groups-rings-fields

[formal ermine]

gave us exercises

[dusty verge]

Probably because they're learning category theory in high school

[formal ermine]

and left

[rotund aurora]

No, I don't do a lot of AG stuff. I am learning about elementary algebraic geometry, but I try to keep it classical in a sense. So no schemes and sheafs

[formal ermine]

I could've just gone home during those two hours

but no, I spent them at school

doing nothing

[void cosmos]

what kind of problems

[lethal dune]

life

[void cosmos]

yo there was this cool ass problem

but i cant remember it exactly

but it was something of

the probability that an element is in Z(G) is 2/3 or something like that

given G IS non abelian finite

[lethal dune]

example of a module with no ass

[void cosmos]

😠

[lethal dune]

you can refine to to say at most 5/8 (if i remember correctly

given in Gallian

[void cosmos]

it was in gaillan?

wow

[lethal dune]

serious question btw

[void cosmos]

whats an ass

[lethal dune]

associated primes

to lazy to define here

[void cosmos]

yea idk about those

[chilly ocean]

Find a three-digit number "abc", where "a" is the first digit, "b" is the second digit, and "c" is the third digit, such that the second digit subtracted from the sum of the first and third digits, added to the sum of the second digit and the first digit subtracted from the last digit, results in (a+b+c)/2 that is similar to a prime number to the power of three.

If possible, find all three-digit numbers where (a+b+c)/2 equals a square number, and add them together

[void cosmos]

what does similar mean

ig u should expand it out

ie write "abc" as 100a+10b+c

and then u can form ur system of equatiosn

@chilly ocean

[frigid lark]

You seem to be projecting something here

[karmic moat]

very smart, here is my new attempt

not at pc cant tex :(

(WTS = want to show)

[dim widget]

I think that's a good strategy, it's important to learn what algebraic geometry actually is, then you'll appreciate the Grothendieckian stuff more imo.

[frigid lark]

Sooo, in algebraic geometry, is 57 a prime?

[south patrol]

Uh

Nice joke lol

[dusty verge]

Ends in seven is a prime

[tranquil parcel]

How big of a field of study is Lattice theory, generally? Lattices here being posets with pairwise supremums and infemums, not just the vector space lattices.

Because sometimes I go to look stuff up while reading from Gratzer's lattice theory text, and I feel I just get vector space stuff most often, or very sparse results from my searches.

[sonic coral]

77 goes brrrrr

[void cosmos]

would dualizing free modules

call them co-free modules

would co-free modules be injective?

[barren sierra]

So I know how to do this directly

However I know that if I can show that Q is injective, then by Baer's criterion, the result also follows

but how would I show that Q is injective?

[void cosmos]

Q is injective iff this

they are equivalent

[barren sierra]

no I know

[void cosmos]

so u start with a set

[barren sierra]

point is I want to know how to just show it's injective in a different way

[void cosmos]

the set of all homomorphisms {f | f:C-->Q}

and order this set with inclusion of domains

like f<g iff dom f subset dom g

[barren sierra]

sure

[void cosmos]

then u get some maximal element by zorn

this maximal element must be the homo f:B-->Q

if it were not

say its a homo f:D-->Q

B is the module ur trying to go from to in the injectivity diagram

pick any element in B not in D call it b

and let I be the ideal { i in R | ib is in D}

this is a left ideal

i cant remember much of the details but then u construct some map

from this left ideal to the module

u then use ur assumption

and define another map from this map that has bigger domain ( u do this by using the element that is in B not in H, i would guess it would be defining a map from H+Rb to Q)

so contradicts the maximliaty of the map "wrt to our order"

where Rb = {rb | r in R}

this is essentially proving the -> partr

the other part is easy you can do it on your own

just consider the diagram with inclusion

hence they are equivalent

now ur proof would be to show that any sequence Q-->A-->B--> 0is split exact

so Q is injective

and by what we said above

or not by what we said above by the other way

Q attains this property

thats it

[white oxide]

does F(1) = F imply F[x] = F? here F(a) denotes phi_a[F[x]], the image of F[x] under the evaluation homomorphism by a

bc you're just plugging in one into every single polynomial with coefficients in F, and my thought process was that you could only get F if you had F[x] = F

[summer path]

F[x] is a polynomial ring, F is a field

so no F[x] is not equal to F

if you evaluate x at k in F, then of course you have F(k) = F

[white oxide]

oh yeah my bad

thanks

[woeful sage]

Woah then F[x] would be like a field valued function

[hot lake]

F(1) here is a very awful and unfortunate notation to denote the F-subvector space of E spanned by 1

It does not mean evaluation in any way, shape or form

[solar shore]

how do i go about proving that $A_4$ is the only subgroup of $S_4$ with order 12?

[cloud walrus]

blanket

[lethal dune]

It involves few steps

First show any index 2 subgroup of S 4 must contain all 3-cycle

[solar shore]

im osrry i was doing something

why do we need to show this??

[frigid lark]

A_4 is generated by 3 cycles

[crude sail]

would someone mind explaining the last sentence to me? i don't understand how $eR'$ and $e'R'$ are isomorphic to $F_{11}$

[cloud walrus]

jonny

[crude sail]

the only ways i could think of checking are actually computing the elements of $eR'$ and $e'R'$ which is really unwieldy, or by finding isomorphisms from $F_{11}$ to those rings, which i don't know how to do. i know there's maps from $F_{11}[x]$ to $F_{11}[\delta]$ and from $F_{11}[\delta]$ to $eR'$ and $e'R'$ (the canonical map from a ring to a quotient, and then the multiplications by idempotent elements are surjective maps too) but i don't know how to use this, or if i even can

[cloud walrus]

jonny

[crude sail]

feels like im overthinking tbh

[hot lake]

There is only one ring with 11 elements

[crude sail]

shit really

ok yeah definitely overthought it, i get it now

cyclic groups of prime order right

i always forget that the group structure still matters

[hot lake]

Here you could check that x -> ex is an isomorphism from F11 to eR'

[crude sail]

i think ill skip that and go with the first fact you told me

thanks for the help i feel dumb for missing that

[sharp sonnet]

~~there is also the ring (without unity) where a*b = 0 for all a, b 🤓 ~~

[sly storm]

Please use the same straightedge and compass that you always use in Euclidean geometry.

[dim widget]

Google is your friend https://scholar.rose-hulman.edu/cgi/viewcontent.cgi?article=1052&context=rhumj

[frigid lark]

Based Lang

[elder wave]

but is this simple and intuitive

[delicate orchid]

:true...:

[dim widget]

It is at least simple, whether or not it is intuitive is arguably a skill issue

[delicate orchid]

brutal....

[sly rain]

If I want to show that $Z/nm \cong Z/m \times Z/n$ if $m$ ,$n$ are coprime, is this a valid way?

Find a mapping $\phi: Z/nm \mapsto Z/m \times Z/n$, show that its well defined and a homomorphism (in general), show that its injective and both sets have the same order (in this case), thus its surjective and bijective.

Most likely there is a nicer way, but I have to review what we have done about cyclic groups first I think.

