groups-rings-fields

[mild laurel]

It's kind of the starting line and one of the theorems that class field theory generalizes to arbitrary number fields

[vestal snow]

Our book says that it is beyond the scope of our book

[mild laurel]

iirc there's a better proof of this

[vestal snow]

Is there a proof which doesn't use the prime thing?

[mild laurel]

yeah kronecker weber is cool bc usually you prove the local version first and then use that to prove the global version of Q

yeah I think so but i'm remembering how it works

hm

so I remember that when d is a prime p, you can say the same thing that the discriminant of x^p - 1 is pm p^p and is a square which means its square root exists in your field

[vestal snow]

Yes that's part b

[mild laurel]

But you can write down that $\sqrt{p^p} = p^{(p-1)/2}\sqrt{p}$

[cloud walrus]

64606131

[mild laurel]

So $\bQ(\zeta_p)$ contains $\sqrt{p^p} = p^{(p-1)/2} \sqrt{p}$ which implies that it must contain $\sqrt{p}$

[cloud walrus]

64606131

[mild laurel]

modulo some signs

[vestal snow]

that makes sense

[mild laurel]

and this probably only works when p isn't 2

you should be able to do such a thing in general too

[vestal snow]

So what's the purpose of having 8d instead of 4d?

Because you will have i in the field if you replace 8d by 4d

[mild laurel]

yea idk, i think 4d should work too iirc

oh

the question requires you to have square roots for both d and -d?

I think that's why you need 8d?

[vestal snow]

If we have a square root of d and i in the field

Can't we just multiply those to product a square root of -d?

[mild laurel]

I know that Q(sqrt(d)) is always embedded in Q(zeta_{4d}) but

hm yeah that sounds reasonable

[vestal snow]

Oh my god

I just realized how easy this was

\sqrt{d^d}=d^{\frac{d}{2}} or d^\frac{d-1}{2}\sqrt{d}

[mild laurel]

yeah

this is what I wrote with p above basically

[vestal snow]

Forgetting basic arithmetic facts seems to be a common thing on this channel

Thanks for the help 13190949

[mild laurel]

no problem 91969759074873114613

[vestal snow]

You need 8d for the case where d is even

so that you have sqrt 2 in your field

and you can divide by it to make d odd

[mild laurel]

but Q(zeta_8) already contains sqrt 2

so if d is even then 4d will already contain sqrt 2

[vestal snow]

Darn you're right

I think the author put it there because he thought what I did

and didn't realize what you said

[mild laurel]

yeah maybe

or maybe the author just wasn't concerned about getting the best possible bound but just something that works idk

[rigid cave]

Why does A_5 have two separate 5 cycles as conjugacy classes?

Why can you split the 5 cycles into two separate conjugacy classes?

[carmine fossil]

Let A be the conjugacy class in S_5
You can write A={x(1,2,3,4,5)x^-1| x is even} + { x(1,2,3,4,5)x^-1 | x is odd}

Elements in the first set are conjugate to each other in A_5

You can write an odd permutation as a product of an odd permutation and a even permutation

[rigid cave]

Okay I think I understand that, but I do not quite get why that should give me two different conjugacy classes for the 5 cycles

Is it maybe because the first conjugacy class is of the form (abcde) and the other one is (abc)(de)?

They are obviously equal to each other but are they of that form?

[carmine fossil]

I mean the latter is odd

Elements in the second conjugate class are still even

[rigid cave]

Oh, right

[carmine fossil]

Let's say the 2 sets share a common element

Let x(1,2,3,4,5)x^-1=y(1,2,3,4,5)y^-1 where x is even and y is odd

Then (y^-1 x) ((1,2,3,4,5) ) (y^-1 x)=(1,2,3,4,5)

But x is in center of (1,2,3,4,5) implies x is in {(1,2,3,4,5)}

That is y^-1 x is even

[viscid pewter]

this doesn't make sense to me tbh

[carmine fossil]

( x(1),x(2),x(3),x(4),x(5) )= (1,2,3,4,5)

x(1) can be 1,2,3,4 or 5

And for each case you get a member of that group

Because x(2) ,x(3)... are uniquely determined by choice of x(1)

[rigid cave]

Let's look at the picture I sent before. What does the representative mean? Is it the element "a" witch generates CL(a)?

[carmine fossil]

Yes

A={x(1,2,3,4,5)x^-1| x is even}, Here (1,2,3,4,5) is a representative

[rigid cave]

Okay, but how can (12345) and (13524) generate two different conjugacy classes?

[carmine fossil]

(1,3,5,2,4) is in the second set

[carmine fossil]

This

[rigid cave]

Okay but how can x be odd in the "post" you just replied to?

[carmine fossil]

You write x as eo where e is an even permutation

(oxo^-1) will be your representative

Which is even

Ok,Not exactly

[rigid cave]

Okay, so the reason behind the two conjugacy classes lies in the fact that you always could choose between an even and odd permutation?

[carmine fossil]

Yes

[rigid cave]

Oh okay! But how can you choose an odd permutation? Odd permutations do not belong to A_5 right?

[carmine fossil]

Let's say x is an odd permutation and let a be even,will x(1,2) be even or odd?

[rigid cave]

odd?

[carmine fossil]

(1,2) is odd

[rigid cave]

oh, then it will be even

[carmine fossil]

So you rewrite that set as {x(1,2) ( (1,2)(1,2,3,4,5)(1,2) ) (x(1,2))^-1|x is odd}

x(1,2) is even

[rigid cave]

okay I think that I get it. But where do you get your x from? From S5?

[carmine fossil]

yes

[rigid cave]

Okay I think that I get it now. You always have the choice to either choose an element in A5 to conjugate or an odd permutation in S5 which you "make" even and then conjugate?

[carmine fossil]

Yes

[rigid cave]

Okay, thank you so much!

[strong valve]

@rustic crown

can you recommend some text for galois theory?

[rustic crown]

maybe check #book-recommendations or #books-old
i've read Aluffi and found it pretty good. many people like Lang.

also heard about jacobson and emil artin galois theory

[strong valve]

Ugh a friend of mine read aluffi and didn't like it much

what's the book name by jacobson?

[rustic crown]

how did we all became numbers btw?

[chilly ocean]

epic funny april fools joke xd

[strong valve]

hihi

[chilly ocean]

"basic algebra" 1 and 2

[rustic crown]

nice

[strong valve]

are the number random?

[chilly ocean]

people with name change permissions can change their name, so long as the result is a number

e.g., my name is a special date

[carmine fossil]

What happens if the new name is not a number

[strong valve]

[chilly ocean]

some people with permissions are just some combination of 420 and 69

[chilly ocean]

it gets changed instantly

[carmine fossil]

Texit powers?

[chilly ocean]

probably

im going to change my name now

see if you can watch it get switched back

[rustic crown]

oooh

[chilly ocean]

do you see it?

[carmine fossil]

How are there people whose nicknames are not numbers?

Yes

[rustic crown]

it changes intantly

[chilly ocean]

bot slow

[strong valve]

why can't i change?

another question

why do so many people here have weeb profile pictures?

