groups-rings-fields

[void cosmos]

fuck

[rustic crown]

if you had such a prime ideal, then its preimage under the map A --> A[a^-1] would give you a prime ideal of A which doesn't contain a. and this contradicts it lies in all primes.

[void cosmos]

is this the localization

[rustic crown]

yep

or simply A[x]/(xa-1)

(if you haven't seen what that is so far)

[oblique river]

Ive always liked that proof, it felt so clever to me

[void cosmos]

where can i learn about localization?

atiyah mcdonald has a whole chapter about it

dont wanna get that deep

im studying for an algebra test

[rustic crown]

so since this ring has no primes under our assumption, only way this is possible is if it was the zero ring! (this is where zorn enters)

[rustic crown]

what do you wanna know it for? if it would be used later in whatever you're reading then it's good to spend quite some time on it.

else reading the definition would be enough for now

[void cosmos]

i am supposed to take a test in the normal undergrad algebra

but i just want to learn it for the sake of the "slick" proof

[rustic crown]

i was introduced to localization from aluffi's exercises lol

[void cosmos]

okayy will just google them or something

anaywanys 1 more question

whats the ascending chain condition?

is it showing that every ascending chain has a maximal element that contains all of them?

or is it just showing an ascending chain exists or what

so for example im supposed to show principal ideals follow the ascending chain condition , so suppose (a1) subs (a2) subs .... then the Ua_i is an ideal ( will prove ) and contains all of them (trivially)?

[rustic crown]

yep, but it's usually phrased in terms of the chain stabilizing

so the chain becomes equality after a while

if every chain of ideals satisfies ACC you call the ring noetherian

[void cosmos]

what type of study is this

[rustic crown]

it was this btw. so if you want a quick intro, you could use this.

[void cosmos]

like if one is interested in solving just these problems in his or her math life

what is this

just algebra?

[rustic crown]

that's a weird way to put it lmao

ACC shows up a lot, it's like a finiteness condition in algebra

you see it quite often if you do number theory or algebraic geometry

[void cosmos]

cool af

tysm

[void cosmos]

ty

[void cosmos]

is this correct

[rustic crown]

it's not true for all rings

[void cosmos]

yes

i thinnk i need

pid

[rustic crown]

this is equivalent to existence of factorizations into irredudibles in a domain?

[void cosmos]

cuz i assumed they are generated by one elements

[void cosmos]

its literally the next problem wow

p smart

[rustic crown]

lol

[rustic crown]

hehe

so yea, for pids that's true.

[void cosmos]

will try to do it

and post out my answer

tyty

[rustic crown]

[solar shore]

still a legend for giving me the right pdf

[chilly ocean]

what do we need exact sequences for?

like its so random

also these random ass diagrams

[rustic crown]

ig in the start it's just a nice notation for book keeping

these things appear automatically as well

the simplest example is when you try to write a generator-relations description of some object

you have some generators, the kernel would be relations, now you can ask generators for these relations, and then relations among those

and keep on messing up your head :3

but all this would be easily described as a nice exact sequence
... R^c --> R^b --> R^a --> M --> 0

which says like you have 'a' generators for M, and the relations among these have b generators, and relations among those are described by c generators

[pastel cliff]

old convo at this point but im not sure i see this

sigma o pi(m) is in N

repost of question

[agile burrow]

pi(m) is in N, applying sigma to that takes you back to M

[pastel cliff]

oh i was reading it backwards woops

[rustic crown]

walter

[agile burrow]

And the you just apply pi to the element to see that it's in the kernel

Det

[rustic crown]

you removed your scarf

[pastel cliff]

[rustic crown]

is it warm again?

[agile burrow]

Yeah, it was 74F yesterday

But today it's down to like 40 again

[rustic crown]

oh

.<

[agile burrow]

So I think I'd rather just stay in

[pastel cliff]

ok i understand most of what you said before this but idk how it's a "breakdown" of an element

sorry for ping

ok we need to define a new map in terms of pi and sigma that gives us an iso

maybe not "in terms of" but using that identity property thing

[agile burrow]

yeah, you can definitely define an iso using sigma and pi

[agile burrow]

the point is that if you have an element m in M, then pi(m) lies in N and m - sigma o pi(m) lies in ker(pi), so this is kind of a natural way to decompose m into something that lies in N and something that lies in ker(pi)

and hopefully it's not too difficult to see how you might recover m from m - sigma o pi(m) and pi(m)

[pastel cliff]

apply sigma to each...?

[agile burrow]

well you can't apply sigma to m - sigma o pi(m) since this lies in M

but yeah, you can apply sigma to pi(m) and add this to m - sigma o pi(m), which recovers m

[void cosmos]

to show x^2+x+1 is irreducible over Z_2[x] can we just do it by bruteforce?

[rustic crown]

yep

that's the easiest

[void cosmos]

yea so just suppose not then its = (ax^n+bx^n-1+....)(a'x^n+b'x^n-1+....)

[rustic crown]

(btw irreducible in Z_2[x] and over Z_2)

[void cosmos]

and then just try a= 0, b=1

b=0=a

a=1 b = 0

and so on?

[rustic crown]

nah

[void cosmos]

oh you meant roots

[rustic crown]

first, is Z_2 the 2-adics?

[void cosmos]

no

[rustic crown]

nice

then it a field :3

[rustic crown]

so yep

[void cosmos]

okay cool

ty

[rustic crown]

[maiden heath]

Hi, I don't quite understand why the algebra $(\mathbb{C}S_n)^{S_{n-1}}$ is a permutation representation of $S_{n-1}$

[cloud walrus]

ArtyLeAardvark

[maiden heath]

I can see how $\mathbb{C}S_n$ is a permutation representation of $S_{n-1}$ but how is it that $(\mathbb{C}S_n)^{S_{n-1}}$ is also a permutation representation of $S_{n-1}$?

[cloud walrus]

ArtyLeAardvark

[maiden heath]

Any help is much appreciated

[tawdry crystal]

I don't understand how h is an Abelian subalgebra

[chilly radish]

h is a subalgebra so closed under bracket, so [h,h] is in h which is contained in m, but it's also in l, so it's in the intersection, but this is a direct sum so it has to be 0

[tawdry crystal]

That makes alot of sense thank you!

[chilly radish]

No problem!

[fervent agate]

I'm struggling with part (c) in this problem. In general, since the order $P$ divides the order of $G$, then [|P|=p^kl] where $k|n$ and $l|m$. I have been able to show as a result of part (b) that in fact $k=n$, so $|P|=p^nl$. Now I am struggling to show that $l=1$.

[cloud walrus]

ecurtiss

[agile burrow]

If l is greater than 1, then it has some prime divisor. Try using that to derive a contradiction

[fervent agate]

Ok I've thought about this for a little while, and I'm not sure what to do with it. My instinct is to either (a) use Cauchy's theorem somehow or (b) show that |P| fails to divide |G/P| or |G|. I've been unsuccessful with both

[next obsidian]

What do you get if you apply Cauchy's theorem?

