groups-rings-fields

[hazy lion]

it's not too bad illuminator

Z_p is the closure of Z

since (-)a is continuous

Z_p a=(Z_p)a =(Z closure)a is contained in closure (Za) contained in closure (ZH) contained in H if a is in H

and if Z_p a is contained in H for all a and H is a subgroup it's clear H is an ideal

[delicate orchid]

ah I think that's what I was trying to get at

very cleaver....

[formal ermine]

what does this prove

[teal vessel]

can't I just say the set G of all generators (of this group) is G={x | (x,202)=1}={2n+1| n in N, n<100, n=/= 50}? I don't feel like writing out all 100 odd non-101 numbers.

[hazy lion]

n.b. famous and easy lemma, a map f:X-->Y is continuous if and only if f(closure K) is contained in closure (f(K)) for all K sub X

[delicate orchid]

if x_n converges to x' in H then ax_n converges to something else in H

[hybrid pike]

So I'm asked to determine all finite groups with exactly two conjugacy classes. Am I correct in thinking that, since one of the conjugacy classes is always the singleton {e} that the other should contain the other n - 1 elements. But since the size of a conjugacy class is a divisor of the size of the group, this is impossible, so there are no such finite groups?

[hazy lion]

generators of Z/nZ are just the elements of the multiplicative subgroup

[delicate orchid]

C_2 has two conjugacy classes

[hazy lion]

exactly phi(n) of them

[hybrid pike]

Oh, duh, if there's only 2 elements then it does work out

[delicate orchid]

there you go

[hybrid pike]

Thanks

[delicate orchid]

good proof otherwise!

[crystal turtle]

average wew proof

[teal vessel]

I mean, yeah, that's basically what I'm saying.

but for the specific case

[delicate orchid]

I have not seen a sequence outside of a colimit in 3 years

[crystal turtle]

[coral spindle]

ewwwww your topology is based off analysis ewwwwwwwwwwwwww

[delicate orchid]

no my topology is based off of functors

and natural transformations

[coral spindle]

Oh you have a closed set? What polynomials does it satisfy?

[coral spindle]

b....bbased.....

[delicate orchid]

the only topological space I need is \mathcal{N}(E(G)) can a brotha relate

[coral spindle]

No sorry :(

[hazy lion]

nerve of (something) of a group G?

[delicate orchid]

that's ok I don't think that is what I meant anyway

unviersal covering space of BG is what I meant

[crystal turtle]

anyways @formal ermine this I think is a clear explanation if you don't understand Wew's

[delicate orchid]

E(G) is the category with Ob(E(G)) = {g \in G} and one morphism between two dudes (given by multiplying by g^-1g')

[delicate orchid]

yeah my proofs universally have a very "stream of consciousness" writing style that the literature PhDs don't take too fondly to

[coral spindle]

Two dudes having a morphism? I'm happy for them

[delicate orchid]

two dudes sitting in a hot tub five morphisms apart cause they're not gay

[formal ermine]

ahh yeah this makes more sense

but what was the book going for with the first thing

[hazy lion]

I explained...

[crystal turtle]

same thing

[hazy lion]

you can't just go straight to Z_p a contained in H for all a in H

you need to use the argument about continuity

[crystal turtle]

god, it would help to read the proof before I start memeing on illum huh

[delicate bloom]

I thought the multiplication part was just to get that it's more than a group, it's an ideal

[formal ermine]

right

[delicate orchid]

oh wait yeah this is really obvious, if the sequence isn't eventually constant it has to go to zero

which is in H

[formal ermine]

wait okay really dumb question but how do we know that it's continuous from that

[delicate orchid]

a function is continous if and only if it commutes with limits

[delicate bloom]

Lipschitz continuity

[formal ermine]

ohhhhhhhhhhhhhhhh

I didn't see that as a function

that makes a lot of sense

[delicate orchid]

yur

[hazy lion]

saw that , you said functor at first

[crystal turtle]

i mean also true

[formal ermine]

thanks guys

[hazy lion]

the function x ---> xa

[formal ermine]

yeee

[delicate orchid]

I said function changed it to functor then changed it back as a little treat for the fans in the back

[delicate bloom]

|f(x)-f(y)|=|a||x-y| with |a|<=1 means |f(x)-f(y)| <= |x-y|, you can think for all epsilon you pick delta=epsilon now

[formal ermine]

wait x -> x|a| or x -> xa?

[delicate orchid]

x -> xa is continous

[crystal turtle]

I hope x -/-> x|a| since they are not in the same set

[delicate orchid]

I don't want to think about interpreting |a| p-adically

[formal ermine]

oh

lmao

ic

[crystal turtle]

nor does that make sense really (i guess they are in the same set tho in Q_p a tleast idfk)

[teal vessel]

"find the number of generators for Z/49000Z"
duh, just phi(49000)

yes, that can be calculated (16800) using prime factors and such, but that seems quite the dull problem lol

[crystal turtle]

yeah

[formal ermine]

,, phi(49000)

[formal ermine]

fuck

wrong bot

,w phi(49000)

[crystal turtle]

"dull problem" yeah that sounds like D&F

[cloud walrus]

Results provided by WolframAlpha

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

Requested by projective

[delicate bloom]

you should think of it as a nonnegative real number yeah

[hazy lion]

x ---> xa of course.

[formal ermine]

yeah ok

[delicate orchid]

x ----------------> xa naturally

[teal vessel]

at least it reminded me how to calculate the totient function

[hazy lion]

we don't want Z_p |a| in H, we want Z_p a in H

cat very sad, was discriminatory against |a|

[teal vessel]

we love permutations of 4 letter words lol

[solar trench]

are there any of the topics here that i'd be able to handle after only learning group theory(part 2 in this book) the book is by gallian

like i know sylow theorem could be covered with only knowledge of group theory but for rest of the chapters i'm not so sure

[delicate bloom]

idk try reading and finding out

[delicate orchid]

all of them except maybe 32 and 33 if you don't know about fields

but yeah

just give it a go

[crystal turtle]

at least the first couple yeah

[solar trench]

oh okay thanks

i was self studying gallian and am quite a bit through part 2 on groups

[delicate orchid]

Crystallographic groups immediately after generators and relators 🤨 interesting pedagogy

[solar trench]

wondered if i could study some part 5 topics before going on to rings

[crystal turtle]

surprised they have Sylow theorems in "special topics"

[delicate orchid]

yur they're fairly fundemental

[high peak]

Hmm a lot of symmetry

[summer path]

how is this characterization of projective modules different from free modules?

is it just that there's no guarantee on uniqueness?

[coral spindle]

Well maybe an example would be good

Consider the projective Z/6-module M = Z/3

There's just a single map phi you need. It'll send 1 in M to 2 in R

So it's a bit like a free module, but there are coordinates missing.

Does that kinda make sense?

But yes in essence it is the uniqueness that causes issues. If the values phi_i(x) were unique with that property then they would be free.

[summer path]

what do you mean by some coordinates are missing?

[coral spindle]

So ok.

R = Z/6, M = Z/3.

My elements will be just the element 1 in M

The map phi will be phi(1) = 1

Wait hold on I just need to think for a sec. I'm making a mistake but I need to check where

[summer path]

phi(1) = 2, phi(2) = 4, phi(0) = 0? or is that not it

[coral spindle]

OK I'm going to write M in a different way to make something a little clearer.

