]

If you think of a group action as moving the elements of X around, then the orbit of x is... the orbit of x

[noble nexus]

groupoids are cool though

yeah its important that the orbits partition the space and break up the action into transitive actions

[copper kestrel]

truth LMAO

[alpine plank]

We should experiment on an undergrad by starting them off with alg top

[noble nexus]

any abstract algebra book that skips over group actions is a bad book

[chilly ocean]

Lmao

[alpine plank]

My bad we should experiment on 5th graders by starting them off with alg top

[chilly ocean]

I started learning some alg top after pointset, with no actual abstract algebra knowledge, but quickly stopped

Not in HS though, cause I'm homeschooled

[noble nexus]

"starting alg top in hs" aka learning a few terms to look smart but not actually knowing anything

[thorn jay]

i agree with you that its natural but, like, thats not helpful

a lot of categorical things like adjunctions are perfectly natural but youre not gonna bombard a first year with those

[chilly ocean]

Fair point

[alpine plank]

Strange

[thorn jay]

monitor the cognitive development

[chilly ocean]

downwardslope.jpeg

[noble nexus]

high schooler: "whats the cartesian product of two sets"
category theory bro:

[thorn jay]

math enjoyer vs mathematician

[thorn jay]

ts 🔥

[wraith cargo]

What's awful is that some people would react this way 💀

[gentle wolf]

could u ask u what's in ur pfp?

looks hella cool

,av blake

[noble nexus]

its a domain coloring plot of a weierstrass p function

or perhaps its derivative I forget

[cloud walrus]

noncommutative's Avatar

Click here to view the image.

[noble nexus]

ya its the derivative

this is the p function

https://en.wikipedia.org/wiki/Weierstrass_elliptic_function this guy

[balmy python]

i suck so bad at group theory 😭

like i'm really bad with finding counterexamples or even having a maturity to prove something

the theorems are easy to understand and i'm able to understand proofs but doing the questions myself is so much harder

[noble nexus]

if you're bad at finding counterexamples and lack some of the intuition id recommend looking a lot of examples in more detail

also while you are following the proofs in the book, really try and pay specific attention to the common tricks that you see over and over

[balmy python]

hmm yeah i'd like to say i do that

also finding isomorphisms are kinda hard

like yeah i know that you should look at mapping similar structures of the groups to each other

[noble nexus]

a lot of proofs in group theory also follow a pattern where every move you make is somewhat "forced" in that there's really not a lot you can do with the information you are given

[balmy python]

but it's a lot harder than that 😭

[balmy python]

ah

[noble nexus]

do you have any specific problems you are struggling with

[balmy python]

many 😭

i've been looking at problem sheet after problem sheet

like i did (i) and most of (ii) with showing that they are subgroups and only the first subset is normal

but i got stuck at showing that the quotient is isomorphic to some well known group

[noble nexus]

from an intuitive perspective, when you quotient out by something you are essentially treating everything in that subgroup as equal to zero

so if you take the transformations of the form ax+b, and you imagine that b is equal to zero

or you treat b as being zero

what do you get?

[balmy python]

ax

which is that second set

ahhh but i don't really understand why

[noble nexus]

yeah so you get the group of transformations of the form ax

but that group is pretty well known

[noble nexus]

the best way to do this is with the first isomorphism theorem

which I assume you have probably seen

[balmy python]

yeah so do i want to construct a homomorphism where the image of the quotient set maps to the identity?

[noble nexus]

this group btw has a pretty simple structure, since if you compose the function x -> ax with x -> a'x you just get the map x -> (aa')x

so its just the multiplicative group of real numbers

[balmy python]

yeah

[noble nexus]

what you would then want to do is construct a homomorphism from this ax+b group into the nonzero reals

show its a homomorphism that is surjective with kernel equal to the translation subgroup

[balmy python]

hmmm

so we want f(x+b) = 0 but f(ax+b) = ax

[noble nexus]

or just f(ax+b) = a

[balmy python]

f(x) = x-b works

[noble nexus]

careful though, because ax+b is not the group element

[balmy python]

wait no

[noble nexus]

the group element is the function that takes x to ax+b

[balmy python]

wait i'm getting mixed up

let me write this down

[tulip otter]

maybe write the functions as $f_{a,b}$ where $f_{a,b}(x)=ax+b$

[cloud walrus]

ali yassine

[balmy python]

woah it's ali again

[tulip otter]

so the elements of the group G are f_{a,b}

[tulip otter]

yooo how are you pentium, is everything alright

[balmy python]

yesss

uni was long today 😭

analysis lecturer gave me some extra questions though

[tulip otter]

same here, reached home at something like 5 pm i think

[balmy python]

alright so we want ker(phi) = H so phi(f) = 0 when f = x+b

[balmy python]

oh that's not bad tbf

i get home like 6 or 7 some days 😭

[merry harness]

6 or 7...

[balmy python]

six seveennnn

[balmy python]

yeah

it's more because i get a shit ton of train delays

[balmy python]

also gker(phi) -> phi(g)

[tulip otter]

what is phi?

[balmy python]

a homomorphism

which i want to find

[tulip otter]

ah so you are trying to find it

i see

[noble nexus]

yea I would suggest using the notation that ali suggested above

using subscripts

[balmy python]

ah okay

[tulip otter]

i actually remember this notation from lang, i saw it not too long ago (he uses T instead of f but it doesnt matter)

[balmy python]

so the map takes $f_{a,b}$ to$ f_{a,0}$

[cloud walrus]

pentium

[noble nexus]

that would probably work but its easier to think about a map from the group just directly to the positive reals

instead of a map into a subgroup

or I guess nonzero reals

[balmy python]

this is confusing

can you explain it more?

[noble nexus]

right so you are trying to construct a map from the group into the nonzero reals with quotient equal to the translations

the nonzero reals are a different group

[balmy python]

yep

[glad osprey]

Just a side note, is G itself isomorphic to a semidirect product of R* and R (where R* acts on R)?

[noble nexus]

yes

[balmy python]

whats a semidirect product

and what's R*

[noble nexus]

nonzero reals

[balmy python]

ah

[noble nexus]

don't worry about what a semidirect product is you'll learn that later

but its basically a way of combining two groups in a more complicated way than just a direct product

[balmy python]

i learnt a direct product i thik

stuff like AxB right?

[balmy python]

ahhh

[noble nexus]

a direct product is like combining two groups but they don't really interact with each other at all (explicitly, the two groups commute with each other)

[tulip otter]

yea a direct product is the cartesian product with coordinatewise product as the operation of the group

[noble nexus]

in a semidirect product you allow the two groups to interact basically and as a result they don't commute (meaning (x,1)(1,y) is not in general equal to (1,y)(x,1) in a semidirect product)

[merry harness]

i don't really understand semidirect products eitehr :(

[noble nexus]

they are often presented in a very confusing way

[merry harness]

are there any good concrete examples

[noble nexus]

but a lot of the examples are quite simple to understand

the best is just the example of the group of affine transformations

[thorn jay]

a semidirect product is a way to define conjugation in a group "externally"

[noble nexus]

of say R^n

[merry harness]

affine transformations, like linear automorphisms?

