groups-rings-fields

[gilded trellis]

but I still don't understand why we only have those two :",

[chilly ocean]

[gilded trellis]

Ledog casually exposing the fact that he is a cat

[wooden ember]

when $\alpha$ is a single cycle take $\alpha=(a_1,a_2,a_3,a_4...)$ and $\beta=(a_1,a_2)...$ then $\beta\alpha\beta^{-1}\alpha$ is necessarily smaller because $a_1$ is fixed

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

(as long as you make sure beta only permutes elements that alpha does)

if you write $\alpha$ as a product of disjoint cycles you should be able to do the same thing

[cloud walrus]

𝓛ittle ℕarwhal ✓

[gilded trellis]

okay ..

thanks btw !

[wooden ember]

this is the way I showed that A_n is generated by 3-cycles over some exercises

first you show that $A_n$ has no subgroup of index $<n$ for $n>5$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

then you show A_n is generated by 3 cycles by proceeding by induction

[gilded trellis]

Yep, that proof also came up whilst I was googling

[wooden ember]

the way to do that is to take $A_{n-1}$ embedded in $A_n$ and to adjoin a cycle that permutes $n$, and then noting that it cannot be a proper subgroup of $A_n$ by what we just showed

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

ah okay i see

[gilded trellis]

Thanks again !

I really do appreciate your time mate !

[steel saffron]

So… just ignoring $ P = (2,1+\delta)$ I take it?

Delta is sqrt{-5} here

[hot lake]

what's wrong with P = (2,1+sqrt(-5)) ?

oh I misread the question

I thought it was asking about showing V tensored with R/P has rank 1

[maiden ocean]

so suppose we have an extension A -> B of dedekind domains and that p in Spec(A) factors into prod q_i^r_i in B

and that the extensions k(q_i) | k(p) are all finite separable

if B/pB is a finite etale algebra over k(p), say the direct product of k1, ..., kn

can we show that the r_i are all 1?

my thought here is that like

[cloud walrus]

Moffz

[maiden ocean]

but im not really sure how to use this to show that the r_i are 1 unless you know that the q_i correspond to the k_i

and that they and have the same degree

each k_i should correspond to a subfield of B containing pB i guess but im not really sure how that helps

Hmm well finite etale algebras decompose uniquely up to isomorphism iirc...

[maiden ocean]

Ignore this.

[next obsidian]

I already was 😊

[weary terrace]

Second attempt

Let $G$ be a finite group and $\rho$ a one-dimensional representation of $G$. Denote $N:=\ker \rho$ and we know that $G/N=\mathbb Z/n \mathbb Z$. Take another irrep of G, $\sigma$ and observe $\text{Res}_N \sigma$. What can we say about $\text{Ind}_N^G (\text{Res}_N \sigma)$?

[cloud walrus]

RiesZ

[novel parrot]

we need a ring homomorphism from A to A[x] yeah?

but the only one i see is the trivial map?

oh wait nvm

[dusty river]

are you doing the first 4 chapters of AM simultaneously

[golden pasture]

who says you cant

wtf why

[upper pivot]

the mistake is using AM in the first place

use an actual good book

[next obsidian]

@scarlet estuary do you recognize this result / know anything about where it’s proven or anything else really? I was told the proof is by k-theory magic. I might not be remembering all the assumptions so if you know it when something is Noetherian or whatever else that’s probably the statement I’m looking for.

A ring A, M a finitely gen projective A module. Then M^(x)n free for some n is equivalent to M^(+)m free for some m

If M is a fg proj module, for any n, you can find a fg projective module P such that M (x) P^(x)n is free

[novel parrot]

kinda

i kinda like AM

i learnt a bit from dummit and foote

but theres just too much info in dummit and foote

for me

typo in part 3 i think

what should it be

sp < sp' makes senseig

[hidden haven]

Seems correct to me. x is in S_p(0) iff some element of A - p annihilates it → the same element of A - p' annihilates it so it is in S_p'(0)

[novel parrot]

ya

[novel parrot]

if m is in A

n n ' in B

how are they equal

they contract the n's twice?

[next obsidian]

Wat

Do you get the proof?

[novel parrot]

no

[next obsidian]

You’re using that theorem that says if an ideal lies over a maximal ideal then it’s maximal

You’ve just localized to make it so n and n’ lie over the same maximal ideal

[novel parrot]

n contracted is in B ?

or is it n cap A

[next obsidian]

n cap A

[novel parrot]

but

n = q^e

[next obsidian]

I mean really it’s in A_p

[novel parrot]

so n'c is in B

[next obsidian]

You have a map A_p -> B_p

And you have n,n’ in B_p

They both lie over m the maximal ideal of A_p

No it’s in A_p

[novel parrot]

uh

[next obsidian]

You’re just pulling the ideal back

I mean A_p even injects into B_p

So it’s a set theoretic intersection with A_p

n\cap A_p = n’\cap A_p = m

So n and n’ are maximal and then they have to be equal

[novel parrot]

what

m is in Ap

uhhhh

[next obsidian]

m = pA_p

It’s the extension of p inside A_p

n and n’ are qB_p and q’B_p respectively

Because q and q’ lie over p you see that q and q’ both don’t intersect A\p so that qB_p and q’B_p are actually prime ideals

Because q and q’ contained p, qB_p and q’B_p will contain pA_p

Thus qB_p\cap A_p is an ideal containing the maximal ideal pA_p, and is thus equal to it

Likewise for q’B_p

Now qB_p and q’B_p are both maximal because they lie over a maximal ideal in an integral extension, but one is contained inside the other so that they’re equal

Now since qB_p = q’B_p by that correspondence for primes inside a localization you have q = q’

[rose axle]

Sometimes category theory is extremely based

Namely I had a moment yesterday where I proved for the first time that Z/a tensor Z/b is Z/gcd(a,b)

And I did it without ever looking at a proof for this before

And I did it a different way than I think most people do

[languid moss]

Can someone help me?

[chilly ocean]

what are your thoughts/what have you tried?

also this kind of thing belongs in #proofs-and-logic

[languid moss]

This from a grad algebra course tho 😦

[chilly ocean]

question still stands

[upper pivot]

produce some original thought and try the problem, or else no one is going to help you

[next obsidian]

@latent anvil it’s your silly proof lol

I’m pretty sure Shamrock did this same proof just to see if he could once

I don’t mean silly in a condescending way to be clear

[rose axle]

It’s a based proof

[next obsidian]

Or actually I’m not sure if this is exactly what he did

[rose axle]

Similar vibes?

