serious-discussion

[timid portal]

Wack

Yeah never actually defined an n-tuple in any context before

[brittle socket]

Either one

Another interesting fact is that you need the function definition of tuples if you want to interpret the arbitrary / infinite cartesian product in a way that similar to how you would interpret the regular cartesian product (in that it is a set of tuples)

[timid portal]

That's kinda cool

[bronze wedge]

Infinite Cartesian products are so much worse than infinite direct sums

[brittle socket]

Why are they so bad in your opinion

They're fine from my view

[bronze wedge]

How do you describe the basis of an infinite Cartesian product?

Direct sums are nice - the basis is the disjoint union of the different bases

But the Cartesian product has no such construction

@brittle socket

[brittle socket]

Wdym by describe the basis of it?

[bronze wedge]

I never said visualize

[brittle socket]

Fyi I have not heard of direct sums yet so yeah

[brittle socket]

Oops typo, corrected it

[bronze wedge]

Describe meaning give a way to write it out

For instance R3

The basis is e1 e2 e3

[brittle socket]

The infinite cartesian product is indeed defined though. Like, it is the set of functions from [so on and so forth] (you get the point). So I don't get what you are getting at though

[bronze wedge]

The Cartesian product is defined - it's just that writing out the basis is a mess

Essentially you are finding a basis for the set of sequences on R

Which you only know exists

But you cant write it out

[brittle socket]

Axiom of choice?

For me, I have no issue with not being able to write it out, as of right now at least. Its just an axiom to me, so I just take it for what it is

Sometimes the AOC can seem quite natural in its formulation imo.
Like you can define the AOC to state that for all relations R, there exists some function f with dom R=dom f. Which I find rather natural and intuitive.

[bronze wedge]

I have no problem with not being able to write it out either - it's just that it's nice to have a specific way to write out the basis

[brittle socket]

Ah true

[brittle socket]

#cats is the best way to spend our data

100% wholesomeness

[midnight iron]

is anyone having any sort of handouts on the topic of matrices and determinants?

[river moon]

look up matrix cookbook

[tender tulip]

okay so random ass shit I noticed

for 2, 4, powers of odd primes, and 2 times powers of odd primes, there’s elements of the multiplicative group that seem to generate the whole damn thing,

why the fuck is said group isomorphic to the additive group n-1

[frail lagoon]

It's a theorem of Gauss that (Z/nZ)* has a generator for those n

But then all cyclic groups of the same order are isomorphic

[frail lagoon]

But it shouldn't be this - it should be that $(\mathbb Z/n\mathbb Z)^* \cong \mathbb Z/\varphi(n)\mathbb Z$

[fathom swallow]

potato

[brittle socket]

\cong

[frail lagoon]

What lol

[brittle socket]

Idk just felt funny to me

[tender tulip]

oh

[tender tulip]

wacky as shit

[frail lagoon]

Why

[tender tulip]

never thought of that before

[frail lagoon]

euler phi of n is (by definition potentially) the order of teh first group

Ah okay

But yeah it's a cool theorem

[brittle socket]

You didn't use \( \)

[tender tulip]

wonder if I can exploit that for galoisy shit

[frail lagoon]

Basically you prove that (Z/pZ)* is cyclic and then from then on it's fairly simple to generalise

[tender tulip]

ye

Can the term primitive root be used for the generators given it’s the roots X^(n-1) - 1 of the finite commutative ring Z/nZ

for example: Z/17Z, then (3)^16 - 1 is a root

[frail lagoon]

Yeah a primitive root mod n is just an integer whose image in (Z/nZ)* is a generator

[tender tulip]

alrighty

this is entirely new number theory stuff to me

[storm sage]

potato is the best

[tender tulip]

lets say we have the polynomial X^5 - 1 over Q. Then we can consider the splitting field which I’ll call Q5. The Q-stabilizers of automorphisms of Q5 is iso to (Z/5Z)* from what I can tell.

[frail lagoon]

No I am not

[bronze wedge]

Then who is?

[frail lagoon]

The fact I am not a maximum does not imply there exists one

[tender tulip]

There exists a maximum for this moment but it continuously changes sporatically so there is no limit

[frail lagoon]

Have you seen any theory of cyclotomic extensions Mizalign

But yes the splitting field of x^n - 1 is a galois extension with galois group isomorphic to (Z/nZ)*

And so yes with n = 5 this is isomorphic to Z/4Z

Basically cause there are 4 primitive 5th roots of unity

[odd narwhal]

cyclotomic extensions are a very well-behaved class of algebraic extensions

very nice

[frail lagoon]

Kummer extensions p cool too

Need to just remember how Galois theory works soon

[odd narwhal]

ey gl with studying lol

[tender tulip]

i haven’t seen anything I’m mostly working on shit I’ve done on paper and read nothing about

[frail lagoon]

Ah fair enough

[frail lagoon]

i mean just been too long ig

But reviewing is always easier yay

[tender tulip]

I’ve proven the fund theorem of galois theory through order-degree inequalities but it doesn’t feel sufficient to me

[odd narwhal]

galois theory comes back quickly (from experience, trust me)

[frail lagoon]

Yeah it's just basically a few lemmas about normal and separable extensions I end up forgetting aha

But I'll be doing it next term and have lots of problems to do so it'll be calm

[tender tulip]

i plan to consider the structure of a commutative domain subring lattice and the automorphism-stabilizer lattice and see what happens to the structure when I progressively add the following conditions:

1. The ring is a field (every ring homomorphism from the ring is mono)
2. For some subfield, every intermediate ring is also a field (algebraic)
3. For said subfield, the extension formed is separable (unable to figure out how to do without calling upon the tensor product)
4. The extension is finite
5. The only subfield fixed by the stabilizer of the aforementioned subfield is the aforementioned subfield (galois)

kind of seeing what restrictions it places

the 2nd condition makes orbits finite for intermediate fields

5th means finitely many intermedite fields when combined with separability afaik

the ultimate result is bijection between intermediate fields and stabilizers of the galois extension but I want to see what conditions ultimately lead to that specifically

idk if there’s a tensor product of fields, but I wonder for two subfields A and B, their intersection being C, if there is a way to quotient it’s direct product ring such that (c,1) ~ (1,c) for any c in C

[bronze wedge]

Fields are rings (which have a tensor product)

[tender tulip]

i originally didn’t really know what I was looking for was a tensor product

i just wanted a way to do the aforementioned equivalence

[tender tulip]

I know this is inefficient but some stuff like this is fun to me

[tender tulip]

can the tensor product of two rings A and B with intersection C be constructed as a quotient ring of A and B’s direct product?

What’s the difference of direct sum and product

it’d have to be over some ideal

[sleek wing]

I'm pretty sure the tensor product is always a quotient of the direct product

cause you just take A and B to be modules over something (not my problem)

[mint canopy]

No not typically

[sleek wing]

am I getting direct product and direct sum mixed up again

the difference is so tiny I always forget it

[mint canopy]

They're the same for modules anyway, doesn't matter

[sleek wing]

ah it's the cartesian product

[mint canopy]

It's a quotient of the free module over the cartisean product

[sleek wing]

yur you take V x W as da basis

[mint canopy]

Exactly

[sleek wing]

so if we take the direct product as the basis

lemme think

I don't think scalar multiplication survives the quotienting

kinda sounds like an apocalyptic event doesn't it

"the quotienting"

[alpine kindle]

it does not

nothing survives the quotienting

[sleek wing]

_ _

[alpine kindle]

it is a massacre

[tender tulip]

o v e r w h a t i d e a l

[sleek wing]

the ideal generated by the multilinear relations?