[cloud walrus]

aabb

[delicate orchid]

this is a perfectly fine way of doing this

[sly rain]

Great, thanks for the reassurance

[karmic moat]

i am curious, is there a nicer way to do that?

[delicate orchid]

nicer? probs not

there are other ways of doing it, for example showing that Z/nm is the internal direct product of Z/m and Z/n

[south patrol]

Well you can prove that it is surjective and injective directly

[delicate orchid]

actually you can invoke a lattice based argument for this which makes me happy

[south patrol]

Nerd

[delicate orchid]

they're just lil dudes with the dots and lines

[karmic moat]

i would like to see lattice based argument :)

(i have no idea what a lattice is other than something which plants can climb on)

[delicate orchid]

it's basically this but rephrased slightly

[south patrol]

Isn't that a trellis

[delicate orchid]

france...

[south patrol]

Or are both intervhangeable

Idk

[karmic moat]

oh wait yeah trellis

oops

[delicate orchid]

decided to check my notes before spouting shite:
two subgroups H_1, H_2 form an inner direct product if their meet is trivial, their join is the entire group, and one is contained within the centraliser of the other

which is just a rephrasing of the usual definition but I find it NIFTY

[sly storm]

Thanks for the link.

[novel parrot]

"or else, that if an irreducible element of R
divides c(f g), then it divides c(f ) or c(g), which is what we will do."

How does that prove that c(f_0g_0) = 1?

Is it because

[dim widget]

Lol this book again

[novel parrot]

sadly

[dim widget]

Yes this is another plausible looking but wrong claim (at least without more context)

[novel parrot]

What should it be?

I know that if an irreducible divides c(f_0g_0) then c(f_0g_0) cant be equal to 1

[dim widget]

Yes but they are not showing that irreducibles do not divide c(f_0g_0)

let's phrase what they're saying in terms of integers

[novel parrot]

They show it below sorry

i didnt include it

Because i dont understand the logical statement that they are using to prove that c(f_0g_0) = 1

[dim widget]

It sounds like from what you posted

that they are saying something like

n = abc,

if p|n with p prime then either p |a or p|b, thus c = +/-1

But this is clearly false.

[novel parrot]

What if that was true for all primes p

then its true?

[dim widget]

Even if it was true for all primes p

consider n = 12 a =2 b = 3 c = 2

If they showed this for all powers of irreducible elements then that's fine.

[novel parrot]

heres the whole thing for more context

i basically dont understand why it implies that c(f_0g_0) = 1

[dim widget]

Yes it does not.

[novel parrot]

rip

[dim widget]

Unless the content of a polynomial can't be squarefree or something?

But isn't the content just the gcd of the coefficients or something?

[novel parrot]

yes

[dim widget]

Yeah this (as far as I can tell) doesn't work because all they have proved is that c(fg) and c(f)*c(g) have the same prime factors

but two numbers can, shocker, have the same prime factors but not be equal.

On the other hand the key idea of how you would actually prove this is in this proof amusingly

[ebon pine]

I don't follow the last statement here, I feel I'm missing something very simple

[cloud walrus]

numbpy

[novel parrot]

Do you know an alternate correct proof for this?

[dim widget]

So the correct proof is that $c(f_0g_0) = c(f_0)c(g_0)$, and the same proof works but it actually gives you what you want.

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

Because knowing that $c(f_0g_0) $ has only prime factors which divide $c(f_0)c(g_0)$ actually tells you that it is trivial

[cloud walrus]

Topos_Theory_E-Girl

[novel parrot]

So i need to prove if f_0 and g_0 are primitive then f_0g_0 is primitive also to complete this proof

right?

[dim widget]

Yes but actually if you just substitute f = f_0 and g = g_0 in the proof they just wrote it already shows this

So they weirdly managed (as they seem to do a lot in these lecture notes) to give a proof which contains all the key ideas of an actual proof but which makes a very simple mistake about divisibility of numbers.

Like a lot of this feels like someone half remembered the right ideas, wrote something down, and didn't spend any time checking if it made any sense.

[novel parrot]

Professor most likely did exactly this lol

[dim widget]

Anyway you are a hero for having the stamina to find all the mistakes in these notes.

We stan a detail-oriented king

and/or queen and/or themperor

[novel parrot]

Thanks lol

Grinding through the lecture notes

And problem sheets

[dim widget]

If $\sigma$ is in N and $(a b c)\sigma(a b c)^{-1}$ then $\sigma^{-1}$ is in N and $\sigma^{-1}(a b c)\sigma(a b c)^{-1}$

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

is also in N

Because N is a subgroup.

[ebon pine]

So just the closure property

[novel parrot]

c(f) is a unit, maybe not 1?

[dim widget]

Yes under inverses and multiplication

[ebon pine]

Thanks

[dim widget]

How you define c(f) when f is primitive is pretty arbitrary since it's really only well-defined as an ideal.

But you could just define it to be 1 wolog

Or any other unit

If you want content to be multiplicative you have to define it to be 1

[novel parrot]

So if the content is a unit, the definiton is to just let it be 1?

[dim widget]

Yeah, again a better definition is to say the content is not a number but an ideal

[novel parrot]

oh ok

Usually we have this equality this f = c(f)f_0 where f_0 is primitive

[dim widget]

yeah I mean you can say that "there is a generator for c(f)" such that that is true

It will be true for any generator

Or just say "okay everything we write about content being multiplicative etc. is only true up to units"

[novel parrot]

Hmm ok

[dim widget]

Just because the gcd is only well-defined up to units etc.

[novel parrot]

So we call a polynomial primitive if its content is a unit, not neccsarily 1

right

Generalise a bit

[dusty verge]

Is every element of a ring uniquely identified by the ideals it belongs to?

The fact that it's true for integers is obvious, but I'm wondering if it extends

[dim widget]

It is not true for integers.

[dusty verge]

Up to additive inverses then

forgot negative numbers existed for a second there

[chilly radish]

If you want to generalise this then it'd probably be up to units

[dim widget]

But even then think of Z_{(p)}

Which only has 2 prime ideals.

But countably many elements.

[dusty verge]

What's Z_{(p)}?

It's not Z/pZ, is it?

[dim widget]

All fractions with denominators not divisible by p.

[dusty verge]

Gotcha

[dusty verge]

Yeah that makes sense

I'm trying to unpack your counterexample, and figure out what the ideal structure is and so on

[dim widget]

Okay let me know if you have questions.

It's a good exercise to do so.

[dusty verge]

Are all the ideals prime? Does it have infinitely many units?

[dim widget]

I think you can work out the second question for yourself

[dusty verge]

Wait, let x and y be in the same ideal every time

[dim widget]

Ah I did not see that you said ideals and not prime ideals

[dusty verge]

Then $y \in (x)$, and $x \in (y)$, meaning $x = \alpha y$, and $y = \beta x$. In particular, $x = \alpha \beta x$, and so $\alpha$ and $\beta$ are units

[cloud walrus]

Halliday

[dusty verge]

Differing up to units should be true by the above

[dim widget]

Yes I completely misunderstood what you were asking.

[dusty verge]

I think I would need a pen and paper to work out the ideals, and I'm not near one

[jaunty glacier]

How would you show that the set of all symmetries of a figure is always a group?

