groups-rings-fields

[vestal snow]

wait wait

what's phi?

[next obsidian]

Since it’s a colimit of im(a)(U)

[vestal snow]

or did you mean a?

[next obsidian]

Yeah

In my notes I had It as phi lol

[vestal snow]

np let's go with phi

[next obsidian]

Show that im(a_x) also satisfies the property of being the colimit

aka can play the role of stalk of im(a)

[vestal snow]

okay

[next obsidian]

Now one thing to note here is that if what I did isn’t cheating

So im a is actually a sheafification of some presheaf

[vestal snow]

yes

[next obsidian]

The adjunction of the forgetful functor and the sheafification functor means it suffices to show that this thing is the colimit of the presheaf version of im a

[vestal snow]

wait

[next obsidian]

I think, I remember asking my prof about this

Yeah let me think about this harder so I can explain what the hell that means

[vestal snow]

so the colimit of presheaf = colimit of sheaf in this case?

[next obsidian]

I think that’s the idea

Since it’s a left adjoint to forgetful which preserves colimits

The only issue I see now is that

[latent anvil]

is this LAPC?

[next obsidian]

Yes but

[vestal snow]

Is it bad that I don't know much about the forgetful functor?

I know it takes away structure

but thats it

[next obsidian]

Only issue i see is that the colimit itself isn’t a sheaf so it’s sort of eh

It’s literally that

It says “okay you were a sheaf, pretend you’re a presheaf”

It makes it “okay” to have a map from a presheaf to a sheaf

Because that map is actually mapping to the sheaf, as a presheaf

Oh okay so

This is what I argued, I guess Shamrock can pipe in if he thinks it’s okay

So I want to show that im(a_x) is the stalk of im(a)_x

What I noted was that, for a presheaf F, F_x is iso to F^+_x

So it suffices to show that im(a_x) is the stalk of the presheaf version of im(a)

Since then the iso to the stalk of the sheaf im(a) gives it a structure as the colimit of that

The thing about forgetful functor was used for showing f_* and f^-1 were adjoints

[latent anvil]

pls do not introduce even more identifications

[next obsidian]

This is the only way I know how to do this

[latent anvil]

"im(a_x) is the stalk of the presheaf version of im(a)" pls

[next obsidian]

I don’t know how you did this

[latent anvil]

I'm just saying track the isomorphism between the stalk and the stalk of the sheafification

[next obsidian]

Oh, sure

Yeah to show the final triangle commutes you need that but

[latent anvil]

I also forget, I might have it typed up

Sorry I was doing analysis

[next obsidian]

First I think you want an abstract iso

Then show it makes the triangle commute

So formerly edible banana, this is sort of why this problem made Shamrock groan lol

It gets really messy, and maybe there’s a nice way to do it but I sure don’t know it

[vestal snow]

Maybe its time to email my homie Cais

[next obsidian]

That might be a good idea haha

Definitely try and think about it

[latent anvil]

who's cais?

[vestal snow]

Oh I will

[next obsidian]

I think hidden in this deluge of text the idea is there

[latent anvil]

the thing is some people will see this convo and be like

[vestal snow]

My algebra prof

[latent anvil]

shamrock and chmonkey are overcomplicating this!

[vestal snow]

and the guy who wanted me to learn AG in 15 days lol

[latent anvil]

oh

That's bad

imho

good luck

[next obsidian]

Shamrock do you remember

That pdf that umm Max sent with the solution

[scarlet estuary]

wow i just read this convo, shamrock and chmonkey are overcomplicating this!

[latent anvil]

oof

I've been owned

[next obsidian]

With the solution to that one Hartshorne problem

[latent anvil]

namington you do algebraic k theory right? So at some point you had to learn basic ag and the basics of sheaves

[next obsidian]

Which was solved by “universally Japanese”

[scarlet estuary]

but uh what are we overcomplicating

[latent anvil]

Add a third chef to the kitchen

It's not overcrowded enough in here

[next obsidian]

Cais was the guy who wrote that pdf

[latent anvil]

oh lol

[vestal snow]

universally japanese?

[latent anvil]

Namington were showing taking stalks commutes with taking images

but stupid

[next obsidian]

Commutative algebra thing haha

[scarlet estuary]

oh ew

[latent anvil]

The text used an =

[next obsidian]

It says that like all finite algebras over it are Japanese

[latent anvil]

we're trying to explain what = means

Lmao Alex

[scarlet estuary]

fuck that i dont want to deal with this at 1 am

[next obsidian]

😏

[latent anvil]

yeah right??

[scarlet estuary]

take my moral support

[latent anvil]

same

[vestal snow]

I will think about it a bit

[latent anvil]

Definitely spend some time thinking about it on your own

like

[scarlet estuary]

= in AG has an implied "up to shit we dont care about"

[vestal snow]

before giving up and playing chess

[latent anvil]

Try to understand wyy the = doesn't quite make sense

chmonkey I got very annoying in manifolds last year

[next obsidian]

My advice is show that the two are iso abstractly first then try and show that triangle commutes. However much progress or lack of progress you make will be valuable

[latent anvil]

Because there's a ton of identifications we make with the tangent space

That are ""canonical""

and every time we got a new one

[next obsidian]

Yeah haha, I talked to someone and he was like “bro that pisses me off”

[latent anvil]

Especially with lie algebras

I was like "okay so is the compatible with the others"

[next obsidian]

Are they?

[latent anvil]

And then lee would be like "wait literally one minute I'm getting there"

yeah ofc otherwise they wouldn't be canonical :^)

[next obsidian]

[vestal snow]

Chmonkey can you send me a screenshot of your notes on this?

[next obsidian]

Yeah

[latent anvil]

Lie(G) can be identified with the tangent space of G at the identity

but also if M is a submanifold of N and p is in M, we have this canonical embedding T_p M -> T_p N

but also lie is functorial so if H is a subgroup of G, we get a map Lie(H) -> Lie(G)

and the rectangle u want to commute commutes

all the identifications are compatible

Manifolds are blessed

amen

I found my solution to the problem

[next obsidian]

Did you show it commutes with inclusions and stuff

[latent anvil]

warning, I havent looked at this in a while

no idea lol

yes it looks like I did

kind of

I showed some kind of niceness of the isomorphism

no i did not

[next obsidian]

Haha me neither!

[latent anvil]

I decided on a different meaning of = that day

[next obsidian]

I didn’t realize that’s what I actually wanted. Same

[latent anvil]

I showed it was compatible with the maps from (im phi)(U) for each U

[next obsidian]

My meaning was “this is defined by universal property so = means it has that universal property”

[latent anvil]

yeah lmao same

[next obsidian]

After thinking for like 6 hours and going

[latent anvil]

ughhh

[next obsidian]

Wtf

[latent anvil]

but obviously it should be about embedding into G_p

right??

