groups-rings-fields

[crystal turtle]

a poset category is (generally) not concrete tho lol

An easier example, rather than a monomorphism that isn't injective, is to find an epimorphism (right cancellable morphism) that isn't surjective. For example, the inclusion of Z into Q is right cancellable in the category of commutative rings with identity, but clearly isn't surjective

[warm wyvern]

this is the only example of a non-surjective epimorphism that I know as well

https://cdn.discordapp.com/emojis/1090593114504314910.webp?size=48&name=anyapat&quality=lossless

oh I got my very active back

nice

no longer will I be bullied by whitenames

[rocky cloak]

Consider the poset category of subspaces of any space. Every morphism is an epimorphism but only the identity is surjective

[crystal turtle]

That's not concrete

[warm wyvern]

that's fine

[rocky cloak]

It has a faithful functor to Set. Or what do you mean by concrete?

[glossy crag]

Really? Is this just some set-theoretical nonsense and hence all rings of some cardinality are isomorphic or something?

[unkempt stream]

bimorphism not necessarily isomorphism jumpscare

[warm wyvern]

wtf does that mean

[unkempt stream]

chunky category

[rocky cloak]

All algebraically closed fields with the same transcendence degree over it's prime field at least

[crystal turtle]

Okay actually you may be right. We can map every object to the singleton lol

Huh, not typically what I think of as concrete

[unkempt stream]

epimorphism & monomorphism not necessarily isomorphism

[crystal turtle]

non-topos moment

[warm wyvern]

oh

[rocky cloak]

You can just map every object to itself...

[crystal turtle]

or that

[rocky cloak]

Aka its underlying set

[glossy crag]

Huh, I need to learn the basics of transcendence degrees, all I know is Luroth's theorem.

[rocky cloak]

How is that not what a forgetful functor should be

[crystal turtle]

no you're right

[unkempt stream]

i avoid them at all costs

[crystal turtle]

I just never think of poset categories as concrete lmao

[glossy crag]

It does feel kind of nebulous, e.g. like cardinalities, but it leads to some interesting statements.

[rocky cloak]

For uncountable field, its the same as cardinality, so you don't need to think about it

(over it's prime field)

[glossy crag]

G*d, I hate cardinalities, maybe that's just my mathematical immaturity speaking

[rocky cloak]

I love that you census god

[glossy crag]

Tetragrammaton, baby

[rocky cloak]

... But not cardinality

[glossy crag]

Although in this case it's more of a triagrammaton

[warm wyvern]

G*d

https://cdn.discordapp.com/emojis/1127339423764328458.webp?size=48&name=clopencry&quality=lossless

[unkempt stream]

SAME

[formal ermine]

why?

[rocky cloak]

Model theory nonsense

[long nebula]

Model theory 🥰

[glossy crag]

And what exactly is the transcendence degree of C over Q?

[rocky cloak]

It is the cardinality of C

Just like any other uncountable excursions of Q

(the cardinality equals the transcendence degree)

[white oxide]

hm my book defined monomorphisms as a homomorphism which is injective as a map fo sets

[rocky cloak]

Your book = bad

[crystal turtle]

I'd say in probably even most categories of (common) algebraic structures that should be true, but that's not the categorical definition

[white oxide]

but hungerford good :(

[rocky cloak]

In any category with a free functor, it's true.

That's a categorical definition at least

[crystal turtle]

I meant mono iff injective

[rocky cloak]

Right, yeah having a free functor only means mono implies injective

Don't know what would be a good characterisation of the converse

[crystal turtle]

Yeah I'm not aware of any characterization like that

[warm wyvern]

that's not the typical definition lel

[molten viper]

(what's the typical definition)

[crystal turtle]

this

[molten viper]

hmmm

[elder wave]

eh i mean

[unkempt stream]

timo!

[molten viper]

Is that equivalent to being injective?

[unkempt stream]

in SET yes

[elder wave]

intro textbooks on both lin alg and AA still define monos as injective homs

[molten viper]

(I really know nothing about category theory yeet)

[elder wave]

sure it's not accurate in the more general sense when introducing categorical language

[molten viper]

But I guess when talking about groups and rings, it's equivalent to being an injective homomorphism

[elder wave]

No

[molten viper]

!

Tell me more

[crystal turtle]

Okay but if they're not going to use it right, then why not just say "injective homomorphism"?

[elder wave]

because monomorphism is shorter and equivalent for some stuff

[molten viper]

Well

[unkempt stream]

less words :troll:

[crystal turtle]

Checkmate textbook writers

[molten viper]

I guess in the context of rings and groups it's equivalent to saying fg = fh => g = h cause that's the definition lel

[warm wyvern]

it is in the category of sets

not in all categories tho, allegedly

[molten viper]

what's it equivalent to in the categories of rings and modules and groups and algebras etc

[unkempt stream]

timo, my man

[elder wave]

it is equivalent

in ring

i was smoking sorry it's late

[warm wyvern]

yea

[elder wave]

did an oopsie with epis

[warm wyvern]

we were talked about this just a bit ago

[crystal turtle]

Ah, I guess it is equivalent in these categories since they have a free functor

[unkempt stream]

does it have to do with being locally small or smth?

[warm wyvern]

they're also concrete categories over Set

you told me this, no?

[unkempt stream]

i.e non-class “sized” homies

[molten viper]

I don't know what a free functor is but like idk if I want that explained to me rn

[warm wyvern]

dw about it lel

[crystal turtle]

https://math.stackexchange.com/a/1644040, having a free -| forgetful adjunction is sufficient apparently

[warm wyvern]

monomorphisms are stupid tbh

the definition is cute

it's cool that it's equivalent to injective most of the time

[unkempt stream]

it’s adjoint to free, essentially like how the forgetful “forgets stuff”, the free functor constructs the most general way to add structure to it

[elder wave]

hello

[warm wyvern]

but actively classifiying monomoprhisms in different categories is just blegh

[unkempt stream]

it is like the “simplest” way to add structure to something

that’s how I view it

[molten viper]

so like, if I give it a set a free functor will give me some more complex structure

[unkempt stream]

yes, sorta

[molten viper]

I'll worry about this in grad school lel

[unkempt stream]

as in it is almost the “most general” way to do so

[elder wave]

you've seen the construction of tensor products right

[molten viper]

I have not

[unkempt stream]

that’s ajoint to hom i thought?

[elder wave]

?

[crystal turtle]

Right, but it's a useful nothing sometimes. Like what's an "injective" homotopy class of maps in hTop otherwise?

[elder wave]

this is not where i was going with it

[molten viper]

what does adjoint mean

[unkempt stream]

tensor algebra I know is an adjoint to a free

[elder wave]

i was thinking of an example of a free construction

[crystal turtle]

Don't worry about that yet lmao

[unkempt stream]

free group over a set?

[elder wave]

...

the construction of the tensor product

you take the free module on the cart product

and mod out stuff

[molten viper]

silly question, does this relate to "free groups"

[crystal turtle]

Yes

[elder wave]

i thought they might've seen that in lin alg

[unkempt stream]

It directly does actually!