[fervent agate]

Not sure. The "somehow" is because I don't know how to apply it, since l having a prime divisor relates to the order of an element instead of the order of a group.

Ah but if |P| has a prime divisor and |P| divides |G/P| and |G|, then those should also have a prime divisor

[next obsidian]

so what proof technique do you think you should apply?

Direct proof, contradiction, contrapositive

[fervent agate]

Contradiction. I'm assuming that l > 1 and am hoping to contradict something related to prime-ness

[next obsidian]

sure, so what does Cauchy's theorem tell you about P if l > 1?

[fervent agate]

If l has a prime divisor q, then I know that both G/P and G have elements of order q, but I don't know what that tells me about P.

[next obsidian]

what does Cauchy's theorem state?

[fervent agate]

It says that if a prime p divides the order of a group G, then G contains an element of order p

[next obsidian]

Right, so if |P| = p^kl and l has a prime divisor q what does this say for P?

[fervent agate]

Oh my god P is a group 🤦‍♂️

[next obsidian]

do you see how to finish now?

[fervent agate]

No but I will think about it!

[next obsidian]

it should be pretty easy from here on out

[fervent agate]

I've thought about it and still don't see how to reach a contradiction. Certainly there is some a in P such that a^q = e. After that I've written a lot of things that might be correct but also aren't helpful.

[next obsidian]

How could a be in P?

What is the definition of P?

[fervent agate]

Hold on I think I got it. a^q = e, but also a^p^k = e, implying that q | p

[next obsidian]

Right

But…

[fervent agate]

So q = p. And this is bad because... (thinking)

[next obsidian]

Maybe just go back and look over everything you’ve done so far and the problem statement

And it should pop out to you

[fervent agate]

p doesn't divide m!

[next obsidian]

🗿

[fervent agate]

😮‍💨

Thank you for your patience. This took... so long. And so many hints. I really appreciate it

[next obsidian]

[primal tusk]

I need help proving that the ideal (5, x) in Z[x] is maximal.

and I know that an ideal is a maximal ideal if and only if R/I is a field for a ring R

So basically I need to prove that Z[x]/(5,x) is a field but im not sure how I should go about doing that

[chilly ocean]

write it as a field you know

got any guesses?

[primal tusk]

Im not really familiar with fields since my class has only really learned about rings

but i know the reals and rationals are fields

and Zmodp is a field if p is prime

[summer path]

you can also show that $\frac{k[x]}{(a,b)}= \frac{k[x]/(a)}{(b)}$ and try to show that RHS is a field in this case

[cloud walrus]

Tubular Cat

[chilly ocean]

cool

[primal tusk]

wait what is RHS?

[summer path]

right hand side, i.e. take k=Z, a=5, b=x

[primal tusk]

oh i guess i should add im not really sure how you show that something is a field

thats really my main issue

[summer path]

think of the fields you are familiar with and try to see if you can find one that "fits" in this case

[primal tusk]

so I know Z mod p for a prime p is a field

but im not sure how to make it "fit" since now i am working with polynomials

[summer path]

this is good

let's just start by considering Z[x]/(x)

do you have any idea as to what this could be?

[primal tusk]

a field

?

just a guess

[summer path]

ok choose some polynomial in Z[x} and then take it mod (x)

see if that gives you a better idea

[primal tusk]

okay so (x) is an ideal in this scenario right?

[summer path]

yes

so say you have some polynomial, like f = x^3 + 4x^2 + 8x + 9, what do you get if you take f modulo x

[primal tusk]

hm

is it 9?

[summer path]

ok good

now do you have a better idea of what Z[x]/(x) is?

[primal tusk]

is it just the integers?

[summer path]

yes

[primal tusk]

ohh cool

[summer path]

so now you have $\frac{\mathbb{Z}[x]/(x)}{(5)} \cong \frac{\mathbb{Z}}{(5)}$, can you figure out this last part?

[primal tusk]

like show that Zmod 5 is a field?

or show that that statement is true?

like show Z[x]/(5,x) = (Z[x]/(x))/(5)

[cloud walrus]

Tubular Cat

[summer path]

this should answer your first question

[primal tusk]

how hard is it to show that Z[x]/(5,x) = (Z[x]/(x))/(5) ?

[summer path]

as for the second, it's a good exercise to show that they are isomorphic

it shouldn't be hard

i'd say give it a try

[primal tusk]

okay ill go for it thank you so much you just made all of this make so much more sense

I didnt realize that the modular stuff was similar to the modular arithmetic with integers

[summer path]

when you're taking R/I, you're kind of thinking as sending everything in I to zero, which makes thinking about modular arithmetic with integers kind of an example case

[primal tusk]

yea its super awesome how this all connects

and how there are prime ideals just like there are prime integers

[primal tusk]

also is this only true with a polynomial ring? Or would any ring have this same property?

[summer path]

oh maybe i should have written R there oops

yeah it works for any ring

[zealous spear]

can someone explain to me what lattices are I tried wikipedia but it doesnt click with me from those pages

[long robin]

https://en.m.wikipedia.org/wiki/Lattice_(order)#:~:text=A lattice is an abstract,greatest lower bound or meet).

[zealous spear]

thank you but like i said wikipedia didnt really do it for me

[void cosmos]

if R is an euclidean domain

what can we say about the norm function?

other than the usual euclidean algorithm

say the function is called N

does N(a)<N(ab) follow from the euclidean algorithm?

[formal ermine]

this is not true if b is a unit

[void cosmos]

okay what do i know

about the norm

properties

i am to prove that if x and y are associates then norm(x)=norm(y)

i can do that but idk specifically what i am allowed to use

[formal ermine]

what does it mean to be associate?

[void cosmos]

differ by a unit

[formal ermine]

yes

[void cosmos]

a =b*u for u unit

[formal ermine]

and the norm of a unit is?

[void cosmos]

have no idea

let me try it

what does the norm satisfy?

in df its only the euclidean algorithm

is there anything else?

[formal ermine]

the proof I'm thinking of involves the euclidean algorithm

[void cosmos]

i got it

tysm

[livid willow]

I have a question. In rings, Z/nZ for any prime n is said to be a field and there is a proof of it. But I cant make sense out of it. Say, we take Z/3Z and its residue classes are {0,1,2}, then how can 2 be a unit here?

