groups-rings-fields

[dim widget]

Similarly for the V_i

[delicate orchid]

so you're looking at the image \rho(g)(V_i)

[dim widget]

Then maybe you would say \rho_U or \rho_{V_i} for the homomorphism form G \to Gl_n

So to be clear g is a matrix \rho_U(g) and Vi is a vector space which is a subspace of the underlying vector space of U.

[tired horizon]

hey guys i have a question in groups theory

what is an arbitrary group

and how do we define it

[dim widget]

You may have to clarify your question... are you asking about the definition of a "group?"

[tired horizon]

no

ik what a group is

but what is an "arbitrary" group

wait

[dim widget]

It depends on context, but probably the author just means "any group"

[dim widget]

Yeah so they just mean that they don't assume anything about the structure of the group except that it is finite.

[tired horizon]

okay so its just a finite group G

[dim widget]

It is just what you would mean in english when you say "an arbitrary person" or "an arbitrary apple"

I mean the phrase "arbitrary group" just means any group, but you can tell from context that they are assuming that G is finite.

So are you confused about why g(V_i) is an N-subrepresentation of U?

[tired horizon]

okay i didnt know that since its not my first language

[dim widget]

Yep, no worries.

[tired horizon]

yes we use finite groups in that case

thanks

[dim widget]

Yes

Yes okay, maybe before you go at it one more question: if W \subset U is an N-irreducible subrepresentation, obviously so is g *W, but what is its character?

I think with these two pieces of information you'll be like 75% of the way there.

[novel parrot]

How to do this question

[south patrol]

Well firstly by induction it suffices to consider the case of an ideal generated by two elements

[novel parrot]

Thats what im trying

so supposing that I = (f,g)

then any function in I is $h_1f + h_2g$

[cloud walrus]

ActiveChapter

[dim widget]

This is a great problem. A good follow up is to ask what the non-principal ideals are like.

[novel parrot]

i havnt done the question yet hah

[novel parrot]

If we fixed h_1 and h_2 then i can see it equaling a single function

but my problem is that for all h_1 and h_2 it has to be that same function?

[hot lake]

You also need to explain why principal ideals are of the form described in the question

[novel parrot]

Yeah

First doing the first bit

Showing its principal

Think i have it

(f,g) = (k)

then h_1f + h_2g = h_3k

for any h_1 and h_2, (h_1f + h_2g)(x) = b

and k(x) = z

we can defin a h_3 that makes (h_3k)(x) = b

since Real is a field

[hot lake]

I have absolutely no idea what you are doing

[novel parrot]

I want to show that all finitely generated ideals are principal

[hot lake]

You want to show that for any f and g, there exists a function k such that (f,g)=(k) right ?

[novel parrot]

no

I want to show that (f,g) = (k) as ideals

So any function in (f,g) has the from (h_1f + h_2g)

(h_1f + h_2g) is a function from Z to R

[hot lake]

What

[south patrol]

Isn't that what zef said oop

Like you need to find k in the first place

[hot lake]

Well I was busy editing

First you need to come up with a suitable k

Then you need to argue why the ideals are equal

[novel parrot]

Couldnt any function that isnt 0 work

[hot lake]

Idk which part you are on

[novel parrot]

this part

[south patrol]

No

[hot lake]

Are you saying (f)=(g) whether f and g are not the zero function

[novel parrot]

For a fixed h_1 and h_2, (h_1f + h_2g)(x) = b_x, say that we fixed k(x) = z, we can define a third function h_3(x) = (b_x)/z

Does this not work

[south patrol]

But why are you fixing stuff likw

You need a generator for the ideal

[hot lake]

So you are just taking k=1 ?

[novel parrot]

any non zero function

[hot lake]

1 is a nonzero function

[novel parrot]

idk just trying stuff? doesnt it work

[south patrol]

It seems overly messy - I'd try to stare at what they gave and think how you can reverse engineer S, for example

[hot lake]

Well you will be in trouble showing that (k) = (f,g) with your choice of k

[novel parrot]

i just showed that any h_1f + h_2g = h_3k

[south patrol]

This is a cool question tbh

[hot lake]

According to however you put your quantifiers that's at best only one inclusion

[novel parrot]

hmm

How should we reverse engineer S

[hot lake]

By looking at what they say the principal ideals look like

[south patrol]

Well, try to think how to deduce S from I_S for example

If that makes sense

[novel parrot]

Intersection of their 0's

[dim widget]

I think maybe you should think of the following alternative approach:

Step 1: Show that I_S is principal

Step 2: Show that if V(I) is the set of all common zeroes of functions in I, then f_{V(I)} (the generator of I_{V(I)} from step 1) is in I.

Obviously I \subset I_S by definition

So that shows that all finitely generated ideals are principal.

Discord got rid of my subscripts 😢

[novel parrot]

Hard to understand

If (f,g) = (k)

k should also be 0 on their common 0's

but how should k act on everything else?

[dim widget]

Well it must be nonzero on everything else otherwise the common zeroes would be larger, after that it doesn't matter.

[hot lake]

Generators of ideals are usually not unique

[dim widget]

Since there are so many units.

[novel parrot]

i was saying this above ;-; Zef Klop telling me im wrong

k can be anything

[dim widget]

So here's a simpler question: what are the units in this ring R?

[novel parrot]

all non zero functions are unit ?

[dim widget]

No

There are lots of units but not that many units!

[hot lake]

Nonzero here is a little ambiguous

[novel parrot]

non zero constant

[dim widget]

Those are units, but there are even more.

[novel parrot]

Functions that have no 0's are unit

[dim widget]

Yes

good

[novel parrot]

My bad

[dim widget]

So that means that after multiplying by a unit, we can assume any function is just valued in {0, 1}

So the only relevant thing about k is where it vanishes and where it doesn't vanish.

[novel parrot]

righty

On the non zero values k could have any values

Thats what i was trying to get at initally but perhaps didnt explain the best

I didnt realise that functions with no 0's are the units

[dim widget]

Yep! And any function which vanishes on S clearly differs from any such k by multiplying by an appropriate function.

So you've basically proved that I_S is principal, now time to prove that any finitely generated ideal contains some I_S!

Hi Joe, so did you try to solve the question about what the character of gV_1 is?

(As a representation of N)

[novel parrot]

Yep, thanks for your help

[dim widget]

It is! that is correct

[novel parrot]

What about this part?

[dim widget]

That is a formal consequence of the first part.

No but if you look at the definition carefully, that ideal has empty V(I)

[novel parrot]

right

[dim widget]

So it cannot be one of the I_S

So if you have shown that every finitely generated ideal is an I_S, you know it is not finitely generated.

Okay so then what is gV_1^{n_1}?

[novel parrot]

@dim widget i understood, thank you very much for your time

[dim widget]

Yes okay so either that is V_1^{n_1} for all g \in G (and then what can we conclude?) or some g \in G maps V_1^{n_1} to some V_2^{n_2} right?

Yep

So what can we conclude about U in that case?

Well we already know gU = U

because U is a representation.

Well if gV_1^n_1 = V_1^n_1 for all g what does that mean about V_1^{n_1}?

