groups-rings-fields

[south patrol]

Well not so much separable as perfect (i.e. all algebraic extensions are separable)

[glad osprey]

Nice proof

[toxic zephyr]

ah right. fermats lil theorem isn't that basically?

[tribal moss]

Yes, that's why it's zero everywhere.

It's also the product of all the x-a (and then a factor of 2), but that fact is easiest to see by noting it has the right roots (by Fermat!) and degree to be that product.

[toxic zephyr]

hmm. i think the main thing kicking my ass about this class ||besides my health|| is that there are these huge relationships between these concepts that don't seem to be emphasized. I struggle to keep track of all of them. like normal extension is basically the same as a splitting field, but they're sort of taught as separate mysterious concepts and eventually it's like a theorem they're actually the same.
not a particularly new thing in math but for some reason I'm struggling with this class especially.

im currently on a journey to connect all these concepts properly and actually figure out what's been going on all quarter 😭

i think separable in particularly confused me because it seems almost all the examples we usually look at are separable. i really only know this one example of a nonseparable extension and it feels very pathological lol

[glad osprey]

Just do what I do, pretend everything is separable

Seriously tho, I think there's a reason it feels pathological, it kind of is, you won't need to work with non-separable polynomials a lot, particularly if this is an intro course to Galois theory

Like, char 0 fields and finite fields are perfect, and those are the fields you will work with the most to begin with

Are you following a book btw? If things seem very mysterious, maybe a different book could help? I can also send you notes from my class if you're interested

[south patrol]

I always find this funny when Galois theory courses spend loads of time worrying about things like separability and then all examples are over Q

[tardy hedge]

Same thing in my class

Had a whole homework just dedicated to it

[glad osprey]

My personal recommendations: either of Aluffi's book are great (if you pick Chapter 0, be sure to skip the algebraic geometry in the middle of the galois theory), and Galois Theory by Cox is also really nice, and finally Galois Theory through Exercises has a really nice selection of exercises with hints and solutions

[tardy hedge]

Man ngl i wasnt a big fan of galois theory

I like learning commutative algebra more

[tardy hedge]

Prob cuz it was hard lol but idk i also didnt like thinking about it all too much

[woeful sage]

Oh that's true, "zero everywhere" only implies that the constant term is zero

[woeful sage]

commutative galois theory when

[tardy hedge]

Abelian extensions and shit

[tardy hedge]

I think id appreciate it a lot more a second time through tbf

[drifting mauve]

So,

I've shown that all primes congruent to 3 (mod4) are irreducible in Z[i].

I now just need to find all irreducible gaussian integers.

I'm quite stuck, I'm kinda thinking that the answer is all associates of primes congruent to 3 (mod4). But I'm sure if this is correct and I'm stuck in how to prove this.

[toxic zephyr]

my course supposedly follows hungerford. i've been having some trouble following it tbh. i would be interested in your class notes if you don't mind sending them (:
my professor posts very like. skeletal solutions and notes. they don't really help much with understanding tbph

[alpine plank]

Not quite, what can you say about 1 + i ? it's norm is 2 so if 1 + i is reducible 2 isn't prime. I believe rather than "finding" all irreducible Gaussian integers you will have an easier time characterizing which ones are irreducible based on their norm.

[drifting mauve]

okay, I'm with you. but how should I go about characterising it basd on the norm?

[alpine plank]

well what did we just observe about 1 + i

[drifting mauve]

well, ig its irreducible, but its not an associate of a prime, p st p = 3 mod 4??

so ig my inital guess is wrong

[alpine plank]

yes yes but what did we use to show that its irreducible

[drifting mauve]

that its norm isprime?

[alpine plank]

yep

[drifting mauve]

really? is that a result? So does the norm of x+yi being a prime number always imply that x+yi is irreducible?

[alpine plank]

prove it!

[drifting mauve]

true.

ok, thank you

I think I've got an idea of direction now

[alpine plank]

Yeah just don't forget to ask what happens when the norm isn't prime

[drifting mauve]

will do thank you

[dire turret]

(maybe wrong channel idk) is there a name for a monotone map between i guess semi-lattices such that f(x v y) <= f(x) v f(y) or the other direction maybe

is this like a lax something

[rocky cloak]

If it's monotone you have
f(x)vf(y) <= f(xvy) automatically. So this would be join-preserving I guess.

[dire turret]

wait do you

lol i didn't realise

thank you

i should probably read a text about this content properly

right of course since f(x) <= f(x v y) and f(y) <= f(x v y), that makes sense

[merry summit]

how would i prove (a+b) mod n = (amodn +bmodn)mod n . while trying to do it i came to a point where i have to use that fact itself to prove any further

[chilly ocean]

how did u use the question to solve question

[rocky cloak]

Write
a = kn + r and b = ln + s for r and s between 0 and n.

Then use that x+n mod n = x mod n

[merry summit]

ya i used that approach, what i was missing was the fact that the sum of remainders r and s can be greater than n, so i substituted (r +s) = nk + m

i was able to prove it now thanks

[glad osprey]

Here are the notes: https://wiki.math.ntnu.no/_media/ma3202/2024v/lecture_notes_latex.pdf They're pretty succinct, but maybe they'll help getting an overview atleast

[cloud walrus]

tm

[spice wagon]

im pretty sure about my proof but just to make 100% sure

[glad osprey]

Is H^G the set of functions from G to H? Is that a group?

[mighty kiln]

Yes if H is a group

[glad osprey]

Oh, right

[spice wagon]

yes with the action $f \cdot g (x)=f(x)g(x)$

[cloud walrus]

tm

[south patrol]

Pointwise multiplication

[spice wagon]

yes exactly

[glad osprey]

The proof looks good 👍 Note that f(xy)g^-1(xy) = f(x)f(y) g^-1(x)g^-1(y) isn't because H is abelian tho, it's just because f and g are homomorphisms

[spice wagon]

$g^{-1}(xy)=[g(x)g(y)]^{-1}=g^{-1}(x)g^{-1}(y)$ because H is abelian no?

[cloud walrus]

tm

[spice wagon]

instead we would have g^{-1}(y)g^{-1}(x)

[tardy hedge]

why in commutative algebra do we care so much about prime ideals?

[glad osprey]

ah, right, good point 👍 g is a homomorphism, but g^-1 isn't unless H is abelian

[tardy hedge]

M(x)_A A_s is considered a right As-module here?

i wouldve liked it to be written as As (X)_A M lol

[thorn jay]

I assume A is a commutative ring?

[tardy hedge]

yea

[thorn jay]

In that case left and right modules structures coincide

As A^op ≈ A

[tardy hedge]

yeah

[thorn jay]

I think it's just customary with a tensor product to write the "modifier" on the right

Like you also do this for homology with coefficients

[tardy hedge]

i like modifying on left 😢

im so distraught

[thorn jay]

pat pat you'll survive

Wait until you have bimodules over a noncommutative ring

[tardy hedge]

idk if u guys know that crank guy who got kicked here i think

he's telling me how he's proving riemann hypothesis

i honestly dk why im engaging him still but he keeps messaging me

idk hes kind of just interesting in the fact that hes throwing around so much jargon yet doesnt know what hes talking about

lol

[woeful sage]

that's basically me

[dire wren]

what i mean is
K_n+1 is K_n extended to include all roots of polynomials with coefficients in K_n

isnt that equivalent?

whats a spliiting feild?

whats f(x_f)

[chilly ocean]

a polynomial,with variable x_f

[tardy hedge]

woahh wew used the Upon the Witnessing emote

[chilly ocean]

absolute cinema

[tardy hedge]

he is a mysterious figure

[chilly ocean]

why so

[tardy hedge]

He is usually not present

[rocky cloak]

The splitting field of a polynomial is the smallest field extension where the polynomial splits (factors as a product of linear polynomials)

[tardy hedge]

but every once in a while, he appears

[chilly ocean]

like catking

[spice wagon]

the canonical inclusion $i :H \to G$ is always defined by $h \mapsto h$ ?

[cloud walrus]

tm

[spice wagon]

H is a subgroup of G

[spice wagon]

bc i saw different ones (by always i mean usually)

[candid patrol]

yes ofc

[jade mason]

You can also embed groups which arent set theoretically subgroups

Like $i: \mathbb R \to M(2\times 2, \mathbb R)$ via $x \mapsto x\cdot I$

[cloud walrus]

Jussari

[jade mason]

But in that case you wanna think of real numbers as diagonal matrices, so that i is again of the form "A -> A"

[rustic crown]

before you prove the statement that jagr wrote, you could only add roots of finitely many polynomials at once. not an infinite collection.

i can start with ℚ, then get ℚ(sqrt2), then ℚ(sqrt2, sqrt3), ... but each time you're making the field bigger you're also making the set of polynomial that needs to have roots bigger. and it's not totally obvious how to resolve this issue.