[molten viper]

oooo

[warm wyvern]

I dunno what that is yet lel

[unkempt stream]

do you know what a free group is

[molten viper]

We kind of mentioned them in my intro to proofs class

I seem to recall it being a way to make a group out of a set

[crystal turtle]

Topological spaces as objects with homotopy classes of maps as morphisms

[unkempt stream]

like say for a set {a,b,c}, the free group, the group of all “sentences” or algebraic combinations of a b c and their inverses, like a^-1ba or cb, or any product of them

Is the “simplest” way to make a group out of that set

[warm wyvern]

I see

[unkempt stream]

ignoring the formality

[warm wyvern]

I dunno about simplest

[unkempt stream]

Adjoints my beloved(s)

[warm wyvern]

I think most general fits best

[unkempt stream]

yeah

[molten viper]

so uh, is the free group what happens if I feed a set into a free functor

[unkempt stream]

Yes

[molten viper]

like it's the "most general" way to make a group from that set

[warm wyvern]

yea

[ivory trail]

it's the "most general" group "on" the elements of that set

[unkempt stream]

yes

[ivory trail]

lots of "quotes" can be made concrete

[unkempt stream]

usually the free monoid is more simpler to explain, but the word monoid is scarier

[warm wyvern]

lmfao

[molten viper]

and I guess we could also make "free rings"

ignoring the formalism for now lol

[ivory trail]

free monoids are encountered in computer science pretty early on

[unkempt stream]

It’s a pain in the ass but yes

even free products

[crystal turtle]

"most general" as in it's easy to map out of. You may have learned that for free groups, you only have to define what it does to the generators to get a homomorphism out of it

[molten viper]

the naturals are a monoid right?

[warm wyvern]

I don't actually think it's productive to learn about free groups from free functors tho lel

[crystal turtle]

Yes

[warm wyvern]

yep

[unkempt stream]

under addition AND multiplication (without 0 for the latter)

[molten viper]

yeah my intro to proofs class did a sort of survey of some group theory concepts

[ivory trail]

naturals are actually the free monoid on a singleton set

[warm wyvern]

I thought monoid only required one operation

[ivory trail]

if my singleton set is {S} then the naturals can be written as 0, S0, SS0, ...

[molten viper]

after we learned how to do induction we just kinda dove into the definition of a group basically

[unkempt stream]

No just two different ways to “forget” it into a monoid

it’s cancellative remove 0 under multiplication

[warm wyvern]

@molten viper serious suggestion

[molten viper]

Hm?

[unkempt stream]

Once I learned adjoints it’s like constant mathematical deja vu

[warm wyvern]

if you actually want to learn the basics of cat theory you should read the couple sections in aluffi chapter 0

[open sluice]

we didn’t get to groups in my proofs class

[warm wyvern]

it's a very good read

[crystal turtle]

Leinster's basic category theory is also a good read imo

Riehl's CTC if you're up to it

[molten viper]

That prof is a character

I tihnk he leans to CS side of things

[warm wyvern]

you should prolly skip the monomorphism/epimorphisms section tho, I don't actually think it's that important

[unkempt stream]

my introduction to adjoints was actually the image functor from Top to Set (or Sigma algs to Set)

[molten viper]

At some point, rn I'm focusing on algebra stuffs

[crystal turtle]

Which functor?

OH

Open sets with inverse image?

[unkempt stream]

May have misworded it

[warm wyvern]

aluffi is an algebra book

[ivory trail]

forgetful functor?

[molten viper]

Well I have recs to cover

[unkempt stream]

almost, something more complicated about it

i had it written down

[warm wyvern]

that's fair lmfao

[unkempt stream]

then I realized it was the same with Hom and the tensor product

and i looked it up

called adjoints

[molten viper]

category theory is an appendix in one of the books I've been recommended @warm wyvern, maybe I'll give it a peak

[warm wyvern]

but if you wanna learn about basic cat theory and maybe free groups and that stuff I really like aluffi

[unkempt stream]

Also I sorta like how the tensor product construction uses both adjoint functors for Ring

you take the forgetful functor from F-Vec to Set, then immediately apply the free functor back to get the vector space you quotient out by

[molten viper]

though uh

maybe I should cover more basic things

like modules

[crystal turtle]

I learned category theory while learning groups and shit

[molten viper]

and whatever the heck a cohen-macaulay ring is

[unkempt stream]

same

[crystal turtle]

You don't strictly need much background, but it helps with motivation

[warm wyvern]

same

[warm wyvern]

my motivation was understanding cat theory memes

best decision ever

[unkempt stream]

Lol

[molten viper]

My motivation will be relevance to my PhD in a few years lol

if it is relevant

[warm wyvern]

that's valid too ig

[crystal turtle]

How it feels to be a monoid in the category of Endo functors 🥴

[warm wyvern]

🙄

[unkempt stream]

i was gonna ask if you where is a proof that, for a finite group of order N of automorphisms of a field F, that F forms a degree N extension over the fixed field

[unkempt stream]

imagine not doing EE

[elder wave]

?

[warm wyvern]

yea I don't understand that either lel

[unkempt stream]

and therefore your study not being remotely benefited by abstract alg

[elder wave]

Galois/Field theory isn't exactly my strongest trait

[unkempt stream]

there’s a proof via Artin’s Lemma but I kinda don’t like it

[elder wave]

there's better people to ask

[warm wyvern]

timo AG arc when

[unkempt stream]

is galois theory technically a part of ag?

[warm wyvern]

no

[elder wave]

was supposed to start but now it's AT again

[crystal turtle]

It's field theory

[elder wave]

after i finish my exams and focus on my thesis

[warm wyvern]

AT has been fun so far tbh

[unkempt stream]

Turns out to be a-finitely generated algebra (as a field) over a field is the same as being a finitely generated module (as a field)

which I’m pretty sure is due to the poly ring being a PID and just, making an explicit integral extension out of it

Noether normalization proofs i have seen are always either

1. Cardinality black magic
2. Weird substitution black magic

[warm wyvern]

😵‍💫

[topaz solar]

I remember coming across an article on like

Free lattice ordered groups

[formal ermine]

noob

[south patrol]

Hm the Nagata proof of Noether normalisation is a sub but I'd not say its black magic like

It's easy for a certain class of polys and its not bad to put stuff in that form by perturbing stuff

[topaz solar]

https://www.ams.org/tran/1985-290-01/S0002-9947-1985-0787955-7/S0002-9947-1985-0787955-7.pdf

https://www.sciencedirect.com/science/article/pii/0021869370900244

Gotta love free things

[unkempt stream]

the one that does X_N |-> X_N + Y^P^N

that one is fucking WILD

[south patrol]

Oh I like it lol

Wouldn't call it wild

[unkempt stream]

i can’t imagine how they came up with it

[cloud walrus]

Irony Incarnate

[wraith cargo]

this is eisenbud proposition 7.16

I assume we get a commutative diagram like so from which we can see this but I can't seem to get this map exactly

[chilly radish]

I don't see why S needs to be complete

[wraith cargo]

This is the setup of a more general theorem that needs the completion of S
But it's just this step I don't understand

Tho I thought it would be usable to show that the map is epic

[long nebula]

What lol how is it a pain

free commutative rings are even better

[rocky cloak]

You can just map the xs to the generators of n. And then you actually need S to be complete if you want to extend this to a map from the power series ring.

[drowsy sand]

wtf is this nerd stuff

[rocky cloak]

nerd stuff? THIS! IS! ALGBRAAAAAA!

https://youtu.be/G7AinOjDiNQ

[prisma bluff]

it’s a feedback loop designed to get more grant funding from the nsf

[unkempt stream]

i see R algebra a lot and I swear different resources say it’s either a ring rhat contains R (in it’s center) or is just a general algebra on the module agony

[coral spindle]

For fields, this is equivalent. If you're only working with fields then there's no difference.

For non-fields, I've only ever seen the second definition used.