Because U is irreducible!

So it cannot have any nontrivial G-stable subspace.

So that means that in that case G transitively permutes the factors of V_1^{n_1} = U.

If g*W = W for all g \in G then W is by definition a G-subspace.

[trim dome]

bro what even is all this

[dim widget]

We only assumed to start that it is an N subspace

[trim dome]

someone pls exlain to me this i have no idea wth is happening

[dim widget]

Don't let it get to you! rep theory is hard.

[delicate orchid]

I'm doing rep theory for my thesis and I still struggle

silly little field of study

[dim widget]

We just let G act on one of the V_i^{n_i}. Basically our conclusion is: if it moves V_i to another V_j, then we don't know anything, but if it stabilizes V_i^{n_i}, then V_i^{n_i} is a G-stable subspace, so it is all of U, and we are in the case that U is just a bunch of copies of a single N-representation

[novel parrot]

Anything outside a maximal ideal is a unit right

[delicate orchid]

if the ring is local, sure

[novel parrot]

in general?

[delicate orchid]

no

[dim widget]

Anything outside of every maximal ideal is a unit

[delicate orchid]

^

counter example, 3 is not in (2) in Z

and is not a unit

[novel parrot]

ok

but R/I is a field if I is maximal, so anything not in I should be a unit? Where am i going wrong

[delicate orchid]

the inverses in R/I are only inverses mod I

again, back to Z

[novel parrot]

right irhgt

[delicate orchid]

2 is self inverse mod 3, doesn't mean 2*2 = 1 in Z

[novel parrot]

so essentailly

if fg = 1 mod I

then fg - 1 is in I

in R

make sense

[delicate orchid]

yes

[dim widget]

Because they are subspaces of U

And G acts on U

I am identifying v \in V_1^n_1 with (v, 0, 0, ..., 0) inside of V_1^{n_1} \oplus ... \oplus V_m^{n_m} = U

Yeah it's inside the a larger vector space on which G acts.

[delicate orchid]

topos just in case u didn't know this server has a bot which will actually compile the latex in messages for you

$v \in V_1^{n_1}$ with $(v, 0, 0, ..., 0)$ inside of $V_1^{n_1} \oplus ... \oplus V_m^{n_m} = U$

[cloud walrus]

Wew

[delicate orchid]

anyway sorry for interrupting

[dim widget]

Oh I did not notice that before.

That would be a lot more helpful

[silent oxide]

yeah texit is pog

[dim widget]

Feel free to join in!

[delicate orchid]

I read the backlog first

[dim widget]

Joe 1 maybe you can repost the question for @delicate orchid ?

[delicate orchid]

oh it's clifford's theorem

kinda

[dim widget]

Yeah it is a difficult question to do under time pressure but I think you are learning a lot from practicing it.

I think for finite index subgroups it is pretty easy to recognize an induction, because for some subgroup H and H-rep W it should look like $\oplus_{gH \in G/H} gW$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

where gW is the representation with the character $h \to \chi_{W}(g^{-1}hg)$

[cloud walrus]

Topos_Theory_E-Girl

[sly storm]

Hello. Can anyone here explain in a simple and intuitive way and in detail why in the hyperbolic plane it is generally impossible, starting with an arbitrary circle of area less than 2\pi, to construct (with straightedge and compass) a quadrilateral with all sides and angles congruent having the same area, and vice versa, that is, starting with an arbitrary quadrilateral with all sides and angles congruent, it is impossible in general to construct a circle having the same area?

[dim widget]

So the idea is now that we have shown that there is some finite set I and a set of irreps W_i of N for i \in I

And a set of integers n_i such that $U = \oplus_i W_i^{n_i}$ for $i \in I$

[cloud walrus]

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

[dim widget]

Yep, so here's a question: does it act transitively?

Hint: use that U is irreducible again.

Have you tried proof by contradiction?

[delicate orchid]

if it doesn't act transitively it has multiple orbits, there's a hint

yus

[dim widget]

Maybe it would help to look at a simple example?

You could write down some representations of D_8, the dihedral group Z/4 semidirect product Z/2

And restrict them to Z/4

And actually do this decomposition we're doing

I dunno why i picked D_8 you should do S_3 instead since it's easier

[delicate orchid]

do C_4 :pack:

[dim widget]

I think it's good to do a nonabelian example because otherwise both types of phenomena never occur.

Irreps of abelian groups are never induced obviously.

[delicate orchid]

yeah I know, hence the ":pack:"

I'm smokin that good stuff

[charred bison]

i asked this earlier but couldnt quite figure it out, if you have that for a field F,
[F[a] : F] is an odd number, show that [F[a] : F[a²]] = 1

[formal ermine]

tower law

F(a) : F = F(a) : F(a^2) * F(a^2) : F

what are the possible values of F(a) : F(a^2)? @charred bison

[charred bison]

must be an odd number and must be lower than [F[a] : F]?

[formal ermine]

typo, meant the left one lol

[charred bison]

hmm let me write it out and see

[formal ermine]

think minimal polynomials

[charred bison]

the minimal polynomial of a over F[a²][x] is x² - a²?

[lethal dune]

[charred bison]

am i wrong?

[formal ermine]

what if a is already in F(a^2)

[charred bison]

then we already get F(a²) supseteq F(a) so they are equal

[formal ermine]

yes so the degree is either 2 or 1

can it be 2?

[charred bison]

right and 2 is not possible since that is odd

got it thanks

[formal ermine]

ye

epic

[charred bison]

does anyone knwo a good reference for field extensions and all that

[summer path]

D&F is nice, also this problem was in D&F lol

Although there's also a book review of algebra books in #book-recommendations

[pastel cliff]

today ii am become algebra

[lethal dune]

Wii

[karmic moat]

is this proof sufficient

for some reason i feel like i'm missing some really important details

[pastel cliff]

same question as yesterday but how do we get that second isomorphism

and is this normal to come up with yourself a common argument with tensors

this is to try and show C \otimes C ≠ C

and ofc it's = C x C instead

[formal ermine]

not sure but that kinda looks like A/a otimes M = M/IM?

[pastel cliff]

proving that is a later question lol

but yeah i guess

[formal ermine]

o

[lethal dune]

what's the dimension as R vs?

[pastel cliff]

2

basis (1, i)

though in my homework is \otimes_Z not R

[lethal dune]

and what about ℂ ⊗ℂ

[pastel cliff]

uhhhhh

4 i guess

i guess that's enough to disprove yea

[lethal dune]

are they tensor prod as Algebras or just modules?

[pastel cliff]

modules im assuming

[delicate orchid]

the complexification of R[x] is C[x], x^2+1 factors as (x+i)(x-i) over C

[dim widget]

But this is as abelian groups not R-vector spaces.

[zealous forum]

If we have a ring A and the nilradical N of A is it true that every element in A \ N is a unit? I find it hard to believe that getting rid of elements x for which there exists n such that x^n = 0 leaves only elements that have inverses.

[south patrol]

This is true if and only if A has exactly one prime

So like, most examples you can think of will work

e.g. nilradical of any integral domain is {0}, but not every integral domain is a field

[glossy crag]

You didn't prove the inductive step, i.e. that F(S)(T)=F(S\cup T) (you use this in the finale line). Proving this is basically the entire proof.