[noble nexus]

is equal to the semidirect product $\mathbb{R}^n\rtimes GL_n(\mathbb{R})$

[balmy python]

ah so i would want the kernel of my map to be the subgroup H

[cloud walrus]

Blake

[balmy python]

so phi(f_{0,b}) =0

[noble nexus]

an affine transformation is a transformation of the form $v\mapsto Av+b$ where $A$ is an invertible matrix and $b$ is a vector

[cloud walrus]

Blake

[merry harness]

GL_n(R) is linear automorphisms right

[noble nexus]

yeah or equal to 1

[noble nexus]

yea

[merry harness]

so what does the R^n bit contribute

[noble nexus]

because the nonzero reals have identity equal to 1

the R^n bit contributes the translations

[balmy python]

but then how do i know what to do from there?

oh wait

[merry harness]

ohhh

[noble nexus]

so if you think about it, every affine transformation consists of a pair (A,b) where A is a matrix and b is a vector

but iits not isomorphic to the direct product

[merry harness]

So what is the difference between $\mathbb{R}^n\rtimes GL_n(\mathbb{R})$ and $GL_n(\mathbb{R})\rtimes \mathbb{R}^n $

[cloud walrus]

qiu

[noble nexus]

because if you take two transformations say $v\mapsto Av+a$ and $v\mapsto Bv+b$ and then compose them, you get $v\mapsto B(Av+a)+b = (BA)v+Ba+b$

[noble nexus]

the other thing you wrote doesn't really exist

or you need to be more specific

[merry harness]

okay

what is the diff between left and right side of \rtimes

[noble nexus]

for a semidirect product you need to specify two groups as well as an action of one group on the other

so in this case, you have an action of GLn on R^n

[tulip otter]

so just a note that this does work, and from here you would need to somehow obtain an isomorphism with the multiplicative group of non-zero real numbers

[balmy python]

the thing is, i did group theory in my first year but didn't really grasp the technique but i decided to pick the extension module so that i can get better at it

but im getting worse and worse

[noble nexus]

and that is encoded in the group structure of the semidirect product

[merry harness]

okay so there's an implicit action group

[tulip otter]

but you can get to an isomorphism directly instead of going this way

[noble nexus]

you have to be a bit careful though, because its not just any action it has to be an action that preserves the group structure

[tulip otter]

you only have to modify the homomorphism you got slightly so that it gives output in the non-zero reals

[noble nexus]

so in this case, what that means is that A(v+w) = Av+Aw

[balmy python]

and because all cosets are of the form {ax+b where a is fixed and b can vary}, we map each coset to it's corresponding a?

[tulip otter]

(remember that a neq 0)

[noble nexus]

you shouldn't need to think about cosets here

because the first iso theorem takes care of that

instead you should just think about a map from your group G to the nonzero reals

[cloud walrus]

Blake

[noble nexus]

if we abstract from this example of GLn and R^n by the way

it makes it very easy to remember the group law for semidirect products

lets denote the transformation v -> Av+a by the pair (a,A)

[glad osprey]

This means the action has to be a group hom H -> Aut_Grp(N), not just H -> Sym(N)? ie. a group action in Grp?

[balmy python]

yeah

[noble nexus]

yes precisely

[tulip otter]

so here the homomorphism you found maps from G to the set
{ax |a in R, a neq 0}. Think about a way to slightly change this homomorphism so that it maps from G to R* instead of {ax |a in R, a neq 0}

[noble nexus]

then the group law is given by (b,B)(a,A) = (Ba+b,BA)

[balmy python]

yeahhh

i mean im pretty decent at everything other than groups

i guess groups has a bigger learning curve

what yr u in?

[balmy python]

why though?

yeahhh

[noble nexus]

so abstractly: If we have two groups $G,H$ together with an action $\rho:G\to\text{Aut}(H)$, then the semidirect product $H\rtimes G$ consists of the pairs $(h,g)$ with the group law $(h_1,g_1)(h_2,g_2) = (h_1\rho(g_1)(h_1),g_1g_2)$

[cloud walrus]

Blake

[noble nexus]

it looks confusing because (a) both groups are written multiplicatively and (b) the notation for group actions is a bit clunky

[balmy python]

such that all translations map to 0

hmmm

[tulip otter]

(to 1)

[noble nexus]

right so, if you have a transformation $f_{a,b}(x)=ax+b$ and you want to pick out a nonzero real number, what's the simplest thing you could pick?

[cloud walrus]

Blake

[balmy python]

1

😭

[noble nexus]

well ok, besides 1

[balmy python]

2

[noble nexus]

we also want it to be a homomorphism

[balmy python]

joke im just ragebaiting

[noble nexus]

you want something like $\Phi(f_{a,b}) = ?$

[cloud walrus]

Blake

[balmy python]

hmmm

a?

[noble nexus]

yep

[balmy python]

but why does a work

[tulip otter]

why does mapping to ax work?

[noble nexus]

so you would need to check its a homomorphism, but that isn't too bad

[tulip otter]

you did this earlier right?

[noble nexus]

and more importantly lets look at the kernel

[balmy python]

oh yeah

that's fine then

[tulip otter]

here

[noble nexus]

If $\Phi(f_{a,b})=a$, then we have $\Phi(f_{a,b})=1$ if and only if $a=1$

[cloud walrus]

Blake

[balmy python]

yeah 😭

[noble nexus]

but $f_{1,b}(x)=x+b$

[cloud walrus]

Blake

[tulip otter]

yea, you basically were more or less done when you chose that map

[balmy python]

yeah

[tulip otter]

but the thing was that its easier if you map to a instead of ax

easier as in you would have to do less work

well only slightly less

[balmy python]

so ker phi = f_{1,b}

[tulip otter]

more precisely, its the set of these functions which is the kernel

[balmy python]

hmmm so if i was to write this formally:

By the first iso thm, there is a bijective homomorphism G/H -> phi(G) where phi: G -> R* where H is the kernel of phi.
let phi(f{1,b}) = 1

phi(f{a,b})
= phi(f{1,b} f{a,0})
= phi(f{1,b}) phi(f{a,0})
= phi(f{a,0})

So it would make sense to define phi(f{a,b}) = a and we have an isomorphism from G to R*

[glad osprey]

Finding the right homomorphism from G is mostly just guesswork, but there is some intuition that should lead you to the right guess: first of all, the elements of G are described by a pair of reals a and b, so a homomorphism from G is probably gonna be some (arithmetic) expression in a and b. Secondly, you know that the kernel should be the elements x + b, so they should map to the identity; this is a sign that the homomorphism should not involve b at all. And finally, the homomorphism you're looking for needs to be surjective, so just mapping to ax + b to 1 isn't good enough. In fact, since you know the elements are described by real numbers a and b, it's likely that the image the homomorphism should be R or some subgroup of it

[balmy python]

yeahhh

thanks!

[tulip otter]

if you want to write a formal answer then i think you should just say let φ:G->R* be a map defined by φ(f_{a,b})=a, then φ is a surjective homomorphism (you prove that it is) whose kernel is H={f_{1,b}| b in R} (you may want to write some steps to show this too). Hence H is normal in G and by the first isomorphism theorem, G/H \cong R*

something like this

you dont need to show your thought process that led you to this particular homomorphism

(you should probably verify this with the prof but i think this will probably be the case)

also define the notation f_{a,b} beforehand since its not defined in the original (given) question

it also depends on how detailed does the prof wants the proof to be

[lavish ferry]

Anyone know of ways to preserve a degree function (in the Euclidean domain sense) of a field inside the ring of polynomials generated by the field?

To be specific, if you have a field $F$ with degree function $\delta: F\setminus{0} \to \mathbb{Z}_{\geq 0}$ and I realise this question… is easy if you you are looking at $\delta(u) = 0$ lol. Then the degree of anything in the imbedding of $F$ in $F[x]$ is the degree of its corresponding element in $F$

[cloud walrus]

Elizabeth

[lavish ferry]

(Where degree is defined of course)

[rapid cave]

wdym by a degree function on a field?

[lavish ferry]

All fields are Euclidean domains and I am meaning that sort of function used with Euclidean domains

I think they also have the name norm function? I learnt it as degree function in my course

[tough raven]

For a field you can just take the function |0| = 0, |x| = 1 otherwise.

[lavish ferry]

Yeag

Are we using the same definition?

[tough raven]

Maybe.

Is your question: you have a norm on the field (which may not be so trivial), you want to extend it to a norm on the polynomial ring?

[lavish ferry]

I wanted a way to do this, but I realised you can do it in a sort of trivial way as I was writing it out. But I am curious on this question you posed

[tough raven]

Actually checking the definition on Wikipedia, the only norm function on a field in the Euclidean domain sense is (scalings of) the trivial one:

For all nonzero a and b in R, f (a) ≤ f (ab).

Because of this.

[lavish ferry]

Ah yeah. I wonder if you omit this?

[tough raven]

I'm not sure one should want to do that.

[lavish ferry]

I’ll need to practice more problems to understand ur wisdom

[tough raven]

I mean, it just seems like the kind of property that would probably be used in a lot of places.

[lavish ferry]

I know some nice properties don’t require it (Euclidean domains are principal ideal domains for example) but it seems useful so idk

Besides it is induced by the existence of a norm without the property right?

