groups-rings-fields

[white oxide]

what's a cool greek letter

other than \varphi and \psi

[hushed hamlet]

xi

[white oxide]

oh facts

[coral spindle]

\varpi

JOKES

[hushed hamlet]

it is, but its also an in my opinion better way to see, why we need to impose that its abelian

[white oxide]

$\phi$ is so fucking basic

[cloud walrus]

okeyokay

[white oxide]

yea i'll check it out later, looked elegant bro

[hollow mica]

wannabe empty set

[hushed hamlet]

also xi is really easy to write actually

[white oxide]

i just scribble a line down lmfao

kinda like a quarter rest

[hushed hamlet]

start with a horizontal bar, then without lifting your pen start writing a c, and then an s

[white oxide]

ah

you do it properly

[hushed hamlet]

ya

i put emphasis on handwriting

even tho mine is only mediocre

[cloud walrus]

Dragonslayer Sharp

Dragonslayer Sharp

[crystal turtle]

"don't real"

[white oxide]

bro said don't real

what the fuck is the first one

[crystal turtle]

Capital Xi in first pic, then lowercase xi and zeta in second

[coral shale]

and here i thought we were discussing math

but no its just some squiggles

[white oxide]

ugh. okay so I defined $\xi: M \to M' \oplus M''$ by $\xi(m) = \bigl(\psi(m), g(m)\bigl)$, it worked, cool. assumed 1 held and defined $\phi: M' \oplus M'' \to M$ by $(m', m'') \mapsto f(m') + \varphi(m'')$. also an isomorphism. I was able to show 2 by letting $\psi = \pi_{M'} \circ \phi^{-1}$. but assuming 2 and trying to show 1, I just took $\xi$ and let $\varphi = \xi^{-1} \circ \iota_{M''}$. the problem was trying to determine an inverse image of $\xi^{-1}(0, m'')$; could I just assume that $g(m) = m''$ by the definition of $\xi$ and $m \in \text{Ker }\psi$?

[cloud walrus]

okeyokay

[hushed hamlet]

do you have like your whole proof written down

i think that would be easier to read

[white oxide]

Yeah, I’m aboutta latex it

Like in 15-20 minutes

And then I have to do two more things in this dumb Lang exercise

But I’ll send the full proof of proposition 3.2

[white oxide]

\noindent 1. First, we prove Proposition 3.2. Suppose $0 \to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0$ splits. Let $\psi$ be such that $\psi \circ f = 1_{M'}$, and $\varphi$ such that $g \circ \varphi = 1_{M''}$. Define $\xi: M \to M' \oplus M''$ by $\xi(m) = \bigl(\psi(m), g(m)\bigl)$. Clearly $\xi$ is a homomorphism (since $\psi$ and $g$ are homomorphisms). $\xi$ is surjective since $\psi$ and $g$ are surjective. If $\xi(m) = 0$, then in particular $m \in \text{Ker }g = \text{Im }f$. Hence $m = f(m')$ for some $m' \in M'$, and $\psi(m) = \psi\bigl(f(m')\bigl) = 0$. But $\psi \circ f = 1_{M'}$, so $m' = 0$, whence $m = 0$.
\
\
We have already seen that $M = \text{Im }\varphi \oplus \text{Ker }g$. Observe that any $x \in M$ can be expressed as [x = x - f\bigl(\psi(x)\bigl) + f\bigl(\psi(x)\bigl)] where $x - f\bigl(\psi(x)\bigl) \medspace \in \text{Ker }\psi$ and $f\bigl(\psi(x)\bigl) \medspace \in \text{Im }f$. Thus $M = \text{Im f} + \text{Ker }\psi$. If $y \in \text{Im }f \cap \text{Ker }\psi$, then $y = f(m')$ for some $m\ in M'$. But then $\psi(y) = \psi\bigl(f(m')\bigl) = m'$, so $m' = 0$ since $\psi \circ f = 1_{M'}$. Hence $f(m') = f(0) = 0 = y$, and $M = \text{Im }f \oplus \text{Ker }\psi$.

[cloud walrus]

okeyokay

okeyokay

[white oxide]

I'm assuming that this submodule is $N = \text{Ker } \psi$?

[cloud walrus]

okeyokay

[white oxide]

I'm too lazy to check right now lol

[steel light]

@balmy belfry Come talk in the big boy channel

[balmy belfry]

Okay

Yay

I made it guys!

a = n^k

b = n^z

Cuz G is cyclic

ab = n^{k + z}

Right

[steel light]

What are a and b?

[balmy belfry]

Random elements of G

[balmy belfry]

Then ab = n^{z + k}

Then ab = n^zn^k

Then ab = ba

Lmfao that’s like the 4th time I proved that and then forgotten how it’s done

Fuck

Is this valid tho

[steel light]

Looks good to me

[balmy belfry]

Grape juice

Well what now

[steel light]

Idk ask more questions

[balmy belfry]

Let’s see

First isomorphism theorem.. no

Yeah I ran out

You ask some

[steel light]

LOL

I have other HW to do

due tom

I cannot

[balmy belfry]

Why don’t you ask for help then

With that

[white oxide]

anyone?

[steel light]

It’s not math

[balmy belfry]

Oh

[balmy belfry]

Ic

How do I give myself studying

And take it away

[steel light]

bot channel

,iam studying

[cloud walrus]

Gave you the studying! selfrole.

[steel light]

…

oops

LMFAO

,iamnot studying

[cloud walrus]

Removed the studying! role from you.

[steel light]

like that amukh

[long nebula]

nice!

[balmy belfry]

I studying role now :D

,Iam studying

[cloud walrus]

You already have the selfroles studying!, do you want to remove them? (y(es)/n(o))
(Tip: use ,iamnot to remove roles without this prompt.)

[balmy belfry]

Sthu texit bot

Nobody even asked

Lol

[cloud walrus]

Session timed out waiting for user response.

Sep 5, 2023, 12:05 AM

[long nebula]

yeah, seems legit

[balmy belfry]

What’s a sub module

[long nebula]

You're not ready

[balmy belfry]

Ok

Can we ask questions about vector spaces here

Where do you ask questions about modules ? I assume here

[long nebula]

But if you know what a subspace of a vector space is

A submodule of a module is the same

[balmy belfry]

Ofc

[long nebula]

[long nebula]

Probably #linear-algebra

[balmy belfry]

But if modules are over rings

And modules are here

And vec spaces are over fields

This is the place

[long nebula]

I know the types of questions you're going to ask

#linear-algebra is a better place for them

[balmy belfry]

Ok

Yeah I know

Linear algebra sounds cooler too

Cuz groups sound boring asf

Algebra sounds funner

[long nebula]

Okay take off-topic discussion somewhere else lol

[balmy belfry]

But I can’t

I’m studying role

[white oxide]

A subspace (if you allow the scalars to be from a division ring instead of any arbitrary ring lol)

[long nebula]

Don't talk here then

[balmy belfry]

Okay

[long nebula]

Why division ring?

Oh

You're talking about vector subspaces lol nvm

[white oxide]

Ye lol

[white oxide]

Bro said groups are boring

[balmy belfry]

no

[long geyser]

bro is speaking facts

[balmy belfry]

i said they sound boring

[long geyser]

imagine having only one binary operation

[balmy belfry]

fr bruh

[long geyser]

can't be me

[balmy belfry]

2 is always better than 1

[long geyser]

can't even multiply by scalars

[balmy belfry]

fr

the fris isomorph theo in vec space is so much better than the group version like bffr

[chilly ocean]

[noble saddle]

the biggest failing of the education system is the pervasive idea that groups are just rings with one less operation

[chilly ocean]

groups are symmetries of spaces and rings are the functions on the spaces

[long geyser]

that is a pervasive idea?