[glossy crag]

If L/K is a field extension K[x]/(f)\otimes_K L is isomorphic as a K-algebra to L[x]/(f), i.e. the larger field absorbs the smaller.

[karmic moat]

hm i thought i included that in the inductive hypothesis

and then X and X_s are disjoint by assumption, so induction hypothesis gives result

maybe i missed something

[glossy crag]

You assumed the result is true for s-1, i.e. F(X_1)...(X_s-1)=F(X_1\cup\cdots\cup X_s-1)

[glossy crag]

That's not the same as this.

[glossy crag]

Besides, idk why you're so adamant about disjointness. Just prove this, then your result follows easily by iteration (or induction, same thing).

[karmic moat]

hm good point

[glossy crag]

I.e. it doesn't matter if you adjoin stuff in stages or all at once.

[karmic moat]

i have disjoint stuck in my head bc partition

so thought it was necessary/important

[pastel cliff]

does surjectivity here just follow immediately from the defn of tensor products of homomorphisms

[glossy crag]

Yes, injectiveness doesn't (because it's not always true).

[delicate orchid]

u basically just asked "is this question lying to me"

[pastel cliff]

[glossy crag]

He might've been thinking if it's just symbol pushing or something more is required.

[pastel cliff]

i was asking if it was as immediate as it seemed lol

[glossy crag]

It is.

[pastel cliff]

i'll think abt injectivity for a bit tho

[delicate orchid]

true

[karmic moat]

i proved this for the case F(u_1, u_2) = F(u_1)(u_2) in a previous problem

i guess i generalize that to all of U?

[glossy crag]

You mean adjoining elements? Yeah, that's a special case of this (X_i={u_i}).

There's really little to prove, just think of the definitions (i.e. F(X) smallest subfield containing F and X)

[karmic moat]

yea

ok i’ll think about it ty

F(X_i)(X_j) is the smallest subfield containing F, Xi and Xj, so equivalent to F(Xi cup Xj)?

[pastel cliff]

is it enough to just say "elementary tensors of N_1 \otimes N_2 arent necessarily spanned by the image of elementary tensors of M_1 \otimes M_2"

that's kinda just restating shit but idk

like it seems clear that it's not necessarily the case

[glossy crag]

Succinctly put, yes, but if you want to go about it properly:
F(S)(T) is the smallest subfield containing F(S) and T. F(S\cup T) contains F and S, therefore it contains F(S) and it contains T, so it contains F(S)(T).
F(S\cupt T) is the smallest subfield containing F and S\cup T. F(S)(T) contains F(S), therefore it contains F and S, and it contains T, therefore it contains F and S\cup T, so it contains F(S\cupt T).

[glossy crag]

Eh, idk. IIRC the problem is there might be unexpected relations between tensors, so stuff that's not 0 in M_1xM_2 might turn into 0 in N_1xN_2. IIRC the prototypical example involved 2Z and Z/2Z or smth like that.

[pastel cliff]

i'll play with it ty for pointer

[karmic moat]

ty!

[pastel cliff]

i can see it all good

[glossy crag]

As always, KC is the answer to all of life's problems

[pastel cliff]

i keep forgetting to check those

[glossy crag]

KC be praised

I just pray to god he releases a textbook (preferably on algebraic number theory) within my lifetime

This might sound embarrassing, but I've unironically written fanmail to him (and another author).

The man has collectively taught me more about algebra than any amount of textbooks of courses.

[pastel cliff]

based

what's the other author

det summon

[glossy crag]

Martin Isaacs, his algebra textbook is godly.

I rate it and Jacobson's BA as the best.

[glossy crag]

Turns out he also knows Russian and is a total dork irl (in a fun way), based on the few videos of him I've found.

[pastel cliff]

this isnt phrased correctly but

do elementary tensors commute

like m times n = n times m

[south patrol]

Nope

But this fact also leads to symmetric/exterior powers and stuff

[glossy crag]

Not literally (i.e. not inside M\otimes N), but M\otimes N is canonically isomorphic to N\otimes M

[fervent agate]

I'm a bit stuck on this exercise. The right-to-left direction is obvious, so I'm working on the left-to-right. Currently, I have the surjective homomorphism $f: F[x]\to F$ such that $f(p) = p(a)$, and I'm trying to figure something out in $F$ and then pull that information back to $F[x]$.

[cloud walrus]

ecurtiss

[void cosmos]

use the fact that given F is a field you have F[x] is an euclidean domain with the norm defined to be degree of the polynomial

@fervent agate are u still stuck

[fervent agate]

Yes

[void cosmos]

do you know what an euclidean domain is

[fervent agate]

Yep. It's the last thing in this chapter, so I only have the definition. I don't know of any special properties they have

[void cosmos]

np

whats the definition?

can u state it

[void cosmos]

okay

now

lets try wishful thinking

[frigid lark]

Maybe show that deg(f) is a valuation

in an id

[void cosmos]

if i were u once i see this expression ( the one i wnat to prov )

prove*

i would immediately guess euclidean algorithm

do u see it?

from b)

[fervent agate]

Sure. I could take literally any other nonzero polynomial and do the division algorithm. Although I would probably want to use a polynomial such that r=0.

[void cosmos]

well ur already forced

the problem wants ut o show that p(x)=q(x)(x-a) for some q(x)

u have it alreaddy but with some r(x) polynomial with the condition on deg r

so apply the euclidean algorithm on p(x) and (x-a)

what do u get

[fervent agate]

I get that p(x) = q(x)(x-a) + r(x), and then I want to show that r(x) = 0.

[frigid lark]

evaluate at a

[fervent agate]

:o

[frigid lark]

and look at the degree of r

[void cosmos]

thats not all what u get

u missed something

[fervent agate]

deg(r) < deg(a) = 1

[void cosmos]

this r(x) has a condition on its degree

yea so this says what really?

and not (a)

[fervent agate]

So r is constant but we need p(a) to be 0, meaning r must be 0

[frigid lark]

ye

[void cosmos]

yup so ur done

and i think by a you meant (x-a)

but yea

good job

[fervent agate]

Thank you!!

[void cosmos]

something to note tho:

u took for granted that given F is a field F[x] is euclidean

u should try to prove that

it would be hard to do it on your own

so look up the proof of Z is an euclidean domain

and try to imitate the argument

[fervent agate]

This is a good point, will do

[void cosmos]

😠

[south patrol]

You don't even need F to be a field for this hm

Like it is easiest to prove this directly imo

[south patrol]

But then again that is also a good exercise

[void cosmos]

😠

[frigid lark]

don't need F to be a field for F[x] to be a ED? What's an example of this? I'm just think that F[x] needs to be principle, so that restricts you quite a bit on your choice of F

[void cosmos]

no

they meant

the problem

[frigid lark]

oh

[south patrol]

No for the conclusion

[frigid lark]

lol

[south patrol]

A[x] is euclidean iff A is a field actually

[frigid lark]

oh nice

[void cosmos]

yo

Hom(-,D) is exact if D is injective

both left and right?

so for example like

suppose A-->B-->C-->0 is some exact sequence

[pastel cliff]

yeah i meant inside

[void cosmos]

with A being injective

then Hom(A,M)-->Hom(B,M) -->Hom(C,M)-->0 have trivial ker/im?