[tender tulip]

i mean I could just look up the universal property and go from there

[sleek wing]

it's in the construction pretty explicitly

[tender tulip]

hm

i haven’t seen the general construction

[sleek wing]

the perils of self-teaching

[alpine kindle]

hol up how are you quotienting if scalar mult and addition aren't preserved

[deep mango]

Yooniversal property

[tender tulip]

was gonna just see if I can match the quotient universal property and the direct product universal property to the TP’s

thought that might be a lot of work

[alpine kindle]

?

literally just use the tensor universal property

it's literally a commie triangle it's super simple

[sleek wing]

which does not survive The Quotienting ™️

one day my internet will work

and on that day I will call ryc names

ah ha!

@deep mango poophead

[deep mango]

[fringe needle]

r00d

Why wait for that

[deep mango]

He doesnt

[alpine kindle]

@deep mango you are a very nice person

[deep mango]

How dare you

[alpine kindle]

[tender tulip]

huh

For all homomorphisms from A into X, B into X such that f(a)g(b) = g(b)f(a) in X, then there is a homomorphism from X into A tensor_C B

[odd narwhal]

yes

[tender tulip]

Category arrow time

AxB is a quotient of AB, so there's an epimorphism t from AB to AxB

[odd narwhal]

.....what is AB

[tender tulip]

direct product

formatting laziness

[odd narwhal]

AxB is the direct product

if you really can't be bothered to use latex, then do A(x)B for tensor

[tender tulip]

alr

[odd narwhal]

also note that bilinear maps out of AxB are not algebra homomorphisms

they in particular don't respect scalar multiplication (If we consider AxB as a direct product of algebras with the corresponding structure(

[tender tulip]

fuck it's rotated wrong

[mint canopy]

,rotate

[mint canopy]

:)

[tender tulip]

thanks

[mint canopy]

no worries

TeXit's pretty good huh

[tender tulip]

C is a "subring" of A and B (with the injections i_a and i_b), and using the UP of AxB then there's p_c from C to AxB

[alpine kindle]

Miz are you becoming a hsct

[tender tulip]

what

[alpine kindle]

high school category theorist

[tender tulip]

… it’s just easier to work with

[alpine kindle]

it's a meme I don't have any problems

[tender tulip]

I’ve only used like a grand total of 4 categories anyway lol

shit. If C has monos into A and B, then sending an element x of c to (x,0) , (0,x) and, (x,x) in AxB are technically all monos

[bright hill]

Set, Grp, Rng and Vec? :troll:

[tender tulip]

Set, Mon, CMon (cancellative monoids), Grp, Rng, R-mod

[alpine kindle]

that's 6

[odd narwhal]

Don't forget about PokeMon

[tender tulip]

perfect

[tender tulip]

i do algebra, not arithmetic

[odd narwhal]

Also if you're constructing t from t_A and t_B then it is not a C-algebra homomorphism

[tender tulip]

Ah

[odd narwhal]

As I said, it doesn't respect scalar multiplication

Because it's bilibear

[tender tulip]

Oh, h

[odd narwhal]

While AxB has scalar multiplication that goes c(a,b)=(ca,cb)

[zealous garden]

Is there a way to build a group structure on pokemon types

[alpine kindle]

um actually R is an algebraic structure

[tender tulip]

no shit i was joking

[alpine kindle]

yes i know lmao

[odd narwhal]

If t is the canonical surjection this still holds

[tender tulip]

monoid structure where we can map pokemon to types, and consider product of types as a set

Commutative semigroup

but that’s boring

how tf do I keep mixing up semis and monoids bruh

[bright hill]

The set of Pokemons is finite so you can technically just slap a cyclic structure onto it

[tender tulip]

i literally swap them by accidwnt entirely in my own shit

[tender tulip]

i’m going to try to construct something different suited to my own purpose

[zealous garden]

I should've asked about a meaningful group structure

[odd narwhal]

If it's different then it's not a tensor product

[tender tulip]

Basically that’s what I was saying

[odd narwhal]

Then don't denote it by (x)

That's bad notation

[tender tulip]

I’m not going to from this point further

[odd narwhal]

Good

[tender tulip]

Not gonna go back yo prior messages and change ut

let C be a subring of A and B, and let D be a ring such that there is an epimorphism t from AxB to D such that t((1,x)) = t((x,1)) for all x in C. Does the D, t pair exist

[dapper badge]

just took a look at my quantum mechanics homework 😵‍💫 I think this is the end of me

[wooden flax]

qm 👀 👀

[sick burrow]

wait wait wait wait

hold up

so you're telling me

MATLAB

the industry standard software

does not have

a fucking dark mode???

[bright hill]

[zealous garden]

Does that mean darq is banned

[bright hill]

...

I'm not darq as in dark

I'm darq as in DarQ

smh

[zealous garden]

dark is a subset of darq

[neat lintel]

Gently molest you into learning moment generating functions

[safe harbor]

has anyone seen Andrew Dotson's new video, his channel might be taking a weird turn

[storm sage]

solution: don't use matlab because proprietary programming languages suck :)))

[eager reef]

Are you doing physics

[dapper badge]

it's the only physics class I'm taking

school has some requirements outside math so I figured this would be fun

it is not

[eager reef]

Isn’t it just

A bunch of func Ana

[dapper badge]

that's what I thought it would be, but I talked to the prof after class the other day and he said he has to put off talking about Hilbert spaces and all the fun stuff to talk about experiments and motivation so as to not scare the physics students

[eager reef]

[dapper badge]

idk the prof is really cool, we just haven't gotten to any actual math yet but hopefully that should start next week

[fickle remnant]

mathematicians don't care about reality

[eager reef]

Fuck them physics students

QM is just maths

[dapper badge]

all of their courses should be catered to my learning style and tastes

[eager reef]

That’s what I’ve heard from people who do it at least

That it makes a lot more sense once you start regarding it as just maths

[eager reef]

Exactly

[fickle remnant]

sean carroll has a paper where he says the physical universe is just a hilbert space

[dapper badge]

I figure that's why I don't understand any of it yet, once we actually set up the foundational math behind everything I'm sure I'll find it much more digestible

I also really want to talk to the prof about Lie group stuff, I've heard it's actually useful for quantum mechanics so it would be nice to see some applications

[eager reef]

Stop overcomplicating maths by calling it physics

[eager reef]

That sounds neat

[fickle remnant]

it seems that instead of going outside academics will just declare social interaction an exercise left to the reader

[cold needle]

it did feel off

[eager reef]

What is it about

[fickle remnant]

'universities in 2050 be like'

[storm sage]

I don't get this video

this is pretty cringy

[bronze pelican]

I'm doing the hw iassigned my students

[dapper badge]

how goes it

[bronze pelican]

Ok

[wooden flax]

unbased

[wooden flax]

watch dis its good https://www.youtube.com/playlist?list=PLOaEOh8qMwDLoBJinaH3p31edODHdlb93

his corresponding book is also decent https://www.math.columbia.edu/~woit/QM/qmbook.pdf

[dapper badge]

looks neat

[wooden flax]

quantum theory for mathematicians

like charge is just an operator on a representation of U(1)

[dapper badge]

so true

I'm becoming unitary

[wooden flax]

lol

[tender tulip]

t((1,x)) = t((x,1))
t((1,x) - (x,1)) =
t((1 - x, x - 1)) = 0
so (1-x,x-1) is in Ker(t) for all x in C

I don’t think {(1-x,x-1), x in C} is an ideal of A x B if C in A and B

[odd narwhal]

That's cute you think you'll see up the relevant math

[dapper badge]

[safe harbor]

yeah I'm hoping his channel doesn't go in a more political direction

[tender tulip]

I still have no damn clue how to construct a tensor product of rings via quotient on the direct product

[wooden flax]

Bro I'm trying to apply to publix not amazon https://blog.publix.com/publix/6-tips-for-getting-a-job-at-publix/

mfw asking for SSN on the short form

[dapper badge]

I don't think the tensor product is constructed as a quotient of the direct product?