[dusty verge]

Define a symmetry

[jaunty glacier]

A function f that when you apply to a figure C, f(C)=C.

at least is what is written in my notes

[dim widget]

A figure is like some kind of shape like a polygon or polyhedron right?

[jaunty glacier]

Yea !

[dim widget]

So f(C) has to take edges to edges, vertices to vertices, etc.

[dusty verge]

So given that, what's f(g(c))?

[jaunty glacier]

Yea like if you rotate a square by 90 degrees, it is the same figure

[jaunty glacier]

g(c) ?

[delicate orchid]

if g is another symmetry of the figure C you can go further
(we're showing the set is closed)

[dusty verge]

let me rephrase, let f and g be symmetries of C

Then is f(g(C)) also a symmetry?

[jaunty glacier]

It is, since f(g(C))=f(c)=C

[dim widget]

yarrr

[jaunty glacier]

But given that, don't I have to show that the composition law is associative for my group?

Or is it by definition

[delicate orchid]

function composition is always associative for any functions, given the composition is actually defined of course

[dusty verge]

damn your edit

[delicate orchid]

see if you can prove that because it's just as easy as doing it generally as it is to just do it for this specific case imo

[jaunty glacier]

That makes sense

thanks to all of you 🙂

[dim widget]

Wait but you haven't proved that it is a group yet

[delicate orchid]

identity should hopefully be obvious as well

all semigroups are groups 😌

[dim widget]

All semigroups are the trivial group 🦾

[jaunty glacier]

just taking the identity (the element that fixes the shape) should work

[jaunty glacier]

true I'm still missing that every element has an inverse

[delicate orchid]

yeah the identity function works

[dusty verge]

The functions have to be bijective

[jaunty glacier]

But are they?

[dusty verge]

Yes

From C to C they are

[jaunty glacier]

How can I be sure of that

Same set

So bijective

[dusty verge]

One to one and onto

[jaunty glacier]

makes sense

[delicate orchid]

well we know they're surjective immediately

[jaunty glacier]

true and we need injective

[dim widget]

Just want to point out that to actually prove this you also need to show something like "the inverse of this function is also a 'symmetry of a figure'"

[delicate orchid]

yur

that's very easy to do though

[dim widget]

Since the definition of symmetry is not just an abstract bijection, but one with very rigid properties

[jaunty glacier]

hmm

I'll try it out and see where it gets me

Thanks @delicate orchid and @dim widget

[dusty verge]

:(

[jaunty glacier]

and @dusty verge

[dusty verge]

:)

[jaunty glacier]

was searching your name<

mb

lol

[novel parrot]

Couldnt b0 or c0 been 0?

[void cosmos]

p^2 must not divide a_0

[novel parrot]

But how does r >0 and s>0 imply that b_0 and c_0 are divisible by p

[void cosmos]

a_0=b_0c_0

u have a prime that divides a_0

[novel parrot]

This is true anyway if r = 0 or s = 0

[void cosmos]

wdym

[novel parrot]

if r = 0 then g(X) = some integer right

g(X) = b_0

[void cosmos]

what is r

oh kay

[novel parrot]

I mean a_0 = b_0c_0 even if r= 0 or s = 0? so we always have contradiction lol?

im probably wrong tho lol

[void cosmos]

wtf is r

b_r is your leading coefficint

in your polynomial g

b_0 is your constant coefficint

in your polynomialg

[dim widget]

So r is the degree of the polynomial g

and s is the degree of h

[novel parrot]

r = degree of polynomial g

[void cosmos]

you meant p^2

or nvm

[novel parrot]

nah p

[void cosmos]

yea

[novel parrot]

p divides both of em

so p^2 divides a_0

[void cosmos]

what question do you have

[novel parrot]

like

Whats the contradiction that they even trying to prove?

[void cosmos]

do you know about ideals

prime ideals?

[novel parrot]

yes

[void cosmos]

it would be better do this more generally and actually easier

okay

[novel parrot]

yeah im looking at the proof in dummit and foote

[void cosmos]

this is not the proof

in df

the one

u posted

[novel parrot]

no

dummit and foote uses prime ideals

[dim widget]

@novel parrot I think you're getting really hung up on this r, s > 0 thing

Since $a_n$ is nonzero mod p we know that $b_r * c_s = a_n$

so if r = s = 0 then n = 0 and this whole statement is trivial

Similarly if either r = 0 or s = 0 then the factorization is trivial.

[cloud walrus]

Topos_Theory_E-Girl

[novel parrot]

this is the proof from df

how do we know that both of the constant terms are 0?

if b_0c_0 = 0 then we can only say that one of them is 0 (in P)

[void cosmos]

if u have say x = 0 mod (p) then x is in p

p as in the ideal p

[novel parrot]

ye

[void cosmos]

so u suppose f is reducible

so u have a non trivial factorization

say f(x) = a(x)*b(x)

[novel parrot]

yes

[void cosmos]

now the hypothesis which ur screenshot is missing is this

[novel parrot]

cant the a(x) or b(x) be just elements of R, just not units?

[void cosmos]

P is a prime ideal in an integral domain R , f(x) = x^n + a_n-1x^(n-1)+.... where each a_n-1 are all elements of P but a_0 is not in P^2

[void cosmos]

good question

u tell me

[novel parrot]

ah

units of R[x] are the units of R

so forced to be non constant

nvm

stupid question

[void cosmos]

not to me ig

anyways

[void cosmos]

u got it?

[novel parrot]

yes

[void cosmos]

now again we suppose f has a non trivial factorization , f(x) = a(x)b(x)

now lets reduce everything mod P here

ie lets look at R/P[x]

u would get x^n = a(x) mod P * b(x) mod P

[novel parrot]

yes

[void cosmos]

why? @novel parrot

where did the rest of the ceofficints go

[novel parrot]

all coeff lie in P

[void cosmos]

cool

now from now on we are in mod P

but i wont write it

[novel parrot]

ok

[void cosmos]

so u have x^n = a(x) * b(x)

then the constant terms of a(x) and b(x) multiplied together must be what

[novel parrot]

a_0

[void cosmos]

yes

what is a+0

a_0

[novel parrot]

= 0 mod P

[void cosmos]

yup

[novel parrot]

b_0c_0 = 0 in P

but that only forces 1 of them to be 0

how to force both

[void cosmos]

notice that P is prime

so our quotient ring here is an integral domain

[novel parrot]

yes

[void cosmos]

so that would mean that both a_0 and b_0 are in the prime ideal P

right?

[novel parrot]

yes

we need c_0 aswell though

[void cosmos]

whats c_0

oh mb

[novel parrot]

idk

Like we need to show that c_0 is in P too

[void cosmos]

a_0 is the constant coeff of a(x) and b_0 is constant coeff of b(x)

cool?

