groups-rings-fields

[nimble folio]

but i cant say I know them

[knotty badger]

You can take a concrete existing construction, and ask whether there’s an alternative description for maps into or out of it

Ok cool, for first iso, you want the universal properties of quotient groups and subgroups

Want me to recap them?

[nimble folio]

Sure

[knotty badger]

So quotients first

Let G be a group, and N be a normal subgroup

One then defines the quotient group G/N - its elements are cosets

With the usual group multiplication

[nimble folio]

I'll take a look at the book, the wikipedia page for universal properties does it no justice

[knotty badger]

Yeah totally

[knotty badger]

The universal property gives a nice alternative description for group homomorphisms out of G/N

Namely, group homomorphisms G/N -> K naturally correspond to group homomorphisms G -> K which send N to the identity

To be clear they can send more than N to the identity too

[nimble folio]

so N gets annihilated

[knotty badger]

Yeah

The way this correspondence works is

Given a group hom G/N -> K

You compose with the quotient map G -> G/N

This gets you a group hom G -> K that annihilates N

The reverse is a little trickier though

You have a group hom $\phi : G \to K$ which annihilates $N$

[cloud walrus]

Pseudonium

[knotty badger]

And you want to define $\tilde \phi : G/N \to K$ by $\tilde \phi(gN) = \phi(g)$

[cloud walrus]

Pseudonium

[knotty badger]

Note that if this was possible you’d have $\tilde \phi \circ q = \phi$ for $q : G \to G/N$ the quotient map

[cloud walrus]

Pseudonium

[knotty badger]

So it really is an inverse

[knotty badger]

Anyway one just checks this is “well-defined” and is a group hom, and you’re good

[nimble folio]

That makes sense

Its starting to all come back

I remember learning about this at office hours last year

[knotty badger]

Neat neat

The other one you need is for subgroups

This one’s a lot more straightforward

Let G be a group, and H be a subgroup

You’re familiar with the inclusion map H -> G?

[nimble folio]

Yes

[knotty badger]

It’s also a group homomorphism

[hidden wind]

right that makes perfect sense

[knotty badger]

Then, here’s the universal property, which gives you an alternative description of maps into H

Group homomorphisms K -> H naturally correspond to group homomorphisms K -> G whose image is contained in H

Does that make sense?

[nimble folio]

wait

so if $\varphi: K \rightarrow G$

[knotty badger]

One direction is composing with the inclusion map

[cloud walrus]

clubsoda14

[nimble folio]

sorry im at work i had to talk to soemone

So if $\varphi:K \rightarrow G$ is a group homomorphism then $\textt{im}\varphi \subseteq H$

[cloud walrus]

clubsoda14
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[knotty badger]

That’s not exactly what I’m saying

What I’m saying is - if you have such a group homomorphism, and its image is contained in H

Then you can shrink the codomain to H

Given a group homomorphism K -> H, you can compose with the inclusion map H -> G to get a group hom K -> G whose image is contained in H

I’m saying you can reverse this process, too

[nimble folio]

ok

that makes sense

[knotty badger]

Lovely!

Oh and one final fact

Do you know that a bijective homomorphism is an isomorphism?

[nimble folio]

Yes

[knotty badger]

Cool, then we have all we need

Ready?

[nimble folio]

Yea

[knotty badger]

So

Let $\varphi : G \to H$ be a group homomorphism

[cloud walrus]

Pseudonium

[nimble folio]

Oh i see where this is going lol

[knotty badger]

You can use the universal property of subgroup, with $\text{Im}(\varphi)$

[cloud walrus]

Pseudonium

[errant wedge]

Ik abt universal property of x in a bunch of diff contexts but how does one define univ properties in general

[knotty badger]

So you can write $\varphi = \iota \circ \psi$

[cloud walrus]

Pseudonium

[knotty badger]

Where $\psi : G \to \text{Im}(\varphi)$

[cloud walrus]

Pseudonium

[knotty badger]

And $\iota : \text{Im}(\varphi) \to H$ is the inclusion map

[cloud walrus]

Pseudonium

[knotty badger]

Is that fine?

[nimble folio]

Yes

[knotty badger]

Cool

There’s two properties of $\psi$ we need

[cloud walrus]

Pseudonium

[coral spindle]

In fact I define it just below that message! Universal properties are properties of objects in categories that specify those objects up to isomorphism. Phrasing things in terms of category theory alone allows us to prove results in a lot of general cases, which is why we study category theory at all.

[knotty badger]

First, do you agree it’s surjective?

[nimble folio]

Yes its surjective

[knotty badger]

Again I think this is a little too limiting

[coral spindle]

OK

[knotty badger]

Cool

The second property we need is

Do you agree that the kernel of psi is the same as the kernel of phi?

[nimble folio]

Uh thats not obvious to me

[knotty badger]

It’s worth checking why, then!

[knotty badger]

Universal properties are just alternative descriptions for maps into or out of your object

It’s not like I used the universal property of subgroup to define the subgroup up to iso, after all

[nimble folio]

I guess I see why the kernel of psi and phi are the same

is it because 0 is contained in the image

[knotty badger]

I think the important part here is that the inclusion map is injective

So clearly if you’re in the kernel of psi, you’re in the kernel of phi, just because group homs preserve the identity m

For the converse, if you’re in the kernel of phi, then you must get mapped to something in Im(phi) which gets mapped to the identity

But the only element which does that is the identity itself

So you must be in the kernel of psi

[nimble folio]

ok got it

[knotty badger]

Cool!

Now we use the universal property of quotient

So we can write $\psi = \alpha \circ q$

[cloud walrus]

Pseudonium

[errant wedge]

Ah sorry, missed that. Thank you. Ive been wanting to study some category theory, not sure where to start text wise lol

[knotty badger]

Here $q : G \to G / \ker \varphi$ is the quotient map

[cloud walrus]

Pseudonium

[knotty badger]

And then $\alpha : G / \ker \varphi \to \text{Im}(\varphi)$

[cloud walrus]

Pseudonium

[knotty badger]

First off - do you agree this is possible to do?

[nimble folio]

Defining the map alpha?

[knotty badger]

Yeah

[nimble folio]

My guess is yes

[knotty badger]

Well, have we applied the universal property of quotient correctly?

[nimble folio]

Yes

[knotty badger]

Why?

[nimble folio]

right here

[knotty badger]

Yep! That’s what the check was for

Ok next - psi was surjective. Do you agree that alpha is still surjective?

[nimble folio]

Yes it would have to be

[knotty badger]

Yep yep

Indeed, whether you write a surjective function f as g o h, g must be surjective

[nimble folio]

yes

[knotty badger]

The final step is to check alpha is injective

For this, I’d suggest showing the kernel is trivial

Note that $\alpha(g \ker \varphi) = \varphi(g)$

By how we’ve defined everything

[cloud walrus]

Pseudonium

[nimble folio]

Yes from a while ago lol

[knotty badger]

So do you see why the kernel is trivial?

[nimble folio]

i dont know if this reasoning is correct

but if it were not injective then $\psi \neq \alpha \circ q$

[cloud walrus]

clubsoda14

[nimble folio]

so the kernel must be trivial

that seems a bit circular though

[knotty badger]

I mean, I think it’s easier to check that

When is $\alpha(g \ker \varphi) = e$?

That’s what it means to check the kernel

[cloud walrus]

Pseudonium

[nimble folio]

oh when g = e

[knotty badger]

Not necessarily

Remember the definition of alpha

[knotty badger]

.

[nimble folio]

oh when g is in the kernel of phi

[knotty badger]

Right - and then?

How do you deduce the kernel of alpha is trivial from that?

[nimble folio]

Uhh

[nimble folio]

from this?

[knotty badger]

From what you said here, which is correct

[nimble folio]

oh

hang on

i lost it

[nimble folio]

im a little stuck on this i keep thinking about it

i know the kernel has to equal just the identity element

[knotty badger]

Right, but which identity element?

In what group?

[nimble folio]

H

[knotty badger]

No

[nimble folio]

G?

[knotty badger]

What is the domain of alpha?

