groups-rings-fields

[hot lake]

your text has the right piece acting on the left piece

and in there

https://cdn.discordapp.com/attachments/496784958430380033/847168376832131092/unknown.png

the left piece is acting on the right piece

so you have to mirror everything

[sinful mirage]

oh,right

okay,so we have [(v_1,z_1),(v_2,z_2)]=[[z_1,z_2]+\rho(v_1)(z_2) - \rho(v_2)(z_1), 0]

how to get from here to here

[hot lake]

no

[(v_1,z_1),(v_2,z_2)]=([v_1,v_2]+rho(z_1)(v_2) - rho(z_2)(v_1), [z1,z2])

[v1,v2] = 0

[z1,z2] = 0 because C is also an abelian lie algebra

and rho(z1) = rho(z1 * 1) = z1 rho(1) = z1 D

[sinful mirage]

right

[hot lake]

so [(v1,z1),(v2,z2)] = (z1 D(v2) - z2 D(v1), 0)

that's how the lie bracket on g works

[sinful mirage]

right,now I see

what is the (0,1) here?

v_1=0, z_1=1?

[hot lake]

it's an element of g

v=0 and z=1

[sinful mirage]

and since sigma is an automorphism,it will give some other artibtrary element v_0 in V and lambda in C

right?

[hot lake]

sigma maps g to g so the image of (0,1) will be some element of g

which we call (v0, lambda)

[sinful mirage]

so v0 need not be in V?

I thought v0,lambda is a pair of (V,C)

[hot lake]

v0 is some element of V

[sinful mirage]

some element v_0 in V and some elemend lambad in C

[hot lake]

and lambda is some element of C

[sinful mirage]

right,then the next steps are clear,just manipulating properties of involutive automorphism

this though,why?

be restricting to V,what does that mean?

if we want to restrict to v,lambda by def =0?

[hot lake]

it means looking at what the map does on V

if you have a function f : X -> Y

and a subset Z of X

the restriction of f to Z is uuuh

f|Z : Z -> Y

[sinful mirage]

right,but why does this imply the formula given?

above we have the adjoint rep,here we have D

how did D come out of nowhere

[hot lake]

because first you need to check that the adjunctions always map V to V

[sinful mirage]

what is an adjunction?

[hot lake]

adj(x) is the map that eats a y in g and spits out [x,y] in g

[sinful mirage]

i found this but i don't understand a word

[sinful mirage]

yes,that's the adjoint representation

of the lie algebra on itself

[hot lake]

yeah that

well not your picture

i mean the adjoint representation

[sinful mirage]

yes,I know how to prove this

but I do not know how D comes out of this

[hot lake]

well adj(0,1) restricted to V is literally D

[sinful mirage]

why?

[hot lake]

because when we say that V is a subspace of g

we mean that for each v in V we have the corresponding (v,0) in g

then (adj(0,1)|V) (v)

[sinful mirage]

right,but in this case lambda=0

this is my confusion

[hot lake]

no

that's not what it means

it means z=0

we have a map from g to g, the restriction of the map to V is going to be what it does on stuff on the form (v,0)

[sinful mirage]

(adj(0,1)|V)((v,0))?

[hot lake]

we identify V with (V x {0}) in g

so then (adj(0,1)|V) (v)

= adj(0,1) (v,0)

= [(0,1 ), (v,0)]

[sinful mirage]

right

but how is this D?

[sinful mirage]

this

[hot lake]

because remember when we worked out what was the lie bracket in g

so [(v1,z1),(v2,z2)] = (z1 D(v2) - z2 D(v1), 0)

we just apply this

[(0,1 ), (v,0)] = (1 D(v) - 0 D(0), 0)

= (D(v),0)

[sinful mirage]

=D(v)|V

right

[hot lake]

= D(v) after the identification of V with Vx{0}

[sinful mirage]

and how could I see the rhs?

lambda D

why is ad(v_0,lambda)|V=lambda D

[hot lake]

you do the same

with adj(v0,lambda)

[sinful mirage]

ok,sec i'll try

[hot lake]

adj(v0,lambda)|V (v) = [(v0,lambda),(v,0)] = ??

[sinful mirage]

=(lambda D(v)-0 D(v_0),0)

=(lambda D(v),0)=lambda D(v)|V

right?

[hot lake]

yeah

[sinful mirage]

okay,and I see we got a contradiction after the computation

[hot lake]

well D is already only defined on V so no need to say you restrict it

[sinful mirage]

but in which step did we use that it has a real form?

where did we use this

[hot lake]

in the step where you said that sigma existed

[sinful mirage]

why?

[hot lake]

because Proposition 1.33 (iii) says that if there is a real form then there is an involutive antilinear automorphism and vice versa ?

[sinful mirage]

ahhhhhhhhhhhhhhhhh

let g be a real lie algebra

ok,now it makes sense

thank you so much for your patience @hot lake

finally now it's clarified

[hot lake]

nice

[sinful mirage]

I should always get my definitions straight

[tender marten]

Hello, I am looking a bit at graded modules and tensor products over them, but I am a bit confused. I know that for example k[X] is a graded ring over some field k, and k[X] is also a graded module over k[X]. The tensor product is defined like https://math.stackexchange.com/questions/1616156/if-m-and-n-are-graded-modules-over-the-graded-ring-r-what-is-the at the bottom of the first answer, but I am a bit unsure how I should interpret this. An element of k[X] can be seen as k_1 + k_2X, is an element of the tensor product then (k_1 x k_1') + (k_2X x k_1') + (k_1 x k_2'X) + (k_2X x k_2X) where the x means some sort of abstract tensor product (not the ring multiplication) or is it something else?

I guess the example is k[X] / X^2 instead of k[X], for k[X] it would be an infinite sum I presume

I find it a bit confusing though, since from the definition, seemingly (k_2X x k_1') and (k_1 x k_2X) are supposed to be in the same component sort of

or am I getting it wrong, and should x just be the ring multiplication?

[sinful mirage]

to be precise,why would we not be able to use iii)? I am thinking now again and we suppose that g) has a real form,we did not specify that sigma(x+iy) acts on g

is sigma always an automorphism on the complexification of g?

right,yes,sorry for bothering

my bad

[tender marten]

Im not sure how this works, but should I ping the helpers in this channel as well?

[chilly ocean]

probably won't help

[tender marten]

alright

well if anyone can help, please ping me 😅

[past temple]

not a specific math question but

i'm planning to review some basic undergrad-level ring theory before i start studying commutative algebra

does anyone have any book recommendations for that other than Dummit and Foote or Artin

[nova plank]

Rotman

Always Rotman

[young gazelle]

How exactly do you calculate this sum? What does that weird expression below it mean?