[next obsidian]

He did some bullshit with a lot of adjunction and showed they represent the same functor

IIRC

[rose axle]

I mean that’s what I did lel

Just phrased it a certain way

[next obsidian]

Yeah I think the method was a little different tho

[latent anvil]

playing vivdeo game

ill find it later

[next obsidian]

I don’t think he ever explicitly defined a new functor?

[rose axle]

Vidja

[next obsidian]

Gaim

[rose axle]

Yeah. If you want you can say it as like the universal properties of free groups mod relations

I just defined it for clarity

[next obsidian]

Chmonkey

FWIW it’s probably good to just prove that R/I (x) R/J = R/(I +J) if you’ve never done it before

It’ll come in handy at some point

I guess this is actually a special case of M (x) R/I = M/IM

[latent anvil]

My proof involved the pontryagin dual

[rose axle]

Tf

[rose axle]

This feels like it should be similar

[latent anvil]

https://twitter.com/grassmannian/status/1379648180170584065?s=19

Brendan Murphy (@grassmannian)

A short proof that Hom(Z/nZ, Z/mZ) = Z/gcd(n, m)Z:

Lemma: if A is a finite abelian group, A is isomorphic to its dual Hom(A, Q/Z)
Proof: Since Hom sends finite direct sums (in the first argument) to finite direct sums, wlog by the classification of finite abelian groups A = Z/nZ

Twitter · Apr 7, 2021, 4:12 AM

[rose axle]

Your proof is like morally similar to mine

Well

It’s just worse

[latent anvil]

Yeah, I think I used a specific dual whereas you used all possible duals

[rose axle]

You can skip the Q/Z garbage and ext

Yeah

[latent anvil]

oh wait, I was spending all this time relating the hom and the tensor

[next obsidian]

Fight between the two meme proofs, which is better

[rose axle]

The proof I give is not meme tbh

It’s correct

[next obsidian]

-_-

[latent anvil]

My computation of the actual tensor is just what Alex said, R/I tensor R/J

[rose axle]

And good

[next obsidian]

As is Brendan’s

[latent anvil]

My proof is also correct

[rose axle]

Yeah but

[next obsidian]

Your proof is “I wanted to see if I could” at best

[rose axle]

The one I give is good and should be the proof given in a good algebra class unironically

[latent anvil]

Disagree

[rose axle]

Chm I only think in terms of universal properties

[latent anvil]

but also, we're proving things which are very different

[next obsidian]

Calling it good is disingenuous

[rose axle]

I could not write down the explicit iso

[latent anvil]

I spent a big chunk of time relating Hom and tensor

[next obsidian]

True

You don’t want the explicit iso

[latent anvil]

You do

Wtf?

[next obsidian]

Okay nvm

[latent anvil]

Yoneda gives you the explicit iso

[next obsidian]

In this case it’s useful

True

[rose axle]

Oh true

Lel

[next obsidian]

If you wanna unravel crap

[rose axle]

Yoneda is so based

[next obsidian]

I just think knowing tensor rules is good

[latent anvil]

I'm a big believer in knowing isomorphisms

[next obsidian]

Because your life can be saved by them

[latent anvil]

Also I stopped gaming because I tilted

[next obsidian]

Yeah Sham this is true

[rose axle]

Agree but

I proved this bc i had never proved this tensor rule

[next obsidian]

A good algebra class

[latent anvil]

went from 19xx mmr to 1895 mmr in like 3.5 hours

Pain

[next obsidian]

Would give you 20 tensor rules

And have u prove them all

[rose axle]

Yes

[next obsidian]

And then you’d GG

[rose axle]

And have you peofe@many of them this way :)

[next obsidian]

And tensor products become easy 4e we

No

[rose axle]

*prove

[latent anvil]

A good algebra class would have already done Fourier theory and pontryagin duals

[next obsidian]

Just use right exactness

[latent anvil]

so you can use my proof

No need to munge with Ext groups

[next obsidian]

A good algebra class would assume you know the tensor product

So there’s no reason to do any of this

[latent anvil]

You also literally don't need to do Ext stuff, it's easy to compute Hom(Z/nZ, Q/Z). I was just trying to maximize meme potential

[next obsidian]

Tfw

Actually hold up

I disagree that Brendan you were computing a different thing

[latent anvil]

Yeah?

[next obsidian]

Faye was computing Tor0 and you Ext0

Those are the same

[latent anvil]

I was proving they coincide

[next obsidian]

Or is that only for the higher ones?

[latent anvil]

And then at the end has a line about what Tor0 is

Chm we literally just computed them

[next obsidian]

I know

But like you can prove for groups

[latent anvil]

and both got Z/gcdZ

[next obsidian]

That abstractly Ext and Tor coincide

So if you did that first

You were computing Tor0 actually

But I forget if that holds for the 0-th groups

[latent anvil]

Right, and we literally just computed those 0th groups

And they coincided

[next obsidian]

Yeah

So?

[latent anvil]

So it does hold for the 0th order groups

[next obsidian]

I thought it’s for all fg ones

But I guess that follows

[latent anvil]

Yeah

[next obsidian]

[latent anvil]

Except for Ext(Z, -) maybe?

[next obsidian]

Why not?

If it’s higher than 0 they’re 0

[latent anvil]

I don't see how it follows from the Z/nZ case

[next obsidian]

If it is 0

Then both end up as just -

[latent anvil]

Wait sorry I meant Ext(-, Z)

[next obsidian]

Oh

True

D:

[latent anvil]

Okay shamrock going offline hf with algebra

[next obsidian]

hf

[latent anvil]

I'm a gamer

[next obsidian]

Oh have fun

I was trying to figure out how the word “if” fit in there

I am addicted to

I use it too much

[rose axle]

I

Am so confused

That y’all don’t think this proof is the correct way to approach it

[next obsidian]

-_-

[rose axle]

At least as a first pass at tensor Product of abelian groups

[next obsidian]

?????