[carmine fossil]

Yes

[strong valve]

i didn;t ask a yes or no question

[carmine fossil]

It's part of internet culture

[chilly ocean]

Yes

[rigid cave]

This relly good for galthery plz read u learn veri much

[golden pasture]

what the fuck

[rigid cave]

I read this myself and me now becom profesor of veri hard math me veri recomend to u

[maiden ocean]

catthink

[latent anvil]

catwiggle

[rustic crown]

catuwu

[oblique leaf]

This is probably dumb but can anyone explain why {1,sr} for example is a subgroup of D_3?

i mean isn't (sr)^2=r^2 so {1,sr} isn't closed ...

[chilly ocean]

Why does it equal r^2?

[uneven folio]

Is r the reflection or the rotation? If it's the reflection, then r^2 = 1.

[oblique leaf]

because (sr)^2=s^2r^2=r^2?

r is the rotation and s is the reflection

wait i thought r^3=1

[chilly ocean]

(sr)^2 does not equal s^2r^2

[uneven folio]

The group isn't abelian, so you would get (sr)^2 = srsr. Not s^2 r^2.

[oblique leaf]

omg

okok

I got it

[uneven folio]

Yeah, you're right that r^3 = 1. Sorry, I was thinking "r is reflection, s is spin."

[oblique leaf]

yea i remember now, (sr)^2=1 no? cuz (sr)=(sr)^{-1} I think

tysm

[sour plume]

r is the radius and s is the sign of the permutation

[maiden ocean]

hmmm

in atiyah macdonald

it says that if A is local with maximal ideal m and residue field M then if M is finitely generated clearly M/mM is annihilated by M (i follow) so it is an A/m module, i.e a k vector space

but then it says "as such it is finite dimensional"

i dont get how this follows

[oblique river]

the images of the generators of M

generate M/mM over k

[maiden ocean]

is the fact that m is maximal here important

[oblique river]

nope

[maiden ocean]

oh

[oblique river]

for any ring A, ideal I, and module M, if S generates M over A, then the image of S in M/IM generates M/IM over A/I

[maiden ocean]

yeah if m = sum a_i x_i for x_i finite then m = \sum (a_i + I)x_i

ok this makes sense

er m + I

[oblique river]

yep

[maiden ocean]

the way he worded it was kinda wack but i get it

[oblique river]

oh wow

names are back

yeah I think the grammar is a bit awkward

[maiden ocean]

hahaha

[latent anvil]

why are names back :(

[next obsidian]

Sad

[latent anvil]

I'm rebelling

We need a :cocatKing: emote to express dissent

[maiden ocean]

:cocatKing: would be useful for me

i support this

[sour plume]

am i name

yes

[chilly ocean]

[maiden ocean]

lmfao

im tempted to make that my pfp

[chilly ocean]

hold on

[chilly ocean]

let me try doing it

but

the crown is not inverted

[latent anvil]

I love that

[maiden ocean]

that would be so good

[chilly ocean]

ok gimme a few mins

might take some careful cropping

that's the bad version

now let me be a bit more careful with the crop job

[latent anvil]

this rocks

[chilly ocean]

it doesn't look too bad when small

[maiden ocean]

this is so good

[viscid pewter]

ok i am 90% sure this proof is wrong but idk how, pls help

trying to prove that if there's a composition series for G, Ni normal in Ni+1 as standard, and also G is solvable, so basically all the composition factors are prime abelian cyclic simple etc.; then Ni is normal in all G also

so induction over Ni: 1 is always normal in G, so we have a base case

then we assume that Ni is normal in G, so there exists a homomorphism f from G to some H where ker(f) = Ni

Ni+1/Ni is prime cyclic, so it's generated by any aNi where a is not in Ni; then Ni+1 is just {a^b Ni}, where 0 =< a < p?? is that right?

then we can just define a homomorphism g from G to some K where ker(g) = 1?? or something like that

and then the kernel of the composition of f and g is Ni+1, i think? so Ni+1 is normal in G, so by induction all Ni in the composition series is normal in G??

[vestal snow]

In characteristic $p$, is there a "good" way to write $\prod_{i=0}^{p-1}(x-(b+i))$?

[cloud walrus]

ɥɔʇᴉB 'ɐuɐuɐB ɐ ǝʌɐH

[vestal snow]

Assume p is not 2

[latent anvil]

Hmm, the constant term is 0 right?

Ah I guess I see

[vestal snow]

I don't think so?

[latent anvil]

Up to reordering this is the product of (x-i)

[vestal snow]

No wait

[latent anvil]

Oh wait sorry banana

I was thinking of Fp

[vestal snow]

b is not neccessarily an element of Z_p

[latent anvil]

Yeah, my bad

Misunderstood the question you were asking

[vestal snow]

If it helps, you can assume that $b^p-b=f(a)$ where $f$ is a rational function which is defined at $a$.

[cloud walrus]

ɥɔʇᴉB 'ɐuɐuɐB ɐ ǝʌɐH

[vestal snow]

Let me just state the problem: Suppose (a,b) are points on the curve defined by $y^p-y=f(x)$ where $f$ is a rational function. The goal is to write $\prod_{i=0}^{p-1}(x-(b+i))$ as something nice like $f(x)-f(a)$ or $f(x-a)$

[cloud walrus]

ɥɔʇᴉB 'ɐuɐuɐB ɐ ǝʌɐH

[mild laurel]

is there motivation for this? Like can you do this in the elliptic case or something?

[vestal snow]

This is related to my research, but I don't know if this is true in the elliptic case or not

The guy who wrote the paper did not explain what was going on in detail

And this is what I think he did

Look at the end of page 2 and the beginning of page 3 here

[vestal snow]

Maybe you can figure out what's going on

He claims that for any $a,b\in k$, $(x-a)^v(y-b)^ug_u^{-1} dx$ (with some condition on v and u) is a basis for the holomorphic differentials. I proved this. From this, he somehow claims the set of v + u is contained in the set of orders at all unramified points

[cloud walrus]

ɥɔʇᴉB 'ɐuɐuɐB ɐ ǝʌɐH

[vestal snow]

This would be true if we somehow knew that given an element $a$ such that the place corresponding to $x-a$ is unramified in $F/k(x)$, there is an element $b$ such that the order of $y-b$ at the place corresponding to $x-a$ is 1.

[cloud walrus]

ɥɔʇᴉB 'ɐuɐuɐB ɐ ǝʌɐH

[latent anvil]

wtf

[vestal snow]

Did I stop making sense?

I might be completely wrong about this

[latent anvil]

No look at the bot

Sorry

They're doing an April fools

[mild laurel]

I'm a lil busy tonight but I'll try to take a look at this tomorrow

[vestal snow]

The idea is this. Given an $a$ ordinary, $y^p-y=\phi(a)$ is defined (where $\phi$ is the melomorphic function at the beginning of the paper). Let $b$ be a root of this polynomial. Then $b+i$ are also roots for all i in $Z_p$. If I somehow get that $\prod_{i=0}^{p-1}(y-(b+i))$ is an element of order 1 at the place corresponding to $x-a$, I can use the triangle inequality to get that exactly one of the $y-(b+i)$ has order $1$ at $x-a$ and all $y-(b+j)$ have order 0.