[next obsidian]

It means all tuples (j1,...,jn)

Such that that expression is true

An example could be like
j1 + j2 = 0

[young gazelle]

Oh, that makes sense

[next obsidian]

Okay, no need to go further on this example :)

[young gazelle]

Thanks!

[next obsidian]

[chilly ocean]

is there an easy way to show ix²+x-i is irreducible over Q(i)?

[gritty sparrow]

Quadratic formula?

[chilly ocean]

yes

show there is no sqrt3 in Q(i)

yea that sounds slightly easier than what i did thank

@chilly ocean Can I dm you?

yes

also is this gucci? Proof finite field has p^n elements for some n
Let un be the finite sequence of elements of F. F is field extension of some Zp cuz finite. So induction at each step adjoin 1 root they obviously algebraic so finite dimension bla bla each step get previous number to some power done

[oblique river]

i think it's much easier than that

if E is a field extension of Fp then E is a vector space over Fp, i.e. E = (Fp)^n for some n, i.e. |E| = p^n

[chilly ocean]

yea ofc wtf was i thinking

[chilly ocean]

if two field extensions have same finite dimension and one is contained in other they're the same right

[oblique river]

yes

that is a linear algebra fact -- if V is n-dimensional and W is an n-dimensional subspace of V, then W = V

[chilly ocean]

yes ofc, but i keep being paranoid that i made some minor mistake and everything is wrong

how to be sure of the things you're doing? but actually justifiedly so, not just being sure and being mostly wrong

[viscid pewter]

git gud

[delicate bloom]

if you're paranoid try to focus on what steps you take and what claims you make. Some things you say will be weaker or stronger than others. Try to focus on the weaker things and ask yourself what about them you doubt.

main thing is don't stay with aimless paranoia, hone it into a precise doubt you have about something

otherwise don't waste time being paranoid, it's just unhealthy

[chilly ocean]

hence it is an endomorphism since F is finite??! what

isn't an endomorphism just any map from F to itself

[oblique river]

i think it should say automorphism

it's surjective

[chilly ocean]

ah yea endomorphisms of finite fields are automorphisms ofc

[viscid pewter]

maybe squaring is an automorphism when you apply it to some things, but not others?

so they just call it all an endomorphism?

[delicate bloom]

like the field F_2(X), it's not an automorphism

[maiden ocean]

Shouldn't we need sigma(L) = L for surjectivity because otherwise the restriction to L is just an automorphism into L and not necessarily surjective?

[oblique river]

i think the point is that all you need is that assumption and you end up getting equality for free

in a special case, imagine that L/k is finite-dimensional, say of dimension n. sigma in injective, and injective is equivalent to surjective for automorphisms of vector spaces of finite dimension

[maiden ocean]

yeah i can see it for the case of finite degree

but i dont see why itd hold if you have infinite extensions

[oblique river]

oh wait it's actually even easier I think. sigma is an automorphism of k_s, so it has an inverse sigma^(-1). both sigma and sigma^(-1) satisfy the given property, that is, sigma(L) \subseteq L and sigma^(-1)(L) \subseteq L

I think these two together give you that sigma(L) = L

otherwise, if there were some element x of L not contained in sigma(L), then sigma^(-1)(x) would not be in L, contradicting that sigma^(-1)(L) \subseteq L

[maiden ocean]

oh lmao yeah

ok that works

[oblique river]

yeah i dont think this is a galois thing after all

just like, "if a divides b and b divides a then a = b"

[maiden ocean]

right

that makes sense

[vestal snow]

what does x |--> y = x^2 mean?

[tough pewter]

a map f where f(x)=x^2 (=y)

[vestal snow]

So the map x |--> y = x^2 simply sends x to x^2?

[vestal snow]

Why even include the y then?

[tough pewter]

probably so they could refer to it as y later on

[latent anvil]

because they want to pass to coordinate rings

on coordinate rings you send the result variable y to the thing it's being set to, ie x^2

[limpid edge]

how would you compare category theory and abstract algebra

is the latter a subset of the former?

[tough pewter]

they're not subsets of each other

I've heard category theory get compared to set theory, but I'm not an expert

[scarlet estuary]

category theory is kind of its own thing (but with a lot of connections to other fields and typically studied to better understand those fields rather than for its own sake)

algebra is only a subset of category theory insofar as its a subset of logic

(ie not at all in practice)

[limpid edge]

aren’t morphisms that are native to abstract algebra generalized by category theory

it seems like category theory had to come first

[scarlet estuary]

sure, and the integers that are native to number theory are generalized by ℝ

but i woulnt call elementary number theory a subset of analysis

[limpid edge]

ah true

[scarlet estuary]

and no, category theory did not come first.

[gritty adder]

it's usually the other way around anyway - things that generalize come later

[scarlet estuary]

general theory of natural equivalences was published in 1945

this paper is credited as the birth of category theory

noether had formalized ring theory and commutative algebra decades ago

and galois used groups a century before then

[limpid edge]

ok yea this makes more sense

what is meant by F_{p^2} = F_{p}[i]

[gritty adder]

you know that there's a unique field (up to isomorphism) of order p^2 right

this is saying that you view that field as Fp[t]/(t^2-1)

[limpid edge]

what’s t

[gritty adder]

an indeterminate

[limpid edge]

t is in F_p?

[gritty adder]

no like you start with the polynomial ring Fp[t], and mod out by an element

[limpid edge]

oh

I’m gonna read https://kconrad.math.uconn.edu/blurbs/galoistheory/finitefields.pdf

and then come back here

[celest mantle]

Shouldnt your polynomial be t^2 + 1 ? As t^2 - 1 isnt irreducible therefore F_p[t]/(t^2-1) isnt a field

[gritty adder]

right, that was a typo

[celest mantle]

ok

[vestal snow]

The field k is is Q right?

Because if I is an ideal of Z[x_1,...,x_n] whose inverse image is zero, then the elements of I are non-constant polynomials (union 0)

And we can write any ideal of Q[x_1,...,x_n] as (f_1(x),..., f_n(x)) where each f_i has integer coefficients

[chilly ocean]

Everytime I see your name I read Have a Banana at first.