[next obsidian]

I think?

[latent anvil]

@vestal snow not a full solution but very close

I showed that = was nice but nice in a different way lmao

Okay I'm signing off

[vestal snow]

@latent anvil Thanks!

[vestal snow]

Any hint for this?

I don't think there was any theorem introduced which could be used to conclude that a given set is open

And since X is an arbitrary topological space, I don't know what the open sets look like

[latent anvil]

okay so one useful way to think about openness is that U is open iff it's a neighborhood of each of its points

What does it mean for N to be a neighborhood of x? It means that N contains some open set containing x

So it's a little circular, but a lot of the time you can show a set is open by looking local to any particular point and showing you can fit in a smaller open set

So let U be the set of points where s_x = t_x

let x in U be arbitrary

Then s_x = t_x as germs

[vestal snow]

Yeah thats what I tried

[latent anvil]

When are two germs equal?

[vestal snow]

WLOG, we can just consider the set of all points such that s_x = 0

[latent anvil]

Sure y

That's a little less general since it doesn't work for F just a sheaf of sets

but I'm nitpicking

so when is s_x = 0?

[vestal snow]

When some restriction of it is 0

[latent anvil]

right

So there's an open set V containing x with s|U = 0

[vestal snow]

Yes

[latent anvil]

what can we say about s_y for y in V?

[vestal snow]

OHHHH

[latent anvil]

haha yeah

[vestal snow]

Yeah

I got until s|U = 0

And I tried intersecting with the original set, but that would need that the original set is open

so circular

But yeah

I didn't think about that

Thanks

So quick question

How would this change if we wanted to do it more generally?

If s_x = t_x

[latent anvil]

same thing

when is s_x = t_x?

[vestal snow]

Uhhhh

idk

[latent anvil]

That's worth reviewing

so what are germs?

By definition

[vestal snow]

I mean that means that s and t map to the same thing in the direct limit

But I don't think that's what you were looking for?

[latent anvil]

No, that's still pretty abstract

Do you know how to explicitly construct a direct limit like this?

[vestal snow]

I want to say that they eventually map to the same thing, but idk

[latent anvil]

that's exactly right

So if you have a big directed system

We can just take the disjoint union of all the members of that system

[vestal snow]

I have seen the construction of the direct limit

[latent anvil]

Oh okay

[vestal snow]

It was in Atiyah MacDonald

[latent anvil]

Then yeah it's exactly the equivalence relation defining the direct limit

That they eventually map to the same thing

[vestal snow]

But it didn't talk about what happens when s_x=t_x

[latent anvil]

at some "finite" step, before we take the limit

[vestal snow]

only s_x = 0

I think A-M used a different definition

[latent anvil]

what was AM's defintion?

Did it take a direct sum?

[vestal snow]

Yeah

And then modded out by something

[latent anvil]

Yeah so we can simplify things in the casw of a direct limit

Im grabbing my copy of AM

To make sure we're on the same pag.

[vestal snow]

pg 33

[latent anvil]

Okay so instead of taking the direct sum, let's take the fiddling disjoint union of all the Mi

[vestal snow]

Okay

[latent anvil]

And then mod out by the equivalence relation (xi in Mi) ~ (xj in Mj) iff there's some k >= i, k >= k such that μik(xi) = μjk(xi)

[vestal snow]

and then define the equivalance relation in the obvious way

okay yeah

[latent anvil]

we define module operations by choosing representatives and pushing forward to some common Mk where they're defined

[vestal snow]

yeah using this definintion

it's pretty straightforward

[latent anvil]

This is iso to the construction in exercise 14 by using exercise 15

[vestal snow]

thanks

[latent anvil]

so if μi(x) = 0 we have μij(x) = 0 for some i, right?

then if μi(x) = μi(y), look at x - y

[vestal snow]

Yes

Wait

what if its a direct limit over sets

[latent anvil]

for some j we have μij(x-y) = 0 since μi(x-y) = μi(x) - μj(y) = 0

[vestal snow]

• might not be defined

[latent anvil]

well how do you define the direct limit of sets?

Presumably you're not using a direct sum, right?

The construction I gave using a disjoint union does have the right universal property in Set

[vestal snow]

No I meant that what if we had a sheaf

of sets

[latent anvil]

Right, I'm asking how you define the directed limit of sets

I define it by taking a disjoint union and modding out by an equivalence relation like above

[vestal snow]

Oh okay

[latent anvil]

That has the right universal property

[vestal snow]

So the construction in A-M doesn't work for sets

[latent anvil]

So no matter what construction you choose it'll be iso to this one

no, it involves a disjoint union

And quotienting by a submodule

neither of those make sense in the category of sets, but coproducts (= disjoint union) and coequalizers (= mod out by equivalence relation) still work

[vestal snow]

okay

[latent anvil]

I find the direct sum definition of a directed limit of modules ugly

and more awkward to work with

[vestal snow]

Got it

I would look at your construction in more detail sometime later, but I think the one I know is gonna work in most contexts for now

[latent anvil]

Sure

[vestal snow]

Working through AG is

really time consuming

[latent anvil]

yeah lol that's why I don't do it

[vestal snow]

Like I spent 6 hours going through just one page

[latent anvil]

Yup

[vestal snow]

Wait

[latent anvil]

I have bad news

[vestal snow]

So where did you learn about sheaves?

[latent anvil]

it gets much much harder after sheaves

[vestal snow]

No pls

[latent anvil]

Chmonkey and I and another friend not on here did an AG study group last summer

then we took an AG class last year

I bailed after the second quarter because I didn't have time and wasn't learning that much

[vestal snow]

tbh I'm kinda learning it just for the application to ANT

[latent anvil]

I know very little number theory

I really like algebra and I really like geometry

but

When you mix them?

Too hard

[vestal snow]

It doesn't help that the authors don't define things sometimes

For example

[latent anvil]

Oh god yeah

You mean how they don't explain what f^#_x is?

It's just implicit

[vestal snow]

Based on the notation introduced up to this point, f^#_x should not have the codomain that is listed here

[latent anvil]

The "obvious map"

yeah ikr

I hated this

okay so

[vestal snow]

Magician and I spent like 3 hours on it

[latent anvil]

apply the inverse image adjunction

[vestal snow]

He was trying to explain it to me

[latent anvil]

and prove a thing about stalks

oh okay

Lol I was gearing up

he is much better at this stuff than me

has been slamming himself into Hartshorne and doing nothing else for about 3 months now

very diligently

[vestal snow]

Everybody gangsta until Hartshorne starts slamming back

I've heard that that book is even more difficult than Liu

[latent anvil]

it is insanely hard

literally made me cry like

once a week

in the math lounge

[vestal snow]

lmao

[latent anvil]

I am a very mentally weak man though so take that with a grain of salt

I'm considering going back to it next month

I have until the 30th for school to start

[vestal snow]

lucky

[latent anvil]

But I also kind of want to finish this chapter of the analysis book I've been reading

Haha absolutely not

[vestal snow]

mine started

[latent anvil]

I'm bored out of my mind

I just managed to reset my sleep schedule

After sleeping from 5am to 3pm every day

It's now 3am, it's happening again

[vestal snow]

10 hours?