[cloud walrus]

ɥɔʇᴉB 'ɐuɐuɐB ɐ ǝʌɐH

[vestal snow]

You can also assume that $a$ is neither a root nor a pole of $\phi$

[cloud walrus]

ɥɔʇᴉB 'ɐuɐuɐB ɐ ǝʌɐH

[next obsidian]

You assumed that H is a p-subgroup haha

|HP| = |H||P|/|H\cap P|

this is still a power of p

you have like

p^kp^n/p^j

|H\cap P| is a subgroup of a p-group

so it's a p-group

Lagrange

|H\cap P| divides |P|

eh?

I mean you've proved Lagrange right?

that for any subgroup the order divides the order of the group

Then that gives it to you yeah?

Okay

In fact even w/o lagrange the formula

|HP| = |H||P|/|H\cap P| tells you that |H\cap P| has to be a power of p

or else it won't even be an integer on the right haha

yeet dab

[mild laurel]

[next obsidian]

Is the buncho meme up

yeah it is

[amber charm]

okay, so i am trying to partition Q_8 into its conjugacy classes, to which i have done, but my problem is that they have asked for it to be done in the notation a^ib^j, where a and b are generators of GL(2,C), any idea how this notation works and if i can transfer the solution i have already into this form?

[viscid pewter]

so it's just like there's an isomorphic representation of Q_8 in terms of some random matrices

https://en.wikipedia.org/wiki/Quaternion_group#Matrix_representations

[amber charm]

oh yeah of course....im just being dumb, cheers!

[chilly ocean]

Hello, I am working on part (ii) of this problem

So I get $uv = a^4+3a^3+10a^2+3a+9$

[cloud walrus]

ɹǝlnƎ ɐƃǝW

[chilly ocean]

I'm just having trouble writing $a^3,a^4$ as a rational-linear combination of $1,a,a^2$

[cloud walrus]

ɹǝlnƎ ɐƃǝW

[chilly ocean]

I know that since $a\in\mathcal{C}$ is a root of $f$, we can write $f(x)=(x-a)\cdot g(x)$ for $g$ is degree 2

[cloud walrus]

ɹǝlnƎ ɐƃǝW

[next obsidian]

You know that a^3 + 2a + 1 = 0

by a being a root of that

So if you expand out uv you get
a^4 + 3a^3 + 10a^2 + 3a + 9

you can actually get rid of the a^4 and 3a^3 term

[chilly ocean]

Ah I see, let me give this a shot

[next obsidian]

I'm gonna go do some other stuff, but I'll leave what I think is the correct calcualtion in a spolier

If you can't figure it out you can just un-spoiler it for yourself

[chilly ocean]

Thank you

[next obsidian]

||a(a^3 + 2a + 1) + 3(a^3 + 2a + 1) + 8a^2 - 6a + 6||

[chilly ocean]

That's amazing you did that so quickly

[next obsidian]

¯_(ツ)_/¯

I just did what you had to do in order to get rid of a^4 and a^3

then just figured out what I had to use up in order to get it

[mild laurel]

Chmonkey is galois theory god

[next obsidian]

Chmonkey is galois theory god

ftfy

[vestal snow]

yes

[next obsidian]

[vestal snow]

I will be looking forward to asking Chmonkey for help on every other AG problem I do over the summer

[next obsidian]

😌

You know the other day I looked at some chapter I exercises

and I knew how to do like half of them just from schemes stuff lol

I was like "oh this is because this is true for integral schemes"

[vestal snow]

I feel that

It's kind of weird how a lot of things that confused me less than a year ago became much easier to understand after learning abstract nonsense

[next obsidian]

[vestal snow]

Like stuff about tensoring being right exact

[next obsidian]

hahaha

that is very based tho

[vestal snow]

Oh it's just a left adjoint functor

[next obsidian]

A lot of stuff can lowkey be seen as just

oh this operation is just tensoring

so you just write it as some like triple tensor product

commute some stuff

do some stuff

tadah

[vestal snow]

Currently I'm on a big high after proving Yoneda lemma without looking at the hint in the book

I'm gonna ride this out for at least a week

[next obsidian]

Just trace the identity

Idk I don't think I ever proved Yoneda

but I did prove the 2-yoneda

so I'm gonna assume that implies normal Yoneda

haha

[vestal snow]

The what?

[next obsidian]

It's for 2-categories basically

haha

I did this for stacks class

[vestal snow]

Thanks for ruining my high chmonkey

[next obsidian]

2-categories suck btw

no

you don't get it

you have to carry around so much data

it sucky

[sturdy marsh]

yoneda works for infinity-1 cats

[next obsidian]

infinity-1

I don't even know what the thing after the hyphen means

[latent anvil]

that's just infinity lol

[vestal snow]

Are 2-categories categories?

[latent anvil]

brofib idoit confirmed

[vestal snow]

Or are they completely different?

[latent anvil]

yes, but also no

[next obsidian]

2-categories are like a category except you also have the data of maps between the morphisms

and these satisfy certain composition things

so think of natural transformations of functors

there's like a horizontal and vertical composition you might have seen

and you can talk about a diagram 2-commuting

[sturdy marsh]

it was supposed to be a ","

[next obsidian]

you can think of it like

say you had a square of functors

and they don't on the nose commute

but there's a natural isomorphism between the two

a 2-commuting diagram would be like that square along with a specific choice of that natural iso

so you carry around a lot of stupid data

and it sucks

[vestal snow]

My advisor was telling me about an undergrad who asked them to advise a project on category theory

[next obsidian]

did he say no

[sturdy marsh]

[vestal snow]

Even though they only really knew linear algebra

[next obsidian]

[vestal snow]

by they I mean the person, not my advisor

[next obsidian]

I imagine ur advisor knows category theory being an arithmetic geometer

haha

[vestal snow]

Haha yeah

He has a very intuitive way of explaining category theory

[sturdy marsh]

category theory

is

so

dry

[latent anvil]

awww

[vestal snow]

Deadass, if he knew the details Mochizuki's "proof", he could probably explain it to most people on the server

[next obsidian]

[latent anvil]

[vestal snow]

(Maybe an exaggeration)

[next obsidian]

"proof"

also most people on the server is a big statement considering like... 99.8% of it are like "I NEED CALC HELP"

xD

I think there's like... 40k members or something

[vestal snow]

All the honorable tag people

[next obsidian]

Ugh I need to finish this proof of Krull-Akizuki

time to go MIA

😭

Today has been so hard to focus

[vestal snow]

Sayonara Kamisama

[next obsidian]

[chilly ocean]

I hope you're not dissing my boy Mochizuki, y'all just wait, I'll be laughing when it is widely acknowledged that mochizukis IUTT is totally correct.

[vestal snow]

I have never read his paper so I can't say

[chilly ocean]

Me neither, but there is no way it is wrong

[vestal snow]

Bro can I have some of your copium?