[rustic crown]

once you prove that statement, then you can construct K_{n+1} from K_n like you say. and now given any polynomial f with coeffcients in the "union", this polynomial already will lie in some K_n[x] as polynomials are "finite gadgets". and so it will have a root in K_{n+1}. this shows that the union is algebraically closed.

[glad osprey]

There's also more complicated examples, like embedding (R, +) in GL_2(R) by x |-> [1 x; 0 1]. But the canonical inclusion is pretty much always just x |-> x

[rustic crown]

here i mentioned a different issue. you might hope to define a directed system on the poset P of non-zero polynomials under divisibility. given f, we can define K_f to be the splitting field of f. and when f | g, we would need to choose a map K_f --> K_g, and in general this choice is NOT unique. and it's not obvious we can make a compatible choice such that the triangle will commute when f | g | h.

[chilly ocean]

semirings what

[spice wagon]

alr thanks guys

[dire wren]

i could just union them all? i mean ik how to add any given root, so i would assign them ordinals, and i could just add each of the roots and at a limit orfinal i could just take union of all previous feilds and extend that

[dire wren]

hm

[rustic crown]

that works yep. just one thing to be aware of is that for each non-constant polynomial you should adjoin all the roots of it. else we need the argument that jagr wrote, which proved that given an algebraic extension L/K such that each non-constant polynomial f in K[x] has at least one root in L, then L is already algebraically closed.

if you adjoin all the roots of the polynomial, then we would need to show this easier statement that given an algebraic extension L/K such that each non-constant polynomial f in K[x] splits completely over L, then L is already algebraically closed.

[tribal moss]

A problem with adjoining an entire layer (or even just finitely many!) roots in one step is that perhaps some of them could be made using other ones, and if you try to adjoin them separately you might not get a field.

For example if we want to extend Q with sqrt(2) and sqrt(8), it's tempting just to do Q[x,y]/(x²-2, y²-8), one unknown per root.
Unfortunately, in the resulting quotient we have 4x²=8 and therefore (2x)²-y²=0 and therefore (2x-y)(2x+y) = 0 -- but neither 2x-y nor 2x+y are in the ideal we quotiented by, so those are zero divisors and we're definitely not getting an extension field.

So I think you need to handle all your polynomials one by one, and for each of them ask, is this still irreducible in the field I've built yet? And then only adjoin new elements if it is irreducible.

[spice wagon]

if we have for example $GL_n(\mathbb{R})$ and we want to study its possible decompositions into direct/semi-direct products, we have to guess them?

[cloud walrus]

tm

[spice wagon]

or there is a general method

[thorn jay]

Normal subgroups come from homomorphisms, so I guess you generally start by looking at group homomorphisms from this group into some other group

Then once you have a since description of the kernel, use the first isomorphism theorem to look for a potential splitting, or decomposition

[spice wagon]

you mean the characterization $H \triangleleft G \iff \exists \varphi \in Hom(G,K)$ such that $H=Ker(\varphi)$?

[cloud walrus]

tm

[spice wagon]

with K another group

[thorn jay]

Ye

[velvet hull]

Schur-Zassenhaus may or may not be useful

[thorn jay]

GL_n(K) has a very nice natural homomorphism to K*, btw

(multiplicative group of K)

Which splits

[spice wagon]

ok so i have to check the ker of an homomorphism from Gln(K) to K* and then i’ll have a normal group

i have to do it 2 times so

with another homomorphism

[thorn jay]

I don't see why

Not if it splits

[spice wagon]

to have two normal groups

[thorn jay]

That's only when it's a direct product

First you should focus on finding a splitting homomorphism

And then either prove or disprove that the image of the section is normal

[spice wagon]

but what do u mean by split ?

sry i don’t study in english lol

if $\pi$ is the canonical projection, $\exists s$ such that $\pi \circ s=id$ ?

[cloud walrus]

tm

[spice wagon]

alright i’ll check that thanks

[keen badge]

\textbf{Definition (Residually Finite Group):}
A group $G$ is \emph{residually finite} if for every nontrivial element $e\ne g \in G$, there exists a normal subgroup $N \trianglelefteq G$ of finite index such that $g \notin N$.

\medskip

\textbf{Theorem:}
Let $H_1, H_2\le G$ be finite index subgroups. Then their intersection
$$
H:=H_1 \cap H_2
$$
contains a normal subgroup $N \trianglelefteq G$ such that $[G : N] < \infty$ and $N \leq H$.

\medskip

\textbf{Is it true that:}
Since every finite index subgroup contains a finite index normal subgroup, we can equivalently define a residually finite group as:
"For every $e\ne g \in G$, there exists a \emph{finite index subgroup} $H \leq G$ such that $g \notin H$."?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[rustic crown]

yea that makes sense

ig one reason to think in terms of normal subgroup is so that we can rephrase it in terms of finite quotients.

G is residually finite if for every non-trivial g in G, there is a surjective map G --> H with H finite group such that g isn't in the kernel.

in other words G --> G^ the map to the profinite completion is injective.

[inner acorn]

What stops us just taking the trivial subgroup {e} as a normal subgroup not containing g?

[fresh bison]

I’m trying to understand the proof of all cosets being the same size. My book says that if $H$ is a subgroup of $G$, $a,b \in G$, and $aH$ and $bH$ are cosets, then $f(x)=ba^{-1}$ is a bijection that transforms elements of $aH$ into elements of $bH$. I understand why this transforms elements, but why is it a bijection?

[cloud walrus]

Anne S

[dull marsh]

Its index

[inner acorn]

Yeah just realised in the inf case, {e} won't have finite index, so ignore me

[inner acorn]

Might be easier to show H and aH are the same size with the bijection f(x) = ax, for any a in G

[jade mason]

you can either directly prove that its injective and surjective, or you can construct an inverse function ( g(y)=ab^{-1}y works as you should check)

[inner acorn]

Huh, that's neat, unlike intersection of residually finite groups, this isn't closed under taking subgroups - e.g. the Baumslag-Solitar subgroup BS(2,3) is not residually finite, but is contained in GL_2 (Q) which is residually finite

Seems useful in that, if you have a residually finite group G and distinct elements g1, ..., gk

You can find a finite group H, and morphism f: G --> H, where f(g_i) =/= f(g_j) for i =/= j

I'll stop now

[rocky cloak]

Being residually finite is closed under taking subgroups. For any element g, just pick N in the larger group, then intersect with the subgroup

[inner acorn]

Yeah, ignore me again, saw a theorem (Mal'cev’s Theorem) which talked about linear groups over char 0 being r.f., but I missed the "finitely generated" piece, so applying the theorem to GL(2, Q) was in error.

Cheers for the corrections tho, proof makes sense to just take N intersect H normal in H, with N not containing h in H (in G) etc...

[rustic crown]

sasha

[inner acorn]

Hoi hoi

How is the det

[rustic crown]

det is

[inner acorn]

Might need to post here more and get my bad habits ironed out xD

Especially when I've not heard of half the things mentioned here 👀

Curious if there's anyone here also looking at Ariki-Koike (or cyclotomic KLR) algebras, and all the crystal / Lie theory stuff involved

Teach me det

[rustic crown]

det doesn't know

[inner acorn]

What kinda stuff does det do

[tardy hedge]

What the flip!

[rustic crown]

det is a tiny eevee ><

me thinking of learning some rigid analytic stuff

[inner acorn]

What has caught your interest in rigid analysis

(iirc that's analysis with p-adics right?)

[rustic crown]

yee me wanna learn about geometric langlands and fun stuff

[rocky cloak]

I attended this years abel lecture. It was (partially) about KLR algebras.

I could not tell you what they are though

[south patrol]

SashaMomo

[inner acorn]

Holla

[rustic crown]

Potato

[south patrol]

Hi

[inner acorn]

I've seen them in the context of representation theory of the symmetric group (or rather the wreath product of Z/mZ with Sn) where you construct so called Specht modules which are indexed by partitions (or rather multipartitions) and can by quotiented out by a radical corresponding to some blinear form to get the simple modules!

What I've described is technically the Ariki-Koike algebras (see attached presentation, notice if r = 1, T_0 = 1 + Q_1 and R a field you get the symmetric group algebra)

[thorn jay]

What are the braid relations doing here

Lmao

[rocky cloak]

Feel like the other relations are weirder than the braid relations

[inner acorn]

Turns out the cyclotomic KLR algebras are isomorphic to these guys as algebras (they have their own definition related to Quantum groups and Catan data), and this was proven quite recently (2009)

[thorn jay]

They are, sure, I guess it's interesting that they pop up

[inner acorn]

Which is cool because the cyclotomic KLR algebras have a relatively simple grading, which is not obvious in the Ariki-Koike definition

[rocky cloak]

Theyre part of the relations defining the symmetric group, so not that surprising if the algebra is related to the symmetric group

[thorn jay]

Mm that makes sense

[inner acorn]

If you just take r = 1, these guys are isomorphic to the Iwahori-Hecke algebras, which give nice solutions to the Yang–Baxter equations

[thorn jay]

❤️

I'm gonna look into that equation with a physics friend of mine soon

[inner acorn]

Small world, then you might be interested in quantum groups and the Hecke algebra

[thorn jay]

Quantum groups seemed to come up yes

Thank you

[rocky cloak]

I was originally planning to do a masters on Yang-Baxter equations and Hopf algebra stuff, but the pandemic changed my plans a bit

Ah, the twists and turns of life amarite

[kind temple]

what did you do instead?