[cloud walrus]

bim

[latent anvil]

G/G^0 is the continuous image of G, so it's compact. But it's also discrete, since the preimage of any point is a connected component of G, which is open (and since G/G^0 has the quotient topology this means the point is an open set downstairs). A discrete, compact space is finite

G/G^0 is abelian because it's a quotient of an abelian group

[minor badger]

so basically the components of G becomes the elements of G/G^0 ?

[latent anvil]

yup, that's pretty much it

The cosets of G^0 are exactly the connected components of G

[minor badger]

great thanks!

[latent anvil]

yup, one way to see this is to think about localization. The prime ideal pulling back to 0 means it's disjoint from S = Z \ {0}, and primes of this form are the same as primes of the localization S^-1 Z[x1,...,xn] = Q[x1,...,xn]

[vestal snow]

Thanks

[minor badger]

shamrock if i may ask, are you a student, phd or what?

[vestal snow]

I would guess a postdoc

at least a postdoc

because he was refereeing a paper the other day

[latent anvil]

wait what

I was doing peer review for a course, maybe that's what you're thinking of?

I'm an undergrad about to finish my 3rd year

[minor badger]

WHAT

[vestal snow]

Well I was wrong lol

[latent anvil]

(banana, I'm chmonkey's classmate)

[vestal snow]

Oh right

Dang I forgot about that

[minor badger]

im the worst D:

[latent anvil]

The comm alg class we're taking has a peer review component

[minor badger]

do you study in germany?

[latent anvil]

nope

In the US

university of washington

[minor badger]

wow

Im on my last year masters and im trash

[vestal snow]

Do you study in germany bim?

[minor badger]

No, sweden

[latent anvil]

oh

I was supposed to study abroad at KTH this year

But the pandemic happened

[minor badger]

Well im located in uppsala, better math here tbh

[latent anvil]

ah nice haha

[minor badger]

You are really good at math

[latent anvil]

thank you

[tender marten]

I asked yesterday about graded modules and tensor products on them, but there hasn't been anyone who was able to help yet, if someone can, then I'd be very grateful 🙏

[minor badger]

If im not mistaken, you could think of k[x] as coefficients, so k[x][y]

where the direct sum is the grading over y

[tender marten]

hmmm so essentially P(x) tensor Q(x) becomes P(x)Q(y)?

err

yes I think that thats what it would turn out to be

[minor badger]

yeah

[tender marten]

I see

[final pasture]

wait you're an undergrad

[next obsidian]

Same as me

[hidden haven]

yal built different

[next obsidian]

I’m built with a plastic lawn chair body

[hidden haven]

[minor badger]

and what are you shika? 😮

[vestal snow]

The open sets are basically just the empty sets, A^1_k, and A^1_k - finitely many maximal ideals right?

[chilly ocean]

shika

[final pasture]

highschool

also just curious, what are you reading hahn banach ?

[vestal snow]

a book

[chilly ocean]

sea?

[final pasture]

[vestal snow]

Ravi Vakil's book on AG

[chilly ocean]

the rising sea

[vestal snow]

yes

[final pasture]

[vestal snow]

This discord has too many goomers smh

Sadly I will be one by the end of the summer too

[final pasture]

what's a goomer ?

[next obsidian]

Yeah the topology is fhat

But why?

I think it’s worth thinking about a bit

When K is algebraically closed it’s a pretty simple application of the Nullstellensatz

But for general K I think it’s a little harder to see

[vestal snow]

Isn't it like gluing the galois conjugates of k

[next obsidian]

I mean can you justify this omegalol

[vestal snow]

And then thinking of the topology as the cofinite topology on it

with an extra punkt

[next obsidian]

Can you justify it tho?

[vestal snow]

Uhhhh

[next obsidian]

Ultimately I think it boils down to

[vestal snow]

How do you mean?

[next obsidian]

Why is I contained in finitely many maximal ideals

When I is non-zero

[vestal snow]

Like prove it?

Oh yeah

V(S) for S not equal to 0 is a finite number of maximal ideals

so it's complement is what I described

[next obsidian]

Yeah, but justifying that is I think not immediately simple

When k isn’t algebraically closed

[vestal snow]

I mean

[next obsidian]

It definitely isn’t true in general rings

Not even true for k[x,y]

[vestal snow]

It's a factorization argument right?

Every element of S must factor as finitely many irreducibles

[next obsidian]

Right, but you need to use that I is principal to make that work

But yeah, that’s the idea

[vestal snow]

Wait what's I?

[next obsidian]

Replace S with the ideal generated by S

[vestal snow]

Oh yeah I meant an ideal S

[next obsidian]

Forget V(S) for arbitrary subsets

[vestal snow]

Bad notation

[next obsidian]

Lmao

Okay well sure haha

[vestal snow]

Actually

[next obsidian]

But yeah, you have to use that it’s principal and then it boils down to irreducible decompositions

[vestal snow]

I don't think you need to use principal

Just UFD

[next obsidian]

I’m not sure that works

For any given element f in S

You know that f is in finitely many maximal ideals with UFD

But there can be infinitely many f in S so how do you handle that?

[vestal snow]

You take the intersection over all f

[next obsidian]

I guess that’s true

[vestal snow]

Thanks for verifying

[next obsidian]

For k[x,y] I guess it isn’t the lack of being a PID that mucks stuff up

It’s that the irreducible factorization tells you which height 1 primes contain it

Then the roots (along with potentially some other stuff when k isn’t algebraically closed) tells you the maximal ideals

[vestal snow]

Wait Chmonkey, how old are you?

[next obsidian]

22

I took a gap year after high school

[vestal snow]

Ah okay

I wish I did

[next obsidian]

Rip

I didn’t by choice

[vestal snow]

Would've tried to grind out AG and AT

Oh? What happened?

[next obsidian]

This year gap year you mean?

I just didn’t get into any school

[vestal snow]

No in HS

[next obsidian]

I only applied to ivies

Lol, I didnt do any math that year

I just worked

[vestal snow]

That does not seem like a wise decision lol

[next obsidian]

It wasn’t

Kekw

[vestal snow]

I applied to 3 schools and only one was an Ivy

(Princeton)

Didn't get in

[next obsidian]

I figured

Since you’re not there rn

Haha

[vestal snow]

Plus it's also really expensive lol

[next obsidian]

Right

I’m gonna go sleepy mode now

💤

[vestal snow]

Night

[latent anvil]

I also applied to Princeton and didn't get in

solidarity

(planning on making up for it by going there for grad school though)

[chilly ocean]

I'm a bit confused about how the normality property of subgroups work. Say if H is normal in G and K is normal in H then it doesn't follow that K is normal in G (just saw a counterexample with S4). But say if K is normal in G and K is in H then K is normal in H too right?

ie. going from the bigger set to the smaller preserves normality, am I right?