I'm going to write M = 2Z/6Z. It's the same module but writing it as Z/3Z obscures the action of the ring a little bit.

[delicate orchid]

M = {0+6Z,2+6Z,4+6Z} if you want to be even more explicit

[coral spindle]

So this time the element that I'm looking at is gonna be 2 in M. The corresponding map phi : M → R ought to be described by phi(2) = 4. Note it's 4! It's a bit weird Then it should be clear that, phi(2)2 = 4x2 = 2 in Z/6Z, phi(4)2 = 2x2 = 4 in Z/6Z. (fixed a bad typo)

So if we wanted to think of these a_i as a 'basis' and the phi_i as functions picking out the coordinates, then there are coordinates 'missing': in the projective Z/6-module 2Z/6, there is no element m with phi(m)=1, for example.

[summer path]

why is 4*2 = 6 in Z/6Z

[coral spindle]

What's 4 times 2

[summer path]

8

[coral spindle]

Oh did I just write 6 instead of 2 mb

[summer path]

lol

[coral spindle]

It should ofc just be 2

[summer path]

ye ok

[coral spindle]

by the formula we need to hold

Anyway you haven't really talked about the substance of my explanation so if you do have any qns about that then ping me or sth idk

[summer path]

why is it that phi(2) = 4 and not 2?

[coral spindle]

Doesn’t satisfy the definition

[summer path]

can you not have phi(2) = 2, so that phi(2)*4 = 8 = 2 in Z/6Z and phi(4) = 4, so that phi(4) * 4 = 4 in Z/6Z?

[coral spindle]

I need to go so I will respond later

[summer path]

ok

i will go take a nap

[next obsidian]

@summer path you can’t map 2 to 2

This is Z/3-> Z/6 right?

[summer path]

ye

[delicate orchid]

defines the unique map from {2} to itself

[next obsidian]

1 has to to go to something which satisfies 3x = 0 because 3 = 0 in Z/3

This forces 1 to go to either 2 or 4

[summer path]

well rather it's 2Z/6 -> Z/6

[next obsidian]

Okay then 2 can go to 2

[coral spindle]

I fixed a bad typo in the message, funnily enough the arithmetic worked out

Might want to check again

[next obsidian]

Anyway uh Boytjie

Why is Z/3Z a projective Z/6Z module?

I do not believe it is

It has torsion lol

[delicate orchid]

any torsion enjoyers?

[summer path]

I thought it is because it's a direct summand of free module

[void cosmos]

i have been

stuck on a problem

[next obsidian]

How is it a direct summand?

[void cosmos]

for 1 hour

[delicate orchid]

if it has torsion it can't be

[next obsidian]

There’s also a characterization for when a quotient of a ring is projective

And it’s when the ideal you quotient by is principal, generated by an idempotent

eg it must be of the form R x S -> R

[summer path]

Z/6 is free module over itself right?

[next obsidian]

Yes

[delicate orchid]

wait chmonkey does it have torsion?

[next obsidian]

1 is killed by 3

[delicate orchid]

3 is a zero divisor

[next obsidian]

Hmm

[summer path]

Then Z/6 = Z/3 \osum Z/2, it's fine?

[next obsidian]

[delicate orchid]

yeah I agree with :3c on this one now

[next obsidian]

What the fuck is (0,1) under that decomposition?

It has to square to itself mod 6

It’s 3

[delicate orchid]

4

[next obsidian]

Hurb

[delicate orchid]

I suppose 4 would be (1,0)

[next obsidian]

Indeed

[delicate orchid]

also chinese remainder theorem lol

[next obsidian]

Okay nvm

Also yeah

[delicate bloom]

#elementary-number-theory

[next obsidian]

😂🔫

[void cosmos]

are you guys done

[summer path]

I thought (1,0) would be 2

[next obsidian]

It doesn’t square to itself

[summer path]

Oh wait

[delicate orchid]

I'm never done I'm on the run I got a gun fly to the sun yuh yuh yuh yuh

[summer path]

[void cosmos]

okay

a ring R is left noetherian iff it satisfies the property that the direct sum of any family of injective R-modules is injective

i proved the --> part

<-- is impossible form e

for me*

[next obsidian]

Lmfao

[summer path]

Is that why phi(2)=4 and not 2?

[next obsidian]

This one is gonna be rough

[void cosmos]

how do u know

so fast

did u see it b4

[next obsidian]

Idk ¯_(ツ)_/¯ I don’t understand

Yes I saw it before

[void cosmos]

its on a past paper

on an exam i am going to be taking

so

rip

[chilly ocean]

just see it then too

[next obsidian]

You show that every ideal is finitely generated by assuming the opppsite

[void cosmos]

yea thats what i had in mind

[coral spindle]

I think the answer is basically 'Chinese Remainder Theorem' lol

[void cosmos]

and then u get some chain

[next obsidian]

and doing some crap taking injective hulls to extend a map and shit

[summer path]

Boy said phi(2) can't be 2, but idk why still

[void cosmos]

ik that (n) is an injective R-module

right?

whenever n is in R

cuz u can extend any map

by sending 1 to n

[next obsidian]

If ultimately the thing wasn’t finitely generated you would end up with an element inside of an infinite direct sum which is non-zero in infinitely many places

[coral spindle]

Like there's a decomposition of 1 in Z/6Z as the sum of two idempotents, and the relevant idempotent here is 4, right?

Yeah there you go

4 is the idempotent that works

[void cosmos]

okay first

can u give any counterexample

[coral spindle]

1 = 3 + 4 describes the isomorphism Z/6Z = Z/3Z (+) Z/2Z

[void cosmos]

if R is not noetherian

[coral spindle]

So that's where it comes from

[void cosmos]

just a direct sum of injective modules

that need not to be injective

i am terrible with examples for injective modules

i can only think of divisible abelian groups

[delicate orchid]

wait... the chinese remainder theorem.... is an idempotent decomposition....

can I do block theory on this

[coral spindle]

Babe yes

[void cosmos]

any hot babes?

[next obsidian]

Let me think of the proof

[void cosmos]

okay king

thank you for yoru time

your*

i literally googled the whole fucking problem

and i just got a paper that describes some results from the theorem

rather than the main theorem itself

and i found some stackexchange post on whatever

[dim widget]

This whole conversation is gold

[next obsidian]

[void cosmos]

this is so fucking c honestly

[coral spindle]

Hehe just saw this

Hehehehehehe

[void cosmos]

yea its just cuz Z/6 is jusut Z/2 and Z/3

as Z/6 modules

both Z/2 and Z/3 are projective Z/6 modules

[delicate orchid]

so true bestie...

[void cosmos]

yea algebra is p hard tbh

i am in semisimple rings and just

everything is collapsing

[delicate orchid]

semisimple rings are the nice ones

[next obsidian]

Okay yes

[dim widget]

The issue is that the representation is not unique, this is a rephrasing of what it means to be a summand of a free module

[next obsidian]

So take an infinite ascending chain I_0 < I_1 < …

[delicate orchid]

"every module is projective" uhhh yes please???

[void cosmos]

ok

[next obsidian]

Let I = U_0^infinity I_n

Let E_n be an injective module containing R/I_n

Consider the map f:I -> (+)_0^infinity E_n

[void cosmos]

what

I to where

whats that

[next obsidian]

I -> R -> R/I_n -> E_n

It’s the map defined like, into the direct sum of all of these

[void cosmos]

u direct summed all of the E_ns

over n?