[rose axle]

If you’ve like touched tensor Hom which you should

[next obsidian]

Because it follows from an easy tensor product rule so you should just do that one, the tensor product rule is more succinct and more applicable to a variety of situations

If you can do a universal property way to do that

[rose axle]

What if you don’t know / have that one

[next obsidian]

Then you should prove that and then use that

That’s I think the “correct” approach

[rose axle]

I mean it’s the same proof to do that

[next obsidian]

To develop a general rule and then apply it

I mean I don’t care how you prove the tensor rule if you can do it with universal properties and representable functors go off

I just think you’ll run into an issue adapting the proof because you said “Bezout’s”

And that won’t work always

[rose axle]

You won’t. That part is unnecessary

And gets wrapped up into

[next obsidian]

I mean maybe that’s just computing nZ + mZ

[rose axle]

nZ + mZ = dZ

Yes it is

[next obsidian]

¯_(ツ)_/¯

I don’t think using representable stuff is the right way to do stuff if there’s a more low tech way to do it in general tho

[rose axle]

Is it not a cool proof of the general tensor rule then?

[next obsidian]

There’s a point you have to

Yeah I mean it’s cool

[rose axle]

I mean idk the regular proof of that

[next obsidian]

Just i don’t think that’s what id as a prof show if I was gonna prove it

You just show M (x) R/I = M/IM

That follows by right exactness

Like

[rose axle]

Right exactness

Feels more high tech than what I did

But maybe I’m braindead for thinking that?

[next obsidian]

You used hom tensor adjunction

Right exactness follows directly from that

Left adjoint means right exact so idk not really

Like if I wanted to demonstrate how you can use representable functors to do stuff this is a reasonable example

If I was just showing the tensor product for the first time I wouldn’t adopt that approach to thinking about it because most people aren’t ready to do that I think

They’re going wtf is t3ns0rZZZ

And then I’d just do the M (x) R/I = M/IM thing

[rose axle]

The first way I would introduce tensor to someone is via tensor hom

That’s how i’d motivate it lol

[next obsidian]

I just don’t think the avg person who is at a point theyre just learning a tensor product is that comfortable with functors

You come from a different background which makes that not the case

[rose axle]

Or bilinear

[next obsidian]

Yeah

Bilinear is how I would go and then you’d eventually show hom tensor to get right exactness

And then I’d use right exactness to derive a lot of properties

[rose axle]

Bilinear and tensor Hom are like one step away

Also

[rose axle]

Failure of the education system 😎

[next obsidian]

-_-

Category brain worms

[rose axle]

Lel

[next obsidian]

I think ppl confuse “best” way with “the ultimately cleanest way” with most appropriate for the time

[rose axle]

Fair

[next obsidian]

You see it when ppl try to help ppl all the time here in the channels

Someone is like help how do I prove __

And someone is like lmfao sheaf cohomology is 0 bruh

And they’re like “what the fuck is a sheaf”

Or something like that

[rose axle]

Yeah but this isn’t what I’d show someone first learning it unless I knew they had brain worms

[next obsidian]

I mean sure

Idk 2 many thoughts for 1 time

I should be thinking about commutative rings

coooooooooommutative ring theory

[latent anvil]

https://twitter.com/GrothendieckG/status/1433619429003341825?s=19

Grothendieck Googling (@GrothendieckG)

least Noetherian ring

Twitter · Sep 3, 2021, 2:34 AM

https://twitter.com/GrothendieckG/status/1433593960912588805?s=19

Grothendieck Googling (@GrothendieckG)

how to check diagram commutes

Twitter · Sep 3, 2021, 12:53 AM

Good account alert

[next obsidian]

I checked if you follow them

Because I thought this was a new Shamrock account lol

[latent anvil]

lmao nah its not me

[runic hemlock]

In a talk, the speaker stated without proof that it's possible to extend the $p$-adic valuation on $\mathbb{Q}$ to a valuation $v_p:\mathbb{C}\rightarrow \mathbb{R}$. Does anyone know of a proof of this fact?

[cloud walrus]

Phorphyrion

[hot lake]

the algebraic closure of the p-adic completion of Q is an algebraically closed field of characteristic 0 with the same cardinality as C, so it is isomorphic to C

and of course there are many isomorphisms possible (basically as many as field automorphisms of C)

so that gives many different extensions of the p-adic valuation to C

[runic hemlock]

well how do you define a valuation on the algebraic closure of Q_p?

[hot lake]

I'm reading something that says if {x1...xn} are roots of an irreducible degree n polynomial over Qp, you can just pick vp(x1) = ... = vp(xn), = vp(x1x2...xn)^(1/n) and you know that last one because x1x2...xn is in Qp

https://wstein.org/129/projects/hamburg/Project.pdf

[runic hemlock]

oh nice thanks

[digital yoke]

I have a bit of a question

let a be some eigenvalue for f: V \to V for any V veector space

so if dim(ker(f - a*I)) = n

does it mean that the minimal polynomial of f must contain (x -a )^n

or could it just have some (x - a)^k where k < n

[hidden haven]

Anything in between should be possible

For example the minimal polynomial for I is x - 1, even though the kernel of (I - 1I) is dimension of the vector space

On the other end you can come up with matrices whose minimal polynomials are higher powers using the Jordan form

[digital yoke]

oh wow

alright, ty

yea idk why I thought there's an upper cap

it's probalby to do with this one assignment problem I'm doing that caps the power

[hidden haven]

Wait by higher powers I only mean upto n

But higher than 0

[digital yoke]

oh

lol

yea ok that seems a lot better

[hidden haven]

Hmm it may not be capped by n actually

It would be capped by the maximum k such that ker ((f - aI)^k) ≠0

[hidden haven]

This condition gives n 1x1 Jordan blocks

So in all, those n dimensions only contribute a single degree to the minimal polynomial

The ker((f-aI)^k) gives a kxk Jordan block with degree k minimal polynomial

And if there are multiple such kernels, the highest power will be the degree of (x-a) (since the minimal polynomial is the lcm of the minimal polynomial of each Jordan block)

[digital yoke]

a

I eed some time to digest this

but I htink I kinda get it

ty

[wooden ember]

im pretty lost with this one, all i can think about is that we've shown before that there a no such groups of order less than 100 so we can assume |G|>100 and then to write |G|=p^a * m with p^a<100 and m>= 100

where G is a minimal counterexample

but I really dont know what approach to take

[wooden ember]

Im thinking I need to show G has a simple subgroup of non prime order <100, then it must be non-abelian, contradicting minimality of G

nah probably being too restrictive

[cloud walrus]

Spamakin🎷

[barren sierra]

I am stuck on this how many part

[storm turret]

Idk about the number of homomorphisms but notice ${i^n | n \in \mathbb{Z}_4} = H$

[barren sierra]

yea

so that's fine

[cloud walrus]

T0lgi01

[barren sierra]

it's the number part tripping me up

[storm turret]

so Homomorphisms preserve neutral elements and inverse elements

[chilly ocean]

I guess there aren't many choices for f(1) for 1 in Z_4

[storm turret]

since f(1+1) = f(1) + f(1) you can get all f(1), (2), f(3) from just your choice for f(1) and therefore the entire homomorphism

[chilly ocean]

f(1+1) = f(1)f(1) *

[dusty river]

You can talk through mute?