[scarlet estuary]

But say if K is normal in G and K is in H then K is normal in H too right?
i think you phrased this wrong

[chilly ocean]

[scarlet estuary]

if K is normal in H, then yes, K is normal in H

[chilly ocean]

I was doing some exercise and came across the fact that N is normal in G and N lies in HN (the set product where H is a subgroup), then the solution said N is normal in HN

[scarlet estuary]

If K is a normal subgroup of G, and H is a subgroup of G containing K, then K is a normal subgroup of H.

is this what you meant?

[chilly ocean]

Yeah they're all subgroups*

But yeah that's exactly it

[scarlet estuary]

thats not what confused me about your statement

But say if K is normal in G and K is in H then K is normal in H too right?

ah wait

[chilly ocean]

Aha

[scarlet estuary]

sorry

i misread lmao

[chilly ocean]

That's fine

[scarlet estuary]

okay in any case

seems you have the right idea yea

[chilly ocean]

Thanks

Sorry for the confusion in my notation

[scarlet estuary]

going from the bigger set to the smaller preserves normality, am I right?
yep, this is the way to think of it

[chilly ocean]

I'd have been easier to use the "triangle" notation hehe

[scarlet estuary]

nah its fine

N is normal in G and N lies in HN (the set product where H is a subgroup), then the solution said N is normal in HN
and indeed, this is true for the stated reason

[vestal snow]

Heck yeah. If I apply for grad school and get accepted, I'll be like you didn't admit my man Shamrock when he applied for UG and decline their offer.

[vestal snow]

Is it possible to do this using the universal property of the topological coproduct or am I being dumb?

[wind steeple]

the coproduct exists bc Rg^op = Aff, but the exercice asks you to describe such morphism

[tough raven]

Is the arbitrary union of closed sets closed in the spectrum of a ring?

[chilly ocean]

idk anything about spectrums of rings except that they're a topology so probably not cuz usually topologies don't have closed for arbitrary unions of closed

[tough raven]

Same

[final pasture]

I don't think so Raghuram, consider the case of Z.
Spec(Z) = {pZ, p prime or p = 0}.
(U_[p prime, p != 2] {pZ}) isn't closed anymore, but it is a countable union of closed sets

[tough raven]

But if they do there might be a very slick proof of this

[tough raven]

You take unions of sets of ideals right? Not ideals themselves?

[final pasture]

yes, sorry

All ideals of Z are of the form nZ, each nZ is contained in finitely many prime ideals (the ideal generated by the prime divisors of n)

So you can't have a closed set that contains infinitely many prime ideals

and "just" a countable union is enough to get one, as I showed I think

But this isn't stuff I know very well, so I may be saying something totally wrong

(also idk why I included p = 1, fixed it)

[tough raven]

Isn't 1Z in Spec(Z)?

[hidden haven]

Whole ring isn't a prime ideal

[tough raven]

oic
All basic closed sets are finite, so all basic open sets are cofinite
So it's coarser than the cofinite topology? So all closed sets are finite or the full space
And there are infinitely many closed points

OK

[tough raven]

Wait really?

ab in Z => a in Z or b in Z

[hidden haven]

Prime ideals have to be proper by definition

[tough raven]

Oh

[final pasture]

I've always seen it defined with the condition of the ideal being a proper ideal, yes

[hidden haven]

And units aren't considered prime

[final pasture]

on Spec(Z) I think it's like literally the cofinite topology

[tough raven]

That would require all points to be closed 🤔

[final pasture]

oh yeah {0} isn't closed, is it ?

[hidden haven]

Yeah it's not

[final pasture]

yeah ok so it's just coarser than the cofinite topology, you're right

[chilly ocean]

Module Theory

[final pasture]

and yeah, the remaining part of the reasoning is what I was saying

but you said it better lol

[red fox]

I have a few questions about the order of operation in group theory. I know that something like r ∘ f would be read right to left: first do an f flip, and then do an r rotation.

1. Is the order always right to left, or does it depend on the author/school?
2. when condensed to a single element of the group, would r ∘ f be written rf or fr ?

[final pasture]

r o f and rf are both common notations

as for the 1 question, it's always right to left AFAIK

@red fox

[red fox]

awesome, thanks

[chilly ocean]

Can anyone give a hint?

I'm trying to derive the equivalent notation of the Eisenstein integers

$a+b\sqrt{-3}$

[cloud walrus]

Laïka

[carmine fossil]

Shouldn't it be Z[w]?

[chilly ocean]

Exactly

My textbook says the Eisenstein integers are of this form

So Z[sqrt(-3)]

[carmine fossil]

I see

[chilly ocean]

There's an alternative notation apparently

The one you suggest

With

where a,b are in Z

I'm trying to understand how these 2 notations relate

[carmine fossil]

There are some differences,tho -1/2+√-3/2 is not in Z[√-3]

If you take Z[√-3] as a ring

[chilly ocean]

Yeah you're right, I double checked, it's actually not quite Z[sqrt(-3)]

$a+b\sqrt{-3}, a,b \in \mathbb{Z}$ or $a-\frac{1}{2}, b-\frac{1}{2} \in \mathbb{Z} $

[cloud walrus]

Laïka

[chilly ocean]

Now that's the correct definition my textbook gives of the Eisenstein integers👍

[carmine fossil]

Yea,That works I think

[chilly ocean]

Now I'm trying to understand how this is linked to the notation that Wikipedia gives

[chilly ocean]

^

[carmine fossil]

Suppose a,b in Z, a+b√-3 can be written as (a+2b)+(2bw)

[chilly ocean]

Ohhhh

I thought there was some geometric meaning

[carmine fossil]

Similarly you get an expression for when a-1/2 in Z and b-1/2 in Z

[chilly ocean]

Thanks

[carmine fossil]

You still have to show all numbers of a+bw form are either a+b√-3 or that

But that direction is easy

[chilly ocean]

I think you meant a+b +2bw

[carmine fossil]

Same thing

[chilly ocean]

is End(R) iso to R? Would f \mapsto f(1) be an isomorphism?

[next obsidian]

What is R here?

[chilly ocean]

reals

End of R as a group

[next obsidian]

Yeah, it’s even better then

Well, what sorts of endomorphisms are you considering actually?

[chilly ocean]

idk, just a question I asked myself since I was doing some problem asking to show end(Z)=Z and end(Q)=Q so was wondering why didn't they ask for R

[next obsidian]

What sort of endomorphisms there?