[rapid cave]

A norm is defined without this property

[rocky cloak]

The point of the norm is to control the remainder when doing euclidean division.

In fields you can always divide, so there's no remainder and the norm becomes sort of pointless.

Though yes you can pick whatever as your norm if you like I suppose.

And whether you can extend this to a norm on F[x] would highly depend on what you pick.

[lavish ferry]

Yes makes sense

[lavish ferry]

Because fields are commutative division rings by definition right?

[rocky cloak]

that is what they are

which means you can divide by stuff

[lavish ferry]

But our notion of division becomes messy with skew fields because you have left and right divisibility?

[silk orchid]

Let phi: A to B be a homeomorphism of rings and let J be an ideal of B.
I can see why phi(phi^{-1}(J)) should be inside of J, but it seems to me that the two should actually be equal, which the book doesn't seem to imply

am i correct?

[lavish ferry]

From set theory this holds only when we have a surjection

[silk orchid]

whoops i see what i did there mb

[lavish ferry]

Consider the zero ring homomorphism

[silk orchid]

thanks

[lavish ferry]

It happens

[silk orchid]

i'm resuming serious maths after maybe a couple months, so my gears are rusty

[rocky cloak]

Less time for oiling gears if you're busy doing math

[south patrol]

This isn't usually a ring map

(Unless you have a non-unital convention)

Or unless by 0 map you mean the map to the 0 ring, in which case your example doesn't work

[south patrol]

Presumably this question also has a typo and should have J instead of B

[silk orchid]

that's indeed the typo that happened, thank you, i'll fix it

[south patrol]

Anyway I think it is nice just to take the example J = B and any nonsurjective map of rings (as Elizabeth suggests originally). Then this is a counterexample working independent of any conventions

[silk orchid]

lol yes i'm not sure how i ever missed that
but ty :)

[south patrol]

Np

[tough raven]

Hello!

[silk orchid]

hii

[lavish ferry]

I mean $f: R \to S$ s.t. r |-> 0. This is a ring homomorphism. I am reading the question again now and seeing that it mentioned homeomorphism instead of homomorphism but aren’t those topology isomorphisms?

[cloud walrus]

Elizabeth

[lavish ferry]

(Assuming S is non-trivial this mapping isnt surjective)

[south patrol]

Only with nonunital conventions I mean

Usually ring maps are required to send q to 1

[lavish ferry]

Oh yes cause you need it to be a multiplicative monoid hom right? Yeah I was meaning a rng hom

[karmic moat]

Do you know about the structure theorem for fg modules over PIDs? It’s used in the proof iirc

[tough raven]

Let H be a non-normal subgroup of a group G. Is there a description of intermediate subgroups in terms of the coset space G/H?

[rocky cloak]

They correspond to quotients of the G-set G/H at least

[candid patrol]

And it preserves the notion of normality

[dull ginkgo]

Because nightmare

It also gives Jordan normal form for fields

Since F[X] is a PID

[sacred wharf]

jordan normal

why learning about basketball

[dull ginkgo]

lebron james ring

lebron james cohomology

[velvet hull]

lebron caught assuming his PID is a ED

[sacred wharf]

he didn't memorize the chain fr

ED --> PID --> UFD --> ID

right ? did i miss smth

[dull ginkgo]

UFD -> GCD -> ID :3

[velvet hull]

All gcd domains are ufds (real)

[dull ginkgo]

Un-noetherians your principal ideal poset

[sacred wharf]

artinian > noetherian

[tough raven]

Please

PID => Dedekind => Prufer => integrally closed domain => integral domain

[dull ginkgo]

100 years of noncomm jacobson ideal spam effective immediately

[elfin wraith]

Bro hasn’t heard of Hopkins-Levisky

[sacred wharf]

Nah noetherian > artinian

Because wth is finitely cogenerated

Finiteky generated as an equivalence actually makes sense

[rocky cloak]

Finitely generated = every sum reduces to a finite sum.
Finitely cogenerated = every intersection reduces to a finite intersection. Makes sense to me

[sacred wharf]

Hmm

What it even mean for infinite sum

Man i hate infinity

[delicate orchid]

what category we in

ewww looks like Ring... PASS!

[sacred wharf]

Dw nobody is offering you a ring

[delicate orchid]

I believe you have failed to account for your mother when performing this calculation

[sacred wharf]

Nah she doesnt like field with one element lovers

[next obsidian]

Who up alg-ing they bra

[rocky cloak]

In the category of modules.

Is there a way to define a ring being Noetherian in the category of rings?

[sacred wharf]

What is a category

[delicate orchid]

do I have brainrot for thinking about category theory when I see "cogenerated" rather than ring theory or something

[next obsidian]

Probably. Will stew upon it

[rocky cloak]

In commutative rings you can probably just use quotients (as regular epimorphisms)

[delicate orchid]

all rings are commutative so that's ok

[next obsidian]

Oh you want in all of ring

[thorn jay]

congruence poset being Noetherian

congruences of R are precisely the fiber products of homomorphisms from R as subobjects of R × R (i.e. fiber product of f : R → S with itself)

[thorn jay]

same thing

lol

[rocky cloak]

Non-Noetherian simple rings would like a word

[thorn jay]

but then the regular epimorphisms wouldnt work either

[rocky cloak]

Yeah, I'm assuming nothing works

[thorn jay]

isnt there an adjunction between Ring and Rng? may be useful

[rocky cloak]

There is, but I'm not sure how it would help

[thorn jay]

right i forgot we want left ideals

[rocky cloak]

I mean taking a ring to its opposite is an autoequivalence, so best case would be determining rings that are both left and right Noetherian

[thorn jay]

yeah i was about to say

R being left noetherian doesn't imply being right noetherian right

[rocky cloak]

It does not

[elfin wraith]

What a world that would be

[rocky cloak]

[R, R; 0, Q]
being an example that is left artinian (and Noetherian), but not right Noetherian

[elfin wraith]

What is that? I don’t think I’ve see that notation before

[rocky cloak]

Matrix where the corresponding elements are taken from R, Q or 0 respectively

R real numbers, Q rational numbers

[elfin wraith]

Ah I see

That’s a very cool example

[rocky cloak]

Basically the left ideals behave like real subspaces of R^2, while the right ideals behave like rational subspaces of R.

[elfin wraith]

Very cool, I think I only knew like quotients of k<x,y> and the twisted polynomial ring

[rocky cloak]

Fun fact: if F/K is a field extension then [F, F; 0, K] is representation finite iff the degree [F:K] is less than or equal to 3.

[elfin wraith]

What does representation finite mean?

[white oxide]

Can somebody please explain to me why this is the goal? If $\overline{\mathfrak{q}2}$ is a prime ideal of $B{\mathfrak{q}_1}$, then its contraction in $B$ is a prime ideal which doesn't meet $\mathfrak{q}_1$. But we want this prime ideal to be contained in $\mathfrak{q}_1$, no?

[cloud walrus]

okeyokay

[rocky cloak]

Has finitely many indecomposable representations

[elfin wraith]

That sounds like a pretty cool result though, I always like a “this holds for all less than [small number]” theorem

[elfin wraith]

Ah ok, very nice

[rocky cloak]

The real result is Gabriel's theorem: a hereditary finite dimensional algebra is representation finite iff it's Gabriel species is a Dynkin diagram.

Then these have species A2, B2 and G2 for degree 1, 2, 3 respectively.

[white oxide]

Never mind I got it mixed up

The prime ideals of B_{q_1} are in correspondence with the prime ideals of B which don't meet B/q_1

[elfin wraith]

Ahhh I see, I came across that like a year ago when I was doing my non com project. I do really want to learn about quivers

[rocky cloak]

Yeah species is like quivers, but for non-algebraically closed fields

[quiet pelican]

Quivers are nice, but I’ve probably forgotten most of what I learned about them now

[elfin wraith]

Oh interesting, I hadn’t realised the definition of them required algebraic closure

[rocky cloak]

The definition doesn't, but if the point is to capture the behavior of finite dimensional algebras they don't work

Like given a finite dimensional algebra you think of the vertecies as representing projectives and arrows as (a basis for) irreducible maps.