[nimble folio]

i thought phi was a map from G to H

[knotty badger]

Yes

[nimble folio]

the domain of alpha is G/ker phi

[knotty badger]

But I am asking about the kernel of alpha, not of phi

Yes, so you need the identity of that group

[nimble folio]

$e \text{ker}(\varphi)?$

[cloud walrus]

clubsoda14

[knotty badger]

Yes!

Or just $\ker(\varphi)$ if you prefer

[cloud walrus]

Pseudonium

[nimble folio]

okay

[knotty badger]

You have this

What can you deduce about $g \ker(\varphi)$?

[cloud walrus]

Pseudonium

[nimble folio]

when $g \in \text{ker}(\varphi)$ then $g\text{ker}(\varphi) = \text{ker}(\varphi)$

[cloud walrus]

clubsoda14

[knotty badger]

Yes!

And that shows the kernel is trivial

Hence, alpha is injective

We already had alpha being surjective, so in fact alpha is bijective

Thus an isomorphism

[nimble folio]

Yep

proved the first isomorphism theorem

[knotty badger]

So, we can write $\varphi = \iota \circ \alpha \circ q$

[cloud walrus]

Pseudonium

[knotty badger]

In other words, every group homomorphism takes the form of a quotient map, followed by an isomorphism, followed by an inclusion

So there’s a duality between the study of group homs, and the study of subgroups/quotients

To me, that’s what the first iso theorem is about

[nimble folio]

Wow

[knotty badger]

E.g. have you heard of simple groups?

[nimble folio]

Yes

[knotty badger]

They have only two quotients

[nimble folio]

0 and itself

[knotty badger]

And using the first iso theorem, this puts very strong constraints on group homs out of simple groups

Because they have to start with a quotient

So a simple group either embeds injectively into another group

Or it gets sent fully to the identity

This is the sense in which they’re “simple” and can’t be broken down further

It’s an all-or-nothing deal

[nimble folio]

Thats really cool

[knotty badger]

I’m glad!

[nimble folio]

We didnt get to talk much about simple groups in my algebra course

[knotty badger]

This is a reason why they’re useful

They’re “primes” in a strong sense

[delicate orchid]

it's like schur's lemma but for groups :thecosmoshumswithatunemostsweet:

[nimble folio]

Yea I know where that leads to

[knotty badger]

This is also why I like the universal property approach

[knotty badger]

Because you get this fact

[chilly radish]

(Schur's Lemma)

Shit

[knotty badger]

And to me that’s cool - it feels like a “classification of group homomorphisms”

[chilly radish]

I was beaten to the punch

[chilly radish]

Le weak factoring system

[knotty badger]

Every group hom consists of forgetting details, relabelling, and then modelling a smaller group inside a larger one

[knotty badger]

Yep yep

[nimble folio]

It was a little hard to follow along ngl

But that wasnt your fault its hard reading this stuff through discord

ill have to go back and review everything slower

[delicate orchid]

if that's weak.... u don't want to see what I got cooking...

[knotty badger]

Of course of course

[chilly radish]

I feel like this kind of perspective is something you pretty much have to work out yourself at some point

[nimble folio]

did we essentially prove the first isomorphism theorem?

[knotty badger]

Yep!

[nimble folio]

we didnt cover that in my group theory course

nice

[knotty badger]

Oh wow what

I thought I was showing you a proof you’d already seen

[nimble folio]

Nope lol

Never seen that before

[knotty badger]

You’re telling me this is the first proof of first iso you’ve seen?

Bruh

[nimble folio]

Yes

[knotty badger]

I wouldn’t have done universal properties from the start if I’d known that lol

[nimble folio]

Wait let me check my ring theory notes

if we didnt prove the first isomorphism theory of rings we definitely didnt prove it for groups

[knotty badger]

It’s honestly a small miracle that you followed

[nimble folio]

Yea I think we did prove it for groups I just forgot

[knotty badger]

Oh ok

Well yeah, the same proof essentially works for all algebraic structures

You use:

1. Universal property of substructure
2. Universal property of quotient
3. Bijective homomorphism is an isomorphism

In other words, every homomorphism takes the form of a quotient, then an isomorphism, then an inclusion

Between groups, rings, vector spaces, modules…

[nimble folio]

got it

silly question but for the proof you just showed me

there'd be absolutely no difference if we had let G and H be modules right

[knotty badger]

Yeah totally

[knotty badger]

It’s always this

[dull ginkgo]

Then there’s continuous/measurable maps which are algebraic in nature but tell you to screw off unless images of open/measurable are open/measurable lo

[knotty badger]

Yeah totally

Except first iso does work for compact hausdorff spaces iirc

Which gives you a hint that they’re algebraic

Essentially because “3” works for these spaces

Whereas it doesn’t work for general topological spaces

[dull ginkgo]

Loc compact or just compact Hausdorff

[knotty badger]

Uh I forget

It’s one of those

[nimble folio]

well i bring that up since the kernel of any module homomorphism is the same as its kernel when viewed as a homomorphisms of abelian groups

so in a sense did the proof you showed me (for groups) also prove it for modules?

[knotty badger]

Yeah, in a sense I’ve proved it for all algebraic structures simultaneously

[nimble folio]

Oh lol

I should probably review the simpler proof before I review the one you showed me

[knotty badger]

Yeah lol

Apologies for that

[dull ginkgo]

Did you show the congruence relation universal alg one lo

[nimble folio]

No thats my fault

[knotty badger]

I would’ve done the simpler proof if I knew this was your first time

[nimble folio]

Nah it wasn't too bad

[knotty badger]

Wait really?

[nimble folio]

I'm familiar with all of the language and logic you used

Im at work and its jsut hard reading through discord

[knotty badger]

Damn wow

Well that makes me feel pretty good

[nimble folio]

Had this been in front of a whiteboard it would have gone easier

could you scrap the notion of groups and prove a "categorical" first isomorphism theorem

[knotty badger]

Oh yeah totally

[knotty badger]

It’s always this

I think it’s called like

[dull ginkgo]

Either you’re talking about monads (horror) or the universal alg idea

[knotty badger]

An epi-mono factorization system?

[nimble folio]

That sounds awful

[knotty badger]

Lol

Yeah I prefer this presentation

[dull ginkgo]

For monoids there is a first isomorphism theorem stated through congruences

for groups, inverses let the congruence class of the identity (the kernel) “represent” the congruence via Quotienting

[nimble folio]

well this came up from trying to understand universal properties

[knotty badger]

It did!

And - do you feel like you understand them a bit better?

They’re just alternative descriptions for maps into or out of your object

[nimble folio]

Yea

[knotty badger]

Yay!!

[nimble folio]

But i was gonna ask is it in good interest to just study category theory

[knotty badger]

I mean I will probably give a biased answer here

But honestly I’m not sure at your stage

[knotty badger]

I think it’s worth just… seeing how many of these you can find

Cat theory helps give a general theory of universal properties

[dull ginkgo]

The very basics yes, like what objects and homomorphisms are, and different examples of objects, and lastly universal properties and natural morphisms but that’s about it at this stage

[knotty badger]

But that’s more useful when you already know a bunch of them

[dull ginkgo]

These ideas can make the “essence” of a lot of algebraic and topological ideas stand out

[knotty badger]

So just, take existing constructions you meet, and see if you can find such alternative descriptions

It’s a fun game honestly

[dull ginkgo]

Like how to properly use things described off their universal property

[knotty badger]

Yeah, a universal property tells you what something “does”

[dull ginkgo]

E.g. polynomial rings, free objects, tensor algebras, etc

[knotty badger]

Not what something “is” though, which is what the construction helps with

[nimble folio]

tensor algebras is why i ask

im studying modules and tensor products right now

are those the same thing?

[dull ginkgo]

Tensor algebra is essentially the “weakest algebra you can put on a module”

It’s called a “free object”, the most general structure of an object you can put on a lesser object

so like

The free group of a set is the set of words to powers and their inverses right

It is the most general and simple group you can make over a set

And all other groups from elements of that set factor through it, i.e are a quotient

[knotty badger]

Yeah this is the other perspective on universal properties - initial/final stuff

I prefer representability myself usually but it’s good to know they’re equivalent

[dull ginkgo]

Tensor algebra is like the “free algebra” of a vector space or module

Well free associative, unital algebra

[knotty badger]

And indeed some situations are more elegantly phrased with initial/final stuff than with representability

Initial algebras of endofunctors come to mind

[nimble folio]

Cool

[dull ginkgo]

Functors are also helpful to understand

[knotty badger]

Hmm at this stage?