As additive groups?

[chilly ocean]

ye

[next obsidian]

I feel as though it won’t be true for R then

I don’t know the answer off the top of my head tho

If you looked at R-linear maps

[chilly ocean]

f \mapsto f(1) works for Z and Q, not sure why wouldnt for R

[next obsidian]

So looking at it as a module

Then that map will be an isomorphism

But there could be group endomorphisms which aren’t R-linear I think

So I feel as though there might be more group endomorphisms

[rustic crown]

iirc this is false when R is non-commutative

[next obsidian]

R is the reals

[chilly ocean]

then its false when reals are non-commutative

[rustic crown]

oh oopsie

[next obsidian]

So if End is R-linear maps it’ll be true

[chilly ocean]

wdym by R-linear

[next obsidian]

But this is asking as a group

[chilly ocean]

ye what's the difference?

[next obsidian]

Linear maps

Like as a vector space

It means f(ab) = af(b)

[chilly ocean]

aight so liek treating R as a field>?

[next obsidian]

Well, no it’s like enhancing the abelian group structure

To also make it a vector space over itself

If we treated it as a field we would want

f(ab) = f(a)f(b)

So really this means that f(1) determines the entire map

Because f(x) = xf(1)

That’s why the map f -> f(1) is an isomorphism

This is why the map you have is an iso of groups for Z and Q

Because for Z, you know f(n) = nf(1)

And for Q you also fix the inverses

[chilly ocean]

ye

[next obsidian]

So like f(m/n) = m/nf(1)

So the question is if there’s a group endomorphism of R which isn’t R-linear

[rustic crown]

and yea there are plenty of those

[chilly ocean]

wait what why

[next obsidian]

Well any R-linear map is a group endomorphism

Cuz it’ll be additive

[chilly ocean]

isn't definition of endomorphism over R to be R linear?

[next obsidian]

No

Well you were asking for endomorphisms of R as a group right?

[chilly ocean]

yes like this

[next obsidian]

Oh that’s not as a group!

That’s R-linearity

That’s as a module

[chilly ocean]

okay yeah

[next obsidian]

Yes if you go to this type of maps then it’s true yup

[chilly ocean]

thatswhy I wasnt following

[next obsidian]

This is true for any ring then

Hom(R,M) = M

By f -> f(1)

[chilly ocean]

wait actually I dont know what is an endomorphism one second its a screenshot from a chapter I havent read

[rustic crown]

oh so ig i was confusing it with invariant base number

[next obsidian]

Yee

That’s really odd

The IBN thing

[chilly ocean]

Ok Im interested whether or not the ring of endomorphisms of the abelian group R is isomorphic to the ring R

[next obsidian]

There’s no way that’s true I think

The map f -> f(1) isn’t even a ring homomorphism

[chilly ocean]

what breaks?

[next obsidian]

Multiplicativity

On one side it’s function composition

[chilly ocean]

ok

like sqrt2sqrt2 = 2 sort of thing right?

[next obsidian]

No

On one side

f•g is the composition of f and g

So it will map to like f•g(1)

But it would need to map to

f(1)g(1) to be a ring homomorphism

[chilly ocean]

wait so why doesnt it fail for Z?

[rustic crown]

but does this show that there isn't an isomorphism... we just showed that the isomorphism isn't of the form f |--> f(1)

[next obsidian]

Because it’s not a ring map

You’re doing the map as abelian groups

[next obsidian]

Right, it could randomly be isomorphic but I find that so hard to believe

For Z the map End(Z) -> Z is as groups

Not as rings

[chilly ocean]

nono as rings

[rustic crown]

i think i can prove that End_Z(R) is not an integral domain

[next obsidian]

Ummm

Okay well

In that case I guess it works...

So

If you look at module maps

So like R-linear maps then it does end up being a map of rings

[rustic crown]

by picking basis of R over Q. pick two basis elements x and y... let f(x) = x and f(other basis) = 0. and similarly g(y) = y and g(other basis) = 0.

then fg is 0 everywhere

[next obsidian]

But if you looked at just group endomorphisms then it won’t work

The reason it works for Z and Q

Is because abelian group endomorphisms

Are exactly the linear maps as well

[blissful ice]

I have a question regarding blow ups of varieties and the points on exceptional divisors (the line where you're blowing up)
so if I have a curve in C[x,y], with a singularity at 0
and, let's say the tangents at that singularity are y=0, y=ix, y=-ix

The consider a blow up at 0. We're looking at points on zariski closure(pi^-1(variety\{(0,0)})) intersection Exceptional divisor
Then, as far as I understand, this will have 3 points, namely [1,0], [1,i], [1,-i]; correct? Well, assuming you identify the divisor with P¹, otherwise all of them have an affine (0,0) in front.
Now assuming that's fine, my question is why. Intuitively it makes sense, but I can't quite get there

So what I've tried is attempted to calculate that blowup, first without closure.
You say that it's equal to {((x,y), [x,y] | f(x,y) = 0 and x != 0} U same with y!= 0 (meaning you split the projective cases in half: One open set where x non-zero, the other when y non-zero. Then you can divide)
then, upon simplifying I get that it's isomorphic to V(x-y-y^3) for x!= 0 and V(yx^4-x^2-1) for y != 0
is there any way to see those tangent singularities from here? The points on the exceptional divisor.

[sinful mirage]

could anyone please help me seeing how the last two lines are equal?

[chilly ocean]

$(a \oplus b)(c \oplus d) = ac \oplus bd$, probably

[cloud walrus]

ττερρα

[sinful mirage]

why is this property true?

intuitively it makes sense that i can't compose different vector spaces

i.e. ad wouldn't make sense

but idk how to prove it formally from the def of direct sum

the 'multiplication' between (a oplus b) (c oplus d) is the composition of homomorphisms,right?

[cloud walrus]

ττερρα

[chilly ocean]

cursed

[sinful mirage]

wait im a bit confused,why v oplus w?

then,strictly speaking this is false?

[chilly ocean]

v, w works fine, too

i wrote v \oplus w to mean (v, w)

[cloud walrus]

ττερρα

[chilly ocean]

there that should match your notation

mega cursed bracketing

[sinful mirage]

yeh,now it makes sense

just a question to be sure,the 'multiplication' between (rho_V(g)+rho_W(g))(rho_V(h)...) is the composition of maps

right?