[void cosmos]

ie exact too?

[pastel cliff]

this would be p easy if it did "commute"

[void cosmos]

so ig the homology grp of this would be like how far is A from being injective in some sort?

is that correct intuition

cuz its 0 if it is

same idea with like derham stuff in geom

[dim widget]

What do you mean by that

[delicate orchid]

Like they were elements of the symmetric space

[pastel cliff]

e1 tensor e2 is not equal to e2 tensor e1 right

[dim widget]

But then it would be false if that was true

No they are not equal

[pastel cliff]

yeahik

proabbly misspoke lol

[dim widget]

I think one useful way to think about tensor products is this:

[pastel cliff]

by easy i meant obvious

[dim widget]

The space of n x n matrices is the tensor product V \otimes V^{dual}

For simplicity if we identify V and its dual with some choice of pairing, we can just write M_n(F) = V \otimes V

Where V is a V-dimensional vector space with a basis.

Given two vectors v_1, v_2 \in V you can always form the Kronecker product or (v_1 \otimes v_2)_{i, j} = (v_1)_i \cdot (v_2)_j

These are what the "simple" tensors look like in the space of matrices, and it's clear that not all matrices are of this form.

I think this is a nice toy model for various basic facts about tensor products.

[pastel cliff]

wdym by ^{dual}

[dim widget]

like V^* or Hom_F(V, F)

[pastel cliff]

hmmm i might have to sit on that for a bit

[pastel cliff]

i have to prove that the tensor prod of fin. generated modules is also finitely generated - is this obvious from just considering the image of the generating set of each module in the tensor?

[ivory trail]

yeah, you can generate the pure tensors and then add them together

(linear combination of generators of A) otimes (linear combination of generators of B) and then distribute

not really image i guess because there is no canonical map A -> A otimes B for modules

[void cosmos]

yo can we show Q is not a projective Z-module from the first defintion only?

like if we suppose it is

can we construct some diagram that it doesnt work

like maybe for example f is the identity map

[ivory trail]

what's the "first definition"

the lifting one?

[void cosmos]

and the exact(?) sequence ZxZ-->Q-->0

yea

so u would have a map that must be the identity map by commutativity

from Q to ZxZ

which is a contradiction(?)

thats my first thought

this doesnt work ig

[white oxide]

ohh ok that helps

thank you

[vagrant zinc]

Guys, how can I prove that every cyclic of length k has order k, this is equal to the identity

[patent girder]

What if you try to consider an n-cycle and work from there

What do we know about the order?

[ebon pine]

Consider the cycle to be (a1 a2 ... ak) then you want to show that applying this permutation to any arbitrary element of the cycle ai returns back ai

[solar shore]

looking at rings for the first time

do the proofs for the properties look alright?

[charred bison]

F is a subfield of C(complex numbers), f(x) ∈ F[x] is an irreducible polynomial(over F). If f(x) has more than one root say a and b, then F[a] = F[b]

how do i show this?

[lethal dune]

What happens if you look at Q and f(x)=x^3-2

[dim widget]

Do you really mean that F[a] = F[b] are equal or do you mean isomorphic?

[charred bison]

yeah up to isomorphism

[dim widget]

If it's up to isomorphism then both are isomorphic to F[x]/(f(x)) as rings. If you are actually asking about equality as subfields of C then counterexamples are generic. And the one @lethal dune gave is fine.

[lethal dune]

You don’t really need to ping me

[dim widget]

I thought it might be educational for you 🙂

[lethal dune]

Again no need to ping

[dim widget]

Oh, sorry 🙂

[lethal dune]

god dammit

[coral shale]

[tender bough]

In a Hilbert space, let 𝑃 be an bounded operator, is it true that if 𝑃𝑃* = 0, then 𝑃 is zero?

where * is the conjugate

[coral spindle]

There is no such exact sequence of Z-modules

Q is not finitely generated

[warm urchin]

you mean adjoint?

[tender bough]

yes

[warm urchin]

then yes ,because ||P*P||= ||P||^2

[tender bough]

thanks!

is this true in any star algebra?

[warm urchin]

I dont think so , i think the ring Z/nZ with identity involution should be a counter example , But it is true over any C*-algebra

This is an answer if you are asking whether xx*=0 implies x=0

[tender bough]

@warm urchin thanks!

[novel parrot]

How should I find the generators of I_S ?

[dim widget]

What is your guess as to what I_S is?

[novel parrot]

A function that is 0 on S

[dim widget]

No sorry, more precisely what is your guess as to what the generators of I_S are?

[novel parrot]

No idea

[dim widget]

What are some functions in I_S?

Besides 0.

[novel parrot]

Does S include the inside part of the circle?

[dim widget]

No it's just the unit circle. But if S was the entire unit disk could you show that I_S is just 0?

[novel parrot]

Maybe after i can try that

f = X^2 + Y^2 - 1 is a polynomial that is 0 on S

[dim widget]

Yes, so my first guess would be that that generates all of the polynomials which vanish on S because I can't think of any others (except for multiples of this one).

[novel parrot]

ok

[dim widget]

So how would you go about that?

[novel parrot]

Supposing we had arbitrary polynomial that vanishes on S and try and see if it is a multiple of f

Is that how you would do it

[dim widget]

Well yes

[novel parrot]

We could do long division by X^2 + Y^2 - 1?

And show that the remainder is 0

[dim widget]

That is the doctrinaire way to do it, but it's a bit complex

[novel parrot]

I have never done division in 2 variables

[dim widget]

Since R[X, Y] doesn't have a division algorithm

[novel parrot]

Yeah

[dim widget]

R(X)[Y] and R(Y)[X] do and sometimes you can use that division algorithm

[novel parrot]

The remainder is forced to be 0 on S too

But i need to show that the remainder is exactly 0

Whats the difference between R(X)[Y] and R[X,Y]

R(X) is a field?

[dim widget]

Yes R(X) is all the P(X)/Q(X) where Q(X) is not zero.

[novel parrot]

How would you do this question?

This way?

Or a better way

[dim widget]

No probably not

But I think this is not an unreasonable way to do this:

R(X)[Y] is a euclidean domain thus a PID, so we know $I_S*R(X)[Y]$ is a principal ideal

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

Since X^2 + Y^2 - 1 is irreducible in R(X, Y), by Gauss's lemma it is irreducible in R(X)[Y] so it generates the ideal I_S

over R(X)[Y]

This means that any P(X, Y) which vanishes on S satisfies P(X, Y) = (X^2 + Y^2 -1)*(Q(X, Y)/R(X)) for Q some polynomial in two variables and R some polynomial in only the variable X.

So then R(X)*P(X, Y) is a multiple of (X^2 + Y^2 - 1)

But it is easy to see from this discussion that the ideal (X^2 + Y^2 - 1) is prime, and also not hard to prove that R(X) cannot be a multiple of (X^2 + Y^2 - 1)

So P(X, Y) is a multiple of (X^2 + Y^2 - 1)

but there are also other ways to prove this.