What functors will they even have met

[nimble folio]

This

[knotty badger]

Ah

Meeting a functor before the first iso theorem, hm…

[nimble folio]

LOL I've seen the first iso theorem

completely forgot the proof though

i've covered groups, rings, fields, and galois theory last year

[knotty badger]

Oh damn

[nimble folio]

the only thing we skipped was module theory so ive been reading up on it

[knotty badger]

You know more algebra than me lol

[nimble folio]

no i do not

[knotty badger]

To be fair that is a low bar

Idk Galois theory

Or field theory

[nimble folio]

me neither LOL

I barely made it through the course

[knotty badger]

No but like I never even went to a course on it

[nimble folio]

Oh

I wasn't a fan of it

Polynomial rings are kind of yuck to me

[knotty badger]

I ain’t an algebraist lol

I’m a physicist

[knotty badger]

Ah I see

[nimble folio]

it was a weird course

it covered rings, polynomial rings, vectorspaces, localization, field theory, and then galois theory

everything after localization kinda just skipped my mind

[knotty badger]

That seems like a huge amount for one course

[nimble folio]

Yes it was very accelerated

it was hard to follow along at a certain point

but atleast i came out of it learning rings and more stuff about vector spaces

[dull ginkgo]

The fucking free Functor lmao

Which I feel like should be explained

Because it helps us know when and how we can go between objects

Like sets to groups

Or groups to sets

And so on

[knotty badger]

Anyway I hope you liked my explanation!

[dull ginkgo]

Which confused me at first and seemed really off until i learned about functors

[knotty badger]

Oh interesting

[nimble folio]

Yea it was great I plan on rereading more thorougly later

ill add a section for it in my notes

[knotty badger]

Aw, I’m honoured!

[dull ginkgo]

Another great example

A ring to its group of units!

[nimble folio]

Thats a functor?

Oh

Duh

[dull ginkgo]

If you have two rings A and B, a map between them, then applying that functor is a group map from A^x to B^x!

[delicate orchid]

a group to it's groupring is another good one, and very related to this one

[dull ginkgo]

Isn’t that the adjoint

Anyway

[nimble folio]

well these are all covariant functors right?

what about an example of a contravariant functor

[dull ginkgo]

So we also have examples of functors of a category into itself (endofunctors)

Examples include Hom into AND out of a module

It has a related functor, later you’ll know it as an adjoint, of the tensor product :3

[knotty badger]

Internal hom?

[dull ginkgo]

I should probably be specific

Homomorphisms of an R module

Fix the module you’re going INTO, that’s a contravariant endofunctor, fix the one you’re going OUT OF, that’s covariant

[nimble folio]

Are you saying $\text{Hom}(A,\square)$ is covariant and $\text{Hom}(\square,B)$ is contravariant

[cloud walrus]

clubsoda14

[nimble folio]

if A,B are modules

[tribal moss]

Yes.

[nimble folio]

Sweet

[dull ginkgo]

Hom and Tensor are probably one of the best intros to adjoints

You’re on tensor products right now, right?

[nimble folio]

No

[dull ginkgo]

Oh

[nimble folio]

Almost though

[dull ginkgo]

Okie, no rush :3

[nimble folio]

im one chapter away

in the book im reading

[dull ginkgo]

Tensor products and how they act “categorically” as introduced in Jacobson made everything click for me

[dull ginkgo]

Epic

[tribal moss]

A map f: X->Y gives rise to a map Hom(Y,B) -> Hom(X,B), so Hom(-,B) is contravariant.

[dull ginkgo]

Which book?

[nimble folio]

introduction to homological algebra by Rotman

I was reading chapter 10 of dummit and foote which is on modules

They did a good job on the basics but once they got into tensor products and exact sequences it was too verbose

they introduce category theory in chapter 1 so i've spent a lot of time on that

[dull ginkgo]

[nimble folio]

its nothing too extreme

they define what a category is, subcategories, covariant and contravariant functors, Hom, natural transfromations

[dull ginkgo]

Yeah I’m just doing Jacobson which is alright

[nimble folio]

and other stuff im not reading

[dull ginkgo]

Did you do rings yet

[nimble folio]

I have done rings yes

I took a course on them at university and reviewed them this summer

[dull ginkgo]

How would you describe the universal property of the free algebra

[nimble folio]

before getting into modules

[nimble folio]

no clue

i've never even heard of that

[dull ginkgo]

Hmm what about

Field of fractions :3

[nimble folio]

Can i read my notes LOL

since we talked about it I just dont remember

[dull ginkgo]

Sure

[frail summit]

Quick question, the abelianization of <x,y| x^2 = y^4k = (xy)^4k = 1> is not cyclic, right?

[nimble folio]

Well its a universal property

I don't know how I'd describe it without saying what the theorem is explicitly

[dull ginkgo]

Try to rephrase your relations in terms of xyx^-1 y^-1 terms

[dull ginkgo]

I was gonna ask do you understand what it’s trying to say

[dull ginkgo]

Also is this not a dihedral group

[nimble folio]

wait just to be sure, do you mean this

[dull ginkgo]

Yeah

[nimble folio]

Oh okay

yea I get the gist of it

[dull ginkgo]

How would you describe that in plain English

[frail summit]

I know

[frail summit]

I don't see any way to express x in terms of exponents of xy, or y, if that's what you mean

[dull ginkgo]

The elements that are being “sent to 0” when you abelianize it are of the form “xyx^-1 y^-1”

So you can try to actually compute the abelianization

Or other commutators for elements in the dihedral group, which are all of the form y^n or xy^n

So try to compute your commutators

[dull ginkgo]

This is an exercise in understanding btw

[nimble folio]

I'm not gonna kid myself

I don't think i can explain it in words

best guess: you can turn a ring into a field

[nimble folio]

using this

[dull ginkgo]

Here’s how i would describe it

[nimble folio]

like Frac(Z) = Q

the universal property is just another way of describing it

like i was explained earlier

[dull ginkgo]

If your ring R injects into a field, then there is a map from the field of fractions into that field extending it

Extending meaning that your map from the field of fractions composed on the embedding of R into Frac(R) is the same as the map from R into the field

[nimble folio]

Yep

[dull ginkgo]

Much like the universal property of a quotient group, where your map from A to B is factored through the quotient

[nimble folio]

Yep yep yep

I see how its related

[dull ginkgo]

Often times trying rephrase it without symbols helped me

[nimble folio]

Yea I need to get better at that lol

[knotty badger]

I find representability helps for this often

[glad osprey]

I have a question about Lie theory - not sure what channel it should go into, but I'll try here: $SL_n(\Bbb R)$ is a closed subset of $M_n(\Bbb R)$, since $SL_n(\Bbb R) = det^{-1}(1)$ where $det : M_n(\Bbb R) \to \Bbb R$ is the determinant, which is continuous. What if we restrict the determinant to the domain $GL_n(\Bbb R)$? Then the determinant is still continuous, which shows that $SL_n(\Bbb R)$ is also closed in $GL_n(\Bbb R)$, correct?

[cloud walrus]

sheddow

[delicate orchid]

I buy it

[agile burrow]

That's correct. But also, the topology on GL_n(R) is inherited as an open subset of M_n(R) = R^{n^2}. The closed subsets of an open subset U of X are precisely sets of the form Z \cap U where Z is a closed subset of X. In this case, Z = SL_n(R), U = GL_n(R), and X = M_n(R). In particular, Z is already contained in U

[glad osprey]

So in this case it is trivial since SL_n(R) is contained in GL_n(R), but in general if A is closed in B and B is open in X, then A would usually not be closed in X, right?

[tribal moss]

Right. We could take X=R, A=B=(0,1).

However, that's not quite the situation we're looking at here.
It is true that: If A is a subset of B, and A is closed in X, then A is closed in B too.