[next obsidian]

Yes

[void cosmos]

okay so this is injective

[next obsidian]

This actually lands in the direct sum

[void cosmos]

so ur seq is exact

[next obsidian]

What

No

[void cosmos]

.

okay

[void cosmos]

wdym,

[next obsidian]

The image of everything actually will have all but finitely many coordinates be 0

Because any x in I is only not in finitely many I_n

[void cosmos]

okay so

slowly cuz im bad

I is the union

of ur ascending chain

so any element of I

would not be present in only finitely many I_ns

[next obsidian]

Yes

[void cosmos]

so the image of that x is?

[next obsidian]

x + I_n

And then we include R/I_n into E_n

This gives us maps I -> E_n which then gives a map I -> Prod_0^infinity E_n

But the image actually lands in the direct sum

[void cosmos]

wdym actually lands

yea i can see the maps now

so u get a map from I to ur infinite direct sum

[next obsidian]

Yes

But you get why it’s the direct sum and not the direct product

[void cosmos]

okay so f(x) is in that direct sum

is it cuz of the map E_n --> direct sum ?

like the inclusion map?

or what :d

[next obsidian]

Do you know what the difference between the direct sum and direct product is?

[void cosmos]

yea

[next obsidian]

Then it’s because each element lies in all but finitely many I_n

So the result f(x) is 0 in all but finitely many places

[void cosmos]

oh yeah

yeah mb i just thought of

that htey are dual thats it i forgot about that

cuz we are

dealing with

infinite

indices

mb sorryh

okay so yeah this f(x) is in the direct sum now

[next obsidian]

Okay so the direct sum is injective and so this map extends along I -> R to give a map f:R -> (+)E_n

[void cosmos]

yea

[next obsidian]

Now consider f(1) = e

[void cosmos]

whats e

[next obsidian]

f(1)

[void cosmos]

okay

[next obsidian]

I just called it e

That’s it

[void cosmos]

ok

[next obsidian]

Whatever, so for n > N, e_n = 0

[void cosmos]

yea

for some N

[next obsidian]

So now, for x in I, f(x) is equal to (x + I_0, x + I_1,…)

By construction

But also f(x) = x•f(1) = x•e

So this means for all n > N, f(x)_n is 0

This means that x + I_n = 0

But this holds for all x in I, which forces I = I_n for all n > N

[void cosmos]

yea

damn

[next obsidian]

But that contradicts the fact that the chain was an infinite ascending chain

[void cosmos]

yea

yea this is insane honestly i would never have got that

gotten

okay so you supposed you have an infinite ascending chain , let I be their union , then any element of x would not be present in (atmost) finitely many I_n, okay so define E_ns to be the injective hulls of R/I_ns , so we have I --> R/I_ns --> E_ns ---> infinite direct sum of E_ns ( which is injective )

so we can extend the map to the whole ring to the direct sum of E_ns

since this is an infinite direct sum f(1) would be 0 for all but a finite number of coordinates ?

is that correct

incorrect? :d

[long geyser]

why is it that if sigma is in Gal(E/k) then its restriction to B is an automorphism of B? is this always true of field isos?

also idgi why lemma A-3.98 lets us deduce the rho map is surjective, here's the lemma in question

also where do they use that E/k is normal, I'm getting KO'd by this proof

[dim widget]

If L/E/K and L/K is galois and E is normal, then the restriction of any \sigma: L \to L to E gives an automorphism of E

This is because \sigma permutes the roots of polynomials with coefficients in K, and by the definition of normality E is the splitting field of a polynomial with coefficients in K, so it is generated by the roots of this one polynomial

[long geyser]

right, if alpha is a root of f so is sigma of alpha, so you have to send E to E
I think I got confused because a result like that was proven in narrower scope

well, makes sense because it only works here because B is normal and normal extensions just got defined

still dont understand how the surjectivity of rho follows from that lemma and I still don't think we've used the normality of E/k yet?

[rocky cloak]

Any polynomial has only finitely many roots and for functions on finite sets injective=surjective

[long geyser]

could you clarify further?

rho is not injective either

[rocky cloak]

Ah, sorry I misread you. Yeah for surjectivity you want to use that you can always extend homomorphisms in normal extensions

You can do this by induction / Zorn's lemma

Proof is similar to how you would prove that algebraic closures are unique

[long geyser]

wait wait, you are spiraling down a path that seems too complicated for this question

I think the logic is that E/k is normal, so it is also normal over E/B (?) now we think of B as base field and we have iso B -> B, so there is an extension E -> E by lemma A-3.98

okay I think this works

[rocky cloak]

Yes exactly

[long geyser]

alright

thanks

[alpine island]

Can you get a "feld" by taking the definition of a field and removing the necessity for a multiplicative identity, like how you get a rng by removing the identity from a ring?

So addition is a group, multiplication is an associative quasigroup

[solar vessel]

if there is no identity

then what is meant by invertibility

ah

quasigroup

[coral spindle]

Well yeah you certainly could define that

And it would still have exactly two ideals if it’s nonzero

I have no idea how well-behaved it would be. The modules over such a thing may be a bit freaky

[south patrol]

i mean isn't that just a semiring

or am i dumb

oh quasigroup lol

[coral spindle]

Yeah

[south patrol]

then that's not really removing it from the definition but doing smth else imitating a field lol

[coral spindle]

It would be nice to find an interesting example of such a thing. I'm struggling to come up with an example in my head.

[solar vessel]

wait are we assuming multiplication is commutative?

if that's the case I think if it's nonzero it's just a field

[alpine island]

This just randomly popped into my head while walking btw, there is no motivation here

[solar vessel]

because like

solving xa=a

and yb=b

you get

xab=ab

and also

my wifi died back on phone

anyway

[alpine island]

Yeah I can't think of anything either

[solar vessel]

yab=yba=ba=ab

[coral spindle]

Nah I think it was discord

[solar vessel]

so x(ab)=y(ab)

and I'm pretty sure ab isn't zero provided a and b aren't

so x=y

i.e. the multiplicative identities for each nonzero element are the same

[coral spindle]

Nice argument. And indeed such a per-element identity exists. Neat.

[solar vessel]

actually not sure sure about this now

[coral spindle]

Well it's required, right?

[solar vessel]

oh

right yeah

[coral spindle]

We require F\{0} to be the quasigroup, so it's by definition

[solar vessel]

nonzero elements quasigroup

[coral spindle]

if indeed we are defining it to be so

[solar vessel]

ok yeah so just fields

[alpine island]

Yeah

[coral spindle]

Now do the noncommutative case ig

lol

[alpine island]

It's like how for a ring:
R, + is a group
R, * is a monoid
Then you turn that into a rng:
R, + is a group
R, * is a semigroup
So in a similar process you take a field:
R, + is a group
R, * is a group
And turn it into a "feld":
R, + is a group
R, * is an associative quasigroup

[crystal turtle]

[alpine island]

Reasonable reaction

[crystal turtle]

Now what about a ringoid or fieldoid: multiplication is a semigroupoid :^)

[coral spindle]

I mean that's basically an additive category lol

just without identities

[rocky cloak]

But an associative quasigroup is just a group right?

Disregarding the empty set I suppose

[alpine island]

No identity

[rocky cloak]

With associativity you can prove that a/a is an identity no?