[latent anvil]

I'm a very well rested boy

just at the wrong time of the day

[vestal snow]

I usually can only sleep 2-3 hours at a time

[latent anvil]

Woah

I can't nap or sleep for a short amount of time at all

[vestal snow]

So I usually sleep, wake up and reddit for an hour, and then sleep again

[latent anvil]

If I fall asleep I'm out for the day

[neat ginkgo]

nailed my algebra 1 exam, thank you so much to everyone who helped me !!!

this exam was one of those where none of the questions are generic or similar to some i've already done

so i really had to work my brain to get it

but i nailed it

[carmine fossil]

Wait,Isn't that an highschool exam?

Not totally sure about the algebra 1 part

[neat ginkgo]

nah its abstract algebra, we just call it algebra here

Algebra 1

(im from eastern europe)

we also call calculus "analysis"

🧐

[carmine fossil]

Ok,The americans are weird

[neat ginkgo]

fancy as fuck

[next obsidian]

Nice job, glad it went well for you

[neat ginkgo]

ty!

[solemn rain]

@neat ginkgo how was the exam

[neat ginkgo]

it was ok tbh

like sometimes we would straight up get questions like:
"a) prove H and K are subgroups of G. are they normal, abelian?
b) prove that f is a homomorphism
c) determine if G/H is isomorphic to K"

which is usually like the braindead problem anyone can solve because

it always turns out to be the first iso theorem

for the homomorphism we were literally given

lmfao

but this time it was a bit more non-obvious

but still pretty easy

[solemn rain]

gj

ur p good man

[neat ginkgo]

actually damn we did have a super easy problem i forgot

[solemn rain]

what was it

[neat ginkgo]

it was some subgroup of the symmetric group S9, and an action from it to X which is literally {1,...9}

lmfao

and we had to find the orbit and stabilizators for every element in X

that was like...

redundant and easy

so kind of annoying really

[solemn rain]

yea

i hate finding shit of for every elements

[neat ginkgo]

ikr

the exam was 3h i ended up finishing in 1.5 tho

lets gooo

[solemn rain]

lmao

gj

[neat ginkgo]

ty for the help btw man

[solemn rain]

i hope u had fun with groups

[neat ginkgo]

legit couldn't have done it without you

[solemn rain]

its the only good math u will ever encounter

jk

[neat ginkgo]

yeah man im 99% sure ill take algebra 2 next year

[solemn rain]

yea

and alot of cool ppl who will help you here

actual good ppl not me

[neat ginkgo]

👼

[solemn rain]

is algebra 2 about rings?

[neat ginkgo]

no clue

[solemn rain]

okay

[neat ginkgo]

i mean algebra 1 was groups rings fields

[solemn rain]

you took rings and fields?

[neat ginkgo]

but i guess all algebra courses are just that lmfao

yeah

[solemn rain]

an algebra is a vector space tho

[neat ginkgo]

actually fk i didnt have fields, because corona cut our last few classes

[solemn rain]

haha nvm

yea

i mean

if ur interested u should pick up a text on galois theory

[neat ginkgo]

i hope algebra 2 gets into galois theory

[solemn rain]

idk how wil lthat help u in ur cs studies

well if algebra 1 is about that then yea algebra 2 should be on galois theory

[neat ginkgo]

but abstract algebra 1 so far made me love and appreciate math so much more man

[solemn rain]

or maybe more group theory

[dawn kiln]

Galois theory comes up a lil bit in cs

[solemn rain]

really

[neat ginkgo]

i love the kind of formal mathematics that tries to tie together different fields of maths

which i assume is the point of category theory

(idk anything about category theory)

[dawn kiln]

Error correcting codes, elliptic curve cryptography and weirdly enough, RAID 6

[neat ginkgo]

i bet id lov ethat

raid 6 LMAO

[dawn kiln]

(well technically more galois fields than galois theory 😂😂)

Yeah ikr

[solemn rain]

yea i think u can know about category theory if u like

know haskell ig?

idk im 0 in this cs stuff

but i mean

[dawn kiln]

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

[solemn rain]

if ur interested

and ur a programmer

check out category theory for programmers

on utube

shold be cool

[dawn kiln]

Scroll down to"general parity" in raid 6

[neat ginkgo]

its weird i was entering college to become a braindead software engineer and i ended up falling in love with like nieche ass mathematics that im almost never gonna use for engineering irl

and theoretical computer science

[solemn rain]

yea

sadly im forced by my parents to become an engineer as well

and engineering here has like 0 theoritical shit whatosever

u just learn what u need

but u can always just study the shit u want

[neat ginkgo]

@dawn kiln damn this is cool

[dawn kiln]

CS at least has a reasonable excuse for studying random pure maths stuff lmao

[neat ginkgo]

^

[dawn kiln]

"yes, I definitely need to study galois fields to implement this new raid 6 server system 😎😎"

[neat ginkgo]

LOL

ikr dude my abstract alg professor was trying so hard to like

show us how useful it is in computer science

its really not gonna be useful to any of us individually

[solemn rain]

yea boys fuck school

gl

[neat ginkgo]

http://www.ams.org/notices/201402/rnoti-p190.pdf

he has this linked on his website

in bold

on our class page

[vestal snow]

Why is the stalk at every point of V here a local ring?

I don't think that this is trivial

Because (O_Y|_V)_y is the direct limit over open subsets of V containing y whereas O_Y,y is the direct limit over open subsets of Y containing y

Note: Ringed topological space means locally ringed in this book

So the second direct limit being local does not imply that the first is local

[vestal snow]

@latent anvil This is an instance of what we were talking about

To prove the iff we need to use that f^#_x is an isomorphism iff f^# is

However, f^#_x does not really mean f^#_x as we had discussed earlier

[latent anvil]

Wait what were we talking about?