[novel parrot]

yes

[void cosmos]

call the constant coeff of f omething else , call it c_0

ok?

and we know c_0 = a_0b_0

[novel parrot]

yes

[novel parrot]

I dont understand this though

[void cosmos]

mod p we get both a_0 and b_0 are 0

[novel parrot]

How can both be in P

Prime only forces one of them

To be 0

yk

integral domain/ prime

xy = 0 => x = 0 or y = 0

[void cosmos]

when u have such a question

[novel parrot]

not always both

[void cosmos]

use contradiction

suppose one of them is 0 but the other is not

[novel parrot]

ok

[void cosmos]

saay suppose a_0 mod p is 0 but b_0 mod p is not

now what happens when u multiply the polynomial a(x)*b(x) mod p

[novel parrot]

we multiply coefficients

[void cosmos]

do it

[novel parrot]

painful

I think i understand though

if b_0 wasnt 0

then a(x) is forced to be 0 mod P right

[void cosmos]

yes

why

tho

[novel parrot]

because we always have terms like b_0a_i = 0

and if a_i = 0 always

then a(x) = 0

so some b_0a_i = 0 and b_0 = 0

so a(x) doesnt be 0

[lethal dune]

https://tenor.com/view/nike-logo-nike-gif-20909840

[novel parrot]

Am i correct

[void cosmos]

no

[novel parrot]

oof

[void cosmos]

grab two polynomials

and try it on paper

with a pencil

only math gods use pen for math

[novel parrot]

i use a pen

lol

pencils are too faint

[void cosmos]

yea

try it anyways

ur conclusion is correct

but u need to see why exactly

[novel parrot]

should i multiply arbitrary polynomials

[void cosmos]

write out the coefficints that u have

dont foil everything out obv

say a(x) = sum(a_ix^i)

everything here is mod p

[novel parrot]

(X^m + ... a_1X) * (X^l + ... + b_1X + b_0)

a_0 = 0

so i didnt include

and m + l = n

[void cosmos]

and u know this is = x^n

now figure out what happens when u have some element that is 0 while the other is not

[novel parrot]

We get terms like b_0a_1X and b_0a_2X^2 and b_0a_3X^3

basically this

[void cosmos]

yes

now if u have the constant term of one 0 and the other not

u will have a term of say x^j where j is less than n

[novel parrot]

But couldnt the inner sum, sum to 0

since terms are like (b_0a_i + (other stuff) )x^j

Feels hard to visiual the terms

[void cosmos]

only the terms that are multiplied with b_0 that is

[novel parrot]

yeah

[void cosmos]

the point is

that when u have a b_0 non zero and a_0 is 0

then u will have terms that have degree less than n which will force the whole polynomial of a(x) to be in P ( this is becuase P is a subring after all )

hence a(x) is 0 mod p

its cuz of that before last coefficint

do u get it?

[charred crescent]

the identity transformation refers to the identity matrix right ?

[void cosmos]

yes

[charred crescent]

thanks 🙂

[novel parrot]

yes

understood

[void cosmos]

u sure?

the whole a(x) MUST BE in P

so a would be 0 mod P all of it

[novel parrot]

Not 100% because i need to become more familiar with how the terms of polynomial multiplication turn out. But i understand overall

having a_0 = 0 forces b_0 = 0

because it all is = x^n

[void cosmos]

we have a_0b_0 in P

[novel parrot]

yes

[void cosmos]

we claim that both a_0 and b_0 are in P

[novel parrot]

yes

[void cosmos]

proof: suppose a_0 is in P but b_0 is not

we know that a(x)b(x) = x^n [ thats the whole point , it is not true in general that ab = 0--> both a and b are 0 so we know the hypothesis is the key obv ]

[novel parrot]

yep

understood

thanks

then a_0b_0 is in P^2

so we get overall contradiciton

[void cosmos]

no

chill

[novel parrot]

[void cosmos]

we are going to show that all of the coefficints of a

are going to be in P

[novel parrot]

ok

[void cosmos]

now when u multiply things out in your paper

u get a_1b_0 + a_0b_1 is in P

[void cosmos]

remember this.

[void cosmos]

and .

[novel parrot]

yes

[void cosmos]

do you see why

[novel parrot]

yes because x has 0 coefficient

[void cosmos]

yes

and thats in P

[novel parrot]

rightttt

[void cosmos]

good

[novel parrot]

so a_1b_0 = 0

so either a_1 = 0

or b_ 0 = 0

if we say a_1 = 0

[void cosmos]

would that imply that a_1 is in P?

[novel parrot]

yes, or b_0 = 0

[void cosmos]

b_0 cant be 0 tho

cuz we supposed that b_0 is not in P

[novel parrot]

yeah

[void cosmos]

thats our hypothesis

[novel parrot]

so a_1 = 0

[void cosmos]

so now we get a_1 is 0 mod p

which means a_1 is in P

now can u keep going

[novel parrot]

then we repeat for X^2 coefficient right

[void cosmos]

yes exactly

[novel parrot]

gotcha

[void cosmos]

good

[novel parrot]

That makes it so much clearer

I can visualise it now

[void cosmos]

yea pen and paper works

work*X D

anyways

where tf were we

[novel parrot]

I understand the rest

just that part of forcing both a_0 and b_0 to be in P

was unclear

[void cosmos]

u got it now?

[novel parrot]

yeah

[void cosmos]

do u see how this goes to this

[novel parrot]

no

i ditched the original proof

df is clearer

[void cosmos]

u should be able to understand it

now

[void cosmos]

no it is not

in df they totally omit this detail

which is not easy

[novel parrot]

Ah

the original proof says that the polynomial is just 0 if both a_0 and b_0 are not in P

gotcha

[void cosmos]

cool

[novel parrot]

makes sense

i was confusing by their working

that "r > 0 and s > 0"

[void cosmos]

u got it now hopefully

[novel parrot]

yeah

makes so much sense now

Thank you

[void cosmos]

np

[jagged plover]

Does this look right?

[sonic coral]

#odes-and-pdes

[dim widget]

Den här borde inte vara här.

[jagged plover]

jag vet inte vart?

kan du snälla hjälpa mig

jag har suttit i 3 h med den utan hjälp

snart kommer jag explodera lol

[dim widget]

@sonic coral är rätt det borde vara i #odes-and-pdes

Om du posta den där kan jag garna hjälpa dig

[jagged plover]

aaa jag har gjort det

[wraith cargo]

I am probably being very dumb but why does the tensor being 0 imply that the gammas are K-linearly dependent

[dim widget]

If the $\gamma_i $were linearly independent over K then so would be the $\beta \otimes \gamma_i$ for any $\beta$, in particular so would the $\beta_i \otimes \gamma_i$.

[cloud walrus]

Topos_Theory_E-Girl

[wraith cargo]

-_-
Intriguing

[dim widget]

You could see the last assertion by, say, taking a K-basis for L and expressing all of the \beta_i in that basis.

[dim widget]

Is this confusing?