[solar vessel]

uh

[alpine island]

ea = a does not imply eb = b

[solar vessel]

I showed above that

an associative abelian quasigroup is an abelian group

basically

unpacking it all

[alpine island]

But without commutativity, that doesn’t work

[dim widget]

[dim widget]

[dim widget]

[void cosmos]

dense subrings are so cool

[rocky cloak]

Let e be such that ae = a. Then we have aeb = ab and dividing by a in the left gives eb = b

So ex = x for all x

[crystal turtle]

[rocky cloak]

Then same argument gives that there is an e' such that xe' = x and then e = ee' = e'

[solar vessel]

[crystal turtle]

[dim widget]

I am torn between this being 🧠 or

[rocky cloak]

I already mentioned the empty set

[solar vessel]

ik

[void cosmos]

yo quick bad question

[delicate bloom]

most of the stuff in this image is garbage

[void cosmos]

let D be a divisoin ring why and V be a vec space over this division ring , why is Hom_D(V,V) artinian?

[rocky cloak]

Last year I was creating problem sets for a group theory class, I felt very cheeky because I had an exercise that was

1. Blabla [alternate definition of group, where you have division but don't require identities] prove that this is equivalent to the normal definition
2. Show that your proof in exercise 1 was wrong by giving a counter example

[south patrol]

I assume V is f.d. ?

[void cosmos]

yes

correct

mb

[south patrol]

Well then same argument as for vector spaces right

[alpine island]

Is it? Can you elaborate?

[crystal turtle]

who cares about quasigroups

[void cosmos]

hmmm which is what ,

that they trivially satisify both chain conditions?

[south patrol]

yes

[rocky cloak]

Finitely generated module over artinian ring is artinian

[south patrol]

dimension in a chain of subspaces can only increase/decrease finitely many times

[void cosmos]

yea

thats what

thats not

what i had in mind xd

its a free D-module so its a direct sum of a noetherian ring ( D )

hence notherian

correct?

[rocky cloak]

Sure, but also artinian implies noetherian (for rings)

[void cosmos]

yea cool.

tysm guys

[lime warren]

do all mods that aren't integral domains solve for every zero divisor in a ring? if I have a ring R how can I pick a mod that will give me every zero divisor? hopefully that makes sense

[void cosmos]

ask them

[dim widget]

no it does not

[lime warren]

how can I find every zero divisor of a ring?

[dim widget]

It depends on the ring

[lime warren]

hm. I see

[dim widget]

in general it is hard

[void cosmos]

if R is left artinian and a dense subring of Hom_D(V,V)

why does it equal Hom_D(V,V)?

i can prove that V has finite dim over D

[delicate bloom]

a cute trick that may help to know is if you have an idempotent (other than 0 and 1) then you have a zero divisor. try showing that

[lime warren]

oh wow, that's super interesting

thank you

[delicate bloom]

if you show what you get,

someone more awake than me might show you some useful things to do with it too

[alpine island]

Welp, this one will be going in the "weird homemade algebraic structures" pile. Next to the 3-way number line and the "field" with >2 binary operations that each distribute in a chain.

[delicate bloom]

recently was thinking of revisiting the concept of a kind of "group" with an n-ary operator inspired by how the group law on an elliptic curve works, this is probably already a thing though

[alpine island]

Yeah, groups with n-ary operations are in that pile too

I need like 2 more years of algebra minimum before I can talk about these seriously

[delicate bloom]

instead of a degree 3 plane curve with 2 pts on it having a line intersect at a 3rd point, instead have a degree d curve with d-1 points on it and make a curve through that, since it'd give you a unique point... but you'd be restricted from what you can actually put in it, so it's kinda garbage

[alpine island]

Actually I'll bring up the 3-way number line again, just for laughs. This came to me while I was biking home from high school: "what if instead of 2 signs (+1, -1) we had 3?" I imagined a Y-shaped number line, with (+1, -1, :1) as the directions.
Addition on this structure turned out to be hard to define, with it ending up being a unitial magma with non-unique inverses. I wanted 1 + :1 = 0 and 1 + -1 = 0 to be both true, but also -1 + :1 = 0.
Multiplication was okay, it's just like the 3rd roots of unity in C.

[coral shale]

the 4 way number line is clearly better

[coral shale]

i mean srsly tho - so your addition takes you away from the origin if same sign, else towards the origin?

thonk

does that work well 😵‍💫

(:1 + 1) + -1 = -1
:1 + (1 + -1) = :1

[delicate orchid]

absolutely no idea what this means ^

[summer path]

What is a :1

Two rows of people in a rollercoaster ride going up?

[delicate orchid]

TRUE

[open sluice]

be like physicists and use red/green/blue instead

or something

[hybrid pike]

im looking for a group isomorphism to show using the first isomorphism theorem that
$\mathbb{R}/\mathbb{Z}, + \cong { z \in \mathbb{C} | |z| = 1}$, can anyone give a hint or help me on my way?

[cloud walrus]

NotAPenguin

[coral spindle]

Hint: think about e^x.

[open sluice]

\left\{ z \in \mathbb{C} \mid \lvert z \rvert = 1 \right}

[formal ermine]

||(and think about what your complex group on the rhs represents geometrically)||

[hybrid pike]

Ah, what if I write $\phi(x) = e^{i \pi x}$

oops

[cloud walrus]

NotAPenguin

[hybrid pike]

Hmm no that won't do

[formal ermine]

yeah not quite

but you're close

[hybrid pike]

$\phi(x) = e^{i \pi x} - 1$ has $ker(\phi) = 2\mathbb{Z}$ so I think just making it $\phi(x) = e^{i \pi 2x} - 1$ should do the trick no?

[formal ermine]

what is the identity in S^1 = { x in C | |x| = 1 }?

[cloud walrus]

NotAPenguin

[formal ermine]

|0| isn't 1, so it can't be 0

because that's not in your group

[hybrid pike]

Right

[hybrid pike]

Is this simply 1?

[formal ermine]

Yes

since it's a subgroup of C*

[hybrid pike]

Alright

I added the -1 because I figured 0 is the identity of R/Z,+

[teal vessel]

step 1. rewrite in seximal and recognize that all generators of subgroups equal to <1> are congruent to +-1 (mod 10), i.e. all numbers ending in 1 or 5. Step 2. use an array to mark multiples to minimize work. We love long problems.

[formal ermine]

your group iso is from R/Z to C*

so the output is in C*

[coral spindle]

It is, but that is irrelevant, 1 is the identity of the second group, and that's the identity that matters when calculating the kernel.

[hybrid pike]

Oh yeah of course

$\phi(x) = e^{i \pi 2x}$ So this is the one I'm looking for then

[cloud walrus]

NotAPenguin

[coral spindle]

Prove it.

[hybrid pike]

Alright, so $\phi(x) = e^{i \pi 2x} = 1 \iff cos(2 \pi x) + i sin (2 \pi x) = 1$. If $x \in \mathbb{Z}$, then we get $\phi(x) = cos(2 \pi x) + 0 = 1$. Also is ${cos(2 \pi x) + i sin (2 \pi x)} \cong {z \in \mathbb{C} | |z| = 1}$

I guess this is kinda just writing out the definitions and maybe not a great proof 🤔

[rocky cloak]

Why is this an exercise. It's just incredibly long for no reason

[cloud walrus]

NotAPenguin

[teal vessel]

I've elected to "write out" the form of the elements as a set with a rule, rather than all of them by hand, as well as define an equivalence class between all generators which generate the same subgroups.