[chilly ocean]

,w copium

I would give some to you if I had it, I don't know that I do

[vestal snow]

I need it for snk finale next week

[vestal snow]

Let $\phi(x)$ be a rational function over an algebraically closed field $k$. Let $a$ be an element of $k$ which is neither a root nor a pole for $\phi(x)$. Then $a$ is a root of $\phi(x)-\phi(a)$. Is it true that $a$ is a degree one root (i.e. $a$ is not a root of $\frac{\phi(x)-\phi(a)}{x-a}$)?

[cloud walrus]

ɥɔʇᴉB 'ɐuɐuɐB ɐ ǝʌɐH

[delicate bloom]

let k be char p, then $\phi(x)=x^p$ gives you $\frac{\phi(x)-\phi(a)}{x-a} = \frac{x^p-a^p}{x-a} = \frac{(x-a)^p}{x-a} = (x-a)^{p-1}$ which has $a$ as a root still

[cloud walrus]

ʎʇᴉsoɹǝW

[vestal snow]

Dang it

[summer geyser]

I need to show that this extension L/K is seperable then the result is immediate, but I'm not really sure how to show this.

[rustic crown]

L/K may not be separable

consider the simplest example K = Q, L = Q(cbrt(2)) and alpha = cbrt(2)

oh sorry

[summer geyser]

Hm then if its not seperable then I'm not sure how to go about it

[rustic crown]

i read separable, but in my head everything was normal

.<

[summer geyser]

ah

[rustic crown]

but yea, the problem doesn't go away

take L = Fp(t) and K = Fp(t^p)

then L/K is a finite extension. With n = p and alpha = t, the result still holds.

[summer geyser]

hm but the hint sort of implies that it's a seperable extension as my notes say that this norm map only applies if the extension is seperable

[rustic crown]

Ah, so there are two nice definitions for norm. One only works when the extension is separable.

the one you're using is probably, try to embed L is algebraic closure of K, and take products of all possible images of alpha

[summer geyser]

whoops wrong picture

I dont think any other definition was introduced thus far

[rustic crown]

yea so there is also this nice definition that look L/K as a vector space. And then multiplication by x is a linear map. the determinant of this map is the norm, and trace is, well the trace.

[summer geyser]

does it still behave in a similar way in the sense that N(alpha^n)=N(a), then N(alpha)^n=a^([L:K]_s) etc. ?

[rustic crown]

lol i was typing exactly that

[summer geyser]

ah :P

[rustic crown]

i'll just paste it anyway

If you know about this, then the problem is very elementary. taking b = norm_{L/K}(alpha) works! As b^n = norm_{L/K}(a) = a^[L:K].

[summer geyser]

ah so not the seperability degree but the normal degree?

[rustic crown]

yep

thinking about it... a is an element of K

what is multiplication by a?

its a*id

in terms of matrices, this is an [L:K]*[L:K] matrix

[wind parrot]

Wait are these stevenhagens notes?

[summer geyser]

hm I see but the only problem I have with this is that this definition wasn't introduced yet so I'm not sure if I can immediately use it

[wind parrot]

If so, look at exercise 31 in that chapter

[summer geyser]

and yeah it is

[wind parrot]

There is the more general definition

[summer geyser]

oh yeah I see it, I guess I can introduce it myself and apply

[wind parrot]

Also since when do these notes have an english translation

[summer geyser]

my uni made a translated copy

in any case, thanks for the help

[rustic crown]

[summer geyser]

Sorry for the ping but I have a quick question, here you chose n=p. but doesn't the question state that for any a in K and any n in positive Z, we have there exists some alpha in L such that alpha^n=a, so you can't straight up choose n=p right?

maybe misunderstanding something

[rustic crown]

it says if you have a finite field extension L/K and an integer n such that there is an element alpha with the property alpha^n = a in K.

Under these hypothesis, you need to show the existence of b

I just gave a concrete example, where the hypothesis is satisfied.

[summer geyser]

ah I see

hm I'm still trying to figure out a way to do this without this generalized norm map because all the theorems keep repeating that L/K is a seperable extension, even for excerise 31 that was suggested earlier (nvm it says any finite extension, but still should be possible without)

[rustic crown]

maybe this could help.

L/K may not be a separable extension, but you can always look at the elements in L which are separable over K. Call it M. This forms field.

you can show that the extension L/M would be purely inseparable

basically this is how you usually relate these two definitions of the norm

[summer geyser]

sorry just trying to digest this

haven't heard of purely inseperable so had to search that up

[rustic crown]

guessed so... (didn't wanna just throw up some new words)

But yea for that particular problem, its much simpler to stick to the determinant definition. (or if you wish, don't call it norm yet, just look at the determinant of the linear map given by left multiplication)

[summer geyser]

yeah I'll probably try to understand this new def

[rustic crown]

this is well defined because choosing a different basis, would just give a similar matrix, and so the determinant is still the same.

[summer geyser]

I see alright, thank you for the help again

[sturdy lotus]

Hiya 😅 , I am by no means a mathematician, but I think this message goes here; I am currently doing measure theory, and started with sigma-algebras. I understand that if you have a set X, and then a family of subsets B, then B⊆ P(X) which fulfils that 1) the empty set, phi, and X pertain to B, 2) that both C ∈ B and C complementary ∈ B are measurable, and c) that the countable union of C_i pertains to B. I am bit unsure of these properties as well as the usage of sigma-algebras; was wondering if somebody could please shed some light on this? I would be most thankful. Also, please do tell me if this is not meant to be here.

[chilly ocean]

the empty set, phi

[sturdy lotus]

sorry, missed a comma

[next obsidian]

You’ll probably get better answers in #advanced-analysis, even though the name literally has the word algebra they’re really most commonly used in measure theory which falls as a subset of analysis

Or at least that’s where it finds it’s home in most people’s mind

[chilly ocean]

Anyone know what this p.1 notation means?

F is a finite field

[mild laurel]

it's just p times 1 probably

[opal osprey]

What is a good introductory book to lie groups and lie algebras that is suitable for someone with knowledge of smooth manifolds?

[mild laurel]

I enjoyed reading Humphreys

[chilly ocean]

i've been told that this is good

havent really read it though

[carmine fossil]

I guess the gtm book is good

[opal osprey]

Yeah, I think he's prolly giving the definition of the characteristic of a field, so it must be p times 1.

[opal osprey]

Hmmmm

I'll give it a look

I hope it has some good exercises

[mild laurel]

You might enjoy looking at Hall's book too

Lie Groups, Lie Algebras and Representations

[opal osprey]

Haven't heard of this one

[carmine fossil]

Referring to this

[opal osprey]

Ah

[mild laurel]

both Humphreys and Hall are GTMs

[carmine fossil]

mb

[opal osprey]

What does GTM mean?