Oh right

I can give some intuition for wjy stalk of O_Y|V at a point is the same as the stalk of O_Y at that point

So germs at p are like equivalence classes of functions defined on some small neighborhood of p

but if you shrink the neighborhood it's defined on you get the same germ

If f is a function (i.e. section of O_Y) defined on W and W' <= W are neighborhoods of p we have <f, W> = <f, W'>

and you can always shrink the domain of definition of a function by intersecting with V

Really this is a statement about directed limits

If you have some system ${X_i}{i \in I}$ (e.g. $F(U)$ for $U$ a nbhd of $p$) and you take a subset $J$ of $I$ (the $U$s contained in $V$) such that for each $i \in I$ we have $i \leq j$ for some $j$, then $\varinjlim{i \in I} X_i = \varinjlim_{j \in J} X_j$

By = I mean there's a naturally defined morphism from the left to the right and it's an isomorphism

Exercise: prove this

[cloud walrus]

shamrock:

[latent anvil]

If you think of a colimit/directed limit like a supremum this should be clear, and that's what colimits look like in a poset category

the property of J which I mentioned is called being a "cofinal" subset of the poset I

[vestal snow]

Yeah that makes sense

Thanks

[vestal snow]

Given a sheaf O_X and a sheaf of Ideals (of the original sheaf's rings) J, is J_x an ideal of O_X,x?

I think I will need this here

Because we get that for every y in V, 1 is in J_y, but we need J_y to be an ideal in order to conclude that J_y is all of O_X,y

[next obsidian]

Oooo so I think I’m qualified to give an answer here

What you want to try and do, consider that being an ideal of A is the same as, considering I as an A-module, that you have an injective map I -> A

Try and show that if you take the direct limit of a bunch of injections like I(U) -> O_X(U), that the map from the stalks is also an injection

I think this should work

Oh this might not work actually

Since you’re taking a colimit and not a limit, you can make that work when you’re taking a limit

[scenic sage]

to get Q[x]/(x^3+1) field, do I need to add to Q

[cloud walrus]

Commander Vimes:

[woven delta]

It is isomorphic to Q(1+sqrt(-3)), yeah

[scenic sage]

hmhm

so i mean

[woven delta]

Or I guess just Q(sqrt (-3))

[scenic sage]

like for Q[sqrt (2)] it becomes then like a+b(sqrt(2))

[woven delta]

Yeah

[scenic sage]

what with x^3+1 then

would i have triple or what

[woven delta]

Every element is of the form $a + b\alpha + c \alpha^2$

[cloud walrus]

Liquid:

[scenic sage]

yes, i was typing that

[woven delta]

Where a, b, c are rational and alpha is a solution to your polynomial

[scenic sage]

wait, but should not i then just plug in third unity root?

[woven delta]

In general if your polynomial is irreducible of degree n your elements will be $a_0 + a_1\alpha + ... + a_{n-1} \alpha^{n-1}$

[cloud walrus]

Liquid:

[woven delta]

So if you just adjoin that polynomial you get one root. Sometimes it happens to be that that one root gives you other roots as well

[scenic sage]

the issue is that expanding x^3+1 as (x+1)(x^2-x+1) will give sqrt(-3) form

but form with third root of unity is also solution

[woven delta]

Wait what

Oh in that case this is bad

[scenic sage]

ok lol i am dumb

[woven delta]

So actually you can't look at Q[x]/(x^3+1) if x^3+1 is reducible

Sorry I should have been more careful about checking that x^3+1 is irreducible

(or rather it won't be a field)

[oblique river]

the red flag should have been when you wrote that Q[x]/(x^3 + 1) is isomorphic to a quadratic extension of Q

[woven delta]

You want Q[x]/(x^2-x+1)

Yeah

This is what happens when I don't think

[scenic sage]

this is problem btw

[woven delta]

Oh okay

[scenic sage]

well looks like cubic root of unity helps but hmhm

[woven delta]

So you know that an ideal (p(x)) is prime in Q[x] iff p(x) is irreducible

And in a PID we have an ideal is prime iff it is maximal

And an ideal is maximal iff the quotient is a field

So it suffices to show that p(x) is irreducible to show that the quotient is a field

For cubic polynomials it's pretty easy to show they are irreducible by just showing they don't have roots in Q

[scenic sage]

lol, i havent studied ideals and like proposition that ring is a field is after this exercise

[woven delta]

Oh ok

I guess you can just do things by hand

[scenic sage]

wait lol
(x^3+1) can be factored in (x+1)(x^2-x+1) and that means that it is not field, yes?

[woven delta]

Yeah

Basically this demonstrates that the quotient is not an integral domain

So it can't be a field

[scenic sage]

oh, lol, i have proved before that it is equivalent to absense of zero divisors

so good

[woven delta]

Yeah a quotient of Q[x] is a field iff it is an integral domain

Which is equivalent to the statement I made above about an ideal being prime iff it is maximal

[scenic sage]

but here i have field, ye?

[woven delta]

Yeah, you can show this by hand pretty easily

[scenic sage]

since cubic root is of two is not in Q

[woven delta]

Or just use eisenstein

Yeah

Exactly

[scenic sage]

by what i then expand Q? by cubic root of two or by complex cubic root of two?

[woven delta]

There is a theorem that tells you that both are isomorphic

But I think it's more conventional to expand by the real cube root of 2

[scenic sage]

ok, ty

[chilly ocean]

Are there any good books to Galois theory that focus on Q only and ignore the complexity that comes from finite fields?

that takes a modern approach*

why what?

Do you have any reason to think it wouldn't be worthwhile for such a thing to exist?

[smoky cypress]

I feel like the only complication fields with nonzero character bring is that irreducible polynomials might not be separable

But that’s not even an issue in finite fields

[stone fulcrum]

Stewart's Galois theory

It saves finite fields until after the "main theorems"

[chilly ocean]

@stone fulcrum +1

I don't doubt it's useful.

But things are useful at certain points of development. Generalization is useful for experts, but not always so much for learners 🙂

[bleak abyss]

Finite fields are actually way easier

Like, the definitions aren't really at all harder in general

Maybe you don't have to worry in principle about separability if you focus on number fields

But there's basically nobody who will have an easy time learning the foundational theory only when you restrict to number fields. Like oh a normal field extension of Q vs a normal field extension of a general field

And then Galois theory of finite fields is something you can learn in one go and then you know everything

[chilly ocean]

I disagree. For the same reason linear algebra is always taught first with matrices with real entries, rather than abstract linear maps over an arbitrary field.

But I'll take a look into Stewart's book 🙂

[bleak abyss]

Not always, and tbh that's an overrated decision

[latent anvil]

Linear algebra is usually taught first to students who aren't interested in math or the general case at all

[chilly ocean]

It depends. Sometimes abstraction can help, because it removes detail. But there's a point where the abstraction itself becomes a burden.