[wooden ember]

fix a generator for G, then send it to any generator for H

[storm turret]

Yeah I used + as any operation I hope you dont start adding the complex numbers

[wooden ember]

gives 3 homomorphisms

and then there's the trivial homomorphism

which sends everything to 1

[dusty river]

It need not go to a generator of H

But deciding where a generator of G goes decides everything

[wooden ember]

oh yeah oop there's also <-1>

so 5 homomorphisms

[dusty river]

isn't it 8

because you can send one to anything

since all elements of H have order dividing 4

[wooden ember]

8?

shouldnt it be 4

send generator to any element of H

[dusty river]

oh wait I thought H was the H_8 group lol

Yeah 4

[wooden ember]

disregard my earlier answers cause i had the misconception they had to get sent to an element of the same order oof

got confused between homo and iso

[storm turret]

Easiest way is taking all 64 functions and checking them for homomorphism :))

[dusty river]

wait H_8 isn't even a thing is it

it is Q_8

[wooden ember]

there are 256 functions

[storm turret]

oh yeah 4^4

[dusty river]

anyway narwhal wtf is that problem you posted

[wooden ember]

ikr

[chilly ocean]

its very

[dusty river]

and why are you studying this kind of group theory

[wooden ember]

im certain the fact that 10000 is the square of 100 is helpful

since we did it for groups less than 100

[chilly ocean]

just use feit thompson

[wooden ember]

lmao no

[dusty river]

is this like part of some course

or self study

[chilly ocean]

just prove feit thompson

[wooden ember]

self study through all of dummit and foote

and i really liked this chapter

[dusty river]

bruhhh

[wooden ember]

oof

[chilly ocean]

iirc the proof is like 300 pages

[wooden ember]

255

it's mad

[dusty river]

imagine studying group theory on its own

😵‍💫

[wooden ember]

i enjoy it so far

[chilly ocean]

groups
rings, fields

[wooden ember]

this chapter introduced a simple group of order 168 and the connection with the fano plane is so satisfying

groups of lie type seem really cool to study

but i agree ring and fields seem cooler

im almost done with groups after these exercises i only have one more small section on free groups

and then to rings i go

[dusty river]

You are literally gonna be like that one swole arm and one thin arm meme

[wooden ember]

i do plan on looking at a more advanced group theory text at some point

probably wont happen ill loose interest before i finish dummit and foote

[dusty river]

doing some research on finite group classifications and having to look up the definition of a sequence

[wooden ember]

and go do something else

[chilly ocean]

groups appear later in the fields but are actually interesting

[wooden ember]

so i dont suppose any of you have an idea for my problem then

[dusty river]

yeah nah I just wanted to make fun of you for even attempting that problem

[wooden ember]

the next problem shows there are no simple groups of order 420 lmao

this is why group theory is useful

[chilly ocean]

lol yeah everytime I go buy groceries I use that fact, its very useful

[storm turret]

I suppose it's just taking all odd numbers whose prime factorisation have two primes with power >= 2 and check they cannot be simple because the rest is covered by other theorems (?)

[wooden ember]

the last part is the hard part lol

[storm turret]

actually thats a lot of number isnt it

[chilly ocean]

just check every case

[wooden ember]

that's basically what i did for groups less than 100 lol

i proved lemmas for a bunch of different forms and took a few special cases aside

that was hell lol

but now i know the only non abelian simple group of order less than 100 is A5

worth it

[storm turret]

next highest is order 120 iirc there aren't that many

[dusty river]

now i know the only non abelian simple group of order less than 100 is A5

[wooden ember]

na it's 168

[dusty river]

it is that ez

[wooden ember]

XD

ig ill wait till i can harass an intellectual like chmonkey on this

or ill post on stackexchange later

[storm turret]

https://groupprops.subwiki.org/wiki/List_of_simple_non-abelian_groups_of_small_order
This has all the ones up to order 8000 I wonder how they expect to solve it

good luck on all the other exercises

[wooden ember]

danke

[dusty river]

my module theory class had an extra exercise to classify all rings of small order

where small was upto 7, but optionally included 8

[wooden ember]

haha

[dusty river]

This wiki is 3 orders of magnitude above that

[chilly ocean]

wait how do you do that

[dusty river]

Use the characteristic map from Z into whatever ring

So the only ring of prime order p is Z/pZ

and the only interesting cases are 4 and 6

For 4, you have Z/4Z or Z/2Z is contained in it

if it is the latter case, then adjoining x and quotienting by a quadratic is the only option

and you can work out which quadratics give isomorphic quotients

6 case is pretty similar, 8 is somewhat hard but you can figure out a pattern for p^3

[chilly ocean]

[dusty river]

but once you know the Z/nZ that is contained you just adjoin variables and quotient by some stuff and you just have to work out what stuff

[wooden ember]

what year you in @dusty river ?

[dusty river]

Starting MSc in 17 days

[wooden ember]

noice

[chilly ocean]

That's good, gl

What field? Logic stuff?

[dusty river]

isn't MSc just generic and you specialize in PhD?

at least the one I am doing is

specialize in second year

when writing thesis

will decide then

[chilly ocean]

Oh okay here you specialize when applying, there arent any courses every msc student has to take

[bleak perch]

Quick question, is the ideal generated by 7 both prime and maximal in Z[x]?