N.b. I mean whether or not R is a field

[rocky cloak]

A ring with a ring homomorphism from R to it's center is also a common definition I guess

[unkempt stream]

i.e an associative (unital?) algebra over R

honestly i don’t like the word “algebra”, which in mant cases is general as fuck and is USED TO DESCRIBE THE FIELD used to define such a specific structure

[rocky cloak]

Yeah, an associative unital R algebra is the same as a ring A with a ring homomorphism R -> A whose image is in the center

The ring homomorphism is just given by r |-> r*1

[unkempt stream]

like a magma on an already existing abelian group , which is on ANOTHER abelian group with a semigroup on it

jesus fuck

which abelian groups already are specific ig

[molten viper]

I believe I've seen an R algebra defined as ring plus a homomorphism from R to the whatever ring

[rocky cloak]

An associative unital R algebra is a monoid in the category of R modules, if you're looking for more equivalent definitions.

[wraith cargo]

Hmm interesting
Eisenbud makes it feel like it's not important where it's mapped to for this part only that it's an epi

[molten viper]

An update from yesterday's exercise

[rocky cloak]

What is it you're building up to?

[molten viper]

hey I have no guarantee the intersection of ideals is an ideal

[crystal turtle]

It is

[molten viper]

I'll make today's exercise proving that

because I can't be bothered to figure out the notation for the next one in the book

[wraith cargo]

We wanna show that R[[x1,...,xn]]-> S is an epi lol just like you said

Lemme send you the proposition and the part that's confusing me

[rocky cloak]

Right, so the power series ring is a sort of "free" complete ring

Not quite in any formal sense I guess

[molten viper]

Even for an infinite intersection?

[crystal turtle]

It should always be an ideal, yeah

[molten viper]

wild okay

hrm

[wraith cargo]

For infinite intersections I feel like it's more nuanced

Tho most of the time that'll be 0 anyways

So you won't have to worry Abt it

[crystal turtle]

Not always tho

[molten viper]

oh well AM just states that the intersection of ideals is an ideal

but I'm still gonna prove it

[wraith cargo]

Huh okay it is an ideal even for infinite indexing sets that's cool

[crystal turtle]

But it shouldn't be too nuanced when you work it out

[crystal turtle]

Yeah AM just states it lol

[rocky cloak]

I feel like it's a pretty general pattern in algebra, that substructures are closed under intersection

[molten viper]

So like, an intersection of subgroups is a subgroup?

[rocky cloak]

Indeed it is

[molten viper]

oh ok this is kinda trivial

[rocky cloak]

Seems they have a typo where g_1 turns into f_1. Don't know if that's your confusion

[molten viper]

I guess I'm thinking of like, closed sets where an infinite intersection isn't necessarily closed

[wraith cargo]

My confusion is where does that map at the start of b. come from
The one from (x1,...,xn)/(x1,...,xn)^2

Or better yet why it's an epi

[molten viper]

I mean open

[crystal turtle]

I mean for open/closed sets it's by definition. This case it's basically just "if x and y are in the intersection, then they are in every (subobject), hence their sum/product is every (subobject) as well, hence their sum/product is in the intersection"

[rocky cloak]

The homomorphism maps xi to fi, and fi generates n. Hence it's epi

[wraith cargo]

Huh okay
That's easier than I thought lol
I thought that he was saying it was more general than that
Also that it would be an epi even if the fi didn't generate n

[unkempt stream]

i mean both A and B are ideals, then they “consume” multiplication by any element. If you multiply some element x in both by any element r, then their has to be in BOTH ideals

never knew how to like formally state that

i have been trying to view ideals as modules lately

hhh

[warm wyvern]

rA \subset A?

[molten viper]

AM writes it as Aα \subseteq α

[warm wyvern]

bru

is that a flipped a wtf

[molten viper]

Its an alpha

[warm wyvern]

oh

[molten viper]

Fake math fan

[warm wyvern]

is alpha the ideal?

[warm wyvern]

now listen here you little...

$\alpha$

[cloud walrus]

chmonkeynumber1enemy

[molten viper]

They denote ideals with gothic letters, for whatever reason

So it should be uh

$\gothic{a}$

[cloud walrus]

Galstaff, Sorcerer of Light
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[molten viper]

Or something

[agile burrow]

mathfrak

[warm wyvern]

$\mathfrak{a}$

[cloud walrus]

chmonkeynumber1enemy

[molten viper]

Yep

[mighty kiln]

Who's am

[molten viper]

Atiyah Macdonald

Which is an old and terse algebra book

They also use $\mathfrak{b}$

[cloud walrus]

Galstaff, Sorcerer of Light

[molten viper]

If they need 2 ideals

[agile burrow]

Sometimes even a c

[molten viper]

Yes!

Yeah AM is 136 pages or so

[crystal turtle]

AM moment

[molten viper]

128

[crystal turtle]

how far are you?

[warm wyvern]

bruh

[molten viper]

Chapter 1 lol

[crystal turtle]

oh lol

[warm wyvern]

and I thought spivak was bad

[molten viper]

10 chapters

From “Rings and Ideals” to “Dimension Theory”

[crystal turtle]

Chapters 1-2 are almost half the book tho

well, more like 30-40%

[molten viper]

About 30%

[crystal turtle]

I'm working through chapter 5 rn, but I've got a lot of exercises left in the earlier chapters

[molten viper]

And it seems chapter 9 is approximately chapter 1 of Eisenbud

[agile burrow]

I think it's a pretty good book

But I have not finished it

[molten viper]

It was the one recommended to me

I really should have started it earlier

[unkempt stream]

associative algebras < abelian pancake stack

[agile burrow]

Very descriptive

[wraith cargo]

tf is one chapter of eisenbud lol
What length is that

[warm wyvern]

lol

commutative algebra just sounds like pain

[unkempt stream]

use currying to make some ungodly description of it

[wraith cargo]

a chapter of eisenbud is like waht 40-50 pages ig

[molten viper]

Maybe

[wraith cargo]

tho the appendix on homological algebra is like 100pages

[unkempt stream]

anyway! A lot of what I’ve read about noncommutative ring theory, noncommutative ring generalizations depends on the existence of modules over that ring that satisfy some property

be it being annihilated or you have it

but like, why.

It feels weird because ideals are technically modules, but the set of all ideals is a set cuz you’re essentially “chosing” them out of subsets of the ring

but to say a module exists that gets annihilated by an ideal or something

you’re saying it exists in a class of R-modules

and classes scare me :troll:

[crystal turtle]

i mean what's the difference tho

[unkempt stream]

an ideal is a module that is “contained” in the ring that it’s over

[crystal turtle]

yeah

[unkempt stream]

I assume there’s modules not iso to an ideal

[crystal turtle]

which exist by free modules over sets with arbitrarily large cardinality

[unkempt stream]

Can many of these module definitions, like primitive ideals for instance, be in the context of purely ideals in a noncommutative setting

Totally not because I want to avoid formal classes

[crystal turtle]

but like what would you need proper classes for?

[unkempt stream]

to say there exists a module

via there exists M such that M in (module class) implies M is …

[wraith cargo]

I don't see how classes can show existence

or do you mean that you assume like axiomatically that such a module exists by some set theory stuff

[unkempt stream]

yes

[crystal turtle]

just use Grothendieck universes if you don't like proper classes

[wraith cargo]

What's the problem with proper classes tho

[unkempt stream]

Paradoxes.

[crystal turtle]

They don't work very nicely/are hard to work with for formal purposes

[unkempt stream]

When trying to form homsets n stuff

[crystal turtle]

homsets are sets tho

[unkempt stream]

needing locally small categories n stuff

[wraith cargo]

hm interesting
I've never really explored how they work I've only heard they're used to formally axiomatize category theory

[crystal turtle]

Yeah you can either work with proper classes or Grothendieck universes for that. Usually (at least for basic stuff) it's just informally treated

[unkempt stream]

I guess all sets are classes, and you can say that a class is a set iff it is contained in a class

[wraith cargo]

I will need to look into this more it sounds interesting

[unkempt stream]

which sounds weird until you just, apply the axioms for saying what classes are sets via construction mayhaps

relation-pilled

[crystal turtle]

still not sure what you need classes for with this? Can't you just say, instead, "module M satisfies property \phi"?