[novel parrot]

So they show that every irreducible is a prime

but the definiton of prime is, if x|yz => x|y or x|z

but they say that x=yz not x|yz

So how does this prove that x is prime?

[dim widget]

Yes it is a typo. It should be x|yz, but the rest of the argument works unchanged. Edit: I read the argument and I think the whole thing is super weird, but the basic idea is correct.

[wraith cargo]

this whole proof is strange

[dim widget]

one is singular and one isn't

Calculate the tangent space at (0, 0) on the RHS.

Okay that makes proving this harder

Singular means there exists a maximal ideal m \subset R where dim(m/m^2) is not dim(R)

If you don't know what dim(R) is then it doesn't matter, just notice that one ring has a maximal ideal where m/m^2 has dimension 2 and the other one doesn't

[agile burrow]

On a more direct level, can't you show the latter isn't a UFD?

[dim widget]

You could also show that C(x) is not isomorphic to C(x)[y]/y^3 - x^2

You could also do that... but I think proving any particular element is irreducible is probably more work than just computing tangent spaces. But that is another valid approach which maybe seems less magical if you don't know algebraic geometry.

[agile burrow]

Fair enough

[rancid totem]

Is there a cyclic permutation of natural numbers?

[wraith cargo]

Yes

Actually no

For integers yes

For natural numbers no

[dim widget]

Well as sets they are the same.

[wraith cargo]

They are equinumerous but they are not the same set

[dim widget]

So you have to ask what they mean by the cyclic permutation I would say.

[dim widget]

What do you mean by cyclic permutation?

[novel parrot]

how do we know that g(X) also divides f ?

how coprime and irreducible implies that

This proof is to show that if a polynomial has n roots then its degree is atleast n

[wraith cargo]

What exactly is the result being proven here

[dim widget]

They show that since r_i(a_i) = 0 that r_i = 0, so there is no remainder. Thus g divides f

[novel parrot]

i get that

but mean g(X) = (X-a_1) .. (X-an)

how do we know that the whole thing divides f

[dim widget]

Because the (X - a_i) all divide f, and they are coprime since the a_i are distinct

I suppose that is a detail that you have to fill in: that (X - a_i) is coprime with (X - a_j) for a_i != a_j

[novel parrot]

And how does coprime implies that g(X) divides f?

[wraith cargo]

Doesn't this follow by definition?
Like if g(X)|f(X), then g(X)h(X) = f(X) and we already know that g(X)(X-ai) = f(X)

[novel parrot]

We know that individually the (X-a) divide f(X)

but we dunno that the multiples of them all divide f(X)

is what im asking

like 3|6 and 6|6 but 18 doesnt divide 6

[wraith cargo]

write f(X) in terms of it's irreducible factors

we know that each (X-ai) divides f(X)

but since they're all coprime they share no irreducible factors with each others

hence their product can be written as a product of irreducibles that all correspond to irreducibles of f(X)

[dim widget]

I think @novel parrot is confused because the book refers to the P such that P(X)(X - a_i) = f(X) for all i by g(X), then uses g(X) to refer to the product of the (X - a_i)

[dim widget]

Yes this is the key point somehow.

[novel parrot]

you are using the UFD property to do this?

[wraith cargo]

yes

actually wait

[novel parrot]

but what if R is not a UFD

[wraith cargo]

is R a UFD?

[novel parrot]

nah

dont think*

[wraith cargo]

Ahh

oh but wait

we're looking at R[X] as a subring of K[X]

[novel parrot]

oh

K[X] is UFD

[wraith cargo]

yeah

[novel parrot]

what do you mean by this?

[wraith cargo]

so you can decompose f(X) into a unique product of irreducibles over K[X]

[wraith cargo]

definition of coprimality in UFDs

like if they had nontrivial gcd

then they have to share some irreducible factors

[novel parrot]

my definiton of coprime was if (a,b) = (1)

as ideals

[dim widget]

I think that is a good definition for this problem.

[novel parrot]

So if an irreducible divides something, then it must show up in its factorisation into irreducibles ?

[dim widget]

Yes, this is almost the definition.

[novel parrot]

i dont understand why coprime is needed

[dim widget]

Because of this example.

X^2 + X + 1 and X - \omega divide X^3 - 1 in C[X], but their product does not.

[novel parrot]

i see

[rancid totem]

Nevermind, it was a silly question

[novel parrot]

Is there a proof that proves this formally

[dim widget]

It is pretty easy, you have to show that (R[X]/((X - a), (X - b)) is the 0 ring if a is not b.

[novel parrot]

Side question - X-a and X-b are only coprime in K[X] right

not neccesarily in R[X]

[dim widget]

Ahhh sorry, true facts.

[novel parrot]

since b - a is not neccesarly a unit in R

[dim widget]

This is not a side question, it's very pertinent.

Yes this proof is incorrect as stated, but the result is correct.

I would stop trusting this book since it seems like the quality is very poor

[novel parrot]

lmao

lecture notes hah

Many typoes

[dim widget]

Either way, so many completely incorrect or just wrong-headed proofs.

[novel parrot]

yeah

[dim widget]

Anyway here is the simpler and more obvious proof:

The degree of a polynomial in R[X] is the same as the degree in K[X], and in K[X] this proof actually works.

[novel parrot]

oh

Ok so i know that K[X]/((X-a), (X-b)) = 0

[novel parrot]

but how does that prove this?

[dim widget]

That shows that (x-a, x-b) = K[X] which means that (x - a), (x-b) are coprime.

[novel parrot]

Yes

but how does coprime imply a|z and b|z => ab|z

[dim widget]

That's one definition of coprime.

But if you like this is the chinese remainder theorem: $K[X]/(X - a) \times K[X]/(X - b) \cong K[X]/((X - a) \cap (X - b)) \cong K[X]/((X - a)(X - b))$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

The last isomorphism is what we get from the two ideals being coprime

So if you look at what it means to be zero on the right hand side and on the left hand side we see that (X - a) and (X - b) divide P if and only if the product divides P.

[novel parrot]

Right

I understand

Tysm!!

[dim widget]

No worries 🙂 I hope you find some better lecture notes!

[novel parrot]

Me too

But finals in a week lol

[dim widget]

Good luck!

[novel parrot]

Thanks

Need to grind another 50 pages of lecture notes

Oh another thing

Two irreducibles are coprime then they cant be associates right

x,y are irred and coprime then x!= yu where u is unit

[dim widget]

Yes

[sly storm]

Hello. Can anyone here explain in a simple and intuitive way and in detail why in the hyperbolic plane it is generally impossible, starting with an arbitrary circle of area less than 2\pi, to construct (with straightedge and compass) a quadrilateral with all sides and angles congruent having the same area, and vice versa, that is, starting with an arbitrary quadrilateral with all sides and angles congruent, it is impossible in general to construct a circle having the same area?

[fleet pelican]

no

[dim widget]

It's not clear what you mean by straightedge and compass on a hyperbolic space. Are they hyperbolic or euclidean compasses and straightedges?