[latent anvil]

I find it weird when a proof based first course on lin alg for math majors sticks to R and C

[bleak abyss]

I mean yeah for non-math majors obviously don't add extra definitions

I just think that sometimes the jump is substantial and sometimes it really isn't

Like okay maybe linear algebra's a weird case because if you zoom in on R there's dot products

So some people present early linear algebra theorems in terms of the dot product

So okay upgrading to general fields means you have to modify your proofs somewhat. I think for math majors the proofs aren't noticeably harder so might as well but there's a biiiiit of a case

Now, Lebesgue measure on R^n vs general measure, that's pointless to me. The theorems are proved in literally the exact same way

So just do 3 extra definitions at the beginning and then call Lebesgue your main example

Calculus on R vs R^n, that's a real jump in complexity

[chilly ocean]

So okay upgrading to general fields means you have to modify your proofs somewhat.

Yeah. This is more or less the sentiment I was chasing.

I'd like to see what the proofs look like in the Q case. It seems like something you could teach to high school kids without a problem.

[bleak abyss]

So yeah it absolutely makes sense to learn it on R first, then R^n. Galois theory? If anything Galois theory of finite fields is the nicer case rather than Q

Since the Galois groups end up all being cyclic

And you have a concrete generator

[chilly ocean]

Is there any convincing "story" you can tell for why you could care about finite fields that would resonate with someone who passed calculus, though? It seems like it's much easier to explain when you're working with Q-polynomials.

[bleak abyss]

I mean even the motivations for Galois theory on Q is kinda like, okay do you like solving polynomials with radicals or doing straightedge stuff, very historical. Really Galois theory is mostly important nowadays for number theory

But I think finite fields in general have a bit of a CS-y type motivation going on

[golden pasture]

crypto

[chilly ocean]

Indeed.

[golden pasture]

honestly we should all use p-adics rationals instead of Q🤔

[bleak abyss]

Plus I think the proof that cyclotomic polynomials are irreducible passes through finite fields?

Tbh I feel like Galois theory should be swapped with rep theory in undergrad algebra

[zinc depot]

notationally, how do i write that (1,1) is a generator for Z2 x Z3, would <(1,1)> = Z2 x Z3 be correct?

[chilly ocean]

That sounds right to me.

[bleak abyss]

Since the traditional motivation for Galois theory is basically all historical and kinda... I don't wanna say expendable but

It's not like oh everyone should know about unsolvability of the quintic. It's cute but it doesn't come up a lot anymore

Nowadays Galois theory is mainly useful for number theory and to a degree algebraic geometry

[golden pasture]

it's so boring when all the classical stuff are actually done tho ><

[bleak abyss]

So I could make a case for it being in graduate algebra, that if you've got significant interaction with algebra you should know Galois theory

But undergrad algebra to me is stuff that every mathematician in general should know. So I'm much more inclined to make analysts learn rep/Lie theory

[golden pasture]

rep theory is a lot more useful

agreed

[chilly ocean]

I should probably take this to #math-pedagogy but this is a long-running feeling/thought I've had is that mathematics doesn't do an especially good job with recording "narrative" for its development. This is in contrast to, say, physics, which has a very strong story to tell.

Galois theory does seem to have a consistent and well-defined narrative (although @bleak abyss is expressing skepticism towards its utility here). But it comes off to me like there's a conflict between the polynomials-and-geometric-constructions and the actual meat of the theory.

wished he knew more about rep theory.

[bleak abyss]

Basically what happens a lot in math is that people stumble across things in super inefficient ways, or they come up with stuff with an eye toward solving a particular problem

But then it turns out there's a much more powerful, and much harder to appreciate, use

So a long time ago people liked solving equations, which I think Abel actually did first, but the one we tend to remember is Galois

So Galois pulled out Galois theory and group theory to hit those problems. Okay great now those are a thing

[chilly ocean]

Yeah. The method shouldn't be strictly based on history. It should be the most useful lie you can tell to get students to understand, appreciate, and remember the subject.

(If it's not clear by now, I've sunk a fair bit of time into learning Galois theory, but never made any meaningful progress on it 😛 )

[bleak abyss]

But then turns out there's a ton you can do with Galois theory elsewhere, and we've kinda forgotten about polynomials now since that story is over, and there's a more efficient way to phrase the theory that's way more conducive to the newer motivation

But yeah how do you feel right now about Galois theory? What do you know somewhat comfortably atm?

[chilly ocean]

It's been a few years, but I was originally learning algebra through Aluffi's book.

[bleak abyss]

Uh oh

[chilly ocean]

Yeah

[bleak abyss]

Aluffi is not good

[chilly ocean]

I realize the error of my ways in retrospect 😛

[bleak abyss]

Weak problems and his category stuff is only good if you don't get carried away

I can tell you the general gist if you'd like?

[chilly ocean]

I eventually moved to D&F which I liked a lot more. but I didn't get to the chapter on Galois Theory

sure!

[bleak abyss]

D&F is better yeah. It's kinda boring sometimes but does everything and it's super clear

But yeah so, nowadays we're trying to understand field extensions

And note the difference in perspective compared to group/ring theory, in groups you have a fixed group and vary the subgroup underneath

Now you have a fixed ground field and you're asking what fields contain it, rather than what it contains

[chilly ocean]

Right. You want to know how you can extend the thing.

And I imagine you're often preoccupied with the intermediate stages between a field and its algebraic closure.

[bleak abyss]

Yeah. So the very first observation we make is that if you have a field K contained in a field L, then L is a vector space over K

Since scalar multiplication is just multiplication

[chilly ocean]

Yep. So Q(√2) is dimension 2 over Q, etc.

[cloud walrus]

Daminark:

[chilly ocean]

Right. So that precludes R over Q, because that's very infinite dimensional, as is Q(π). So you're really focusing on even a tiny portion of the field extensions between k and k^bar.

[bleak abyss]

Well, careful

R isn't in Q bar

[chilly ocean]

Yes. I'm aware of that.

I meant only that R is infinite dimensional over Q.

as would be Q^bar.

as would be Q(π).

[bleak abyss]

Yeah

[chilly ocean]

And implied in this is that the whole point of doing these extensions is that you're thinking about polynomials in the background.

That each extension you make is really done with the aim to bring new solutions to previously insoluble polynomials.

(aka, to reduce a previously irreducible polynomial)

[bleak abyss]

Well, I won't make strong claims about what the aim is here. But if nothing else you see that these two ideas are linked

You can take an irreducible polynomial and give it a root, this is the classic ring theory trick of k[x]/(f)

[chilly ocean]

Sure. That could be one possible aim, and maybe a good one to highlight.

And yep. The trick then introduces the new equation f = 0 into the resulting field.

