groups-rings-fields

[latent anvil]

It's 3am here

[woven obsidian]

Haha np

[latent anvil]

maybe chmonkey can help

[woven obsidian]

I'm willing to believe that, I can google it later otherwise

Ok but then we have one inclusion

[latent anvil]

Well we have an inclusion

But it's an inclusion of a maximal ideal

And (x-a, y-b) cap k[x, y] is an ideal of k[x, y] too

[woven obsidian]

Ah right

So we're done

[latent anvil]

Not quite

[next obsidian]

I think you can prove it for rings maybe

[latent anvil]

You need to check (x-a, y-b) cap k[x, y] isn't all of k[x, y]

[next obsidian]

and consider F[x,y] as F[x][y][

[latent anvil]

But that's pretty easy

[next obsidian]

and do it in two steps

[latent anvil]

Yeah I was thinking of something similar

But yeah @woven obsidian we're pretty much done

[woven obsidian]

I guess no element except 0 in k is in the ideal

[latent anvil]

Yeah exactly

If it contained a unit, we'd have 1 = (x-a) p + (y-b) q in F

But this is absurd

[woven obsidian]

Yeah, thanks a lot for the help

[latent anvil]

Np

This stuff is great but can be hard to wrap your head around

[woven obsidian]

We didn't do much ideals in my last algebra course, so this book will be a challenge

[latent anvil]

Ah yeah

Is this Reid?

Undergrad comm alg?

[woven obsidian]

Yeah

[latent anvil]

We used it in my comm alg class too

It was good

But definitely worth reviewing some ring theory stuff first

[woven obsidian]

Yeah I'm looking at Durbin at the same time, for the more basic stuff

[golden pasture]

by any chance anyone knows where the word 'characteristic' of a ring/field comes from?

[solemn rain]

its weird

i asked this question b4 and i just got a trash answer

cuz theres also a characteristic subgroup which has a totally diffdefinition

but same name so meh

[chilly ocean]

I think it's often very difficult to track down those kinds of things.

If it were me, I'd be satisfied with making up a reasonable etymology.

In this case, it's characteristic in the sense that it's a property of the ring/field. And it's one that happens to be easy to see on the surface.

[golden pasture]

ig it is quite a natural one to think off but still feels unanswered

[chilly ocean]

Yeah. I would say it's unlikely to find a good answer.

Unless you really dig hard for it. I wouldn't count on the Internet to provide.

[golden pasture]

time to stare at old manuscripts

[carmine fossil]

Probably something to do with hilbert

Or dedekind

Or it's characteristic because it is such a easy property to define

[golden pasture]

Steinitz’s groundbreaking 150-page work “Algebraic theory of fields” of 1910 initiated the abstract study of fields as an independent subject

it seem to have came from this

page 75 of https://www.springer.com/gp/book/9780817646844

[carmine fossil]

Any idea whether electric "fields" are related to fields?

[golden pasture]

they prob originated differently

[woven obsidian]

I'm having some trouble understanding how to conclude the "Moreover" part from the proof

I guess the point is that any (f1,f2) without common factors is of the form described in the statement, but I'm not sure how to show this

[next obsidian]

so do you see one direction?

Namely that any ideal m of that form is maximal?

@woven obsidian

[woven obsidian]

Two such polynomials should not contain any common factor so by the proof of the theorem they should generate a maximal ideal I think

[next obsidian]

well really we should think about it like this

Z[x]/(p,g) can be calculated by quotienting in steps

so first you quotient by p

so you look at F_p[x]/(g)

except the image of g inside of F_p[x]

then if g is irreducible this is a field

[woven obsidian]

And then that should be a field

[next obsidian]

yup

[woven obsidian]

Ok that's a nice way of thinking about it

[next obsidian]

So the other direction is harder I think, but I have a method which might work, but let's try and figure it out

so first thing is a maximal ideal must contain some element not in Z

the proof is, if m is maximal then it's the kernel of some map Z[x] -> F for some field right?

suppose the kernel is contained in Z

then Z[x]/m is actually (Z/m)[x]

which isn't a field

Actually I don't think you even need to consider that it's a kernel for that haha

[woven obsidian]

Hmm ok

[next obsidian]

Now the next idea would be to try to show it contains a prime number

My goal is to show it has a polynomial, then a prime number

Show that there needs to be a polynomial which is irr mod that prime

[woven obsidian]

Ok makes sense

[next obsidian]

then if p, g are those, we know that (p,g) subseteq m

but since we already showed (p,g) is maximal we conclude the two are equal

I don't know if I can actually manage to do this, but we can try haha

[woven obsidian]

The proof does establish that any two polynomials without common factors generate a maximal ideal doesn't it?

[next obsidian]

I remember actually reading a pdf for this and if I recall it's sort of complex

I don't think so

take like x - 1 and x + 1

[woven obsidian]

They generate the whole space?

[next obsidian]

I'm pretty sure

Like I know coprime ideals generate evertying

but I guess a better example is x + 1 and x + 2

you can see that 1 is in there

For x - 1 and x + 1 I struggle to see how you could get 1

I don't think you can

Oh I think you get 1

So consider that if you quotient Z[x] by (x - 1) you get Z

so if you do it in steps again

The image of (x + 1) under this map is just 2

Actually huh

F_2???