[chilly ocean]

yeah

[sinful mirage]

[chilly ocean]

unless there's some fancy representation theory thing it could mean

all i know about rep theory is the definition

[sinful mirage]

[sinful mirage]

how could I continue from here?

in our class the tensor product was defined only in a basis

[chilly ocean]

gorup

[sinful mirage]

ah

yikes

i'll correct sec

I should show that the 2 are equal somehow

also there's an extra bracket in last line

[hot lake]

isn't that completely trivial :/

I think the hard part is checking that their definition makes sense

[sinful mirage]

does this seem ok?

and then doing brackets properly I get that they really are equal

[sinful mirage]

what do you mean by this?

[hot lake]

well they give the definition of the linear map on the pure tensors

it may not be completely obvious that it can be linearly extended to a linear map

depending on what you have done before about tensor products

[sinful mirage]

why does the map have to be linear?

a representation is not linear a priori

[hot lake]

[sinful mirage]

it's just a homomorphism?

only lie algebra reps need be linear

[hot lake]

it's a morphism into GL(V)

[sinful mirage]

yes

[hot lake]

GL stands for general linear group

[sinful mirage]

but the group G does not have a linear structure,which the map can preserve

or?

[hot lake]

V is a vector space

[sinful mirage]

I thought only lie algebra reps need be linear

V is a vector space,but G is not

[hot lake]

I mean rho(g) has to be a linear map

not rho

[sinful mirage]

ahhhhhh

yes

ok,right

a linear map tensor a linear map is not linear by def?

[hot lake]

well it's at least a lemma that you may want to prove in the chapter about tensor product, that you can just define linear maps on pure tensors and nothing will go wrong

but I'm probably splitting hairs xD

[sinful mirage]

okay,will do that in the appendix

I didn't do the appendix yet

[sinful mirage]

no,I think it is important indeed,because I try to make a self-contained script

[hot lake]

the pure tensors generate the tensor product space, but they aren't exactly a basis of it

[sinful mirage]

any suggestion is very welcome

[hot lake]

but yeah this is ok

I can already tell you are going to have so much fun with the symmetric powers and exterior powers and other schur functors

[sinful mirage]

symmetric and exterior products were fun,I had some homework related to them

schur functors I never heard

I solved this

the proof of frobenius formula though was homework and we didn't do it in the tutorial cause we were told it's too difficult

[hot lake]

now don't do the decomposition of V^3 into irreducible representations of GL(V)

[sinful mirage]

I actually still wonder how one can find irreps of a given group

our teacher told us we will not touch that subject and we will always be given the irreps and we just need to use the complete reducibility theorem to decompose reps into irreps

but we will never search the irreps of a given group

[hot lake]

well if the group is finite you can just look at the regular representation and stare at it until it breaks into all the irreducible pieces

[sinful mirage]

we did show that all irreps are subreps of the regular rep

but how does this help?

finding them explicitly

[hot lake]

actually you're right if you don't have the character table you can't really do the projection thing and get all the irreducible pieces

[sinful mirage]

is $V^{\star}$ a common notation for dual?

[cloud walrus]

ProphetX

is $V^{\star}$ a common notation for dual?

ProphetX

[daring ibex]

V^* is most common, some ppl use V'

V^\star would be recognized as the dual but too much effort to write it

[chilly ocean]

write the whole Hom (V, k)

[latent anvil]

some folks also use $V^\vee$

[cloud walrus]

shamrock

[daring ibex]

lang does this

i think it's kinda dum

V* is good enough

[old lava]

V* is good

because you can also write V** easily

imagine writing

$V^{\vee\vee}$

[cloud walrus]

F[x]-module

[hidden haven]

isnt \vee easier to write than *

[old lava]

it looks uglier

is the point

also V** is also easier to write

in latex and just in regular old text

[hidden haven]

I meant in hand writing lol

[old lava]

which is where I do most of my math writing

[hidden haven]

asterisks 🤮

[old lava]

asterisks are easier to write than a superscript \vee imo

like yes, technically more difficult strokes

but easier to distinguish

I doubt most people have good enough handwriting to distinguish V from \vee

in handwriting

[hidden haven]

fair

[old lava]

so I would say V* is better for handwriting as well

[latent anvil]

You don't want to handwrite $v\in V^\vee$

[cloud walrus]

shamrock

[old lava]

$u, v, \nu \in V^\vee$

[cloud walrus]

F[x]-module

[wispy scaffold]

cursed

[latent anvil]

$u, v, \nu \in V^\vee\oplus\mathcal{V}$

[cloud walrus]

shamrock

[wispy scaffold]

[old lava]

$u, v, \nu, \mathfrak{v} \in V^\vee \oplus \mathcal{V}$

[cloud walrus]

F[x]-module

[old lava]

writing mathfrak characters by hand

is when you're at the peak of handwriting ability

[golden pasture]

ikr T_T

[old lava]

some math student should bring like a fountain pen to their test

and write mathfrak letters

[daring ibex]

lmao

[prisma ibex]

That’s me

[daring ibex]

I just want to make some small checks, so that i know im not just being paranoid and i actually understand this

1. yes
2. It's iso to G
3. it's iso to G * A

[hidden haven]

2 is wrong I think

[daring ibex]

why?

[hidden haven]

Through one path, A will map to identity, through the other, to the image of the homomorphism A -> G

So the square won't commute

[daring ibex]

Yeah and it has to commute so it forces the image to become an identity

[hidden haven]

Yes

I mean no

That homomorphism is some given homomorphism

[daring ibex]

hmm i see

[hidden haven]

It won't always be the zero homomorphism

[daring ibex]

so we can't even form the free prod

[hidden haven]

You can, the pushout is always a quotient of the free product

3 is also wrong btw

[daring ibex]

im glad i asked then

[hidden haven]

(these are called special pushouts because the pushouts turn out to be really nice)

[daring ibex]

So is 3. just G or something

[hidden haven]

Yes

With G->G identity and A->G the same as the given map

[latent anvil]

huh G *_A 1 is interesting

Is it ||G mod the normal closure of the image of A in G||?

[hidden haven]

Yes should be

At least that's what I remember from alg top

[latent anvil]

many such cases

[daring ibex]

I am actually kinda confused

one moment

why is 2. not G again?

[latent anvil]

Many such cases

[daring ibex]

lol

[latent anvil]

Mirza have you tried an example?

[hidden haven]

What are the 2 pushout maps?