[unkempt stream]

i think i just don’t know how to formalize a “property”

[delicate orchid]

Why does everything need to be formalised

[wraith cargo]

I assume classes have a similar axiom schema for properties

[crystal turtle]

yeah I mean I don't think that needs to be super formalized

[wraith cargo]

or grothendieck universes

[unkempt stream]

constructivism honestly

to say how you can construct it

[wraith cargo]

Oh okay thank you for being up front about your problems
Saying they exist is the first step to recovery

[delicate orchid]

Then you do not need formal classes to define a module?

[wraith cargo]

choice :3

[unkempt stream]

LMAO

[unkempt stream]

It’s not socially acceptable to talk people out of their religion

[crystal turtle]

constructivist moment

[open sluice]

since when do mathematicians care about social acceptance

[unkempt stream]

i’ve actually never seen anything counter intuitive about choice.

[crystal turtle]

next we'll become ultra-finitists

[unkempt stream]

people mention shit like well-ordering theorem and I’m like, yeah, as sets why not, doesn’t mean that well ordering’s gonna work with whatever structure you use to define it

[delicate orchid]

People tend to mention banarch-tarski more than the well ordering theorem

That one is weird

[unkempt stream]

banach tarski isn’t counter intuitively even remotely

[delicate orchid]

Now you’re being disingenuous

[solar vessel]

real

but what is the discussion at hand

[warm wyvern]

bet

[solar vessel]

oh

yeah I agree

choice is based

GCH is more based

[unkempt stream]

Well to say shit has volume you need to add some kind of structure to the set, banach tarski is basically on the fucking group structure of the sphere, not the fuckin volume

[warm wyvern]

assuming choice, this result is true

[delicate orchid]

What

The elements involved in b-t are line segements which union to give the sphere

[unkempt stream]

That’s like saying because [0,1] has the same cardinality as the whole real line that a 1 inch metal rod has the same volume as an infinitely long one.

[delicate orchid]

SO(3) acts on this set

[crystal turtle]

Okay but choice is also equivalent to being able to choose an element from an infinite product of (nonempty) sets
Which is obviously true

[wraith cargo]

throwing shade is it
and yet when I asked you to imagine a set with cardinality between N_0 and N_1 you told me it's "impossible" and that I was "spreading a disease from which mathematics would one day recover"

[warm wyvern]

banach tarski doesn't talk about the group structure of the sphere at all lel

[unkempt stream]

I meant group structure of the SO(3) action

[delicate orchid]

It uses that

[wraith cargo]

isn't choice necessary to defined the whachama call it set that isn't measurable with lebesgue measure

[delicate orchid]

To produce two spheres of identical volume from one

[unkempt stream]

Pretty sure b-t uses a free subgroup of SO(3) and uses that algebraic freeness to define the sets

[warm wyvern]

also, without choice unmeasurable sets don't even exist

[warm wyvern]

vitali set?

[wraith cargo]

ye

[unkempt stream]

WHAT

[solar vessel]

this result is interesting but not counter intuitive

[warm wyvern]

is it not?

[solar vessel]

no

[warm wyvern]

it is for me

[unkempt stream]

i mean at first glance i’d assume nonmeasurable sets exist without choice because, wtf should every set be measurable

not abstract alg anymore 😦

well besides vitali sets

direct product of all the cosets of R/Q bruhaps

[wraith cargo]

yeah vitali assumes choice

[warm wyvern]

if you don't assume choice you're never encountering a set you can't measure

[delicate orchid]

Bruhaps indeed

[warm wyvern]

in practice, even when assuming choice, you're very likely to not encounter a non-measurable set but still

lmao

[solar vessel]

I think people consider banach tarski nonintuitive because they're indoctrinated from a young age to think area/volume etc are self evident things

[unkempt stream]

do nonmeasurable sets imply choice for sets of continuum cardinality

[delicate orchid]

The real problem with this conversation is if you think about anything non-intuitive enough it becomes intuitive

[warm wyvern]

also, god I'm so glad no set theorist is here to flame me for this

[open sluice]

the outer measure of a disjoint union is not the same as the sum of their outer measures
that’s a bit unpleasant to me

[unkempt stream]

To describe a proposition as self referential is self referential

agonizing

[crystal turtle]

Imagine rejecting axiom of choice

But not axiom of pairing

[delicate orchid]

I reject all of em I’m just gonna do my own thing

[warm wyvern]

anywho

anyone that thinks aoc is intuitive is fucking somoking

later nerds

[delicate orchid]

Aoc itself is intuitive it just leads to weirdo stuff

[solar vessel]

it leads to interesting stuff

[warm wyvern]

fair

[unkempt stream]

still stuck on Noetherian condition on prime ideals implies noetherian condition on all ideals WITHOUT Zorn’s Lemma

driving me mad

[solar vessel]

let the choice flow through you

[unkempt stream]

it’s for hilbert basis theorem, from which I can see only needs finite choice

but the prime ideal route needs full choice

[delicate orchid]

Noetherian rings have primary decomp

Le circular argument has arrived

[unkempt stream]

My route of trying to prove Hilbert basis through morphisms is basically showing that for a strictly increasing chain of prime ideals, that the sequence of intersections with R must be strictly increasing

so therefore the prime ideal chains must stabilize in all of R[X] given they do in R

[delicate orchid]

I buy it

The only “additional” prime ideal in R[x] is (x)

So intersecting with R is like removing the contribution of (x) to the chain of ideals

[unkempt stream]

R need not be integral

[delicate orchid]

Gulp

[unkempt stream]

We only need zorn’s to say every ideal is in a prime ideal (because every one has a maximal)

[delicate orchid]

Yeah it’s easy with zorn

[unkempt stream]

wonder if we can find a weaker form using finite choice

[delicate orchid]

I thought we were trying to ignore choice

[unkempt stream]

I.e just every ideal in a prime

well you’re working with sequences of ideals

so you need a weak form of choice otherwise spooky scaries

[agile burrow]

oh, are you assuming that every ascending chain of prime ideals terminates?

[delicate orchid]

I think I’ve misunderstood the set up here

[unkempt stream]

Assume R is noetherian

thus every ascending chain of prime ideals stabilizes because all chains of ideals must

Well, for prime ideals of R[X], if A is strictly contained in B, then their intersections with R preserve that

so every ascending chain of prime ideals in R[X] stabilizes

Using the leading coefficient form of hilbert basis proof doesn’t need the full power of choice

while this one does to extend it to all chains of ideals in R[X]

if you can prove any ideal has a prime ideal containing it without using that full power of choice then I’m happy

[agile burrow]

Ok I think I just don't understand what you're asking lol

[delicate orchid]

“”Prove the thing without using choice or I will CRY””

[unkempt stream]

THERE’S A WEAKER PROOF WITHOUT CHOICE, WHY DOES THIS BITCH NEED IT

I want to avoid the leading coefficient map because it’s not a morphism

i just want to kind of get insight into the structure of ideals in R[X]

[agile burrow]

Lol it's a morphism in a different category

[unkempt stream]

Because like noetherian-ness, being jacobson is also inherited

[unkempt stream]

in semigroups, yes

[warm urchin]

If that was the case this would mean the hahn-banach implies choice , which is not true

you can prove lebesgue non measurable sets exist using just the hahn banach , and the hahn banach is weaker than choice

[unkempt stream]

i’m just seeing if there is some structure of ideals in R[X] inherited from R that causes it

[delicate orchid]

Look at the spectrum then

[agile burrow]

But isn't that exactly what the leading coefficient proof shows?