[woven obsidian]

Hmm

[next obsidian]

Oh duh

[woven obsidian]

I see now though

[next obsidian]

Here's the proof lol

you have 2 in there

so it's equal to like (2, x + 1)

which is of the criteria from before

But (x + 1, x + 2) is equal to (1)

[woven obsidian]

Yeah

I see that the proof supposes it's a prime ideal too

[next obsidian]

I mean

[woven obsidian]

Which (x+1,x+2) isn't

[next obsidian]

a maximal ideal is always prime

[woven obsidian]

Yeah

[next obsidian]

Point being is coprime polynomials gives weird stuff

I think

Anyway, I want to figure out why a prime ideal is in any maximal ideal

[woven obsidian]

I'm guessing the author thinks the moreover part is a trivial consequence of the proof

[next obsidian]

I believe this is the idea

OOO

Nice

Consider the inclusion of Z in Z[x]

for any prime ideal of Z[x] p, p\cap Z is prime in Z

You can frame this in terms of ring homomorphisms

for any ring hom f:A - > B, f^{-1}(p) is prime when p is prime in B

Consider that p\cap Z is just the inverse image of p under the inclusion i:Z -> Z[x]

So the point being is m\cap Z is prime, so equal to (p) for some prime number

So that means p is in m

[woven obsidian]

Ah right

[next obsidian]

So now we just need to show that there's some polynomial which is irreducible in F_p

or like, its image is

Also it should be clear that no other prime exists besides this p, if that ends up mattering

because of Bezout's lemma if you had p,q prime they're coprime so you get 1 as some Z-linear combo of p and q

i.e. there exists a,b such that ap + bq = 1

[woven obsidian]

Yeah that's true

[next obsidian]

So if that ever ends up mattering

¯_(ツ)_/¯

Okay so, here's my idea

Maybe this is the kernel of a hom time?

Suppose m is maximal, then it's the kernel of f:Z[x] -> F

then we know m contains (p)

so we can factor f as like

Z[x] -> Z[x]/(p) -> F

by the universal property of quotients

[woven obsidian]

Right

[next obsidian]

Then Z[x]/(p) = F_p[x] right>?

So now we can look at the kernel of F_p[x] -> F

since this is a field, it has to be equal to (f) for some irreducible f in F_p[x] right?

[woven obsidian]

Yeah

[next obsidian]

Like, it's a PID so it has to be generated by one thing, prime <==> maximal so the generator is prime <==> irreducible

So now f has to actually be the image of something in m

And the reason is ummm

Oh it's surjective right? the map Z[x] -> F_p[x]

now if f maps to 0

Okay so take something in Z[x] mapping to f

call this g

Then you know that g maps to f, which maps to 0

In the composition Z[x] -> F_p[x] -> F

but that's the same as saying g maps to 0 in Z[x] -> F

so it's in the kernel, so it's in m

[woven obsidian]

And this g is our polynomial?

[next obsidian]

Yup

since its image, f is irreducible in F_p[x]

So from what I argued before, (p,g) subseteq m, but both are maximal so they're equal

[woven obsidian]

Thanks for the help, I'm not used to these kinds of arguments

[next obsidian]

No worries

I think this is the first time I came up with this argument all by myself so I'm happy to have gone through it

When I was reading the same book, I think I just remember that I saw the classification of maximal ideals in Z[x] before

and I just shrugged it off without thinking about it more

[woven obsidian]

Yeah I haven't really studied anything similar before

Starting to wonder how this course will be if this is in chapter 1 haha

[next obsidian]

FWIW the first chapter is like

harder than the rest

IMO

It's like "here's the stuff about general ring theory you should know"

ludicrously hard problems

Atiyah Macdonald did the same

I guess A-M's exposition wasn't as hard

but god the chapter 1 problems

were crazy hard

Once you start to develop the machinery they do a lot of the heavy lifting

[woven obsidian]

That's very nice to hear :)

But that's also why I want to make sure I understand everything from this chaptet

[eager willow]

I read an uncited claim that said for a maximal ideal $m \subset R$, that the multiplication map $m\otimes m \to m^2$ is an isomorphism whenever $m$ is a flat $R$ module. What is meant by $m^2$? The product of ideals $mm = m$? If so does this simply follow from looking at the exact sequence $0 \to m \to R \to R / m \to 0$ under the $m\otimes _ $ functor?

[cloud walrus]

datorangeguy:

[cloud walrus]

datorangeguy:

[next obsidian]

There’s a result that a module F is flat iff for all ideals I, the map I (x) F -> IF is an isomorphism

Take I, F = m and the result follows

Note: hard to prove without Tor. With Tor it isn’t that bad, without Tor my proof used the snake lemma like... 3 times?

For these purposes, it’s basically as you said, the map is injective by flatness of m, and the map m (x) m -> R (x) m = m is just given on simple tensors a (x) b by multiplying the two. This clearly maps into m^2, and is surjective almost by construction

I don’t see any reason this is limited to maximal ideals, it should hold for all ideals

[eager willow]

Oh right m^2 is not all of m.

Following the exact sequence we get m(×)(R/m) is isomorphic to m / m^2 ? That's pretty cool

[latent anvil]

yes, this is also an instance of a more general fact about tensors

If M is an R module and I an ideal of R, then M (×) R/I = M/IM, where IM is the submodule generated by all products ix with i in I and x in M

[chilly ocean]

quaternions

[smoky cypress]

How do I show that $\sum_{k=0}^n\frac{x^k}{k!}$ is irreducible over $\bQ$ for all $n$?

[cloud walrus]

Whoever:

[latent anvil]

hmmm

This seems hard

[smoky cypress]

I found a proof

https://kconrad.math.uconn.edu/blurbs/gradnumthy/schurtheorem.pdf

That uses prime ideal factorizations to show a more general result

Don't know what prime ideal factorizations is yet so rip me

@latent anvil

[olive mirage]

I have not looked at the link, but I'd try to use Bertrand's Postulate

(and Gauss's Lemma)

[vestal snow]

If the multiplicative sets S, S' are such that S is a subset of S', then there is an injection of A_S into A_S' right?

Not looking for proof, just a yes or no

[latent anvil]

no, not an injection

Take S = {1} and S' = A

A_S = A but A_S' = 0

[vestal snow]

I am using the definition that multiplicative sets are closed under multiplication AND do not contain 0

Is it an injection now?

[latent anvil]

No, take S' containing a zero divisor but not zero

[vestal snow]

Yeah you're right

[latent anvil]

A = Z[x, y] /(xy) and S' = {x^n}

[vestal snow]

What about an injection from A_f into A_fg?

[latent anvil]

hmm

This says that if f^r g^r x = 0 then f^s x = 0 for some r

Doesn't smell like it's always true

Geoemtrically A_fg is the intersection of the open sets A_f and A_h

*A_g

But I'm not sure what injectivity of rings maps says on schemes

oh no it doesn't

Take my example from before

Are you okay with a case where fg = 0?