[glad osprey]

Hmm, it seems like being the preimage of a closed set is stronger than just being closed. Like, if $A = f^{-1}(1)$ where 1 is closed, then we can always restrict f to an open set B, which would show that A is closed in B, wouldn't it?

[cloud walrus]

sheddow

[glad osprey]

Right, got it 👍 just got confused for a sec

[tribal moss]

These concepts can be different in general topological spaces, but in metric spaces in particular (and more generally in "perfectly normal" spaces) every closed set is the preimage of {0} under some continuous function X->R.
Being a metric space is obviously inherited by subspaces, so if your topology on M_n(R) is the usual topology on R^n², then you get the same property of GL_n(R).

[glad osprey]

That makes sense, thanks 🙏

[chilly ocean]

Indeed its different in general, it even has a name: a subset of a topological space that is preimage of {0} by some continuous X->R is called a zero set

and there's also cozero sets

[sly crescent]

Is SL_n(Z_p) closed in GL_n(Z_p)?

[hollow tartan]

Hello to prove [G:K] = [G:H][H:K] index is multiplicative and K < H < G. The book proves it using cosets but my proof using Lagrange theorem was simpler, is it a valid proof it would just be |K| = p, |H| = p+m and |G| = p+m+n, which simply leads to (p+m+n)/p = {(p+m+n)/(p+m)}{(p+m)/p}

[glad osprey]

I guess you would have to choose a topology on Z_p to answer that, I don't think there is a standard one

[sly crescent]

Yes there is

The p-adic valuation

[glad osprey]

Oops, I thought you meant Z/pZ

[coral spindle]

The determinant is continuous (so I would hope!) and SL_n(Z_p) is the preimage of the closed set {1}

[sly crescent]

What about GL_m(Z_p)xGL_n(Z_p) in GL_(m+n)(Z_p)?

[delicate orchid]

I'm scared

[hollow tartan]

Sorry for the ignorant question but by Z_p you mean positive characteristic?, if so is SL_n(Z_p) a manifold?

[tardy hedge]

True

[arctic trail]

discrete topology?

[arctic trail]

should be

they seem open

open subgroups are closed

so it should be closed

[rotund aurora]

how do you define the inclusion? two diagonal blocks of sizes m and n?

[delicate orchid]

yeah

[arctic trail]

that's how I assumed it

[delicate orchid]

it's like S_n x S_m included in S_m+n

[sly crescent]

Are you ker?

[arctic trail]

ker?

[delicate orchid]

Schur-Weyl moment

[rotund aurora]

The map that sends a matrix to the (i,j) entry is continuous, the inverse image of {0} is closed, you are intersecting this closed sets

[sly crescent]

I used to know a user with the same pfp whose username started with ker

Oh, they’re someone else

[arctic trail]

I just stole it off wikipedia

[arctic trail]

not even that

it's the sum of open (additive) subgroups so open

[rotund aurora]

didn't he ask for closed?

[sly crescent]

Yes

[arctic trail]

open subgroups = clopen subgroups in topological groups

[rotund aurora]

yes, but what I said seems more straightforward

[arctic trail]

maybe

also the regular topology on p-adics is really like

forgiving

[delicate orchid]

it's as close to trivial as you can get without being boring

[rotund aurora]

I mean what Mecejide asked is true if you replace Z_p by any Hausdorff topological ring

[arctic trail]

man, matrices over non-commutative and non-division rings

💀

what if you wanted to calculate the determinant of a matrix but god said 'non-commutative and non-division ring'

[rotund aurora]

Z_p is a non-division ring

[arctic trail]

it's commutative

[sly crescent]

Dieudonné:

[rotund aurora]

you said "or" tho

[arctic trail]

shit I meant and

[delicate orchid]

GL_2(M_2(Z))

[arctic trail]

that's why I mentioned division rings

[arctic trail]

At that point

wait

[delicate orchid]

:trollface:

[arctic trail]

yeah GL_4(Z)?

[sly crescent]

I think so

[delicate orchid]

I think so too

[sly crescent]

How about

[arctic trail]

M2 isn't a commutative ring

nor a division ring

is even GL2(M2(Z)) well defined lol

[sly crescent]

GL_2(Hurwitz integers)

[sly crescent]

Yes

[arctic trail]

the heck are those

[arctic trail]

How?

[delicate orchid]

Well yeah of course it is? It's the invertible endomorphisms of some geezers

[arctic trail]

Ah

I guess that makes sense

I was imagining it as a matrix group

[sly crescent]

You don’t need determinants to define invertibility

[arctic trail]

I know

I was thinking of it specifically as det != 0, rather than thinking of it as invertibility, that's why it confused me.

[delicate orchid]

that's fair

[sly crescent]

Quaternions a+bi+cj+dk where a, b, c, d are either all integers or all half of odd integers

[delicate orchid]

oh that one is a classic

[arctic trail]

that's specific

where the heck do they come up

[sly crescent]

I’m not really sure

[dull ginkgo]

I think the Octonions do in that form

[nimble folio]

found another universal property

[cloud walrus]

clubsoda14

[naive remnant]

I'm currently doing this brilliant.org lesson that is explaining why half of the configurations of the 15 puzzle are impossible to solve using group theory. They ask "What's a good way of describing a sequence of moves [using transpositions between two elements]?" However, I'm confused by their solution. They say that the best representation is "A product of transpositions (i, j) where i is between 1 and 15, and j is always 16". Their reasoning is shown in the last image I uploaded. What confuses me is how these 15 operations are transpositions, since moving most elements into the empty slot seems to require moving a lot more elements than just 2. The definition of a transposition is that it only permutes 2 elements, right?

Can anyone help me understand how this operation of moving an element into the blank slot is a transposition?

[rocky cloak]

Any move on the sliding puzzle is given by swapping "16" with one of its neighbouring tiles.

For example in the given picture, siding 12 down will swap 12 and 16.

So every move on the puzzle is given by such a transposition

[naive remnant]

Hmm, I see that. But they are claiming that there are 15 possible transpositions (one for each numbered tile). It appears to me that if it was just swapping with neighboring tiles then there would only be around 2 transpositions at any given time?

[barren sierra]

Quite stuck on this problem

I've shown that p_i(x_i) is indeed a generator but any sort of algorithm as described eludes me

I mean the crux of it is that I don't know how to algorithmically determine if ${1, x_i}$ is linearly independant in $k[x_1, \ldots, x_n] / I$ if I have a Groebner basis $G$ for $I$.

[cloud walrus]

Spamakin🎷

[barren sierra]

once I know how to do that then I can do this problem

This is easy if I can choose the Groebner basis to have a particular monomial order but the key is that I can't choose, I'm just given the basis

[rocky cloak]

"at any given time" being the key phrase there

[crystal vale]

To show there is only one left Z-module for abelian group M, we need to show there is only one ring homomorphism between Z and End M, and here they defined ring homomorphism with additional condition that unity maps to unity.

Now if f and g are ring homomorphism between Z and End M, since f(1) = g(1) so both are the same on the generator thus f = g.

Is it correct?

[shell mesa]

Can anyone explain to me the notation and meaning of the last part here? I understand the definition of p and R inside the braces, but I wonder about R^3 x SO(3), and perhaps specifically the cross notation.

[kind temple]

going from 2 to 3, it’s just the cartesian product of the underlying sets and the group multiplication is given coordinate wise, coming from the group multiplication of R^3 and SO(3), respectively

so like, (x,A)(y,B) = (x + y, AB)

[shell mesa]

Ah, right, thank you!

[kind temple]

in general, when you see a cartesian product of groups, that’s what it means. it can slightly be different/ambiguous in the infinite case if written with ‘x’ notation, but it’s all the same idea: the group multiplication works coordinate wise.

these are called the direct sum and direct product if you are interested.

[shell mesa]

Yes, very much so! I'll read up on this, thank you for the pointers

[knotty badger]

The other main product on groups is the semidirect product iirc

I realised recently that it’s a left adjoint so that’s nice

[noble bronze]

Hes conveniently cropping out the sentence which follows that one. I think most semigroup theorists would agree that 3-nilpotent semigroups are uninteresting. But nilpotent semigroups, and nilpotent monoids in particular, are very interesting (note, almost all monoids are 3-nilpotent monoids as well). I think boy is under the assumption that semigroup theory and (semigroup) structure theory are the same thing, Which is not true. In fact, most of the work done on semigroup theory is studying the varieties/pseudovarieties/quasivarieties they generate. Nilpotent monoids are absolutely crazy in this sense. Even though their structure is "simple", they are a great source for generating non-finitely based/non-finitely related semigroup varieties. In the monoid signature they can also be used to exhibit varieties with continuum many subvarieties.