[<x>]={ n | n generates <x>}
just because I can.

so instead, I'm using the minimal member of the EC as a representative, and writing out the form of each subgroup, i.e.
[<3>]={6n+3 | n in N, n < 48}
<3> = {3n | n in N}

[rocky cloak]

I feel like at that point it should be clear enough you understand it that you don't need to write anything down

[delicate orchid]

just a sublattice of Z under gcd innit or whatever

[teal vessel]

yes, but my teacher decided I should do every example in the textbook, with only minor modification to instructions. (autodidact joke)

[delicate orchid]

well just... don't

[teal vessel]

why not? I have little better to do.

[delicate orchid]

you could be reading about more exciting things like uhhhh

completion of rings

[dim widget]

Don't listen to new user Wew Lads Tbh, he's still getting used to the server

[teal vessel]

I could. There's still some structure I'm teasing out of this one tho. It's getting me back into using seximal.

[delicate orchid]

sex

[teal vessel]

not yet. math first.

like how to describe the set {4, 20, 28, 44} as a rule. now it's a little mini side puzzle because I'm dumb and want symmetry in my writing

me: counting up generators and realizing I only have 47 [insert panik meme]
also me: 0 can't be a generator [kalm]

[summer path]

That's a lot of generators..

[teal vessel]

yeah, for all the subgroups of the cyclic group of order 48

solution: ibid

[delicate orchid]

holy tedium

literally just relabelling

[dim widget]

yeah this is really dumb

but it's good to use this exercise as a chance to unlearn the notation $Z_n$ for the group $\mathbb{Z}/n\mathbb{Z}$

[summer path]

[cloud walrus]

Topos_Theory_E-Girl

[mighty kiln]

Z/nZ > Zn

[teal vessel]

I mean, the text only just introduced Z_n as an isomorphic group to Z/nZ as a way to make things more convenient. Thusfar it's all been Z/nZ

[summer path]

Z/n > Cn > Zn

[dim widget]

it's more convenient because then if you forget one piece of notation you can still use the other

genius

[summer path]

I'm not sure where I'd put In along there but maybe between Cn and Zn

[dim widget]

going to tell my kids this is what error correcting codes was

[mighty kiln]

Z/nZ >〈x|x^n〉> Cn > Zn

What's In

[summer path]

What rotman uses

[mighty kiln]

Bruh

[summer path]

I had to check the glossary of terms

Xd

[formal ermine]

Z/nZ > Z/(n) > Z/n > C_n

it's not even worth mentioning Z_n

[summer path]

"Integers mod n" thanks rotman

[mighty kiln]

Z/nZ > Z/(n) > Z/n >〈x|x^n〉> Cn > Zn > In

[delicate orchid]

ranking <x|x^n> higher than C_n

all of the first three are also used for the ring

[mighty kiln]

We use Z for (Z, +) and (Z, +, ×) anyways

[summer path]

Wow you really hate In that much?

[delicate orchid]

in fact Z/(n) is literally the notation for quotienting by an ideal

[mighty kiln]

Ye which is pog

[delicate orchid]

overloading notation is not "p"og

[summer path]

Z/pog

[mighty kiln]

Abuse of notation
Overloading notation

[south patrol]

I like Z/n

[formal ermine]

hi potato

[south patrol]

Exactly matches what we say, has 0 ambiguity and is the quickest to write besides Z_n which has other meanings and isn't obviously a quotient

[south patrol]

Hi!

[crystal turtle]

he's finally learned

[crystal turtle]

also the correct notation, ofc

[high peak]

[alpine island]

Yeah basically. It was an intuitive way to do it at the time, I was picturing a sort of "pulling" motion. Where you pull on the end of one of the lines, and the other two get dragged through 0 onto that line. So :1 + 2 = 1

[alpine island]

No lol

[minor wraith]

any tips for b?

proving that it's infinite

[teal vessel]

what is the form of 1/x for a number in Z adjoin root 2?

you'll find you might be able to perform some algebraic manipulations through rational conjugates

of course, you have to ensure the end result is a polynomial in root 2 with only integer coefficients, but that simply means you need a particular form for your free variables a and b in your element (a+bx) using x as a substitute for root 2

[delicate orchid]

alternatively, you can use the fact that you need N(x) = \pm 1 for it to be a unit and then show that N(a+b\sqrt(2)) = a^2-2b^2 = \pm 1 has an infinite amount of solutions

[rotund aurora]

where can I find proof that the number of abelian groups of order at most x is asymptotic to Cx+O(sqrt x) where C=zeta(2)zeta(3)...?

there is the paper by Erdös and Szekeres which is short, but it's in German

[minor wraith]

how do you show the "has infinite number of solutions" part

[hollow mica]

#advanced-algebra

the last sentence of the #advanced-algebra channel description says that group theory is allowed if it’s beyond what’s in a typical first course

[rotund aurora]

💀

[teal vessel]

as opposed to a poorly defined homomorphism. (I know, but it's funny)

[formal ermine]

just learn german, ez

[rotund aurora]

translate it, its just 7 pages

pls

🙈

[delicate orchid]

I've found an easier way

recall that the product of two units is a unit, and that 1+sqrt(2) is a unit (inverse given by -1+sqrt(2))
so we can just take the family (1+sqrt(2))^n to be our infinite set of units

[formal ermine]

send it and I'll translate it

[rotund aurora]

https://www.renyi.hu/~p_erdos/1934-05.pdf

[formal ermine]

alright

give me like

20 mins

nvm this is gonna take longer

on page 3 now

[formal ermine]

page 4

[rotund aurora]

Theres quite some people that can potentially beneffit from what you are doing, so take your time 🙂

[formal ermine]

almost done with page 6

[hollow mica]

bro is a miracle worker

[formal ermine]

@rotund aurora https://illuminator3.github.io/On the amount of abelian groups of a given order and a related number theoretic problem.pdf let me know if this is what you were looking for / if you find any typos

[jovial quiver]

A is a ring, and S is a multipliciative subset. If I have two S^-1 A-modules M and N, any A-module homomorphism M --> N is automatically also an S^-1 A-module homomorphism no?

[south patrol]

Yes, in fact I guess this is just functoriality of S^-1(-)

[jovial quiver]

oh so true thanks

my proof uses the actual construction of S^-1 M, I wonder, could one show functoriality using only the universal property?

[south patrol]

Hm functoriality is part of the universal property right at least to me

Anyway you can just do this by hand with the actual construction as you say yeah

It's probably clearest lol

[white oxide]

I get how you can find $y_j \equiv (\text{mod} \medspace \alpha_j)$, but how can you find $y_j$ such that $y_j \equiv 0 \medspace (\text{mod} \medspace \alpha_i)$ for $i \neq j$ if the process only describes how to find what it says above similarly if that makes any sense

[cloud walrus]

okeyokay

[rotund aurora]

yeah, thank you!

will read it tomorrow tho

Honestly, when I asked I was half joking and didn't expect you to do it lmao

I suggest tho that you put the names of the authors and say that this is a translation of the original paper or something.

[rocky cloak]

So a1 + a2 = A, so you can find b1 in a1 and b2 in a2 such that b1+b2 = 1. Then b2 = 1 - b1 is 1 modulo a1 and 0 modulo a2. Similarly you can find b3 that is 1 modulo a1, but 0 modulo a3, etc.

Then the product b2 * b3 * ... will be 1 modulo a1, but 0 modulo ai for all other i. Then choose this product to be y1.