[south patrol]

it's wrong anyway

[noble saddle]

among undergraduate students who have taken algebra 1, I feel like yes

[south patrol]

really?

aren't groups usually taught before rings anyway

[noble saddle]

at least, I sure have heard it expressed fairly often

[crystal turtle]

who the hell has that opinion

that's a pretty bad idea of groups

[noble saddle]

yes, agreed

[topaz solar]

I have literally never heard this

Wholly nonsensical, anyhow, since not all groups are abelian

The symmetries/functions perspective doesn’t matter much, its completely ill typed

[south patrol]

(Also the other axioms and usually existence of 1 I suppose)

Although, to be fair, an abelian group is just a ring without the monoid structure*
*internal to Ab

[topaz solar]

Especially without identity

Since then like

Yep take an arbitrary Ab & zero multiplication it

[last spoke]

Let's maybe crop my uni outta that

[chilly ocean]

doxxed!!!!

[rotund dragon]

we love some selfdox

[last spoke]

Oh I'll doxx myself plenty, come find me bitches

Anyways that was my test today

[crystal turtle]

[void cosmos]

yo one time

someone called my number

at 5am

and he went yo bro i found your number on doxedbin

and ur fucked

so i was like ?? whats that and stuff

[last spoke]

Oh that was me

[void cosmos]

turns out the post literally had my name , some shirtless pics of me and where my country is

thats it

lmfao it was so funny

and my number ofc but nobody seemed to care sadly

like it was that only one guy

😦

[void cosmos]

yea its from this server 100%

[void cosmos]

i got it

let R be a simple ring and let M be any left ideal in R

consider D = Hom_R(M,M)

this is a division ring ( every map here has trivial kernel and full image )

now consider Hom_D(M,M)

R is isomorphic to Hom_D(M,M) with the isomoprhism r --> phi_r = ra for a in M

once we get the details we have Hom_D(M,M) is R

then yeah M is a left ideal , consider it as a left R-module

then i think M is like 1-dimensional?

and if so u then just like the matrix corresponding to the Hom and we are done

or better consider it as a right R-module

not left

does this work

[white oxide]

why is $p^n(y) \in (x_1)$? is it because $(p^n)$ is an ideal and hence $p^ny \in \text{Ker }\varphi$ where $\varphi(\bar{a}) = \bar{a}\bar{y}$ and hence $p^ny \in (x_1)$?

[cloud walrus]

okeyokay

[white oxide]

can any gs pls explain why s = r implies that y has the same period as y bar

<@&286206848099549185>

[hollow mica]

bro wasted no time

[white oxide]

care to help out big man?

[hollow mica]

idk modules sorry

[agile burrow]

Does period mean the same thing as order?

[white oxide]

im not sure

here's the definition

[crystal turtle]

I take it R is a PID?

Information like this is generally quite helpful to include

[white oxide]

yeah

[white oxide]

the c is throwing me off i think

i understand that the period of y is the generator m of the kernel a maps to ay

ok so

the least such element such that ay is in (x1) or in other words in the kernel

is p^r

and this is an ideal

hence if r = s

then p^r = p^s

so like we have to show that

(p^s) = (p^n)?

nvm i don't get this shit at all

[crystal turtle]

it looks like, if s < r, then you'll find some other representative y' of \overline{y} such that the period of y' is the same as \overline{y}

like it looks like that might be the next step in the proof, not super sure of this tho

[white oxide]

fucking lang man

okay

i'll consider that thanks

how is one supposed to read like 25 pages of lang per week and do 7 hard ass exercises rip

i should prob drop out of this course tbf

[crystal turtle]

since \overline{y} has period p^n by assumption, p^n is a generator of the kernel of the map a --> a* \overline{y}. In particular, that's why p^n * \overline{y} = 0 in E/(x_1), or in other words, why the representative of the coset y satisfies p^n * y in (x_1)

idk if that makes sense

[white oxide]

okay thanks, i'll probably return to this later

i've been trying to do math at night but apparently it still doesn't work

thanks

[crystal turtle]

same

[pliant haven]

what does the =1 exactly mean here?

is it like, its normal state?

[peak root]

Here sigma is the permutation of the set {1,2,3,...,n}, which means it is a bijection from {1,2,3,...,n} to {1,2,3,...,n}. So sigma(n)=1 means this permutation sends n to 1 in the set {1,2,3,...,n}.

[pliant haven]

ooh i see, thank you

[coral shale]

(no one saw anything)

I think the usage of word "isomorphism" to refer to a bijection of sets is confusing in this context

[peak root]

Sorry it's my bad. I should have say bijection. English is not my first language and I just couldn't find the correct word.

Updated

[coral shale]

Nice. My explanation would be:
A permutation is a bijection from a set to itself, hence the function notation

[delicate orchid]

woah!

[white oxide]

ladies and gentlemen

i don't need the structur etheorem for this problem right

[topaz solar]

I don’t think so?

Is PID even essential here hurmm

[white oxide]

ok good

[quiet pelican]

Doesn’t this work for any domain R?
Do we even need finitely generated?
And like
The proof is the same with those looser assumptions

[topaz solar]

Yeah that’s what I’m thinking

x in M -> nx = 0 for some n ye?

[quiet pelican]

Yes

[topaz solar]

Does it even have to have all zero divisors gone? (I mean ofc otherwise the fractions are scuffed but you know what I mean)

[white oxide]

what would a map from M to Q even look like? I'm assuming it's determined by the coefficients in any linear combination which make up an element m of M since the coefficients are in R

idk

[topaz solar]

Aight just making it wasn’t some weird coefficient thing, as opposed to some n thing

[topaz solar]

Consider, \phi(rm) = r \phi(m) yeah?

[white oxide]

yuh

that was the first thing i thought of

cuz like

Ya know

[topaz solar]

This is the definition

[white oxide]

torsion

well yes

definition of r-module homomorphism

[topaz solar]

Consider: multiplicative inverses

[white oxide]

hrmmmm

$\varphi(r^{-1}rm) = r^{-1}\varphi(rm)$

[cloud walrus]

okeyokay

[white oxide]

:kek:

[topaz solar]

Yes

[topaz solar]

What about n= n*1

[white oxide]

hmmmmmmmmm

ready for this? $\varphi(m) = \varphi(m + 0) = \varphi(m) + \varphi(am) = \varphi(m) + a\varphi(m) = (a \cdot 1)\varphi(m) = a\varphi(m) = \varphi(am) = 0$, for $a \in R$ such that $am = 0$

[cloud walrus]

okeyokay

[white oxide]

🤯

[topaz solar]

huh

[white oxide]

lmfao

[topaz solar]

why is phi m + a phi m = a phi m

[white oxide]

idk

[topaz solar]

Well I mean yeah since it’s all 0 but

That’s what you’re trying to prove

[white oxide]

oh oops

i forgot the distributive law

[topaz solar]

real

Anyhow, ye am = 0

And also

[summer path]

"Trick the grader" distributive law

[topaz solar]

what about 1/a phi (am)

[white oxide]

fax

[white oxide]

hm ok

[white oxide]

ye

i thought of this earlier but like

this is equivalent to showing that the exponents for every element of M are invertible right

or am i TRIPPIN

[topaz solar]

huh

That’s

Being a field?