[unkempt stream]

the spectrum i thought is about primes

[unkempt stream]

mainly I want to try to prove it via the universal property

[delicate orchid]

🧍‍♂️prime ideals

[unkempt stream]

i did so for the proof that if every prime ideal in R is an intersection of maximals, then the same holds for R[X] using evalution morphisms from the universal property

no use of degrees or anything of polynomials

Just the evaluation

[warm urchin]

you need to reject weaker assumptions too for that to happen iirc you can construct non measurable lebesgue sets with weaker assumptions than choice

[open sluice]

the illusion of choice

[warm wyvern]

shush, james, I know

[unkempt stream]

And their kernels n stuff

[warm wyvern]

and I thought no set theorist could figure me out. Yet I got called out by a fellow analyst

shame

this is what folland says for a more accurate description

[agile burrow]

How do you propose passing from an ideal in R[x] to an ideal in R using just evaluation?

[unkempt stream]

evaluations from R[X] to R are always epi, obv cuz they fix R

[delicate orchid]

Evaluate x to 1 :letrollface:

[unkempt stream]

Evaluate x to literally anything in R

evaluating to 0 gives the maximal R-disjoint ideal (X)

[agile burrow]

Sure, but to extract meaningful information about the polynomials themselves, it seems like evaluation forgets too much in general

[delicate orchid]

Yeah 0 is a better shout

[unkempt stream]

i forget specifically but I did show via rhe universal property that R[X] is the monoid alg of R[N] and essentially every element is a polynomial

[delicate orchid]

Evaluating to something in R is just quotienting by some (x-a) doodad

So I do agree with walter if I must have a serious opinion

[unkempt stream]

it was kinda nutty but used powers of (X) and some ideals quotiented by higher powers of (X) via lattice iso

and that R[X]/(X) is iso to R

[delicate orchid]

Yeah maybe we can think of these mofos as lattices that could be funny

[unkempt stream]

yeah lol

[unkempt stream]

it was simultaneously the most annoying and enjoyable/enlightening universal prop exercise I’ve done

[chilly ocean]

[unkempt stream]

i think it was taking (X)^n to (X)^n/(X)^n+1 and using the (forgetful’d) map from that ideal to R

which gives the coefficients in the module sense

absolute insanity but whatever :troll:

[agile burrow]

Ok so like, if you restrict to polynomials of degree less than or equal to d, then the leading coefficient map is a morphism of R-modules, right?

[unkempt stream]

yes?

it also shows R[X] is iso to it’s associated graded ring I think

anyway gtg for a bit

[agile burrow]

Ok lol

I was just gonna say that this gives you an ascending chain of ideals in R so I don't see what you were saying about the leading coefficient map not being a morphism

[unkempt stream]

Not really

descending i think

[agile burrow]

I'm thinking let I_n be the image of the leading coefficient map on polynomials of degree less than or equal to n, where if the degree is strictly less than n then you send it to 0

[unkempt stream]

I could try some weird diagonalization technique

[agile burrow]

And then I_n should be contained in I_{n+1} as n ranges over natural numbers

[unkempt stream]

the full proof of the universal property of the polynomial ring in the reverse was showing that the kernel of the evaluation map of 0, call it J, is principal, constructing powers of that kernel, and constructing J^n/J^n+1 via quotients and isomorphism theorems, then using the principality of J^n to construct the nth coefficient map, and then doing the full isomorphism

it was my first universal property exercise i did

after learning the def and applying it to R[X] lol

[agile burrow]

ok

but does what I said make sense

[unkempt stream]

yes

[agile burrow]

So then you get the ascending chain of ideals in R and you can just follow through with the rest of the proof in leading coefficients

I just don't see what bothers you regarding that proof

[unkempt stream]

the proof that the image of an ideal under that map is an ideal and all the other stuff making sure it preserves strict inclusion n stuff

[agile burrow]

You can explicitly verify those by hand though

[unkempt stream]

it’s fairly annoying

it’s “almost” linear

i.e there exist a b such that f(ax + by) = f(x) + f(y)

and f(xy) is not necessarily f(x)f(y) afaik if R is not integral

like for example f((ax + b)(cx + d)) = f(acx^2 + (ad + bc)x + c)) = ad + bc if ac = 0

which you have to do some ideal properties at that point to prove the image is still idelic

it’s a pain

[agile burrow]

Wait it shouldn't be a ring homomorphism though, it's a module homomorphism

[unkempt stream]

If that

[unkempt stream]

the ax + by is used to prove the image is an ideal

[agile burrow]

I don't get it. As we've set it up, you just add the leading coefficients if you add two polynomials together

And similarly if you multiply a polynomial by an element of r, the leading coefficient gets multiplied by r

So it's a module homomorphism

[unkempt stream]

Fuck i was thinking about the general leading coefficient one

not the one that sends polynomials of degree n to it’s leading coeff

[agile burrow]

Yeah I mean I think you need this family of maps to get a reasonable ascending chain of ideals in R

[unkempt stream]

it’s just, starting from rhe universal property to get to here is a LOT

i wanted to see if there’s a shorter route

another attempt of mine was kind of considering the ideal (X_1…X_N) in it’s multivar poly ring to be the “archetypical” finitely generated ideal, as any finitelt generated ideal is that ideal’s image under some evaluation

it works by appending X being just another finite var poly ring but it’s annoying

[white pivot]

can someone explain this answer? https://math.stackexchange.com/a/605260/1196218

they say that a function is surjective since Q(sqrt(2)) is the smallest field containing Q and sqrt(2), but i dont see how that proves it surjective

Here is what they say:
$f$ is surjective, because $\mathbb{Q}(\sqrt{2})$ is the smallest field containing $\mathbb{Q}$ and $\sqrt{2}$. Now the set
$$
{a+b\sqrt{2} : a,b\in\mathbb{Q}}
$$
is another field (check this) that contains $\mathbb{Q}$ and $\sqrt{2}$. Hence, $\mathbb{Q}(\sqrt{2}) \subset f(\mathbb{Q}\times \mathbb{Q})$.

[cloud walrus]

MyFavoriteAccount

[delicate orchid]

it's fairly clear, if we have some a+bsqrt(2) then it's the image of (a,b)

hence it's surjective

[white pivot]

why do they say "another"? Isnt Q(sqrt(2)) = {a + b*sqrt(2) | a,b in Q}

[delicate orchid]

the only nuance I can see in this is the fact that sqrt(2)^{-1} is still in that set

otherwise I agree with you, it's unneeded fluff

[topaz solar]

I don’t really get why they bring up fields here tbh

[delicate orchid]

yeah like literally just do this

[topaz solar]

I mean obviously Q(sqrt 2) is a field but like

Show sqrt(2)/2 = 1/sqrt(2)

[delicate orchid]

I really think this is crux of it

yeah

[topaz solar]

it’s so extra to bring up that field line imo

[rocky cloak]

I mean, you would have to show that all the other elements have inverses as well.

[topaz solar]

Ah fair

[rocky cloak]

It's not very difficult, but you have to check it

[topaz solar]

But presumably you’ve already checked that since a + bsqrt(2) is given as a representation of things

[rocky cloak]

representation of things?

[topaz solar]

as in they’re all equal to a + b sqrt(2) for any element in it

Not rep theory type of representation

[rocky cloak]

I'm not sure what you're saying

[topaz solar]

Well, when they asked the related question in #proofs-and-logic I asked what the elements looked like

And they already said they were all a + b sqrt(2)

So that’s a step that’s already done

[rocky cloak]

I see, I'm missing some context then

[glossy crag]

Say I have two monic separable irreducibles f and g of degree n (over a field K with alg. closure C) and I know that for every root b of g there is a root a of f with K(b)=K(a). I want to show that I can order the roots {a_1,...,a_n} and {b_1,...,b_n} of f and g in such a way that K(a_i)=K(b_i), how do I do this? I clearly need to use Aut(C/K) and how it permutes the roots, but I can't see a clean way of doing it.