Then define the other ys similarly

[void cosmos]

let D be a division ring

V be a vec space ( with infinite dim )

consider Hom_D(V,V)

define I = { f in Hom_D(V,V) | dim(img(f)) < infinity }

why is this an ideal

only when dim(V) is infinite

dim(img is just the rnak btw

rank*

[south patrol]

I'm confused what the question is

[void cosmos]

.

[south patrol]

when dim V is finite this is clearly an ideal

[rocky cloak]

If V is finite this is just the whole ring

[south patrol]

^

[void cosmos]

yes

[rocky cloak]

(which is an ideal)

[void cosmos]

why is this an ideal when dimV is infinite sorry

[south patrol]

Wait that's a different question again

[void cosmos]

i meant i am asking only in the inifnite dim case :d

[south patrol]

Oh

[void cosmos]

mb

bad phrasing

okay the question is

prove that I is an ideal

given that dim V is infinite

[south patrol]

Well

[rocky cloak]

So im(fg) is contained in im(f)

[south patrol]

note that im(f + g) is contained in im(f) + im(g)

Like basically the way I would view this is like

[void cosmos]

is it cuz like i can just take off the f out of the linear combinations?

f being the ring element

[south patrol]

not sure what you mean

[void cosmos]

i just wrote out some linear combinations ( elements of the span )

[rocky cloak]

So it being a right ideal should be clear. And g(im f) is the image of a finite dimensional space

[south patrol]

But this is hopefully clear

[void cosmos]

yes

[rocky cloak]

So it is also a left ideal

[void cosmos]

is g your ring element?

[south patrol]

there are two rings at play

f,g are maps V -> V

in both of our notations

lol

i like how we have given complementary contributions to an answer

[void cosmos]

what i just tried was suppose g in Hom_D(V,V) and f is in I , the span of g(f(V)) would be just be linear combinations of g(f(elements)) which is the g( of something finite dim )

thats what jagr was saying i think

[south patrol]

yes

[void cosmos]

cuz like

g(f(x1))+g(f(x2)) is just g(both

cuz g is linear uk

lol

its so simple lmfao im so stupid

xd

[south patrol]

this is also like standard linear algebra if you pretend D is a field

fortunately everything still works

that's how i am thinking about it anyway lol

[void cosmos]

so this statement : 2- If D is a division ring, V is a D-vector space, and R is a dense subring of homD(V, V) , then R is a simple ring

is clearly false

with the ideal i gave right?

[rocky cloak]

Is there anything in linear algebra that doesn't also work over division algebras?

[void cosmos]

regardless of the dimension of V

[rocky cloak]

Surely there must be something

[rocky cloak]

What does dense mean here?

[void cosmos]

a dense subring of Hom_D(V,V) is a subring that has the property that whenever U is a linearly independant subset of V and A is any subset of V then u have an element in that subring such that it sends each element to the other

( assuming ofc they have the same number of elements )

Hom_D(V,V) is a dense subring of itself

cuz take any linearly independant subset , consider the basis that contains it

[rocky cloak]

Yeah, you probably need finite dimensional for that to be true

[void cosmos]

and then just define the map manually

[void cosmos]

by true you mean the statement i wrote above right

?

[rocky cloak]

Yeah

[void cosmos]

yea thats prob what i meant

by infinite dim

cuz i had this problem in mine

its a prove or disprove problem

[south patrol]

I imagine stuff with determinants would be messed up lol

for example

[rocky cloak]

At least it is true that End(V) is a simple ring, haven't checked the dense stuff

[south patrol]

and related tools

[void cosmos]

given V is finite dim?

or both

lol

sorry

or yeah

[void cosmos]

.

so to show this is false

i would assume that V has infinite dim

and then this ideal wouldnot be the whole ring

it would be proper

so im done

correct?

[rocky cloak]

Yes

[void cosmos]

and also i get Hom_D(V,V) not artinian

by the wedderburn theorem?

[rocky cloak]

I'm not sure how you would apply wedderburn, but if V is infinite dimensional then it is not artinian yes

For any subspace U, {f | Im f < U} is a right ideal. Then just make an infinite chain of subspaces.

[void cosmos]

< is subspace right?

[solar vessel]

yes

[rocky cloak]

Yeah

[void cosmos]

cool

tysm

i got it

suppose R is simple left artinian , then R is primitive

just a quick proof check guys

proof: suppose R is simple left artinian , then there exists a minimal left ideal call it J , we claim that J is faithful as an R-module.

proof: Ann(J) is a (two-sided) ideal of R ( since J is an R-module ) , so we have either Ann(J) is 0 or R

if its 0 then we are done , if Ann(J) is R then that implies Rj = 0 for all j , if J is the 0 ideal we are done , if not then rj is 0 for all nonzero j for all r , giving that j is contained in the zero ideal

which is a contradiction so Ann(J) must be 0?

is it gucci? i am bit skeptical as i literally did not use it being artinian other than having a minimal left ideal

and also

quite a bad questio

or nvm

[green flint]

elgato do you have to come up with the group yourself?

else you could just write down the group presentation for the dihedral group

[mossy lintel]

it's an exercise from one of my ag book

[green flint]

what you wrote should be fine, I'd say

[wicked harbor]

I'm stuck on (ii)... I have found a basis for $\mathbb{Q}(\theta)$, but it's really messy and even then I can't seem to write $i\sqrt{5}$ in that basis... Any tips?

[cloud walrus]

sunnyside1

[wicked harbor]

Maybe I should show that $[\mathbb{Q}(\theta)(i\sqrt{5}) : \mathbb{Q}(\theta)] = 1$?

[cloud walrus]

sunnyside1

[next obsidian]

I mean doing this is equivalent

I don’t think there’s a clever way to do it, you probably just have to stare at the element and figure out how to get the element

Alternatively you can show that Q(theta) is degree 6 so you’d want to write down theta’s minimal polynomial and then…

But that sounds painful

[wicked harbor]

Yeah, I just had that idea... We know that $[\mathbb{Q}(\theta) : \mathbb{Q}] = 6$, and so if we can show that $[\mathbb{Q}(\theta)(i\sqrt{5}) : \mathbb{Q}] = 6$, then it would follow that $[\mathbb{Q}(\theta)(i\sqrt{5} : \mathbb{Q}(\theta)] = 1$.

[cloud walrus]

sunnyside1

[delicate bloom]

in principle it's straight forward, subtract isqrt(5) from both sides -> cube both sides -> factor out sqrt(-5) -> divide out what's left

[wicked harbor]

I've already shown that the degree of the minimal polynomial of $\theta$ over $\mathbb{Q}$ is 6

[cloud walrus]

sunnyside1

[wicked harbor]

It's pretty scary though

[delicate bloom]

what I described solves part ii

[wicked harbor]

I've shown that $\theta$ has minimal polynomial $x^{6} - 5x^{4} - 4x^{3} - 25x^{2} + 20x + 129$ over $\mathbb{Q}$, which is irreducible over $\mathbb{Q}$ by reduction mod 5... But then the basis for $\mathbb{Q}(\theta)$ over $\mathbb{Q}$ is ${1, \theta, \theta^{2}, \theta^{3}, \theta^{4}, \theta^{5}}.$.. And that's scary

[cloud walrus]

sunnyside1

[stone bane]

do PDFs form a group under convolutions?

[eager willow]

Over what domain? I'm not sure if there can always be an identity.