[wraith cargo]

Ahh

that makes sense

[wraith cargo]

Nah I wasn't seeing it at first but now it makes sense thanks

[dim widget]

Nw 🙂

[rotund aurora]

Let f be a monic polynomial in Z[x]. What is the ideal (f, f') ?

where f' is the derivative

[fleet pelican]

what do you want to know

[rotund aurora]

What is the ideal (f,f')

[fleet pelican]

It is the ideal generated by those two polynomials

[rotund aurora]

what's the intersection (f, f') cap Z

[dim widget]

It's the ideal generated by those polynomials intersected with Z :kappa:

[rotund aurora]

ok so

I think (f, f') contains the discriminant

which is an integer

so I guess (f,f') cap Z is the ideal generated by the discriminant, or maybe not. That's what Im kinda asking

[dim widget]

Haha yeah the image of (f') is the different ideal (or inverse different idr which is called which) of Z[x]/(f)

[rotund aurora]

what's the different ideal

[dim widget]

The norm down to Z is the discriminant of the order Z[x]/(f)

The different ideal is the dual of Z[x]/(f) under the trace pairing inside of Frac(Z[x]/(f)). I am assuming that (f) is irreducible just because that's what I'm familiar with.

The relevant thing is that the norm of the inverse of the different ideal is the discriminant.

But if you are just interested in which primes divide the discriminant then yes (f, f') \cap Z has the same support as the discriminant of the order Z[x]/(f).

[rotund aurora]

differents are cool

D_(K/F) is not defined previously, is that a typo and they meant to write D_(K/F)=(alpha)^{-1} o[alpha] ?

[dim widget]

What is the complementary module?

[rotund aurora]

after that they use $\mathfrak D_{K/F}$ for the different

[cloud walrus]

Croqueta

[dim widget]

They must define D_{K/F} somewhere otherwise the proposition has no content.

Anyway from context it is the different of O[\alpha]

probably they reserve \mathfrak{D}_{K/F} for the different of the maximal order O_K

[rotund aurora]

D_{L/K} is defined in the appendix

lol

[dim widget]

Ah makes sense. How is it defined that this proposition is not a tautology? Just as (f')?

[rotund aurora]

$D_{L/K}(\alpha)=f'(\alpha)$

[cloud walrus]

Croqueta

[dim widget]

Yeah makes sense

One thing to keep in mind by the way: not every ring of integers O_K is of the form Z[\alpha] for some \alpha \in K (although most examples you encounter will be).

So you cannot always compute the discriminant this way.

[rotund aurora]

ik

[dim widget]

Have you seen the cool obstruction to a ring of integers being monogenic which comes from the F_2 points?

[rotund aurora]

no

[dim widget]

So if O_K = Z[\alpha] then there is a surjection Z[X] \onto O_K so Spec(O_K) is closed inside of A^1_Z

but then if you tensor both sides with F_2 you see that O_K \otimes F_2 is a closed within A^1_{F_2}

But

A^1 only has 2 points over F_2

So if you can engineer a cubic field with 2 totally split for instance, it is automatically not monogenic.

[rotund aurora]

what is Z[X] here

[dim widget]

Because then $\text{Spec}(O_K) $has too many points in the fiber above $\text{Spec}(\mathbb{F}_2)$

[rotund aurora]

like polynomials

?

[dim widget]

Z[X] is polynomials over Z yes

X is an indeterminate

[rotund aurora]

ah yeye lol

[dim widget]

$O_K = Z[\alpha]$ iff $Z[X]$ has a surjection onto $O_K$

[rotund aurora]

you either use latex bot or dont

[cloud walrus]

Topos_Theory_E-Girl

[dim widget]

Latex bot is for the weak

https://www.youtube.com/watch?v=fBDifUjNzbQ

[cloud walrus]

Topos_Theory_E-Girl

[rotund aurora]

what does it mean Spec(O_K) to be closed inside A^1_Z. Guess it means its an algebraic set, but how?

[dim widget]

Ah okay I forgot that you didn't know about schemes

So Spec(A) is closed in Spec(B) iff B surjects onto A

That's it

[rotund aurora]

yes

[dim widget]

It is then the "algebraic set" cut out by the equations defined by the generators of the kernel

[rotund aurora]

and if you have a surjection R -->S then you have an injection Spec(S)--> Spec(R) right

or im smoking

[dim widget]

Yes but I'm claiming even more, it's actually a closed immersion

If you like you can think of Spec(Z) as an infinite number of points (one for each prime, and then a mysterious one for (0) which corresponds to Q), and $Spec(Z[X])$ is an $A^1_{F_p}$ at each prime p, and an $A^1_Q$ at $(0) $ all glued together in some mysterious way. For now you can think of $Z[X] \twoheadrightarrow O_K$ as meaning that $O_K \otimes \mathbb{F}p$ is a closed subvariety of $\mathbb{A}^1{\mathbb{F}_p}$ for all primes p.

[cloud walrus]

Topos_Theory_E-Girl

[rotund aurora]

in this context, you can view numbers as functions right

[dim widget]

Yes exactly, and they vanish wherever their prime divisors are, etc.

And the order of vanishing is the multiplicity of that prime divisor.

Anyway maybe this was too much to go into, but I think it's very cool.

But the different stuff is what you were originally asking about.

[rotund aurora]

I am interested in the analogy between number fields and Riemann surfaces

[dim widget]

Yes rings of integers are very much the same as riemann surfaces from a geometric point of view

Both one dimensional, both have the same ramification theory, elements of a number field are like meromorphic functions, there is a Riemann Roch theorem for both

[rotund aurora]

I don't really know about the analogy yet, but how strong is it? Like do you get a functor { number fields } --> { Riemann surfaces } ?

[dim widget]

No not a functor, it is not that literal of an analogy. Number fields are much more difficult to classify than Riemann surjaces.

It's more of an analogy where the same structures tend to pop up on both sides.

[rotund aurora]

👍

[dim widget]

The analogy is actually closer with curves over finite fields though

[white oxide]

can somebody give me a hint for this question? I feel like it has to do with the characteristic of R, or the fact that R contains a subring isomorphic to Zn.

i'm trying proof by contradiction but for the moment i seem to be stuck

[frigid lark]

What happens if you quotient out a prime ideal

[novel parrot]

How do i decide if I is free?

[dim widget]

If it has an R-basis with no nontrivial relations.

Usually you have to use something cleverer to figure out if a module is free, but for this exercise just the definition is fine.

[novel parrot]

The Ideal I can always be spanned right

by some set of elements

only linear independence is a possible issue?

[dim widget]

Yes every ideal has some possibly infinite set of generators.

So there is a surjection from some free module onto I

[novel parrot]

i see

linear subspace means submodule right

in the context of a field

[wraith cargo]

Why does the last line of this proof imply that d is a unit?

[hot lake]

g is monic and df divides g in A[x] so the leading coefficient of df (which is d) divides 1 in A, so it's a unit

[wraith cargo]

Ahh true

thanks

[novel parrot]

the annihalor of 1 is <X - 1> and annihalor or t,t^2,t^3 is <(X-1)^2> right

because (X-1)^2 * t = (X-1) * 1 = 0

[hot lake]

are you saying (X-1)².t² = 0 ?

[novel parrot]

yes

[hot lake]

what's (X-1). t² ?

[novel parrot]

0?

could be wrong

[hot lake]

what's X. t²

[novel parrot]

(t+1)^2

[hot lake]

yes

so now what's (X-1). t² ?