Actually wait

Take f = 1

g = x

Then my example from before works

@vestal snow

[vestal snow]

Hmmm

[latent anvil]

Generally localization is weird

It will add things

But it can also collapse things

[vestal snow]

Can you explain the last line of the proof?

we get that a/f^n = 0 in A_gf

[latent anvil]

gf is not p

Because f not in p and g not in p

[vestal snow]

Ah yes

That's an important thing I didn't use

[latent anvil]

So eventually your element is 0

In the colimit

So we get injectivity

[vestal snow]

I don't think you need to deal with colimits

[latent anvil]

But φ is a map out of a colimit?

I'm not sure how you'd avoid dealing with colim A_f here

[vestal snow]

My bad you're right

I was thinking of something else

So the idea here is that anything that maps to 0 in the codomain eventually becomes 0 in the colimit right?

[latent anvil]

Yup

like formally

Take an element of the kernel

That can be represented as the equivalence class of an element of A_f for some f

[vestal snow]

Yeah I get the argument now

Thanks

[solemn rain]

any help?

in the ring A[x] prove that the jacobson radical is equal to the nilradical

[latent anvil]

What is A here?

any jacobson ring?

[solemn rain]

any ring

commutative

with identity

[latent anvil]

This doesn't seem true to me but I've been unable to find a counterexample

[solemn rain]

its an exercise for me in atiyah macdonald

number 4

the problems are super hard

for me

tbh

[latent anvil]

Yeah this first section is brutal

I did this problem at some point

[solemn rain]

im getting fucked badly

[bleak abyss]

Hmm, so we're saying a maximal ideal in k[x] must contain every nilpotent polynomial

[solemn rain]

i only did 1 problem on my own

[bleak abyss]

And a polynomial being nilpotent implies the coefficients are nilpotent

[solemn rain]

yes

i didnt know how to prove this for jack shit

the author just hitned at induction and i was just put off

im now doing the zariski topology thing

which looks eaiser

does 0 and 1 in number 15 mean the ideals generated

or the subsets {0} and {1}?

obv subsets right?

[bleak abyss]

You mean exercise 2?

[solemn rain]

yea 15 in 2

[latent anvil]

So I feel like you want to use 2i and 2ii

[solemn rain]

ofc subsets right?

[bleak abyss]

Oh oh

[latent anvil]

Oh lol other people said that didn't they

[bleak abyss]

Lmao

LMAO

Okay

Lol

[solemn rain]

?

what

im scared

[bleak abyss]

With that hint

[latent anvil]

yes it means that

[bleak abyss]

This problem is an actual freebie

[latent anvil]

Or just {0} and {1}

[solemn rain]

yea im sorry for that stupid question

if an ideal has a unit then its just the whole ring

and 1 is a unit so lmfaoa

[bleak abyss]

Oh I was talking about problem 4

[solemn rain]

i left it for u guys

i gave up

the problems are just too hard for me

i cant do single one of them

[latent anvil]

The hint should help

[bleak abyss]

Viburnum's hint gives it away

Just think carefully about problems 1, 2i, and -2

[latent anvil]

dude they're too hard for everyone, I haven't met anyone who's been like "oh yeah ch1 of AM was a breeze"

which is to say

[solemn rain]

i literally cant solve any of them

[latent anvil]

Do them

[solemn rain]

i solved number1 by pure luck

showing if x is nilpotent 1+x is au nit

[bleak abyss]

mo2men this is how you get good lmao

You gotta power through

Here's the hint

Look at problems 1, 2i, and 2ii

[latent anvil]

lmao -2

Good bit

[solemn rain]

i couldnt even do those

i only did 1

[latent anvil]

yeah cuz they're hard

[bleak abyss]

Use them even if you didn't do them

[solemn rain]

u know what?

im going to do it

[bleak abyss]

Don't be afraid to use theorems

[latent anvil]

those problems could take hours

[bleak abyss]

4 is easier than 2

[latent anvil]

They're hard

[solemn rain]

i cant do 4

[latent anvil]

or days

[bleak abyss]

Think more

[solemn rain]

maybe somethings wrong with me

anyways im trying again

but after i finish the zariski topology tghing

its def unfolding so wont take ami n

a min*

[bleak abyss]

:0

It won't take me?!

[solemn rain]

?

[latent anvil]

dox????

[solemn rain]

??!

[latent anvil]

@ mods

[bleak abyss]

Whoops

[solemn rain]

u guys are fucking scaring me

wtf

oh amin

lmfao

fuckign nerds

[latent anvil]

@bleak abyss hey someone doxed an admin

[bleak abyss]

😠

[latent anvil]

Not cool

[bleak abyss]

Who

Dares

[solemn rain]

dark

daminark

lmfaaao clever

i just got it now

anyways back to am

[latent anvil]

It was either mo2men or daminark

Not sure

[bleak abyss]

If only we knew

Then I could ban one of them

And yeah mo2men everyone has that revelation eventually

And they're like ohhhhhhhhhh

But yeah you continue, tbh I should've done Atiyah-Macdonald or something

I had various opportunities to just take some time and be like

Aight doing some problems now

But I missed that window so instead I'm reading Matsumura

[scarlet estuary]

wait what revelation

[latent anvil]

That sloth king daminark is an anagram of I Am Lord Voldemort

[scarlet estuary]

oh i thought it was "dami is a nerd"

but i figured that no one can truly understand the full scope of dami's nerdhood

[solemn rain]

is this correct proof of iii:

[latent anvil]

2iii?

Or 15iii?