But if you're not over an algebraically closed field, then the endomorphism rings of your projectives can have different residue fields (some finite dimensional division algebras over your base field) and then just counting irreducible maps doesn't work cuz you get a different answer of you consider them going out of the domain or into the codomain.

[elfin wraith]

Very interesting, I really wish I could’ve taken some amount of rep theory

[thorn jay]

grahhhh amazing theorem

gotta be up there as a 10/10

[thorn jay]

Erdmann has a very good book on algebras and representation theory, in particular building up to Gabriel's theorem

it's an undergrad book so doesn't go particularly deep but still very well written

it also gives a nice bigger picture for e.g. group representations that Serre doesn't give

[crystal vale]

if w is a primitive cube root of unity, and i want to find unit of Z[w], how do i find?

i can say 1, -1, w,-w, w^2, -w^2, w + w^2, -w-w^2, they are units

units elements

[delicate orchid]

surely a norm argument proves that these are the only ones, although w+w^2 is -1

[crystal vale]

ah i thought about norms, but then i got lost, can you tell me how do i define norms on that ring?

[delicate orchid]

it's a subring of C

[south patrol]

Tbf norm doesn't always agree with the norm from C tho. Tho stuff is still good here

[delicate orchid]

yeah but it's trivial to check it still works

[south patrol]

Ig I mean like you need to know the norm is an integer

[delicate orchid]

you are right, I more meant it as a hint to "maybe see if the obvious norm works"

[south patrol]

Ah ok

[delicate orchid]

is this even true? norms are R^+-valued

[south patrol]

I mean the norm in the sense for number fuelds lol

As in in the algebraic / number theory sense

[delicate orchid]

no clue what that means

[south patrol]

Like product of conjugates of your element

[delicate orchid]

ah I see

[south patrol]

If you start with an algebraic integer then this gives you an integer

Here fortunately all is good

[delicate orchid]

yeah cause potentially we could've had rational norms which is ew

[south patrol]

Just thought may be worth amplifying

[crystal vale]

but i think if norms take value in Z, then we can comment about inverse of an element, if a is unit then N(a) should be 1 or -1, right? i think norm usually takes value in N, but as complex entries it can take in R^+

i don't see how norm on Z[w] will be help me here

[delicate orchid]

you can't have -1 cause norms are positive valued, otherwise yes

[delicate orchid]

you listed exactly every single element with norm 1

although 1 and -1 were listed twice

[alpine plank]

Norm in the number theory sense can be negative valued

[delicate orchid]

considering I didn't know the definition of that until 5 minutes ago, it's kind of obvious that's not the sense I'm using "norm" in

[alpine plank]

I can't read my bad

[delicate orchid]

A normed ring. A ring with a norm.

a norm with a ring? maybe

science cannot yet answer these questions

[crystal vale]

if a + bw + cw^2 is an unit then | a + bw + cw^2 |^2 = 1, right? but | a + bw + cw^2 |^2 = a^2 + b^2 + c^2 + ab + ac - bc

[delicate orchid]

am I missing something here? The only complex numbers with norm 1 lie on the unit circle. You have 1, w, w^2 in the intersection of the unit circle with Z[w] and nothing else

1 = |1| = |aa^-1| = |a||a^-1| <=> |a| = |a^-1| = 1 as |x| >= 0 for all x

[crystal vale]

|a||a^-1| = 1 doesn't imply |a| = 1

[alpine plank]

The last step is sus

[crystal vale]

because |a| can be any positive real number

[alpine plank]

Example: In Z[root2] (1+root2)^n is a unit for all n and it can have arbitrarily large euclidean norm

[delicate orchid]

am I losing my mind? this is the absolute value of a complex number

this is $\sqrt{Re(z)^2+Im(z)^2}$ how on earth is that ever negative

[cloud walrus]

FIELD WITH ONE ELEMENT LOVER

[crystal vale]

but in case of Z[w], i verified that the norm takes only non negative integer, therefore we can say N(a)N(b) = 1 implies N(a) = 1

[delicate orchid]

it's the length of a line

[delicate orchid]

ok cool

[alpine plank]

a times 1/ a = 1 for every positive real number a

[delicate orchid]

these are integer multiples of roots of unity

[alpine plank]

Yeah I meant you have to specify the norm takes integer values

[delicate orchid]

it's a lattice in C who's intersection with R is Z like

I don't understand where I'm going wrong here

[south patrol]

I mean you have to prove those are the only ones

Ig clear from a picture though

Lol

[crystal vale]

how do i find the all integers solution of a^2 + b^2 + c^2 + ab + ac - bc = 1?

[quiet pelican]

Find an explicit basis in which that quadratic form has form x^2 + y^2 + z^2 (which you can do by like Gram Schmidt using that as your form) then it’s easy

[crystal vale]

i don't know about quadratic form

[quiet pelican]

So you can rewrite your equation as like x A x^t = 1 where x = (a, b, c) and A has 1s on the diagonal, and every other place is +-1/2 and A is symmetric

Then you can orthogonally diagonalise A by a theorem of linear algebra

That then gives you a new basis in which your equation looks like yy^t = 1 for y = Bx, for some orthogonal matrix B

And that equation is trivial (and I suspect B ends up having only integer coefs, or at least something nice to give a criterion when x is integer in terms of y)

[sacred wharf]

yo yo yo

a submodule of a free module is only a free module if it is finitely generated

?

[south patrol]

Being free is essentially irrelevant to being finitely generated

E.g. over a PID A, every submodule of a free A-module is free

Or e.g. over a field every module is free

[rocky cloak]

Submodules of free modules are usually not free (finitely generated or not), but the result is true for PIDs

[knotty badger]

Subgroups of free groups are free

[crystal vale]

I = ( x ), right? So K[x]/I is just isomorphic to K( as a field isomorphism)

[rocky cloak]

I = (x) seems a little too big

[crystal vale]

I = (x) as an ideal

I mean K[x] is PID so ( x^3 + x^2, x^6 ) = ( GCD(x^3 + x^2, x^6) ).

And i found that GCD is x

Oh x^2

[sacred wharf]

ohh i see

yes i meant pids

[warm ember]

is the ring of algebraic integers UFD

is it PID and is it euclidean domain

[velvet hull]

of an arbitrary number field?

[warm ember]

of Z

[velvet hull]

so you mean Qbar?

[warm ember]

Z bar

[rocky cloak]

It's sort of a PID in that it is a Bezout domain

Which is like a non-Noetherian pid

[warm ember]

its def a UFD tho rifht

[warm ember]

basically that means every ideal is either principle or infinityl generated right

[rocky cloak]

Well, you can't necessarily factor things into irreducibles.

Like 2 = sqrt(2)^2, but also cuberoot(2)^3, etc

[warm ember]

oh

ok ty

[tough raven]

Is there a finite group G with a Hall subgroup H and a subgroup N such that H ∩ N is not a Hall subgroup of N?

Note that HN must not be a subgroup of G, and in particular N must not be normal, or the conclusion does hold.

[signal hedge]

This is with respect to D_10 (or D_20) on a dodecagon.
I am doing the reflections for "bead-to-bead" I'll call it. 10 beads with 3 colors, trying to find how many colorings are possible with burnsides lemma.
With respect to the reflecting line 0,5 (yellow) - my thoughts are that this is a 3^6 option as 0, 1, 2, 3, 4, 5 can be any of the three colors, but 6,7,8,9 are only given no choice (3^0).
Does that sound right? (this is a homework problem)

[sacred wharf]

So

If I want to find the SNF of a mxn matrix

I first begin by trying to find RREF of V with respect to M
And then with this RREF I transpose it and try and find RREF with respect N or smth

and then i transpose back and i got it

idk

[white oxide]

As a sanity check, this is true because if we have an ascending sequence of subgroups then it stops at G and descending sequence stops at {e}?