[novel parrot]

no

oops

(t+1)^2 - t^2 = 2t + 1

[hot lake]

yes

and now you can compute (X-1)² . t² = (X-1). ((X-1).t²)

[novel parrot]

yeah i was wrong xd i treated t^2 = t * t and not f(t) = t^2 XD

[hot lake]

I'm honestly not sure what's the difference

[novel parrot]

not sure either 🙃

but i did it wrong

but i know where i went wrong

[dim widget]

I think maybe you're thinking that R[X] is acting by ring homomorphisms, but it's obviously not, it only preserves the addition.

[novel parrot]

yep

did this

[hot lake]

ah the sum of ring morphisms isn't a ring morphisms

[dim widget]

Yes but also on a more basic level: scalar multiplication is not a ring homomorphism unless the scalar is idempotent.

[hot lake]

ah yeah

[proper moth]

Trying to prove |cos x−cos y| ≤ |x−y| for all x, y ∈ R. - using the mean value theorem
Can I get a hint (which doesn't give away the solution too much). Tried a couple things and not quite sure

[south patrol]

Well what does the mean value theorem tell you about cos x - cos y

[proper moth]

that between two points x,y there is a point with its derivative equal to the slope of x and y?

[maiden ocean]

This probably is better relegated to #real-complex-analysis than here

[south patrol]

That is also true lol I didn't even realise

[proper moth]

solved it. & yup! 🤦 (I've been asking a few group theory Qs recently, so instinctively clicked on this channel)

[formal ermine]

where is the sylvester matrix used other than to define the resultant

[dusty verge]

Can you use Turing machines to prove that the group given by a presentation is undecidable?

Feels like there'd be a way to twist it

After googling it, the answer is yes. Neat

[rain otter]

can someone help me with this, Im trying to understand how 4 principal ideals (2), (3), (1+sqrt(-5)), (1-sqrt(-5)), in Z(sqrt(-5)) can be expressed as products of 4 other ideals in Z(sqrt(-5))

my first thought was to pick the 4 ideals to be
a = (2, 1+sqrt(-5))
b = (2, 1-sqrt(-5))
c = (3, 1+sqrt(-5))
d = (3, 1-sqrt(-5))
but that didnt really help

[barren sierra]

Ok can someone check my work on this?

Suppose $A = 0$, the matrix of all $0$'s. Then the left action of $p(x) \cdot v = p(A) \cdot v$ is essentially just scalar multiplication by $k$ and so our module is $M = k^2$ with scalar multiplication. We can write this as $(k \times \{0\}) \oplus (\{0\} \times k)$ which are both simple submodules of $M$. Thus fixing $A = 0$ yields a semisimple ring.

[cloud walrus]

Spamakin🎷

[barren sierra]

I'm not at all comfortable with semisimple rings (absolutely 0 intuition) so it may be basic but I want to make sure. I think it's right so if it's wrong I want to know for sure

[wide shard]

\documentclass[12pt]{article}
\usepackage{tikz}

\begin{document}

\begin{center}
\begin{tikzpicture}[scale=0.2]
\tikzstyle{every node}+=[inner sep=0pt]
\draw [black] (8.3,-12.7) circle (3);
\draw (8.3,-12.7) node {$q_0$};
\draw [black] (23.2,-11) circle (3);
\draw (23.2,-11) node {$q_1$};
\draw [black] (11.28,-12.36) -- (20.22,-11.34);
\fill [black] (20.22,-11.34) -- (19.37,-10.93) -- (19.48,-11.93);
\end{tikzpicture}
\end{center}

\end{document}

[cloud walrus]

m64sky

\documentclass[12pt]{article}
\usepackage{tikz}

\begin{document}

\begin{center}
\begin{tikzpicture}[scale=0.2]
\tikzstyle{every node}+=[inner sep=0pt]
\draw [black] (8.3,-12.7) circle (3);
\draw (8.3,-12.7) node {$q_0$};
\draw [black] (23.2,-11) circle (3);
\draw (23.2,-11) node {$q_1$};
\draw [black] (11.28,-12.36) -- (20.22,-11.34);
\fill [black] (20.22,-11.34) -- (19.37,-10.93) -- (19.48,-11.93);
\end{tikzpicture}
\end{center}