[white oxide]

wait what

R a PID tho

[topaz solar]

Things are invertible

[white oxide]

and a in R

wat

wait why

integral domain is not a division ring right

[topaz solar]

Ah wait we’re not making this a Q module that way oop

But yeah things in Q are invertible

\phi(m) is in Q

[white oxide]

ohhh

ah

ok that makes sense then

[topaz solar]

So x\phi(m) is in R for some x ye?

[white oxide]

and yea i guess R is embedded into Q anyhow

yup

ok gotcha

[topaz solar]

And that’s integral

[white oxide]

proof should be easy now

hopefully

[delicate orchid]

|R| <= |Q| u heard it here first folks

[white oxide]

is it really that? $m \in M$, $a \in R$ such that $am = 0$, $\varphi \in \text{Hom}_R(M, Q)$ arbitrary, $\varphi(m) = \varphi(a^{-1}am) = a^{-1}\varphi(am) = 0$

[cloud walrus]

okeyokay

[delicate orchid]

if it's anything like the proof for the tensor version of this (which is should be they're adjoint) then yeah probably

[white oxide]

damn bro that's crazy

[topaz solar]

Well, you can’t just do 1/a ig since it’s not as a Q module, but that’s the idea

[white oxide]

foock

[cloud walrus]

Dragonslayer Sharp

Dragonslayer Sharp

[topaz solar]

It falls out from here

[delicate orchid]

so true bestie...

[topaz solar]

That’s how the tensor one goes, m (x) 1 = m (x) n/n = nm (x) 1/n = 0

[delicate orchid]

can we do this directly through the adjoint

[topaz solar]

directly

[topaz solar]

||na/b = 0 my child||

[delicate orchid]

yeah DUH

[topaz solar]

this is way nicer than the adjoint

But

[delicate orchid]

yeah but adjoint is cooler

[topaz solar]

Uhhh Hom(A (x) Q, B) ~= Hom(A, Hom(Q, B)) no?

[delicate orchid]

Hom(X,Hom(Y, Z)) = Hom(X \otimes Y, Z)

if we wrote them the wrong way around just pretend we're in R-mod^op or whatever

[topaz solar]

Issue is, that’s Q -> B

[delicate orchid]

pain and suffering

[tough raven]

M is not a module over the field of fractions so you can't apply a^{-1} to elements of the domain.
What you do have is
φ(m) = a^{-1} a φ(m) = a^{-1} φ(am) = a^{-1} φ(0) = a^{-1} 0 = 0,
choosing any non-zero a such that am = 0.

(That is, the idea is the same but keep the a^{-1} applied “outside the φ”.)

[white oxide]

so in this case it would just be $\frac{\varphi(am)}{a}$

[cloud walrus]

okeyokay

[white oxide]

ah

[topaz solar]

I mentioned this oop, after realizing it’s not tensor shenanigans

[cloud walrus]

Dragonslayer Sharp

[delicate orchid]

like the tensor one :breakfast:

wait

our ring is a domain right

[topaz solar]

Yes

[delicate orchid]

ok good

[topaz solar]

Otherwise Q is scuffed

[delicate orchid]

it's not THAT scuffed

[topaz solar]

Since, ya know, dividing by zero divisors bad

[tough raven]

Total field of fractions. 🥱

[topaz solar]

Disgusting

[tough raven]

Of course, who knows if a is invertible anymore?

[summer path]

[white oxide]

sorry how is a^{-1} \varphi(am) not the same as applying a^{-1} to elements of the domain? since \varphi(am) is in Q

[topaz solar]

Bro you can’t apply a^-1 to m

[tough raven]

Keep in mind that Q is the codomain here.

[topaz solar]

it ain’t a Q module chief

[delicate orchid]

I simply do not care. Multiply them.

[tough raven]

I'm saying here that a^{-1}m and a^{-1}am do not make sense for m in M.

[tough raven]

Extension of scalars lessgo

(inb4 the extension of scalars to Q is 0)

[white oxide]

ah gotcha thanks

[delicate orchid]

coextenstion of scalars

[tough raven]

Hmmm.

[delicate orchid]

has anyone actually ever performed a coextenstion of scalars

[tough raven]

Yes.

[topaz solar]

Poor soul

[delicate orchid]

I refuse to believe it

[tough raven]

There is a right adjoint to restriction of scalars, just like extension of scalars is the left adjoint.

[white oxide]

i need a hint!

😹

[quiet pelican]

||Every homomorphism sends the torsion elements to 0||

[white oxide]

bro got it instantly

[topaz solar]

I mean yeah

[quiet pelican]

I literally have she/her pronoun role

Please don’t call me that

[white oxide]

bro is gender neutral

but aight gotchu

[topaz solar]

consider \phi: M —> Q

[delicate orchid]

everything is gender neutral because we are not FRENCH with their stupid le la lalalala leellelele

[topaz solar]

Apply earlier problem

[tough raven]

IIRC, for R -> S, you send a left R-module M to Hom_R(S,M), where S is viewed as a (R,S)-bimodule.

[delicate orchid]

yeah it follows pretty quickly from the previous problem

[topaz solar]

Literally immediate

[white oxide]

:(

[topaz solar]

Enough clowning

[delicate orchid]

I love how u can just quotient by the torsion elements... R-mod is so nice...

[topaz solar]

What’s ker phi

my child

[quiet pelican]

Come on tablet
Die so I can do some category theory

[white oxide]

oh wait.... if I can show that Hom_R(M/M_tors, Q) is zero then I'm done!

wait nvm

M not torsion

oh wait....

[quiet pelican]

||What you want to show is that every element of Hom(M, Q) factors through the projection onto M/M_tors||

[white oxide]

thanks dudet i'll take a look if i'ms tuck

oh wait

[topaz solar]

Yes

[white oxide]

if M is torsion then it's trivial

holy shit

i'm a math god

[topaz solar]

but yeah what if M isn’t torsion

[delicate orchid]

very categorical way of thinking about it, I'd just construct an explicit iso lol ||which is basically the same thing actually - fair point||

[topaz solar]

What’s ker phi

[topaz solar]

I’d think about it that way tbh, iso theorem moment

[quiet pelican]

I mean #groups-rings-fields message

[delicate orchid]

I hate category theory. All groups are S_n and all categories are FinDimVect

[topaz solar]

Sylow in shambles

[tender kestrel]

Hello brilliant people! I need to decide between taking either differential geometry or topology class for this semester. I'd only take one before I graduate and might go to different school for masters where I'm not sure math class would be so available to engineering student. I want to study Lie groups and apply to robotics.

[white oxide]

I appreciate being called brilliant

[tender kestrel]

well okey what do you think?

[topaz solar]

Idk just take one and read a book on the other gg

[white oxide]

$\varphi \mapsto \varphi'$ where $\varphi'(m + M_{\text{tors}}) = \varphi(m)$

[cloud walrus]

okeyokay

[tender kestrel]

well, I think that's probably how I'll go about. Maybe overthinking too much lol

Thanks!

[void cosmos]

anyone

[rocky cloak]

This is not necessarily a division ring, for example of M equals R this just equals R^op

Maybe you want to assume R artinian and take M to be a minimal left ideal

[void cosmos]

yews

yes

but R is finite right?

or does that say nothing

[rocky cloak]

Yeah, if it's finite it's artinian, but you still need to assume M is minimal

[void cosmos]

yea ofc

[rocky cloak]

Also M won't be 1-dimensional in general. Can be n-dimensional for any finite n

[void cosmos]

so its simple as an R-modu;e

with n being the mulitplicity ?