[rocky cloak]

Finitistic dimension conjecture

[inner acorn]

Why is there no distinction between Nakayama's conjectures lmao

Was just googling/reading that the Finitistic dimension conjectures implies Nakayama's conjecture

but it's the homological one

not the blocks of the Hecke algebra one

[rocky cloak]

Could call it Nakayamas selfinjectivity conjecture maybe

Or the dominant dimension conjecture

[inner acorn]

Small world u.u

[rocky cloak]

,rotate

[inner acorn]

[rocky cloak]

I made this nice diagram of which homological conjectures imply each other

(NC is the Nakayama conjecture)

[inner acorn]

That's neat, I might have to steal this idea lol

[thorn jay]

Based

[rocky cloak]

The idea of making a diagram of implications? I think that's an idea we can all share

[velvet hull]

Terrible thought, but if I made a diagram of implications inspired by your diagram of implications wouldn’t there be an implicit 2-implication from your diagram to mine

[thorn jay]

Seems like FDC is a really strong statement

[rocky cloak]

Yes, and there are quite a few stronger conjectures that imply FDC that have been proven false.

So it's kinda at that sweet sspot

While simultaneously being a pretty early one (from 1960)

[thorn jay]

Gotta love those conjectures

[inner acorn]

There are a lot of algorithms computing the Mullineux map, and a few conjectures which would imply better algorithms (currently working on proving, and making NO progress with, such a conjecture from Jacon & Lecouvey) - would be useful to have a visual

yoink

[next obsidian]

Hi Sasha Momo!

[inner acorn]

Hoi hoi

[tardy hedge]

Wow ive never seen sasha momo

Hi sasha momo

[thorn jay]

Haiii, I've not seen you either

[elfin wraith]

What is the deal with Yang Baxter? There’s someone at my uni who’s very interested in set theoretic solutions and in my quantum programming class we did some stuff with it using string diagrams but I’ve never really understood the motivation beyond “something about physics”

[rocky cloak]

I never found out, but something with physics sounds about right

[elfin wraith]

Very fair

I should look it up because it seems like an odd relation, but a lot of people are interested in it so there must be something

[thorn jay]

Something about particles on a string switching places and interacting, and Yang-Baxter gives a restriction on how the particle states can change, based on some symmetry condition

Iirc

And also something to do with statistical mechanics apparently

I like set-theoretical Yang-Baxter cuz it gives rise to a neat algebraic structure, kind of two shelves braided together type beat

[elfin wraith]

Braces?

[thorn jay]

Hmm, no more general

It's a universal algebraisation of the set-theotetical Yang Baxter equation

Damn, universal algebra being a somewhat useful framework

[spice wagon]

i just got introduced to resolvable groups but what does it means concretely ?

[coral spindle]

Wym "concretely"

The origin, in some sense, of solvable (I think this is what you mean!) aka soluble groups is Galois theory. It is a theorem that elements of a field extension are expressible as radicals iff the Galois group is solvable. So there is a great motivation.

It turns out solvable groups are important and nice in other ways. So there you go.

[spice wagon]

ah okayy because they introduced it with the definition with subgroup sequences etc but how can we see that « intuitively »

[coral spindle]

You cannot.

[spice wagon]

idk how to say 😭

ah

[coral spindle]

Or more accurately

Just get used to it.

[spice wagon]

okay let’s wait for the galois theory lol

[velvet hull]

well the very rough reason why

is when you take a square root, you are kind of like adding a cyclic group

if you squint at it

[coral spindle]

(I think they are asking how they can intuit the definition, not why it's equivalent to the extension being solvable)

[spice wagon]

wtf 😭

[velvet hull]

mmmh, I interpreted as them asking for intuition for why solvable group iff solvable polynomial

but I guess they don't know that theorem yet

[velvet hull]

so forget what I said lmao

[spice wagon]

yes 🥲

465 pages left until i’ll discover it

[velvet hull]

it'll go by faster than you think

[south patrol]

Which book is this lol

[spice wagon]

it’s a french book from Jean-Marie Arnaudiès

[coral spindle]

I can only imagine you are exaggerating a little bit

[south patrol]

I feel like I hadn't read like 100 pages of algebra when I did galois theory

[spice wagon]

there is too many definitions in group theory 😩

[coral spindle]

The hard part of algebra is coming up with the correct definition

[spice wagon]

and also to visualize what we’re manipulating

[cloud walrus]

tm

[spice wagon]

is it the right definition?

bc i saw different ones

some definitions say that H_{i+1}/H_i must be abelian, monogenous

[south patrol]

It's equivalent for finite groups

For arbitrary groups one usually uses abelian rather than cyclic in the definition

(And says solvable rather than resolvable)

What do you mean by monogeneous

[spice wagon]

generated by a single element but not necessarily finite

idk how we say in english

[south patrol]

Well that's just cyclic

[spice wagon]

cyclic is when the group is finite no

[south patrol]

Nah that's "finite cyclic"

At least in English anyway

[spice wagon]

ahh okayy

[spice wagon]

i’ll write that in my notes thanks

[tribal moss]

It's confusing because an infinite cyclic group doesn't "cycle around".

[south patrol]

This is true

[velvet hull]

it cycles... to the point at infinity and then back (real)

[tribal moss]

It cycles because it embeds in the circle group.

[dim widget]

Ah yes <e^ei> my beloved

[south patrol]

Old Macdonald had a farm

[tribal moss]

Hmm, it's surprisingly difficult to google up discussion of whether e/pi is rational or not. Plenty of yapping about e+pi, e·pi, e^pi, etc. though.
Is there a simple proof of its irrationality and that's why nobody deigns to mention it?

[dim widget]

Wouldn’t that be (ei)^{ei + 0}?

[south patrol]

My apologies.

[south patrol]

Hm

[dim widget]

I doubt it’s known

[south patrol]

I would guess it is irrational but not proven yet to be so

[dim widget]

I think it would imply that \pi + e = (r+1)\pi and \pi e = r\pi^2

[south patrol]

It is sort of funny for this reason

[dim widget]

So we would know both of these are irrational if we knew it was rational

[south patrol]

Like it can't be easily provbly rational

[dim widget]

I guess that doesn’t really help

[south patrol]

Ofc id assume it's irrational but unknwn

[dim widget]

But also we don’t know \pi e is irrational

So it’s hard to imagine

[south patrol]

Hard

I wonder what a proof would look like

[dim widget]

Idk transcendence theory proofs are always bizarre

[tribal moss]

Someone proves Schnauel's conjecture and this drops out as a corollary.

[south patrol]

Who is it who said like

Hmm when you start caring bout transcendence you stop caring about number theory

Or smth

Lol

It feels a shame that pi was never proven to be irrational in my undergrad

Feels like a cute thing

[thorn jay]

how would you even begin with smt like that

[dim widget]

If I knew I would be writing an annals paper instead of talking to you ;p

[thorn jay]

hey that's fair enough

[wintry sluice]

with respect to what norm is Zp euclidean to (p is prime)?

[dim widget]

p-adic valuation

but it's almost never useful to know it's a euclidean domain

[white oxide]

Is every subset of non-units in a commutative ring contained in a maximal ideal? Can this be proved with something like Zorn's lemma

[dim widget]

No and thus no.

like in Z/6 you have 2, 3 which are not contained within any maximal ideal

[white oxide]

Damn

[glad osprey]

thank god you didn't say no and yes

[dim widget]

math survives for today

[thorn jay]

I mean, if you reject Zorn's lemma..

[glad osprey]

I always work in ZFE (zermelo-fraenkel set theory with the axiom of explosion) 💪

[next obsidian]

This happens only when you’re local

[rocky cloak]

The set of non-units being contained in an ideal is called being a local ring

[delicate orchid]

I work in KFC

[tardy hedge]

Is it because (2,3) = (1)?

[dim widget]

that is one way to see it, but with a 6 element ring there are many ways to argue

[trail rose]

Let H be an arbitrary subgroup of G. Is it true that F_n(G) \cap H is contained in F_n(H)? Here F_n stands for the nth Fitting subgroup in the upper Fitting series

[delicate orchid]

it's definitely true for n = 1 lol uhhhh

I guess the question is if H/F_n(G) \cap H = HF_n(G)/F_n(G) is nilpotent? If we can reduce it down to that then yeah it's pretty clear

and we can reduce it down to that, given a nilpotent chain 0 < H_1 < ... < G we must have H_i is contained in the ith fitting group. Moreover, HF_n(G) <= G implies that HF_n(G)/F_n(G) is isomorphic to some subgroup of G/F_n(G), and subgroups of nilpotent groups are nilpotent, we should be able to construct a nilpotent chain of H containing F_n(G) \cap H, and hence it is in F_n(H)

[spice wagon]

for p a prime number and G a group of cardinal p^2, show that G is abelian

should i show that the center of G=G?