Sorry

[carmine fossil]

[mild laurel]

graduate texts in mathematics, it's a book series that Springer prints

[opal osprey]

AAAAAAH

OK

I know the book series

Just didn't know people called it this way

I will take a look at Hall's then

Because people have already recommended this one

[next obsidian]

Yellow book

[opal osprey]

I was first thinking of reading Serre's book on the topic

[next obsidian]

mmmmmm

just like my name

[opal osprey]

It is fairly short

[mild laurel]

oh I didn't know Serre had a book on the topic

[opal osprey]

I was taking a look at the content table

Looks nice

[wind parrot]

Serres book is based

[mild laurel]

All of Serres books are short lmao

[opal osprey]

I mean

[wind parrot]

He does Lie groups over p adic fields lmao

[opal osprey]

It is short

But the content table is like

Really dense

There's lots of deep stuff being covered in few pages

So that's one of the reasons I didn't want to read this one at first

[elder valley]

is there any test for elements of F_p^2 having square roots, similar to the legendre symbol for F_p?

[next obsidian]

All Serre books are like

1 line is actually 5

lol

[opal osprey]

I think it would be better to just read a more introductory book with a slower pace

Before reading Serre's book

[opal osprey]

Lmao

Looks like Atiyah's introduction to Commutative Algebra

It's really dense

But anyway

Thanks for the suggestions

[chilly ocean]

thank you

[mild laurel]

not sure what you mean, the legendre symbol is just a function to {-1,0,1}. You can define the same thing for F_p^2. Are you asking about an easy way to compute it?

[elder valley]

the value of the legendre symbol indicates whether the input has a square root in F_p, and so can be used to test if an element is a quadratic residue. i'm wondering if there's an analogous test for F_p^2

[mild laurel]

https://en.wikipedia.org/wiki/Quadratic_reciprocity#Imaginary_quadratic_fields

it should just be this extension I think

[elder valley]

i was reading that. doesn't number field mean you're working specifically over an extension of the rationals though?

[mild laurel]

yes?

in the same way that F_p^2 is an extension of F_p

so for any two odd primes p,q, you should be able to find an imaginary quadratic field such that both p,q remain inert in that extension

[elder valley]

i think the assumption there is that it's a number field, so it doesn't necessarily apply to finite fields

[mild laurel]

so that the quotients are F_p^2 and F_q^2 respectively

I mean, quadratic reciprocity is a statement about the integers really

[elder valley]

integers mod p, yes

i probably didn't phrase my question very well. i'm just looking for a way to test whether an element of F_p^2 has a square root that's also in F_p^2

in F_p, the test is exactly the legendre symbol computation

[mild laurel]

I think it should be the polynomials over finite fields section maybe?

if you take f to be monic, irreducible, degree 2

then that ring mod f is isomorphic to F_p^2

[elder valley]

oh good catch, that might be it. not sure why i didn't look at that section 😅

that may be exactly it, i'll look into it

it doesn't say how to compute that symbol however

followed one of the references and found this gem

[chilly ocean]

what book is that

[elder valley]

Algorithmic Number Theory: Efficient algorithms

[mild laurel]

they give a reciprocity law so I think you can just use that

[latent anvil]

This seems rather hard

[chilly ocean]

quite the trivial exercise

[next obsidian]

SMH

this person is just ripping off of Lang's famous exercise

in the first two editions of Algebra

[chilly ocean]

my algebra professor's advisor was lang. i wouldn't be surprised if he also wrote a book and did this

[small bison]

i think so cause you can view restriction of scalars as both a tensor and a hom

and then you just apply tensor hom adjunction

yeah i forget exactly how it goes but i think it's something like tensoring against A

should be the same as restriction of scalars

[opal osprey]

I think that what young smasher says makes sense

Restriction of scalars is the left adjoint functor of the extension of scalars

Which is given by some tensor product

You can prolly do this without actually knowing anything about category theory tho

[magic owl]

you know all of the maps and stuff pretty concretely

it should be a (potentially annoying) straightforward computation

[opal osprey]

Can't we just take an injective function f : M -> N where M and N are A-modules and prove that F(f) : F(M) -> F(N) is also injective?

[magic owl]

Like let 0->A->B->C->0 be exact

[opal osprey]

Where F is the restriction of scalars functor

[magic owl]

show that the image of that SES is an SES under restriction

probably im on mobile rn

sorry

I think you need to be careful because it is not clear that generators are still generators when you remove some of the scalars. For example, think of C as a complex vector space, and restrict to R. The element 1 was a generator before but doesn’t suffice over R

But i mean really you should think about what I said before

just show the inclusions of all the images and kernels

it should be follow your nose

In particular i think you should find that nothing changes.

||if f(a) = 0 then f(a) is gonna be 0 no matter what your scalars are||

Can it?

Remember that like

underneath all this structure

you still have set maps

and those don’t change

under this functor

so like, basically nothing changes except for generators lmfao

But i think it is worth working it out in detail if its confusing you

Maybe I can convince you that if the SES you care about is

0->A->B->C->0

surjectivity at the end and injectivity at the beginning won’t change for set-reasons

so the only thing that a priori might go wrong is in the middle

fwiw i am not convinced the categorical arguments above suffice

although one can sum this up with forgetful functors should be exact

[magic owl]

Can you expand on this, I can’t think of why this is true. Extension of scalars is a tensor but I think restriction is just a plain old forgetful functor

Oh, maybe it is a hom

Yeah it should be

But I don’t think its both

[small bison]

i had to look it up in my notes

but view B as an (A,B) bi-module

and it $B \otimes_B - \simeq \operatorname{Res}_\alpha$

[cloud walrus]

ɹǝɥsɐɯs_ƃunoʎ

[small bison]

where \alpha is your ring homomorphism

it's trivial in light of tensor hom adjunction

[magic owl]

pls stop

its trivial just on a basic ring level

Like

okay

Think about this

let F be the restriction functor

and f be a ring map

what is F(f)(x)?

by defn

okay

so the kernel is?

why

I think you are confusing yourself by thinking too hard

Write down, naively, what the conditions are for that exact sequence to be preserved

and then try to prove them purely on a set theoretic level

nothing

yeah

you’re not missing anything

When all you are doing is forgetting structure

you have no new structure

to worry about

[latent anvil]

Here's a question for you slim

If you're interested

Ah wait I misread your original question

I thought you were just asking if it was a functor at all

[opal osprey]

I am sorry for the bad quality picture

So, if you have a ring homomorphism between two comutative rings with unity

I called it lambda

Which goes from B to A

It induces a functor between Mod_A and Mod_B

And what it does

Is that for a given module M over A

F(M) has the same abelian group structure

As M

So it's just M, the difference is that you get a multiplication by scalars given by * : B × M -> M where (a,v) gets mapped to λ(a) v

So the underling sets are the same

F(M) and M are the same sets

Now

If you have a module homomorphism f : M -> M' between two A-modules

You naturally get that this same function f

Is also a homomorphism between B-modules

Since if you have a scalar b in B and a v in M, f(b v) = f(λ(b) v), but since f is a homomorphism of A-modules f(λ(b) v) = λ(b) f(v) = b f(v)

So this restriction of scalars

Basically does nothing to the sets or the homomorphism of A-modules

So if you have that for example, f : M - M' is an injective homomorphism of A-modules

F(f) : F(M) -> F(M') is also trivially going to be an injetive homomorphism of B-modules

Because since F(f) and f are the same as functions

f being injective implies F(f) is also going to be injective

And so on

With this

You naturally see that if you have a short exact sequence of A-modules

This functor is naturally going to give you a short exact sequence of B-modules

Because since F(f) and f are the same as functions

F(g) and g are also the same as functions

And so on

Therefore, Images and kernels are going to be the same

So you just get another short exact sequence

But now of B-modules

Sorry for just giving a long ass explanation

But this is basically the reason why the restriction of scalars induced by a homomorphism of rings is an exact functor

I thought it would be a harder question, but it's all about how you construct this functor from a given ring homomorphism.