[velvet hull]

well it's because its finite

there are finitely many submodules so every chains has to stabilize

[white oxide]

Yeah

Does this follow from Sylow

looks like Sylow

[sacred wharf]

Finitely generated is equivalent to being noetherian

[sacred wharf]

this follows from sylow + abelian

[white oxide]

I forget the relevant sylow theorem

[velvet hull]

finitely generated module over noetherian ring implies noetherian as a module

[wraith cargo]

Finitely generated ideals not finitely generated module

[white oxide]

matter of fact I don't remember any of them just have to do with prime order groups

[sacred wharf]

ohh

yea

[sacred wharf]

So there are 3 sylow theorems

The first one is that a group of order of the form p^n*m where p doesn't divide m must have at least one group of order p^n

[white oxide]

Mm okay

I guess it's time to review

[sacred wharf]

bro needs anki

[white oxide]

anki for theorems is crazy

[sacred wharf]

y

[white oxide]

Idk I just always viewed anki as a way to learn vocab lol

actually i'm too lazy to review it just for one example 😹

blackbox time!!!

[sacred wharf]

u have to remember it lil bro

[white oxide]

it's been two years since I've learned ts and I'm not reading allat just for one example

Here, what is G?

Oh it's from the previous example

[tardy hedge]

Lol

Ya lil bro

[white oxide]

😂😂😂

[rocky cloak]

G is the p-sylow subgroup of Q/Z, but the fact the subgroup of order p^n is unique isn't really related to Sylow.

It follows pretty immediately from just describing what the elements of Q/Z are though.

[sacred wharf]

does the last line use correspondence theorem ?

[rocky cloak]

Sure. It's saying that a subgroup of G/C is a subgroup of G containing C yeah

[sacred wharf]

What does finite theorem of abelian groups say succintly

[rocky cloak]

That every finite abelian group is a product of cyclic groups.

Or even product of cyclic groups of order prime powers

[noble belfry]

Can someone give me a shit ton of group theory questions which use group actions / Sylow theorems / semidirect products

[elfin wraith]

Looking up group theory quals would probably bring up a lot of stuff on that

If not, can’t go wrong with Dummit and Foote

[sacred wharf]

here the key idea is that

[rocky cloak]

Pick an integer, prove / disprove there's no simple group of that order.

Count the number of groups up to isomorphism.

[sacred wharf]

The key contradiction idea here is that if there is non nilpotency than F is non empty ?

[sacred wharf]

r u a prof in algebra ?

[rocky cloak]

I'm a soon to be PhD in algebra

[sacred wharf]

ohhh

i thought you were like a 45 year old professor

[rocky cloak]

Give it a decade or two and I'll reach 45 soon enough

[sacred wharf]

are you doing the phd at NTNU ?

[elfin wraith]

Is your defence next week? I know you said it was pretty soon

[rocky cloak]

It is yeah

[candid patrol]

Damn

[elfin wraith]

Exciting times

[rocky cloak]

Well, next next week I guess

[candid patrol]

Same 🫣

[elfin wraith]

I’ve got an interview for a PhD scholarship that week, it’s like the passing of the guard

Except I’m much worse at maths

[rocky cloak]

Next generation to carry the mantle

[elfin wraith]

Someones got to pretend like noncom rings matter

[rocky cloak]

The real noncommutative rings were the friends we made along the way.

[elfin wraith]

I do actually need to brush up on all that stuff before my interview, also pray the 2 academics there don’t really know a lot of that stuff

Someone I know did it last year, spoke about wanting to do Lean and got Kevin Buzzard as his interviewer

[sacred wharf]

to prove that kernel --> normal

Is this the idea:

p(gkg^-1) = p(g)p(g^-1) = p(k_1)

so gKg^-1 is a subset of K

and I believe that should be enough to prove it is equal to K

[kind temple]

i think its more clear to write p(gkg^-1) = p(g)p(g)^-1 = 1

but yea, then gKg^-1 = K

[sacred wharf]

rhomomorphism is when ?
given unity in R
is when phi(ra + rb) = rphi(a) + rphi(b)
otherwise it is
phi(ra) = rphi(a)
and
phi(a+b) = phi(a) + phi(b) =

[tardy hedge]

Ts whole time jagr not a phd but he knows more algebra than my professors

[tardy hedge]

Yea that first one is like equivalent

if phi(ra+b) = rphi(a)+phi(b) ur good broski

[sacred wharf]

only works if r has unity right ?

that is for equivalency

u better not give me that 'all rings have unity '

[tardy hedge]

Yeah

I havent really thought too much about non unital modules tho

[sacred wharf]

aight second isomorophism theorem is AB/B is isomorphic A/B\cap A

let me try and remember the proof

i think it's some stupid composition map

a --> aB
ok the kernel is just a/a n b

right

A/B = AB/B

or

hm

[thorn jay]

wdym "rhomomorphism when"

this is just the definition of an R-module homomorphism?

[sacred wharf]

that is what i'm asking

[sacred wharf]

ph(ra+rb) = rphi(a) + rphi(b)
this is equivalent to the definition of rhomomorphism iff unity ?

[thorn jay]

uh, i guess

[thorn jay]

AB → A/A ∩ B given by ab ↦aA ∩ B is well-defined and has kernel B

[sacred wharf]

idk what you said but

i'm using the method where apparently you do:
A --> AB --> AB/B
and you look at the kernel of this map apparently it's just A n B

i don't get what is AB/B

so that is just all aBB ?

the fuck is that 😭

[thorn jay]

do you know what AB is

[sacred wharf]

all finite sums of the elements in the form ab

?

[thorn jay]

okay bro are we doing rings modules or groups

[sacred wharf]

groups

[thorn jay]

we know B is normal, right?

[thorn jay]

so how can one speak of "sums"

[sacred wharf]

yes

[sacred wharf]

oh um

all fininte multiples of ab ?

ig

i think

[thorn jay]

not even! because B is normal, AB is actuall just { ab ∣ a ∈ A, b ∈ B } as this will already by closed under multiplication

[sacred wharf]

well they are equivalent

[thorn jay]

sure, but im telling you that its not needed to think of it as product of elements of the form ab

[sacred wharf]

ok but the map A --> AB --> AB/B
it's a--> aB right

aB --> aBB

[thorn jay]

no

we just talked about this AB doesnt have elements of the form aB

the total map is given by a ↦aB

[sacred wharf]

wym total map ?

[thorn jay]

composition of A → AB → AB/B

[sacred wharf]

ok so where does a map to

in A --> AB

?

[thorn jay]

inclusion, bro

[sacred wharf]

oh

a goes to a

[sacred wharf]

cn u confirm this ?

[thorn jay]

wdym

[sacred wharf]

so

a maps to a

in AB

since we know a is in AB

[thorn jay]

yesah thars what inclusion means

[sacred wharf]

yes

okay

thanks

[tardy hedge]

a maps to a*1 if u want to make more sense of it ig

Actually idk the context so i might be wrong rn

[sacred wharf]

nah ur right

[twilit wraith]

happy halloween yall i have a spooky problem from a algebra prelim at my uni

If G is a group of order 504 containing an element of order 21, then there exists a subgroup in G of index 8

ive made good progress on it but i havent quite finished it

maybe i should explain my progress so far

[kind temple]

is this not straight up the Sylow theorem? ||504 = 2^3 * 3^2 * 7, so there is a Sylow 2-subgroup of order 2^3 = 8||

[twilit wraith]

Let g in G such that |g| = 21

The number of Sylow 7-subgroups is 1,8, or 36

so if theres 8 were done

as the normalizer of the group generated by g^3 has index 8 just by virtue of g^3 being a sylow 7-subgroup

or the normalizer of any of the sylow 7 subgroups i suppose

there cant be 36 sylow 7-subgroups as then the normalizer of the group generated by g^3 would have order 504/36 = 14

but <g> normalizes <g^3> and has order 21

if there is one sylow 7-subgroup then we can just semidirect product it with one of the sylow 3-subgroups to get a subgroup of order 63 and thus index 8

i think thats good enough but idk

[twilit wraith]

ooh i realized i meant index 8

btw is my argument correct?

im most skeptical about the semidirect product part

[paper kite]

technically there's no need for the SDP
you can argue that: if there's 1 7-sylow, then it is necessarily normal. the quotient group has order 72, and thus has a 3-sylow of order 9.

the preimage of this subgroup under the projection map is a subgroup of order 63

[twilit wraith]

oh interesting

yeah i suppose so

[paper kite]

in general (as i understand) you can't just take two subgroups H, K in G and form a H sdp K type of subgroup

the underlying set would be HxK, but it's not obvious how you can identify that as a substructure of G

in some sense

[twilit wraith]

well its because we know that the 7-subgroup is normal

or maybe thats still not enough

[paper kite]

uh

[twilit wraith]

also its disjoint from any sylow 3 subgroup besides identity

[paper kite]

this is like a short exact sequence of groups i believe

and SDPs are a specific case of this

let N be the normal 7-sylow, and Q the quotient group

wait never mind

[twilit wraith]

wait so does that mean im good

bc i thought the way SDPs work are what gives them the ability to be identified as a substructure of G

[paper kite]

SDP is for two groups though

what's the plan for SDPing two subgroups

the multiplication law is already determined since there's already a larger group

[twilit wraith]

i see

[thorn jay]

should be enough

[crystal vale]

If my ring R[X] is ED, then is that result valid?