[candid patrol]

Yes

It is one of the classics proofs

[spice wagon]

[spice wagon]

par inclusion et argument d’ordre j’imagine ?

[candid patrol]

Tu es français haha ?

Tu m’as cramé à mon anglais 😓

Sorry let’s speak english

[spice wagon]

nn jte connais lol

we say order in english for card(G) ?

[rocky cloak]

A nice chain of arguments might be that the center of a p-group is nontrivial (use for example class equation)

The center can't have order p because G/ZG is never cyclic unless G is abelian

[candid patrol]

De où ?

[spice wagon]

maths en direct

[candid patrol]

T’as pas MED dans tes servs en communs, t’as leave ?

[spice wagon]

why it’s never cyclic ?

[spice wagon]

ui

[candid patrol]

Reviens le serv est bien

[spice wagon]

never

[glad osprey]

G is abelian iff G/ZG is cyclic, pretty well-known result, you should try to prove it

[candid patrol]

Wrong

[spice wagon]

quand ils disent cyclique c’est pour monogene

l’anglais c’est bzr

[candid patrol]

By cyclic you mean finite + monogen ?

[glad osprey]

French and their terminology is in their own little world smh

Yes, by cyclic I mean the definition the entire rest of the world uses

[spice wagon]

in french cyclic is necessarily finite

[knotty badger]

Does it even need to be finite

For the proof I mean

[candid patrol]

Of course

[knotty badger]

I don’t remember finiteness being used in the proof is the thing

[glad osprey]

I mean no, I don't mean finite

[candid patrol]

Same, i was wrong mb

But G/Z(G) is monogene iff G is abelian is true

[spice wagon]

ça existe en anglais monogene ?😭

[candid patrol]

Cause we don’t have don’t use that G/Z(G) is finite

For the proof

[candid patrol]

J’espère mdr, mais viens mp

[glad osprey]

Monogene isn't a term that is used in english

[candid patrol]

Oh mb

When G = <g>, you say that G is cyclic then ?

[knotty badger]

Yes

[candid patrol]

😵

[delicate orchid]

I'm invading france

[spice wagon]

it reminds me smth

[glad osprey]

france's terms of surrender is to literally change their terms

[vivid kestrel]

So let R be a Ring with principal ideal I = (p) and assume we have elements a and b in R/I such that ab = 0, can we recover from that a factorisation of p

[delicate orchid]

well from this you know that ab = rp for some r in R, and that r depends on both a and b in general

if you had precise control over this r then maybe it's do-able (like when R = Z)

[rocky cloak]

It's possible for p to be irreducible without (p) being a prime ideal. So then you would have a and b with ab = 0 in R/I without any (nontrivial) factorization of p existing

[south patrol]

Real

[vivid kestrel]

oh i see

thanks yall

[trail rose]

Why do we need H/(F_n(G)\cap H) to be nilpotent? F_n(H) is not necessarily the top Fitting quotient

this is my progress, and after this i get stuck

[delicate orchid]

I tried to do it inductively and also got stuck

[vagrant zinc]

Name of book?

[trail rose]

this is notes i am writing

[spice wagon]

you sure about this ?

[south patrol]

Yes

common exercise

[spice wagon]

I searched and found only this implication: If G/Z(G) is cyclic, then G is abelian

[south patrol]

Well the converse is trivial

If G is abelian, what is Z(G)

[spice wagon]

G

[elfin wraith]

The quotient should then be pretty easy to generate

[spice wagon]

its the classes to the left or right of G

[glad osprey]

Do you remember how quotient groups are defined? What are the elements of G/G?

[spice wagon]

yes sure its gG so G

[glad osprey]

yep, so what group is G/G? (I'm assuming by the emote that you didn't quite understand the converse)

[spice wagon]

abelian ?

like its G which is abelian

[glad osprey]

no, we're proving that G abelian implies that G/Z(G) is cyclic. Which group is G/G isomorphic to? Or even simpler, what cardinality does it have?

[spice wagon]

to {e}

which is cyclic

[glad osprey]

yep, nice

[spice wagon]

oki i'll try to do the other sense

[spice wagon]

can this be generalized to any p-group?

[candid patrol]

No…

[spice wagon]

😢

[candid patrol]

H8 which is a 2-group is not abelian

Same for D8

[spice wagon]

wtf is H8 😭

quaternion?

[candid patrol]

Quaternion group yes

[spice wagon]

ah

didn’t study yet quaternions

[candid patrol]

That’s a hard group to study

Omg my english lvl 😵‍💫

[spice wagon]

dihedral is the worst i think

[candid patrol]

Clearly not

[spice wagon]

cleerly?

😭

[candid patrol]

Edited 🤫

[spice whale]

i mean
as far as nonabelian groups go, H8 is one of the most abelian ones

[candid patrol]

Why ?

Z(H8) has only 2 éléments

[knotty badger]

i think they mean some inequality

about the number of pairs where ab = ba

[spice whale]

the proportion of pairs of elements that commute is maximal over all finite nonabelian groups

[knotty badger]

mhm

[candid patrol]

Still a non-abelian group but allright

[spice whale]

yeah I'm mostly kidding

[spice wagon]

jpp

[keen badge]

Let $\iota : K \to L$ and $\tilde{\iota} : \widetilde{K} \to \widetilde{L}$ be two field extensions (i.e., injective field homomorphisms).

We say the extensions are isomorphic if there exists a pair of field isomorphisms $(\lambda : K \to \widetilde{K}, ; \mu : L \to \widetilde{L})$ such that:
$$
\mu \circ \iota = \tilde{\iota} \circ \lambda
$$
That is,
$$
\mu(\iota(k)) = \tilde{\iota}(\lambda(k)) \quad \text{for all } k \in K
$$

Let $\iota : K \to L$ and $\tilde{\iota} : K \to \widetilde{L}$ be field extensions.

Is the following statement true?

$$(K \xrightarrow{\iota} L) \cong (K \xrightarrow{\tilde{\iota}} \widetilde{L}) \iff L \cong \widetilde{L} ; ?$$

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[jade mason]

I think you can consider something like K=L=L'= k(x) with i: K->L as the identity and i': K -> L' which sends x to x²

[south patrol]

"Injective field homomorphism" is a bit redundant

[south patrol]

Yeah

[thorn jay]

I love injective embeddings

[spice wagon]

in rings and fields theory, do symmetries, permutations etc.. play as important a role as they do in group theory?

[dim widget]

symmetries are important in all of math

too vague of a question

but they are very important in field theory as well

[spice wagon]

ah

i’ll have to get back to symmetric group 😥

[south patrol]

I would say it is more like lol

Symmetries arise in mathematics and that is how groups arise usually

[spice wagon]

yh i really hate this part of groups theory 😭

[thorn jay]

We study a group by studying how it can be interpreted as symmetries of some object, while we often study objects by considering it's symmetries

So everywhere symmetries play an important role, but just a little different

[spice wagon]

alright i know what i have to do..

[quick laurel]

working on a galois theory problem where im asked to find the galois group of the polynomial $x^4 +3x^3 - 5x^2 + 3x + 1$
i have the fact that if $f(\alpha)$ is a root, then $f(1/\alpha)$ is also a root

[cloud walrus]

rubixcyouber

[quick laurel]

my vague understanding is that any automorphism in the galois group Gal(K/Q) will permute the roots, so since we are permuting 4 things, we are going to be finding a group living inside of S_4

letting the roots of the quartic be ${ x_1, x_2, x_3, x_4}$ where wlog we have that $x_1 x_2 = x_3 x_4 = 1$, i have tried applying an arbitrary $\sigma \in \text{Aut}(K/Q)$ to the relations, getting us $\sigma(x_1) = \sigma(x_2)^{-1}$, but im stuck here

[cloud walrus]

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

[quick laurel]

i guess since $x_1, x_2 \not\in \mathbb{Q}$, we have that $\sigma$ must be like
x_1 <-> x_2 and x_3 <-> x_4

[cloud walrus]

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

[vagrant zinc]

Nice

[dim widget]

okay so you know that \sigma(1/\alpha) = 1/\sigma(\alpha) for all \sigma \in Gal(f) so then every element of Gal(f) commutes with the permutation \alpha \to 1/\alpha of the roots

what does that tell you?

[quick laurel]

the permutation $\alpha \to 1/\alpha$ is in the center of the galois group?

[cloud walrus]

rubixcyouber

[dim widget]

it might not even be in the galois group a priori I guess

but what does it tell you about the possibilities for the galois group?