[rocky cloak]

I guess unless the domain is discrete you'll need something like the dirac delta distribution to get an identity. Then you have to deal with convolutions of distributions, which sounds scary.

[formal ermine]

I did on the main page

[hybrid pike]

Hi, my algebra textbook introduces division rings as a ring where each element has a multiplicative inverse, and then adds a quick note that this means that these rings have no zero divisors. I wasn't initally convinced this was obvious, so I tried to make a proof. Could someone check it and see if what I wrote isn't complete nonsense?

Suppose R is a nontrivial division ring with a zero divisor. Then take $x \neq 0 \in R$. Suppose x is a zero divisor. Then there is an $r \neq 0 \in R$ such that $xr = 0$ or $rx = 0$. Suppose $xr = 0$ (the other case is proven analogously). Because x has an inverse $x^{-1}$, we can write $x^{-1}xr = 0 . x^{-1} \rightarrow r = 0$, which contradicts the choice of r. Therefore x cannot be a zero divisor.

[cloud walrus]

NotAPenguin

[wooden ember]

Yeah that works, as you’ve shown generally units cannot be zero divisors

[hybrid pike]

Cool, thanks

[void cosmos]

4a)

is the faithful R-module going to be V?

with action phi o v = phi(v)

suppose cuz let f be in Ann(V)

or idk

how is the Ann(V) not just whole of R?

cuz just take one subset to be the element and one to be just {0}

wait

wtf im so stupid ig

f(v) = 0 --> v = 0

rihgt?

cuz its linear?

tf

im confused tbh any help?

i just find it instanly trivial cuz with Ann(V) we have fV = 0

which just implies f = 0 and thats it

but surely thats wrong idk why

every subring of Hom_D(V,V) is 1-fold transitive ig

[rocky cloak]

Yes, f being in the annihilator means exactly f(V) = 0, which means f is 0. So V is always a faithful module. You need the transitivity part to prove that it is simple

[rocky cloak]

No, D itself is a subring for example viewed as diagonal matricies.

[void cosmos]

Lmfao i forgot about simplicity lol

okay let me try it out

Thank you sm

[void cosmos]

the jacobson radical is literally fucking wild as fuck

[rotund aurora]

If R is noncommutative and M and N are R-modules, can you define the product of M and N?

[rocky cloak]

Indeed we can. You do it in exactly the same way as you would if R was commutative

[rotund aurora]

yeah I realized

ig tho, things can collapse no?

[rocky cloak]

How do you mean?

[crystal turtle]

You might run into some issue defining internal product of submodules

[rotund aurora]

mmh

[crystal turtle]

But external product (i.e. categorical product), should work the exact same

[rotund aurora]

so you are saying, take the set theoretic cartesian product MxN. The group structure is clear, now for the R-action, do r(m,n)=(rm, n)=(m, rn) ?

or you dont let the r get inside?

[rocky cloak]

r(m, n) = (rm, rn)

Just like usual

[rotund aurora]

lol right

[rocky cloak]

I don't see why. They're still just abelian groups

[rotund aurora]

I was reading abt tensors, confused things xD

[rocky cloak]

For tensor products it's important that one is a left module and the other is a right module

[crystal turtle]

true, I was thinking of how you may run into issues like with internal product of noncommutative subgroups, but the fact that addition is commutative I guess should prevent issues

makes sense tho

[rotund aurora]

oh ok so the tensor product of R-modules M and N, M right and N left, R noncommutative, is not an R-module itself

[void cosmos]

truee

do u have any other questions king?

[crystal turtle]

[timber blaze]

why is r-2 even?

[void cosmos]

if R is a ring and I is a left regular ideal then I is contained a left regular maximal ideal

is this trivial? ie showing that the maximal ideal that contains a regular ideal must be regular and thats it?

regular means that for all r there exists e such that r-re is in the ideal

[void cosmos]

strong induction assumes not just P(n) but all P(k) for k<=n

[rotund aurora]

I don't understand why we need to take a right and a left module to define the tensor product

[timber blaze]

what

nvm

🧠 🔨

[rotund aurora]

like let M and N be left R-modules, R highly non-commutative. Consider the free abelian group F over MxN. Consider the subgroup S generated by (rm,n)-(m,rn), (m+m', n)-(m,n)-(m',n), (m,n+n')-(m,n)-(m,n') where r in R, m,m' in M and n,n' in N.

Define F/S to be the tensor product of M and N

whats wrong with this

[south patrol]

hm i mean it won't give you the desired structure right

e.g. note rsm (x) n = m (x) srn

which is why it makes more sense to just make one of them a right R-module so you have the correct compatibility

[rotund aurora]

ah wait ye Im dumb

[rotund aurora]

in the category of abelian groups, the injective objects are precisely all the divisible groups?

[rocky cloak]

That is correct

[rotund aurora]

are there any other categories where the injective objects have a nice characterization?

[rocky cloak]

Nonunital rings don't necessarily have maximal ideals right?

[rotund aurora]

honestly I'm having trouble proving Q is injective. I think I'm not getting the right idea

[rocky cloak]

Over a finite dimensional algebra A, injective modules are exactly direct sums/summands of DA (the dual of the regular module).

[rotund aurora]

ah wait I got it

also proved the characterization

[rotund aurora]

what's "the dual of the regular module"

[rocky cloak]

The regular module is just the algebra A viewed as a module over itself. The dual of a (right) module M, is the set of linear maps from M to the underlying field k. Which becomes a (left) module with the action

(af)(m) = f(ma)

[void cosmos]

this is my whole problem..

like idk

cuz the proof assumes it

like where did we use regularity in this zorn application

its so confusing

[rocky cloak]

I guess there saying that an ideal J that contains I is equal to R iff it contains e. And ideals not containing e is certainly closed under unions of chains.

Basically R/I becomes a unital ring with unit e, and then things proceed as normal (at least if R is commutative)

Then in the noncommutative case I guess that's "morally" what's going on

[rocky cloak]

Over a (finite) group algebra a module is projective iff it is injective.

[rocky cloak]

In the extorthogonal complement of a cotilting module, the injective objects are sums/summands of your cotilting module. In particular over a Gorenstein ring the injective objects in the category of Gorenstein projectives are exactly the projective modules.

(probably need to add in Noetherian and finitely generated everywhere in this statement)

[rotund aurora]

[void cosmos]

okay can u go on with that

if I is regular then R/I has identity

now what

it has a maximal ideal

in the form K/I

for some ideal K in R

[rocky cloak]

Yeah unital rings have maximal ideals

[void cosmos]

does K have to be maximal?

[rocky cloak]

Yes there is an order preserving bijection between ideals in R/I and ideals containing I

[void cosmos]

omg yeah obviously

bro ur super smart lmfao

okay

can it be our candidate

no right cuz why would I be contained in K

[rocky cloak]

Because K comes from an ideal of R/I

[void cosmos]

yea and thats it

by the same iso theorem

lmfao so cool

this proof is so much better

[white oxide]

what do they mean by I is a maximal ideal w.r.t the property listed? as in I is a maximal ideal in the ring R that has that property, or that I is the biggest such ideal that has that property?

oh well i assume it's the latter since every maximal ideal is prime

[south patrol]

Well sort of the latter, though not " the biggest"

A maximal element of the poset of ideals I such that ...