[latent anvil]

Take 2Z mapping into Z

[daring ibex]

there's the 0 map and a morphism A -> G

We're trying to form G *_A 1

[hidden haven]

Yeah those are the maps whose pushout you're taking

But the pushout itself is a tuple of 3 things

The pushout group with the 2 maps that complete the square

[daring ibex]

oh oh

[hidden haven]

So when you say the pushout is G that isn't a complete description

[daring ibex]

yeah okay okay makes sense

So this forces both pushout maps to be the trivial morphisms

Like 0 map

[hidden haven]

Assuming the group is G, yes

But it's not, because then the square doesn't commute

Unless A->G is trivial

[daring ibex]

Doesn't it commute tho

like both ways you're getting a trivial map

[hidden haven]

Ohh ok

Then it is not universal lol

Try to prove universality

[daring ibex]

you mean the second square? Which maps the group homomorphically into another group in two ways?

Doesn't that one work too? One sec

[hidden haven]

Yeah, like the definition of pushout is "a tuple such that that square commutes and for any other tuple there's a unique map from the universal one to that thing such that whole diagram commutes"

So take a second tuple with a commutative square

[daring ibex]

le mouse drawing has arrived

[hidden haven]

Yez

[daring ibex]

So the second square does commute

[hidden haven]

You don't know what f_1 and f_2 are

[daring ibex]

Oh wait we need \phi_1 to be trivial too

[hidden haven]

But it won't be, in general

[daring ibex]

yea

[hidden haven]

Right, so the problem here is that in the pushout, you're sending too many things to identity

You need to send im(A) to identity

But as is usual in universal constructions you should do anything extra

So from this try to guess the pushout first, then prove it

[daring ibex]

hm i see

However if we did have A -> G was trivial then we'd get that G *_A 1 \cong G, right? or is that crankery too

[hidden haven]

Yeah

Then it kinda reduces to problem 3

Because you might as well replace A with the trivial group (any diagrams that commuted with A there, will remain commutative with this change and vice versa)

[next obsidian]

[daring ibex]

Okay another confusing thing, i can describe a pushout group with respect to $\alpha_1, \alpha_2$ over $A$ as $G *_A H = \langle S_G \cup S_H \mid {\alpha_1(a) = \alpha_2(a) \mid a \in A} \cup R_G \cup R_H \rangle$. Being a little lax about $\alpha_1(a)$ and $\alpha_2(a)$ not being in the same group, and with the understanding we're talking about their inclusions into the product. Then we have $G *_A 1 = \langle S_G \mid {\alpha_1(a) = a \mid a \in A} \cup R_G \cup R_H\rangle$

[cloud walrus]

c(o)hmirza

[daring ibex]

This is correct, right?

[hidden haven]

Yez

[latent anvil]

Good idea

[hidden haven]

But you can also get a more usable construction in this case

[daring ibex]

Yeah yeah i know the free prod and quotient by the smallest normal sugrop generated by \alpha_1(a)\alpha_2(a)^-1

smH

[latent anvil]

is an hmirza a mirza up to homotopy?

[hidden haven]

More usable in the sense you won't have to worry about free product etc

[daring ibex]

Yes up to isomorphism

[hidden haven]

Try to do it for the case where im(A) is normal in G

[daring ibex]

So then doesn't this force this free prod to be defined no matter what \alpha_1 is?

[hidden haven]

Yes

[daring ibex]

in the quotient we just turn it into the identity

[latent anvil]

yup

[hidden haven]

Pushout

[daring ibex]

Yeah pushout

[hidden haven]

They always exist for groups

[daring ibex]

So then G *_A 1 really is G then i think

[hidden haven]

What are the 2 maps?

[daring ibex]

It's canonical inclusion

[hidden haven]

You saw that 0 maps don't work

[daring ibex]

there's a cringe notational thing

g \mapsto x_g and you formulate everything above in terms of x_g as elements of the pushout

Been trying to avoid it since i hate the way it looks

[latent anvil]

what's the presentation you just found for G *_A 1 mirza?

Say G is presented by <S|R> and we have this map α : A -> G

[daring ibex]

<S_G | R_G \cup {\alpha_1(a) = 1 | a \in A}>

[latent anvil]

This isn't right

[hidden haven]

Alpha_1(a) = 1

[daring ibex]

oh yeah

doh

[latent anvil]

oh

Wait I am now more confused

should it be like α_G(a) instead of a?

A isn't a subgroup of G

[daring ibex]

Ye i said before i'm being a little lax with this right here

[latent anvil]

oh I see

anyways yeah moldi pointed out what I was failing to communicate

[daring ibex]

But it's with the understanding that you'll just include everything into the pushout

[latent anvil]

But it's not an inclusion

[hidden haven]

Include everything into free product you mean

[latent anvil]

I guess I don't get why you write α_1(a) but then just a

[hidden haven]

[latent anvil]

seems asymmetric

[daring ibex]

Okay so let's try being more precise

[latent anvil]

sure

[daring ibex]

We'll use this presentation instead

[latent anvil]

I don't care about writing g for the word g in the free product/pushout

[daring ibex]

for G_1 *_A G_2

[latent anvil]

What I'm saying to be careful about is writing a for the image of a under the map A -> G

[daring ibex]

It's equivalent i think but this is a little more precise

[daring ibex]

yeah

[latent anvil]

okay so

What's the presentation of G *_A 1?

[daring ibex]

So we have

$G *A 1 = \langle {x_g \mid g \in G } \mid {x{\alpha_1(a)} = x_1 \mid a \in A } \cup R_G\rangle$

[cloud walrus]

c(o)hmirza

[latent anvil]

Yes, I agree with this

Where x_1 is the identity element of the free product

Or whatever

Oh I guess that's enforced by R_G

anyways, do you see why this isn't just G?

[daring ibex]

nop

[hidden haven]

You have the same generators, but extra relations

[latent anvil]

well you're taking G and adding in more relations

[daring ibex]

oh wait

tfw literal genius

[upper inlet]

pepega

[latent anvil]

yes?

[daring ibex]

[hidden haven]

So can you find a way to nicely represent this group without generators and relations?

[daring ibex]

G * 1/N where N is smallest normal subgroup containing that whole thing

[latent anvil]

Wdym "that whole thing"?

[daring ibex]

and that's just G/N

[frigid bough]

The topological intuition here is that G describes some holes, and A is sewing in some discs that trivializes the holes I guess

(just for another perspective)

[daring ibex]

the whole set for the x_\alpha_1(a) crap

[latent anvil]

Sure

[hidden haven]

ie I'm(A) ⊂ G

[latent anvil]

So inside of G this is just the image of A

[daring ibex]

okay so this makes it painfully obvious that this isn't G

wtf am i doing these days

[latent anvil]

have you thought about this in the context of Top?