(just using the structure of S_4)

[quick laurel]

im assuming the spliting field over Q so i think this implies it is Galois

[dim widget]

No I mean the permutation \alpha \to 1/\alpha is not necessarily in the galois group

it may not be an automorphism of the splitting field Q(x_1, x_3)

[quick laurel]

i see

[dim widget]

but.... it is an abstract element of S_4

[quick laurel]

namely a transposition ?

[dim widget]

no it's a product of two transpositions

it is (12)(34) using your notation x_i for the ith root

[quick laurel]

oh yeah because you need to apply it to both groups of the roots

[dim widget]

anyway so what's the centralizer of that?

[quick laurel]

of what? the group generated by (12)(34) ? apologies

[dim widget]

yes

which is the same as just the centralizer of that element

[quick laurel]

i forget, is it all products of two transpositions ?

one moment

its not because (34) is also in there

as is (12)

i dont think any 3 cycles are in the centralizer

all products of two disjoint transpositions are as (12)(34) is one

(1423) and its inverse are also in there

so C = {e, (12), (34), (12)(34), (13)(24), (14)(23), (1423), (1324) } and we can stop because 8 | 24

is this the Galois group ?

as an aside, if anyone has any recommendations for galois theory that are less dense than dummit and foote that would be much appreciated

[dim widget]

well that’s the question isn’t it?

the galois group has to be a subgroup of this

[dim widget]

You should prove rigorously that this is the whole centralizer (however you are correct that it is)

[dim widget]

And you know that it must act transitively on the roots

[quick laurel]

could you explain what "transitively" means here

[dim widget]

Like any element can be sent to any other

Using some element of the group

So just one orbit of roots

[quick laurel]

I see

[fresh bison]

I haven’t made it to Galois theory yet, so I can’t comment on that part of the book, but I’ve found undergrad Hungerford (Abstract Algebra:An Introduction) to generally be easier to read than Dummit and Foote. It’s also pretty cheap (you can find an used older edition for less than 10 US dollars)

[ripe crest]

It's very natural to consider commutative monoids as 'Z+'-"modules", yes?

[dim widget]

Not modules but yes there are maps [n] sending any x to x^n for a commutative monoid

[twilit wraith]

Hello, today is my first day of abstract algebra and I am asked on my homework to show that there is a unique group of order 4 such that no elements of the group have order 4.

im confident ive found the group, which is <a,b : a^2 = b^2 = e>, but i am wondering if there is a more elegant way than checking the cases of |a| WLOG

the simple way that ive identified is to note that |a| cannot exceed 4 were there to be only 4 elements in the group, and from there to just check |a| = 1 and |a| = 3

[ripe crest]

Yes, I don't know the name of that structure, that's why I just went with "module"

Do we need Z+, or does this work with N as well...

It works with N ....that makes sense...

Given we have abelian groups as Z-modules

[next obsidian]

Yes there is but not based on day 1 of abstract algebra

Once you develop some basic theory you can say |a| has to divide |G| so it gives you 1,2,4

[twilit wraith]

i see

do i really need to check cases then

[next obsidian]

I mean

Yeah

[twilit wraith]

damn

i guess it cant be too bad

[next obsidian]

It’s easy

I mean it’s a 4 element group, the multiplication table has what, 64 choices or something

[twilit wraith]

i know that |a| < 4 as otherwise we would have at least a^0 through a^4 being in the group

[next obsidian]

But once you say one element is the identity, even less

27?

[twilit wraith]

maybe

well if |a| = 1 that just means a = e

[next obsidian]

My point is that 4 is such a small number I think point is for you to just get used to doing cases and bashing some stuff out

Get used to it, finite group theory has a lot of cases

[twilit wraith]

fair enough

[next obsidian]

Based on orders of things

[twilit wraith]

im just worried im gonna forget a case

i know what would happen if i had 4 "distinct" elements a through d

they wouldnt actually be distinct

if i had 3 then i would still end up with one of them being the identity

[next obsidian]

Idk just, don’t forget a case

[twilit wraith]

yeah

[next obsidian]

I think the point of this is for you to just flounder

And try some shit and discover ways to reduce the work you gotta do

So you should just grab some paper and start writing

And once you are convinced you handled all cases, maybe try to edit and see if you can reduce that work down to some smaller subset

Until you’re happy

[twilit wraith]

sounds good

[twilit wraith]

just to check though, am i right that, for all generators of a group, their orders add to the order of the group?

[next obsidian]

Idk what that sentence means

[twilit wraith]

i might just be wrong then

i was thinking that if i have 3 different generators, id have to have a situation where one of them is order 2 and the others are order 1

[next obsidian]

Idk what you mean 3 different generators

Like 3 elements which collectively generate?

Or 3 elements which each are a generator

[twilit wraith]

im not knowledgeable enough to know how this is different

ig as an example

i think the dihedral group has 2 generators right

[next obsidian]

it is generated by 2 elements

[twilit wraith]

i see

[next obsidian]

But I wouldn’t say it has a generator

Cuz that means one element gets you everything

[twilit wraith]

ohh i see the confusion

ok then suppose a group of order n is generated by some number of elements

then the order of those elements must sum to n

[next obsidian]

And I mean based off what I think you’re trying to say, this just tells you that’s not true right?

[twilit wraith]

oh right

[next obsidian]

2 + n ≠ 2n often times

[twilit wraith]

true

[next obsidian]

And besides your proposal is also nonsense

Because if it was true for some set of generators {x_i}

You can just throw in an extra element, change the sum, and that still generates

At the very least you’d need to say a minimal set of generators or something

[twilit wraith]

oh ur totally right

man i having some getting used to to do

[next obsidian]

I just say it’s total nonsense so you can start to realize how you can gut check an jdea

[twilit wraith]

yeah i know, im not offended at the proposal that i said utter nonsense

especially if its true that i did lol

[next obsidian]

It wasn’t a bad thought, but if you realize “wait I can add in more stuff” you’ll immediately realize what you’re suggesting is kapoot

[twilit wraith]

cant afford to have a progress-preventing ego in math in my opinion

plus its ok to be really wrong sometimes

[next obsidian]

Yeh

Here’s NL as a xenomorph

I’m taking a shower

[twilit wraith]

lol

[tardy hedge]

Wat

Who’s NL

[next obsidian]

tm…

[tardy hedge]

I dunno

[spice wagon]

😥

[south patrol]

No

[stone elbow]

Hi all, I need intuition for why ideals of a PID are free modules. I get that they are b/c they’re generated by a single linearly independent element but then it has to deeply affect how we think about quotient rings. Again, I can see logically that such a quotient is either 0, A, or a field, and in the latter case it’s just a torsion module so a version of the rank nullity theorem is preserved. But I just don’t know how to grok these ideas so that I’m completely at home with them

[next obsidian]

You’ve like, described the entire thing

Idk what else you want to say

[stone elbow]

E.g. if an ideal is completely free as a module, it means that the residue field is determined by a dimension 1 free module automorphism - which is super unusual

[velvet hull]

it is exactly because the ideals are principal. and a single element is free, with respect to itself (when viewed as a generator for a module)

[stone elbow]

It’s like, my rational brain gets it, but my monkey brain insists that ideals must be “smaller”

[velvet hull]

how so?

[stone elbow]

Because I always thought of residue fields as information “missing” from the ideal

But it looks like the correct way of thinking is that it’s determined by the ideal’s “position” within the ring

So the internal structure of an ideal really just amounts to its dimension, I guess?

It’s like my intuitive mental picture of what an ideal even is turned out to be wrong and I’m struggling to form a new one

[velvet hull]

the quotient of PIDs can be more than just 0, A, or a field, if that's what you meant

[stone elbow]

Right, sorry I assumed that the ideal is prime for some reason

But even for primes, if all non-zero primes are the same as modules, and their embedding amounts to specifying the residue field plus some extra bit of data, what would that data be?

[velvet hull]

when you consider ideals as modules you forget the internal multiplication

so the inutition for ideals, when considered as objects for taking quotient rings, do not carry over 100% to modules

especially since when you include the multiplication it fails to be a free algebra, I think

[stone elbow]

This makes sense

[velvet hull]

so there is more data that is missing here

[next obsidian]

It isn’t even a ring

[velvet hull]

also that lmao

[stone elbow]

Another unrelated question: what are the applications of an associated Lie algebra of a filtered group? Is it a useful viewpoint on finite nilpotent groups?

[delicate orchid]

hell of a jump to go from quotients of PIDs to (what I've gathered from a shit ton of googling) a highly technical niche modular representation construction

I'm gonna say "yes, it is useful" though purely because F_p[G] for G nilpotent has most of it's structure focused on the modular reps of the p-Sylow of G

[thorn jay]

parkour

[keen badge]

Can't I take then λ to be the identity, and μ to send x to x^2?

[keen badge]

yeah lol

[jade mason]

mu isnt an isomorphism then

[keen badge]

Can you think of some relation(s) of the fields, to conclude when the extensions are isomorphic?

other than definition?