[arctic trail]

Is this the special Euclidean group?

[arctic trail]

I don't think this is the direct product

[shell mesa]

Yes, indeed and SO(n) is the special orthogonal group

[rocky cloak]

Yeah, SE is the semidirect product of R^n and SO, so I guess they just meant to describe the underlying set.

Sort of bad communication if you ask me...

[coral spindle]

These ought to be definable by polynomial equations in the entries of the matrices, so yes.

[knotty badger]

I’m so happy I understand semidirect products now

[coral spindle]

I'm not sure why you're saying I'm conveniently cropping the following sentence, which is this:

One needs to yank out these weeds to find the interesting semigroups.
But in any case, it's cool to know that you can see some cool stuff in the varieties n-nilpotent monoids generate.

[arctic trail]

man, if I'm honest I think semidirect products are harder to understand than group extensions

understanding them through split group extensions to me is simpler

[knotty badger]

I understand them as left adjoints

[tough raven]

They probably meant the sentence after that one, which looks like it might start with "On the other hand".

[coral spindle]

It says nothing about 3-nilpotent semigroups at all, it's not relevant:

(On the other hand, semigroups have wider applicability than groups, especially in Computer Science)
This is just true.

Groups just don't show up as much in CS as semigroups and monoids do, ofc they're more applicable

[chilly ocean]

Is every non-0 prime ideal of Z[sqrt(5)] maximal?

More generally, if R an integral domain such that its integral closure is a Dedekind domain, can R have a non-0 prime ideal that isn't maximal?

[rotund aurora]

yes

[chilly ocean]

why?

[rotund aurora]

I mean you could try proving it, but just note that ||a finite integral domain is a field||

[chilly ocean]

I see

[chilly ocean]

I'm guessing this is probably false in general then

[rotund aurora]

In general, there is the concept of an "order" O which is defined to be a subring of O_K which contains an integral basis of length n=[K: Q]. Then O is noetherian and has dimension 1 (dimension 1 meaning that all nonzero prime ideals are maximal)

[rotund aurora]

integral extensions preserve Krull dimension https://math.stackexchange.com/questions/2421885/why-does-an-integral-extension-over-a-ring-has-the-same-krull-dimension-as-the-r

(one of the comment references Atiyah-Macdonald, go there I guess)

[chilly ocean]

I see

[tribal furnace]

I thought so, too

[chilly ocean]

Is there a way to prove this without using the fact that the multiplicative group of the field of order p is cyclic of order p-1?

[rotund aurora]

Yes

[chilly ocean]

How?

[rotund aurora]

Do you see why this is the same as determining the splitting behaviour of x^2+1 mod p? Splitting behaviour means whether it has no roots, two distinct roots or two equal roots

[chilly ocean]

of course

[rotund aurora]

Well so there are many proofs that -1 is a quadratic residue mod p iff p=1 mod 4 that dont use primitive roots

[chilly ocean]

I wasn't able to think of any

For example for the p=4k+3 case, I tried looking at the equation x^2+1=(4k+3)l, but this probably won't be enough since it doesn't use that 4k+3 is prime

[rotund aurora]

How did you prove it with primitive roots? I dont think its really a shortcut

[chilly ocean]

Let g be a generator for F_P^*

then g^(p-1) = 1

if we have x^2 = -1

then x^4 = g^(p-1)

so p-1 must be divisible by 4

[rotund aurora]

And the other direction?

[chilly ocean]

If p-1 is divisible by 4 then g^((p-1)/4) is a solution to x^2 = -1

(g^((p-1)/4))^2 = g^((p-1)/2) which is not equal to 1

so it must be equal to -1

[rotund aurora]

This is the easy direction btw, just note that if x^2=-1 and p=3 mod 3 then (x^2)^(p-1/2)=x^(p-1)=-1

@chilly ocean how Euler originally proved it is by showing that if p=1 mod 4 then there exist integers x and y such that p divides x^2+y^2

and just so you know, it took him several years, so this is by no means trivial

[chilly ocean]

that's was really ugly and difficult proof right

[chilly ocean]

why x^(p-1) = -1 in end?

[rotund aurora]

because (p-1)/2 is odd

I believe you can also use Wilson's theorem

another proof exploits the symetry of the number of solutions of x^2+y^2=1 mod p

there are many ways

[rotund aurora]

no? The proof is sensational

[chilly ocean]

I remember reading Cox's book primes of form x^2+ny^2 and they presented a proof at beginning for it and it was really unintuitive and hard to follow

[rotund aurora]

yeah, Cox's book is a great reference

[chilly ocean]

maybe I'm just dumb but I don't know how anyone would come up with a proof like that

The proof using cyclicity of F_p^* seems so much more intuitive and easier to find to me

[rotund aurora]

well, because it is not easy. But you can do similar stuff for other similar equations (I guess just quadratic forms)

[rotund aurora]

as I said, it took Euler several years...

[chilly ocean]

It's hard to me to understand why you find it sensational, because to me it looks like random manipulations of terms that somehow magically end up working in the end

[primal beacon]

groups didn’t even exist as such at the time

[rotund aurora]

Euler discovered primitive roots much later

[chilly ocean]

Right, I mean its more intuitive if you already know about basics of cyclic groups

[primal beacon]

the fact that F_p^x is cyclic is not obvious as well

[chilly ocean]

true

I'd really like to understand and appreciate this proof but. It's hard when I have no idea what is the motivation for each step. For the proof using group theory at least you can break it down into many smaller steps that I can understand on their own. Is something wrong with me?

[primal beacon]

have you looked a bit more at cox ?

(the theory of binary quadratic forms etc)

and you know about dedekind rings ? or just rings of integers or number fields

[chilly ocean]

you mean dedekind domains? i know about those

[chilly ocean]

yeah a bit

[primal beacon]

so the descent step is essentially working with factorisation in Z[i] (a dedekind domain) but without this theory

(when you know the equivalence between ideals and binary quadratic forms)

the insight was probably « this theory exists » and descent arguments are an effective way of proving that, already this requires Euler’s knowledge of bqf and primes

for the reciprocity step it was much harder explicitly because Z/pZ wasnt considered as a field or a group, if it was the proof would just be an effective construction of the primitive g!

as Euler tries to find x^2k+1!=0

[chilly ocean]

What does the assumption p | x^2+y^2, gcd(x,y)=1 correspond to in terms of factorization in Z[i]?

[primal beacon]

write n=x^2+y^2 then p|n, but x^2+y^2=(x+iy)(x-iy) now it happens that either p is prime in Z[i] or splits as (a+ib)(a-ib), if p doesnt split then p|(x+iy) or (x-iy) so p|x and p|y (Z[i] is a pid )

(i ommited p=2)

so that your assumption happens only if p splits

and the splitting happens if p=1 mod4

once you know that Z[i] is a pid with galois theory a you can derive this splitting behavior easily

[chilly ocean]

why p | (x+iy) implies p | x and p | y?

[primal beacon]

Z[i] is a pid

so p appears in the factorisation

so once you develop you get. (x+iy)=p(x’+iy’)

[primal beacon]

(the assumption is that p is prime in Z[i] above)

[magic dew]

how does it work, if for example x = p+1, y = p+i, (x+yi) = p + 1 + ip - 1 = p(i+1) and x' = 1, y' = 1

[primal beacon]

x and y are integers there

[magic dew]

ahh ok

[dull ginkgo]

What is more intuitive is every finite subgroup of a field’s multiplicative group is cyclic (due to the minimal polynomial of the group :3)

[naive remnant]

Hmm, okay. Is it not weird that some transpositions can't be performed at any given time? Shouldn't I be able to apply any transposition I want whenever I want? It feels strange that I would only be able to apply 2 out of the 15 transpositions at any given time.