[glossy crag]

This implies that symmetrically, for every root a of f there is a root b of g s.t. K(a)=K(b) (using Aut(C/K)), but it's not helping me order the roots nicely 1to1.

[rocky cloak]

Don't you just order them? Like pick one b to be b1, then pick a1 to be the a such that K(b) = K(a)?

[glossy crag]

I mean what if there are repetitions

[delicate orchid]

then either ordering is fine?

[glossy crag]

Wdym

[delicate orchid]

if K(b_i) = K(a_i) = K(a_j) and K(b_j) = K(a_j) then K(b_j) = K(b_i) so either ordering works?

[white oxide]

could I get a hint to show that the requisite series of subgroups has abelian factors? my process was that if I could show that $H_iN \trianglelefteq H_{i + 1}N$ then each quotient group would be isomorphic to $H_{i + 1}N/H_iN$ (by the third isomorphism theorem), then the problem would reduce to showing that $H_{i +1}N/H_iN$ is abelian. but I'm stuck on trying to show that $H_iN \trianglelefteq H_{i + 1}N$ lol

[cloud walrus]

okeyokay

[topaz solar]

If you want no repeats, for the a_i, you’d just need to use like, if K(b_i) = K(b_j) = K(a_i), then you want another a_j with the same extension right?

[glossy crag]

I don't understand what you mean by this. E.g. n=4, b_1->a_1, b_2->a_2, but then it turns out that K(b_3)=K(a_2), what do I associate to b_3?

[topaz solar]

Once you have this just exhaust em

[white oxide]

I let $a_{i +1}n_1 \in H_{i +1}N$ and $h_in_2 \in H_iN$, and so it reduces down to showing that $a_{i + 1}n_1h_in_2n_1^{-1}a_{i + 1}^{-1} \in H_iN$

[cloud walrus]

okeyokay

[white oxide]

upon which I have no clue

[delicate orchid]

there will be some a_3 ~ b_3 (by assumption), so a_2 ~ a_3

so it doesn't matter

equality is transitive

[topaz solar]

What he’s worried about is he wants each b_i to have a distinct a_i

[glossy crag]

I still don't get it.

[delicate orchid]

well that just isn't always possible

[topaz solar]

Mhm

Kinda hard if you have repeated roots

[glossy crag]

To make it even simpler, what if n=2 and the roots are a,a' and b,b' and it turns out that K(b)=K(a) and K(b')=K(a)

[delicate orchid]

then K(a') = K(a)

by assumption

[glossy crag]

Why?

[delicate orchid]

because K(a) = K(b) = K(b')

and K(a') = K(b')

[delicate orchid]

it really is this simple

[glossy crag]

Why K(a')=K(b')?

[topaz solar]

K(a’) does not have assumed equality with K(b’)

[delicate orchid]

what

ok I have no fucking idea what's going on

what is a' if not the associated root to b'

[topaz solar]

There’s a term of a_i with K(b) = K(a_i)

[white oxide]

nvm I'm slow I just need to show that H_{i + 1}N is abelian ignore all this

[topaz solar]

He wants a functional assignment where they’re all given them

[delicate orchid]

we have two sets

with an equivalence relation on their cartesian product

you may pick any two representitves

and label them with the same index

[topaz solar]

I wonder how you got the symmetric bit, but if you have that just zig zag

Start at b_1

Grab your guaranteed a_1

Then do an a_2 term

Which has at least one b_2.

[topaz solar]

Might not immediately get uniqueness but (linked reply) might be an idea for that

[glossy crag]

If a is any root take e.g. b_1, then it has to it some a_i, then if x\in Aut(C/K) moves a_i to a (C/K is normal and f is irreducible) you have K(xb_1)=xK(b_1)=xK(a_i)=K(xa_i)=K(a) and xb_1 is a root of g.

[topaz solar]

now zigzag and remove choices after taking them

[glossy crag]

Hm, lemme think about it.

[rocky cloak]

K(a') is isomorphic to K(a) so there is an automorphism of C which maps K(a) to K(a'). This will necessarily map K(b) to K(b'') where b'' is some conjugate of b. Thus the relation is symmetric

[topaz solar]

Ye, just use the symmetric argument you had earlier

[glossy crag]

This "removing" choices bit is bothering me, it's not clicking.

[rocky cloak]

Seems you already figured that out

[topaz solar]

This condition is pretty much just saying we can exhaust our roots

If b, b’ both have a as a possible choice for them

then we’d want an a’ which is also possible

then associate b & a, b’ & a’

[glossy crag]

Yeah I see how the case n=2 works, I'm just having trouble formalising it for n arbitrary

[topaz solar]

Pulling marbles out of hats & not replacing them

[glossy crag]

Maybe I should do induction lol

[topaz solar]

Well, b, b’, b”, all go to a, so we want two more congruent ones to a

Etc

[glossy crag]

I'm feeling particularly thick today, it's just not clicking

Maybe i just need to take a break

[topaz solar]

Something something root permuting on either side

[glossy crag]

Yeah I know, it's the something something that is not clicking

[topaz solar]

K(b) = K(a), and we can apply that element of the Galois to K(b)

Where it send a?

[void cosmos]

why

[topaz solar]

For larger numbers of roots, what about larger numbers of permutations

[void cosmos]

are noetherian rings important in geometry

oh its taken sorry

[topaz solar]

Not sure if you might run into a repeated root issue, but that should be fine?

[glossy crag]

There are no repeated roots, the polynomials are separable.

[topaz solar]

Ah right

peabrain

[glossy crag]

@topaz solar you can stop now, I'll come back to this later with a clear head.

Thanks

[rocky cloak]

Definitely

[void cosmos]

yea i meant

why

[void cosmos]

.

is it because

I^n can be thought of polynomials

[rocky cloak]

Because polynomial rings are Noetherian

[void cosmos]

I is an idea

yea but like

[rocky cloak]

And quotients and localizations preserve being Noetherian

[void cosmos]

where is the geometry

tho

[delicate orchid]

the variety

[void cosmos]

where is the square

and the shapes

triangle

[delicate orchid]

you'd know this if you read the first page of any alg geo book

[rocky cloak]

A shape can be understood by looking at the functions defined on that shape

[topaz solar]

Circle (radius 1) is roots to x^2 + y^2 -1

[delicate orchid]

ok maybe not page

chapter

[void cosmos]

what about triangle

just some lines ig

wher do rings come tho

[glossy crag]

Man I just CAN'T let it go

What if it's a situation like this (above roots of g, below roots of f)

The final root of f has to have a root of g associated to it

what if this happens

How do you then order the roots in the desired way

[delicate orchid]

the question is, can that happen

[glossy crag]

True, but then why shouldn't it?

[delicate orchid]

that's the mystery

[glossy crag]

https://www.youtube.com/watch?v=ZqGC4-n3cu8

[topaz solar]

Try permitting b_2 -> b_3, then since a_2 is in K(b_2), and K(a_2) = K(b_2), what do we have as the image of a_2 under that permutation

I’d certainly expect a_3 and K(a_3) = K(b_3), but that’s left to be shown

But might be useful to use how b_2 and a_2 both give the whole extension, and b_3 gives the whole extension K(b_3)

[delicate orchid]

ah right now I see where my mistake in my understanding was

[topaz solar]

Yeah we want a linear ordering of the roots

[delicate orchid]

s-s-s-ss-scary

[topaz solar]

real

[white oxide]

help i'm actually stupid if Hi is normal in Hi+1 and N is any normal subgroup of G why is HiN normal in Hi+1N help

[topaz solar]

But yeah, consider b_2 and a_2 as polynomials of each other maybe?