[rustic crown]

because 2 * 2 = 1 in Z/3Z

[formal ermine]

k is invertible in Zn iff k is coprime to n, see eg fermat's little theorem

[livid willow]

Oh okay

I will have a look at Fermat's little theorem

[formal ermine]

what's a group theoretic proof that Zp^\times is cyclic? I can only think of a field theoretic proof

[rustic crown]

use structure theorem

:p

[formal ermine]

ah I see

ok thanks

[rustic crown]

(the standard basic proof uses this lemma that if there is an element of order n and an element of order m, then tehre is an element of order lcm(n,m))

that's like part of the structure theorem but done directly by hand

so you have an element of maximal order w.r.t divisibility

but you have to input some field theory at some point, to say that x^(maximal order) - 1 can't have too many roots over a field

like basically you have to use somewhere that p is a prime lol

[formal ermine]

I want a purely group theoretic proof

else I could just say Zp iso Fp

F^\times cyclic

done

[rustic crown]

maybe check some keith conrads expository paper

he would have given like 5-10 proofs of this

but a small generalization of this statement is that any finite subgroup of (the multiplicative group of) a field is cyclic

so any proof there should be sort of easily generalized to this

and ig that would make it harder to avoid some property of fields

[simple valley]

I'm trying to come to come up with a way to enumerate all submodules M of the Z-module Z^2 such that Z^2/M has order 28. Would appreciate ideas on how to approach this

The only Z-modules of order 28 would be Z/28 and Z/2 x Z/14 but idk if that helps

[next obsidian]

Isn’t there also Z/4Z x Z/7Z?

[woeful sage]

mniip caught asking for help

[simple valley]

By CRT that's iso to Z/28

[next obsidian]

Ah yeah

Lol

So the first thing is to count how many M there are of the form M’ x M’’

This should be fairly easy

[simple valley]

I think all M have to be iso to Z^2

[next obsidian]

The issue comes to determining how many are not of this form

[simple valley]

But how do they embed?

[next obsidian]

Yeah

[simple valley]

Automorphisms of Z^2 kinda mess up the picture

[next obsidian]

Actually I think this is like a matrix thing

Quotients of Z^2 should be something like

Some sort of rational canonical form thing?

[simple valley]

Yeah I've been thinking about that

[next obsidian]

Anyway sorry, I can’t be of too much help

I think you’re right about the Z^2 thing tho

M ≈ Z^2 I mean

[simple valley]

So if we take the two generators of M and put them as rows into a matrix

We can do elementary row ops on it

Is there a row ops related canonical form

If we have {{a, b}, {c, d}} then we can do extended Euclidean algorithm on gcd(a, c)

Like if we do egcd don't we get numbers x and y such that ax+cy=gcd(a, c) but ay+cx=0?

No I don't think so, but we do get a SL(Z) matrix that takes {a, c} to {gcd(a, c), 0}

So I think we get {{gcd(a, c), p}, {0, q}} where p,q can be arbitrary (?)

Further canonicalizing we can take p mod q

So it suffices to consider upper triangular {{a, b}, {0, c}} where ac=28 and b in [0, c)

[rustic crown]

i would try to use some structure theorem stuff here. by that we know there exists some basis {u, v} of Z^2 such that the basis of the submodule is {au, bv} with a dividing b. from what you said we have two options a = 1 and b = 28 or a = 2 and b = 14.
so the ig start with the submodules generated by {e1, 28e2} and {2e1, 14e2} and see what all you can get using element of GL(2, Z).

[simple valley]

Aut(Z^2) is SL_2(Z) x C_2 though innit

[rustic crown]

it doesn't split, does it?

we have 1 --> SL(2, Z) --> GL(2, Z) --det--> {+-1} --> 1

(lemme read what you doing lol)

[simple valley]

Oh wait

Does GL mean specifically invertible?

As opposed to det /= 0

[rustic crown]

oh yep

(that's what i used it for at least)

right now i'm trying to see how GL(2, Z) acts on the submodules where the quotient is iso to Z/28Z.

the stablizer subgroup is this, {g in GL(2, Z) | A^-1gA in GL(2, Z)} where A is our [1 0; 0 28]

[simple valley]

What's A^-1 then?

[rustic crown]

that over Q

[simple valley]

Ah

[rustic crown]

we want to describe the cosets of this nicely >.<

idk how to

wait, so if i multiply that out, i get [a 28b; c/28 d] so the only condition is 28 divides c

[simple valley]

I think any matrix can be brought into upper triangular form by acting with GL

[rustic crown]

so the stablizers are matrices in GL(2, Z) which look like
[* *;0 *] (modulo 28)

[rustic crown]

yea right

[simple valley]

And the way GL acts on the space of triangular matrices is that any off diagonal element is taken mod the diagonal element somewhere below it

[pliant forge]

det

[simple valley]

And the occasional situation where we get a block triangular matrix and can permute the blocks

[rustic crown]

okie i think i'm done

the above observation let me to consider the set function GL(2, Z) --> (Z/28Z) which sends a matrix [a, b; c, d] to c mod 28. this is ofc a werid map but the nice thing is that the fiber above 0 is exactly that stablizer subgroup.
now i noticed that stablizer subgroup has diagonal entries which would be units mod 28, so this value c mod 28 should only be considered up to units.
so consider the composite
GL(2, Z) --> Z/28Z --> {1, 2, 4, 7, 14, 28}
this is surjective and an entire coset of the stablizer would be sent to a single thing.
so we do get a nice factoring

so in particular a nice set of coset representatives here would be

[1 0]
[i 1]

where i varies in {1, 2, 4, 7, 14, 28}

if i do this for A = [2, 0; 0, 14] then stuff would be mod 7 instead of 28, so in that case we should have the coset representatives as {1, 7}

this looks nice enough that there is probably a good theory behind this :p

so the answer to your initial question would be that there are 6+2=8 submodules of Z^2 of index 28

[gleaming fable]

Hello, I wish to ask
So we have something called a "law of external composition" which is an operation "£" such that
£ : E £ F ---> F
Whereas E and F are ensembles
Right?
Now Let's assume the corp F
(F, &, £)
And the group E
(E, &)
And £ is a law of external composition
Now for E to be a vectorial space one of the conditions is that :
(a £' b) £" x = a £" (b £" x)
Now my teacher marked a little F under £' and a little E under £" and said that those two operations are different
When I tried to ask him he was busy, so basically why are they two different operations? Does it just mean that £' 's product is in F and £" 's is in E? Or are they really completely different operations (like one of them might be multiplication and the other addition)?
Sorry for my bad terminology again btw TvT

[rustic crown]

[gleaming fable]

Haha sorry

[rustic crown]

so corp = field?

i'll read (F, &, £) as (F, +, *)

and E is an abelian group, so i would read (E, &) as (E, +)

what are £' and £''

ig the condition shoudl be (a * b) * x = a * (b * x)

[gleaming fable]

I guess?
For F to be a corp
(F, &) must be a commutative group
And £ must :
-be assosiative
-be distrubutive to &
-admit a neuter element
-its neuter element is different than &'s
All elements in F are inversible except for &'s neuter element

[rustic crown]

(which language is that btw?)

[gleaming fable]

French 💔

[rustic crown]

oh sorry >.<

please restore your heart

i don't want you to die

[gleaming fable]

My english sucks sorry

[gleaming fable]

It's okk

[gleaming fable]

Uh also one other question
Can F be like
(F, +, *)
While E is
(E, ○) where ○ is a different operation? Does it still work?