[bleak abyss]

-i

[solemn rain]

(just one side for now): let x be in V(UE_i) --> x is a prime ideal that contains UE_i --> the intersection must be contained in x aswell --> E_i is contained in x for every i in I --> x is in intersection(V(E_i)) for i in I

no thats the topology stuff

i just wanna make sure

15iii

[latent anvil]

I don't understand the last implication

okay yes I get it now

[solemn rain]

okay okay

[latent anvil]

The intersection must be contained in x is a bad way to phrase it

[solemn rain]

i do the otrher side and last one

yea yea

sorry

okay i do other side and last one and now i know what spec(A) means

then i do the hard boys

[latent anvil]

Well

Kind of

[solemn rain]

kind of

what

[latent anvil]

This is the easy half of Spec A

The hard part is the sheaf

[solemn rain]

oh i think that comes later in AG

[latent anvil]

The topology isn't bad to write down

[solemn rain]

the word sheaf

wtf theres an exercise asking me to draw

lmfao?

[latent anvil]

yeah dude draw

[solemn rain]

idk how to do that i dont know topology

[latent anvil]

lol it's not asking for real pictures

These exercises are kind of silly

okay so

what are the primes of C[x]?

[solemn rain]

wait

b4 that

i did the other side :

let x be in the intersec(V(E_i)) --> x is in V(E_1) and V(E_2) ..... --> x is an ideal that contains E_1, E_2 .... --> x contians the union --> x is in V(UE_i)

correct?

hence both equal by containment or wahtever

V(a intersects b) = V(ab) where a and b are ideals

let x be in V(a intersects b)

i want to show x contains ab

so let z be in ab --> z = cd for c in a , d in b --> c is in x and d is in x --> x is an ideal tho --> z is in x

thats first containment

okay i can do the other1

for number 4n ow

so the jacobson radical is the intersection of all maximal ideals

and maximal ideals are prime iddeals ( cant remember how to prove that lmfao im bad )

can i say therefore the jacobson is a subset of the nilradical?

tag if help

[golden pasture]

yes

i would send a drawing of like spec C[x,y] for lolz but may be spoiler

[solemn rain]

okay

so now the other side

let me think

[winter vigil]

Hello, here is my attempt to prove the uniqueness of the division algorithm

Suppose there exists $s,s',r,r' \in P(F)$ such that, $q=sp+r$ and $q=s'p+r'$, where $p,q \in P(F)$. Subtracting both equation, we obtain $p(s'-s)=r-r'$ , rearranging $p=\frac{r-r'}{s'-s}$ . Since $p \in P(F)$, $s-s'$ is divisible by $r-r'$, hence we could write $r-r' = Q(s'-s)$ where $Q \in P(F) \setminus {0}$.

However, $deg p > deg r$ and $deg p > deg r'$. So $deg p > deg (r-r')$. The only case where $p=\frac{r-r'}{s'-s}$ holds is when p is the zero polynomial. Thus, $0=r-r', r'=r$. From
$r-r' = Q(s'-s)$ we can conclude that $s'=s$, as $Q \neq 0$.

[cloud walrus]

Otoro:

[winter vigil]

<@&286206848099549185>

[knotty mason]

@winter vigil if sp + r = s' p + r'

you can rewrite it like this: (s-s')p = r'-r

now you want to show r'-r = 0

using the fact that 0 <= r',r < p

and the fact that p divides the LHS so it divides the RHS

I would avoid doing the division p = (r-r')/(s'-s)

[winter vigil]

Oh what would be the difference ?

May I ask why would we avoid division ?

Oh, I'm sorry I haven't learnt rings yet. This is a chapter about polynomials in axler's linear algebra

So what would be the changes do I have to make to my proof ?

[knotty mason]

i outlined what to do

my advice was to go straight from (s-s')p = r'-r to showing that r'-r = 0

rather than dividing

the LHS tells you that we have 0 or p or 2p or 3p ..

the fact that r' and r are small lets you conclude that we have 0

[chilly ocean]

@knotty mason is right. And since r-r'=0, you can conclude that s-s'-0 proving uniqueness

The fact that $r-r'$ is a multiple of $p$ implies $r-r'=0,\pm p, \pm 2p, \pm 3p,\ldots$. But remember that each of $r,r'$ are such that $0\leq r,r'\leq(p-1)$. Hence, $r-r'=0$ is the only appropriate choice.

[cloud walrus]

Mega Jeff:

[winter vigil]

Okay, thank you

[lavish sigil]

is there any well-established name for the semidirect product $\mathbb{Z}\rtimes\mathbb{Z}/2\mathbb{Z}$, where $\mathbb{Z}/2\mathbb{Z}$ acts on $\mathbb{Z}$ by $n\mapsto -n$?

[cloud walrus]

gustavn64:

[fair obsidian]

Maybe something like the affine transformations? On a vector space, if you consider maps of the form f(v) = Av + w for some invertible A and vector w, I think those are the affine transformations.

This is like that but on Z (as the identity and n to -n are the only automorphisms of Z)

[latent anvil]

I think some people wanted to read matsumura with me

Maybe @maiden ocean @bleak abyss

I'm going to do that now

[maiden ocean]

nice

[latent anvil]

However I don't know if I should read commutative algebra or commutative ring theory

I'm waiting for comm alg to download

Probably comm ring theory since that's what chmonkey has been reading

[bleak abyss]

Commutative ring theory seems better™️

The only thing it doesn't have are "Excellent rings"

[latent anvil]

Yon says some homological methods are missing but I'm not planning to get that far that it matters

On this read

And I'll cover those in my classes in spring anyways

[kindred mist]

How do I show that a finite algebraic extension is simple? Any hints are appreciated (I tried to read this in the one book and I did a bunch of googling and none of that helped)

Note I do not want to assume separability

[next obsidian]

Are you assuming the field is characteristic 0?

[kindred mist]

nope

I guess I can do it by cases?

char 0 and char p

[latent anvil]

Is this true?

Let K = Fp(x, y)

consider K(x^(1/p),y^(1/p))

Isn't this algebraic but not simple?

[next obsidian]

Is it finite?

[latent anvil]

Yeah it's a degree p^2 extension, right?

[next obsidian]

I want to say yes

Like x^k/p y^j/p

Makes up a basis

[latent anvil]

Well yeah we also adjoined two elements which are degree p

I want to say every element of that extension is degree p over K

[next obsidian]

What’s the result in the char 0 case called?