These 15 "transpositions" just don't feel like swapping two elements of a set. It feels like they neccesarily do more than that.

This is my original question for anyone who wasn't around last night:
#groups-rings-fields message

[chilly radish]

No, because you aren't just permuting things around with no restrictions, to perform a transposition you need the board to allow that move

If the "16 tile" isn't next to the thing you want to swap it with you can't perform that move

[naive remnant]

Oh okay. I guess that makes sense. Thank you! So the "16 tile" is also going to be moving around as I perform these transpositions, right?

[chilly radish]

Yea

Every move is just a sequence of "swap 16 with tile x", "swap 16 with tile y now" etc

[primal beacon]

yeah the point was that you need group theory to do that and Euler didnt have group theory neither field theory !

[naive remnant]

Got it. I guess it was just counterintuitive because transpositions I've encountered before could be performed anytime, no matter the state. For instance, the transpositions that swap (1, x) for the set {1, 2, 3, 4} don't have this kind of restriction on which transpositions are currently available.

Thanks for your help!

[tough raven]

Hmm, maybe you can model this with a groupoid. Specifically, consider the groupoid with one object for every position in the grid and generated by one isomorphism between adjacent positions. For each object, take the set of configurations with the hole at that position. Each generating isomorphism induces a map between the respective sets. So this is a "representation" of the groupoid.

In particular, we can look at the group of automorphisms of a single object and this group meaningfully acts on the set of configurations with the hole in a given position.

Is that group the free group on 15 generators?

[urban geyser]

Just a quick clarification: There can exist a bijection from $GL_{n}(\mathbb{R}) \to \mathbb{R}^{\cross}$, but these two groups are not isomorphic since $\mathbb{R}^{\cross}$ is Abelian and $GL_{n}(\mathbb{R})$ is Non-Abelian for $n>1$; in other words, there can exist a bijection between them, but not a homomorphism, and therefore no “bijection homomorphism” (aka isomorphism). This means that any map from $GL_{n}(\mathbb{R}) \to \mathbb{R}^{\cross}$ , such as the determinant, cannot be isomorphic.

[cloud walrus]

thimg

[coral spindle]

Yup

N.b. This argument requires n>1

Of course if n=1 then the groups are certainly isomorphic :)

[urban geyser]

yiss :3

thank u

[urban geyser]

Just out of curiosity, would another example of a bijection between objects of different dimensions be space-filling curves?

[chilly ocean]

space-filling curves are usually not bijective

a space-filling curve from R into R^n cannot be bijective iirc

you can still have a function from R into R^n that is bijective tho, but it won't be continuous so it won't be a curve

[rustic jolt]

does red contradict purple here

[chilly radish]

You can also have a continuous space-filling curve, it just won't be injective

[arctic trail]

man I love minkowski

[tribal moss]

The red claim is right about "made into a ring with pointwise operations", but that is not what "group ring" usually means, and the claim that "elements of G multiply as they do in G" is not true under that definition.

[rustic jolt]

would that mean the book is wrong? its a really popular book so i am confused

rotman homological algebra

[tribal moss]

Yes, that screenshot looks desperately wrong.

[knotty badger]

Yeah it contradicts itself lol

It talks about pointwise multiplication and convolution on the same page

[rustic jolt]

is there a better book on the same topic?

[tribal moss]

Instead of the red box, the right definition would be something like:
The multiplication in $kG$ is the unique $k$-bilinear map satisfying $f(\delta_g,\delta_h) = \delta_{gh}$.
The convolution definition then follows from that.

[cloud walrus]

Troposphere

[knotty badger]

Aw that is neat!

Using the universal property of the tensor product I see~

Ok I guess strictly you don’t need to phrase this in terms of tensor products

You can stick with bilinear maps

[tribal moss]

As far as I'm concerned, "the universal property of the tensor product" is that each bilinear map V×W -> U arises from a unique linear map V tensor W -> U, and I don't think I'm using that here.

[knotty badger]

Yeah yeah mb

I’m, uh, understandably eager to see universal properties everywhere lol

So I saw “bilinear” and “unique extension” and kinda just did a short circuit

[tribal moss]

I think the appropriate abstract nonsense for what I'm doing is that $\mathrm{Free}_k(A) \otimes \mathrm{Free}_k(B) = \mathrm{Free}_k(A\times B)$ -- where $\mathrm{Free}_k$ is the free functor $\mathbf{Set} \to k\text{-}\mathbf{Mod}$ -- and then using the universal property of free modules.

[cloud walrus]

Troposphere

[knotty badger]

Yeah! That seems about right

Since free modules are all about bases

[knotty badger]

Aluffi maybe?

[nimble folio]

im trying to prove the first isomorphism theorem for groups using the universal property of subgroups and universal property of quotient groups. Heres the proof I've written up so far:

[cloud walrus]

clubsoda14

clubsoda14

[nimble folio]

Am I allowed to use the universal property of quotient groups before showing that ker{varphi} = ker{Phi}? My guess is yes because the map Phi:G ----> im{varphi} does have ker{Phi} as a normal subgroup of G.

[knotty badger]

Ah that’s what you meant by using the universal property of quotient first

Then yeah this is fine

Nice diagram

[nimble folio]

Thanks

Do you mind helping me understand why ker{Phi} = ker{varphi}? seems a little weird that Phi(k) = 0_H immediately implies k is an element of ker{varphi}

Its not obvious to me

[knotty badger]

Ah so

Phi(k) = 0_H implies (iota o Phi)(k) = iota(0_H) = 0_G

So that shows that k in ker Phi implies k in Ker varphi

Conversely, suppose k in Ker varphi, so varphi(k) = 0_G

Then you have (iota o Phi)(k) = 0_G

So iota(Phi(k)) = 0_G = iota(0_H)

But iota is injective

So Phi(k) = 0_H

[urban geyser]

||bookmark for myself to read this later||

[knotty badger]

Thus k is in Ker Phi

[nimble folio]

Thank you! I am at work so when I have the chance I'll read this more carefully

[knotty badger]

Essentially the same proof (and diagram) works for any algebraic structure that has:

1. Universal property of substructure
2. Universal property of quotient
3. Bijective homomorphism is isomorphism

[nimble folio]

Yea

I'd love to prove this for an arbitrary category C but quotients are defined a little weird

[knotty badger]

Oh with coequalizers you mean?

[nimble folio]

What is a coequalizer? I've been meaning to take notes on them

[knotty badger]

Yeah I wouldn’t suggest defining the quotient categorically

You should define the quotient just using normal set theory

However, once you’ve defined it, you can still show it satisfies the universal property

[knotty badger]

What I like about this is that all of 1, 2, 3 are useful outside the first iso theorem

[knotty badger]

Colimit of a diagram whose shape is two parallel arrows

[nimble folio]

Oh I think I see what you mean, I could say let C be any category and let $A \in \text{obj}(C)$, then you can define quotients on A easily

[cloud walrus]

clubsoda14

[knotty badger]

Yeah I think people often assume universal properties are only useful for defining things up to unique isomorphism

This is just not true

Even when you have a concrete explicit construction, it can still be useful to investigate whether maps into or out of your object have a nice alternative description

This establishes a universal property for that construction

It’s like a little shortcut you can take to define a map into or out of your object

And it’s just useful to know a lotta shortcuts

[nimble folio]

That's really cool

Category theory is kinda fun

[knotty badger]

I’m glad! People don’t, uh, usually think that

But I do!

[nimble folio]

I take this back actually

This seems really hard

[knotty badger]

Might be useful to focus on a specific example first

Say, rings

So, you want ring homomorphisms R/I -> S to naturally correspond to ring homomorphisms R -> S which send I to 0

Here I is an ideal of R

[chilly ocean]

For arbitrary categories you need to define internal equivalence relations

[knotty badger]

That’s if you want to define the quotient categorically

[chilly ocean]

and not all epimorphisms are quotients.. things are more complicated

[knotty badger]

I’m not suggesting that

I’m suggesting defining the quotient using regular ol’ set theory

[nimble folio]

What do you mean by this?