What's the pushout of X and the one-point space along a map A -> X

[hidden haven]

Does Mirza know top?

[daring ibex]

only bottom

Hm is this like wedge sum stuff, i'm not good at top

[latent anvil]

no worries

[hidden haven]

Not wedge sum, that would be there if these were pointed top spaces

[latent anvil]

just an interesting visualization

[daring ibex]

i cee i cee

yeah i should start doing more top too

am being lazy

[latent anvil]

¯\_(ツ)_/¯

alright shamrock time 8)

Say you have a ring $A$ and you know that
$R_1$: if $\mathrm{ht} \mathfrak{p} \leq 1$ then $A_\mathfrak{p}$ is a regular local ring
$S_2$: if $\mathrm{ht} \mathfrak{p} \geq 2$ then $\mathrm{depth}A_\mathfrak{p}\geq 2$

[cloud walrus]

shamrock

[daring ibex]

! Undefined control sequence.
<recently read> \marhrm

l.63 ...hrm{ht} \mathfrak{p} \geq 2$ then $\marhrm
{depth}A_\mathfrak{p}\geq 2$
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., \hobx'), type I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.

[cloud walrus]

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

[daring ibex]

new copypasta

[latent anvil]

I'm supposed to show that A is an integrally closed domain

I've shown it's reduced

[upper inlet]

that is the worst copypasta I've ever seen

srry

[daring ibex]

wdym it's great

😌

[latent anvil]

I've handled the case where depth R < 2

So depth R >= 2

ah fuck I'm mixing up R and A

Whatever

chmonkey told me to try and show R is integrally closed in its ring of total fractions

Progress on that is not so good

I was thinking that if t = α/β is integral then consider B = A[t], we know that A_p = B_p for any prime p of height <= 1

idk maybe I need to learn local cohomology stuff

so depth B <= depth A

Vibes tell me depth B = depth A

yes

[next obsidian]

You don’t need local coho

Try showing the claim based on the dimension of A

You can use the last hw

[next obsidian]

S_2 also says it’s Cohen-Macaulay for all lower height primes!

[latent anvil]

I used the last homework already

[next obsidian]

:O

[latent anvil]

I don't remember exactly where

[next obsidian]

Chogpamp

[latent anvil]

I think that's how I know depth A >= 2

[latent anvil]

Which claim?

[next obsidian]

I think you could use the hw for that yeah

That integrally closed in the field of fractions

[latent anvil]

Ah I see

[next obsidian]

Like you can prove it for dim 0

Cuz like

Omegalol

[latent anvil]

Well I know dim A >= 2

[next obsidian]

and now whenever you reduce dimension

You can assume it’s integrally closed

Hell you could even aaaume it’s an integral domain

[latent anvil]

ah sure I see

so if I eg localize

Nice

[next obsidian]

Right

💡

[daring ibex]

💡

[next obsidian]

Hi Mirza

[daring ibex]

hi chomky

[latent anvil]

wait uhh isn't there like a thing

like a thing thing

[tough raven]

What is G * A? Doesn't it have to be over some group mapping into G, A?

[latent anvil]

about extending past codim 2 singularities

hmm

@next obsidian did we cover this in lecture?

probably doesn't matter, I can cite whatever on the hw

[next obsidian]

I mean, yeh we did but how do you want to use that tho haha

[tough raven]

Category theory FTW

[latent anvil]

hmm

I was thinking you could look at an integral element and argue that it's not singular in very many places

[next obsidian]

This

Above my brain

But integral element is good yes

So maybe secretly what you said is right

[latent anvil]

I mean I already had the "take an integral element" thought lol

you can find some finite superring B = A[t] and say B_p = A_p for any prime p ≠ m

[next obsidian]

Yes... yes

AND THEN WHAT

What does this say about B/A

[latent anvil]

I mean it says its support is {m}

OH

So it has m as it's only associated prime

I hadn't made that connection yet

[next obsidian]

[latent anvil]

So it's depth 0

which seems important

[next obsidian]

Doesn’t the HW then like

Finish it?

[latent anvil]

umm let me see if I see why

the cokernel is annihilated by m

Yeah okay

Sure

[final pasture]

poor cokernel

[latent anvil]

neat!

[next obsidian]

Now for integral

Remember that djekcndoxjdks

[latent anvil]

lol

[next obsidian]

I know you won’t want me to tell you

But I have in the far distant past

Given you the key

flys away

👼

[latent anvil]

lol

Yeah I see, I'd thought about supp(B/A) but the induction didn't occur to me

[next obsidian]

Right, the induction is pretty shmooving

[latent anvil]

clever

oh I ordered a new laptop

Excited to be a gamer

[vestal snow]

Try titanfall 2

[next obsidian]

What laptop

[latent anvil]

I'll dm

[vestal snow]

I proved the hint but I don't know how to go from there

I said that if D(f) does not interesect the set of closed points (i.e. maximal ideals), then f is in every maximal ideal of A

[latent anvil]

The nullstellensatz should tell you that the intersection of all maximal ideals of A is the nilradical

[gritty sparrow]

The ring is jacobson

[gritty sparrow]

Ye

[vestal snow]

How?

[latent anvil]

good question, I'm trying to think about a simple argument for why this is true for non-algebraically closed k right now

[vestal snow]

Maybe you could look at the hint?

I'm pretty sure Vakil has some other solution in mind

[latent anvil]

sure

Ah sure

So take your D(f)

This contains some point p

Let me back up. If the ring A_f is nonzero then it has some maximal ideal m. Let p be the preimage of this under A -> A_f

Then A/p is a sub k-algebra of A_f/m

By the thing, A_f/m is finite dimensional over k

then the field of fraction of A/p will be contained in A_f/m, so it'll still be finite over k, so p is closed

Alternatively, since A_f/m is finite over A it is integral, so A/p is an integral domain which is integral over its subring k, and this implies A/p is a field. Thus p is a maximal ideal

Does that argument look good to you @vestal snow ?

[vestal snow]

Yeah

[latent anvil]

The argument I had in mind for algebraically closed k is that I(V(J)) = r(J) for any ideal J of k[x1,...,xn], which (if you unravel definitions) says that the radical of any ideal J is the intersection of all maximal ideals containing J

passing to the quotient by J we see that the nilradical and Jacobson radical of a fg k algebra coincide

I'm not immediately sure how to extend this to non algebraically closed k, probably you can just tensor with the algebraic closure?

[vestal snow]

Why does this imply p is closed?