I proved that its prime, so my assumption says its maximal cause it is irreducible

[dusty river]

look at the quotient

maximal iff quotient is a field

[bleak perch]

is it being irreducible also that that it is a field?

or is that circular

[dusty river]

no, irreducible doesnt imply that the generated ideal is maximal

counterexample: ||your example lol||

or like x in Z[x] is irreducible, but (x) is not a maximal ideal

[bleak perch]

lol, okay I se

see*

[barren sierra]

I got what you meant lol

[bleak perch]

Another question, how would I Reduce $$I = x^{37} + (100! + 1)x^9 + (2^{201} - 2)x^3 + 101!$$ ? to show that Q[x]/I is a field? I assume I have to use Eisenstiens criterion but the manipulation of it is not clear to me

[cloud walrus]

freshzak187

[dusty river]

eisenstein works

[bleak perch]

I assumed so lol but idk how to apply it

[dusty river]

hints are

no transformations are needed, it is directly applicable

and there is only one prime which can divide both the degree 9 term and the constant term

too high and you don't divide the constant term, too low and you don't divide the degree 9 term

After that it is just verifying that the eisenstein criteria are met

stronger hints: ||wilson's theorem, fermat's little theorem||

its a packed problem

[bleak perch]

definitely haha

thanks for the hints

[urban acorn]

it does for a PID which Z is (EDIT: disregard, I didn't notice it was Z[x])

[urban acorn]

(edit: disregard this) Assume I is an ideal of Z properly containing (7), let x be some element of I - (7), since 7 doesn't divide x and 7 is prime, by Bezout's identity you can write 1 = 7a + xb for integers a, b, meaning 1 in I, and therefore I = Z

LMAO I just noticed we were talking about Z[x]

my bad

[topaz leaf]

Anyone here into Group Representation Theory?

[lethal cipher]

are there any concerns regarding the last implication?

That's definitely a lot more clear

[south patrol]

Couldn't you also just multiply the original xy = yx by x on left, y on right to get between the two statements

[lethal cipher]

No, because we are not assuming that xy=yx in the second implication

[south patrol]

I just mean $xy = yx \iff xyy = yxy \iff xxyy = xyxy$

[cloud walrus]

potato

[lethal cipher]

Ah, I see what you mean. That would have been quicker for sure.

[topaz leaf]

yeah, a lot of algebraic equalities (especially early group/ring theory) can be proved with iffs and clever manipulation

[south patrol]

and imo don't be scared to just multiply stuff through

[lethal cipher]

For sure, I just didn't see it

[south patrol]

Like a crappier version of what I did would be say like

yeah nws!

like to say suppose (xy)^2 = x^2 y^2. Then xyxy = xxyy, so yxy = xyy by multiplying on left by x^-1 and so yx = xy etc

[lethal cipher]

I am gonna hold out though because my proof still gets the job done, but now I know about these tools for the future; that way I can discover these things on my own next time.

[south patrol]

yup sure, sounds good

:)

[languid moss]

Part A

[wooden ember]

The last argument works just by rearranging: $xyxy^{-1}x^{-2}=e$ iff $xyxy^{-1}=x^2$ iff $xyx=x^2y$ iff $yx=xy$ (step by step)

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

Oh I misunderstood this is your answer isn’t it

Well yeah it works

But yeah potato’s answer works better

[languid moss]

Anyone 😦 @upper pivot

[frank fiber]

in a ring, if 1+a is unit, then a is nilpotent?

[upper pivot]

Not necessarily. Take -1 in Z

There is a similar condition for the Jacobson radical which is probably what you are thinking of

If 1+ab is unit for all b then a is in the Jacobson radical

[frank fiber]

yes, i want to prove radical jacobson is in the nilradical in a polynomial ring

[upper pivot]

It’s probably just easier to use the definition as being intersection of maximal and primes in that case

[hidden haven]

You can characterize nilpotent and unit polynomial based on their coefficients

[frank fiber]

yes, a polynomial is nilpotent if all his coeficients are nilotent

and a polynomial is unit if his first coeficient is unit and the others coeficients are nilpotent

[hidden haven]

So if for a polynomial p, 1+pq is a unit for all q, then you have to prove that p is nilpotent

This follows from those 2 statements

[hidden haven]

Ah I see why you wanted this

Well you can avoid it by choosing q such that nice things happen

[frank fiber]

i think q=0, then the first coeficient of pq is 0 and 0 is nilpotent

[hidden haven]

Yeah that just tells you that 1 is a unit then, which doesn't give anything

[frank fiber]

q=x

[hidden haven]

What does that tell you?

[cloud walrus]

Or x1

[frank fiber]

Let $p(x)=a_0+a_1x+...+a_nx^n$ in the radical jacobson of A[x] then $1-p(x)q(x)$ is unit for all $q(x) \in A[x]$, taking $q(x)=x$ we have $1-xp(x)=1-(a_0x+a_1x^2+...+a_nx^n)$ is unit , then $a_0,...,a_n$ are nilpotent and $p(x)$ is nilpotent, so $p(x)$ is in nilradical

[cloud walrus]

Or x1

[hidden haven]

Yep nice

[frank fiber]

thanks

[wooden ember]

asking this again in case someone can answer it now, any hints on approaching this question?

[hot lake]

probably just more of the same thing that you did for order < 100 ?

[wooden ember]

but for order <100 i basically went through a lot of different forms of order and took a few special cases apart

definitely not the way to do this one, the order is too big

the fact that 10000 is the square of 100 must be important though: im thinking the fact that you can write |G|=ab with a<100, b>=100 must help

(and we can assume |G|>=100 since we showed it is true for <100)

where G is a minimal counterexample

G is a counter example so it must be simple, G/H isnt defined?

or do you mean coset space

i dont use that notation

[hot lake]

uuuh

[hidden haven]

I cannot count past 100 so larger simple groups are uncountable and we know from model theory that ever uncountable simple group has a countable simple subgroup

[hot lake]

you saw nothing

[wooden ember]

lmao

[upper pivot]

i see everything

but also this problem seems very

[wooden ember]

that's what they said yesterday

[hot lake]

yeah the problem looks like it wants to inflict maximum pain on the reader

[wooden ember]

i refuse to believe there is no elegant way to do this

other than feit-thompson lol

[hot lake]

just prove the feit thompson theorem smh

[wooden ember]

too fast for you

[hot lake]

I am getting too old now I am slow and half of what I write is nonsense

[hidden haven]

Just do proof by likely existence of elegant proof from presence in d&f smh

[wooden ember]

haha

[hidden haven]

Are these song lyrics

[hot lake]

the zef song

[wooden ember]

i thought the bigger brains would know but i guess not

[upper pivot]

use sage or smth ez

[wooden ember]

i think ill ask on stackexchange in the end

will let yall know if i come across a nice proof

[urban acorn]

check all cases with a computer like a chad

[wooden ember]

i wish your username was true then id have a vacuously true result

[urban acorn]

lmao

[gritty sparrow]

Bruh use the classification of finite simple groups

[wooden ember]

[wooden ember]

btw did you figure anything out regarding that question i asked yesterday

[gritty sparrow]

Oh sorry, I didn’t think too much about it

[wooden ember]

lmao it's fine was just wondering

[gritty sparrow]

But here are some thoughts I had:

You can’t just quotient the ring C[[X]] by the maximal ideal containing a power series t to get roots, because the equivalence class of X still might not be a root. Once I noticed that I basically stopped thinking about it

Not much progress I suppose

[wooden ember]

ill understand that once i study more ring theory lmao

[novel parrot]

i dont get thisd

ive understoof

taking a max in Bp, contracting it will give the max in Ap (p exteneded) and then contracting it will give just p

but how will inverse beta give q that contracts to p

[next obsidian]

You’re just pulling the ideal back along both ways

If you contract q into A

It’s the same as if you contracted n into A_p (giving m) and then contract that into A giving p

[novel parrot]

okay

m

its the same because they are all injections?