For example, we know that F(x)=F(α) for some transcendental α in K, and am trying to prove(if that's even correct) that i:F->F(x) isomorphic to i':F->F(α)

[rocky cloak]

So a good example might be L equal to the real numbers.

There are no nontrivial automorphisms of the real numbers, so there would be no automorphism taking Q(pi) to Q(e) or anywhere else for that matter.

As for relations, if you have extension
L/K/F such that L/F is algebraic and normal, then any embedding K/F into L/F is conjugate by an automorphism of L/F. So here your conjecture would basically hold.

Then if you take F to be a prime field, then automorphism of L/F is the same as automorphism of L, so then it would hold on the nose.

[keen badge]

but this is not connected to what I said with F(x)=F(α) right?
Because my question is when the embedded fields are equal

[rocky cloak]

It's just a more convenient shift in perspective.

If L and L' are isomorphic. And you have embeddings
i: K -> L and i': K -> L'
then you can just compose i' with an isomorphism L' -> L.

And now the question is about two embeddings of K into L

[keen badge]

ohh ok

[rocky cloak]

So you're question here is if the canonical isomorphism F(x) -> F(a) takes F to F?

That should be clear from the definition, since it's defined to take F to F and x to a

[keen badge]

thank you!

I am trying to read Ian Stewert's "Galois Theory" book(I think it's the 4th edition) and he tries to classify field extenstions via this definition, so instead of saying when fields are isomorphic, in this chapter, he focuses a BIT more about when field extensions are isomorphic

so he tried to show that the extenstions F(α)/F and F(x)/F are isomorphic via isomorphism between F(x) and F(α) but no where did he explain why the identity is enough, so thank you(I do skim the book, and not trying to read it all, I just read the theorems and stuff, and not the "words" so maybe I did skip it somewhere)

[mortal spindle]

Not sure where to out this. If I want to compute (x+y+z)^n mod (xz-y) is there an easy way to do this?

I would apply binomial theorem on y and x+z but things get messy

[velvet hull]

well it's equal to (x+xz+z)^n

and that's about the best you can do

[mortal spindle]

😠

[rocky cloak]

What does "compute" even mean here? Just listing the coefficients?

[mortal spindle]

Yes

This works fine

I just wish it wasnt messy by hand

Like multinomial theorem is just going to look messy?

[rocky cloak]

Well, seems mostly annoying and pointless to me, but good luck

[mortal spindle]

Trying to relate it to generalizing pascal triangle

Instead of sum of directly 2 above what if im looking at sum of directly 3 above or n above

In case of 3 each coefficient choice can be made into a path on the tree of the triangle similar to pascals triangle

So you have x being left y being middle and z being right

So its the same as looking at coefficients of (x+y+z)^n mod (xz-y^2)

I believe this logic checks out for verifying the coefficients of the triangle and generalizing just means the ideal is going to get more complicated

[keen badge]

If K subset L are fields, then a monomorphism, i:K->L is a field homomorphism such that i(K)=K(equal, not isomorphic)?

[dim widget]

that would be a homomorphism of K-algebras vs a field homomorphism

you have to decide from context what is meant by that notation

but if no context was given I would assume that i:K \to L can have any image not necessarily within the copy of K you have already singled out

[keen badge]

so you would assume that a monomorphism is just an injective ring homomorphism between the fields?

[dim widget]

absent any other information that's the definition

it depends on what category you're working in

or what objects you are considering

[south patrol]

Well maps of fields are automatically injective and monomorphisms (in the category of fields) so there's a considerable amount of redundancy here

[dim widget]

yes, which as potato mentions is any homomorphism between fields

[keen badge]

It's not related to this, but is there a category of all categories?

[dim widget]

yes

[keen badge]

lol

That's funny as hell.

someone just said "Let me solve Russell's paradox"

[dim widget]

a monomorphism i: x \to y is supposed to satisfy that if f: z \to x, g: z \to x then i \circ g = i \circ f iff f = g

it's pretty easy to check that when your category is sets with some algebraic structures (like rings, fields, etc.) this is implied by (and in fact equivalent to) being an injective map

[keen badge]

yes, so if we are over fields, then are you saying it is enought that i(K) isomorphic to K which happens trivially

[keen badge]

I just don't see why that is true

[dim widget]

and for fields any map is injective

[keen badge]

and thus can't understand how there might be field monomorphisms i:K->L where K is not a LITERAL subset of L(and not isomorphic to a subset)

[dim widget]

well they won't exist unless there is an abstract field F within L such that F \cong K

by definition

[keen badge]

that's what I am saying

[dim widget]

but if you fix ahead of time your F it may be that i(K) \neq F

[south patrol]

But with the caveat that we are using category in two different ways here, lol

[keen badge]

I don't understand why IF there is an abstract field F within L s.t. F\cong K then there exists a monomorphism

There will be an embedding, yeah, but a monomorphism, I am not sure

[dim widget]

uh

[south patrol]

What is the difference lol

[dim widget]

well I just explained to you that injective maps are monomorphisms

[keen badge]

I am cooked, aren't I

[dim widget]

and you didn't seem upset about that claim

[keen badge]

if you are talking about this

then I don't understand why it is true for injective maps

[rocky cloak]

You're not convinced that embeddings are monomorphisms?

[keen badge]

because we want the composition to be EQUAL and not isomorphic

[glad osprey]

What is your definition of embedding?

[keen badge]

no.

[rocky cloak]

I'm curious how you're defining them without it being obvious

[dim widget]

okay that's a reasonable subtlety to bring up, but if i: S \to T is an injective map of sets, and f, g: X \to S, then if i \circ f = i \circ g then literally for every x \in X f(x) = g(x)

so there is not so much wiggle room here

[keen badge]

Given two fields F,L. An embedding i:F->L is a ring homomorphism.

[keen badge]

And a monomorphism is this

They are the definitions that I work with

Maybe I am wrong?

[knotty badger]

There's a general result that faithful functors reflect monos and epis

the forgetful functor from fields to Set is faithful

[rocky cloak]

So what you're unsure about is why an embedding must be injective?

[knotty badger]

so an injective function is always a monomorphism

[keen badge]

no.
Why an embedding must be monomorphism

[knotty badger]

and a surjective function is always an epimorphism

[keen badge]

this is true because ker(i) is an ideal

[mortal spindle]

Can you do something with the iff part of the statement

[keen badge]

By this definition:

a monomorphism i: x \to y is supposed to satisfy that if f: z \to x, g: z \to x then i \circ g = i \circ f iff f = g
Why an embedding is a monomorphism?

[mortal spindle]

Like algebraically

[keen badge]

what do you mean

[knotty badger]

It's because embeddings are injective

[rocky cloak]

Say
f,g: X -> F
are two different functions.

Then there is an x such that f(x) does not equal g(x).

Then i(f(x)) =/= i(g(x)) because i is injective

[knotty badger]

so they're a monomorphism in Set

and then faithful functors reflect monos

so the original morphism is also a monomorphism

[keen badge]

What you said abt faithhful functors I dont know what that is.
I only know the definition of a functor and category and that's that practically.
I have learned abt functors for an hour(in 3 different classes)

[keen badge]

yes, ok

[knotty badger]

ah so a faithful functor is one whose action on homsets is injective

essentially if Uf = Ug then you must have f = g

[keen badge]

so that means that an embedding is not necessarily a monomorphhism

[knotty badger]

(up to considerations about domains/codomains)

[rocky cloak]

It's just a very fancy way of saying "a homomorphism is in particular a function"

[keen badge]

ohhhh it's like a group that acts properly discountinuous on a set

[knotty badger]

um idk what that means unfortunately

[keen badge]

exactly that, lol

[knotty badger]

the proper definition is that if Uf = Ug and dom f = dom g and cod f = cod g, then f = g

when U is a faithful functor

you can verify that the forgetful functor from Fields to Set is a faithful functor

[keen badge]

But never mind that now

[knotty badger]

indeed all forgetful functors you'll meet are

[keen badge]

ok

[dim widget]

I don't think all of this stuff about category theory is important

[keen badge]

so why doesn't it mean that an embedding is a mono IFF the smaller field is a literal subset of the bigger one

[knotty badger]

i only mentioned it because they were bringing up monomorphisms etc

[dim widget]

one should just verify by hand that i: R \to S being a monomorphism of rings is implied by it being a monomorphism of sets

then win

[rocky cloak]

Because you can have embeddings that are not literally subsets...

[knotty badger]

imo it's easy to verify that in this example but not necessarily illuminating, hence why i prefer the "faithful functors reflect monos" argument

but i do have the brainworms so

[keen badge]

Let's look at an example?
F->F[x]/(f(x)) where f(x) is an irreducible polynomial over F

(I am ok with some other examples, if you prefer)

[dim widget]

let's look at an example

Q(i)

[knotty badger]

(btw by "faithful functors reflect monos" i mean that if U is faithful and Uf is a mono then f is a mono)

[keen badge]

ok, with what other field?