[rustic crown]

okie so £' : F x F --> F and £'' : F x E --> E right

[gleaming fable]

Yes

[rustic crown]

yea it all works, we just overload the same notation so that it's easier to read

[gleaming fable]

Aha okk

[rustic crown]

these operations need to satisfy all the axioms which tell you how any two of these relate with each other

[gleaming fable]

So

[rustic crown]

like only thing you need to keep in mind is that "multiplication distributes over addition" which is why we use those symbols

[gleaming fable]

the only thing I don't understand is why are F's operations not the same as E's operations

[rustic crown]

cause E and F are different sets. when will you call two operations the same?

i would only call them same if the two function F x F --> F and F x E --> E are "equal"

you can always use the same name for two operations if it doesn't cause any confusion

for example when i write (a * b) * x

the first * is for F x F --> F

but the second star is F x E --> E

i would need to distinguish the two operations (+), (*) : F x F --> F as they're between same stuff

[gleaming fable]

But for example if the operation * is
a*b = (a+b+3)/5
And we take two elements from F
c and d
So c*d = (c+d+3)/5
And now we take x from E
Is c*x also equal to (c+x+3)/2 or does it have to be a different operation? Or by saying different operation you just mean that the results are in different sets?

I hate how discord uses *

[rustic crown]

just do \*

[gleaming fable]

Wait Ill write again

Done!

[rustic crown]

it's up to you how to define the two operations on F, one operation on E and one action F x E --> E as long as they satisfy all the properties

so c*x need not be that

c+x need not even make sense

also the way you define * won't make F a field/corp

[gleaming fable]

Hmm

So the set in crucial in the definition of an operation?

[rustic crown]

yep

for example 1 + 2 is 3 in Z but [1] + [2] = [0] in Z/3Z

[gleaming fable]

Didn't know that
Thank you for explaining mate 😁

[rustic crown]

operation is just a function

but writing +(a, b) is annoying and hard to read

also writing associativity this way would be super weird

[gleaming fable]

Ok I think I get it

[rustic crown]

+(a, +(b, c)) = +(+(a, b), c)

[gleaming fable]

Oh also one more thing

For fields

Let's say (F, +, *) is a field/corp/ chicken idk

[simple valley]

that's odd because the number I get is 53

the submodules generated by {1, k} and {0, 28} where k = 0..27 are all different submodules are they not?

[gleaming fable]

One of the conditions is that all elements of F should be inversible by * except for the neuter element of +
But if +'s neuter element is by chance inversible, does that cancel that F is a field completely?

[rustic crown]

i think not... but lemme confirm again

[coral shale]

u can prove this only occurs when 0=1

and that this will also only occur when there is only 1 element

[simple valley]

the submodule generated by {1, k} and {0, 28} contains exactly the points (x, y) where kx=y mod 28

these are the same set only if k = k' mod 28

[gleaming fable]

Oh
So it can't happen

[coral shale]

can depending on your definition of field

[rustic crown]

GL(2, Z) --> Z/28Z --> {1, 2, 4, 7, 14, 28}
this is surjective and an entire coset of the stablizer would be sent to a single thing.
ah idk why i assumed the factored thing would be bijective...

it's just surjective

[uneven sundial]

can anyone help me with this?

this is the version of gauss' lemma from the book which i assume im supposed to use

[tribal moss]

Do you have the concept of "content" of a rational polynomial available?

(Your exercise would seem to make more sense as a stepping stone to Gauss' lemma than as a consequence of it).

[uneven sundial]

its not in the book but i think i get it

[median pawn]

this is how normal extensions seem to be defined for number fields

but we also define L to be normal over K (where [L:K] < infty) iff L is closed under taking conjugates over K

how does the more general definition specialize to the first one?

[lethal dune]

This one just says K is a splitting field of a polynomial, hence prod of their irreducible polys. Any conjugate is again the root of the same poly hence lies in the same field K

[rustic crown]

could you confirm this once more, i did some stuff and now i think it should be 28(1+1/2)(1+1/7) + 7(1+1/7) which is 56

[simple valley]

$\begin{pmatrix}28&\&1\end{pmatrix},\begin{pmatrix}14&\&2\end{pmatrix},\begin{pmatrix}14&1\&2\end{pmatrix},\begin{pmatrix}7&\&4\end{pmatrix},\begin{pmatrix}7&1\&4\end{pmatrix},\begin{pmatrix}7&2\&4\end{pmatrix},\begin{pmatrix}7&3\&4\end{pmatrix},\begin{pmatrix}4&1\&7\end{pmatrix},\begin{pmatrix}4&2\&7\end{pmatrix},\begin{pmatrix}4&3\&7\end{pmatrix},\begin{pmatrix}4&4\&7\end{pmatrix},\begin{pmatrix}4&5\&7\end{pmatrix},\begin{pmatrix}4&6\&7\end{pmatrix},\begin{pmatrix}2&1\&14\end{pmatrix},\begin{pmatrix}2&2\&14\end{pmatrix},\begin{pmatrix}2&3\&14\end{pmatrix},\begin{pmatrix}2&4\&14\end{pmatrix},\begin{pmatrix}2&5\&14\end{pmatrix},\begin{pmatrix}2&6\&14\end{pmatrix},\begin{pmatrix}2&7\&14\end{pmatrix},\begin{pmatrix}2&8\&14\end{pmatrix},\begin{pmatrix}2&9\&14\end{pmatrix},\begin{pmatrix}2&10\&14\end{pmatrix},\begin{pmatrix}2&11\&14\end{pmatrix},\begin{pmatrix}2&12\&14\end{pmatrix},\begin{pmatrix}2&13\&14\end{pmatrix},\begin{pmatrix}1&1\&28\end{pmatrix},\begin{pmatrix}1&2\&28\end{pmatrix},\begin{pmatrix}1&3\&28\end{pmatrix},\begin{pmatrix}1&4\&28\end{pmatrix},\begin{pmatrix}1&5\&28\end{pmatrix},\begin{pmatrix}1&6\&28\end{pmatrix},\begin{pmatrix}1&7\&28\end{pmatrix},\begin{pmatrix}1&8\&28\end{pmatrix},\begin{pmatrix}1&9\&28\end{pmatrix},\begin{pmatrix}1&10\&28\end{pmatrix},\begin{pmatrix}1&11\&28\end{pmatrix},\begin{pmatrix}1&12\&28\end{pmatrix},\begin{pmatrix}1&13\&28\end{pmatrix},\begin{pmatrix}1&14\&28\end{pmatrix},\begin{pmatrix}1&15\&28\end{pmatrix},\begin{pmatrix}1&16\&28\end{pmatrix},\begin{pmatrix}1&17\&28\end{pmatrix},\begin{pmatrix}1&18\&28\end{pmatrix},\begin{pmatrix}1&19\&28\end{pmatrix},\begin{pmatrix}1&20\&28\end{pmatrix},\begin{pmatrix}1&21\&28\end{pmatrix},\begin{pmatrix}1&22\&28\end{pmatrix},\begin{pmatrix}1&23\&28\end{pmatrix},\begin{pmatrix}1&24\&28\end{pmatrix},\begin{pmatrix}1&25\&28\end{pmatrix},\begin{pmatrix}1&26\&28\end{pmatrix},\begin{pmatrix}1&27\&28\end{pmatrix}$

[cloud walrus]

mniip

[simple valley]

non-diagonal:
27 of the form {{1, k}, {0, 28}},
13 of the form {{2, k}, {0, 14}},
6 of the form {{4, k}, {0, 7}},
3 of the form {{7, k}, {0, 3}},
1 of the form {{14, k}, {0, 2}};
diagonal:
diag(1, 28), diag(2, 14), diag(4, 7)

total 53

[coral shale]

nani

[rustic crown]

okie maybe i'll tell my reasoning as well

so let X be the set of submodules of Z^2 such that the quotient is isomorphic to Z/28Z

from structure theorem, i get that GL(2, Z) acts transitively on the set. from this we also see that SL(2, Z) also acts transitively, because changing the basis from {u, v} to {v, u} doesn't change the submodule.