[zealous garden]

omg no

[tender tulip]

I was told it can be

[dapper badge]

I've always seen the explicit construction as a quotient of the free module on A x B as a set

[zealous garden]

it's a quotient on the free product iirc

[tender tulip]

I thought the free product of rings doesn't exist

[zealous garden]

What is the free product btw, does it just turn every possible pair into its own basis vector or something

[dapper badge]

are you using free product in place of coproduct?

[maiden bear]

wew made a mistake

he admitted that

[dapper badge]

you can define the free product of two algebras over a ring

[maiden bear]

it's a quotient of the free module over the cartesian product

[tender tulip]

hm

[dapper badge]

i forget if that's only for commutative algebras

[maiden bear]

wait i misread

you're asking for the TP of rings not of modules

mb

[dapper badge]

but yes, for commutative algebras you can construct the tensor product of the underlying modules which is explicitly a quotient of the free module with basis given by the cartesian product

[tender tulip]

Any ring extension CAN be viewed as a module algebra

[dapper badge]

then there's a natural multiplication structure on this tensor product given by (a \otimes b) (a' \otimes b') = aa' \otimes bb'

I think for noncommutative algebras you have to take a quotient of the tensor algebra

[tender tulip]

Can every R-algebra over a ring R be viewed as a ring containing R

[mint canopy]

Typically, yes. Some definitions of an R-algebra vary lol

What's the definition you're using rn?

[tender tulip]

I was wondering what the general one was

[mint canopy]

So uh, yeah people use several

[tender tulip]

not sure if R-algebra is just an algebra on a module with R as it's ring

THe module's abelian group might be independent from R

[mint canopy]

The most specific one is when R is a field, and we usually require that the field be centrally embedded

[tender tulip]

F in Z(R)

?

[mint canopy]

Yeah that's right

Now I've also seen the definition of an R-algebra as a ring that is also a (left) R-module

And then the multiplications must be compatible

[tender tulip]

I was reading a paper before resorting to the wikipedia article because confusion

[mint canopy]

This does mean that R will be embedded in the ring, but not necessarily centrally

[tender tulip]

looking for a universal property

[mint canopy]

Fair enough!

[tender tulip]

formatting moment

[alpine kindle]

yo tensors

based

[dapper badge]

right, if you look at like any finite ring R as a Z algebra then there isn't an injective map from Z to R

[tender tulip]

Essentially I'm trying to find a universal property of the tensor product of commutative rings A and B over common subring C

Using ring homomorphisms

R-algebra homomorphisms I assume are just ring homomorphisms from extension rings that fix R

f(ax + by) = f(a)f(x) + f(b)f(y) = af(x) + bf(y)

[dapper badge]

yes, R-algebra maps are just R-linear ring homomorphisms

So to make sure I understand, you want to find a universal property of $A \otimes_C B$ in the category of rings?

[fathom swallow]

walter

[tender tulip]

Yes

Basically from there existing a monomorphism from C to A and C to B

I know there's a monomorphism from A and B to the tensor product

... but what about C

a |-> (a,1) [a in A] and b |-> (1,b) [b in B] are monomorphisms from A and B to the tensor product respectively

[dapper badge]

If you want a C-linear map, just send 1 -> 1 \otimes 1

[tender tulip]

but if x is in C, then (x,1) = (1,x) so essentially the square diagram we make commutes

Let me latex

what's the enclosed cross called in latex syntax

[fickle remnant]

ocross

otimes

or something

[tender tulip]

@dapper badge

[dapper badge]

nice latex

[tender tulip]

i love drawing those diagrams

but yeah

[dapper badge]

but for a general ring, the maps C -> A and C -> B need not be monomorphisms?

[tender tulip]

C is a subring of A and B and that’s how I represent that

[dapper badge]

ah right, I forgot about that

[tender tulip]

yeah

[dapper badge]

I mean, if you consider the category of commutative C-algebras as a subcategory of all rings, then this is just the tensor product of C-algebras which already has a nice universal property as a coproduct

idk what you would want to say more generally than that

[tender tulip]

what I mentioned isn’t a universal property afaik

[dapper badge]

what did you mention

[lean pawn]

topology masters is anyone here?

[tender tulip]

it’s just a property, not universal

[dapper badge]

idk what property you're referring to

that it's a coproduct?

[odd narwhal]

I don't think it's true that the tensor product is a coproduct in R-Alg

[dapper badge]

is it not for commutative algebras?

[odd narwhal]

Oh ok apparently this is true in commutative Algebras

Yea you're rifht

It's not in R-Alg tho

Generally

[dapper badge]

right, then you have to do a quotient of the tensor algebra or something

it's bigger

[untold sapphire]

bruh

[odd narwhal]

Ye

[untold sapphire]

that sounds atrocious

yeah that makes sense tho

[odd narwhal]

Oh ok it's obvious why it's true in commutative R-Algs

[untold sapphire]

shin i actually didn't learn this until i studied hartshorne and this is necessary to explain why the ring of functions on the product of varieties is the tensor product of rings

[odd narwhal]

It's cuz maps fron A and B determine a unique map from the tensor product by multiplying (the images always commute cuz comm algebras)

[odd narwhal]

Hmm yea that makes sense

There's a duality going on there

Ok that's nice I didn't know this

But it makes total sense in retrospect

I guess this shows CRing is not additive

Truly a cringe Category

[dapper badge]

very much so

running out of vegetables

but it's been raining all day and I don't want to walk to the grocery store

[tender tulip]

is this still accurate though

[dapper badge]

yeah, that's a commutative diagram

[tender tulip]

But it doesn’t imply it’s a coproduct, just that it’s a commutative diagram

it also doesn’t give any hint to how to construct it

[dapper badge]

but we've been telling you how to construct it

[tender tulip]

free product of R-modules i thought COMES from the tensor product

[dapper badge]

take the free module over C on the basis A \times B and quotient out by the bilinear relations

idk what you mean by free product of R-modules

[tender tulip]

”the free module” ????

[dapper badge]

a free module over a ring R is a module with a basis

for this construction, you take the module which contains a copy of R for each element of A \times B

but yeah, you put the natural multiplication structure on it (a \otimes b) (a' \otimes b') = aa' \otimes bb'

And then you would have to explicitly show that this is a coproduct in the category of commutative C-algebras by showing that it satisfies the universal property

but that isn't particularly difficult to do

oh yeah, @odd narwhal if I learn group/galois cohomology this semester, can we talk about brauer group stuff sometime?