And the result naturally follows

[cursive temple]

In fraleighs abstract algebra, im supposed to prove that ${1, y, \ldots, y^{p-1}}$ is a basis of $\bZ_p[y]$ over $\bZ_p[y^p]$, where $p$ is a prime and $y$ is an indeterminate.

[cloud walrus]

ɐddɐԀ

[cursive temple]

I cant see how this isnt obvious, as the maximal degree of a polynomial over $\bZ_p$ is $p-1$, and so i feel like we dont even need the entirety of $\bZ_p[y^p]$, but the constants suffice.

[cloud walrus]

ɐddɐԀ

[cursive temple]

im probably missing something gigaobvious

[mild laurel]

Is Z_p Z/pZ here

@cursive temple

[cursive temple]

yes

[mild laurel]

The maximal degree of a polynomial over Z_p isn't p-1

Two polynomials are equal if their coefficients are all equal

Not if they take on the same values at all elements

x^p and x are two different polynomials

[cursive temple]

why is that distinction made

[scarlet estuary]

because we're not actually evaluating the polynomials

when we talk about polynomial rings

we're talking about the polynomial itself, not its evaluation function

analogy: $\sum_{n=1}^{\infty}\frac{1}{2^n}$ and $\sum_{n=1}^{\infty}\frac{(-1)^{n+1} + 1}{\sqrt{2} + 1}$ both converge to $1$

[cloud walrus]

uoʇƃuᴉɯɐN

[scarlet estuary]

but its fair to call them "different series", right?

[cursive temple]

sure

[scarlet estuary]

from an algebraic perspective, we often care about the abstract behaviour of polynomials rather than what happens when we evaluate them

oops typo

the denominator of the second sum should be raised to the n

$\sum_{n=1}^{\infty}\frac{(-1)^{n+1} + 1}{(\sqrt{2}+1)^n} = 1$

[cloud walrus]

uoʇƃuᴉɯɐN

[mild laurel]

maybe one reason is that when you consider these as functions on some field extension of F_p, like F_p^2, you'll get different functions in that case

so you don't necessarily want to equate the polynomials just because they agree on F_p, because you might extend the domain

and I mean, there's no distinction here really. This is just how the definition of R[x] for any ring R works

it's just that, over the real numbers, two polynomials in R[x] are equal if and only if they take the same values everywhere, so there's no distinction for some rings like the real or complex numbers

[cursive temple]

interesting

[mild laurel]

There are probably some algebraic geometry things to say here too

[cursive temple]

its always cool when your intuition gets challenged

[mild laurel]

It's just a weird consequence of considering polynomials over finite things

This is also one of the reasons classical algebraic geometry only works over algebraically closed fields, so you don't run into these issues

[cursive temple]

can you expand on what issues you run into

[mild laurel]

This isn't rlly my specialty but it's basically what we've talked about

In algebraic geometry, you try to study spaces by looking at the functions on them

But on Z_p like we noticed, the functions on Z_p aren't Z_p[x] cause some of these give us the same functions on Z_p

The study of schemes is one way you can fix these issues

[cursive temple]

Hmm

Sounds v cool

I hope to get to shafarevic at some point

As a sidenote, the problem is super simple now that i got the defns sorted out

[warm flame]

can someone help me out with proving the order of the dicylic group to be 4n?
given that its generated by $a=e^{\frac{i\pi}{n}}$

[cloud walrus]

˙ʎɹɹɐq

˙ʎɹɹɐq

[warm flame]

now just need to apply those ideas rigorously to prove the order

[cloud walrus]

˙ʎɹɹɐq

[carmine fossil]

All elements in the dicyclic group are either of the form a^k or a^k j for some k in {1,2...2n}

Assuming you know DC_n=(j,a|a^(2n)=1, aj=ja^-1 , a^n=j^2)

[steady axle]

I know how to find roots of matrices over $\mathbb C$ if they are diagonalizable(thats the trick). But is there an easy way to tell if a real matrix has root (which has to be a real matrix) when you know that the real matrix is diagonalizable over reals but dont want to diagonalize it?

[cloud walrus]

ʇɹǝq

[steady axle]

The specific question I had in mind was does there exist $B\in M_2(\mathbb R)$ such that $B^3=A$ where $$ A = \begin{pmatrix} 19 & 2019 \ 2019 & 1 \end{pmatrix}$$

[cloud walrus]

ʇɹǝq

[thorn delta]

Since A is diagonalizable, you should be able to just compute the cubed roots of the eigenvalues. cube roots of reals always exist, so the cube root of A exists

[carmine fossil]

This might be useful

[thorn delta]

is what i said incorrect?

[carmine fossil]

,w factorize (x-19)(x-1)-2019*2019

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by DrunkenDrake#1521

[carmine fossil]

Ok,Wa is drunk

But,yea that works here

[thorn delta]

nvm about what I said. (possibly?) Misremembered theorems about symmetric matrices.

[steady axle]

I think what you said is correct. I somehow had square roots in mind and was worried what if one of the diagonal entry is negative.

[thorn delta]

alright, cool. the thing i was hung up on was whether real symmetric => real diagonalizable or only complex diagonalizable. I'm pretty sure real diagonalizability holds, but I couldn't quickly find a reference for that.

[sturdy marsh]

yeah it's real diagonalizable

[terse crystal]

Real symmetric matrixes always have real eigenvalues

And all normal matrixes are diagonalizable where normal means that

AA^h =A^hA where A^h_ij is the conjugate complex number of A_ji

[sturdy marsh]

If you're invoking the usual spectral thm for normal operators, there is more to prove

I assume by real diagonalizability, kxrider wants the diagonal matrix to be conjugate to the symmetric matrix over the reals

but this is not hard to prove if you know RCF

[terse crystal]

As a matter of fact any square matrix A can be factorized as A=UDU^h where U is an unitary matrix(U^h is the inverse of U), D is an upper triangular matrix with eigenvalues of A sitting at the diagonal. A is diagonalizable if and only if D is diagonalizable if and only if DD^h=D^hD if and only if A is normal.

If A is a real matrix such that all eigenvalues are real (for example real symmetric matrixes)you can discuss this in R (A=PDP^t where P is an orthogonal matrix)

[chilly ocean]

Hello, I was wondering if someone can help me out on a problem. I'm on part (iii) of this problem.