Let A = {{1, 0}, {0, 28}} be matrix corresponding to the "trivial" submodule. i wanna see when g * A is also the same submodule. this happens if one can find a matrix x in GL(2, Z) such that g*A = A*x, i.e. A^-1gA in Gl(2, Z). since this matrix automatically has det 1, so stabilizer of [A] is exactly the matrices in SL(2, Z) such that A^{-1}gA has integral entries. which happens precisely when the third entry in the matrix is divisible by 28.

we wanna find index of this subgroup. we can do this in a little extra generality. G = SL(2, Z) and H = matrices which look like {{*, *}, {0, *}} mod n. and from some theory of the congruence/hecke subgroups it's known that this has index n * prod(1+1/p) for p dividing n.

for the submodules whose quotient is Z/2Z⊕Z/14Z, the matrix A would be {{2, 0}, {0, 14}} so the condition A^-1gA having integral entries is same as the third entry of g being divisible by 7.

[simple valley]

I think you're just overcounting the diagonal matrices

[rustic crown]

wait so why is diag(1, 28) and diag(28, 1) considered same?

[simple valley]

oh

it's not

I've mentally pre-quotiented Z^2 by D8 because such is the subject at hand

[rustic crown]

i dunno of a nice list of coset representatives of that stablizer subgroup, but at least the index has a nice expression

[formal ermine]

the subrings of Zn are all of the form Zk where k divides n?

and all of them are ideals?

[coral shale]

this part is generally false

show that if an ideal contains a unit, its the entire ring

[formal ermine]

huh wait

[coral shale]

This is in the context where your rings are defined to have identity - ie. subrings must also contain the same identity

if thats not the case, make it clear in Q

[formal ermine]

https://math.stackexchange.com/a/2035307/943349 this answer says otherwise

[coral shale]

likely handling this

but this is implicit from the question

i think

[pastel cliff]

what did i walk into

[coral shale]

because under the other supposition isn't the question trivial

im half asleep so do correct me if this is bs

[pastel cliff]

[coral shale]

yeah there we go, so the question becomes trivial if you take rings with multiplicative identity

There is only the non proper subring and thats it

So the MSE question can be assumed to apply to the other context

[formal ermine]

ah ok thanks

[coral shale]

So back to this.... taking rings with no mul id

I think so??? Im not familiar with this

So for example, taking Z_10, {0, ..., 9}

{0, 2, 4, 6, 8}
{0, 5}

These should be the 2 subrings

which are also the ideals

so yes to both, I believe is the answer

=====
If you're learning rings for the first time I recommend you take it in the context where they're commutative and have multiplicative identity - that's how its almost always taught afaik

[formal ermine]

not learning rings for the first time just having a brain fart

[median pawn]

i'm not sure i fully understand the remark about well-definedness here; could someone elaborate?

which Z^n's are pairwise non-isomorphic? there's just one

[coral spindle]

They're saying that if Z^n is isomorphic to Z^m, then n = m.

[median pawn]

oh lmao

thanks lol

[pastel cliff]

wont ping walter but i think this is the right start

idk why this proof has had me so stuck for so long

[void cosmos]

6a is just basically solved right?

construct the chain by inclusion

this has a maximal element with the union of the ideals

hence by zorn this has a maximal element

[rustic crown]

walter

[pastel cliff]

det

[chilly ocean]

waltuh

[pastel cliff]

,av walter

[cloud walrus]

walter#1555's Avatar

Click here to view the image.

[pastel cliff]

,av TTeppa

[cloud walrus]

Couldn't find a member matching TTeppa!

Feb 18, 2023, 7:09 PM

[pastel cliff]

,av TTerra#5291

[cloud walrus]

TTerra#5291's Avatar

Click here to view the image.

[agile burrow]

this is a good start. I will remark that when I said pi is a projection from M onto N, I meant that kind of literally since pi is surjective. With that said, it feels a bit off to say that sigma projects pi(m) back into M - it can be thought of moreso as an inclusion since sigma is injective

[pastel cliff]

abstract-chillgebra

not enough to show that it's actually an isomorphism but the bijection is obvious right?

[agile burrow]

The bijection should be clear from our last conversation, yes

and in fact, the bijection we discussed ends up being the isomorphism

it should be straight forward to see that the map is a homomorphism

[void cosmos]

now 6b is done by just saying suppose x is in I_a then x=f(I) + b(a) for f,b in R

but I is an ideal so f(I) = I?

is this right?

[pastel cliff]

"this is clearly a bijection" moment

[rustic crown]

nit pick, that may not be the maximal element of the chain. it's the least upper bound.

[void cosmos]

yes

my bad

least element that is maximum

[rustic crown]

(upper bound?)

[void cosmos]

what

[pastel cliff]

supremum?

[void cosmos]

do u want me to say its the sup

yea i get it ig

[rustic crown]

what does least element that is maximum mean

[pastel cliff]

not much wbu

[void cosmos]

smallest upper bound

[agile burrow]

upper bound and maximal aren't interchangeable

[rustic crown]

yee that's what i asked if by "maximum" you meant "upper bound" :p

[void cosmos]

yea my bad guys

logicians moment

[rustic crown]

[agile burrow]

I'm not sure how this shows I_a is principal

[void cosmos]

idk

this looks like a coset to me

[gentle pendant]

The ideals you need to be principal are principal because of the maximality of I.

[pastel cliff]

is it okay to think of normal subgroups of modules

bc a module obv has it's ring as a part of it

[agile burrow]

modules are in particular abelian groups, so every subgroup is normal

[void cosmos]

oh lmfao

cuz if not then it would be a bigger ideal which is not maximal

right?