[odd narwhal]

Sure of course. I might know more about it by then too. You might also wanna ask ng or yamin

I'll at least hopefully finish the group cohomo chapter of weibel by then

[dapper badge]

I should probably look at CSA stuff too but idk how much time I'll have

idk how people manage to read during the semester, it gets so busy :(

[tender tulip]

does there exist a monomorphism from The tensor product to the direct product or vice versa

[dapper badge]

I don't think so?

[tender tulip]

(a,b) to (a cross b) or vice versa

[dapper badge]

you'd either want natural maps A <- A \otimes B -> B or A -> A \times B <- B

But I don't see either of those

[odd narwhal]

This generally is not going to be a map of algebras

[tender tulip]

I don’t want it to be a map of algebras, i want it to be a map of rings if it can be

[dapper badge]

.

[odd narwhal]

Why

[dapper badge]

In general I don't think there would be a monomorphism from the direct product to the tensor product because the tensor product can be smaller the direct product

[burnt dune]

@velvet dagger hey

if i have two generating sets

and they generate quasiisometric metric spaces

what can we say about the two groups generated from the two generating sets

do you know

[tender tulip]

is there a ring monomorpism from the tensor product to the direct product

[velvet dagger]

They're two generating sets for the same group

[burnt dune]

oh

no thats not what i meant

but ur right

okay reverse it

like

lets look at the geometry first

like

we cant extract a group

from a metric space right

like u cant go backwards

i just thought it would be really really really really cool

if you can know things

algebraic things

from the "induced" metric spaces especially if they are quasi isometric

[pure sun]

no. basically any example you write down is a counterexample.

Z/2Z (x) Z/4Z is isomorphic to Z/2Z

but there's no ring map at all from Z/2Z into (Z/2Z) x (Z/4Z)

let alone a monomorphism

[tender tulip]

alrighty

[alpine kindle]

tensor product and direct product are usually completely different

[pure sun]

yeah they basically have no relation to each other

the universal property does give you a map from A x B into A (x) B

but it's definitely not a monomorphism

wait nvm i think i have the arrows backwards lol

[tender tulip]

There is naturally a morphism between a coproduct and a product

But one is the the direct product of rings is, the other is the coproduct of R-modules (morphisms that fix R)

[pure sun]

the direct product in the category of R-algebras is the usual product though

[tender tulip]

good point

[pure sun]

ok i seem to have gotten myself mixed up

[tender tulip]

hm?

[pure sun]

i dont think this is true

the arrows don't go the right way

[tender tulip]

This is how I remember it

For a coproduct, morphisms out of the multiplicands imply a morphism out of the coproduct

For a product, morphisms into the multiplicands imply a morphism into the product

[pure sun]

yes

that is correct

so you can map out of the coproduct; let's say you want to map that to the direct product, you need maps from the "multiplicands" into the direct product

which don't exist

[alpine kindle]

it's bilinear tho, not linear

[pure sun]

we're talking about maps of rings

[alpine kindle]

oh

uh

[pure sun]

and yeah that's why i got confused

it's not actually true

i was thinking of the linalg situation and got the arrows backward

[alpine kindle]

wait do you just tensor rings as Z algebras?

[tender tulip]

Obviously, there’s morphisms into the coproduct, and out to the product for the multiplicands so by the universal properties, there should be one both ways actually, right

[pure sun]

yeah; more generally you can tensor any two R-algebras together to get another R-algebra

[pure sun]

no?

let's say you want to map from $A \coprod B$ to $A \prod B$. There are two ways you could achieve this. one way is using the u.p. of coproduct, and to use that you would need maps from A and B into the product, which don't exist. the other way is to use the u.p. of the product, and to use that you would need maps from the coproduct into A and B, which also don't exist

and there's no hope of finding a map from the product to the coproduct, since mapping out of the product isn't really a "thing" from a u.p. perspective

well okay i take that back

there are maps they just aren't nice categorically

e.g. you can project A x B --> A and then follow that with the canonical map A --> A \otimes B

and you can similarly do that for B

but you can't do much more than that

[tender tulip]

hm

alright

i get that

[pure sun]

the arrows just don't line up

yoiu could get lucky in some categories, e.g. there might happen to be morphisms A --> A x B and B --> A x B so you could try to get a map from the coproduct into the product

and those maps do exist in the category of R-modules, for example

(but coproduct in R-mod is not tensor product)

thsoe maps do not exist in the category of R-algebras

[tender tulip]

Alrighty

moving on

separable extensions

[pure sun]

[tender tulip]

I jotted that shit down in my notebook

[pure sun]

i cant imagine what more you have to think about separable extensions lol

i feel like we went through literally everything

like a month ago or whenever

[tender tulip]

I added a lot of revisions to my shit

but the one thing rhat’s stumped me is separable extensions

[pure sun]

what do you want to say about them

[tender tulip]

The notion of an algebraic extension is generically saying that for extension F/H, any morphism from H[X] to F that fixes H has nontrivial kernel. This statement is equivalent to saying that every intermediate ring is a field

or if you’re categorically deranged, a ring being sandwitched between monos between F and H implies any morphism out of the ring is mono

[pure sun]

why are you so compelled by that last fact

[tender tulip]

But like, i can’t find an interpretation for Separable extensions along these “categorical” lines.

[pure sun]

i have literally never used that fact

outside of the first time i proved it as an exercise

[tender tulip]

it’s more important for something else I’m doing

[pure sun]

you shouldn't be thinking categorically here

the category of fields is awful

[tender tulip]

no non-monos

[pure sun]

you're not going to get any valuable statements out of it

if you really like talking about monos

you can say that an extension is separable iff the number of monos into an algebraic closure is equal to its degree

[tender tulip]

i just want to see if I can prove fundamental theorem of galois theory WITHOUT even TOUCHING the polynomial ring

[pure sun]

if you look specifically at the category of algebraic extensions of a field F

well nvm idk what im saying

[tender tulip]

personal challenge I guess

[pure sun]

the standard proof of the fundamental theorem of galois theory also doesn't reference polynomial rings

you just work directly with field extensions

[tender tulip]

how do you define separable extensions then

[pure sun]

i just told you

the number of maps into the algebraic closure of the base field is equal to the degree of hte extension

[tender tulip]

need not be finite, right

[pure sun]

okay fine let's make it finite

L/K is separable if every a in L is separable over K

[tender tulip]

i don’t want to impose the finiteness condition

[pure sun]

an element a in L is separable over K if K(a) is separable over K

and now K(a)/K is finite

I'm not imposing any finiteness condition

[tender tulip]

this sent later than I expected it to

[pure sun]

let L/K be any algebraic extension. this extension is separable if, for any a in L, the number of maps from K(a) into \bar{K} is equal to the degree of a over K

just like algebraicity, separability is something that can be checked on elements

[tender tulip]

degree of a over K is the degree of it’s minimal polynomial, correct

[pure sun]

yes

equivalently, degree of K(a)/K

[tender tulip]

alright

[pure sun]

again, if you don't want to reference polynomials, you don't have to

you can just say "degree of K(a)/K" which is just some vector space dimension

no polynomials in sight

[tender tulip]

alright

[tender tulip]

my insane ass just designated it as the order of an orbit

Nvm

Just now realized that’s wrong

[pure sun]

so, the degree of K(a)/K is equal to the degree of the minpoly, which (if it is separable) is equal to the number of conjugates, i.e. the degree of the orbit of a under Gal(\bar{K} / K)

[tender tulip]

the reason I’m doing all of this is because of how the separability and finiteness conditions impact the lattice structure between subfields and subgroups of the auto-group. The things at a glance don’t seem to coincide and I want to see how adjoining conditions to specific extensions impacts the structure of the lattices

[pure sun]

if the extension is galois then all subfields are separable so there's nothing exciting to say

well more generally subfields of separable extensions are separable

so in that case, "separability" isn't going to tell you anything becuase everything is separable

if the big extension is not separable, then its automorphism group is going to be "too small"

so there will be too many subfields

[tender tulip]

For alg F/H and action of Aut(F) on F, if the only field fixed by the stabilizer of H is H, then the extension is normal and separable is what boggles me a bit

[pure sun]

but what is true

[tender tulip]

Specifically seperability.