Or ya know, a similar variation thereof since it’s a field and inverses and all

[delicate orchid]

let x be in Hi+1N
xHiNx^{-1} = xHix^{-1}xNx^{-1} = HiN

think this works

I made up several inclusions of subgroups in my head

[topaz solar]

Yeah I think that should work

[white oxide]

ah right thanks

[topaz solar]

What these lil permutations applied to the relevant fields should show is that this doesnt happen

b ~ a, and b’ ~ a implies we get another a’ ~ a, I think?

And then successively yoink them out till you exhaust them, giving a compatible linear order on both

[glossy crag]

Imma take a break now, try again later I think. Thanks for the continued interested @topaz solar

[unkempt stream]

localization at a multuplicative set has the ideals that are outside of the multiplicative set, right?

i know units lie outside any proper ideal

i know very little about multiplicative sets besides the definition

[rocky cloak]

I think I have a proof:

First consider an equivalence relation on the bs with b ~ b' if K(b) = K(b'). Let B be the set of equivalence classes and do the same for the as creating A.

Since each b has a corresponding a, we get a map from B to A, and by how we constructed the equivalence this must be injective. Symmetrically we also get an injection from A to B so they have the same size hence this is a bijection.

Applying automorphisms shows that the equivalence classes in B have the same size n/|B|. Similarly for A, so you can just map the bs arbitrarily within each equivalence class.

Hopefully there is a shorter argument, but this works at least.

@glossy crag

[rocky cloak]

Yeah, that's right. The ideals of S^-1 R correspond to the ideals of R that don't intersect S.

[topaz solar]

I mean, yeah that should work, the part he was hung up on was the size of each class so I was trying to point (poorly) at being able to do that by the automorphisms

[vague granite]

Formally how would one go about showing 1+x+x^2+x^3+x^4 is irreducible over Q[x]

The only way I can think of is to verify that each strict subproduct of distinct linear factors (over C[x]) isn't an element in Q[x]

But that would require 4C2+4C3=12 checks

[crystal turtle]

multiply it by x-1

(x^4+x^3+x^2+x+1)(x-1)=x^5-1

Which we know the roots of 🙂

(in C)

[unkempt stream]

there’s a few different ways

[vague granite]

I'm not sure how that implies it's irreducible

(1+x^2)^2 is not irreducible despite having no real roots

[unkempt stream]

One method is eisenstein’s criterion but i haven’t thought about where the terms are ALL one

there’s a really… weird way

[vague granite]

I'm not sure Eisenstein's criterion works here

[crystal turtle]

ah you're right

[vague granite]

There's no prime which divides one...

[unkempt stream]

show that (1 + x + x^2 + x^3 + x^4)(x - 1)= x^n - 1

NOOOOO WAIT

[crystal turtle]

Still reduces the number of checks to (4 choose 2), no?

[unkempt stream]

if a poly divides rhat

then it must be cyclotomic

[crystal turtle]

And this allows you to work with roots of unity in C

[unkempt stream]

right?

[vague granite]

What and how?

[crystal turtle]

There's something with this, I just can't remember the exact argument lol

[unkempt stream]

I can think of a galoisy method

[vague granite]

It's an exercise in a pre problem sheet on a course on galois theory so I think that terminology is out of the scope haha..

[crystal turtle]

Certainly it doesn't have a linear term. You just need to check different combinations of multiplying (x-\zeta_i)(x-\zeta_j) for 5-th roots of unity \zeta_i and \zeta_j

[vague granite]

Oh right, yes I mentioned that, and I think it's 4C2+4C3

so 12

still seems unreasonably tedious for a simple exercise

[topaz solar]

You know the roots to x^5 - 1 right

Just argue on how no product is rational or wtv

gg

[vague granite]

That's 12 products...

[unkempt stream]

so clearly galois theory itself is out

lol

[vague granite]

Yes

[topaz solar]

Well, think in terms of the angles

[crystal turtle]

WEll when you multiply out any pair like that, you will get x^2 -(\zeta_i + \zeta_j)x + \zeta{i+j}

[topaz solar]

Since roots of unity

gg

[crystal turtle]

At least one of them will be complex

[vague granite]

This is not a mathematical argument

I can't even begin to decipher how that would work

[topaz solar]

it is, because any product of the linear factors has to multiply to a thing with coefficients in Q right

[vague granite]

Yes

[topaz solar]

and the roots are gonna be like

e^wtv

so just add those things together, and you won’t get a rational

Unless you do em all

[crystal turtle]

^^

[topaz solar]

And that’s exactly the constant term

So it has to be rational in order to get a reducible thing

[vague granite]

Oh right

[topaz solar]

Ye

[vague granite]

You only have to do 2 cases

(x-zeta)(x-zeta*)

that's nifty actually

[crystal turtle]

yeah with symmetry of the roots in C

[topaz solar]

Don’t have to check everything, literally just say it’s 5th roots of unity and you can’t get a rational polynomial

[vague granite]

But you do have to check something

[topaz solar]

That’s what we do by saying it’s roots of unity

That’s the check

[vague granite]

Namely that (x-zeta)(x-zeta*) isnt int Q[x]?

[topaz solar]

obviously it’s not

Adding them together isn’t rational

[vague granite]

Okay but x^3+x^2+x+1=0 has roots of unity despite being reducible

[topaz solar]

But that’s 4th roots

Not 5th

5th roots you can’t add and get a rational like that

[vague granite]

Okay but you need to prove that no?

I don't see a priori how that's obvious

[topaz solar]

If they’re not opposing roots, the product isn’t even real

And if they are, they’re neither 1/2 nor -1/2 real part

[vague granite]

I don't know what you mean by opposing roots

[topaz solar]

Done

[topaz solar]

Symmetric across the real line

[vague granite]

Every real polynomial has symmetric roots about the real line

?

[rocky cloak]

Do you have the concept of minimal polynomial, and how Q(a) = Q[x]/f(x) when f is the minimal polynomial of a?

If so there is a very short solution

[vague granite]

Yes that's allowed

This is after a course on rings and modules

[rocky cloak]

Right well you have a map Q[x] -> C, mapping x to a fifth root of unity. Show that the image is the same for all choices of primitive roots. Thus they all have the same minimal polynomial.

(this also uses that Q[x]/f = Q[x]/g implies f=g)

So the minimal polynomial has degree 4

[unkempt stream]

i think I know a way.

would you like to hear.

[topaz solar]

I feel like just a quick argument on adding roots of x^5-1 is easier imo but oh well

[vague granite]

yeah sure

[unkempt stream]

so.