\end{document}
```Compilation error:```! LaTeX Error: Can be used only in preamble.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.57 \documentclass
                   [12pt]{article}
Your command was ignored.
Type  I <command> <return>  to replace it with another command,
or  <return>  to continue without it.```

[formal ermine]

@wide shard #latex-testing

[wide shard]

this is abstract algebra

[summer path]

that is not a question

nor an answer

[formal ermine]

nor abstract algebra for that matter

[summer path]

well i can't tell what that is because there is not enough context lol

[wide shard]

oh this channel is used for questions and answers

[summer path]

but i don't really want that context

if all there is is some tikz diagram

[wide shard]

i thought it was used for just discussions or just abstract algebra

[summer path]

it is used for abstract algebra

[formal ermine]

@wide shard do you know what abstract algebra is

[wide shard]

yeah

[summer path]

but this is not abstract algebra

[barren sierra]

if you're just trying to figure out tikz stuff + play with bot go to #latex-testing

[summer path]

look, if you have something to say, say it

[rain otter]

@wide shard my brother in christ that is not abstract algebra

[summer path]

otherwise go to #latex-testing

[barren sierra]

let him cook and ask a question ig

[wide shard]

yeah bro

[formal ermine]

🧑‍🍳

[wide shard]

let me cook

[barren sierra]

(also if anyone has any ideas about my question above 🙃)

[wide shard]

what does that mean btw

[formal ermine]

I can cook minute rice in 58 seconds.

[wide shard]

lol

[barren sierra]

just ask your question if you have one

[rain otter]

same 💀

[summer path]

what does what mean

[wide shard]

i can cook biryani in 10 minutes

[barren sierra]

instead of clogging the channel

[wide shard]

let him cook meaning

[summer path]

ya know what

[barren sierra]

and further burying my question

[summer path]

just go to #chill

[rain otter]

theres no hope for us my guy, dw

you can ask on mse and get deleted if you want

[wide shard]

whoops

[hot lake]

what is stopping you from multiplying a,b,c,d together

[rain otter]

wdym

I tried combinations

like I tried a*b but it didnt seem to help

or I messed up somewhere

[hot lake]

you are thinking that your 4 principal ideals might be expressed as products of some of a,b,c,d, right ?

[rain otter]

My idea was that if we do that then for example a * b = (2)

[rain otter]

ye

[hot lake]

so what happened when you computed a * b

[rain otter]

I got an expression, which didnt seems to be (2)

but maybe i made a mistake

lemme send pic

I was thinking maybe the first 4 terms in the final step were all equivalent?

but honestly im not sure

[hot lake]

that's uh probably not the best way to go about it

[rain otter]

I see

and also it should be Z\sqrt[-5] not 5

but thats beside the point

[hot lake]

you should try to get a set of generators for ab

[rain otter]

I think this is where my weak background in abstract algebras shining through

hm

[hot lake]

also uh

2*(1+sqrt(-5)) is not 2 + sqrt(-5)

[rain otter]

lovely

ok but I just put a factor of 2 before the sqrt(-5) in the last step

at least thats easy to fix

ah fuck it, its too late for this anyways, im gonna go to bed and look at it tomorrow

thanks for the help anyway

[hot lake]

also if you want to show this is equal to (2)

you can show that it's included in (2)

and then show that it contains 2

[rain otter]

and (2) is included in it?

hm yeah

that would be a good approach

[wraith cargo]

The last sentence of this proof is confusing me
I get that N_{L/K}(\beta) and \Tr_{L/K}(\beta) are integral over A and elements of K, but I can't see why they're elements of B?
It seems more realistic to me that they're integral elements of some field over A

[dim widget]

Yes that is right. But did you find a non-semisimple example (that is arguably harder)?

[wraith cargo]

Also this is the referenced proposition

[barren sierra]

not yet, stuck on that part

[dim widget]

They have to be integral over A because B is integral over A

[wraith cargo]

yeah I mean I understand why they're integral I don't understand why they're elements of B

[dim widget]

Beta is in B, so all of its conjugates must also be in B (Edit: images of B under various embeddings of L into \bar{K})

Since integrality is independent of the choice of embedding (or invariant under automorphisms fixing A if you look at everything in the algebraic closure)

[wraith cargo]

Aren't the conjugates elements of the algebraic closure of L?

Ah wait

The minimal polynomial is Monic

So the roots are restricted to B subset L

[dim widget]

Weren't you worried about B not being integrally closed in L?

[dim widget]

I guess I thought after this comment that was your confusion.

[chilly radish]

this isn't necessarily true tho. Like take e.g. B=Z[cbrt(2)] in L=Q(cbrt(2)), its conjugates are certainly contained in an isomorphic copy of B inside the algebraic closure, but they aren't inside of L even, since they involve complex roots of unity

[dim widget]

Yes but having them live in isomorphic embedded copies of Z[sqrt(2)] is enough to deduce that the products are integral over A, and contained in Z[sqrt(2)]

More precise is to say "contained in the image of B under some embedding" but I was being lazy

[chilly radish]

I think that's the part that irony incarnate is confused about

that the product is indeed contained in B

[dim widget]

Ah I see, then perhaps it's better to say that the product is integral and obviously in K.

That's a simpler way to end the proof anyway

[chilly radish]

honestly yea

either way, the product ends up being integral over A and in L. The only elements of L which are integral over A are elements of B anyways. If you have non-trivial denominators then you're gonna end up not satisfying a monic polynomial with coefficients in A

[dim widget]

This isn't obviously true from the problem statement

I thought this was what they were worried about.

But there's no reason that B must be integrally closed in L

Unless I'm misunderstanding what they mean by "integral extension of domains"

[chilly radish]

hmmm ok yea you're right

my bad

[chilly radish]

this is the best way to think about this

[dim widget]

Yes I think so, I realize now that my proof of the norm and trace being in B goes through proving this.

[white oxide]

i know that if i quotient by a prime ideal i get an integral domain, and also the ring is finite and Zn is an integral domain iff n is prime and is hence a field?

although i'm not sure if it would automatically be iso to Zn

[dim widget]

Do you know a theorem about finite integral domains??

[white oxide]

nah

hmmm

well i suppose maybe we could regard R/N as a finitely generated abelian group under addition

since it's a ring

and perhaps then it's isomorphic to Zn which is an integral domain iff n is prime and hence is a field

??

idk if that works

oh ya

every finite integral domain is a field

oops

[dim widget]

So proposition: all finite integral domains are fields. Can you prove this?

[white oxide]

forgot about that

[dim widget]

Yep

[white oxide]

wait would my argument still work tho

or nah

[dim widget]

Nah

[white oxide]

rip

[dim widget]

Because you don't know what the multiplication is like

[white oxide]

ahh true

[dim widget]

Structure theorem for FGAB's will not say anything about whether it's a domain or not

E.g. F_p[x]/x^2 is just (Z/p)^2 as an abelian group just like F_p \times F_p

[white oxide]

ohhh that makes sense

got it got it

thank you!

[dim widget]

nw

[white oxide]

so my book gave Z2 x Z3 as an example, since i tcontains a subring isomorphic to Z2 and a subringg isomorphic to Z3, but how is this possible? wouldn't imply that if such a ring exists that has a subring isomorphic to Zm and one isomorphic to Zn, then that ring would have characteristic m and characteristic n for m neq n? but i thought the characteristic of a ring is defined to be the least such integer n such that n x r = 0 for all r in R, and hence a unique integer value?

[south patrol]

There it doesn't say anything about the characteristic though

Like for Z2 x Z3 the characteristic is 6

[dim widget]

The characteristic of the ring is the order of the multiplicative identity. Here is an example that will really bake your noodle: $\prod_{p \text{ prime}} \mathbb{F}_p$ has characteristic $0$.

[cloud walrus]

Topos_Theory_E-Girl

[white oxide]

ah true

[white oxide]

damn

what does \mathbb{F}_p denote here

just any field of order p or Zp?

[dim widget]

both, there is only one field of order p as it so happens

[dim widget]

.

[white oxide]

oh yea true

[formal ermine]

I suppose the profinite integers have characteristic 0?

[dim widget]

Yep, they're even torsion free.

[prisma ibex]