[quiet pelican]

Possibly not if it’s an ED by something other than the degree function
But I’m not sure that’s actually possible
Actually I think I vaguely remember a result that if R[X] is a PID, then R is a field, so it’s the same iirc?

Yeah that’s right

So the result for R[X] an ED (and in fact even a PID) is equivalent to the result for fields

[crystal vale]

Yes R[X] pid implies R is field

I see

I got it

[quiet pelican]

Like suppose r isn’t a unit and consider (r, X)
Then this is generated by some a \in R[X]
As R is a domain (it’s a sub ring of a domain), we have r = ab, so a is of degree 0
Also X = ac, where c must have degree 1, and hence c = a’x + b’
So a is a unit, as aa’ = 1
But (r, X) \cap R = (r) as anything multiplied by X can’t contribute in degree 0
So (r) = (1), done

[crystal vale]

And I did this by repeatedly applying division algorithm

[quiet pelican]

Yeah that’s how I’d do it

[crystal vale]

Now i want a hint for a question, if R is a commutative ring with unity, I have to prove R[X] has infinitely many maximal ideals.

[quiet pelican]

Hint: ||quotient by a maximal ideal of R, showing that it suffices to prove it for a field||

[crystal vale]

So you mean first I have to show for F[X]

[quiet pelican]

Yeah where F is a field
(You could also choose to work on the reduction to the field case first)

[crystal vale]

Okay so for a field, it is enough to show that F[X] has infinitely many irreducible elements, and if they have finitely many p1,..,pN then we can generate one more irreducible element distinct from p1,..,pN that is p1...pN + 1

[quiet pelican]

Yeah
(Well technically the irred is something that divides p1…pN + 1)

[crystal vale]

But we can say p1 ..pN + 1 is irreducible right?

But what if it is a unit element?

[rocky cloak]

The product of subgroups HK is called semidirect if H and K don't intersect and one of them is normal in HK

[quiet pelican]

Consider the highest degree term

[crystal vale]

I am dumb

[quiet pelican]

We can say it’s divisible by an irred not in our list (by UFD)

[crystal vale]

Yes

So it must be irreducible

[quiet pelican]

We must get an irreducible

[crystal vale]

Yes

So we are done for the field

[quiet pelican]

Yup

[crystal vale]

Okay R has maximal ideal M, R/M is field

And R/M [X] is isomorphic to R[X]/ M[X]

[quiet pelican]

(I’m about to disappear to walk to lectures for 30 mins)

[quiet pelican]

Yeah

[crystal vale]

Oh

[quiet pelican]

2 words and you’re done

[crystal vale]

I got it, thank you

[crystal vale]

Correspondence theorem, right?

[quiet pelican]

Yup

[crystal vale]

[tardy hedge]

Im not actually sure how that argument works

Is it like p1..pN +1 has a factorization into irreducibles with factors that arent p1…pN

It has at least one factor that is distinct because F[x] is a ufd?

[quiet pelican]

Yeah

Actually I’d move the “because F[x] is a UFD” up a line but like yeah

[crystal vale]

I think, for 1, it is not principal ideal

For 2, Z[X]/P is isomorphic to Z/3Z [X] /[ < ( X+1)^2(X+2) > ]

So it is not an integral domain, right?

[quiet pelican]

It’s non-principal because any generator would have to divide 3, so would be 1 or 3
Neither works

And that argument works for “not integral”

(Assuming the factorisation is right)

Actually I don’t think that factorisation is right

-1 isn’t a root

Mod 3

I think it should be integral

The polynomial is irred mod 3

[tardy hedge]

I forget why we can argue this sort of way, you sort of considered the quotient one at a time or something right?

[quiet pelican]

Yeah
It’s essentially the “computational” half of the correspondence theorem

Ie (R/I)/(J/I) \cong R/J

Where J contains I

[tardy hedge]

Just trying to understand what details matter here: so after using that then it comes down to checking if that polynomial is prime, (so that it generates a prime ideal). Since Z/3[x] is a ufd, then we can just check if its irreducible, bc irred -> prime?

(I think i just want to understand better ufds, prime elements etc)

[wraith cargo]

yesh

[crystal vale]

Why? The polynomial is x^3 + x^2 + 2x +2 so 1 is one root

[quiet pelican]

X^3 - X^2 + 2X + 2

Note the - before the X^2

[crystal vale]

Oh sorry i didn't see that

[sacred wharf]

Alright I want to prove third isomorphism theorem:
Conditions necessary:
H,N are normal
N belongs to H
Define a functioon gN --> gH
Well defined?
suppose gN = g1N
gN --> gH =/= H or H
Suppose gN maps to H
then: g belongs to H
gn = g_1
gn belongs to H
so g_1 belongs to H
so g1H = H
Suppose now that gN maps to gH =/= H
g1N maps to g1H
pick element in g1H such as g1h1
we know gn_1 = g1
g1h1 = gn_1h_1 which is clearly in gH
Symmetrical argument to show that gH = g1H
So we have well defined.
We will now show homomorphism
p(gg1N) = gg1H = gHg1H = p(gH)p(g1H)
Done!
Kernel of this is clearly H/N
So using first isomorphism theorem and we are done!

Can someone evaluate the proof ?

Conditions necessary: $H, N$ are normal, $N \subseteq H$.

Define a function $gN \mapsto gH$.

Well-defined?
Suppose $gN = g_1N$.

$gN \mapsto gH \neq H$ or $H$.

Suppose $gN$ maps to $H$.
Then $g \in H$.

$gN = g_1N$ implies $g_1 \in gN$, so $g_1 \in H$.

Hence $g_1H = H$.

Suppose now that $gN$ maps to $gH \neq H$.

$g_1N \mapsto g_1H$. Pick an element in $g_1H$, such as $g_1 h_1$.

We know $g_1N = gN$, so $g_1 h_1 = g n_1 h_1$, which is clearly in $gH$.

Symmetrical argument shows $gH = g_1H$.

So we have well-defined.

We will now show homomorphism:
$$p(gg_1N) = gg_1H = gH g_1H = p(gN) p(g_1N)$$

Done!

Kernel of this is clearly $H/N$.

So using the first isomorphism theorem, we are done!

[cloud walrus]

mq

[sacred wharf]

Is this correct ?

I guess froma. first glance, I think we can remove the proof by cases that is simply look at gN --> gH

this is proof of third iso theorem

[crystal vale]

If R is a ring then what is the definition of R-algebras and sub algebra?

[quiet pelican]

You can either define it as an R module that is a ring such that you have associativity between the R-action and the ring multiplication, or as a ring with a morphism from R
Subalgebra is just a subset that is a sub ring, and a submodule (by the first definition) or a sub ring and contains the image of R (by the second defn)

[crystal vale]

So when I said A is R-algebra if there is f : R -> A homomorphism, do I need that f(R) should be in the centre of A.

f: R -> A homomorphism gives me A as R -module structure

But i think there is some bilinear notion

Can you tell me in details?

[thorn jay]

yes, f(R) should be in the centre of A

(in particular you want R to be commutative)

[thorn jay]

yes, you want to have a bilinear map m : A (x) A -> A which is associative and a unit 1 : R -> A (i.e. an algebra of the monoid operad)

[tardy hedge]

The diagram/ theorem it references only shows id = phi o i, but dont you need to also justify id = i o phi?