[pure sun]

is that if E/F is normal but not separable

then the fixed field of E under Aut(E/F)

call it K

has that K/F is purely inseparable

and Aut(E/F) = Aut(E/K)

[tender tulip]

hm

[pure sun]

(uh wait let me make sure that's true)

okay yeah i think that's true

[pure sun]

yes. what this is saying is that if Aut(F/H) fixes more than just H, then the extension is either not normal or not separable

[tender tulip]

so what interested me is how, we can generally think of automorphisms of abelian structures, and form group actions and lattices, and sometimes find where fix(stab(H)) = H for some substructure. But for fields, with F/H being finite and separable, then a bijection is formed for all fields that contain H in the lattice, and all subgroups of the stabilizer. This specific fact originally made me want to find the specific constraints on this lattice for it to occur

[pure sun]

basically, Aut(F/H) "should" permute the roots of irreducible polynomials, and the size of the orbit of an element should be equal to the degree of its minpoly, i.e. the number of conjugates

so if the orbit is too small, there are two reasons why that could happen

[tender tulip]

finiteness of the extension and separability of the extension implies finite stabilizer, thus finite subgroups

[pure sun]

first, maybe not all of the conjugates of the element x are in F. so, x has lots of conjugates, just not inside F. that means F isn't normal over H

[tender tulip]

I think

[pure sun]

second, maybe x just doesn't have enough conjugates at all, i.e. the minpoly doesn't have as many roots as it should. that means that x isn't separable

[pure sun]

idk what you mean. the automorphism group of a finite extension is finite and so all subgroups are finite

[tender tulip]

Yeah, that’s what I was saying

but more specifically that there are finitely many subgroups

[pure sun]

again... finite groups only have finitely many subgroups

[tender tulip]

i know. That’s the ultimate fact I was going for

[pure sun]

are you actually interested in galois theory

or are you just interested in lattices

[tender tulip]

in a way what galois theory does to them

like I KNOW the FToGT because i’ce worked through it so I want to see it more in depth

[pure sun]

galois theory doesn't "do anything" to lattices

[tender tulip]

i’m thinking of a better way to phrase it. I just want to look at it through a different perspective

[pure sun]

every field extension has a lattice of subfields, and its automorphism group has a lattice of subgroups. there are maps going in both directions. ftogt gives necessary and sufficient conditions for those maps to be bijections

that is the perspective

[tender tulip]

the condition is finiteness and that fix(stab(H)) = H for subfield H, or for some subgroup, stab(fix(G)) = G for subgroup of gal(F/H)

[pure sun]

i dont think you actually need finiteness; it's also true for infinite extensions, you just have to replace "subgroup" with "closed subgroup"

which, in some sense, is already in the finite version, since all subgroups of the finite galois groups are closed

also, I was going to say that the condition was "normal + separable"

which is equivalent to the things you wrote

you can investigate directly (as I have already tried to start explaining) the direct link between normal+separable and the fix(stab(H)) = H condition

[tender tulip]

only when algebraic, correct

otherwise normality and seperability mean nothing

[pure sun]

yes, so why would you even bring up that condition?

[tender tulip]

if there was anything more to it

[pure sun]

this entire conversation has been about algebraic extensions because these conditions are meaningless for nonalgebraic extensiosn

[tender tulip]

yeah, idrc about normality right now, just separability

[pure sun]

which makes no sense for nonalgebraic extensions

so if you care about separability, then you should only be talking about algebraic things

which is what we have been doing

for the past 30 minutes

[tender tulip]

For A/B/C, A/C being galois implies A/B being galois, so if Fix(stab(C)) = C, then Fix(Stab(B))= B

[pure sun]

i feel like im having a stroke

yes that is correct

but why is it relevant right now?

i honestly don't know what your goal is right now

it seems like you're just listing true facts

[tender tulip]

just seeing if my assumptions are correct

[tender tulip]

this i gotta prove though

just the fix/stab part

[tender tulip]

ah, that makes sense

[pure sun]

the way to prove it is to prove that that condition implies galois and then use galois theory things

[tender tulip]

another thing I think is true is that if a is conjugate to b (same min poly) then their simple ring F(a) is iso to F(b)

wait a second it’s literally the same quotient

F(a) iso to F(b) iso to F[X]/<min(a)> = F[X]/<min(b)>

[pure sun]

im going to go now

https://en.wikipedia.org/wiki/Galois_connection

if you like lattices, read this

[fair mural]

is mizalign not reading books again

[velvet dagger]

Is "again" an appropriate term to describe a continuous process?

[arctic grove]

Lmaoo

[neat lintel]

why are there so many trolls in the help channels all the time lol

what do they gain 💀

[open aspen]

amusement

[neat lintel]

tbh tho

sometimes i can't even tell

if the teacher just gave them some really weird question

and they're actually genuine

[tender tulip]

Let A/B be an algebraic field extension.

An element x in A is separable iff the degree of the minimal B-poly of x is the same as the number of conjugates?

I assume minimal polynomials of separable elements cannot have multiple roots

[tender tulip]

bro you just posted that in #math-discussion

[velvet dagger]

Mizalign by def yes

I'm gonna ping this person

@steel cloud read #❓how-to-get-help, in particular do not post your stuff in discussion channels and do not multipost

[tender tulip]

For an extension A/B with separable element x in A the following are equal:

1. The number of B-conjugates of x
2. The degree of x's min B-poly
3. The orbit of x under Aut(A/B) (or Stab(B) in Aut(A))
4. The number of monomorphisms from A to any splitting field of x's min B-poly
5. Degree of the extension B(x)/B
6. Number of isomorphic intermediate fields to B(x)

I know 6. and 4. are equivalent by literal definition of the image.

Now I just gotta choose which one to use lol

[velvet dagger]

Uh, hold on wait

Something's bugging me about this

So if we're in characteristic 0 then everything is separable right?

[tender tulip]

ye

[fathom swallow]

Sloth King Daminark

[tender tulip]

Oh. It has no other conjugates in that field.

[velvet dagger]

Yeah

[tender tulip]

shit.

[velvet dagger]

Sorry fam

[tender tulip]

I'm trying to think of the best way to define separability without referencing the minimal polynomial

[tender tulip]

This implies the extension's automorphism group is trivial, correct?