[next obsidian]

No

This is just… a thing

If f = g then f^-1(S) = g^-1(S)

[novel parrot]

ok

[next obsidian]

Here f and g are the two maps
A -> B -> B_p

And A -> A_p -> B_p

[novel parrot]

ahhh

that makes it soo much clearer

Thank you

[stone fulcrum]

This the one you're asking about?

That's just in an intro group theory course?

[chilly ocean]

It's an exercise from D&F

[novel parrot]

[stone fulcrum]

I think your mute isn't working

And what okay D&F I know you want to bore the reader but yike

[dusty river]

Ledog may not be silenced

[barren sierra]

ok so suppose I have a set G with an associative operation

and suppose that for all g, h in G

I know there is a solution to the following two equations

g * x = h

and x * g = h

so if I have g = h

I know there are solutions for g * x = g and x * g = g

so identity for g

but how do I know that for different g

such an x isn't different?

My goal is to prove that G is a group

so I know that there is a solution to

x * g = g
y * h = h

how can I show x = y (cause identity is unique)

[topaz leaf]

so you want to show that there is at most one identity element? what would happen if you had more than one, and tried to multiply them together?

[barren sierra]

well the thing is

doesn't that only work with left and right identity?

how do I do this to show that left identity is unique

@topaz leaf

[mild laurel]

The group axioms tell you that there exists some e such that eg = ge = g for all g in G right? You can use this to show that if xg = g, then x = e and similarly, y = e

[topaz leaf]

honestly, not sure rn, i don't have a notepad to scribble on 🤷

[barren sierra]

I don't know that it is a group

that is what I am trying to prove

[mild laurel]

Oh I misread

I don't think this is possible. You can definitely have sets G with associative operations that have different identities for each element

[barren sierra]

I don't think I mistyped above but just sending the question here again

the "show first" imo is misleading

cause I think you have to have unique left and unique right identity

as a preliminary

[mild laurel]

Ah I see what you're saying now

[barren sierra]

and that's where I'm stuck

(x * g = g) and (y * h = h) => x = y

that's what I need to show

at least I think I need to show it?

[mild laurel]

No, I don't see why you need to show that first

If you show the hint is true, then if xg = g, then the hint says that gx = g too. Using the hint again lets you show that x = y

[frank fiber]

if gx=h and h=xg have a solution for every g,h then you can take h=g so gx=xg=g, this is, x is the identity

and if you take h=e, then gx=e=xg, this is, g has inverse

@barren sierra

[hot lake]

yeah zoph is right the hint says "for some g and h" if you had to show first than there is a left inverse and a right inverse it wouldn't have said that

now if I wanted to prove what he hint says I would get g * e1 * h = g * h = g * e2 * h, so to deduce e1 = e2 from that I would like multiplications to be injective

not sure how to get that though

[frank fiber]

are there polynomials which are not unit in A[x] but are units in the ring of formal power series?

[barren sierra]

yea exactly

[barren sierra]

the issue is

I don't think we can say that if h =/=g

then h * e_1 = h and g * e_2 = g => e_1 = e_2

[barren sierra]

why not?

[urban acorn]

fun fact: all finite simple groups are actually just the prime cyclic ones

[urban acorn]

yes. i.e. (x+1)(1-x+x^2-x^3+x^4-...) = 1 in Z[[x]]

[mild laurel]

Read my other message, you can derive that from the hint

I agree with Zef that I'm not sure how to prove the hint though

[barren sierra]

I'm so lost

hint says "if we have left and right identity, show they're the same"

which is easy

however

I can't do that

without showing unique left and unique right identity

if h =/=g then h * e_1 = h and g * e_2 = g => e_1 = e_2

[mild laurel]

If you show the hint is true, then if xg = g, then the hint says that gx = g too. Using the hint again lets you show that x = y

Oh that's what you're saying

[barren sierra]

now you see where I'm stuck?

[mild laurel]

Yes I agree that showing the uniqueness of left identity is one way that you can prove the hint through

It basically boils down to what Zef said though, you need to know that the solutions to the equations are unique in general

Is there some assumption that G is finite here? Because I'm not sure this statement is true otherwise

[barren sierra]

no there isn't

[barren sierra]

this is the full question

[topaz leaf]

okay i'm back - what progress have you made?

[barren sierra]

none cause my question hasn't really been answered

I've answered the whole question assuming we have unique left and unique right identity

however I do need to prove this assumption some how

which was my original question

[topaz leaf]

thanks for the edit

[barren sierra]

lol

[topaz leaf]

Assume you have some number of left identities (l_i) and some number of right identities (r_j)
then l_i = l_i * r_j = r_j
(where the left equality holds because r_j is a right identity, and the right equality holds because l_i is a left identity)
so all left identities are equal to all right identities is unique

[barren sierra]

ohhhhh

ok so that's how I proved l_i = r_i

but extending that reasoning is slick

ok cool

[topaz leaf]

https://tenor.com/view/phew-relieved-sigh-gif-6021378

i feel like that guy that's just walked in, scribbled a couple of lines on the board and saved the day

[barren sierra]

🙏

[topaz leaf]

So I take it this is the reasoning:

1. there is some right identity for some element g (same for left)
2. there is at least one right identity that works for all elements g (same for left)
3. all these right and left identities are equal

[mild laurel]

I don't see how this logic is enough. The question is whether a left inverse for g and a left inverse for h necessarily have to be the same right?

[topaz leaf]

inverse? or identity?

[mild laurel]

Identity sorry

[topaz leaf]

i haven't done that yet (not sure if @barren sierra has either) but i suspect you can show this using properties i) and ii) give in the question

[barren sierra]

well doesn't this show that they necessarily have to be the same?