[latent anvil]

And so it can't be simple

primitive element theorem

[kindred mist]

Ok, let me reprhase then. I want to show that a finite algebraic extension of the form k(c_1, ..., c_n) is simple

[latent anvil]

this is the usual counterexample

my example is of that form, isn't it?

[kindred mist]

yeah I think it is, lemme grab the book page

to make it clear what theorem I mean

sorry about that

[kindred mist]

So if we restrict to characteristic 0, then I am looking to prove this "primitive element theorem" then I suppose

[next obsidian]

Any finite extension is algebraic anyway so

[latent anvil]

Yes, in char 0 separable is automatic

[next obsidian]

Shamrock’s example probably works

[latent anvil]

works for what?

[next obsidian]

And if you buy Wikipedia’s thing then that’s a counterexample

A counterexample in char p

[latent anvil]

yeah you just take the pth power of an arbitrary term (sum of ai x^i/p y^j/p) and use the freshman's dream

[next obsidian]

Yeah

[latent anvil]

To get that the pth power is in K

[next obsidian]

Oh wait

Lmao this is exactly the example they give

[latent anvil]

oh lol

I didn't read your screenshot

[kindred mist]

ok so use separability to prove it for characteristic 0 then

got it

[next obsidian]

Yup

[latent anvil]

yee

Algebraic and separable are equivalent in char 0

or over Fp

[next obsidian]

Uhhh wait what?

In particular, finite separable extensions are simple.

Oh nvm

Right, we worked over Fp(x,y)

Not Fp

[latent anvil]

yup

Don't ask me to prove it over Fp I forget why it's true

Probably obvious™

[fair obsidian]

I believe the multiplicative group is cyclic, so you can pick a generator

[latent anvil]

Oh lol I can just appeal to the classification can't I

I know what all algebraic extensions are

[next obsidian]

Because?

Finite algebraic ones

You mean?

[latent anvil]

yes but we know what the algebraic closure is

It's the union of all finite ones

Actually no that's not exactly what I mean

Separable means the minimal polynomial of every element is separable

But this is "local"

It reduces to finite algebraic extensions

[next obsidian]

Right

[latent anvil]

so we just need to know that Fp^n/Fp is always separable

[kindred mist]

I was in an AG lecture today and someone used the phrase "a fixed algebraic closure of k" but this confused me because k bar is unique, right?

[next obsidian]

Sure and that’s like

Nope

[latent anvil]

no countable

It's weird okay

So

It's unique up to isomorphism

But there's lots of weird isomorphisms

[next obsidian]

Isn’t there a good example like if you take an algebraic closure of R

You can get C in the usual way

But you can do some other weird crap

[latent anvil]

okay not in that case

Gal(C/R) = 2

[next obsidian]

Does that actually

[latent anvil]

and that's not as convincing

[next obsidian]

Okay

[kindred mist]

Gal was Aut_k(L) right?

[latent anvil]

yes

I'm trying to think of a good analogy

[kindred mist]

is it not unique no matter the characteristic?

[latent anvil]

someone compared it to choosing a basepoint when you calculate the fundamental group

you have to choose a basepoint, but it doesn't really effect the outcome up to isomorphism

[next obsidian]

affect

[kindred mist]

ok I see what you mean

[next obsidian]

#owned

😏

[latent anvil]

But it kind of does, because you have to pick a path to show π1(X, p) ≈ π1(X, q)

[kindred mist]

so like Pi_1(s^1) = Z but the equivalence classes are technically different with a different basepoint

[latent anvil]

And that depends on the homotopy class of the path from p to q

right, if I choose a different basepoint

And my path between 1 and that basepoint winds around the circle three times

I'm identifying the fundamental groups in a kind of weird way

[kindred mist]

yeah

so there is sort of an analogous link with different k bars then eh?

[latent anvil]

similarly there are a lot of automorphisms of "the" algebraic closure, and so lots of isomorphisms between two algebraic closures

Yeah

[next obsidian]

The algebraic closure is always like some colimit?

[latent anvil]

kind of chmonkey

I don't think I ever got the details entirely right but

[bleak abyss]

Probably a colimit of algebraic extensions

[latent anvil]

If you choose all the finite extensions

algebraic is dumb dami

you're including the algebraic closure

[next obsidian]

I’m trying to get for free any algebraic closure is isomorphic

Haha

[kindred mist]

colimit sounds like something category theoretic, lemme look this up real quick

[latent anvil]

so if you take all finite ones but remove the isomorphisms

Like

You choose a canonical way of embedding all possible finite extensions into one another

You can then take a colimit

But if u have multiple embeddings

You're going to like try and coequalize them

And get fucked up

[bleak abyss]

Oh isn't it a thing that like, if L is an algebraic extension of k, then a homomorphism from k into an algebraically closed field extends to a homomorphism out of L?

[latent anvil]

Yeah that's true

[bleak abyss]

So yeah if L is another algebraic closure

You apply this

[latent anvil]

basically an algebraic closure/extension is the limit of "all" finite extensions

Terms and conditions apply

like inside your actual fixed algebraic closure everything includes into everything else uniquely

[bleak abyss]

You get an embedding and the image should be surjective since otherwise it'll be a smaller field which is algebraically closed and algebraic

[latent anvil]

You want to reproduce that structure in the colimit

wait dami I'm not sure what you're saying right now

Are you proving uniqueness of algebraic closure?

@countable you don't need to learn about colimits right now

I can't ping your nick lol

I mean you can if you want to but the colimit thing was just a question chmonkey was asking

@next obsidian category theory would give you a unique iso

And the whole point is we don't have that kind of canonical structure

[next obsidian]

Yeah but setting it up in a way to express it as something universal seems kinda hard

[latent anvil]

That's why we need to make these weird choices about how all the finite extensions sit inside one another

[next obsidian]

Yeh

[kindred mist]

I see

I guess I am just looking for examples of different alg. closures after fixing a field

so idk say if I fix Q or something

[bleak abyss]

Oh I thought he was asking it specifically in service of uniqueness

[latent anvil]

Oh he was but I think he wanted it for free

Like

By magic

[next obsidian]

Yes

[latent anvil]

Countable what do you mean by different?