[velvet dagger]

Yeah

[tender tulip]

hmmmmmmm

[velvet dagger]

The sort of tricky situation in your case with the 6 points thing is

If you have a field extension K/F

You tend to think of a few kinds of objects

You either think of automorphisms of K fixing F

Or you think of maps from K into an algebraic closure of F

[tender tulip]

you can replace the algebraic closure with just an element's splitting field and it'll be fine too, right and just consider it element-wise

[velvet dagger]

So when you say "an element" you need to specify this element generates K

Which you can always find, this is something called the primitive element theorem

[tender tulip]

I thought iff there is finitely many intermediate fields

[velvet dagger]

I'm only looking at algebraic stuff for now

[tender tulip]

also "finitely many intermediate" fields means up to iso right

[tender tulip]

ah

[velvet dagger]

Not up to iso, in general when you're dealing with one object inside of another you actually care about how things sit inside

(e.g. I don't think of Z as having 2 subgroups up to iso, rather I actually care about the difference between 2Z and 10Z)

[tender tulip]

oh.

[velvet dagger]

To elaborate on why

2Z is isomorphic to 10Z

But Z/2Z is not isomorphic to Z/10Z

[fringe needle]

You care more about the embedding than the subgroup itself

Same for the fields

[tender tulip]

alr

Ultimately, my goal is to phrase it with automorphisms

Because the weaker condition of galois-ity (with algebraicity) being fix(Aut(F/H)) = H can be phrased WITHOUT EVEN TOUCHING any idea of a minimal polynomial

[velvet dagger]

The tricky thing about separability by itself is that without normality in principle you can get a smaller automorphism group than the degree of the field extension

[tender tulip]

That whole bullshit is what annoys me. Idk how to say using purely automorphisms (stabilizers and fixed fields) and algebraicity wether an extension is separable

you can tell when it's separable and normal, and just normal, but not separable

[velvet dagger]

I might have my brain on the blink right now but this might be kind of a dead effort

[tender tulip]

yeah.

[velvet dagger]

Like here's the way I picture it

[fringe needle]

The only theorem I can think of in that direction (not using min poly) is characterizing perfect fields

[velvet dagger]

In general we have two inequalities

[fringe needle]

In terms of characteristic and K^p

[velvet dagger]

Let's say E is an extension of F

And let K be an algebraic closure of F containing E

Then normality says that every embedding E->K has image lying in E

And separability says that there are [E:F]-many embeddings

Together this tells you that [E:F] = |Gal(E/F)|

Does that picture kinda make sense?

[tender tulip]

yeah

[fringe needle]

In which situations do we have this but not separability?

[tender tulip]

I've seen that before

[velvet dagger]

Like it's sorta saying we have two inequalities

[tender tulip]

My entire goal of the endevour is to see if there's "another way to do it"

[velvet dagger]

|Aut(E/F)| \le |Embeddings E->K| \le [E:F]

And the point is that separability is that the second condition is an equality, normality is that the first is

[fringe needle]

So from this we should expect normality is required to link separability to automorphisms?

[velvet dagger]

At least using the size of the field extension

[tender tulip]

If A/B is algebraic, then the orbit of any element by Aut(A/B) is finite

Orb(x) = [Aut(A/B) : Aut(A)] actually by orbit stabilizer i think

Let A/B be an algebraic extension, x in A, with minimal poly p(X) in B[X]
let f be in Aut(A/B), then, and p(x) = 0, so f(p(x)) = 0, and p(f(x)) = 0, and granted the minimal polynomial has only finite roots (less than it's degree), then the orbit of x is finite.

[Aut(A/B) : Aut(A)] ≤ deg(min(x))

[velvet dagger]

Do you want something which I'm not totally totally sure is correct but might be and if so I think it's what you'd want?

[tender tulip]

sure

[velvet dagger]

So let E/F be a separable extension, let's say generated by x

[long epoch]

I dreamt about this

[velvet dagger]

Or hmm

This might not be what you'd want sadly

I was gonna say

Consider other roots of the min poly that are also in E, separable might be equivalent to Galois group acting transitively on these guys

But this still references minimal polynomial in some capacity because you're asking for the roots of it

[tender tulip]

Now I'm wondering when Aut(F/G) is a normal subgroup of Aut(F)

[velvet dagger]

First off eww G

Second... if everything in sight is Galois

Then you have the main theorem of Galois theory which would say that this holds iff the smaller field is itself a Galois extension

[fringe needle]

Apparently there’s a neat separability condition using a bilinear form called the trace form

[velvet dagger]

Oh hmmmmm

[fringe needle]

Namely separable iff trace form is non degenerate for finite degrees

[velvet dagger]

This is interesting actually lemme think about why

[fringe needle]

Yeah I’ve never seen that before either

[tender tulip]

,, $Aut(F/H) \trianglelefteq Aut(F) \Leftrightarrow F^{Aut(F/H)} = H \
\forall g \in Aut(F/H), f \in Aut(F) [f^{-1} \circ g \circ f \in Aut(F/H)]\
f^{-1} \circ g\circ f(x) = x \forall x \in H \
g(f(x)) = f(x) \
\forall x \in H, f \in Aut(F)[ f(x)\in F^{Aut(F/H)}]

terrible spacing and error moment

huh

HMMMMM

yeah I get it now

[neat frost]

,, Aut(F/H) \trianglelefteq Aut(F) \LeftRightArrow F^{Aut(F/H)} = H \
\forall g \in Aut(F/H), f \in Aut(F): f^{-1} \circ g \circ f \in Aut(F/H)\
f^{-1} \circ g\circ f(x) = x \forall x \in H \
g(f(x)) = f(x)

[fringe needle]

Am I thinking about this dumb Dami? Eigenvalues of the multiplication endomorphism only exist if the element lies in the base field right?

[neat frost]

Puts it in align

[velvet dagger]

Hmm, that sounds right?

[fathom swallow]

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

[velvet dagger]

M_x is multiplication by x

Then xy = M_x(y) = \lambda y

[fringe needle]

Ahh I don’t like when eigenvalues don’t exist I get confused

[velvet dagger]

lambda the eigenvalue

Divide by y, oh wait gg

This will likely be choosing the right basis sorta thing

[tender tulip]

idk where to go from here

[velvet dagger]

Like first let's start off with E/F separable. By primitive element theorem E = F(alpha)

So we have the basis (1,alpha,alpha^2,...,alpha^{n-1}) where n = deg(E/F)

[tender tulip]

why are we using primitive element here

the extension need not be finite, just algebraic

[velvet dagger]

I don't care about transcendental things

At this point I'm doing something different

[fringe needle]

The theorem I brought up is for finite extensions

[tender tulip]

is a galois group always normal in the automorphism group of the extension field

[fringe needle]

I have something dami i think

Like suppose separable extension L extending K

And a in L

Take all the other non zero roots of its min poly

Take their product call it b

Then the multiplication morphism associated to ab has non zero trace

So trace form is non degenerate

[tender tulip]

hm

There's also something with the tensor product i think

[fringe needle]

That should be one direction if I didn’t do it wrong

[velvet dagger]

Oh wait a second

[tender tulip]

I think if $L/K$ is separable then for any algebraic extension $V/L$ the ring $V \otimes_K L$ has no nilpotent elements

[velvet dagger]

So so so

[fathom swallow]

Mizalign

[tender tulip]

i read that in one of the papers I went over about tensor product of algebras

or something like it

[fringe needle]

Huh that’s neat

[velvet dagger]

So let's say f is the min poly of alpha

f(x) = x^n + c_1 x^{n-1} + ... + c_{n-1}x + c_n

[tender tulip]

I also want to avoid tensor product because that's, once again, constructing a ring and using it's universal property to show properties of what you're working with

[velvet dagger]

Then the trace of alpha is c_1

[fringe needle]

It is?