@gentle pendant

[gentle pendant]

a bigger ideal that is not principal

[void cosmos]

not bigger but i meant like

contains

yea mb

meant principal

but I is the biggest that is not principal so contradictino

cool

ty

[rustic crown]

it's not the biggest, it's a maximal >.<

[gentle pendant]

https://tenor.com/view/lol-laughing-cat-funny-gif-5670070

[void cosmos]

wdym

whats a maximal

[rustic crown]

maximal = nothing is bigger than it

biggest = it's bigger than everything

[gentle pendant]

you shouldn't say THE biggest when talking about a max in a poset

[void cosmos]

yea

i meant biggest in this

category

in this genre

of ideals

that are not principal

is this correct

[gentle pendant]

it's more the "the" that is misleading/wrong

in "the biggest"

[void cosmos]

so it shoulld be biggest ideal such that its not principal

would this be better

[gentle pendant]

then that's just incomplete grammar lol, simply call it a maximal element of the set

[agile burrow]

I think the phrasing "maximal with respect to being nonprincipal" as in the problem statement is the most descriptive

[void cosmos]

cool

tysm

1 more thing

in 3, every prime ideal is maximal so this is a field --> now what? does this follow by just vacuous truth that fields have no nontrivial ideals

[agile burrow]

it's not true that every prime ideal is maximal

[void cosmos]

in a PID it is

[agile burrow]

no it isn't

(0)

[void cosmos]

every nontrivial ideal?

[rustic crown]

lol we nitpicking so much today :p

[void cosmos]

yes

but ig its funny for you boys

lmfao

[chilly ocean]

in some cases it's necessary

[void cosmos]

yeaa i am actually going to tkae an exam in this shit

so i need to like actually write down stuff

so im grateful

now is what i said enough

[rustic crown]

yep, if the ideal was (0) then you get the same pid back

else you get a field which is again a pid since only ideals are (0) and (1)

[void cosmos]

yea

got it

ty

1 more last thing sorry boys

[agile burrow]

If you want, you could try to prove this without invoking that nonzero prime ideals are maximal

[rustic crown]

so pir + domain?

[void cosmos]

can someone walkthrough with me the proof that given R is a ring ideals of M_nxn(R) are of the form M_nxn(J) where J is an ideal of R?

[rustic crown]

say I is an ideal in M(n, R)

[void cosmos]

okay

[rustic crown]

and let J = stuff in the 1,1 entries

[void cosmos]

damn ur going to do it like instantly without even looking shit up

u guys are op tbh

[rustic crown]

lmao, these are standard things :p

[void cosmos]

[void cosmos]

what does this mean

[pastel cliff]

they really are

[void cosmos]

they really are in what sense? coming up with the arguments yoruselves

or knowing the argument cuz u saw it b4

anyways

[rustic crown]

like all elements in R which appear at the corner left entry of some matrix in I

[pastel cliff]

both

[void cosmos]

coming up with it yourselves is not easy for me tbh but ig im stupid

[pastel cliff]

you n me both

[void cosmos]

you mean in the i,i entries?

like the diagonal entries?

[rustic crown]

nu, 1,1

i'm looking at a weird set

[void cosmos]

just 1,1?

the rest are what

0s?

[rustic crown]

yea but the matrices are varying

I is an ideal in M(n, R)

[void cosmos]

okay

[rustic crown]

J = {M[1][1] : M in I}

sorry for using programming notation

:p

[void cosmos]

im literally a cs student

and idk wtf this is

XD

okay

so

oh you meant 2d arrays

[rustic crown]

yeee

maybe i shoudl have used [0][0]

[void cosmos]

yea

[rustic crown]

but whatever

[void cosmos]

so these are the matrices in I such that what?

have entries only in 1,1 and else is 0?

[rustic crown]

this is a subset of R

[void cosmos]

ohh

you just wrote out M[0][0]

[rustic crown]

yea lol, weird thing to do

[void cosmos]

for M ranging in the ideal

[rustic crown]

but it's not that weird you'll see in a moment

[void cosmos]

okay so you took out every top corner element

these are elements in R

[rustic crown]

yep

first see that J is an ideal in R

[void cosmos]

lemme think of it

yea it is cuz the closure follows from writing out the matrix multiplication when we are in I

cuz I is an ideal so it must be closed under elements from M_nxn(R)

right?

[rustic crown]

yep

[void cosmos]

okay col

J is an ideal in R

[rustic crown]

now show this
r is in J, then the matrix with r at 1,1 and 0s everywhere else would be in I

[void cosmos]

yes

[rustic crown]

you know there is one matrix with top left corner = r, show you can kill all other entries

[void cosmos]

so u like

build a correspondonce

ig

[rustic crown]

you already said the correspondence... J <--> M(n, J)

[pastel cliff]

if i have a direct sum of Z arbitrarily many times... this is still finite right?

[void cosmos]

1 sc plz

[void cosmos]

why

[pastel cliff]

it's a y/n question

[void cosmos]

i dont see that

[void cosmos]

no

[rustic crown]

lol

[pastel cliff]

[rustic crown]

use some nice matrices from M_n(R) to kill other cloumns and rows

I was an ideal, so it will absorb multiplication by stuff

[pastel cliff]

it's not finite, just a finite number of non-zero indices

[rustic crown]

wait, so let me ask you again how did you show that J is an ideal

say a and b are in J, so you know there are matrices A and B such that A_{1,1} = a and B_{1,1}=b

and say r in R

[void cosmos]

J is an ideal because when doing matrix multiplication inside I M[1][1] is forced to be in I cuz its an ideal

[rustic crown]

show that r*a in J and a +b in J

yee just making sure you saw it, do you see both of these things above?

[void cosmos]

not like 100% ig

i want to like make a matrix that would multiply the entries of A by just r

diagonal 1 probably idk im abd with matrices

[rustic crown]

yep diag{r, r, r, r...} will do the job

its just r times the identity matrix

[void cosmos]

yea so this is an element of Mnxn

[rustic crown]

yep

[void cosmos]

so thats it ik r*a must be in R

or no

not in R

[rustic crown]

J

[void cosmos]

yea

[rustic crown]

and a+b is easier

[void cosmos]

yea same idea just use the fact that I is an ideal

okay

[rustic crown]

yea just consider A+B in I

[void cosmos]

okay so now J is an ideal

in R

[rustic crown]

okie so back to the other claim

[void cosmos]

now what do we want to show

I = M_nxn(J)?

[rustic crown]

say M is a matrix with M_{1, 1} = r

[void cosmos]

ook

[rustic crown]

yee

show that I contains the matrix with 1,1 entry r and 0 elsewhere

by killing all other rows and columns

[void cosmos]

where is M

[rustic crown]

in I

[void cosmos]

okay so this r is in J

[rustic crown]

yep

[void cosmos]

well

this doesnt use any like ring theory ig

im sure there are matrices that can do this

like

so like i need two matrices

[rustic crown]

yea you only need to know what a (2-sided) ideal is

[void cosmos]

so

okay so i want just 1 element

in some column in some row

then this will kill all elements except this row

then what XD

so first matrix is

daddy[i][j]=1 else = 0 for i=1,j=1

or wait

idk im bad with matrices

[rustic crown]

i already used the letter M lol

[void cosmos]

let me try it

[rustic crown]

use some different one

it's 1 when i=j=1 and 0 elsewhere

then see that daddy * M * daddy does the job

what did i just write >.<

[void cosmos]

okay it does

now what did we do

and why

[rustic crown]

now show that you can transport this r to any other entry

fix i, j. show there exist a matrix N with N[i][j] = r in J and 0 elsewhere

once you done with this you can see that M(n, J) would be a subset of I

[void cosmos]

then instead of daddy[1][1]=1 just choose daddy[i][j]