[bleak abyss]

Well the work involved in showing it's a colimit of stuff is the content of this proof too

[latent anvil]

Because they'll all be isomorphic

yeah I agree dami

[kindred mist]

I mean different as in not equal sets

[bleak abyss]

Either way you need to know that you can extend homs

[latent anvil]

Oh

[bleak abyss]

And then conclude in a line

[latent anvil]

Then take an algebraic closure K of k

Look at K × {0}

We do all field operations on the first component

[kindred mist]

would we say that K x 0 is an algebraic closure of k though? as opposed to k x 0 ?

[next obsidian]

I mean

K x 0 is an algebraic extension of k

And is algebraically closed

[latent anvil]

okay so take (K × {0})\(k×{0}) union k

Not exactly chmonkey

What does field extension mean?

[next obsidian]

Wat

[latent anvil]

does it mean set theoretical containment?

[kindred mist]

I thought extension normally meant that it must contain k as sets

[next obsidian]

No

[latent anvil]

Or a choice of field homomorphism

lol

[next obsidian]

Okay well

[latent anvil]

we have two answers in chat

[next obsidian]

Just redefine all the k x {0} elements as their k counterpart

And then nothing changes

[latent anvil]

Countable suppose F/k is some field extension and α is an element of F. Let m(x) be the minimal polynomial of α. Would you say k[x]/(m(x)) and k(α) are isomorphic extensions of k?

[kindred mist]

so is it correct to say that $\mathbb{C} \times 0 \times ... \times 0$ is an algebraic closure of say $\mathbb{Q}[i]$?

[cloud walrus]

𝓒𝓸𝓾𝓷𝓽𝓪𝓫𝓵𝓮:

[latent anvil]

No

Any algebraic closure of Q[i] is countable

[kindred mist]

ok then what about if we replace Q[i] with the reals?

[latent anvil]

then yes I would say that

Assuming you define the field structure on C × 0 ×... ×0

[kindred mist]

in the natural way, alright

[latent anvil]

okay so about my question above

Would you say k(α) and k[x]/(m(x)) are isomorphic extensions of k?

[kindred mist]

are you asking me?

[latent anvil]

Yup

[kindred mist]

I'm just used to an extension being a field that literally contains the subfield, but I see the natural way that k[x]/(m(x)) would contain k (I think...)

by sending like c to c bar I guess

[latent anvil]

Sure, I'm just trying to point out that k[x]/(m(x)) doesn't actually contain k

But we refer to it as an extension of k all the time

[kindred mist]

ok that is what I was not sure about (I mean after applying an isomorphism)

[latent anvil]

Like this is one of the most important basic facts about field extensions

That you can adjoin an element by modding out by its minimal polynomial

But you can't state it very well unless you generalize the notion of field extension a little

[next obsidian]

Shamrock do you feel like isomorphic field extensions should mean they’re isomorphic as like in the coslice way or whatever.

I think that’s the most natural way to explain it

[latent anvil]

an extension of k is a field F with a field homomorphism k -> F

That's what I would say

that's what it is, isn't it?

like by definition

[next obsidian]

I mean that adds nuance, so the way they’re iso matters

Since it has to be compatible with the embedding

Anyway, this is not important rn

[kindred mist]

is it literally just a homomorphism? Does it not have to be surjective or anything?

[latent anvil]

well countable, what property does every field homomorphism have?

[kindred mist]

nonzero implies injective

[latent anvil]

Oh

[kindred mist]

(if the map is nonzero I mean)

[latent anvil]

In my brain field and ring homomorphisms have to preserve 1

So it's just "injective"

but yeah I would say an extension is a field with a nonzero map k -> F

[kindred mist]

ah ok I wasent assuming that but then yeah

that would make more sense to assume nonzero I see

because I guess we wouldnt want 0 to be an extension of a field

[latent anvil]

And then a homomorphism between extensions has to preserve these embeddings. So if $i : k \to F$ and $j : k \to F'$ a homomorphism $φ : F \to F'$ of extensions of $k$ should satisfy $j = φ \circ i$

[cloud walrus]

shamrock:

[latent anvil]

So if you embed an element of k in F and apply φ, it's the same as embedding it in F'

[kindred mist]

so in the familliar case of like Q and Q[i] then we just naturally use inclusion for this then I guess

[latent anvil]

Right

Actually Q is special

If Q embeds into a field at all, it does so in a unique way

[kindred mist]

Ah yeah

just by basic properties

[latent anvil]

Right, it has to send 1 to 1

And Q is generated by 1 and field arithmetic

@next obsidian do I need to read the appendices

Of matsumura

[next obsidian]

I didn’t

[latent anvil]

Kk

[next obsidian]

I think it’s like here’s tensor product facts, here’s homological algebra facts we know

Like snake lemma, and some stuff about Ext and Tor

[latent anvil]

Yeah I figured

[next obsidian]

I glanced at it

[kindred mist]

So is it true that: a field extension is not unique as a literal set up to equality iff we take this generalized (what to me is "generalized") definition of extension? as opposed to the definition of extension that requires set containment, right?

if you get what I mean

(I can state that more precisely if it is unclear)

[next obsidian]

If you mean it in the first way

Like requiring set containment

What’s your notion of equivalent field extensions?

[latent anvil]

I'm not sure what unique as a literal set up to equality means

like

The field F with a map k -> F is unique

[kindred mist]

unique object in the category Set

[latent anvil]

Anything equal to it is equal to it

I don't know what that means either

[kindred mist]

Here lemme restate then

I meant closure unique not extension

sorry about that

Is algebraic closure unique if we use the very specific definition of field extension that requires literal set containment --- this is my question

[latent anvil]

No

I gave an example above

(K×{0}) \ (k×{0}) union k

[kindred mist]

not up to isomorphism though (unless Im still misunerstanding)

I mean that another algebraic closure would have to equal a fixed k bar

[latent anvil]

You can always take a generalized extension ι : k -> F and (F\ι(k)) union k will be a literal extension of k which is isomorphic as a generalized extension to the one we started with

I'm not sure what you mean by the last comment

Two algebraic closures might not be set theoretically equal

They will be isomorphic

There are lots of isomorphisms to choose

And it's not clear which one is best

[kindred mist]

yes I know but I mean if we enforce the further condition that they are

then is it unique?