Huh I didn’t know that

[velvet dagger]

Just use the basis

[fringe needle]

Makes sense though

Yeah

[velvet dagger]

(1, alpha, alpha^2, ..., alpha^{n-1})

Then the first column will be, call it e_2

Where e_i is the col vector whose ith entry is 1

[fringe needle]

I never really did anything with the multiplication endo is all

[velvet dagger]

Second column will be e_3

etc

Last column will be alpha^n written in that basis

[fringe needle]

Yeah I get that

[velvet dagger]

So I guess really the trace is -c_1

But then here's the kicker

If alpha is not separable

Then the min poly is really a function of x^p

Where p is the characteristic

[fringe needle]

Hmm right

[velvet dagger]

In general if you give me an irreducible polynomial, it's separable iff it's coprime to its derivative

But then its derivative is lower degree so the only way it's not coprime is if derivative is 0

[tender tulip]

Because which is easier:

F/G is algebraic

1. Every intermediate ring is a field
2. Construct the poly ring, use it's universal property, show every homomorphism guaranteed by this universal property has nontrivial kernel.

Yeah these are equivalent, but it feels like trying to explain Zorn's Lemma using the Vector Space Basis existence theorem

[velvet dagger]

So (a) we gotta be in char p, and (b) you can't have terms that aren't powers of x^{p^n} or else the derivative is not zero

But if we're a function of x^{p^n}

[fringe needle]

That doesn’t really shed any light on the automorphisms side of things though

[velvet dagger]

Then -c_1 would be the coefficient of x^{p^n - 1}, which is 0

So consider F(alpha) as an extension over F. If alpha is inseparable we're in char p, so Tr_{F(alpha)/F}(1) = dim(F(alpha))

But that's deg of min poly of alpha

[tender tulip]

let me restate what I am actually looking for:

How can we define separability WITHOUT constructing or proving the existence of any “external” rings/fields to the extension (e.g. further extension fields, polynomial ring, field tensor product ring).

[velvet dagger]

Which is a power of p, aka it's 0 mod p

[fringe needle]

At least what we did with the bilinear form doesn’t use external rings or fields

[velvet dagger]

And then we just showed Tr_{F(alpha)/F}(alpha) = 0

So actually

Trace map is straight up fuckin 0

Or wait

No no no

Hold on

[fringe needle]

That’s not the trace we want dami

[velvet dagger]

It's not clear that trace of alpha^2 and all is 0

[fringe needle]

We’re looking at trace of ab for all b

[velvet dagger]

That's the trace I want idk about what trace you want

[tender tulip]

i still wonder if a lot of galois shit can be extended to division rings in some regard

[velvet dagger]

Well, what I gave shows what you gave

[fringe needle]

At least the theorem i mentioned uses that trace

[velvet dagger]

If I can show that like

Trace is just the 0 function in the inseparable case

That hard carries what you're talking about lol

[fringe needle]

Yeah fair enough

[tender tulip]

i do not know what trace is generally

[fringe needle]

Should give the other direction yeah

[fringe needle]

Trace of an element is the trace of the multiplication endomorphism

Defined by phi_a(x) = ax

[tender tulip]

ah

i’m just going to get back to separability later

[fringe needle]

What are you learning from mizalign?

[tender tulip]

Many different papers, excerbts from D&F

[fringe needle]

I see

[tender tulip]

i printed a LOT of shit out

[fringe needle]

I should do the d&f galois stuff at some point cause the bit of Galois theory I did in school was very bare bones

Like I’m not at all comfortable with working with embeddings and shit

[tender tulip]

i’ll ask again (sorry)
If L/F is galois, then Aut(L/F) is normal in Aut(L) right

[velvet dagger]

So Aut(L) is Aut(L/prime field) right?

[tender tulip]

? Just the automorphism of the extension field

[velvet dagger]

Yes but like

[fringe needle]

You need F/prime field to be galois right? Or something like that

[ocean marten]

guys i'm kind of confused, but is the reason the GCF of 6x^4 - 14x^2 is 2x^2 is bc when u change the minus sign to plus the 14x^2 becomes positive?
i just need confirmation, thanks

[velvet dagger]

Automorphisms send 1->1

So they fix the prime field always

[fringe needle]

I can never remember fundamental theorem

[velvet dagger]

Narwhal should be orrect

If not then we're both incorrect but in that case we're vibing with each other and that's what's most important

[fringe needle]

Lmao facts

[tender tulip]

i mean, sure. I was thinking arbitrary char.

[velvet dagger]

Prime field doesn't mean p

[tender tulip]

Q

[velvet dagger]

If char 0 then the "prime field" is Q

[tender tulip]

i was thinking Q, or F_p^n

[velvet dagger]

Okay look

[fringe needle]

The point is for fundamental theorem you can say that Aut(K/F) is normal in Aut(L/F) for F<K<L given some galois condition on the intermediate extensions

[velvet dagger]

If F is a field

Then F has a prime subfield

Which is either F_p or it's Q

The automorphism group will always and forever fix that subfield

[tender tulip]

I guess then yes

[velvet dagger]

So Aut(L) is basically Aut(L/prime subfield)

[tender tulip]

yes.

i like viewing it as a stabilizer of the group action by Aut(L) but that’s a bad way of interpreting it

[velvet dagger]

Now, at minimum if shit's finite over the prime subfield then whether Aut(L/F) is prime in Aut(L) = Aut(L/prime subfield) depends on whether F is Galois over the prime subfield

[fringe needle]

normal*

[velvet dagger]

Err yeah

Well hold on separable is given in the case that things are finite over the prime subfield

Because char 0 or finite field

So get fucked I'm still right

[tender tulip]

so generally Aut(F/L) is not normal Aut(F/F_prime)

[velvet dagger]

Yeah

[fringe needle]

No lmao I meant Aut(L/F) normal in Aut(L)

[velvet dagger]

Oh oh

[tender tulip]

ah

[velvet dagger]

Sadge-ittarius Miz

Also Narwhal

Regarding trace form stuff

[tender tulip]

i wonder for transcendential extensions A/B/C that A^Aut(A/C) = C implies A^Aut(A/B) = B is still true

[fringe needle]

Yeah?

Ew transcendental extensions

[tender tulip]

or just, any triple-extension of fields

[fringe needle]

I don’t think it’s true?

[tender tulip]

I still haven’t proven this for algebraic ones w/o using transitivity of separability and normality

[velvet dagger]

I'm thinking like, if alpha is an inseparable element, does the same hold for its powers?