[dim widget]

u: Q(i) \to Q(i) is an embedding by sending u(i) = -i

so you see even though Q(i) is abstractly a subset of Q(i) the image of u is not the inclusion of that subset

new example

[rocky cloak]

Or Q(i) to Q[x]/(x^2 + 1)

[dim widget]

Q(cuberoot(2), \zeta_3)

[keen badge]

so that means that u is NOT a monomorphism, although it is an embedding, and that one is a literal subset of the other?

[dim widget]

Q(cuberoot(2)) embeds into this in three different ways that have very different images

[dim widget]

no

[rocky cloak]

Why would it not be a monomorphism?

[keen badge]

Here! This example

[glad osprey]

Embeddings are injective homomorphisms which are equivalent to monomorphisms in the category of fields, so they are literally the same

[keen badge]

Q(i)->Q[x]/(x^2+1) defined by (a+bi)-> ?

[dim widget]

a + bx

or a - bx

[keen badge]

so a+bi->[a+bx]

ok

why is that a monomorphhism

[dim widget]

because it is an injective map of sets

as I have now said like 10 times

[keen badge]

...

By definition why that is

[mortal spindle]

I think they want to be walkthrough with the definition

[rocky cloak]

.

[keen badge]

okk

I think I understand now

[rocky cloak]

f =/= g implies if =/= ig, which is the definition of monomorphism

[keen badge]

yes ok

[mortal spindle]

I guess its not going to show its a monomorphism if you give explicit maps into Q(i) since you need to show its true for every map

So thats why you are getting this explanation I guess.

[dim widget]

the argument given applies not just for every field map K \to F but for every map of sets S \to F

[mortal spindle]

Yeah thats the point

But i dont know if this was explained properly

you did say it

[keen badge]

so if I have categories C,D where C is a sub-category of D, then a monomorphism of C is a monomorphism of D, that is what you are saying(at least when C=SETS and D=FIELDS)

[mortal spindle]

But i dont know if you said the last part + info about concreteness

[dim widget]

Sets are not a subcategory of fields

[keen badge]

damn

[dim widget]

or at least not in any natural way

[mortal spindle]

Speaking from personal experience its something that needed to be reiterated for me since its unfamiliar for sure

[keen badge]

At least I understand now why injective = monomorphism in FIELDS

[dim widget]

but I'm saying that if you have a category C where the objects are "sets with extra structure" and the morphisms are "morphisms of sets which have some extra properties"

[mortal spindle]

Well look at this

[dim widget]

then if you can check that f \neq g iff i \circ f \neq i \circ g for every map of sets

[mortal spindle]

#groups-rings-fields message

[knotty badger]

(faithful functors reflect monos)

[mortal spindle]

Thats pretty much it

[dim widget]

then you have checked it for every map in the category C

[mortal spindle]

And psued here is just telling you the more general reason

[knotty badger]

mhm

[keen badge]

What is the definition of reflect here? that there exists a bijection between all the faithful functors and monos?

[knotty badger]

it means that if U is a faithful functor and Uf is a mono, then f is a mono

[keen badge]

ohh ok

that is obvious

Ok, thank you all

[mortal spindle]

Let it simmer

[keen badge]

(I did like it tho that I made 4 people write at the same time)

[knotty badger]

you should've seen my category theory hot takes back in the day

[keen badge]

If y'all dont mind sharing, what is your knowledge and (approximate) age?
Like... 20's/30's/40's/50's and undergrads/grads/doctors/profs?

I just want to understand who are the people in this community

[slim kayak]

Zoomers mostly

[knotty badger]

it's in my bio~

[glad osprey]

Tbh I'm not sure how important it is to understand monomorphisms for field theory, or even most other parts of algebra. Do you even need to think about monomorphisms until you start using non-trivial category theory?

[mortal spindle]

Also what I was looking for is here already

https://en.m.wikipedia.org/wiki/Trinomial_triangle

[mortal spindle]

No but its definitely a basic concept

So Id say its important if you continue

[glad osprey]

You can just use injectivity, and for field theory in particular you don't even need to care about that because everything is injective

[keen badge]

As I said here a few hours ago, I am trying to understand isomorphism between field extensions, and in the book it says that we can think of an extension as a monomorphism i:K->L.
And by the definitions I am familiar with, and by the view I had before this convo, I thought(and in this particular context, still is) that it means that K literal subset of L

[knotty badger]

so, subset inclusions are monomorphisms, but not all monomorphisms are subset inclusions

[glad osprey]

I see you can just mentally replace monomorphism with injection in this case imo

[knotty badger]

however every monomorphism differs from a subset inclusion by an isomorphism

cause of the isomorphism theorem

[keen badge]

And because if
i:K->L
j:K->tilde(L)
Are field extensions, then the extensions are isomorphic iff there exists an isomorphism mu:L->tilde(L) with mu(K)=K

[dim widget]

i don't think that's really how it works

[knotty badger]

wait did i make a mistake

[dim widget]

there's no measurable difference between the inclusion of subset and a random monomorphism

it's just changing labels

[knotty badger]

isn't that what i said here

[dim widget]

from the categorical perspective

no it's similar to what you said but not the same

[knotty badger]

i don't quite understand what you're saying then

[dim widget]

there is no canonical way of looking at a category and saying "this is a subset inclusion"

[knotty badger]

i mean if you know your category is the category of fields then there is right

[dim widget]

that's fair

but i don't think from the categorical viewpoint it's at all a useful thing to think about.

[knotty badger]

sure but i'm not married to the categorical viewpoint or anything

i'm happy and willing to use other viewpoints

[slim kayak]

Sometimes it helps to scrape off unnecessary context

[dim widget]

I don't even think from the perspective of field theory it's useful to split hairs about the difference between i: K \to L being a monomorphism vs. K already being thought of as a subfield of L

[knotty badger]

idk i haven't done much field theory

[mortal spindle]

So the (x+y+z)^n mod (xz-y^2) is satisfied by (1+x+x^2)^n

[knotty badger]

though for what it's worth treating isomorphic things as equal usually led to confusion for me when learning cat theory

[mortal spindle]

Is there a quick way to recognize the two

Like what if the ideal was xz-y^3+x

[dim widget]

but labeling the inclusion i: K \to L is exactly not thinking of isomorphic things as equal

[mortal spindle]

Would it be possible to reduce the unknowns?

Or is this just algebra strictly

[knotty badger]

what i mean is not distinguishing between a general mono and a subset inclusion

[dim widget]

but that's a different issue

[mortal spindle]

The way I see it is probably common guesses for what works lol

[keen badge]

Me sleep now
Goodbye good people

[dim widget]

this is not a sentence that makes sense

that's why people aren't responding

[knotty badger]

now im more confused..

[mortal spindle]

Yeah I dont know the right vocabulary but to make it make sense making the substitutions for 1,x,x^2 help in simplifying and calculating the coefficients of the original polynomial

Are there some heuristic methods in reducing the number of unknowns similar to what is shown above

[dim widget]

I don't really get what you're trying to accomplish

[mortal spindle]

Before I continue do you understand what I said above?

[dim widget]

to be honest I am struggling even with the english

[mortal spindle]

Can you let me know what part specifically?

[dim widget]

why don't you just try to state what problem you are trying to solve as clearly as you can

and why you have gotten to the point of trying to calculate various coefficients of a 3 variable polynomial modulo some random ideals

[mortal spindle]

Previously I stated how I reduced finding coefficients of the polynomial modulo some random ideals corresponds to counting number of paths for the trinomial triangle( its like pascals triangle but sum of 3 above)
I am generalizing. But before that I want you to understand what I said above as to not waste both our times.

Im scared of putting in effort to be misunderstood so Im trying to be incremental. This is only way I see respectful of our times

[glad osprey]

I feel like sometimes in field theory you want to distinguish between set inclusion and an embedding. For eaxmple Q[x]/(x^3 - 2) is isomorphic to Q(cbrt(2)), but the latter has a canonical inclusion in C, while the former has 3 embeddings in C none of which is more "canonical" than other. This is not a category theoretic thing tho, more a pedagogical one I guess. I remember being confused by this when I first learned it

[mortal spindle]

The generalization isnt just multinomial triangle, but multinomial on different boundary polygons

[knotty badger]

This is part of why I prefer to have different levels of equality I can work with

[dim widget]

I would argue that you shouldn't do anything with fields that is sensitive to how you label them as sets

[mortal spindle]

I cant find much on this but there is some info about multinomial triangle out there

[knotty badger]

Rather than being forced to work with the weakest one

[dim widget]

exactly for this reason that it just causes confusion

[mortal spindle]

The reason anyone might care is because its useful for partitioning algorithms in CS

[knotty badger]

I actually think limiting yourself also causes a lot of confusion though

[dim widget]

what do you mean number of paths for the trinomial triangle? which paths?