[topaz leaf]

my statement is true if l_i (and r_j) are left (right) identities for ALL elements

[barren sierra]

ohhhhh

yea you're right

shit

[topaz leaf]

so I think @mild laurel is asking about step 1 to step 2

[barren sierra]

yea

[topaz leaf]

and i think you can show that using properties i) and ii) given in the question

[mild laurel]

Yeah idk I feel like this isn't true

The properties basically say that left multiplication by any element g is surjective and so is right multiplication by g

If you replace surjective with injective, then the statement isn't true since you can take the positive integers under addition and this isn't a group

[barren sierra]

yea

idk if I agree that it isn't true but at the very least there is nothing saying that we have an identity that works for all elements

[mild laurel]

Okay nvm, maybe this works. So if you assume that xg = g, then for any h, you know there's some y such that gy = h. So if we take xg = g and multiply both sides on the right by y, you get that xh = h. In other words, x is a left identity for all elements

And you can do a similar thing for right identity, and then apply what Zain said

[barren sierra]

ohhhh

ok

well I mean if I know there is a left identity and a right identity that works for all elements

I don't care about elements that are an identity for only some elements

[mild laurel]

I mean, this is how you show that an identity for some elements is an identity for all elements

[topaz leaf]

yeah, going from step 1 to step 2

[novel parrot]

how does 5.13 work?

how is Ap integrally closed by 5.12

what are they letting ring B be?

S is being A- p and A-m i guess?

[dusty river]

Yeah

A is integrally closed in K

C is the image of A in the inclusion from A to K, you can just all it A to make this easier to see

The integral closure of A in K is itself

Therefore the integral closure of A_p in K_p is A_p

K_p = K = field of fractions of A = field of fractions of A_p

So A_p is integrally closed

[novel parrot]

ok so localizing K has no effect

right?

[dusty river]

Yeah, it's like you've already inverted all non zero things, so adding formal inverses to any non zero things (and then quotienting by reasonable relations) now is not gonna have an effect because they are already all there

It will only have an effect if S contains 0

because then you are inverting something that hasn't yet been inverted

[novel parrot]

right yep

and

those fp and fm mape

maps

are from Ap into Ap/Am right

trivial

maps

[dusty river]

No they are what you get when you localise the map A → K

or rather A → C

[dusty river]

but here I said it is easier to just think of C = A and A sitting in K

so really f is just identity when you view it like that

and you are localising and localisation of identity is identity

ok well we don't a priori know f is identity

but f is inclusion into integral closure

so fp is not Ap to Ap/Am

it is from Ap to integral closure of Ap in K

[novel parrot]

the f maps are just embeddings into K

?

[dusty river]

yeah

[novel parrot]

i didnt understand what you said

[dusty river]

embedding of A into its integral closure inside K

[novel parrot]

ok

but we know integral closure of Ap and Am is Ap and Am

[dusty river]

then fp is the embedding of Ap into its integral closure inside K

[dusty river]

we only know that when proving b → a or c → a

[novel parrot]

sorry you mean a -> b and c ?

but now we want to go from b and c to a

[dusty river]

You are saying that Ap is integrally closed

that is the hypothesis in b

right now I am assuming none of a,b,c

I am just doing in the main set up

[novel parrot]

ok

[dusty river]

f is embedding of A into its integral closure in K, then fp is embedding of Ap into its integral closure in Kp = K (this is exactly what 5.12 says but I have used embedding instead of actual having things be contained in each other, but it works out exactly the same)

[novel parrot]

ok so i understand that Ap is integrally closed iff it maps onto the closure in K

[novel parrot]

ok got it!

[gilded trellis]

Quick question. So there is a question that asks me to prove that the rest and quotient of the euclidean division of two polynomials A, B in Z[X] (with B unitary) are also in Z[X]. The hint stated to do induction on deg(A) for deg(A) >= deg(B) (the other case being trivial).

But since B is unitary..

and Z[X] is a ring,

can't I consider the quotient Z[X]/(B) ?

then [A] = [R] (R being a representative of degree less than that of B), so A-R is in (B), i.e. A - R = QB with Q in Z[X]

and then conclude by uniqueness ?

Am I missing something here ? Is there something stopping me from doing so ?

[mild laurel]

Why does R exist?

[gilded trellis]

B is unitary

so [B] = 0

wait ?

why wouldn't it exist ?

if : $$ B = \sum_{k=0}^{b = deg(B)}{a_k X^k} $$

[cloud walrus]

Der Gegenstand ist einfach.

[gilded trellis]

then since B unitary, : $$ X^b = \sum_{k=0}^{b - 1}{-a_k X^k} $$

[cloud walrus]

Der Gegenstand ist einfach.

[gilded trellis]

so basically we can always have R by replacing the X^k for k >= b in A with the above expression

and its degree would be less than that of B

[mild laurel]

Yeah sure, but this is basically the same as the induction proof

So there's no real need to go through the whole R/(B) stuff

[gilded trellis]

I see

:", I was actually struggling with the induction proof so that's why I resorted to this, but now I can also see how to do it with induction

thanks !

[novel parrot]

i can imagine that x^n = phi

but not 100% sure

like $\phi^m + \cdots + a_n = (x^n)^m + \cdots + a_n$

[cloud walrus]

ActiveChapter

[novel parrot]

because phi is a map we need to actually input something make it equal to the RHS

[dusty river]

You're right

And there is something like that in A[x1,...,xn]

[novel parrot]

so are we saying its isomorphic

rather than equal

[dusty river]

Well ok so the equation you wrote isn't exactly what's happening

since phi isn't x^n, but multiplication by x^n

[novel parrot]

ye

[dusty river]

what you want to prove is that if a polynomial in phi is 0, then the same polynomial in x^n is 0

and the way to turn a map given by a polynomial in phi into a polynomial in x^n is to evaluate it at a point (a polynomial in phi is really an endomorphism of A[x_i's])

[novel parrot]

ok so its gonna be 0 for everything in A[xi]

[dusty river]

yes

[novel parrot]

makes sene

[dusty river]

it is the 0 map\

[novel parrot]

yea

ty

[novel parrot]

wht does it mean by coefficients of min poly of x are polynomials in xi

[topaz leaf]

x is the root of the minimal polynomial
the x_i are all the conjugates of x (i.e. the other roots of the minimal polynomial)
if you expand out (t-x1)*...*(t-xn) then you get the minimal polynomial, right? well, each coefficient a_j is just sums and products of the terms x_i

[novel parrot]

oh rogjt

righty

then is it saying that each a_i is integral over a ?

[topaz leaf]

that's what the sentence says, yes

[novel parrot]

but ehmm a_i is in a yes?