[formal ermine]

4 day streak!

[dusty verge]

You can use $ab = gcd(a, b)lcm(a, b)$ to get that result specifically.

[cloud walrus]

Halliday

[dusty verge]

To get the above, you can use the fact that if $\displaystyle gcd(a, b) = \prod_{\substack{p_i^{k_i}||a \ p_i^{j_i} || b}}p_i^{min(k_i, j_i)}$ and $\displaystyle lcm(a, b) = \prod_{\substack{p_i^{k_i}||a \ p_i^{j_i} || b}}p_i^{max(k_i, j_i)}$

[cloud walrus]

Halliday

[dusty verge]

I think that does it, I might be off on some details though

Wait yeah, because then their product is just $ \prod_{\substack{p_i^{k_i}||a \ p_i^{j_i} || b}}p_i^{min(k_i, j_i) + max(k_i, j_i)} = \prod_{\substack{p_i^{k_i}||a \ p_i^{j_i} || b}}p_i^{k_i + j_i}$

Huh, why isn't the bot compiling this?

[dusty verge]

Actually, I think this explains it right? You can't put a straightedge on a hyperbolic plane because a hyperbolic plane is all curvy and shit

[dim widget]

I mean it's just a question of what you mean, the hyperbolic plane is just the plane with a strange notion of distance. So you could ask for either compass or either straightedge

[dusty verge]

I know I was just making fun of the guy, because no one seemed to like his question

Although it does seem at least a bit galois theory-y

[dim widget]

AFAIK you can actually square the circle with a euclidean compass and straightedge on the hyperbolic plane when the angles are constructible

[dusty verge]

That's actually pretty cool

I wonder if there are geometries where $pi = 3$

[cloud walrus]

Halliday

[dusty verge]

Other than euclidean geometry obviously

[solar shore]

hey y'all

a question's asking to describe the subrings of the ring of integers

is that not just the subrings $n\mathbb Z$?

[cloud walrus]

blanket

[solar shore]

also the trivial subring 0 and Z itself of course but

[south patrol]

Uh

[solar shore]

am i missing any?

[south patrol]

Are nonunital rings allowed here

If so then sure

[solar shore]

i have no idea, this is my first day with ring theory haha

or second day i guess i started looking at this stuff yesterday but didnt get too far, im doing some exercises now

[south patrol]

It depends on the convention of the text you're reading basically but I assume you don't need rings to have a multiplicative unit here in which case yeah that's correct

[solar shore]

ah right yeah

the text emphasized heavily that rings need not have unity

[dim widget]

Then you should also ask if the multiplicative unit needs to be different from 0 🤔

[solar shore]

hmm there were some exercises dealing with subrings of $\mathbb Z_n$ and they did have unity different from 1

[cloud walrus]

blanket

[solar shore]

like this one

the book said that the subrings $n\mathbb Z$ didn't have unity so im assuming they don't have units

[cloud walrus]

blanket

[solar shore]

is that a right assumption?

[dim widget]

Yes of course, but I was thinking about whether (0) should count.

As a ring with unity

[solar shore]

ah

[dim widget]

Which is another thing that people quibble about sometimes

[solar shore]

hmm

well from what i know so far, unity's the element 1 such that a1 = a for some a in the ring

[dusty verge]

Let's see, they have to be an additive subgroup, so you take the smallest positive non-zero integer N, and by induction, it contains NZ

[solar shore]

if its the trivial subring, then 00 is 0, no? i would then say it has unity

[dim widget]

Yeah, I think there is no objective argument one way or another, it's just a convention 🙂

[solar shore]

ahh okay

[dusty verge]

If you have M also in the subring, bezout's lemma gets you GCD(M, N)

[dim widget]

Some people don't like to include it because it's weird

[dim widget]

they have already solved the problem

[dusty verge]

I'm trying to give justification that those are the only subrings

[solar shore]

i do appreciate the justification

[dusty verge]

To justify this, let $n$ be the smallest non-zero element of the subring, and consider $gcd(|m|, n)$ for any non-zero element of the subring m

[solar shore]

the book lacks in explicitness sometimes so ill take all explicit forms of statements

[cloud walrus]

Halliday

[dusty verge]

Basically, because of Bezout's lemma, you can always find a way to get $gcd(m, n)$ for any $m, n$ in the subring, but if $n$ is the smallest non-zero in it, $gcd(m, n) = n$, meaning $n$ divides every element of the subring

[cloud walrus]

Halliday

[dusty verge]

If you're unfamiliar, Bezout's lemma just says there always exists integers $x$ and $y$ such that $xm + yn = gcd(m, n)$ for any integers $m, n$.

[cloud walrus]

Halliday

[solar shore]

right yeah the linear combination

got that one

[dusty verge]

Yeah

[solar shore]

also i do have a question about notation

so for group theory, we used $\langle \rangle$ notation to denote a cyclic group/subgroup

[cloud walrus]

blanket

[solar shore]

is the same notation used for rings?

[dusty verge]

Not really? I think you're actually getting kinda close to the idea of an ideal

[solar shore]

ah

well i will encounter those when i do, its in two chapters for me hehe

[dusty verge]

Basically cyclic subrings aren't really a well defined idea cause of the two operations.

[solar shore]

oh i see

oh right i keep forgetting multiplcation

[dusty verge]

An ideal is an analogy to divisibility, kind of

[solar shore]

im still stuck in the group mindset, there wwas one exercise that messed with me for a bit because i just forgot about the second binary operation

[dusty verge]

You'll get used to it

Don't worry rings are generally easier than groups

Groups can go hard

[solar shore]

gotcha

oh is it okay if i asked for one more clarification before i go back to doing more exercises?

[dusty verge]

go for it

[solar shore]

so it was these two questions that were messing with me

for part a, i said Z_6 with a = 3 sop a^2 =9 mod 6 = 3

b i said z^4 so a = b = 2, but ab = 2*2 mod 4 is 0

and then c i said z_4, a = 2, b = 1, c = 3 so that 2 * 1= 2 * 3 mod 4 = 2

[dusty verge]

Yeah that's all correct

[solar shore]

alright

and then for 7, i said that if we had z_p where p was prime, then its set of units is every element except 0

so it means that they all have inverses

oh too small

sec

\begin{enumerate}
\item $a^2 = a \implies a^{-1}a^2 = a^{-1}a \implies a = 1$. The equation is also true for $a = 0$.
\item $ab = 0 \implies a^{-1}ab = a^{-1}0 \implies 1b = 0 \implies b = 0$. Similarly, if $b^{-1}$ is the multiplicative inverse of $b$, then $abb^{-1} = 0b^{-1} \implies a1 = 0 \implies a = 0$ so that either $a$ or $b$ must be 0.
\item Let $a$ be a nonidentity element. Then $ab = ac \implies a^{-1}ab = a^{-1}ac \implies 1b = 1c \implies b = c$.
\end{enumerate}

[cloud walrus]

blanket

[dusty verge]

If you want a cheesy answer, you could also just point out that the units form a group

[solar shore]

oh yeah thats true

[dusty verge]

Which means that the units are closed under multiplication, and have cancelation

[solar shore]

let me add that as a separate answer that's smart LOL

but is the answer okay?