[knotty badger]

Like

A ring is a set

And you know how to take quotients of sets

You use equivalence classes

This is how the quotient ring R/I is defined

I don’t think you need to alter anything about that definition

[nimble folio]

Ah I see

[knotty badger]

But once you have that construction

[knotty badger]

You can still establish this

By R/I here I really do mean the specific construction with equivalence classes

[dull ginkgo]

me when the field of fractions map is epimorphic but not surjective

[knotty badger]

Yeah epimorphisms are either surjections or like

“Dense inclusions”

[chilly ocean]

there's even naturally occuring concrete categories with surjections that aren't quotients iirc

like Top

[knotty badger]

Yep

This is why I wouldn’t suggest defining quotients categorically in this case

[long obsidian]

In an integral domain A if I have a nonzero product $Q_*=\prod_{j=0} Q_j$ then does this imply that each factor $Q_j$ is nonzero? I feel like this is just the zero product property but im doubting whether this is valid reasoning...

[cloud walrus]

HausdorffT1

[chilly ocean]

What happens if one of the factors is zero?

[long obsidian]

I think this would imply the entire product is zero...

[chilly ocean]

Why are you not sure?

[arctic trail]

the heck

that notation is cursed

[arctic trail]

man

I like my categories algebraic

and with surjective epimorphisms

[delicate orchid]

the capital letters made me think it was a categorical product I ain't gonna lie

[arctic trail]

I don't even understand the question

is it a finite product in an integral domain?

[delicate orchid]

yeah

gotta be

[arctic trail]

man, all of those letters

gotta be the worst choice of letter

[delicate orchid]

if it was a topological integral domain I wouldn't even bother

[knotty badger]

Well I ain’t an algebraist

[arctic trail]

fair

[knotty badger]

I’m a physicist!

[knotty badger]

Why that reaction

[arctic trail]

flashbacks to physics in undergrad

[knotty badger]

Sure

[delicate orchid]

flashback to undergrad physicists

[arctic trail]

man

[delicate orchid]

ug physics student don't be a crank for 30 seconds challenge

[knotty badger]

I mean I loved physics in undergrad obviously

[topaz solar]

Something something prime over the image

[arctic trail]

My physics course was so cursed cause like

[knotty badger]

Ooh is that the algebraic intuition?

Dense inclusion is my topological intuition (from like categories of hausdorff spaces)

[topaz solar]

More the logic intuition since prime here is as in prime model, but you’re determining all of Q from Z after all, and there’s not really a smaller field containing it

[arctic trail]

it was the first class to introduce tensors and functionals. And the introduction of tensors could've been avoided cause like we only used them in contexts where we could use matrices lol. The professor spent 30 minutes calculating a Jacobian matrix.
Who thought it was a good idea to include Classical mechanics (Newtonian, Lagrangian, Hamiltonian), Special relativity and quantum mechanics in a single semester

[knotty badger]

I see I see

Interesting!

[arctic trail]

and bet none of the professors explanations made any actual mathematical sense

[topaz solar]

Yeah it’s prime in the same sense that Q is prime

[knotty badger]

Yeah that’s not really the point in physics

[arctic trail]

if you introduce an object, at least introduce it well

[knotty badger]

The standard of truth is “agrees with experiment”, not “is rigorous mathematically”

[arctic trail]

just sayin'

[knotty badger]

You don’t need to necessarily know precise definitions

[arctic trail]

I'm talking about mathematical objects

[knotty badger]

You focus more on what something “does” rather than what something “is”

For physics at least

[topaz solar]

Every char 0 field has Q inside, so prime here is like categorically initial wrt embeddings (which is just initial for fields) of things containing it

[knotty badger]

Mhm mhm, epimorphisms can always be thought of as initial arrows

And conversely monomorphisms can be thought of as final arrows

Or if you prefer representability, epimorphisms are about representing subfunctors of Hom(X, -), whereas monomorphisms are about representing subfunctors of Hom(-, X)

[arctic trail]

All of the operators in quantum mechanics were assumed to have additional conditions based off the usefulness of theorems of Functional Analysis

[knotty badger]

Yeah exactly

Because making those assumptions agrees with experiment

[arctic trail]

my mans did not say that. He did not say 'this is an assumption which has experimental evidence to support it' he would always say 'it is not physically admissible that it's not' as if anyone there had any intuition for that 💀

[knotty badger]

Yeah physically admissible is just what tends to crop up in experiment

Idk I had intuition for it

[arctic trail]

man, physics was mandatory I'm just mad cause I never wanted to take it and I even considered Erasmus to not have to take it 😭

[knotty badger]

Idk what that is

[arctic trail]

switching university for a semester essentially, where the credits transfer

[knotty badger]

Bruh

That’s a little excessive

I mean I disliked algebra in my degree but not to that extent

[arctic trail]

I heavily despise physics, and my GPA (equivalent) was near a rounding value so it made sense to me at the time to ask that

[knotty badger]

Ok lol

[arctic trail]

Mathematical physics seems interesting though

[knotty badger]

Maybe, I’m more into physics physics

[cloud walrus]

thou_art_an_egg

[clear fiber]

Let R be a ring and let I and J be ideals in R which are not equal as sets. It is possible for R/I to be isomorphic to R/J? If so, what is an example?

Not a homework problem, just something I wondered about

[topaz solar]

Yes

[hollow tartan]

What would be a characterization? Either I c J or J c I?

[next obsidian]

Neither of these have any implication

Take an infinite product of Z and then the ideal which is just Z in one component and 0 everywhere else has the quotient isomorphic to an infinite product of Z

But these ideals are incomparable

[toxic zephyr]

alright this is probably a really stupid question.
for a group, if i want a subset N to be defined in such a way that the cosets xN preserve the group structure (xN)(yN)=(xy)N, then it's not that hard to show that you need that for every n in N, we have gng^-1 is in N for all g. hence a normal subgroup.
for a ring, if we want to define a subset I such that the cosets x+I preserve the ring operations: (x+I)(y+I)=xy+I and (x+I)+(y+I)=(x+y)+I, then we can similarly show that we need I to be closed under addition and for xi and ix to be in I for all i in I and x in R (closed under multiplication by the ring). hence a two sided ideal.
my really stupid question is this: why would it be obvious that we need to define ring cosets via x+I instead of xI like in a group? apart from xI just clearly not working when you try to make it preserve addition. is it something stupidly simple like addition has "priority" over multiplication or what?

[knotty badger]

I like to think of ring cosets as equivalence classes under congruences

That requires this x + I definition

A congruence on an algebraic structure is an equivalence relation which is stable under the operations in that structure

[toxic zephyr]

so, let's say i'm stupid and naive (not a stretch). i'm familiar with groups and normal subgroups and quotient groups. does this congruence property hold for groups, or why, when i decide to move up to a ring with a new addition operation, would i decide that this normal subgroup/quotient group idea would generalize better through the lens of congruence?

[knotty badger]

Yes so

A congruence on a group determines a normal subgroup - it’s the equivalence class of the identity element

A congruence on a ring determines an ideal - it’s the equivalence class of 0

This turns out to be true more generally - the thing you want to quotient by is the equivalence class of 0 under the congruence

[toxic zephyr]

can you write this symbolically? i'm having a hard time connecting gng^-1 in N to that

[knotty badger]

I think it’s more natural generalisation of the idea of “quotienting” from set theory

So, a congruence on a group G is an equivalence relation R on G that respects the group operation

Meaning that if xRy, then zxRzy and xzRyz for any z in G

If you look at the equivalence class of the identity element under such a congruence, it always determines a normal subgroup of G

[toxic zephyr]

could you walk me through that? i'm not quite making the connection yet

[knotty badger]

Ok so, given a subgroup N of G, you can define an equivalence relation by xRy iff x^-1 y is in N

If N is normal, then you should check this equivalence relation is additionally a congruence on G

The normality assumption gets used

Once you’ve done that, it might be easier to show the reverse

[toxic zephyr]

oh interesting. like how xRy iff x-y is in I (or y-x i guess idk). i see. i suppose the key is that it's the invertible operation? so that's why it isn't x^-1y in I for a ring

[lone niche]

For another point of view, if you are willing to accept that ring homomorphisms are the correct structure preserving morphisms, then kernels of ring homomorphisms are ideals.