[pastel cliff]

another quick question - im being asked to show that $\Z[x]/\ang{2x+1}$ is isomorphic to $\Z \oplus \Z/2 \oplus \Z/2 \oplus \dots $ as a $\Z$-module - does the "as a Z module" part refer to thinking of the polynomial ring as a Z module or as the direct sum? or both since theyre supposed to be iso

[void cosmos]

=1

[cloud walrus]

sebbb

[rustic crown]

not quite

basically i'm using that row/column operations correspond to matrix multiplication by left or right

[void cosmos]

yea why cant we just do row/column operations

[rustic crown]

we cannnnn

but i was wring the exact matrix so that you see how we use the 2-sided ideal-ness

[void cosmos]

yeaa

[rustic crown]

killing the ith column is same as multiplying by diag(1,1, ..., 0, ..., 1, 1, .1) on the right where 0 is on the ith spot

[void cosmos]

ohh so i need to like

conjugate without inverse

[rustic crown]

you can do row column operations to bring that r to any entry

[void cosmos]

daddyMdaddy

okay

[rustic crown]

lol

and daddy just kills all columns at once when you multiply on the right

[void cosmos]

yea

so this is the correspondonce

[rustic crown]

so you see that M(n, J) is a subset of I right

[void cosmos]

not really

not a subset

these are matrices and these are not

but ig i see how u would go from one to the other

[rustic crown]

wait mb mb

i should have said something before that

so say M is a matrix in I

if then by row/col operations, you can bring any enetry to the 1,1 enetry

so J actually contains all enries of all the matrices in I

[void cosmos]

yes

[rustic crown]

so you immediateyl get one direction

[void cosmos]

why cant we just say

[rustic crown]

I contained in M(n, J)

[void cosmos]

like why didnt wee just define J = {a | a is an entry in I}

[rustic crown]

you can

doesn't matter

[void cosmos]

okay coilk

col]

why contianed tho 😦

olh

oh M(n,J)

yea

[rustic crown]

the other direction is why we did the computation above

[gentle pendant]

the ideal property is just more trivial by extracting from an entry

or I guess both are trivial but anyway lol

[rustic crown]

yea lol

(i was gonna say that and realized your second statement)

[gentle pendant]

🐱

[rustic crown]

so to finish we need to show that M(n, J) is also inside I

[void cosmos]

okay so first suppose I is an ideal in M_nxn(R) then say J = {a | a is M[1][1] for some M in I} , J is an ideal and the matrices of elements of J is a subset of I

correc?

or is it the other way around

yea I is a subset of matrices of J elements

ig

[rustic crown]

the last thing is still left to be shown

so say M is a matrix in M(n, J)

[void cosmos]

okay so take any element in I , row reduce kill multiply do whatever but basically we can extract any entry and put it in J

right?

[rustic crown]

consider the matrix M_{i, j} which has M[i][j] on the (i,j)th entry and 0 elsewhere

[void cosmos]

ok

[rustic crown]

yep

[void cosmos]

yea and its better to say multiply cuz it justs uses the fact that we are always going to be in J cuz its an ideal

[rustic crown]

sum of these M_{i,j} is M and each M_{i,j} is in I :3

[void cosmos]

so now we now matrices from elements of J contains I

right?

know*

[rustic crown]

okie we're at different pages >.<

i'm proving M(n, J) subset of I

you're thinking I subset of M(n, J)

[void cosmos]

ok

u go fist

first

[rustic crown]

okie lets do again

I ideal in M(n, R)

J = {a in R | a appears somewhere in some matrix in I}

[void cosmos]

ok

[rustic crown]

then we see that if a in J, then then matrix with a on (i, j) and 0 elesewhere is in I

[void cosmos]

why is that

[rustic crown]

cause we kill other rows and columns, and then transport that a to (i, j)

all by row/column operations

which we can do because ideals absorb multiplication on both left and right

[void cosmos]

oh

so a is in J --> a is an entry in some matrix

then destroy erase improve this matrix

[rustic crown]

[void cosmos]

okay

[void cosmos]

then whatr

[rustic crown]

from this we see that J is an ideal lol

[void cosmos]

ok

[rustic crown]

(i'm just changing the order of the proof a litlte)

okie now we need to prove I = M(n, J)

maybe you tell me now why both inclusions follow

[void cosmos]

okay so suppose A is a matrix in I then each entry is in J so why wouldnt A be in M(n,J)?

[rustic crown]

it would be

:3

[void cosmos]

okay

so now one is done

[rustic crown]

yep

[void cosmos]

now suppose A' is in M(n,J)

[rustic crown]

for this easier implication we didn't need this observation

[rustic crown]

we'll need it for this

[void cosmos]

1 sec

[rustic crown]

yee take your time

[void cosmos]

now A' has entries from J , so each entry is in some matrix in I

[rustic crown]

yep

[void cosmos]

okayy so like

we literally ressurect this matrix that its in I

so for example A' has like some entry call it a_i,j

then for each a_i,j we know there exists the matrix with the entry a_i,j and 0 everywhere in I

by daddy

[rustic crown]

yee

[void cosmos]

so thats it? we just show that A' = sum of these matrices?

and this sum is in I cuz I is an ideal too

[rustic crown]

yee

[rustic crown]

but you will have to assume that that a_{i,j} was in the (i,j)th entry

which you can do

[void cosmos]

yea yea

i will just enumerate them

no1 can tell me nothing

okay cool

tysm for having the patience with me

ur amaizng

[rustic crown]

np

[void cosmos]

is this ssupposed to be easy

to figure out on ur own lmao

[rustic crown]

it's just messy

not really hard

[void cosmos]

lmfao ig yeea

[rustic crown]

if someone gives you a few pointers and you sit down and do some computations then you can def do it on your own

[void cosmos]

all is it saying is if u have one element all u need is closure under additon and multiplicatoin so u make a matrix out of this element

or if u have a matrix*

so the statement M_nxn(F) has no nontrivial ideals is true

given F is a field

right?

[rustic crown]

yep

you call such simple rings

[void cosmos]

bro atiyah mcdonald has none of this shit

[rustic crown]

this is was an aluffi exercise

[void cosmos]

like the exercises were actualy easier

[void cosmos]

damn that hint

[rustic crown]

i remmeber grinding that like 2 years ago

[void cosmos]

cool ass exercise

it would have taken me atleast 2 hours

i am taking an exam in this

any tips on how to study

lmfao

i am just grinding dummit and foote's and trying to revise what i knew and forgot

[rustic crown]

ig just study stuff up to a point that you can answer all these in your sleep :p

[void cosmos]

i can answer some of then in my sleep;

them*

[rustic crown]

see that's progress

[void cosmos]

but i always had trouble with actually doing things by hand like actually doing examples

[rustic crown]

yea those require some practice to get comfortable

[void cosmos]

i can piece some thoerems together to get something new

[pastel cliff]

imma interrupt

[void cosmos]

but thats about it

ig

[pastel cliff]

hello det

[void cosmos]

the textbook putnam and beyond really helped with that tho

[pastel cliff]

if i can bother you a moment, just a nudge would be nice

[rustic crown]

that's usually enough for a start, rest you pick up as you study or think about it ever again

[pastel cliff]

im assuming something something first iso

[delicate orchid]

true...