[bleak abyss]

And then if you give me finite field extensions, well each element happens to be the root of some polynomial. And if you take the set of polynomials which kill an element, that's an ideal in k[x]

There's a unique monic generator of that ideal, this is called the minimal polynomial

But there is a bit of a tension here

Let's consider the case where our base field is Q. If we want in principle we can think of everything happening inside C

[chilly ocean]

sure

[bleak abyss]

If we do the Q[x]/(f) trick to get a field extension, well in C you could be talking about multiple things

For example, let's say I want to adjoin a root of x^3 - 2

K = Q[x]/(x^3 - 2) gives me one thing. But then in C that polynomial has 3 different roots, so I can embed K into C in 3 different ways

[chilly ocean]

When you say embed, you're not talking about a ring map K → C though, right?

[bleak abyss]

Yes I am

[chilly ocean]

ok

I'm misremembering something a bit, but yes, that sounds good.

[bleak abyss]

Note that being a ring homomorphism out of a field automatically implies that you're injective

[chilly ocean]

So there should be three such maps.

Yes. That mathfact, I do remember properly

having to do with something something kernels are trivial because fields.

[bleak abyss]

Yup

[chilly ocean]

Let me work this out really quickly

because I'm missing something

[bleak abyss]

Sure thing

[chilly ocean]

Analogously R → C embeds in two different ways, intuitively due to the fact that you can adjoin i or -i.

[bleak abyss]

Well, more that C embeds into C in two different ways

[chilly ocean]

ah

ok

er

[bleak abyss]

R[x]/(x^2 + 1)

But yeah you have the idea. So we can either take the identity C->C, or complex conjugation

[chilly ocean]

Right. I get for C → C that there's two field maps.

But R → C, there should only be one... I think

[bleak abyss]

Yeah

[chilly ocean]

So that means i misunderstood what you said earlier

[bleak abyss]

I didn't say earlier that Q embeds in multiple ways

I said Q[x]/(x^3 - 2) does

[chilly ocean]

ah

ok

👍

So we've extended Q to Q(∛2)

[bleak abyss]

So in the C case, the fact that the polynomial x^2 + 1 doesn't really know the difference between i and -i is basically the statement that we have multiple embeddings. In other words, we have automorphisms of C sending i to -i

[chilly ocean]

and what you said amounts to the fact that our "generator" for the solutions (the x in Q[x]/(f)) might end up being any one of three of the "concrete" solutions in C.

yeah.

[bleak abyss]

Yup, and in fact if we add on all the roots of x^3 - 2 to Q, then we have automorphisms swapping them around

[chilly ocean]

And this is where the group theory comes in

because you see that field extensions permute the roots. But if you play around, you'll find the permutations are sometimes constrained.

and sometimes they aren't.

[bleak abyss]

Yup, and then the magic of Galois theory is that the group is related big time to the structure of field extensions

[chilly ocean]

In our example, you have three items, but I think you can only "rotate" them in the plane.

[bleak abyss]

Yeah, so that's kinda what's going on here at large

[chilly ocean]

As a tool, it seems to me still that -- unless you have a preoccupation with field extensions for their own sake -- we're still mostly either thinking about the nature of the polynomials or of the fields involved.

[bleak abyss]

I think that's one of those things where, before people used to care about the polynomials more

Now I think people have started caring more about the fields, since in stuff like number theory

You want the arithmetic of higher fields

[chilly ocean]

That sounds reasonable enough to me.

stares at his book on algebraic number theory he never finished.

Well, I'm glad at least everything you described was the parts I actually understood from the classes and books I've gone through.

[solemn rain]

alot ig

u can do intersections of fields

like algebraic topology

[chilly ocean]

Algebra is a major artery of mathematics.

[solemn rain]

or algebraic number theory

or u can study like more shit in rep theory

or specific tools like homological algebra ( for AT )_

[chilly ocean]

To the converse, there's very little you could do without algebra 🙂

[solemn rain]

category theory ig

[scarlet estuary]

uh

wdym by category theory

[solemn rain]

xD

[scarlet estuary]

you absolutely need algebra for category theory

[solemn rain]

?

yea

i meant as something u can do higher stuff in

with algebra

[scarlet estuary]

in theory you could be a half-decent analyst without it maybe

[solemn rain]

not something u dont need alg for

[scarlet estuary]

category theory is severely, er

overrepresented

online

to put it kindly

[solemn rain]

iidk

sometimes i jusut thinnk ppl think its cool to hate on it

it just attracts alot of newer guyus

and they get trapped tahts it

but nothing else

really thats just my honest take

yea i agree with viburnum do more algebra

uptill galois theory and some like rep theory maybe

[scarlet estuary]

if youre interested in serious research in (pure) math

definitely try and get exposure to both algebra and analysis

at least at the undergraduate level

lots of fields are interdisciplinary or sit at the intersection, or at the very least you'll get a better idea of what kinds of things youre interested in

plus it lets you know what kinds of tools are appropriate to answer what kinds of questions

like a course in functional analysis will give you an intuition for when analytic tools can be useful in the study of vector spaces

(since in practice functional analysis is mostly the study of vector-spaces-(or-modules)-with-analytic-structures)

[golden pasture]

i heard that it's more because so many algebraic xxx uses the same foundations that have a huge amount of prereq that result in all these foundations being represented quite a lot cuz of lots of qns like how do i learn/understand xxx

meanwhile analysis type areas feel more like mostly disjoint areas without a huge chunk of prereq so qns are generally all in different places and also you dont get answers :p

[leaden finch]

can soemone help me with proofs?

[solemn rain]

sure

just ask

[leaden finch]

im having a hard time constructing this proof

i have no idea how it relates back to the theorems or defintions

divisbilty and divison algorithm right now

[woven delta]

You can do the forwards direction using Euler's theorem if you know that lol

[leaden finch]

i dont know it yet haha

one second, can we do the backwards direction togetehr

[leaden finch]

i got stuck i know it relates to theorem 1.2

[chilly ocean]

What book is that? That looks familiar.

It seems kind of stupid to start off with the wellordering axiom. (aka, "Common sense phrased in a way to confuse students")

[leaden finch]

thomas hungerford

anyone know of a good textbook for modern algerbra?

[vestal snow]

Can someone explain what the proof of uniqueness would look like?

Say (F, f_i) and (F',f'_i) (using f instead of \phi because its easier) satisfy the given property

I feel like you have to show that there is an isomorphism from F to F' which behaves well with f_i and f'_i

Like there must exist an isomorphism between F and F' such that it does not matter if we go into F_i directly, or if we first go to F' and then go to F_i

[latent anvil]

yeah so I think you'd want to show you can glue maps between sheaves first

I think that should be true? This is like the sheaf hom right

[vestal snow]

What do you mean by gluing maps betweeen sheaves?

[latent anvil]

Let F, G be sheaves on X and suppose {Ui} is an open cover of X

Suppose we have sheaf maps φi : F|Ui -> G|Ui for each i

Let i, j be arbitrary and let V = Ui cap Uj. Suppose φi|V, φj|V : F|V -> G|V are equal

then there's a unique (in the sense of actual equality) sheaf morphism φ : F -> G

Does that make sense?