[harsh gale]

I believe proof is essentially the same

[elfin wraith]

Isn’t the definition of an isomorphism commutative?

[tardy hedge]

Yeah i thought something with if left inv exists the right one does too

Thats true in Set right

[harsh gale]

$r\alpha(1)(m) = \alpha(r)(m)$ and $r\alpha(m)(r) = \alpha(m)(r)$

[cloud walrus]

Pyramid

[knotty badger]

what do you mean

it's entirely possible to have a left inverse without a right inverse

[harsh gale]

The idea is that $r \in R$ is essentially just a 1 in R, so it's no better than just action of r

[knotty badger]

and vice-versa

[cloud walrus]

Pyramid

[tardy hedge]

In R-mod do they imply each orher

[knotty badger]

how

this seems very false to me

[tardy hedge]

Idk thats why i asked

[knotty badger]

it's not even true in Set

nor in Ab

[tardy hedge]

Oh

[knotty badger]

unless i am misunderstanding your statement

[rocky cloak]

I don't know what theorem 8 says, but it's not hard to prove that R(x)N = N

[crystal vale]

So using this bilinear map I can define multiplication operation on A and then I can make A as a ring?

[tardy hedge]

Yeah ik u can show it directly easily, this was just instead showing it using universal property of S(x)RM with the map id: M->M

[thorn jay]

yes

a x b will then just be m(a (x) b)

[rocky cloak]

Using universal property would be what I had in mind. But I guess it can be shown directly from the relations too

[elfin wraith]

Yeah I worded this incredibly poorly lol, I was just meaning that the statement about it being the id is the definition of being an isomorphism no?

[rocky cloak]

N satisfies the universal property of R(x)N, hence N = R(x)N

[knotty badger]

can you clarify what precisely you are stating

[elfin wraith]

A morphism $f:X\to Y$ is said to be an isomorphism if there exists another morphism $f^{-1}:Y \to X$ such that $f^{-1}\circ f = \text{id}_X$ and $f\circ f^{-1} = \text{id}_Y$. Im saying the fact its the identity either way around is just what it means to be an isomorphism

[cloud walrus]

Nope

[knotty badger]

this is true

but you need both ways for it to be an isomorphism

[crystal vale]

And what is the morphism notion between two R-algebras?

[quiet pelican]

A ring morphism such that the obvious triangle (with vertices R, and your two algebras) commutes

[crystal vale]

And if I have R \subset S, then I can say S is R-algebra, right?

[quiet pelican]

In the commutative case, yes
In the non-commutative case you may require R is in the centre

[crystal vale]

Yes

[noble nexus]

tbh the definition involving ring homomorphisms I think is just confusing even though it's useful

you should just think of it like an algebra over a field, which is a vector space with a compatible multiplication

[quiet pelican]

It’s kinda more like
The definition you get by asking “well what does it mean for R to be a subring”

[noble nexus]

in the same way an algebra is a module with a compatible multiplication

[sacred wharf]

Is r algebra like r rated algebra

[crystal vale]

[quiet pelican]

Algebra is algebra

[noble nexus]

so A is an R-module together with a bilinear product A x A -> A such that r(ab) = (ra)b = a(rb) for all r in R and a,b in A

if A is unital then you get a homomorphism from R into the center but not always

by mapping r to r1

[thorn jay]

you don't require your algebras to be associative?

[noble nexus]

oh yeah that too

Technically you don't have to

If you want lie algebras to be algebras

[thorn jay]

I will call that a magma over a field tyvm /s

[noble nexus]

but nobody got time to write "associative algebra"

[thorn jay]

I'd call that a nonassociative algebra

[quiet pelican]

You have algebras and not-necessarily-associative algebras

[crystal vale]

So R -> R[ab^-1] gives natural morphism similar for R -> R[X]/(bX-a).

So I think if I define h : R[X]/(bX-a) -> R[ab^-1] by f(x) -> f(ab^-1), which is well-defined.

And it commutes the triangle

Is it correct?

[rocky cloak]

Yes, but the thing to prove here is that h is an isomorphism.

[crystal vale]

Morphism is clear, surjective is also, I have to show injective

If for f, f(ab^-1) = 0 it implies f is divisible by (bx-a), right?

[rocky cloak]

Over K at least

[crystal vale]

I see

So I have to show it is divisible by bx-a over R

[rocky cloak]

Actually this isn't true. If a and b have a common factor it doesn't work. And if R isn't integrally closed like R = k[t^2, t^3] with a/b = t it doesn't work

[crystal vale]

So you mean this question is not correct?

[rocky cloak]

Yeah

[crystal vale]

Okay

Let's assume they are co-prime

[rocky cloak]

Meaning R is a UFD? Then it should be true yes

[crystal vale]

Yes

I proved 1 =>2 =>3

So for 3 => 1, we can write f(x,y) = f(a,b) + ( x-a, y-b).

Now, if f(a,b) is not zero then it is an unit element then - f(a,b) is an unit element in R, but -f(a,b) belongs to (x-a, y-b)R in R, so (x-a, y-b)R cannot be maximal in R

Is there any mistake?

Here ring required 1 ≠ 0, therefore f cannot be any constant polynomial

[near tapir]

I’m trying to prove that every ring R has contains exactly one subring isomorphic to Z/nZ with n being the characteristic.
I’ve written:

Let f be the unique homomorphism from Z to R defined by f(k) = k(1_R). Then it follows by the definition of the characteristic that ker f = nZ.
By the First Isomorphism Theorem, Z/nZ is isomorphic to the image of f and the image of a homormorphism is a subring.

How do I argue that f being unique makes the subring unique?

[tribal moss]

If there's any homomorphism Z/mZ -> R, you can compose it with the canonical map Z -> Z/mZ, and the composition must be what you have just shown is unique.

[tardy hedge]

What is this worksheet from?

[tribal moss]

(And since Z -> Z/mZ, the image Z/mZ -> R is the same as the image of the composed morphism.

[crystal vale]

Do you want, I can share with you

[tardy hedge]

Ya sure

[near tapir]

So if S is a subring of R isomorphic to Z/nZ, then compose it with the canonical maps Z -> S and S -> R. Then since f is unique S is the image of f?

[tribal moss]

Yeah.

[near tapir]

Thanks.

I realized I should say that Z is a PID when arguing for ker f = nZ

[rocky cloak]

The implications also work when f is constant, but I guess they're not so interesting.

Your proof looks correct, though you probably want to justify why f(a, b) is in (x-a, ...

[crystal vale]

Because we already saw that f(x,y) = f(a,b) + (x-a, y-b) so it implies -f(a,b) = (x-a,y-b)R + ( f(x,y) ).

Maybe I am making some notation abuse here

(x-a, y-b)R = (x-a, y-b) + (f(x,y))

So -f(a,b) +( f(x,y) ) in (x-a, y-b)R

[rocky cloak]

Yeah, if you've already shown it that's fine I guess

[crystal vale]

How do I check K[x,y] / ( x^3 - y^2 ) is PID or not?

[fresh gate]

is it correct to say that for groups commutativity is equivalent to * being a homomorphism from G^2 to G?

[rapid cave]

Its isomorphic to k[t^2, t^3]

Should be easier to work with I think

[quiet pelican]

S = K[x, y]/(x^3 - y^2, y^2) = K[x, y]/(x^3, y^2) is not a principal ideal ring as (x, y) is not generated by a single element
If (x, y) in R = K[x, y]/(x^3 - y^2) were generated by a single element f, then (f)R = (x, y)R, so applying the projection map, φ(f)S = (x, y)S by surjectivity, contradiction
I think that should work

[noble nexus]

If the multiplication map $m:G^2\to G$ is a homomorphism, then $xy=m(x,y) = m((1,y)(x,1)) = m(1,y)m(x,1) = yx$

[cloud walrus]

Blake

[crystal vale]

Sorry I don't get it what do you mean by projection?

[quiet pelican]

Quotient map

[crystal vale]

Quotient with?

[quiet pelican]

(y^2)

No, the map R -> S

[crystal vale]

but I don't see what's a contradiction here

Is S something nice ring?

Do we have to show (x,y) is not principal ideal in S?