[dusty verge]

I had it drilled into me to not use \implies like that by a very mean british professor, but other than that yeah

[solar shore]

ah LOL

"mean british professor" i love it

[dusty verge]

He's actually not that bad, but I sure hated him at the time

[solar shore]

LOL

do i have to do cases for these?

where m can either be greater than, less than, or equal to 0

[dusty verge]

How does the book define $m \cdot(ab)$?

[cloud walrus]

Halliday

[dusty verge]

Also, the first implies the second, and also the third so long as you already have $(-1)(a) = -a$, which should be by definition of multiplying an element of the ring by an integer

[cloud walrus]

Halliday

[dusty verge]

You could avoid cases by using some trickery with something like $SGN(m)$ to mean the sign of $m$,

[cloud walrus]

Halliday

[solar shore]

they define it ab + ab + ab m times

i think its because they dont want ambiguity with mab?

[dusty verge]

Ugh, that's annoying imprecise

[solar shore]

i was a bit confused as to why they did that because... mab makes sense as long as i know m is an integer

and ab are ring elements

[dusty verge]

You can probably get away with calling it $SGN(m)\sum_{k = 1}^{|m|}ab$, right?

[cloud walrus]

Halliday

[dusty verge]

And noting that you define the empty sum to be Id_R

[solar shore]

sgn is the sign function right

[dusty verge]

yeah

That way you don't need to do cases, which are annoying to do

[solar shore]

if i didnt use sgn would i then proceed by cases?

[dusty verge]

Yeah

[solar shore]

i forgot we do have access to (-1)a = -a, that was one of the properties we proved

[dusty verge]

But basically you just induct on distributivity for the first, and then use the first to prove the other two

[solar shore]

gotcha

okay i think i got it now

thank you!

i think ill stop here after to take a break and study something else :p

[karmic moat]

is this proof sufficient?

[solar shore]

sorry to bother but which environment do you use for the problem and proposition?

ive been using t color box for stuff but ive been really wanting something like that...

[dusty verge]

You claim that $\mathbb{K}_1$ and $\mathbb{K}_2$ are the splitting field for $p_1$ and $p_2$, but that's not necessarily true right?

[cloud walrus]

Halliday

[dusty verge]

And since the degree if the extension is potentially infinite, I think it might not work beyond that, although I'm super rusty on this stuff, so take that with a pinch of salt

I also have not looked at the proof beyond that point, since I was trying to remember what the hell any of it meant lol

Oh yeah, it doesn't break it, but they're definitely not necessarily the smallest fields such that ...

[karmic moat]

by assumption, i have that K1 and K2 are algebraic closures

[dusty verge]

In $\mathbb{Q}(\sqrt{2})$, the polynomial $x^2 - 2$ factors, but that's not the algebraic closure of $\mathbb{Q}$.

[karmic moat]

mdframed

[cloud walrus]

Halliday

[dusty verge]

The algebraic closure is the smallest field such that EVERY polynomial factors, not any given one.

[karmic moat]

true

hmm

[dusty verge]

And iirc, an algebraic closure can have potentially infinite degree, right?

[fleet pelican]

I think it needs to be infinite. No?

[dusty verge]

Either way, I don't think it actually matters for your proof, since you're just claiming that $p$ splits, the fact that you say it's smaller doesn't matter

[karmic moat]

yeah it does

[cloud walrus]

Halliday

[dusty verge]

infinite size, not necessarily infinite degree though. Degree is the degree of the smallest polynomial that you add the roots to the field to get to the new one

[formal ermine]

how do I form integral extensions of Z

[solar shore]

ok ill look into that, thank you

[dusty verge]

Like the degree of $\mathbb{C}$ over $\mathbb{R}$ is $2$, since $\mathbb{C} = \mathbb{R}(x^2 + 1)$ or whatever the notation is

[cloud walrus]

Halliday

[fleet pelican]

Don't you have irred of every degree for Z and C? I guess not if you include C and R

[formal ermine]

R(i)

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

[dusty verge]

Makes sense lmao. I know so much more algebra than I did when I learned galois theory that that notation seems like it actually makes sense now lol

[fleet pelican]

I guess yeah in geo you wouldn't want to assume that kind of thing

[dusty verge]

Yea, I think the degree of C over Q is infinite

[formal ermine]

indeed

C is bigger than Q's algebraic closure

[dusty verge]

Oh it is?

[fleet pelican]

Q bar is countable,

[dusty verge]

Right right transcendentals

[fleet pelican]

tfw no monic poly satisfies you 😿

[formal ermine]

uwu?

I guess I could just find a finitely generated Z module lmao

[dusty verge]

What's an integral extension?

[dusty verge]

I find $x - \pi$ very satisfying thank you

[cloud walrus]

Halliday

[formal ermine]

every element is integral

[dusty verge]

You mean root of poly with intege coefficients?

[formal ermine]

no

[fleet pelican]

Monic

[dusty verge]

gotcha

Right of course lol

[formal ermine]

R --> S is integral iff every element in S is integral over R i.e. is the root of a monic polynomial in R[x]

[dusty verge]

Multiply by lcm of denominators

duh

What's an extension of Z?

[formal ermine]

Z --> R

[dusty verge]

That has not helped me

[fleet pelican]

Z[x]/f(x)

[dusty verge]

Okay, sure

[formal ermine]

non algebraic extensions count too

[fleet pelican]

Like what?

[dusty verge]

How is a non algebraic extension integral?

[formal ermine]

R

[dusty verge]

That sounds like a contradiction in terms

[formal ermine]

that's not what I meant

[fleet pelican]

Ew

[formal ermine]

Z --> C is a non algebraic extension but still an extension

[dusty verge]

Oh gotcha

[formal ermine]

we have this proposition

[dusty verge]

So it's a ring that contains Z as a subring?

[formal ermine]

for an extension R --> S and s_1, ..., s_k in S. we have s_1, ..., s_k are integral over R iff R[s_1, ..., s_k] is finitely generated as an R mod

[formal ermine]

yes, Z embeds into it

Z --> R

[dusty verge]

R is thick R?

[formal ermine]

any R

it's not as easy as just taking any algebraic extension

[dusty verge]

Oh okay

[formal ermine]

neither Z[1/2] nor Q are integral over Z

though Z[sqrt(2)] and Z[zeta_3] are

I think

we need to take a number field K

and then its ring of integers will be integral over Z

yeah that would make sense

[dusty verge]

what does integral mean in an arbitrary ring, is it just $n \cdot id$?

[formal ermine]

but is the other direction true as well

[cloud walrus]

Halliday

[formal ermine]

like is every integral extension of Z the ring of integers of some number field

[next obsidian]

No

[formal ermine]

ah ok lol

thanks chmonkey

another mystery solved

[next obsidian]

For one it can be way too big

Take an algebraic closure of Q

And form Z’s integral closure in it

This is called the absolute integral closure, and generally is denoted R^+ for an integral domain R

This is hella infinite-dimensional

[formal ermine]

wdym with this? ring of integers?