[latent anvil]

that they're equal?

[kindred mist]

equal yes

[latent anvil]

if two algebraic closures are equal, they are equal

[kindred mist]

like if we include that, and the set containment thing

[latent anvil]

what's the that?

I'm confused by what exactly you're asking

If you have two algebraic closures K, K' of k

k is a literal subset of both

and (???)
then they're (equal? isomorphic?)

[kindred mist]

then they need not be equal right?

after the k is a literal subset part

[latent anvil]

Yeah we can still have K ≠ K'

[kindred mist]

ok that was my question

[latent anvil]

Right, I posted an example earlier

[kindred mist]

sorry I wasnt sure exactly how to phrase it

[latent anvil]

okay so take (K × {0})\(k×{0}) union k
@latent anvil

Let K be some algebraic closure of k

Literally containing k

Then you can turn the set above into an algebraic closure of k too

Add/multiply elements like you were actually in K

[kindred mist]

Thanks

For some reason I can do commutative algebra just fine and then when I try to do Galois theory it is always slow going

[latent anvil]

It's certainly a different flavor

I think these kinds of issues come up in commutative algebra too

Or at least in algebraic geometry

[kindred mist]

I mean its probably healthy to know a good bit of Galois for com. alg.

does not seem super important to me until AG though

[latent anvil]

Oh sure I meant the non uniqueness stuff

I was thinking about defining the structure sheaf on a basis of distinguished opens

we say stuff like O(D(f)) = A_f

But this only makes sense up to isomorphism

[kindred mist]

It might just be a slight, not really linguistic but maybe mindset barrier with me and Galois idk

[latent anvil]

sure

[kindred mist]

I have seen structure sheaf, and I take it O is your functor, but can you remind me of this A and D?

[latent anvil]

oh so like

Let A be a commutative ring

The sets D(f) = { p in Spec A : f not in p } form a basis for Spec A

They're the "distinguished opens"

You can define sheaves on a basis of open sets instead of on all the open sets

[kindred mist]

then the rest of the mappings should just be naturally induced right?

[latent anvil]

and so one way to define the structure sheaf of Spec A is by O(D(f)) = A_f

Well it's more about defining O(U) for U not a basis element than the restriction maps

The problem with this "definition" is that we can have D(f) ≠ D(g) for different f, g, so A_f and A_g are set theoretically different rings

but they're isomorphic, and you can make it all work because there's a big family of isomorphisms between all the possible A_f's and A_g's with D(f) = D(g)

I'm really just rambling at this point

It's not that related to your original question

@open torrent hi

[kindred mist]

This reminds me of how sheaf cateogry was sort of described as 'encompassing' com. ring theory on the wiki page

(although what I said is sort of a sloppy restatement of what they said)

[leaden finch]

can someone help me with this one

[karmic nebula]

From

#❓how-to-get-help

Rule 3

Stick to one channel and don't post the same question in multiple channels. Please don't ask for help in other channels if no one is responding in the one you have posted your question in.

[knotty mason]

@leaden finch you can use mod 5 arithmetic for this

check what possible values n^5 has mod 5 for n=0,1,2,3,4

[leaden finch]

how do you do mod 5?

[knotty mason]

mod 2 is {even, odd}

mod 5 is {0,1,2,3,4}

the system works by adding and then reducing, 4 + 3 = 7 = 7-5 = 2

[leaden finch]

i did this

they all came out to be true

[latent anvil]

@next obsidian I'm already stuck on a matsumura problem

[next obsidian]

Oof

Which one?

The benefit to Matsumura problems is solutions are in the back so if u crack you can look at it, and I have definitely done that before after a while

[next obsidian]

Why should Z[Z/p^nZ] be an interesting ring? It seems really cool

[solemn rain]

i dont know mana

look at its prime ideals

[wintry yacht]

Hi. Can someone help me to prove this?
f:(A,+,) -> (R,(+),()) is a group homomorphism and S <= A then prove that f(S) <= R

[carmine fossil]

Can you use texit,pls?

[wintry yacht]

I don't know how to use it

[carmine fossil]

What do you mean by <=?

Belongs?

[wintry yacht]

lesser or equal

like a subset

[carmine fossil]

S is a subset of A,right?

[wintry yacht]

yes

[sharp sonnet]

what have you tried

[wintry yacht]

nothing for now, but I think that there is a theorem who should prove that

[sharp sonnet]

well, you want to show f(S) is a subgroup of R

there is a so-called subgroup criterion that makes that easier

[carmine fossil]

You want the solution?

[wintry yacht]

yes

[carmine fossil]

$Take A,B \in f(S).$
$ A=f(a),B=f(b)\exists a,b;

AB^{-1}=f(a)f(b)^{-1}=f(ab^{-1})
\implies AB^{-1} \in f(S)$

[sharp sonnet]

this is the greatest help of all time

[carmine fossil]

nothing for now, but I think that there is a theorem who should prove that
@wintry yacht prove that

[cloud walrus]

DrunkenDrake:
Compile Error! Click the reaction for details. (You may edit your message)

[wintry yacht]

@carmine fossil I think a and b should belong to group A

[carmine fossil]

Yes

They belong to subgroup S

[wintry yacht]

I mean the lowercase a and b

[carmine fossil]

Yes

[wintry yacht]

ok but is that enough?

[carmine fossil]

Yes

[wintry yacht]

thanks then

[carmine fossil]

Did you prove the subgroup criterion?

[wintry yacht]

No

[carmine fossil]

Do it

[wintry yacht]

ok

[wintry yacht]

@carmine fossil

then I guess that proves what I had to prove

[carmine fossil]

Not that one

How do you determine whether a subset with the group operation,is a subgroup,in general?

[wintry yacht]

Well by proving the subgroup criterion

but I think that it's not needed here because we have a group homomorphism

[carmine fossil]

Just do that

[wintry yacht]

Why?