[mortal spindle]

Similar to how binomial triangle has a bunch of uses

A graphic might be helpful

[dim widget]

that's very interesting

[knotty badger]

It’s how it happened for me, at least

Once I stopped worshipping the principle of equivalence, a lot of previous confusions resolved themselves

[mortal spindle]

https://math.stackexchange.com/questions/2834476/the-relationship-between-the-pascals-triangle-sequence-and-the-binomial-number

This is the case for pascal/binomial triangle

You can do same for trinomial triangle and so on

And different boundary polygons not just triangles

[dim widget]

why don't you just explain what you're trying to count?

[mortal spindle]

And its important for algorithms in CS vaguely

[dim widget]

is it just the number of paths in this tree up until a certain depth?

or are you meaningfully using that this comes from the trinomial triangle etc.

[mortal spindle]

Its used for trinomial triangle too yea

[dim widget]

no I mean

please explain concretely what you are trying to count

[mortal spindle]

The number of paths to leaves in the tree

So if you look at middle leaf

You can count number of paths

And that corresponds to the coeff

[dim widget]

I see

[mortal spindle]

True for any d weighted tree and this helps with modeling unrooted trees in the polygonal boundary case

But i feel like im far from doing polygonal boundary so im just working with multinomial triangles

[dim widget]

okay but then doesn't exactly the same argument tell you that you are looking for the coefficients of (1 + x + ... + x^{k-1})^n?

for the k-ary generalization of pascal's triangle

[mortal spindle]

Well yes but that isnt entirely obvious to me how you get there because for pascals triangle you really just have (x+y)^n where x,y correspond to traversing left and right from the root

So the most intuitive case for d weighted trees is the sum of d unknowns to the nth power so (x_1+x_2+…+x_d)^n mod(equivalent relations)

The k-ary generalization you mention satisfies those equivalent relations

Which for example would correspond to how going left and then right on the trinomial tree is the same as going middle twice

So xz=y^2 where x is left,z is right and y is middle

Now im asking a general question about if people know methods of reducing unknowns so that I have something to read in case of polygonal boundary

Because relations could be more complicated

And sorry I know it sounds unclear but thats because I dont have good vocabulary to concisely describe this that well

[dim widget]

I think I get it now

I'm just thinking about the problem for a little bit

and how to respond

[mortal spindle]

🙏🏽 thanks

Also you dont have to read this immediately but to explain different boundary polygons this could be as simple the truncated pascals where your right side is all 0s instead of 1s

[dim widget]

okay for your specific problem: we're looking at these graphs which start with 1 node, then that gives k nodes, then the first of those k nodes connect with k nodes below them and every node shares k-1 nodes with its neighbor right?

this is with the usual boundary conditions (i.e. 1's on the sides of the graph)

[dim widget]

If you're trying to count paths in this graph you should prove by induction that, labeling the nodes in the obvious way, the number of paths from the root to the leaf labeled by x^j on row n + 1 is exactly the coefficient of x^j in (1 + \dots + x^{k-1})^n

and the proof is exactly the same as for the pascal's triangle graph

[dim widget]

from this perspective what's going on is that you can always set x_1 = 1 without loss of generality as long as the unknowns and relations are homogenous (which I think for trees they more or less have to be if I understand you correctly)

then x_1 = 1, x_1x_3 = x_2^2 forces you to get (1 + x + x^2)^n

[mortal spindle]

Ok you pretty much understand it perfectly. The next step forward for me is to play with the different boundaries/polygons and see if the relationships and variables being homogenous thing stays true

But its sure to always be something acyclic for sure

But thats most the context I guess

I was really just asking a tangential question about reducing the unknowns of ideals in general for computing purposes

[dim widget]

yeah well one general takeaway is if you are counting coefficients of a polynomial in n variables which is homogenous, that's the same as counting coefficients of a polynomial in n-1 variables that is not (or at least not necessarily homogenous)

[mortal spindle]

I didnt know this

[stone elbow]

I blame Serre for putting it in Chapter 2 of his Lie algebras book 🙂

The first 2 chapters are like a grand tour of every single conceivable way to arrive at a Lie algebra known to man

[chilly ocean]

ㅡㅡㅡㅡㅡ
I think it doesn't need to be Neotherian ...why it shud be a neotherian

[dim widget]

Otherwise what if you have $k[x_1, \dots, x_n, \dots]/(x_i^2 = x_{i-1})$ localized at $(x_1, \dots, x_n, \dots)$?

[velvet hull]

Although every element to some power is eventually in the ideal, theres no guarantee that the maximum power among all elements is bounded

[dim widget]

the radical is $(x_1, x_2, \dots)$ but no power of the radical is contained within the ideal $(x_1) $

[cloud walrus]

Topos_Theory_E-Girl

Topos_Theory_E-Girl

[velvet hull]

Otherwise for a very rudimentary construction, consider <x1, x2^2, x3^3, x4^4, …> in k[x1,x2,…]

[chilly ocean]

radical^'gcd of orders of every elements of A in A/a' is in a

[velvet hull]

Sorry, what? That’s hard to understand

Are you saying it’s true in A/sqrt(I)? because that doesn’t sound right either

Because the nilpotent ideal is also not guaranteed to have an n such that Nil(A)^n = 0, which sounds like is what you’re describing

[chilly ocean]

Why not@velvet hull

[velvet hull]

But it’s the exact same answer here, just because every inidivual nilpotent element has finite index, doesn’t mean that the index of every nilpotent element is bounded

[rocky cloak]

I'm guessing you mean lcm, not gcd.

In which case the lcm of an infinite set can be infinity.

The bound is also not correct on the dot anyway, since the power of an ideal can contain products of different elements

[quartz wind]

is this correct?

[velvet hull]

presumably you meant to type M/S. but yeah it's just an application of 1st iso

[quartz wind]

ah yeah my mistake. thanks

[south patrol]

I wouldn't say 1st iso is needed

Just like it is immediate, as you have written

[thorn jay]

1st isomorphism theorem would've been useful if you had a general surjection

[south patrol]

But then you could just write the same proof with that surjection replacing the quotient map aha

as in f(a) = sum r f(b_i)

[thorn jay]

You still could use it if you rreaaallyyy wanted to

[spice wagon]

if we want to say that A an integral ring has a fraction field K, we denote the fraction field as the pair (K,i) with i the canonical injection or only the field K?

[south patrol]

Usually just K

[velvet hull]

I write Q(A)

or A\{0}^-1 A if youre a psychopath

[south patrol]

Frac A is common

[dim widget]

I would write Frac A

[spice wagon]

so Q is the field of fractions of the ring Z unique up to an isomorphism?

yeah i saw the notation frac A

[dim widget]

well that's obviously true

[velvet hull]

it's unique up to unique isomorphism (that admits the base domain)

[south patrol]

As is any field of fractions

[dim widget]

but in general by construction the fraction field comes with a canonical map from A \to Frac A

[spice wagon]

yes but i wanted to know if we write the canonical injection

[dim widget]

no you don't write it

[spice wagon]

yeah

[south patrol]

Usually no, it is implied

By perhaps slight abuse but it's standard

[dim widget]

but I guess if something you were doing was really sensitive to it you could write it

[south patrol]

I mean also technically like

You can either fix a certain construction of the fraction field or have it as something defined up to unique iso

But the same is true of, say, products of things

It is common to omit the "extra data" from notation

[spice wagon]

yeah i see

[south patrol]

But that is slightly fancier

[velvet hull]

the canonical map is universal in the sense that it factors through any injective embedding of the int domain into any field

in that sense it closely resembles the universal property of the quotient map, which factors through any map that contains the normal subobject in its kernel

(category theory nonsense ^)

or, in human words, the field of fractions is special because it is the unique smallest field that retains all the information from the original ring

and if you embed that ring into any bigger field it will contain a copy of that field inside the bigger field

[spice wagon]

we can prove it’s the smallest ?

[dim widget]

category theory foucault be like why do limits resemble limits resemble limits

[velvet hull]

you just think about the construction

for every d in D, d has to be invertible in K. so K has to contain 1/d as well

but then that's exactly the field of fractions

[spice wagon]

ill write that lol thx

and another question

if we have a ring A, K a commutative field and f : K ->A, when we write ker(f), do we mean the set of x such that f(x)=1 or f(x)=0?

the neutral for addition or the multiplication

[velvet hull]

0, always

[dim widget]

all fields are commutative and all such ker(f) are zero

[velvet hull]

0 is much stickier than 1

is the way I think about it

the point of a kernel is that it's supposed to be a black hole that eats up anything that touches it

but then that's what 0 is

0*x = 0

equally, I * x is a subset of I

[spice wagon]

in my course it is stated that if you have a non-commutative field, it’s called a division algebra

[south patrol]

Commutative field

[dim widget]

well ker(f) would be zero for a division algebra too

[south patrol]

Tell me you're reading French without telling me

Jk