[rocky cloak]

So you end up proving that R is the nxn matrix ring over D, then M will be n dimensional

This is also happens to be the multiplicity of M in R in the sense that R is isomorphic to M^n as a left module

And yes M is simple

[void cosmos]

yea got it

thank you king of kigns

[white oxide]

I had an oopsie moment while tutoring today

If W is the direct sum of a bunch of vector spaces are their pairwise intersections trivial or is there entire intersection trivial or is the intersection of all of them except one and that one trivial

😭😭😭

[rocky cloak]

So if you have A+B+C direct, then you will have that the intersections

(A+B) n C, (A+C) n B and (B+C) n A

Are trivial. In particular this also holds pairwise

[white oxide]

ah okay

Thanks

[rocky cloak]

Yeah, so the intersection of one with the sum of all the others is the thing that needs to be trivial.

Alternatively you can think of it iteratively. I.e.

A+B+C = (A+B)+C

So you want A and B to intersect trivially and you want (A+B) and C to intersect trivially, and that will be enough for it to be direct

[delicate orchid]

think of direct sums as disjoint unions of sets

[rocky cloak]

Might be dangouos, since if sets pairwise intersect trivially, then they are disjoint. Which is not true for vector spaces and direct sums

[delicate orchid]

I've sat here for 5 minutes now and cannot for the life of me figure out why that isn't true

is it an edge case with infinity?

[rocky cloak]

Consider 3 lines in R^2

Or infinitely many if you prefer

[delicate orchid]

I'm still not following, am I just getting my wires crossed that the forgetful functor Vect -> Set is additive?

ah yeah the underlying set of a direct sum is the product of sets

ok now I buy it

[rocky cloak]

Consider the 1D subspaces spanned by [1, 0], [0, 1] and [1, 1]

Yeah, forgetful functor preserves limits not colimits

[delicate orchid]

been working in too many categories with biproducts....

[rocky cloak]

Additive categories my beloved

[south patrol]

Every good forgetful functor has both left and right adjoints

[crystal turtle]

categories with biproducts

except we know the only good categories are the topological ones~ /s

[delicate orchid]

yeah, like Cat

[crystal turtle]

categories are just glorified topological spaces

[white oxide]

ah that makes sense thanks

why is this true? underlined in blue!

nvm

i guess it makes sense

[crystal turtle]

[white oxide]

cuz when you multiply it by p^s then you get p^r

which is an exponent for E

[white oxide]

W or L rhyme tho

can smbdy pls explain why we must have n + r - s \leq r

or how p^r being an exponent for E implies it

oh

well i don't rlly see how it implies it

wait

oh yea because ap^{n + r - s} is in (x1) and is not zero

?

oh yea and if it was greater than r then it would be zero which contradicts the fact that it's nonzero

or am i smokin

can somebody explain how this implies that the above term is a representative for ybar

is it because p^{s - n}cx_1 is in (x_1) and thus we have y + (x_1) where y is an original representative for ybar

and it's in x_1 because it's less than n

here, why can we conclude that $a_iy_i = 0$ for all $i$? $a_i\bar{y_i} = 0 \implies a_iy_i \in (x_1)$, which doesn't imply that $a_iy_i = 0$

[cloud walrus]

okeyokay

[white oxide]

because the kernel in $E/(x_1)$ is the elements sent to $(x_1)$ and not to $0 \in E$

[cloud walrus]

okeyokay

[coral shale]

looks like an independence condition

it says let the y bars be independent

in lemma 7.6 premise

[white oxide]

hm okay thanks

[crystal turtle]

that depends

[terse crystal]

We can’t know how many when the group is not given. Say it has m many, but we do know that m=1 mod p

By sylow

[crystal turtle]

for instance, C_4 (or Z/4) has 1 element of order 2, but V_4 (C_2 x C_2) has 3

so it depends on the structure of the group, but you should be able to determine this at least for certain types of groups. For instance, it'll be easier for cyclic groups

[south patrol]

Hm saying m=1 mod p by Sylow doesn't seem the best reasoning unless I'm confusing something?

[crystal turtle]

yeah that's for sylow P subgroups, right?

not necessarily p subgroups/ele's of order p

[terse crystal]

The proof of sylow by bourbaki didn’t say anything about the power right, p^r | |G|, the proof didn’t require r be maximal

[crystal turtle]

iirc it's for maximal exponent

[south patrol]

I think that is true though yes that is a good point

It is often just stated with maximal but holds more generally

[crystal turtle]

oh okay

[terse crystal]

Yeah. At least the proof by bourbaki didn’t require that. The excellent proof using group action one

[south patrol]

Wait 1 second

[crystal turtle]

but this is still not good reasoning

Because that counts subgroups

not elements of order whatever

[south patrol]

Well there is an elementary argument anyway lol

[crystal turtle]

For instance, Z/p has 1 p-subgroup, but p-1 elements of order p

[terse crystal]

Oh, yeah my bad… sorry, yeah not the same thing… a p order subgroup have p-1 many …

[south patrol]

I mean each distinct subgroup of G of order p will have p-1 elements of order p

and these are pairwise distinct

so we have n(p-1) where n is the # of such subgroups lol

But then hm # of such subgroups

I don't think there is any nice way to enumerate them right idk

[crystal turtle]

well if we don't know it for cyclic we definitely don't know it in general but it seems like we should be able to say something at least?

[terse crystal]

Yeah… can’t tell us how many, only to show that there exist, at least p-1 many…

[crystal turtle]

[terse crystal]

At least it’s a way. Right? equivalently hard😂. You know how many p order elements iff you know how many p order subgroups…

[terse crystal]

q is also a prime?

Yes I think so, assuming this q is also a prime

This result is good, we can show it has p-1 many order p elements without it though

(Because U(Z/qZ) is cyclic of order q-1, A cyclic group of n elements, any divisor d of n, there exists exactly one subgroup of order d)

(Actually if you are talking about U(Z/mZ) in general we always can calculate number of elements of certain order. U(Z/mZ) is a very good group, we know how it can be factored into cyclic groups)

[sonic coral]

how would i go about showing that all the non identity elements of H(F_p) for p>2 prime has order p. H is the heisenberg group of size 3. I know that H(F_p) has order p^3 but how would i eliminate the elements having order p^2 or p^3.

[white oxide]

yo

could i please get help with this problem

i have no clue where to start lol

offering payment of $20 for solution

[chilly ocean]

Let K be a splitting field of x^4+8 over Q. What is a basis of K
?

[white oxide]

is this a hint

[chilly ocean]

No unrelated

[white oxide]

lmao

[terse crystal]

You have complex eigenvalues $\sqrt{a}i$ and $-\sqrt{a}i$ now right?
When you have $A(x+yi)=(r+si)(x+yi)$ for complex eigenvalue $r+si$ and complex eigenvector $x+yi$. Notice that right now you have $A\begin{pmatrix}x&y\end{pmatrix}=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}r&s\-s&r\end{pmatrix}$

[cloud walrus]

Cogwheels of the mind

[terse crystal]

Consider basis of F[x]/(p(x)) where p(x) Is irreducible

[white oxide]

Thanks for the help, I gave up tho cuz it’s late at night and the problem set is due soon

I’ll just lose a few points it’s all good

I’ll prob return tho and follow ur hints

[rocky cloak]

Do you know the structure theorem?