(x^5 - 1)/(x - 1)

right

well, let x = t + 1

[vague granite]

Right that's quite neat

[unkempt stream]

((t + 1)^5 - 1))/t okay

well, for binomial coefficients

Remember, for 1 < n < N

N divides nCr(N,n)

[rocky cloak]

Yeah, this also extends to arbitrary cyclotomic polynomials, so pretty neat yeah

[unkempt stream]

so for that poly in t, 5 divides all the terms except for the leading one

[rocky cloak]

Eisenstein baby

[unkempt stream]

and 5^2 does not divide the constant term, which is 5 ofc

So to recap

your x^4 + x^3 + x^2 + x + 1

set x -> x + 1

[vague granite]

That's a very clever way of introducing Eisenstein into the problem

Sounds like a reusable trick

[molten viper]

(It is)

[unkempt stream]

you get x^4 + 5x^3 + 10x^2 + 10x + 5

use le eisenstein

thank you dummit & foote

it was for the case of 7 lol

just generalized

[crystal turtle]

ah, that's the general argument, I remember seeing that in class lol, I just forget it

[south patrol]

Yes indeed this works for the *p^n) th cyclotomic polynomials (p prime, n arbitrary)

Oh that's been said, though it was said "all" cyclotomics when really it's just for these ones

[unkempt stream]

I got reminded because of the Febronious for the other prime fields

[south patrol]

Febronious

😭

[vague granite]

Someone call a medic

Cause you butchered his name

[unkempt stream]

idk

[topaz solar]

Some names are hard to remember

Like the uhhh, Semen Kusakatdze-something guy

Hang on lemme just look up the book

Semën Samsonovich Kutateladze

Ain’t no way I will remember that name’s spelling

[toxic zephyr]

this is technically a number theory question but I'm guessing there's a relatively simple abstract algebra proof for it (that I'm struggling with bc my algebra is not strong).

let q<p be odd primes such that p=q^2k+1. prove that if x^q=1 mod p then x=y^q mod p for some y.

should I try to do this with a primitive root/generator?

[wraith cargo]

x^q implies that x has order q in the group (Z/pZ)^x
I think that's a good hint

[toxic zephyr]

$(g^\ell)^q=1\implies q^2k\mid q\ell\implies qk\mid \ell$

$\implies g^\ell=g^{cqk}=(g^{ck})^q$

[cloud walrus]

nixxy nilpotent

[toxic zephyr]

I'm being lazy with notation but does that do it?

[wraith cargo]

uhhh if you can assume that (Z/pZ)^x is cyclic then yeah that works

[toxic zephyr]

yeah we can. alright thanks 👍

[wraith cargo]

yeah sorry if my hint was a bit weird I didn't know if you could assume this so I was going for some subgroup nonsense

[crystal turtle]

any finite field has a cyclic group for it's group of units

iirc

[white oxide]

could I get a hint to show that if f: $V \to V'$ is a linear transformation of vector spaces over a division ring $D$, then ${f(x) \mid f(x) \neq 0, x \in X}$ is linearly independent in $V'$? already showed that it spans $Im \text{ f }$

[cloud walrus]

okeyokay

[white oxide]

So I wrote $r_1f(x_1) + \dots + r_nf(x_n) = 0 = f(r_1x_1 + \dots + r_nx_n)$

[cloud walrus]

okeyokay

[white oxide]

trying to see something from this and I'm blanking F

[wraith cargo]

yeah but showing cyclicity isn't very trivial and it took like until gauss for that to be shown

and I know that sometimes like intro NT classes take a while to prove that

[topaz solar]

What is X

[white oxide]

my bad the basis of V

[wraith cargo]

tho yea this is true it's cyclic when n=1,2,4, p^k or 2p^k

[topaz solar]

Couldn’t I map like (0, 1) to (1, 0) and (1, 0) to (1, 0)?

[south patrol]

Lol that statements seems false to me

[topaz solar]

With X = {(0, 1), (1, 0)}

[south patrol]

Oh I was xonfused too cause I use V' as notation for dual space kek

[topaz solar]

Basis for 2 dim

[south patrol]

Oh I guess this is basically first iso theorem

[topaz solar]

Well x is taken from X

So it seems false

[south patrol]

But only after certain modifications

[topaz solar]

yeah as is it’s uh

[white oxide]

tryna verify the first half of this theorem

[south patrol]

I mean just take projection onto an axis and V = R^3, V' = R

[topaz solar]

you did not have the correct statement originally

[south patrol]

For example

Yeah lol

"There exists a basis" rather than given arbitrary X as you seemed to imply

[topaz solar]

Yeah homie

[white oxide]

o

[topaz solar]

Take a basis for ker f

Extend it

[south patrol]

Amd so yes this is basically first iso

[topaz solar]

Yep

[white oxide]

wdym by this

[topaz solar]

Do you have that any vector spaces have a basis

I’m particular, so you have that any linearly independent set can be extended to a basis?

[white oxide]

well i have every linearly independent subset of V is contained in a basis, that's basically equivalent right

[topaz solar]

Yeah that’s the same

Basis of ker f is obviously independent

[unkempt stream]

left module?

[white oxide]

yea

[white oxide]

how do we know this basis exists, could we just take the identity or something which is a linearly independent subset and then presumably extend that to a basis of Ker f

[topaz solar]

Well, do you have that every vector space over division ring has a basis?

@white oxide

[white oxide]

oh right

[unkempt stream]

is this like cardinal arithmetic for the basises

or are you assuming it’s a finitely generated module

[topaz solar]

No?

Well here it’s probably assumed finitely generated since otherwise ya know

Cardinal arithmetic

[unkempt stream]

i don’t know SHIT about cardinal arithmetic

[topaz solar]

Basic of ker f -> extend it -> done

[topaz solar]

Accept choice and just take the bigger one

[white oxide]

thanks you a g

[topaz solar]

gg

[white oxide]

why is not ${x \in X \mid f(x) \neq 0}$ since a basis of $Im \text{ f }$ should be $x \in X - Y$ where $Y$ is a basis of Ker$f$ right

[cloud walrus]

okeyokay

[white oxide]

well I guess it has to be of the form f(xi)

by the definition of image

but I'm still confused shouldn't it also be in V

and not V'

[unkempt stream]

i wonder if notions of galois theory work for division rings

[white oxide]

oh wait you just use the isomorphism $V/Ker$f$ \simeq Im\text{ f }$ right

[unkempt stream]

some automorphism bs basically

[cloud walrus]

okeyokay

[south patrol]

Wait you already have that

Bruh lol

[white oxide]

I'm confuzzled

[south patrol]

I assumed this problem was a lemma for proving that lol

[white oxide]

nah I didn't know how to show that the given set was a basis of Im f so I'm using the actual theorem to show it is 😹

[unkempt stream]

specifically if

Is the fixed point subring of division R under the automorphisms of R a division ring, and finitely generated as a module (or an algebra)?

honestly a massive amount of galois theory can be deduced from [F : F^G] = |G|

[glossy crag]

Good god, he's done it!

This was driving me insane, thanks so much @rocky cloak

[white oxide]

jagr as always the goat

[unkempt stream]

actually

it’s well known that for a field F, F is finitely generated as a module over F^G by |G| elements, where F^G the fixed point field under the automorpgism subgroup G if G is finite

and therefore it’s finitely generated as an algebra

by Noether normalization shows that if F is a field is finitely generated as an algebra over a field H, then F is finitely generated as a module (because you can make it into an integral extension)

is there a general class of rings such that the same fixed point theorem holds, but instead it’s a finitely generated algebra of degree |G|

[wraith cargo]

how does this follow?

[unkempt stream]

every element is equal to the image of a linear polynomial where the evaluation sends the variables to the basis of the module extension

so if x = x_1b_1 + x_2b_2…
Then L(X_1…X_N) = x_1X_1 + x_2X_2… is sent to x when X_n is evaluated to b_n

does that follow?

[wraith cargo]

ah I found the result showing this

[unkempt stream]

?

being a finitely generated algebra over a field is weaker than being a finitely generated module

[wraith cargo]

sorry I was looking throught Eisenbud
I wasn't sure if finitely generated as an R-module implies finitely generated as an R-algebra but that implication is true

[unkempt stream]

WHEN R Is a field

[wraith cargo]

generally

[unkempt stream]

If R is jacobson then all finitely generated algebras that are a field are finitely generated modules tho

zariski’s lemma

but I’m wondering if that [F^G : F] = |G| holds for more general classes where the degree instead denotes degree of an algebraic extension

[wraith cargo]

yeah
more generally you have the iff statement that S is finitely generated as an R-module iff S is generated as an R-algebra by finitely many integral elements