[pastel cliff]

,av wew lads

[cloud walrus]

Wew Lads Tbh#0864's Avatar

Click here to view the image.

[pastel cliff]

jesus

[delicate orchid]

me irl

[void cosmos]

thats literally me

irl

[pastel cliff]

he's just like me

[rustic crown]

can you identify Z[x]/(2x+1) as a subring of Q?

[void cosmos]

yo is it ||Z[1/2]|| @rustic crown

yea

[rustic crown]

yes

[void cosmos]

ik this from my crypto class who would have thought

lmfao

[rustic crown]

wait is that even true? the left side doesn't have torsion elements

[delicate orchid]

is it supposed to be 2x-1?

[rustic crown]

which you see from this

[pastel cliff]

[rustic crown]

that doesn't change it though

[void cosmos]

a ring has identity iff R/I has an identity where I is a proper ideal

this is false right?

[delicate orchid]

you're right they're iso but it's easier for me to think about it as a minus

[rustic crown]

lmao

just think ||Z[-1/2] = Z[1/2]||

[pastel cliff]

is actually broken then

[pastel cliff]

this isnt immediately obvious to me

but probably polynomials w fraction coefficients? something like that?

[void cosmos]

yo sebb its similar to like how u could think of C as some R[x]/(x^2+1)

like field extensions

[rustic crown]

oh i'm saying subring of Q and not Q[x]

can you find a morphism Z[x] --> Q with kernel (2x+1)?

[pastel cliff]

polynomials with -1/2 as a root?

[delicate orchid]

close

that would be the kernel

so what would the map be with that as the kernel

[rustic crown]

lol

[void cosmos]

yo 6 is false right

[rustic crown]

what are equiv SES

[void cosmos]

u can loike

its like when u can make a diagram with one on top one of the bottom

with mapping each one

and everyting is commutative

[delicate orchid]

oh the funny 5 lemma

[void cosmos]

yea

[pastel cliff]

adjoined 1/2 then

so Z[1/2]

[rustic crown]

and all vertical maps being iso?

[void cosmos]

2Z isnt 3Z tho so this is false

[void cosmos]

exactly yea thats why its false

right?

[delicate orchid]

2Z is iso to 3Z lol

[void cosmos]

how

[delicate orchid]

x -> 3/2 x

inverse x -> 2/3 x

[rustic crown]

nZ is iso to Z

as abelian groups

[delicate orchid]

oh wait yeah I assumed they're groups whoops

[void cosmos]

yea mb those are rings

[tribal moss]

Without identity?

[rustic crown]

yee that was what you had to do lol

[void cosmos]

with

[delicate orchid]

3Z does not contain 1

[tribal moss]

But 2Z and 3Z don't have identities.

[void cosmos]

yea mb lmao

[rustic crown]

are we looking at SES in Ring?

[tribal moss]

Or perhaps in Z-Mod?

[pastel cliff]

wait so is this an L

[delicate orchid]

I mean regardless it doesn't seem to work anyway cause I don't get how you'd get an iso from Z/2Z to Z/3Z?

[rustic crown]

oh my eyes ignored this >.<

[void cosmos]

yea sorry boys its not specified

[rustic crown]

what's an L

[void cosmos]

but i just immedaitely thought 2Z cant be isomoprhic to 3Z (i can prove this) and thatsi t

[delicate orchid]

no boss... u basically did it I think

[pastel cliff]

like is it not true

[tribal moss]

We can't even begin to ask that question before we know which category we're working in.

[rustic crown]

yee, it's false

[void cosmos]

Ring

[void cosmos]

cool ty

[tribal moss]

You mean Rng?

[void cosmos]

yea here Rng

mb

[rustic crown]

i was rplying to sebb >.<

[pastel cliff]

oh

[delicate orchid]

I suppose but in any sensible category the fact that they're different cardinalities would stop an iso from existing

[pastel cliff]

wait im confus

[rustic crown]

but ig yours is flase also so nvm

[void cosmos]

cool

sorry for the confusion boys and girls

[tribal moss]

OK, that's fair. Apologies, I was still stuck on the left end of the sequences.

[pastel cliff]

i can see a map from Q -> Z oplus Z/2 oplus Z/2 ...

[delicate orchid]

no worries boss

[pastel cliff]

so i can kinda see how it would be true

[rustic crown]

(0, 1, 0, 0, ...) is an element of order 2 in that infinite direct sum. but the other ring is a subring of a field, so can't have 2-torsion elements

[delicate orchid]

this is the right idea, I think

[pastel cliff]

wew says it can be proven but det says no

[delicate orchid]

it's my idea anyway

go with det over me

[rustic crown]

det confused

which problem we talking about lol

[pastel cliff]

$\Z[x]/\ang{2x+1}$ is isomorphic to $\Z \oplus \Z/2 \oplus \Z/2 \oplus \dots $ as a $\Z$-module.

[cloud walrus]

sebbb

[delicate orchid]

Q is definitely torsion free as a Z-module det is right

it's projective innit

[pastel cliff]

"as a Z-module"

[rustic crown]

so you see that the first ring is isomorphic to Z[-1/2] right?

[pastel cliff]

fuck

yeah

[rustic crown]

and you see that it's torsion free right

because submodule of Q

[rustic crown]

(lol why bring that here xD)

[pastel cliff]

i see that it's torsion free bc no elements of it can be taken to zero

but ig the submodule thing is obv

[delicate orchid]

gonna throw something stupid out there just to see what happens - aren't infinite direct sums weird?

[pastel cliff]

i confus by them yeah but idk

[coral shale]

what does weird mean

uwu

[delicate orchid]

yeah nvm I just remembered why they're weird - doesn't make a difference here

[tribal moss]

They're "weird" in the sense that they're not products :-D

[delicate orchid]

elements of infinite direct sums can only have finitely many non-zero "coordinates" iirc

[coral shale]

i just think of them 'like' infinite vector spaces

[rustic crown]

so that's false right, it's flat but not projective

[coral shale]

and linear combinations

and stuff

[rustic crown]

idr projectives over pid

[delicate orchid]

it's defo projective as a Z-module it's a divisible group

I'll google

[rustic crown]

flat iff torsion free was prlly true

that was injective!!!

[delicate orchid]

https://tenor.com/view/dog-puppy-looking-sussy-staring-gif-26025556

[rustic crown]

and Z[1/2] isn't even divislbe

[delicate orchid]

blunder...

[pastel cliff]

what defn of torsion do you mean

[delicate orchid]

rm = 0, m and r non-zero, I imagine

[rustic crown]

(non-zero divisor)

[delicate orchid]

all rings are integral domains 😌

[rustic crown]

.<