[white oxide]

Yeah

[rocky cloak]

And you see how it applies?

[white oxide]

No

I don’t even know what Jordan form is

[rocky cloak]

You don't need the Jordan form

A module over R[x] is a vector space where x acts by some matrix

If that matrix is A, what do you think the resulting module should be?

The module will be ||R[x]/(x^2 + a)^2||

This has a cannonical basis ||1, x, x^2, x^3||,

||Expressing A in this basis|| is the R-canonical form

[white oxide]

thank you, i'll probably return to this problem later. my brain literally turns off at night and i can't do any form of math lol

[summer path]

On the other hand, my brain turns off during the day

[timber blaze]

Why

[topaz solar]

Well, ht is in H, and t^-1 k is in K

Does this make sense @timber blaze ?

[timber blaze]

yea i get that but why cant we do this with any t not in h intersection k

[topaz solar]

well, if t isn't in both, then ht and t^-1k wont be in H, K respectively

[timber blaze]

aha

thanks 👍

[hybrid pike]

My abstract algebra exam went well and I couldn’t have done it without help from the people here so ty all ❤️

[lime badge]

Quick question: Let R be an integral domain with field of fractions K and let M be a finitely generated R-module. The tensor product $V = M \otimes_R K$ is a $K$-vector space. I need to show $\text{dim}_K V$ equals the rank of $M$ as an R-module

[cloud walrus]

Eternal Way

[lime badge]

My intuition is just that I take the generating set of M and tensor by 1 but I wanted to make this more precise

[rocky cloak]

What is your definition of rank?

[lime badge]

Rank of a finitely generated module M is the max number of linearly independent of elements of M

[rocky cloak]

I see, so what you want to prove is that {m} is linearly independent iff {m(x)1} is

[lime badge]

We want to prove {m \otimes 1} is a maximal linearly independent set of V right?

[rocky cloak]

Yes

[lime badge]

Ok perfect, thanks!

[coral spindle]

Congrats!

[lime badge]

So showing the set of {m otimes 1} is linearly independent over K is not hard. As for showing it generates V = M otimes_R K, isn't this obvious by the structure of the tensor product?

[rocky cloak]

I wouldn't say that it's obvious, since {m} doesn't necessarily generate M.

What you can do is prove that this is a maximally linearly independent set, and then linear algebra gives you that it's a basis

[lime badge]

Wait, we're assuming M is a f.g. module with generating set {m_i}

[rocky cloak]

No, for example take R=Z and M = Z(+)Z/2

Then a maximally linearly independent set consist of the single element m = (1, 0)

Which doesn't generate M

[lime badge]

When my book says "generate", it merely means that there's a surjective homomorphism between M and the free-module on some indexing set A

[rocky cloak]

Indeed, that's the usual definition of generates

But a maximally linearly independent set need not be a generating set

As per the above example

[lime badge]

I agree. But I was trying to show that {m otimes 1} is a basis for V = M \otimes_R K over K

[rocky cloak]

Yes, when over a field maximally linearly independent implies basis implies generating

[lime badge]

Right, and every element of V can be written as $\sum_i k_i m_i \otimes k'_i$, where $k_i, k'_i$ are elements of $K$.

But $\sum_i k_i m_i \otimes k'_i = \sum_i k_i k'_i m_i \otimes 1 = \sum_i \tilde{k}_i (m_i \otimes 1)$

[cloud walrus]

Eternal Way

[rocky cloak]

That's true, but is something one would need to show

It's not immediately obvious

[lime badge]

Why does it not immediately follow from how the tensor product relations are defined?

In the sense that it should follow from the construction we use to show the existence of the tensor product in R-mod

[rocky cloak]

So the elements of the tensor product are sums of things of the form m(x)k for m in M and k in K.

So if m_i generated M, it would follow. But since there can be m not in the span of m_i, you need some argument for why m(x)k is in the span of m_i(x)1

For example in the above example if m is the element (1, 1) in Z(+)Z/2. How do you know it is in the span of (1,0)(x)1?

[lime badge]

But aren't we assuming {m_i} is a generating set for M? {m_i} isn't necessarily a basis, but that shouldn't matter for this

[rocky cloak]

No, that's not what we're assuming

We are assuming m_i is a maximally linearly independent set

[lime badge]

Ah damn

[rocky cloak]

Like in the example M = Z(+)Z/2, the maximally linearly independent set does not generate

[lime badge]

My bad. So I've showed {m_i otimes 1} is independent from the fact that {m_i} is independent. To show {m_i otimes 1} is maximal, I suppose we can do a proof by contradiction. I suspect proving the backwards direction here will help me with this right?

[rocky cloak]

That should work yeah

[lime badge]

And here, are you saying {m} is linearly independent over R iff {m otimes 1} is linearly independent over K or are both to be considered lin indep over K?

[rocky cloak]

Yeah {m} linearly independent over R iff {m otimes 1} linearly independent over K

[lime badge]

Ok, I showed this.

To show maximality of {m_i otimes 1} for V over K, we can do a proof by contradiction and suppose it isn't. Then we can add an element v of V to {m otimes 1} to complete it to a basis. v can be expressed as a linear combination of m' otimes 1 with coeffs in K, where m' are not necessarily equal to the m in {m otimes 1}.

So {v, m otimes 1} is in particular linearly independent over K. But then this implies {m' otimes 1, m otimes 1} is linearly independent over K which implies (by what we proved) that {m, m'} is linearly independent over R, contradicting the maximality of {m} in M

[rocky cloak]

Nice

[lime badge]

Thanks a lot for your help!

As usual 😄

[south patrol]

I believe a fancy way to do this is to show that taking exterior powers commutes with extension of scalars

Lol

[rocky cloak]

How do you detect the rank from the exterior algebra?

[lime badge]

I have another question. Let $k \subseteq k(\alpha) = F$ be a finite simple field extension. I need to show that $\alpha$ is separable over $k$ iff $F \otimes_k F$ is reduced as a ring.

Now its obvious that it wants me to do something like $F \otimes_k \frac{k[x]}{f(x)} \cong \frac{F[x]}{f(x)}$, where $f(x)$ is the minimal polynomial of $\alpha$ and use CRT to split up the factors of f(x), but my concern is that f(x) does not necessarily split over $F$

[cloud walrus]

Eternal Way

[south patrol]

Okay so tbh I don't have a proper answer lol it seems more tedious than I realised, but it should be the case (i suppose is the case) that for a module over an integral domain, the highest non-zero power is the size of the largest linearly independent subset

One of the directions is easy

But yeah tbh this doesn't seem like it'll be particularly pleasant to prove so this is a worse method nvm lol

[rocky cloak]

Hmmm, but something like (Z/2)^n
It should have the same exterior algebra over F2 and Z right

[south patrol]

Just thought it'd be a nice structural result but not applied to this specific case

[rocky cloak]

But the rank over Z is 0, while dimension over F2 is n

[south patrol]

Sure point taken

mb

[rocky cloak]

Would have been nice though

[south patrol]

Wait why is this the case

I guess sure like the Z action would factor through to give a Z/2 action

[lime badge]

Ah actually, I just need to argue that $F \otimes_k F \cong F[x]/(f(x))$ is reduced. If it wasn't, then it would have a non-zero nilpotent element $g(x) + (f(x))$ such that $g(x)^r \in f(x)$ for some power r. So $f(x)h(x) = g(x)^r$. But I can say alpha is a root of g(x) since f vanishes at alpha.