[quiet pelican]

Yes

(Not that hard)

[crystal vale]

Okay

Can I do quotient again?

[quiet pelican]

Yeah

If it helps

[noble belfry]

Guys

If I know that a subgroup H of G contains all p-Sylow subgroups for p!=2

How can I show that H is normal

Or is that not true

[rapid cave]

if G is of size 2^n H can be any subgroup

which most likely won't be normal

[noble belfry]

Let me specify H has odd order

[rapid cave]

so H contains all p-sylow subgroups

[noble belfry]

Yeah

[rapid cave]

then H would be all of G, hence normal

[crystal vale]

Why H needs to be G?

[vapid vale]

it doesn't need to be

H = (123), G = S_3

[crystal vale]

I don't think it needs to be

[crystal vale]

Yes

[rapid cave]

it doesn't contain 2-sylow subgroups...

[vapid vale]

assumption is p != 2

[crystal vale]

I think, containing all p-sylow subgroups, maybe for particular p

[vapid vale]

you can just show it contains all sylow subgroups then its actually generated by all of them, which is straightforward

[rocky cloak]

C3 inside A4 isn't normal

[vapid vale]

oops

[rocky cloak]

I guess it doesn't contain "all p-sylow subgroups"

In that it contains only one of the 3-sylow subgroups

[vapid vale]

lol yeah nvm'

[crystal vale]

For v, i think generators are { x1-a1,.., x_n -a_n }, right?

[rocky cloak]

But I'm guessing you mean it contains a p-sylow subgroup for all p

[noble belfry]

Nah what I mean is

It contains every p Sylow subgroup for all p except for p = 2

Where it contains none of them

[vapid vale]

there are other 3-subgroups though

[noble belfry]

Yeah

[vapid vale]

anyway this should work i think

[rapid cave]

[noble belfry]

I was thinking of something along this line yeah

[rocky cloak]

Anyway, your group will just be the product of all the Sylow subgroups and they're closed under conjugation

So normal

[noble belfry]

How do I show that it’s the product of all the sylow subgroups?

[rocky cloak]

Well, it contains the product as it contains all of them, and it's contained in the product as every element has odd order.

[noble belfry]

Got it

[rocky cloak]

I guess I'm assuming the group is finite

But sylow subgroups don't behave so well in the infinite case

[noble belfry]

Wait so

Oh yeah the second part is the important one

Why is every element with odd order inside the product

[rocky cloak]

Take an element g of order n, then (g) = Z/n = prod Z/p^k for primes dividing n.

p is not equal to 2, so you don't need to take anything from the 2-sylow subgroup

[noble belfry]

I get this now

In my words i would say that <g> has its own Sylow p subgroups which are all normal in <g> which implies <g> = P1 x P2 x … and all of those are inside some P Sylow subgroup of G so we have that every g of odd order is inside the product

I think 💀

[noble belfry]

I want to show that if take the union of one of every p Sylow subgroup in a group, it generates the group

I’m still thinking how to do this though

[rocky cloak]

Think about what the order of the generated subgroup can be

[noble belfry]

Oh it has full order

Because there’s trivial intersection

Like |HK| = |H|*|K| / | h intersection K|

So the order of the product of the Sylow subgroups is |G|

Wait this feels fallacious

[rocky cloak]

So the thing I was thinking about is that H is a subgroup of the group generated by H and K

[noble belfry]

This is only for subgroups but we’re not guaranteed P_1 P_2 is a subgroup

[noble belfry]

Yeah

[noble belfry]

We know <P_1 P_2> is a subgroup but then we’re not guaranteed trivial intersection

[rocky cloak]

Trivial intersection of what?

[noble belfry]

With P_3

[rocky cloak]

And why do you want that?

[noble belfry]

So I can do |HK| = |H||K|

[rocky cloak]

But that only works when HK is a group anyway

[noble belfry]

Oh

This is smart

I just got what you were pointing at

By lagranges we have that p^r | <the generated group>

For every prime p

So it must have full order

[noble belfry]

Damn it’s cool how you immediately saw that

[rocky cloak]

Took me about 16-14 = 2 minutes

[noble belfry]

I need to do way more algebra problems to prepare for my midterm

My mind didn’t even go to that

[sacred wharf]

sudo vs pseudo

[tardy hedge]

Lol

[thorn jay]

23 * 60 + 16 - 23 * 60 - 14 = 2

[rocky cloak]

quick maffs

[thorn jay]

https://tenor.com/view/quick-maths-mans-not-hot-gif-11163522

[crystal vale]

i have to prove if R is infinite comm. ring with unity and it has finitely many units then R has infinitely many maximal ideals, any hint?

[south patrol]

Minus 1 that's 3 quick maffs

[south patrol]

Iirc: adapt Euclid's proof that there are infinitely many primes

[rocky cloak]

Chinese remainder theorem can be useful.

[sacred wharf]

i can just hear the british accent in my head

[sacred wharf]

wait we can prove there are infinite primes

[noble belfry]

Show TFAE:

1. Every p-Sylow subgroup is normal in G
2. for x,y \in G, if x and y have relatively prime orders, xy = yx.

=> consider <x> which is the product of the p-Sylow subgroups in <x> of the primes dividing |X| and the same for y. There is no intersection (since they have no shared prime divisors) and they are commute since x = p_1 p_2 ... p_n and y = p_1' p_2' ... p_n' for different n. We know every p-subgroup is contained in a p-Sylow subgroup (which are normal) and we know normal subgroups with trivial intersection commute, so xy=yx.

any tips on the <= direction?

i know that if x isn't in a P-sylow subgroup, then xPx^-1 = P

but if x has order p^k then im at a loss

its not generally true that if x has order p^k then xPx^-1 = P

[tardy hedge]

WTF

[rocky cloak]

You already proved that the group is generated by its sylow subgroups, so you can assume x is in P if it has order p^k

[tardy hedge]

Black magic bro

[sacred wharf]

bro did the clever -a + a = 0

[tardy hedge]

Brilliant… simply brilliant…

[noble belfry]

why...

[tardy hedge]

A Brilliant Mind I’d say…

[noble belfry]

master of analysis

[rocky cloak]

Like you pick one p-sylow subgroup for each p (and you can let P be one of them). Then every element of your group is a product of elements from those subgroups

[noble belfry]

oh

genius

yeah that makes it immediate

damn theres too many possible moves to make in math 💀

[tough raven]

Oh, this is equivalent to G being nilpotent, right?

[noble belfry]

Yeah

[elfin wraith]

I never really understood nilpotency for groups, like the significance of it. It was kinda just a thing in my group theory class

[sacred wharf]

nilpotency in Rings as modules > nilpotency in groups

[elfin wraith]

I guess from that second point it’s that these groups are reasonably “almost abelian”?

[elfin wraith]

But tbh, my general feeling is that often stuff in group theory are just “things”…

[thorn jay]

(finite) solvable at least has a nice interpretation as being those groups built out of cyclic/abelian groups

nilpotent algebras and congruences appear to have pretty nice descriptions in universal algebra though (a group is nilpotent iff it is nilpotent in the sense of universal algebra)

[thorn jay]

these are nice properties

[elfin wraith]

Ah ok I also didn’t know property c), that is a nice thing to know I guess

Idk, I just have this feeling that so much of group theory is just here’s an adjective, what groups fit under it. And I’m just not sure I care that much about that lol

I’m sure that’s not at all a fair assessment, and likely just very indicative of what I’ve seen

[thorn jay]

no that's fair

it's how a lot of pure algebra seems sometimes

I'll see papers about all these cool algebraic structures with cool properties but motivation to care about them feels lacking

[tardy hedge]

When tensoring an R-mod hom with M(x)R - and M is also an (S,R)bimod, the result can also be viewed as an S-module hom? S-scalars should pass thru cause its just the identity map on the M part

[twilit wraith]

Maybe theres already an answer but it measures how abelian the group is

Also the equivalences for it are nice

G is nilpotent iff every sylow subgroup of G is normal

And G is the direct product of those sylow subgroups

[south patrol]

Yes, indeed the construction can be viewed as a functor from left R modules to left S-modules

[balmy python]

Hello group theorists

I have a question about a specific equivalence relation