[neat ginkgo]

you can use the main homomorphism property to do it

[carmine fossil]

That criteria provides a convenient way to deal and prove things like this theorem. And if something makes your life easier,you should definitely use it. Which means you should prove it

[wintry yacht]

I know that theorems need to be proven

Thanks guys for help

[neat ginkgo]

damn

i knew i would pass the exam but i didn't expect to have 93% 😎

wtf i love rings now

[smoky cypress]

Find a ring R such that RxR=R

Or show that such ring doesn't exist

@neat ginkgo

[neat ginkgo]

{0} x {0} = {0}

[smoky cypress]

Lol

Let's assume 0≠1

[neat ginkgo]

😭

[next obsidian]

this is EZ PZ

[sour plume]

is "all convergence is uniform" a new take on acab

[next obsidian]

No, it's what complex analysts believe

[scarlet estuary]

all modules are algebras

[latent anvil]

@smoky cypress infinite product of Z maybe?

[smoky cypress]

yeah

Now find a countable ring

[latent anvil]

Can I use Löwenheim-Skolem for this lol

Consider a language with two constant 0, 1 and binary function symbols *, +, f. Let A be the set of ring axioms (commutative ig) along with axioms staring f is an isomorphism, so

• for all x, y, z, w, if f(x, y) = f(z, w) then x = x and y = w
• for all x, there exists y, z with x = f(y, z)
• for all x, y, z, w, f(x+z, y+w) = f(x, y) + f(z, w)
• for all x, y, z, w, f(xz, yw) = f(x, y) f(z, w)
• f(1) = 1

Models of A are then rings equipped with an isomorphism A × A -> A. There's finitely many axioms and I showed there's some model, so there's a countable model

we love existence proofs

@smoky cypress do you have an explicit example?

[smoky cypress]

countable product of (Z/(2)), and you take the elements that have finitely many nonzero entries

@latent anvil

[latent anvil]

Oh I thought of that but my rings need to have 1

[smoky cypress]

True

I guess you can just add a 1

[latent anvil]

oh like freely?

sounds plausible but I'm worried it would fuck the product thing up

[smoky cypress]

That is true

Hmmmmm

Yeah that's a good point

[latent anvil]

I thought about this but stopped there

You might be able to get it to work

the model theory above proves there's some example though

just gotta find it

[smoky cypress]

Lmao high tech theorems

[latent anvil]

look I have a hammer for producing countable for examples from uncountable examples

you're telling me this isn't a nail??

[smoky cypress]

[vestal snow]

Has anyone read Algebraic Function Fields and Codes by Stichtenoth?

If so, are there any prereqs from AG?

[leaden finch]

can someone check my work

[solemn rain]

a local ring has no idempotents other than 1 and 0

proof: suppose A has the unique maximal ideal M

and e is idempotent

now what i dont get it

[latent anvil]

think about it lol

[solemn rain]

yea yea i got it

i just wanna check

so

[latent anvil]

what does e being idempotent mean

[solemn rain]

e^2 = e --> e(e-1)=0

[latent anvil]

how can you use that

[solemn rain]

so now

[latent anvil]

yes

[solemn rain]

i know by purer

pure luck

that u cant be both zero dviisor and unit

so now

ik e is not = 0 or 1 so neither is 0

now im stuck here

if e is a unit then e is in M

non*

nonunit*

and e is idempotent

so 1-e is a unit

i proved that b

4

but im still stuck somewhere

[latent anvil]

okay, so why is 1-e a unit?

[solemn rain]

cuz e is idempotent

no]

thats e+1

i think cuz if e is in the maximal idea?

ideal*?

can i use the jacobian shit?

jacobson*

i know the jacobson of the ring is just M right?

[latent anvil]

you could

Yes

[solemn rain]

and any elmeent in the jacobson 1- it is a unit

yea yea cool

okay

[latent anvil]

so if e is not a unit 1-e is a unit

[solemn rain]

yea

[latent anvil]

how did you get that e was not a unit?

[solemn rain]

cuz its as zero divisor

?

[latent anvil]

yeah, you have the equation e(1-e) = 0

[solemn rain]

yea

fucking smart ass i am

[latent anvil]

and you know e and 1-e are not zero

[solemn rain]

yea

yea i got that

so now 1-e is a unit

[latent anvil]

right

[solemn rain]

is 1-e in the maximal ideal

[latent anvil]

think about the equation e(1-e) = 0

[solemn rain]

no right?

if 1-e is a unit its not a zero divisor

[latent anvil]

yes

[solemn rain]

and if its not a zero divisor

how can e(1-e) be 0

right?>

[latent anvil]

yes

[solemn rain]

so what now

e must be 0

?

why cant be e be 1

[latent anvil]

the equation e(1-e) = 0 tells you both e and 1-e are zero divisors

[solemn rain]

yea

[latent anvil]

right?

You concluded this for e earlier

But it applies just as well for 1-e

[solemn rain]

yes

[latent anvil]

You're assuming e ≠ 0,1, so neither term in the product is zero

[solemn rain]

yes

so if e is a zero divisor --> e is not a unit --> e is in M --> M is jacobian --> 1-e(1) is unit

1-e is unit

am i rightr?

jacobson*

[latent anvil]

Yeah that logic tracks

[solemn rain]

now whats wrong with 1-e being a unit

what contradiction is there

1-e is not a zero divisor

ur smart af lmfao

yea the equation

says that e must be 0

now

right?

the equation e(1-e) = 0 tells you both e and 1-e are zero divisors
@latent anvil

so e must be 0 here

cuz 1-e is not a zero divisor

neither e is lmfao

am i right

[latent anvil]

yes

[solemn rain]

so e must be 1 or 0

[latent anvil]

yes

[solemn rain]

cool

fucking dizzy problem

tysm

[latent anvil]

here is a slightly simpler proof

[solemn rain]

okaay

[latent anvil]

e(1-e) = 0 implies e and 1-e are zero divisors and thus not units

so M contains both e and 1-e

So M contains 1 = e + (1-e)