How can I use this to argue a contradiction with alpha being separable?

[cloud walrus]

Eternal Way

[halcyon peak]

consider the equation fh = g^r in a larger field where all polynomial splits, then all roots of f are roots of g, this can only happen when alpha is not sep, otherwise u get f divides g since all roots are distinct, which leads to \bar{g} = 0

conversely if alpha is not sep, assume chark = p, and f = g^{p^k}, it's easy to see \bar{g} is the nilpotent element

[lime badge]

So you're saying fh = g^r implies all roots of f are roots of g. Could you clarify why it contradicts the separability of alpha? If all roots of f are roots of g, g is in the ideal (f(x)) and so represents the 0 element in F[x]/(f(x)) after all

Or is that what you're saying? I guess I'm not seeing why this contradicts f having no multiple roots over its splitting field

[halcyon peak]

yes

if it has multiple roots then u cannot deduce f divides g

like f = (x-a)^p and g = x-a

[lime badge]

Ah ok, thanks!

And conversely, suppose $F \otimes_k F$ is reduced and assume by contradiction that alpha was not separable over $k$. Then it has multiple roots, so $f(x) = (x-\alpha)^t h(x)$ for some $t > 1$ and some h(x) that's relatively prime to $(x-\alpha)^t$. By CRT, one of the factors of $F \otimes_k F \cong F[x]/((f(x))$ is $F[x]/(x-\alpha)^t$ and this is clearly not a reduced ring

[cloud walrus]

Eternal Way

[alpine island]

I'm gonna start calling GCD domains GCDomains

[crystal turtle]

greatest common domains

[hollow mica]

has to be R

[spare totem]

I don't have lot of experience with analysis

I am wondering how can I approach this problem

by the way, deg is defined as deg(a + bi) = a^2 + b^2

[quiet pelican]

What
The
Why
That has better, more standard notation

[spare totem]

I do know that deg(r) will always be lower than deg(m)

[quiet pelican]

Also, without further restriction (I suspect there is one, it’s pretty standard) it should be 0

[spare totem]

so idk, probably 1

[spare totem]

oh can you explain why

[quiet pelican]

With the standard one, this is right

[spare totem]

uhhh

[quiet pelican]

Let q = 0, r = n \in \bZ and let m = 1
Then deg(m)/deg(r) = 1/n^2

If we require deg(r) < deg(m) which is standard, then it is 1

[spare totem]

oh okay okay

Thank you

I have this one more question, sorry for being so nosy about this

How do I talk about permutations and dihedral groups

having multiplication and addition operations

This group theory course (some online MOOC) was second course (but first for me) so I am probably missing something

[terse crystal]

What is the ring in those cases?

R[G], you gave us different G, but didn’t give us R

[quiet pelican]

The group ring is \bZ[G] where the multiplication is given by distributing and multiplying the basis vectors as if they were elements of G

[quiet pelican]

I’m going to assume over \bZ with no further context here

[terse crystal]

I see, makes sense

[spare totem]

OH

Thank you so much

[quiet pelican]

I suspect you get the same answer for any integral domain

(Well, the 3 non-integral ones are non integral for any R, not even necessarily integral, not sure about the integral ones)

[terse crystal]

Do we have R[G] isomorphic R \otimes Z[G] , otimes over Z?

[quiet pelican]

Yes I think we should
Yes

[terse crystal]

thanks… nvm, tensor product of two integral domains over Z is not necessarily an integral domain. I thought I wrote them as a tensor product then it is integral…

[chilly ocean]

Hello

Does someone knows more groups classified by the order? Like a group of order p is cyclic or group of order p^2 abelian, group of order pq with p<q and q is not congruent with 1 mod p is cyclic or group of order p^2 q with p<q and q non congruent with 1 mod p is abelian

This type of classification, not for groups with order less than 15 or something like this

[delicate orchid]

add more than two prime factors and everything goes to hell but uhhh maybe look at Burnside's p,q theorem

[chilly ocean]

What an inverse of 3 in F_9?

[topaz solar]

Is 3 in F_9

[south patrol]

There isn't one

3 = 0 in F_9

The issue is presumably that F_9 is not just Z/9Z, unfortunately

In fact, the fact that 3 does not have an inverse in Z/9Z is a proof that the latter is not a field lol

[chilly ocean]

How can F_(p^2) be a field? p and 2p don't have inverses

[chilly radish]

it's of characteristic p

Both of those are equal to 0 in that field

Did you read what potato said

[south patrol]

[chilly ocean]

Textbook says that any finite field is isomorphic to F_(p^n)

[south patrol]

Yes

[chilly ocean]

How many elements in F_9

[crystal turtle]

[wraith cargo]

determining the number of elements in F_(p^n) (field with p^n elements) is an NP-hard problem

[chilly ocean]

U said "with p^n elements "

[wraith cargo]

yes I did
So that means that F_9 has 9 elements

[chilly ocean]

Please tell me that every 9 elements explicitly

[wraith cargo]

I don't know if there's an explicit description of all the 9 elements

what you can do explicitly is see how the field is constructed

[chilly ocean]

What does F_(p^n) mean....

[crystal turtle]

Presumably you are reading a textbook. Have you read in the textbook how this field is actually constructed?

If not, that might be a better place to start than on discord

[south patrol]

Well we can give explicit descriptions for all 9 elements in particular models of it lol

but it can be constructed in various ways

and ultimately ig not that important since all isomorphic

[crystal turtle]

sure, you can give such a field structure on any set with 9 elements

[south patrol]

Yeah

[crystal turtle]

it's just not meaningful what the elements are if you do that. And the usual construction won't have a particularly enlightening description of the elements unless you actually just work it out

[south patrol]

As a group it is just Z/3 x Z/3, but you have to give it a different multiplication

[void cosmos]

why is an artinian primitive ring simple?

[void cosmos]

is htis even true

primitive rings have 0 jacobson cuz 0 is the annihilator of something

so artiniain primiitve rings are just semisimple ones no?

[sly crescent]

0, 1, -1, i, 1+i, -1+i, -i, 1-i, -1-i

Addition and multiplication work modulo 3 for the real and imaginary parts

[terse crystal]

People have classified groups of order p^r with r up to 7… I don’t know any reference but I am sure you can find many paper about r=3 case…

[hollow tartan]

Why is this true? |G| <= |Z|/4. For finite non abelian groups Z is the center. I was looking at the bound of the prob that 2 elements in a non abelian group commute

[crystal turtle]

Z is the center of G?

[hollow tartan]

Yes

[crystal turtle]

Z is subset of G. How is its cardinality larger

[hollow tartan]

Other way sry

[crystal turtle]

hmm okay a bit more reasonable, |Z| <= |G|/4. Under what conditions is this true?

nonabelian in general i guess?

I see

[south patrol]

Do you understand what they said before

[crystal turtle]

Yeah so if G/Z is cyclic, then G is abelian iirc

Which would be the contrapositive of the third line

[south patrol]

Yeah that is a nice problem to do

Not too hard

And then every group of order <= 3 is cyclic so |G|/|Z| = |G/Z| >= 4

[crystal turtle]

In particular, we know what groups of order 1, 2, and 3 look like

damn

got sniped

[hollow tartan]

Oh okay it's basically the point of non commutativeness

[crystal turtle]

it's exercise 3.1.36 in dummit and foote. I knew i had seen this before lol

[lusty marlin]

Does there exist a 'nice' characterisation of all algebraic numbers that are the root of some polynomial with non-negative integer coefficients?