[next obsidian]

Everything integral over it

[formal ermine]

ah right integral closure

I saw it somewhere in the upcoming lecture notes

I assume it's the same thing as algebraic closure just for integral elements?

[next obsidian]

I mean I said what it is

It’s the ring-theoretic analogue

[formal ermine]

ye

ok

[next obsidian]

Even if you restrict to finite extensions I don’t think it’s true

The ring of integers of Q(sqrt(d)) for d ≠ 2 is Z[sqrt(d)] if d = 1 mod 4 (or was it 3 mod 4 lmfao) and Z[(1+sqrt(d))/2] otherwise

Take d such that it’s the latter

[formal ermine]

that's what I was working on earlier today lol

alg nt is fun ong

[next obsidian]

If Z[sqrt(d)] was the ring of integers of some number field, it would have to be the ring of integers of its fraction field which is Q[sqrt(d)]

But this ring is not the ring of integers

[formal ermine]

why would it have to be the roi of its ff

[chilly radish]

C over R

[next obsidian]

In general you can show that if you have an extension L\ Frac(R) then the fraction field of the integral closure of R in L is L

Frac(R^~) = L

But in this case it’s even easier because of degree stuff

[formal ermine]

I see

thanks chmonkey

[dusty verge]

Yeah, I think my observation breaks your proof actually, since I believe you end up proving instead that two splitting fields for the same polynomial are isomorphic

[karmic moat]

yeah that makes sense

oop

s

[dusty verge]

Hmm, I think the general idea is still mostly good, though

What if you instead look at a chain of isomorphisms, ordering them by inclusion on the domain and codomain

[lethal dune]

you're overthinking

[formal ermine]

what does the upside down capital pi mean

[lethal dune]

Hints:

1. || Every embedding into an algebraic closure can be extended to any algebraic extension. So given 2 alg closure, being alg, it can be extended to a map K1 → K2. ||
2. || nontrivial field morphisms are injective ||
3. || Show your map K1 → K2 is surjective by pulling back the polynomial and pushing the root ||

[lethal dune]

disjoint union in most cases

can also denote coproduct in arbitrary category

[formal ermine]

uhhh

I think coproduct?

,, \mathcal{O}(U) = \set{s : U \to \coprod_{\mathfrak{p} \in U} A_{\mathfrak{p}} | s(\mathfrak{p}) \in A_{\mathfrak{p}}}

[formal ermine]

$U \subseteq \operatorname{Spec}{A}$

[formal ermine]

A_p is localization at p

I assume this is coproduct?

[delicate orchid]

looks like it

[lethal dune]

no it's disjoint union

this the espas etale contuction

aka space of sections

[formal ermine]

is it

this text says it's the sheaf of rings on Spec A

is it the same thing?

[lethal dune]

well this is more accurate

what I said is a more general construction, what you have is more specific. your O here is the structure sheaf of the ring

[formal ermine]

wait so is it disjoint union or coproduct

[dusty verge]

por que no los dos?

[lethal dune]

disjoint union

[dusty verge]

Coproducts are a generalization of a disjoint union

Although the difference sounds like it matters here

[formal ermine]

me no know cat theory ):

[formal ermine]

oke thanks

[lethal dune]

it does

you have coproduct in the category of Rings, namely tensor product

[dusty verge]

cat theory is super cool in theory, but then hell in actuality

[formal ermine]

yeah I tried working my way through category theory in context but it was just too painful

[void cosmos]

meow

[dusty verge]

It's like "Hey lets look at math things not by the actual composition, but by whether they stand in relation to other things"

[formal ermine]

I like abstract nonsense but not THAT much abstract nonsense

[formal ermine]

yoooooooo moamen

[rotund aurora]

category theory without context is just cringe

[formal ermine]

what's good

[dusty verge]

I'm usually good at levels of recursion

[lethal dune]

it's better if you do cat theory before staring any kind of AG

like sheaf / schemes

[formal ermine]

I just do both at the same time

has worked with comm alg too

[lethal dune]

not necessarily but makes a lot easier

[formal ermine]

the method I'm doing right now is fun and that's all that matters

[lethal dune]

sure

but how do you define a morphism of sheafs?

[formal ermine]

gosh darn it

not every commutative diagram is cat theory

[lethal dune]

who gives a sheaf

[formal ermine]

good one

[dusty verge]

What's a sheaf?

[lethal dune]

such bullsheaf

[formal ermine]

a local presheaf

[dusty verge]

Sheaf doesn't sound nearly enough like fuck for that joke to work

[formal ermine]

a presheaf is like

you take a topological space

and you assign some data to every open set

e.g. abelian groups

[lethal dune]

a functor from open(X)^op → Ab

[rotund aurora]

@formal ermine how would you motivate sheafs

like I could make a gazillion structures with compliacted data

why do u care about sheafs

[formal ermine]

no idea, I've started reading schemes just about now

it's like

[rotund aurora]

ok xD

[formal ermine]

I've finished the chapter on sheaves like a couple of days ago lmao

haven't "used" them yet

[rotund aurora]

I mean, I intentionally refuse to learn about sheafs and schemes for now

[dusty verge]

Open as in open sets, what does op mean, and what is Ab here?

Ab abelian groups?

[formal ermine]

[formal ermine]

yeah idk I have nothing better to do

I'm procrastinating my alg nt homework lmao

[dusty verge]

I like math because it's like the textbooks keep trying to make me give up on understanding things

And just do it all computationally

[rotund aurora]

this sounds interesting btw, like every time I see a functor it's pretty cool

[formal ermine]

every time I see a functor I cry<

[dusty verge]

Functors are sick

[formal ermine]

every time I see a right exact but not left exact functor I break into tears

EVERYMODULEISFLATEVERYMODULEISFLATEVERYMODULEISFLATEVERYMODULEISFLATEVERYMODULEISFLATEVERYMODULEISFLATEVERYMODULEISFLATEVERYMODULEISFLATEVERYMODULEISFLATEVERYMODULEISFLAT

[dusty verge]

Functors are homomorphisms except between branches of math

[lethal dune]

but you do a lot of AG stuff

[rotund aurora]

functors are homomorphisms between diagrams

[lethal dune]

without using scheme language?

[formal ermine]

your mum is a homomorphism between me and her bed

[void cosmos]

yoo @formal ermine

nothing much

[formal ermine]

yooo

[void cosmos]

today is rest daay so i can do math

[formal ermine]

epic

[dusty verge]

Functors being homomorphisms between diagrams is a formal extension of the basic intuition I mentioned

[formal ermine]

I had a stressful school day today

[void cosmos]

school?

[lethal dune]

nice

[void cosmos]

ur a highschooler?

[formal ermine]

yeh

[void cosmos]

wow

why is that

[formal ermine]

idk what you call it in english but I had like an hour where I didn't have class

[formal ermine]

I was born at a very young age

[void cosmos]

it's gap

[formal ermine]

yeah

gap

[void cosmos]

i meant why was it a stressful day

[formal ermine]

and then the next lesson

was supposed to be math

the teacher came in

said "I have something to do"