[knotty badger]

Yes exactly, ring multiplication is generically not invertible

You could try and circumvent this by saying xRy iff y in xI

But then this is no longer guaranteed to be an equivalence relation

[lavish nexus]

Prove that x^(p^n)-x+1 is irred over F_p only if n = 1 or n=p=2

If α is a root then α+a is a root for all a in F_p^n

So the polynomial splits in F_p(α)

the only maps in Gal(F_p(α)/F_p) consist of sending α to α+a

there are p^n choices of a, so the size of this group is p^n

and if this polynomial is irred then any splitting field also contains F_p^n

We get [F_p(α):F_p^n]= p^n/[F_p^n: F_p] = p

So F_p(α) = F_p^(n+1) and for this to have F_p^n as a subfield we must have n | n+1

This gives the n=1 case

I don’t see how to get the n=p=2 case

especially since F_8 does not contain F_4

[lone niche]

since n=p=2 is a small number you can check it by hand

i.e. solve (x^2+ax+b)(x^2+cx+d)=x^4-x+1

[lavish nexus]

I mean from my argument F_p(α) would be F_8 and F_p^n would be F_4

so if n=p=2 my argument is wrong somewhere

[lone niche]

here

Gal(F_q/F_p) where q=p^n has size n.

wait nvm idk if F_q is the splitting field of x^(p^n)-x+1

[lavish nexus]

F_p(α) has p^n distinct roots of that polynomial

[lavish nexus]

This step is sus

[F_p(α):F_p^n] = p is definitely true because it’s a hint
Then in fact F_p^(α) = F_p^(np)
but F_p^(α) is the splitting field of a polynomial of degree p^n

Then p^n = np

so we get that n= 1 or n=p=2

But I definitely got the Galois group size wrong

[F_p(α):F_p^n]= |Gal(F_p(α)/F_p)| / [F_p^n: F_p] = p
|Gal| = np

but why np

No it is actually p^n it’s just that I don’t get [F_p(α):F_p^n] = p this way

[rotund aurora]

The Galois group definitely has size p^n. The roots are of the form alpha+x where x in Fp^n

[lavish nexus]

yeah

[rotund aurora]

Then, if the polynomial were irreducible, Fp(alpha) would have degree p^n, and hence it would be the splitting field

[lavish nexus]

Now as to Gal(F_p(α)/F_p^n)

maybe I can get it directly

[rotund aurora]

It may be larger actually

G(Fp(alpha]/Fp^n)=p^n

Because the roots are alpha+x with x in Fp^n, and x is fixed. So they are all translations of alpha

[lavish nexus]

It’s still Galois over Fp^n so it should be the same as [F_p(α):F_p^n] which according to the hint is p

[rotund aurora]

It is not p

alpha-->alpha+x where x in Fp^n looks like an automorphism to me that fixes Fp^n

I am saying this assuming Fp(alpha) is the splitting field

[lavish nexus]

G(Fp(alpha]/Fp^n) has to be cyclic

but α -> α+ x has order p

doesn’t matter the x

some of them must be the same but how

So this thing has order p

[lone niche]

if you proved the hint, then I think we are done

since every extension of a finite field is normal, then deg(x^(p^n)-x+1)=(p^n)=np

[lavish nexus]

yeah I still find the fact G(Fp(alpha]/Fp^n) = Z/p bizarre

[rotund aurora]

Aren't you done? You know Fp(alpha) has degree p^n, but also that Fp(alpha) over Fp^n has degree p, and Fp^n over F^p has degree n. Therefore, np=p^n

[lavish nexus]

Yes I am technically done

It’s just that G(F(α)/F_p) and G(F(α)/F_p^n) seem to have the same form of maps and somehow one is Z/p^n the other is Z/p

[primal beacon]

this is not an automorphism no?

1->1+x!=1

[lavish nexus]

1 is not a root to that polynomial

[primal beacon]

oh alpha was a root i should’ve read the whole thing

[arctic trail]

hmmm

[arctic trail]

Take a random non trivial ring R. Take R^2 = direct product of R with itself.
Take I = R+0 and J = 0+R

[lone niche]

You replied to the wrong message

[rain grove]

Started with rings today, and I have a question here:

K is ring without unit. Show that Z x K is ring with unti if we define:

But for example $nx$ is a product of element in K and in Z

[cloud walrus]

OHHELLNAH

[rain grove]

but that is not defined?

[coral spindle]

Good observation! The book is likely wanting you to realise that you have a way of converting integers to an element of any ring

But really what this is is a homomorphism phi : Z → K with phi(1) = 1

The rest follows from addition and inversion: phi(n) = phi(1 + ... + 1) = phi(1) + ... + phi(1)

So really they should say phi(n)x or use some other notation for that, but this is what they mean.

Note that this homomorphism may not be injective: for example, think of what happens when K = Z_5 or something similar.

[tribal moss]

Note that K was specified to be a ring without unit here, so the whole point is that this phi may not exist.

We can still multiply ring elements by integers simply by interpreting it as repeated addition, though.

[tribal moss]

We can set
$$ nx = \begin{cases} \underbrace{x+x+\cdots+x}{n\text{ times}} & \text{if }n>0 \ 0 & \text{if }n=0 \ -(\underbrace{x+x+\cdots+x}{|n|\text{ times}}) & \text{if }n<0 \end{cases} $$
This automatically gives you a multiplication with $\bZ$ for any (additively written) abelian group -- here, in particular, the additive group of $K$.

[cloud walrus]

Troposphere

[rain grove]

Ohh yeah ofc

But I mean, we could define this in any weird way we want right?

Can i do nx = x+x+x+...x (13 times) if n=0 and then for n>0 and n<0 just adjust for 13

[chilly ocean]

You could but what's the point of that?

[chilly ocean]

It is more natural to define it this way and it has nicer properties

[mighty kiln]

It's an unstated convention

[chilly ocean]

For example your definition won't satisfy (m+n)x = mx+nx

[chilly ocean]

In fact, if you want (m+n)x = mx+nx and 1x = x, then it will imply this

[mighty kiln]

Whenever the group operation is denoted by + one automatically assumes that nx means x added to itself n times

And similarly x^n when group/monoid operation is denoted by ×

[rain grove]

okk ty

Why $xy = 1 \Rightarrow yx = 1$ is not enough for commutative ring? Im thinking: if xy = k then not necessarily yx = k but just saying that doesnt convince me and i can't find an example

[cloud walrus]

OHHELLNAH

[chilly ocean]

I don't understand what you are asking

Are you asking if there are non-commutative rings where xy=1 implies yx=1 for all x,y?

[rain grove]

$xy = 1 \Rightarrow yx = 1 \iff xy = yx$ for all x,y

[cloud walrus]

OHHELLNAH

[rain grove]

This is not true

why?

[chilly ocean]

What noncommutative rings do you know?

Do you know linear algebra?

[rain grove]

Yeah so non-commutative would be matrices in R

[chilly ocean]

Yes, nxn matrices in R (or in any other field) form a noncommutative ring whenever n>=2

There's a well known theorem in linear algebra..

Suppose XY are matrices and XY=1

Then YX=1

Have you heard of this theorem?

[rain grove]

Yes

[chilly ocean]

Well, there you go then

Any ring of nxn matrices over a field R for n>=2 is a counterexample

[rain grove]

ohh dam okk

[knotty badger]

I think this may also work in any integral domain..?

[chilly ocean]

What also works?

Integral domains are commutative so they can't give a counterexample to OHHELLNAH's claim

[knotty badger]

Which matrices aren’t lol

[chilly ocean]

Do you mean non-commutative domains?

[knotty badger]

yeah nvm, I meant like

YX is a nonzero idempotent

I was trying to think of in which rings you can deduce from this that YX = 1

[tribal moss]

No, that would wreak havoc for the distributive law in the resulting ring structure on Z×K.
You need to have a multiplication where (n+m)x = nx+mx.
And you also want 1x = x.
These two conditions, taken together, force the "repeated addition" interpretation.

[glad osprey]

$xy = 1 \implies yx = 1$ just means that every element commutes with its multiplicative inverse. You can't write $xy = 1 = yx$ for all x, y, as that would require $\wedge$ instead of $\implies$

[cloud walrus]

sheddow