[wraith cargo]

This would require a notion of an integral closure of a semiring that I'm not sure exists

[icy bear]

is $(a^{-1})^{n} = (a^{n})^{-1}$ only true for abelian groups?

[cloud walrus]

gabi the ancient

[crystal turtle]

Always

Even for nonabelian

[icy bear]

oh

Imma try proving it

oh it's actually kinda obvious damn

ty ryx

what's n(a) supposed to be?

[coral shale]

possibly order

[icy bear]

I supposed it's just an integer that changes with a but idk

[coral shale]

huh weird

oh right yeah yeah

n is a variable

that depends on a

weird notation

H = {a in G | exists n > 1 : a^n = e}

[icy bear]

I think the point is that the integer power is not fixed for all elements

[coral shale]

in words - H is the elements with finite order

[icy bear]

what's order

[crystal turtle]

Bad way of writing it but yes, elements of finite order

[coral shale]

Also, note this statement holds more generally. G need not be abelian

[crystal turtle]

No

I don't think so

[coral shale]

eh really?

ill have a think

[coral shale]

n = ord(g) is smallest positive n that makes g^n = e

[crystal turtle]

You might be able to come up with a counter example with matrices? Idk infinite nonabelian groups

[coral shale]

im lost ryx lol

i feel like the axioms hold - i rechecked kek

[icy bear]

oh

ty

[crystal turtle]

||I don't think in general if nonabelian that a and b having finite order implies ab does too||

[coral shale]

oh am i being dumb

shit ur right

[crystal turtle]

This is true for finite nonabelian groups for obvious reasons

[coral shale]

yh u can make a free group

[icy bear]

:0 first time I had to use (a little little bit of) number theory to solve a group theory problem, nice

[coral shale]

<a, b | aa, bb>

abab...abab is never the Identity

[coral shale]

im curious how

[icy bear]

I mean now I'm scared of having done something dumb but

I think I did actually

shit

or not

[coral shale]

just show, lets see

[lusty marlin]

What do you mean by 'require a notion of'? Using such a notion seems to just be another way of writing the set I described.

[icy bear]

my argument to say that it's closed, that is, (a.b)^k = e for some k is that using the least common multiple, you can multiply both n(a) and n(b) so that they become equal, so since the group is abelian, you can factor the multiple exponent

[coral shale]

nahhh...

[icy bear]

but I'm not sure of something

[coral shale]

ohhh

wait

[south patrol]

Or just n(a)n(b)

[coral shale]

ok yeah fairs

but like potato said

god this na nb stuff

[icy bear]

[south patrol]

Why sully lol

[icy bear]

idk I hate it when I over complicate my arguments without realising there's a much easier way

[coral shale]

eh its fine.

that was a decent idea

[south patrol]

Oh dw I mean tbf your thing is optimal

[crystal turtle]

Their least common multiple is the actual order of ab

[icy bear]

:0

[crystal turtle]

But just taking na nb is easier

[south patrol]

Assuming n(a) is the order of a ryx

xd

[coral shale]

well theres a few reasons

yeah

why that notation is bad

n(a) need not be unique, and isnt

[crystal turtle]

Wait I don't think so lol

[south patrol]

would be way easier to just say in words lol subsetof finite order elements

[icy bear]

order wasn't presented in the book yet

[coral shale]

thats kindof cursed

[south patrol]

Oh

[coral shale]

why would u not bring it up at the start

[south patrol]

Interesting this'd be before that lol

[icy bear]

this book is full of foreshadowing exercises and I think that's one of those

[coral shale]

also, to note: if g^n is never e, then we say "the order is infinite"

[icy bear]

I'm reeally in the start of the book yet

[crystal turtle]

Even if, take a = 1 and b = 3 in Z/4. Each has order 4 but they add to id

So nevermind that I just said some dumb shit

[coral shale]

im too lazy to think

[south patrol]

Yeah sure fair

[coral shale]

multiplying and adding gg

[icy bear]

this is an exercise of the subgroups section, it's really in the beggining

[coral shale]

its a pretty hard exercise

if its in the beginning

but alright

[icy bear]

nah

[coral shale]

i mean like - bringing up the notion of infinite groups this early lol

even if implicit

[crystal turtle]

Why must groups be finite

[coral shale]

im used to seeing finite groups introduced first

before moving on

because we do order, lagrange, etc

[icy bear]

it's all mixed up, most groups presented are finite but sometimes infinite groups appear

[crystal turtle]

This is true under some conditions like relatively prime order, at least for cyclic groups. I don't feel like thinking about it more though

[white oxide]

what they mean by "the most general combination of sets under which it will work"?

[open sluice]

you want associativity to hold for as many sets as possible

I don’t know the context but it sounds like there are some sets this does not hold for

so try and come up with a general rule that permits associativity

[white oxide]

hm ok

ty

[coral steeple]

Background: we proved that if a,b are positive integers, then aZ+bZ=dZ for some integer d. In my proof, I showed this constructively by defining $d=\operatorname{min}(a\mathbb{Z}+b\mathbb{Z}\cap{1,2,3,\dotsc})$. We then defined gcd(a,b)=d.

[cloud walrus]

person2709505

[coral steeple]

I would like to know if there is a more self-contained solution possible than this:

I know it is possible to show that $g:=\prod_{p\textrm{ prime and }p\mid a,b}p$ is the largest integer that divides $a,b$, so we obtain $d\leq g$. Supposing $d<g$, use division with remainder to see that $g=qd+r,; 0\leq r<d$. Rearranging, $g-r=qd$, so $g-r\in aZ+bZ$. Since $r$ cannot be in $dZ$ by definition of $d$, $g$ is in $dZ$ only if $r=0$, i.e. $g=d$. The proof now reduces to the classical Bezout's identity.

[cloud walrus]

person2709505

[coral steeple]

I have been thinking about this on and off for months now lmao

And if this belongs in #elementary-number-theory lmk

Oog this doesn't even work nevermind

[white oxide]

here, do we have to use the fact that Z is a noetherian domain?

[coral spindle]

You can if you want

It is a convenient way to prove that

[white oxide]

but it's not necessary right because i don't remember the argument at all lol

or wait no the euclidean algorithim has to terminate

[south patrol]

Hm I wouldn't both to consider the noetherian property

[white oxide]

yea i guess the exercise is assuming that i don't know it

which tbf i'm not that familiar, something something ascending chain condition of ideals

any ascending chain of ideals is finite there we go

i'm a little bit confused about the nature of the hint tho since it doesn't apply to any arbitrary ideal in Z

hm

[rocky cloak]

I guess you want to use the fact that the natural numbers are well ordered.

[white oxide]

huh okay

i'll consider that

[rocky cloak]

The hint will give you that any finitely generated ideal is principal. Then to rule out infinitely generated ones you can use that natural numbers are well ordered.

Here you could also use that Z is Noetherian I suppose. Though I feel like you would use this exercise to prove that fact, so...

[south patrol]

Tbh this is a bit silly imo lol like the easiest method doesn't use the Euclidean algorithm

[white oxide]

so i would have to extend it by induction? because the hint only shows that it's true for ideals which are finitely generated by 2 elements right

that was my thought

[south patrol]

The hint I would give (which is really not much of a hint): If I is an ideal and I = aZ, what is a way to characterise a in terms of I? Think about this lol

[south patrol]

Well (a,b) is simply aZ + bZ so the induction would be virtually immediate

[rocky cloak]

Yeah, you would use induction

[rocky cloak]

How can you do this without using the Euclidean algorithm?