[vestal snow]

I'll try proving it

but yes

intuitively it does make sense

[latent anvil]

Do you see why this is useful for the sheaf gluing problem?

for establishing uniqueness I mean

[vestal snow]

Let me think one sec

Okay

So this would help because then we would get a sheaf isomorphism between F, F'

because each \phi_i in this case is an isomorphism

So that would probably imply that \phi is also one

[latent anvil]

yeah you'd look at φ'_i ° φ_i^(-1)

[vestal snow]

Though it would need to be proved

[latent anvil]

If I understand the notation correctly

[vestal snow]

yes

[latent anvil]

Err no sorry

ψ'_i ° ψ_i^(-1)

That gives an isomorphism F|Ui ≈ F_i ≈ F'|Ui

Right?

[vestal snow]

one sec

Yes

[latent anvil]

What I'm trying to say is F and F' are both locally isomorphic to the same thing

And the family of local isomorphisms is coherent

[vestal snow]

I don't know what locally isomorphic or coherent means 😦

[latent anvil]

oh coherent is a nonsense word that mathematicians like to use

it means "plays well with one another" or something like that

The isomorphisms aren't just random isomorphisms, they play well together somehow

It's not really a well defined term

By locally isomorphic I mean you have a cover where F|Ui ≈ F'|Ui

[vestal snow]

I think I understand the problem now

[latent anvil]

There's a formal definition of coherent sheaves which is not what I was talking about to be clear

[vestal snow]

I'm probably gonna skip it because it seems like verifying a lot of details

But I think I have the intuition down

[latent anvil]

Yeah it does

I found this problem annoying

But also

Skipping problems like this is why I gave up on AG

because I didn't have a solid foundation

[vestal snow]

I get that, but I feel like its fine for now

I proabably would study the subject in a year or so more rigorously.

Right now I just need enough of it to understand research problems in functional and number fields

Most of which is in chapters 6 and 7 of Liu iirc

[vestal snow]

For all f,g in A, r(f) = r(g) implies A_f = A_g

How do I prove this?

[latent anvil]

What's r? Nonvanishing set?

And also what do you mean by =?

[vestal snow]

r means radical

and by equal I mean isomorphic

[latent anvil]

It suffices to show A_f is isomorphic to A_{fg} by symmetry. There's a map φ : A_f -> A_{fg} because the canonical map A -> A_{fg} sends f to a unit (1/f = g/fg) and I think that map should be an isomorphism

and the result says that A_f and A_fg must be isomorphic since r(fg) = r(f) cap r(g) = r(f)

[vestal snow]

I dont understand your last line?

[latent anvil]

Your problem implies that if r(f) = r(g) then A_f and A_fg are isomorphic

[vestal snow]

Isn't that sufficient

[latent anvil]

Yes

but I'm saying the conclusion implies it

[vestal snow]

Because A_g will also be isomorphic

[latent anvil]

Exactly, so instead of your problem we can show that A_f and A_fg are isomorphic

[vestal snow]

I meant that I dont get why you need to use this r(fg) = r(f) cap r(g) = r(f)

[latent anvil]

And this feels better to me because there's a natural map A_f -> A_fg

Because I was to say your problem implies A_f ≈ A_fg

And your problem only applies if the two things you're localizing wrt have the same radical

So I'm showing f and fg have the same radical when f and g do

To show that the thing you're trying to prove is equivalent to A_f and A_fg being isomorphic

[vestal snow]

Is there a more direct way?

By proving that the map induced from A_f to A_g is an isomorphism

[latent anvil]

What is that map?

[vestal snow]

A to A_g takes f to a unit

[latent anvil]

Oh, then yes

It just felt clearer to me that that was the case in A_fg

But if you see why f is a unit in A_g then use that map instead sure

[vestal snow]

Yeah I was wondering how to prove injectivity

Say g^m = af and f^n = bg

[latent anvil]

So injectivity says that if g^r a = 0 for some r then f^s a = 0 for some s

Right?

[vestal snow]

Uh I don't think so

x/f^k goes to xa^k/g^{mk}

[latent anvil]

If φ(x/1) = 0, then x/1 = 0 in A_g, so g^r x = 0 for some r

[vestal snow]

I don't think we should use a

[latent anvil]

Sure

[vestal snow]

because we already fixed it above

[latent anvil]

I'm not sure what you meant by this:

Say g^m = af and f^n = bg
@vestal snow

[vestal snow]

Oh I was just writing out r(g) = r(f) in terms of elements

[latent anvil]

Ah I see

Sure

[vestal snow]

there exists m,n,a,b such that

[latent anvil]

oh duh

Okay I see how to do this now

suppose φ(x/1) = 0

Then x/1 = 0 in A_g

So g^r x = 0

So g^(mk) x = 0 for some k

Just multiply g^r by g until you get to a multiple of m

Oh but hm this only gives a^k f^k x = 0

Not f^k x = 0

Oh jsut go the other way lol

If g^r x = 0, then b^r g^r x = 0

So f^(nr) x = (f^n)^r x = (bg)^r x = b^r g^r x = 0

so x/1 is zero in A_f

Does that proof of injectivity make sense? I got a little turned around so understandable if not

[vestal snow]

I think it does

Let me write it out one sec

[latent anvil]

It suffices to only look at elements like x/1 in the kernel since f is a unit

[vestal snow]

Lol now that I write it down I'm a bit frustrated that I couldn't do this

[latent anvil]

dude it's no problem

Math hard

[vestal snow]

Yeah stuff like this happens from time to time

Yes

AG very hard

[latent anvil]

I felt like an idiot when I was like "was it how did you get f = a g^r"

Like it didn't occur to me to use the assumption r(f) = r(g)

AG very hard

It's been too long since I've done commutative algebra (3 months)

The plan is to start again on Tuesday

[vestal snow]

Good luck with that

Matsumura?