people in the literature like calling things like Q_p or Z_p "mixed characteristic": they are characteristic 0, but have char p residue field

it's good terminology, but it's a little funny when you first encounter it

[white oxide]

probably a stupid question, but why couldn't we consider the elements 1, B, B^2, ... B^n+1? wouldn't that show that the deg(B, F) is at most n + 1?

[formal ermine]

isn't it explained in the text between those two highlights?

[agile burrow]

I mean you could, but certainly if the degree is at most n, then it's also at most n + 1

[solar shore]

could someone give a hint for this?

im struggling on where to start with it

[formal ermine]

geometric series

[coral spindle]

A more straightforward hint would be to remember the difference of two squares.

[formal ermine]

oh I didn't see the 2 there lmao

[solar shore]

the next question considers if a was nilpotent, show that 1 - a has an inverse

LOL

[formal ermine]

kek

[solar shore]

but the difference of squares helped

so (a + b)(a - b) = a^2 right?

since a is a unit, then a^2a^(-2) = 1

[formal ermine]

yeah

[south patrol]

Noic

[solar shore]

so a + b's inverse is (a - b)a^(-2)

[formal ermine]

yip

[solar shore]

okee thank you

[barren sierra]

I'm stuck on finding an example that is not semisimple.

[coral spindle]

Keep at it. If you know about the Jordan normal form you may be able to land on an example quickly.

[barren sierra]

Oh I wasn't even thinking in that direction

lemme refresh myself on that

[white oxide]

wait sorry i'm confused wouldn't that contradict the fact that deg(B, F) < deg(a, F)

[coral spindle]

I said if you know it. I will emphasise that you don't have to know it.

Just keep trying matrices.

Btw, the Jordan normal form classifies all finite-dimensional k[x]-modules where k is algebraically closed.

[agile burrow]

No? "At most n + 1" means "less than or equal to n + 1"

[barren sierra]

I guess part of the issue is that I'm not sure how I'd show something isn't semisimple

[coral spindle]

Then you should try showing that

[white oxide]

yea but what if it is n + 1

deg(B, F) = n + 1 which is not less than or equal to deg(a, F) = n

[agile burrow]

It can't be because you can show the stronger condition that the degree is less than or equal to n

[white oxide]

so wouldn't it be incorrect to say that deg(B, F) is at most n + 1 if it can't be n + 1

sorry i'm confused this is my fault

[coral spindle]

No

You are incorrectly thinking that "at most n+1" means that there is a case in which the value is n+1, when this is not how this term is used in mathematics.

If what you were thinking was the case, then it would be incorrect to say "x is at most y" unless the two were literally equal, which is just not useful at all.

"6 is at most 7" is a correct use of the term in mathematics

[white oxide]

ah huh ok i guess i better review my logic then

thanks guys

[barren sierra]

yea but how. For example how would I show that something cannot be written as a sum of simple modules?

[coral spindle]

Exercise: show that Z/p^2Z is not the sum of simple Z-modules.

Trying to do this will help with the problem at hand

[barren sierra]

former has elements of order p^2 but the latter cannot?

[white oxide]

i'm struggling to see how the preceding sentences before consequently justify 1, sqrt(3) being a basis

can somebody help me pls

[solar shore]

how do i start with this

[white oxide]

i understand that {1, sqrt(3)} is a basis, but that's because the irreducible polynomial for sqrt(3) over Q(sqrt(2)) is of degree 2, but how do we know that

[white oxide]

<@&286206848099549185>

[analog zephyr]

guys

need some help here

i've asking some question to ChatGPT then i get this answer from one of those question

We have a collection of normal subgroups $G_n$ of a group $G$, and we form the group $G_N$ as the direct product of the groups $G_n$. There exists a natural homomorphism $\phi$ from $G_N$ to the group formed by the direct products of the quotient groups $G/G_n$. The homomorphism $\phi$ is defined by sending each element of $G_N$ to a tuple of the corresponding cosets in the quotient groups. Moreover, $\phi$ is surjective and its kernel is given by the elements of $G_N$ whose components lie in the subgroups $G_n$. By the homomorphism theorem, it follows that $G_N$ mod the kernel of $\phi$ is isomorphic to the image of $\phi$, which is the direct product of the quotient groups $G/G_n$.

[cloud walrus]

brey

[analog zephyr]

The homomorphism $\phi$ is defined as follows: for $a = (a_1, \ldots, a_n) \in G_N$, we define $\phi(a) = (\overline{a_1}, \ldots, \overline{a_n})$, where $\overline{a_i}$ is the coset of $a_i$ in $G/G_i$. In other words, $\overline{a_i} = a_i G_i$. To show that $\phi$ is a homomorphism, we need to verify that $\phi(ab) = \phi(a)\phi(b)$ for all $a,b \in G_N$. If we write $a = (a_1, \ldots, a_n)$ and $b = (b_1, \ldots, b_n)$, then $ab = (a_1b_1, \ldots, a_nb_n)$. Therefore, $\phi(ab) = (\overline{a_1b_1}, \ldots, \overline{a_nb_n})$. Now, $\overline{a_ib_i} = a_ib_i G_i = a_i G_i b_i G_i = \overline{a_i} , \overline{b_i}$ for each $i$, since $G_i$ is normal in $G$. Hence, $\phi(ab) = (\overline{a_1} , \overline{b_1}, \ldots, \overline{a_n} , \overline{b_n}) = \phi(a)\phi(b)$. Therefore, $\phi$ is a homomorphism from $G_N$ to $\prod_{i=1}^n G/G_i$.

[cloud walrus]

brey

[white oxide]

is it just the intersection of the bases??

like how do we know deg(sqrt(3), Q(sqrt(2)) = 2

<@&286206848099549185>

[wraith cargo]

Sqrt(3) is not in Q(sqrt(2)), so the minimal polynomial can't be of degree 1

x^2-3 is a polynomial with coefficients in Q(sqrt(2)) with root being sqrt(3) so we can conclude that [Q(sqrt(2), sqrt(3)) : Q(sqrt(2))] = 2

[wraith cargo]

What is Z here?

[solar shore]

it might be a typo it should be Z_n i think

[wraith cargo]

Ahh

[solar shore]

idk 🤺

[wraith cargo]

Then show that the set satisfies the axioms for a ring
And argue why it only has n^2 elements

(here addition and multiplication are the regular matrix addition and multiplication)

[empty rose]

...do you have a question about something in here?

[white oxide]

wait sorry why not

[solar shore]

when proving that something is a ring, do i go through all the ring axioms?

[frigid lark]

sometimes

Very often you can just prove that it's a subring

or a quotient ring

[solar shore]

i have not gotten into quotient rings in my readings but i will consider the subring idea

i think because i havent had an exercise that said to check all the ring axioms, i might do it that way just to get some practice with it

[tulip hawk]

How do we define addition for tensor product to make it into a module

[frigid lark]

Addition in $A\otimes_R B$ is just induced from addition in $A\times B$, i.e. component wise. If R is commutative, then $A\otimes_R B$ is an R module.

[cloud walrus]

Parrot Tea

[frigid lark]

Read the construction of a tensor product as a quotient space from here

https://en.m.wikipedia.org/wiki/Tensor_product

[sly rain]

If I have established $B \cong C $ and I need to show that $ A \cong B $, but showing $A \cong C$ is easier for me.
I think i get that im allowed to just show $A \cong C$ and it should follow that $A \cong C \cong B$.
Whats the "good practice"/justification here? If I where to explicitly justify this I would say something along the lines of that a composition of two bijections is bijective hence $A \cong B$.

[cloud walrus]

aabb

[sly rain]

Can I just treat them like equal signs in that case? Like $A \cong C \cong B$ implies $ A \cong B $?

[cloud walrus]

aabb

[carmine fossil]

Well isomorphisms work like that yes

If f is a iso from A to C and g is a iso from C to B, fg is clearly a hom from A to B. It's also iso because g^-1 f^-1 is inverse

[dim widget]

I think this is kind of misleading: the way you word of it makes it kind of sound like $A \otimes B$ has the same underlying space as $A \times B$. Instead maybe it's better to say that $(a + a') \otimes b = a \otimes b + a' \otimes b$ and $a \otimes (b + b') = a \otimes b + a \otimes b'$.

[cloud walrus]

Topos_Theory_E-Girl

[wraith cargo]

That is proved in the sentence before the word consequently

[novel parrot]

Every module has a spanning set right

We could take the spanning set to be just every element of the module correct?

[wraith cargo]

I don't think this is true?

are you thinking of the spanning set as finite

maybe that should've been a good thing to ask beforehand tho lol

[novel parrot]

possibly infinite

nvm tho because i only seen the proof that all vector spaces have a basis in the finite case

[wraith cargo]

yeah not all modules have a basis

take Q as a Z-module

[novel parrot]

the last part where it says that x <= m for all x in P

since P is partially ordered, we cannot compare every x with m ?

so what does it mean?

Or is it saying for this m, we can compare every other element?