I get that $\frac{u}{v} = \frac{a^2+1}{a^2+3a+9}$

[cloud walrus]

ɹǝlnƎ ɐƃǝW

[chilly ocean]

I do get that fact that since $a$ is a root in the complexes, then $f(a) = a^3+2a+1=0$, but not sure how to use this in $u/v$

[cloud walrus]

ɹǝlnƎ ɐƃǝW

[delicate bloom]

do you know the euclidean algorithm?

[chilly ocean]

Yes

[delicate bloom]

then you know that there are polynomials of the form A(x) and B(x) such that A(x)*v(x) + B(x)f(x) = 1

so you can then use that to get A(x) because when x=a you have f(a)=0 and so you have A(a)v(a)=1

that makes A(a) the inverse of v

[chilly ocean]

Where are A(x) and B(x) coming from?

[delicate bloom]

from the euclidean algorithm

[terse crystal]

You can just assume that v^-1=b_0 + b_1 *a +b_2 *a^2 which multiplied by v equals to 1 then calculate those b using a^3=-1-2a

[rich pecan]

can i ask a quick question? or should i wait a bit

[chilly ocean]

Go ahead and let me think through this @rich pecan

Thank you for asking

[rich pecan]

https://math.stackexchange.com/questions/413813/why-mathbbrx-x21-cong-mathbbc

i'm not sure how to modify the argument when the polynomial has more than 1 complex root

like R[X]/(x^2+2x+1), for example

which i think should still be isomorphic to C?

[mild laurel]

No it won't, that's not a field because that polynomial isn't irreducible

[rich pecan]

oh wait right i'm stupid

x^2-x+1

haha

[mild laurel]

Yeah it'd still be true, I'm not really sure what you're asking

x² + 1 also has more than one complex root

[rich pecan]

i'll think about it some more, thanks though!

[vestal snow]

@mild laurel Were you able to look at the problem from yesterday by any chance?

[mild laurel]

Yeah I thought about it a lil but don't really have any ideas

You might want to try math overflow or something honestly

[vestal snow]

Thanks, I'll do that

[weak plinth]

When will I be unmuted

[chilly ocean]

Topkek

If you swear on your soul that you're not a troll, I could relay to to mods

[rigid cave]

Can someone check if I am insane? If G is a group with subgroups G_i and H then G_i intersect H is a subgroup of G_i, right?

[chilly ocean]

Totally

[rigid cave]

Okay, thank you so much!

[opal osprey]

Guys

I have a little question

This first image is just to give a bit of context

But my question concerns the last theorem

Can I make it more general?

Like, if M is a faithful A-module, N is an A-module and u,u' are elements of N such that there exists v an element of M where u tensor v = u' tensor v. Then with these conditions, we have that u=u'

Idk if this is true

But I can't find any counterexamples

If I could prove that in these conditions

The module generated by v, Av, is also a faithful module

Then it would easily follow from this theorem I have already proved

But I don't really know in general if a submodule of a faithful module is also faithful

This is prolly false, but I can't prove nor find a counterexample

So I would like to ask for a counterexample of a submodule of a faithful module that is not faithful, it is false, or for a proof that a submodule of a faithful module is also faithful, if it is correct.

[vestal snow]

I can't think of a counterexample, but I don't see how this can be generalized easily

Just to make sure I'm getting this right, you're saying that if M is faithful and m (x) n = m' (x) n for some n in N, then m = m' right?

[opal osprey]

Yup

[vestal snow]

Let me write this out

[opal osprey]

Where m(x) n is actually m(x) tensor n

[vestal snow]

Its been a while since I've worked with tensors

[opal osprey]

That's what a faithful functor is in general

[vestal snow]

I'm using (x) to mean the tensor symbol

[opal osprey]

F : C -> D is a faithful functor between categories iff for every morphism f : a -> b where a and b are objects of C, F(f)=F(g) => f=g

What the tensor product does to a morphism as a functor is taking a morphism f and tensoring it with the identity

[opal osprey]

Oooh

Ok

If N is a generator of M, then M faithful => N faithful as well

That's the best I could do

[vestal snow]

Try Z_10 (x) Z_6

as Z modules

[opal osprey]

Hmmm

Ok

I will think a little bit about this case

[vestal snow]

Ah shucks neither of them are faithful though

[opal osprey]

Yeah

That's what I was about to ask

[vestal snow]

I definitely don't think that this is true

You are using the fact that (u-u') (x) v = 0 for all v

And unless you have some additional hypothesis on the generators of N, I doubt it would work out nicely

[opal osprey]

Every free module over a ring A is faithfully flat

Oh btw

All my rings are commutative and with unity

So I am trying to find some examples of nice free modules

I mean

If N = 0, ofc that 0 is not faithful

So let's suppose N≠0

[terse crystal]

Am I correct

[opal osprey]

Ok

Let me try to grasp this

A=Z+Z is a faithful A-module, because A is a free A module and every free A-module is faithful.

So I got the first line

e=(1,0) and t=(0,1) are elements of A

Now we want to prove that the A-submodule generated by e is not faithful

So you consider the A-submodule generated by t

and consider eA tensor tA

This is going to be zero, because for for every z in eA and for every z' in tA

We have that there exists an x in A and a y in A such that z=ex and z'=ty

So we have that

z tensor z' = ex tensor ey

But since e=e*e

And you are taking the tensor product with respect to A

e is a scalar

So ex tensor ty = e²x tensor ty = e(ex) tensor ty = e (ex tensor ty), but since e is a scalar, e (ex tensor ty) = (ex tensor e(ty)) = (ex tensor (et)y), but et = (1,0)×(0,1) =(1×0,0×1) = (0,0).

In the end, we have that ex tensor ty = ex tensor 0×y = ex tensor 0 = 0.

So indeed eA tensor tA = 0, but tA ≠0, therefore, eA can't be a faithful A-module.

Sorry

I have to go through all the steps to make sure I understand

And it seems correct

[terse crystal]

Okay, thanks

[opal osprey]

I am really bad at thinking about examples damn

Nah

I think it should be the other way around lmao

Thanks

[terse crystal]

Me too actually,I even skipped examples while reading sometimes 😂

[delicate bloom]

is this polynomial irreducible for all $n$? $$f_n(x) = \frac{(-x-1)^n + x^n + 1}{(x^2+x)^{(n \mod 2)} (x^2+x+1)^{(2n \mod 3)}}$$

[cloud walrus]

Merosity

[next obsidian]

Instead of looking for a specific example that's more complicated, simply see that if you take v = 0 then this is just vacuously true

that u (x) v = u' (x) v

so u = u' for all u,u' in M

[vestal snow]

Every module has exactly one element. Checkmate algebraists

[next obsidian]

If faithful

[vestal snow]

Can you then use that every free module is faithful to conclude that every module is the 0 module?

[next obsidian]

🧠

[vestal snow]

I solved algebra guys

Where's my fields medal?