[latent anvil]

Yup

chmonkey swears by it now

[vestal snow]

I liked that book too

Well, I only read the first two chapters and the appendix on homological algebra

[latent anvil]

If I have time I'll try to do some Hartshorne problems during September

But probably not during the quarter 😢

Homological algebra is so great

Chmonkey and I are taking a course on it in winter

[next obsidian]

You will do Hartshorne during the quarter

[vestal snow]

I'm so glad the Putnam was moved to a later date

[next obsidian]

Because I will text you

[latent anvil]

We had some during last spring

Lmao

I'll do some even if I never open the book

[next obsidian]

It’ll probably be pure commutative algebra by the time I do tho lmfao

[latent anvil]

Don't ask me for help when you're reading actual papers

[vestal snow]

I have so much stuff. There's no way I'd be able to prepare for it

[latent anvil]

That's why I don't do the Putnam

[next obsidian]

That’s why I have forsaken the Putnam

[latent anvil]

Lol

Other things are more interesting to me and it takes a lot of time to prep

[vestal snow]

Haha some problems can be wacky

[latent anvil]

Also Alex that's not quite true

You forsook it by sleeping through the exam

[next obsidian]

Only because I didn’t care enough

[latent anvil]

uhuh

[vestal snow]

The Putnam coordinator (who happens to be the same guy who wanted me to do AG in 15 days) basically recruits math majors to take it kinda like Army recruiters at the mall

[latent anvil]

lmao

Sadly you cant bait him with updog in response like you can with army recruiters

I'm 0 for 3 but someday I'll get one

[vestal snow]

I have never actually been asked by any recruiter

Probably because I'm 100 pounds

[carmine fossil]

Some putnam questions are really easy

[vestal snow]

Maybe the really old ones?

I don't think I've ever solved one in less than 3 hours

[carmine fossil]

Yea,I guess so

[steady axle]

i have a silly question

why cant we calculate any cohomology group using free resolution ...0 -> Z -> Z ->0

[latent anvil]

Oh I meant asked by text

@steady axle wym?

Are we working with abelian groups?

[steady axle]

yes

[latent anvil]

How would you use this resolution to calculate cohomology?

Like, say I give you a cochain complex C

how is this free resolution useful in computing H^i(C)

[steady axle]

hmm

what i thought was we can using standard resolution and any projective resolution gives same cohomology groups

[latent anvil]

Okay so I think you're thinking of a different meaning of cohomology than me

You calculate derived functors using a projective resolution, and sometimes we refer to derived functors as cohomology (eg sheaf cohomology or group cohomology)

Could you define cohomology for me?

[steady axle]

i am talking about group cohomology here (dont really know what sheaf cohomology is)

[latent anvil]

Okay

Please specify that lol

What's your definition of group cohomology? Is it in terms of Ext?

[steady axle]

Could you define cohomology for me?
yes. take projective resolution of Z then apply functor Hom(_,A) and then ker / im

[latent anvil]

Okay yeah that's what I was thinking

You're resolving Z over the ring Z[G]

Not as an abelian group

[steady axle]

What's your definition of group cohomology? Is it in terms of Ext?
yes so it is Ext^n(Z,A)

[latent anvil]

Yeah but it's Ext_{Z[G]}(Z, A), right?

Not Ext_Z

[steady axle]

i should probably add subscript Ext^n_ZG

[latent anvil]

exactly

Z isn't free over Z[G]

[steady axle]

i dont understand why does it have to be? ( we are talking about free resolution of Z)

[latent anvil]

Because 0 -> Z -> Z -> 0 isn't a free resolution

Because Z isn't free

[steady axle]

ooh hmmm

now many things make sense

[latent anvil]

Nice

[steady axle]

actually i remembered that this was used when G={1} so it didnt matter there ! but cant use it everywhere (because as u pointed out Z wont be free)

[latent anvil]

So just a note, if you say cohomology unqualified it usually means the groups H^i(C) = ker(di)/im(d_{i-1}) for a cochain complex C

Which is why I was so confused earlier

[steady axle]

right so i have to specify cochain complex

[latent anvil]

I mean you just need to say group cohomology

Because that's something different

[steady axle]

thanks

i evidently did not learn it very properly

[bleak abyss]

Weibel if you're fine with your theorems only being true up to homotopy

[steady axle]

Ya. Weibel is considered std txt on homological algebra afaik

[solemn rain]

can homological algebra be taught for its own sake

or is it just uysed as a tool for AT

[next obsidian]

Yes to first

No to second

[solemn rain]

is it col

[bleak abyss]

Tbh knowing very little of both homological and commutative algebra

I think homological is cooler

[latent anvil]

agreed

Homological is just very sick

[woven obsidian]

I'm having some trouble understanding why a maximal ideal must be of this form

I.e the general form

[latent anvil]

Which part in particular are you having trouble with?

@woven obsidian

[woven obsidian]

How they conclude that m = the intersection

[latent anvil]

Do you see why m is contained in that intersection?

[woven obsidian]

I guess I'm having some trouble moving from k[x,y] to F[x,y]

Because the elements of F are also "polynomials"

[latent anvil]

My suggestion is to forget about that

F is some abstract field

[woven obsidian]

Hmm ok

[latent anvil]

we don't know what it is

[woven obsidian]

And k a subfield

[latent anvil]

But we have a map k[x, y] -> F with kernel m

Yup

So a map out of k like this has to send x to some value a and y to some value b

[woven obsidian]

Yeah

[latent anvil]

And really it's just φ(f(x, y)) = f(a, b)

[woven obsidian]

Then I would like to think X-a and Y-b lie in that kernel

[latent anvil]

well not exactly

[woven obsidian]

But that's the wrong space

Yeah

[latent anvil]

So if f is in the kernel m

Then f(a, b) = 0 in F

[woven obsidian]

Why is that?

[next obsidian]

x maps to a, y maps to b

so f(x,y) maps to f(a,b)

if in kernel must map to 0

[woven obsidian]

Ok I think the second line is what I'm having a hard time understanding

Hmm

So we map into F

[latent anvil]

I think the thing I said earlier about f being an abstract field is making things more confusing, sorry

[woven obsidian]

Ok so we just use homomorphism property?

[latent anvil]

Kind of

We use the fact that it's a k algebra homomorphism

[woven obsidian]

That would make sense

[latent anvil]

So φ(a x^n) = aφ(x)^n

[woven obsidian]

Yeah ok, that makes a lot of sense

So our homomorphism is determined by where X and Y maps

[latent anvil]

Yup! This is the "universal property" of k[X, Y]

It's the free k algebra on two generators

Okay, so we have a map k[x,y] -> F

It's given by evaluating at (a, b) for some a, b in F

Now if f is in k[X, Y]

and f(a, b) = 0

Then thinning of f as a polynomial with coefficients in F, we have f in (x-a, y-b)

Because if an element of F[x, y] is zero at a point it's contained in the corresponding maximal ideal

For any field F

[woven obsidian]

Ok, we proved that for one-variable polynomials

I guess you can do the same with two variables?

[next obsidian]

any number

[woven obsidian]

I meant this part

[latent anvil]

Yeah pretty much

[woven obsidian]

Then thinning of f as a polynomial with coefficients in F, we have f in (x-a, y-b)
@latent anvil

[latent anvil]

If f(a, b) = 0

[woven obsidian]

Yeah

[latent anvil]

Then f in (x-a, y-b)

It's some kind of multivariable division algorithm

Embarrassingly I don't remember it rn