[fringe needle]

When you say inseparable do you mean not separable or purely inseparable

I have French terminology

[velvet dagger]

I think "not separable" cuts it here

[fringe needle]

Well at the very least you can say that trace of a^k is still 0 by additivity right

No nvm

I should look over my definitions before I speak lol

[tender tulip]

how should I go about proving this in the algebraic case

hopefully without polynomials and just using the subring-is-field property if I can

i’ll try contradiction first

[fringe needle]

Oh doesn’t F_p(t) / F_p(t^p) answer your question @velvet dagger

t is inseparable but t^p is very much separable

[velvet dagger]

Well that's not a problem for me because that's in the subfield

[fringe needle]

Yeah that’s fair

[velvet dagger]

Like I've shown that Tr(1) and Tr(alpha) are both 0

Okay so

If you give me a tower of field extensions

Then well you have tower law right?

So if I know trace is 0 for totally inseparable extensions that should cut it

[fringe needle]

Well haven’t you already shown that?

[velvet dagger]

Yup

So yeah we know it for any non-separable extension

Because we know it for totally inseparable ones

[fringe needle]

Could you explain the tower law argument?

Not sure I got it

[velvet dagger]

So if E subset F subset K

Then Tr_{K/E} = Tr_{F/E} o Tr_{K/F}

[fringe needle]

Hmm right

[velvet dagger]

I think I remember this from Neukirch

[fringe needle]

So then you use separable - purely inseparable decomposition or something?

[velvet dagger]

Yea

[fringe needle]

Neato

I think that’s the first time I see that decomposition be useful

[velvet dagger]

Good stuff

[pure sun]

This is not true btw

Writing down counterexamples is kinda hard but I think C(t)/C is one.

There are intermediate fields that C(t) is not galois over

[tender tulip]

alr

I wonder if the whole intermediate-ring-field shit has something to do with the fix(Stab(B)) = B thing

possibly images of subfields under automorphisms or the like

[pure sun]

No

Because the “whole intermediate-field-ring-shit” is a property of algebraicity

[tender tulip]

it's equivalent.

[pure sun]

And general algebraif extensions dont satisfy the other condition

I said this yesterday but idk why you are so compelled by the intermediate field thing. Its really not anythint special

[tender tulip]

For alg extensions A/B/C, C^Aut(A/C) implying B^Aut(A/B) isn't always true?

[pure sun]

That is true…

[tender tulip]

how do I prove that (without polynomials)

[pure sun]

I feel like we had this exact conversation yesterday

You can prove the fundamental theorem of galois theory without reference to polynomial rings

[tender tulip]

That was about separability

[pure sun]

A is always galois (separable and normal) over C^Aut(A/C), so if that’s equal to C then it means that A is also normal and separable over any intermediate field, like B

[odd narwhal]

You can't really avoid talking about polynomials when normality and separability are properties determined by minimal polynomials

[pure sun]

I mean you kind of can

A finite extension is separable if the number of embeddings jnto the algebraic closure is equal to the degree

[odd narwhal]

Ok sure

[pure sun]

And it’s normal if it’s fixed (as a set, not pointwise) by all automorphisms of the algebraic closure

But i agree that those arent the optimal phrasings

[odd narwhal]

Sure, but in practice this is just restating that min polys have the correct properties

Maybe moreso for normality

[pure sun]

Of course any two definitions should be equivalent haha

[odd narwhal]

Well yea I just mean that trying to avoid Polynomials purposefully seems like an exercise in futility

[pure sun]

Im just saying that you can try to reason about separability without referring to polys

Oh yeah

I 100% agree

Just because you can try to reason that way doesnt mean you should lol

[sick burrow]

Geometry 😠

I love proving obvious facts in annoying ways

So fun so fun so fun so fun oh boy am I having fun yet

[zealous garden]

prove the pythagorean theorem algebraically

[sick burrow]

Twivial

I feel like a bad tf2 soldier main

I can't do anything without my black box

[ancient flame]

can you prove pythag through the distance formula

[velvet dagger]

You guys want a proof that n choose k is equal to n choose n-k?

[ancient flame]

not really

[zealous garden]

how do you get the distance formula without pythag

[ancient flame]

metric spaces?

[velvet dagger]

Problem is, Pythagorean theorem is the statement that you should define the distance between two points as l^2 norm

Basically that "irl" distance is l^2 rather than l^7 or something

[zealous garden]

what kind of proof

[ancient flame]

oh ok

is the only proof of pythag that pictoral proof?

[velvet dagger]

The kth betti number of the n-torus is n choose k

[ancient flame]

there have to be others

[velvet dagger]

Apply Poincare duality

[zealous garden]

I know an easy one with the dot product

[ancient flame]

ohhhhh ok

carry on with your combinatorical bs

:p

[velvet dagger]

Again the fact that dot products correspond to "irl" geometry is a consequence of Pythagorean

[sick burrow]

combinatorial

[ancient flame]

oh I thought there was another c there

[ancient flame]

bruh

[sick burrow]

Me when topological combinatorics

[velvet dagger]

There are lots of proofs of it

[zealous garden]

what

[ancient flame]

oh I saw some on wikipedia

[velvet dagger]

But not the linear algebra constructions

[sick burrow]

What does proving the Pythagorean theorem even mean then

[velvet dagger]

If you start from the definitions in analytic geometry and define distance

Then it's trivial by definition

[ancient flame]

bruh

[sick burrow]

Right but saying it gives you the "true" distance

[zealous garden]

I start with the dot product of orthogonal vectors being zero and equal to the norm squared for a vector with itself

[sick burrow]

Isn't the L^2 norm the only one induced by an inner product

[zealous garden]

that's enough for pythag

[velvet dagger]

No

[sick burrow]

The only L^p norm that is

[velvet dagger]

Wraith: the point is you're trying to say "why is d(x,y) = sqrt(blah)" the distance between my fist and gmod's face for just saying bruh without having a real response

Okay that's true now gamma

[sick burrow]

Yeah that's what I meant

Obv

[ancient flame]

bruh

[zealous garden]

this message has only confused me

[velvet dagger]

Obv ofc

[sick burrow]

So that's sorta the reason why the L^2 norm is the "true" distance

[velvet dagger]

Wraith here's the thing

[sick burrow]

But like is the statement that distance in the actual world is L^2 even a statement in math

And not a statement in physics

Like what do we even mean here by Pythagorean theorem

[velvet dagger]

Honestly I don't know how you like, go down into the woods here

It's subtle

I'm fairly certain they mostly worked with some slightly more rigorous version of the "irl idea" of length

[sick burrow]

You mean in the original sense of the theorem

[velvet dagger]

Yeah

[sick burrow]

The way the Greeks would have understood it

[velvet dagger]

Like I'm fairly certain you have the notion of two things being congruent

[sick burrow]

Congruence was a thing yeah

In the sense of "you can take one to the other by a rigid motion'

[velvet dagger]

And I'm not sure how it's defined but from it you can at least talk about something being twice as long as something else

[sick burrow]

Hold on I have a copy of Elements somewhere

[zealous garden]

do you need pythag for a metric tensor

[sick burrow]

Pretty sure we went over the Pythagorean theorem in its original form in my geometry class

[velvet dagger]

Looking this up on Google

It seems like you have to take as axiomatic that the area of a 1x1 square is 1

In this setting

So yeah they had to take some reasonable "irl" notions for granted before they could do things and eventually hit Pythagorean theorem

And once you make the definition of things in coordinate geometry

Pythagorean theorem is what tells you "Yeah you should probably define the distance this way"

[zealous garden]

yeah because we had to take a notion for granted