[delicate bloom]

I'll give it to you once you answer my question @vestal snow

[vestal snow]

Consider the module of polynomials with coefficients in Q. This module is 0 by the previous result. Thus, your polynomial is the 0 polynomial

[latent anvil]

wow Mero this looks like shit

You're welcome :+1:

[delicate bloom]

Thanks and GFY 👍 😭

[weak plinth]

<@&268886789983436800> unmute me

Thank you In advance

[vestal snow]

Why are you muted?

[scarlet estuary]

okay no longer muted

[vestal snow]

Why was he muted?

[scarlet estuary]

trolling

and evidently not intending to reform.

a bit more context: they went on a long rant against other users and were clearly being intentionally inflammatory

[delicate bloom]

weird, how was he able to talk if he was muted

[scarlet estuary]

i wouldve unmuted tomorrow (or if they DMed probably) but considering instead they exploited a loophole with the role system in order to publicly mod ping

twice

nah

[scarlet estuary]

its how roles in discord work

advanced role has to let you talk in advanced channels regardless of permissions

coulda just deleted advanced role but

i forget

[delicate bloom]

ah I see

[scarlet estuary]

its a dumb override system

"can talk here" takes precedence over "cant talk anywhere"

[golden pasture]

yea afaik you need to set cant talk here for such overwrites?

i remember having to jus manually set every category

for a muted role to work properly

[covert vector]

one clunky patch is to make it so muted people can't see the advanced channels

it "works" but feels strange

can't post in a channel you can't see

[carmine fossil]

So muted is like a forced ,iam studying?

[opal osprey]

How does u tensor v = u' tensor v for v=0 implies u=u' for all u,u' in N?

Considering v is in M and M is faithful

Also

I had to rewrite it

Because u,u' supposed to be in M and not the other way around

Just like the previous weaker theorem

So the question would be let M be a faithful A-module, and u,u' be elements of M such that there exists v≠0 for which u tensor v = u' tensor v, is it true that u=u'?

Hm

So like

[opal osprey]

One of the questions in the homework sheet was to find conditions such that this would be true.

I think that looking for faithful modules is just a waste of time

Because I can only prove something weaker

But I have another idea

Is this correct?

I used the fact that if A is a local ring and M,N are finitely generated A-modules, then M tensor N = 0 implies M=0 or N=0

Which we have already proved

[terse crystal]

Correct

[left sand]

hey about the last question i asked about basis of fractions

how would u show that if I have 2 poly over a field $F$ then write $\frac f g = q+r$, then split r into a lin combo of $t^i/p(t)^j$ where $p(t)$ is irreducible and $i < j$, that the ${t^i/p(t)^j}$ is independent?

[chilly ocean]

assume it isn't

[left sand]

yes, and there are coefficient nonzero

so that it evaluates to 0, but how do i reach conclusion

[cloud walrus]

Anticipation

[terse crystal]

i is smaller than the degree of p not j

[left sand]

ah right

yea but that was jsut typo, doesnt help w/ the problem

[terse crystal]

Writing now

[terse crystal]

New to latex ,I just send photos for now

[left sand]

ah that fine, what is Qk?

[terse crystal]

Wait at the end of the second row , the sum is zero

[terse crystal]

The row below

[left sand]

oh i mean why do you write sum fk Qk, what does that come from?

[terse crystal]

Σf_k Q_k =0

[left sand]

ohh

smart, thanks alot!

[terse crystal]

NP

[terse crystal]

Actually I made a mistake...

The degree of f_k isn’t necessarily smaller than the degree of P_k, but smaller than the degree of P_k^r_k
So replace all P_k in the last three rows with P_k^r_k then it’s correct.

[left sand]

but doesnt it still work? since that also mean p_k divides it, and clearly p_k doesnt divide LHS otherwise f_k(t)/p_k(t)^r_k would be reducible

[terse crystal]

Too many mistakes so I rewrote it...

[terse crystal]

Oh... yeah so never mind😂

[left sand]

nice thx

[opal osprey]

Suppose that I know that for a given module $M$ over a commutative ring $A$, $\dfrac{M}{mM} \neq 0$, for every $m \in \text{Spec}_{m}(A)$, i.e for every maximal ideal the quotient, $\dfrac{M}{mM}$ is non-zero. Can I conclude from this that $M \neq 0$?

[cloud walrus]

MisterSystem

[opal osprey]

Idk if I can just take the contrapositive of Nakayama's lemma

Because here M is not necessarily finetely generated

[oblique river]

sorry but... what?

If any quotient of M is nonzero

then M is also necessarily nonzero

as M surjects onto M/mM

did you accidentally switch "zero" and "nonzero"?

[opal osprey]

Yeah

Exactly

I thought of this for a few more seconds and yeah

It can't be that M=0

Because then mM=0 and therefore M/mM = 0 for every maximal ideal m

The question was really dumb lol

[oblique river]

or even "for any maximal ideal m"

just knowing that M/mM \neq 0 for a single m

is enough to conclude that M is nonzero

[opal osprey]

It's just that I am starting to learn commuative algebra, and since I have seen that Nakayama's lemma has a similar flavor to what I was trying to prove

I thought about something more complicated

Than what is really happening lol

I am still getting used to the concepts

So this kind of thing happens lmao

Thank you anyway

[oblique river]

start simple :)

[opal osprey]

Ok

Now that you have said it

My question was badly formulated

I really meant the other way around

I have that $M$ is an $A$ module over a commuative ring with unity, and $\dfrac{M}{mM} = 0$, for every $m \in \text{Spec}_{m} A$.

[cloud walrus]

MisterSystem

[opal osprey]

Can I conclude that M=0?

[golden pasture]

probably by some form of nakayama yes

[oblique river]

no

you can't

[golden pasture]

oh

[oblique river]

unless M is finitely generated

[opal osprey]

But M is not finitely generated

:/

[oblique river]

for example, M = Q and A = Z

[golden pasture]

oh right nakayama assumed fg rite

[opal osprey]

Is this a counterexample?

[golden pasture]

ah right

yea

[oblique river]

you tell me :)

what are the maximal ideals of Z

and what happens when you look at mQ

for those maximal ideals

[opal osprey]

Spec Z = mSpec Z

[oblique river]

well that's not true but

[opal osprey]

So mSpec Z = {pZ, where p is prime}

[oblique river]

Spec Z includes {0}

[opal osprey]

Oh yeah

You are right

[oblique river]

but yes, the maximal ideals of Z are just the ideals generated by the primes

now, does pQ = Q?

[opal osprey]

I forgot {0}

So

Hmm

Yeah, take any rational number

a/b

Where a and b are integers with b≠0

Therefore

a can be written as p^n*q where p is prime, q is an integer and n is natural

so a/b = (p^n*q)/b = p^n(q/b)

And a/b is in pQ for some prime

[oblique river]

for some prime?

(for all primes)

all of this is to say that, yes, Q/pQ = 0

[opal osprey]

Hmmm

Nice

Yeah

I still need to get some intuition with this

Because it's hard for me to come with examples

[oblique river]

examples come with practice

i know a lot of examples just because i've worked with them a lot