[white oxide]

oh yea oop mixed up the def of biggest and maximal

[south patrol]

But yeah

It's a general phenomenon it seems that maximal elements of posets of ideals often end up being prime lol

[long geyser]

the phrase "with respect to" wouldn't make sense either if it wasn't referring to the maximality

[south patrol]

Though in this case this is different

[white oxide]

could I have a hint for this please? I'm assuming that $xy \in I, x \notin I$, and $y \in I$ (for contradiction). I'm also assuming that $I$ is contained in some ideal $J$ with the property that $J \cap S \neq \emptyset$. Now I was able to show that $y \notin I, J, S$ and that $x \notin J$ but $xy \in J$.

[cloud walrus]

okeyokay

[rocky cloak]

I'm not sure I'm following your strategy, but what you should do is let I be maximal wrt I\cap S = Ø, then assume xy is in I with x not in I and try to prove that y is in I

[chilly ocean]

is this really from an intro to abstract algebra class

[rocky cloak]

The first thing you should note is that Rx + I is a bigger ideal, so must intersect S

[white oxide]

ye sorry, i'm assuming for contradiction that y is not in I

[white oxide]

nah it's from a grad algebra class

[white oxide]

okay thanks, i'll note that

[chilly ocean]

[rocky cloak]

If you also assume y is not in I, then fiddle with some elements and use that S is multiplicatively closed

[white oxide]

ok thank you

[toxic zephyr]

we covered stuff like this in my second level abstract algebra course

[chilly ocean]

I'll be taking my first and only algebra class this fall

[white oxide]

sorry could i have another hint, i've been playing with it but i'm stuck again

here's some of the stuff i got so far (went back to trying a direct proof): xy is in I implies that xy is in Rx + I, so xy = rx + i for some r in R and i in I

x not in I implies that there's some r' in R such that r'x is not in I

okay

these are all definitions basically

so i've gotten nowhere lol

i guess i'm trying to find an explicit element in the intersection of Rx + I and S

[rocky cloak]

So if neither x nor y are in I then both Rx + I and Ry + I intersect S.

So there are r, r', i, i' such that rx + i is in S and r'y + i' is in S

Then my hint is that S is multiplicatively closed

[white oxide]

okay so I'm guessing that I have to show that xy is in S somehow which will contradict xy being in I

what I would do is multiply rx + i and r'y + i' and use the fact that they're in S and try to isolate xy, but S is not necessarily closed under addition so that wouldn't work

oh wait i think i can show it's in I which will complete the proof

[rocky cloak]

So you don't need to show xy is in S, just that some element of I is in S

That will give you your contradiction

[white oxide]

oh right duh that whole entire expression is in I

OHHHHHHHH

okay thank you so much, you basically carried lol

jagr always the hero

[south patrol]

Potentially dumb q

Let A, B be abelian groups and suppose that A (x) F is iso to B (x) F for F = Q and F = F_p for all p

Does it follow that A is iso to B? I assume not, but it does seem to hold for all f.g. groups by the classification

And then what happens if we change it to like suppose f: A -> B is a map and that f (x) F is an iso for all such F, is f an iso

Hm

I mean I have no idea how you'd prove the first to be true and would conjecture that it is false

But 2nd I have not as much intuition for

[solemn dew]

https://en.wikipedia.org/wiki/Base_change_theorems#:~:text=In mathematics%2C the base change,following natural transformation of sheaves%3A&text=is a sheaf on X.

[rocky cloak]

Can't you do something like A=Z/4 and B=Z/2? That even has a map A -> B

Even if you replace Fp with Z/n you still have that tensoring with Q/Z is 0 on everything.

[south patrol]

oh Q/Z is very good then lol

i'm silly

thank

[somber sleet]

hey guys, can somebody explain to me how to find the galois group of $\mathbb{Q}(\sqrt{2}, \sqrt{3})/ \mathbb{Q}$

[cloud walrus]

damn_guuurl

[somber sleet]

What I know until now, is that $Gal( \mathbb{Q}(\sqrt{2}, \sqrt{3})/ \mathbb{Q})$ has order 4, but how do I know if it is isomorphi to $Z_4$ or $Z_2 \times Z_2 $

[cloud walrus]

damn_guuurl

[shy fossil]

find out the minimal polynomial

[somber sleet]

I did

[shy fossil]

what was it?

[somber sleet]

I mean I have 2, $X^2 -2$ and $X^2 -3$

[cloud walrus]

damn_guuurl

[shy fossil]

so it's the splitting field of (x^2-2)(x^2-3), right?

[somber sleet]

yeah exactly

which I know is also galois, since it is normal separable and algebraich

[shy fossil]

now just look at the conjugations of the roots

[somber sleet]

well I mean I have the set of roots ${+-\sqrt{2}, +-\sqrt{3}}$

[cloud walrus]

damn_guuurl

[shy fossil]

right

so look at the map fixing sqrt3 and sending sqrt2 to -sqrt2

and the map fixing sqrt 2 and sending sqrt3 to -sqrt3

except the identity the only other element is sending sqrt2 to -sqrt2 and sqrt3 to -sqrt3

you have your 4 elements

[somber sleet]

is it possible to change them? like sending for example sqrt(2) to sqrt(3)?

[shy fossil]

sqrt 2 and sqrt 3 are not conjugates

[somber sleet]

can you go further into this argument? Cause I have everything you wrote before

the only thing which I don't see, is why the switch is not possible

[shy fossil]

i would leave it to you to check that the map you propose actually doesn't leave Q fixed

[somber sleet]

Oh okay haha, then it is surely something easy, thank you for helping me through it, gonna come back if I have further questions

[shy fossil]

sure

np!

[somber sleet]

I don't know why, but I can't never recognize the group operations

I somehow don't see the properties

[shy fossil]

do you see them here though?

there's a very general algorithm in dammit my foot though

[somber sleet]

the galois group operates on the field extension right?

[somber sleet]

?

[shy fossil]

of finding galois groups of polynomials

so if you get the minimal polynomial, you can work out it out

[somber sleet]

what I mostly know is that the action has to be transitiv, when f is a minimalpolynomial(irreducible), but how do I know if it is free or something else?

[shy fossil]

you don't that's the point of the algorithm

you just calculate a bunch of numbers and find the groups

[somber sleet]

hahah okay well

[somber sleet]

is this because if we square it we have a contradiction to the identity, right?

[shy fossil]

yes

[somber sleet]

I'm sorry for asking such simple questions, I'm trying to understand it thouroughly

[shy fossil]

it's completely fine dw

we were all once in this phase

[somber sleet]

is there a way on how to order field extensions? like normal implies algebraic, finite implies algebraic.. are there some obvious statements?

[shy fossil]

You can look up the chain in wiki I guess

It's not very enlightening though

[hybrid pike]

How do I solve problems like "What ring is isomorphic to $\mathbb{Z}[X]/(2)$"? When no polynomials are involved I can sometimes see the answer more easily but I get stuck on these questions.

[cloud walrus]

NotAPenguin

[hybrid pike]

Is there an intuitive way to understand these quotient rings that I'm missing? Right now I try to think of the cosets of (2) in $\mathbb{Z}[X]$ but I find that hard to visualize, especially with polynomials involved.

[cloud walrus]

NotAPenguin

[delicate orchid]

Z[X]/(2) is "the same" as taking Z[X] and then setting 2 = 0

[formal ermine]

beat my wife to it

[delicate orchid]

child you have to stop saying that

[hybrid pike]

So that would essentially filter out all the terms with even coefficients?

[formal ermine]

you're just jealous

[formal ermine]

no