[topaz leaf]

the statement says "x is algebraic over the field of fractions K of A"
edit: i think i misread

[novel parrot]

x was integral over a in A

but that also implies its algebraic over the frac field

or am i misundertstanding

so then my next question is why/how are we using 5.14 to show a_i is integral over a

[topaz leaf]

ah no, a_i come from the minimal polynomial over K

[novel parrot]

oh right

so initialy

we dunno if those conjugates were in B

or A

[topaz leaf]

[novel parrot]

but the last part

is it lying is rad(a) or rad(a^e)

5.14 says r(a^e)

nvm got it

[topaz leaf]

[novel parrot]

uhh

i didnt get it

so we take an x in B

x is integral over a so satisfies a polynomial with coefficients in a

xi are the other roots

in L

so why is min poly of x different

"The coefficients of the minimal polynomial of x over K are polynomials in the x_i""

if the min poly was different, how do we know it will be in x_i

[topaz leaf]

1. Clearly x is algebraic over K.
2. Let L be an extension field...
3. Each x_i satisfies the same equation....
4. The coefficient of the minimal polynomial....
5. Since A is integrally closed....

@novel parrot which statement are you struggling with

[novel parrot]

part 4

the min poly for x in K

is not the same one as the initial relation over a right?

[topaz leaf]

x is integral over A, and K is the the field of fractions of A
so the min poly [of x over A and of x over K] should be the same, no?

[novel parrot]

so coefficients are automattically integral over a by what you said now?

thats confusing me

[topaz leaf]

i made a small edit, are we still saying the same thing?

[novel parrot]

yes

but if thats the case

then im confused

$t^{n} + \cdots + a_1 t + a_0 = 0$, $a_i \in \bold{a}$

[cloud walrus]

ActiveChapter

[novel parrot]

thats the min poly over a yeah?

ai is already integral over a

am i wrong?

[topaz leaf]

i don't think so
but then you wouldn't need 5.14

[novel parrot]

yes thats what im thinking ...

so whats going on?

[topaz leaf]

honestly, i don't know either

sorry

[novel parrot]

its fine lol

someone else will help

[topaz leaf]

have you got a supervisor or a friend or someone else to ask about this?

[novel parrot]

nope not right now

[topaz leaf]

okay
let me know if you manage to figure it out

[novel parrot]

sure

[novel parrot]

<@&286206848099549185> can anyone help?

still havnt figured it out

[celest mantle]

what's your question again ?

[novel parrot]

so x satisfies a polynomail over the ideal a in A

but over the quotient field, we have a min polynomial for x and its conjugates

firstly, why are these conjugates also satisfying the same integral dependence polynomial over ideal a in A

is the min polynomial for x over K the same as the integral dependence polynomial for x?

@celest mantle

nvm i figured it out

[celest mantle]

i think it just comes from the fact that the minimal polynomial of x divides all polynomials that admits x as root, so P(x) = pi(x)Q(x), and then you do the trick with the conjugates

[novel parrot]

yes

[celest mantle]

oh well

congrats

[novel parrot]

thanks though 😄

[celest mantle]

sorry for the late response

np

[topaz leaf]

so what's the tl;dr?

[novel parrot]

so we got x in B which is integral

consider the min polynomial for x in K

they arnt necessarily the same

but in K, the min polynomial for x will divide that integral dependence for x

so each x_i is integral over a

so coefficients of min polynomial are too

then rest follows by previous result

[topaz leaf]

i see
do you happen to know an example for this?

[novel parrot]

hmmm

not really

f(x) would be our integral dependence over a

[topaz leaf]

alrighty

[novel parrot]

[topaz leaf]

@novel parrot may i ask your background? (nth year undergrad etc. etc.)

[novel parrot]

2nd year undergrad

you?

[topaz leaf]

going to start my phd in a few weeks

i had a few years off though so i'm still a lil rusty

[novel parrot]

sounds fun

what is the phd about?

[topaz leaf]

group theory (/group representation theory if you've heard of that)

[novel parrot]

group theory yes

not group representation theory

no idea about that

[topaz leaf]

in a nutshell, group representation theory looks at studying groups by looking at group actions on vector spaces

[maiden ocean]

are you asking why this is true

Bp2 is the extension of p2 in B, which is the integral closure of A in B because B is integral over A, hence the radical of Bp2 (which of course contains it) is the integral closure of p2 in B

So y in Bp2 is in the integral closure of p2 in B, hence integral over p2

[novel parrot]

yes

but

why is B the integral closure of A

it could be something smaller no?

[maiden ocean]

B is integral over A

so by definition the set of elements of B integral over A has to be all of B

[novel parrot]

right

ok

[novel parrot]

nvm

[maiden ocean]

[novel parrot]

sorry i gotta question

how does showing that

prove it

p2 is a contraction of prime in Bq1

[maiden ocean]

p2 being the contraction of a prime ideal in Bq1 is the base case

for m = 1, n = 2

[novel parrot]

yh

[maiden ocean]

Then they are inducting

i mean if you have an arbitrary $\mathfrak{q}_1 \supseteq \ldots \supseteq \mathfrak{q}_m$ and $\mathfrak{p}_1 \supseteq \ldots \supseteq \mathfrak{p}n$ then you'll get $\mathfrak{q}{m+1$ by just ignoring the first $m - 1$ terms of each sequence and the last $n - m - 1$ terms of the chain of ideals in $A$

[cloud walrus]

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

[novel parrot]

yea i get thaat

part

the induction

but not the part afterwards

specifically

[maiden ocean]

p2 being a contraction of a prime ideal in B_{q1}?

[novel parrot]

how having a prime in Bq1 contracting to p2 shows that a prime in B will contract to p2

[maiden ocean]

prime ideals of the localization B_q1 will be prime ideals of B contained in q1

[novel parrot]

😮

yah

god

okey

my bad

the thing is i knew that at the beginning then got confused

[maiden ocean]

just think of quotients cutting out all prime ideals that dont contain an ideal and localization cutting out all prime ideals not contained in the ideal

so if you want to go up you take quotients

if you want to go down you take localizations

[novel parrot]

but in localizations

[maiden ocean]

(i mean in general like going up as in passing from an ideal to some ideal containing it and going down as passing from an ideal to some ideal contained within it)

[novel parrot]

giving a prime ideal that contains some of p

will not be 0 in the localzation at p

[maiden ocean]

What do you mean by that

[novel parrot]

Like

[maiden ocean]

like its image under the localization map A -> A_p