[south patrol]

Well I would merely use the fact that Z is Euclidean and to show Z is Euclidean doesn't actually need the Euclidean algorithm right, only the division algorithm

Anyway this has the advantage of not needing to worry about # of generators lol

[rocky cloak]

Oh yeah sure, I was just using "the Euclidean algorithm" and "Euclidean division" interchangeably here

[south patrol]

Ah okay

[rocky cloak]

But yeah you only need Euclidean division

[south patrol]

Yeah the way the question is described implies they mean the actual algorithm

I believe

[white oxide]

so if it were the case that if I were infinitely generated, then I must be Z?

[rocky cloak]

Well Z is not infinitely generated (generated by 1)

[white oxide]

ye but like

well i guess you could have countably many elements generating the ideal excluding 1

cuz i'm just trying to rule out the case that proper ideals of Z are infinitely generated

[rocky cloak]

I mean you can always add redundad generators. There isn't really a distinction between proper and non proper ideals there

All ideals are finitely generated, so in particular the proper ones are

[white oxide]

oh i didn't know this, i should probably prove it beforehand then

[rocky cloak]

I'm confused, all ideals of Z are generated by a single element. That's the exercise you're doing

If it's generated by a single element that's finitely generated

[white oxide]

ya but isn't that assuming that's the case before actually proving the claim if that makes any sense

sorry i'm probably confused

[rocky cloak]

Sure, I'm just saying that Z is not infinitely generated

[white oxide]

ye

ohhh

so proper ideals can't be infinitely generated

?

[rocky cloak]

Like in general they can, but the statement "if I is infinitely generated, it must be Z" is strange

[white oxide]

yea i suppose

yea

[rocky cloak]

Okay, I'm sorry. I threw you off I guess

[white oxide]

nah that's my fault, thank you

[white oxide]

ah because if we're assuming that $(a_1, \dots, a_{n - 1}) = (d)$ then $(a_1, \dots, a_n) = (d, a_n)$ which is principal by the base case?

[cloud walrus]

okeyokay

[south patrol]

Yup

[white oxide]

sick

wait this seems trivial but i'm still confused on why no ideal of Z is infinitely generated

assuming that we don't know that Z is a PID

is it just because Z is generated by 1?

oh wait any ideal which is infinitely generated must be contained in (1) which is finitely generated oops

[topaz solar]

[white oxide]

wit

wait

[rocky cloak]

Since the natural numbers are well ordered any ideal has a minimal positive element

Are you able to use that to prove it is principal?

[topaz solar]

This feels way easier than, ya know

Invoking Euclidean shenanigans

[rocky cloak]

You still have to invoke Euclidean shenanigans

[topaz solar]

Significantly less that what it seems like they were suggesting

In the problem hint

[white oxide]

i'll attemtp

[white oxide]

Well the thing is

Oops

Replied to wrong person

[white oxide]

I already proved Z is a PID under the assumption that every ideal is finitely generated

[rocky cloak]

An interesting similar example is Q as an abelian group. For any p and q rational numbers there is an integer x and rational r such that

p = xq + r

With |r| < |q|.

With this you can prove that any finitely generated subgroup of Q is principal, but there are infinitely generated subgroups of Q

(in particular Q itself)

[delicate orchid]

Z is infinite so if an ideal is infinitely generated it must be (Z) :iwon:

[south patrol]

ye thi is what i was saying lol

it weird

[white oxide]

lmao this doesn't work right

[topaz solar]

Just ||bezout that mf, and/or alternatively see that min N \cap I is very generator lookin?||

[south patrol]

Ye

[wet zodiac]

where should i start

[delicate orchid]

well what's the first map that comes to mind

actually no - what do you think S^-1R looks like, keeping the hint in mind

[wet zodiac]

it should look like uhh

since S is closed under multiplication it should look like a ring RxS with addition defined as (r,s)+(t,u)=(ru+ts,su) and multiplication as (r,s)(t,u)=(rt,su)

[delicate orchid]

uhh lemme think for a sec

[wet zodiac]

with the same equivalance relation we have as in the hint

[delicate orchid]

yeah that's right

[delicate orchid]

so?

[wet zodiac]

[delicate orchid]

consider the fact that the underlying set is RxS

[wet zodiac]

wdym

[delicate orchid]

so there's a pretty obvious set map from R into RxS

which you can show is a ring homomorphism

[wet zodiac]

R -> RxS via phi(r)=(r,s) for some fixed s?

[rocky cloak]

You may want to adjoin a certain nice element to S, before doing the construction

[delicate orchid]

oh whoops I thought we already had that!

yes, definitely - my apologies

[wet zodiac]

what

[rocky cloak]

What is a very nice element of R when it comes to multiplication?

[wet zodiac]

1

but why adjoin 1 to S

oh wait a minute

[delicate orchid]

yeah wait a minute

1 being in S doesn't follow from being multiplicatively cloesd

[rocky cloak]

You don't have to adjoin 1, it just makes it easier

[white oxide]

oh wait i got it

consider min{I \cap N} = k

[delicate orchid]

who tf cares about multiplicatively closed sets that aren't monoids bro on the 9 circles of heaven

[white oxide]

if g is in I such that g is not a multiple of k write g = kq + r where k > r >= 0

but this implies that r is zero since k is minimal

i believe

[rocky cloak]

You got it!

[delicate orchid]

I actually do not know how localisations work without a 1 lol I'll leave this to the experts

[rocky cloak]

|| sr/s = r/1||

[delicate orchid]

||but there is no 1||

[rocky cloak]

You may have a problem since they never assume S to be nonempty though.

[rocky cloak]

Exactly, and this shows you don't need 1

[delicate orchid]

but you ||wrote a 1 in!||

||how can you "cancel" the fraction without a 1 in S?||

[rocky cloak]

Yes ||things with 1 can also be written without||

[wet zodiac]

i will get ice cream and return to this problem

i need sugar

[white oxide]

wait yea

wtf

why did i need euclidean algorithim for this shit

[delicate orchid]

the non-empty thing is far more worrysome though

I'll admit I just don't understand how you're supposed to do it without 1 but doing it with the empty set seems suspcious

maybe you shouldn't have definitions be exercises in books you know what a crazy concept

[rocky cloak]

||pick any s in S, then the map is r |-> sr/s||

[delicate orchid]

hmmmmmmmmmmmmmmmmmmmmmm

[rocky cloak]

I mean you can localized with respect to any set just using universal property, but then the construction won't be of the form RxS

[delicate orchid]

I suppose that's the important part of the localisation ig

do you still get all the nice exactness properties with a general set

[rocky cloak]

I mean the construction will just be take the multiplicatively closed set (with 1) generated by S, and proceed

So yeah, should have the same nice properties

[chilly ocean]

If H is a p sylow subgroup of a finite group G and K is a subgoup of G then always H intersected with K is a p sylow subgroup of K?

[quiet pelican]

Consider G = S_7, H = <(123), (456)>, K = <(123), (457)>
Then K is the sylow p subgroup of K, but H intersect K is not K
More generally, if H and K are distinct sylow p subgroups with non-trivial intersection this fails

[chilly ocean]

Thx

[crystal turtle]

For which p is this?

[quiet pelican]

3

[delicate orchid]

pretty clearly p = 3

[crystal turtle]

...

Duh

I was just checking it micose knew