groups-rings-fields

[woeful sage]

an elementary school child can learn elements of naive set theory

just as they learn basic arithemtic

you don't need to sit through a first order logic book

[worn tinsel]

Mh maybe I'm approaching the learning process from the wrong angle, I mean I don't get stuck learning stuff, simply it is hard to judge what to learn next

[woeful sage]

That's why there's this server

You can ask people here

[worn tinsel]

And it is harder when different authors use absolutely different set of axioms or notations, happens a lot in logic

[woeful sage]

well 99.999% of mathematics uses ZFC axioms so you don't have to worry about that

notation is a different story

you just get used to it

[worn tinsel]

well thanks for the suggestions then, so after the relations and functions part you're supposed to go through abstract algebra? I've tried to see a curriculum from an university or something but it is not that clear

[woeful sage]

not exactly

you can go through either abstract algebra or real analysis

[worn tinsel]

never heard of the second one, what would be more fitting if I'm more interested in the structural part of mathematics and computer science in general?

[woeful sage]

real analysis is very necessary for stuff like analysis of algorithms (I think )

You analyze algorithms as the input size tends to infinity, (you look at their asymptotic behaviour) and a major part of analysis is making sense of infinity

But this is getting off-topic this is #groups-rings-fields

[worn tinsel]

Oh maybe I'm already learning that, but I didn't knew it was named real analysis, I was reading that from CLRS

[woeful sage]

you can go through either, if you choose to go through analysis, I'd recommend "Understanding Analysis" by Stephen Abbott

[worn tinsel]

Well if I had to define by the lack of knowledge I have, but atleast knowing what amazed me more, it is more about the foundational fields and pure math, will that be more fitting going through abstract algebra?

[worn tinsel]

I'll save that one too then, ty

[woeful sage]

flip a coin! they're both equally foundational pure math

[worn tinsel]

Okay I don't want to offtopic with it, I'll give it a try then, didn't know that would be consider equally foundational, thought it was more about abstract algebra and category theory.

Well thank you again for all your suggestions, they suffice more than what I was searching for! I'll start with groups, GN.

[stone elbow]

I need a hint on how to prove that $\mathbb{Z}[i]$ is a Euclidean domain. It's trivial to show that it's a domain using the the usual complex norm. I struggle coming up with a division algorithm though.

I tried defining $c, d \in \mathbb{Z}[i]$ such that $a = c + bd$, $|c| < |b|$ by first defining $|d|^2 = \lfloor |a|^2 / |b|^2 \rfloor$, then defining the argument such that the distance between $a/b$ and $d$ is minimal among elements of $\mathbb{Z}[i]_{|d|^2}$. But so far I can't find a bound for $|c|/|b|$, my naive attempt has led to a bound that goes to infinity as $|a|/|b|$ gets large. Am I on the right track, generally? Should I try another rounding function?

[cloud walrus]

a-square

[stone elbow]

Specifically, if we let $|a|/|b| = u \in \mathbb{Z}^+$, $\varepsilon = |a|/|b| - \lfloor |a|/|b| \rfloor$, we get that $-\pi/4 \leq \operatorname{arg}(a/b) - \operatorname{arg}(d) \leq \pi/4$, and this leads to $|c|^2/|b|^2 \leq (u - \sqrt{\frac{u - \varepsilon}{2}})^2 + \frac{u - \varepsilon}{2}$. By taking $\varepsilon = 0$, we get $|c|^2/|b|^2 \leq (u - \sqrt{u/2})^2 + u/2$, which diverges when $u \to \infty$.

[cloud walrus]

a-square

[rocky cloak]

I would start with normal division of complex numbers.

Like given x = a + bi and y = c + di, then x/y is some complex number. Then you can round the real and imaginary part to closest integer, say z = e + fi.

Then
x = (x/y)y = zy + (x/y - z)y
what's the norm of (x/y - z)y? Use that the real and imaginary part of x/y - z is small

[amber burrow]

when does group theory stop being like, just about group theory itself

going through dummit and foote and im realizing more and more that the group theory chapters are just about classifying groups

ok my original question was phrased wrong

i mean like, when do you actually do stuff with groups, instead of just classifying them

[thorn jay]

in fields that aren't group theory :P

group theory is about, well, studying groups

applications are in e.g. algebraic topology with the fundamental group, homotopy groups, and even homology groups (or modules)

[rocky cloak]

And I guess computations in Galois theory is a place where classifications of finite groups is quite useful

[thorn jay]

ah, yes, true

[velvet hull]

i would also mention combinatorics, should one (somehow) want to purse it

[chilly ocean]

the most elegent two line proof was using lattices

[lusty marlin]

What is this proof?

[chilly ocean]

taking gaussian integers as lattice point that was the idea

[lusty marlin]

I don't see how this leads to a proof that ℤ[i] is a Euclidean domain though.

[chilly ocean]

for that see neukirch algebraic number theory chapter 1 page 1

[obsidian whale]

wow that is a very clean proof

[lusty marlin]

Thanks

Very nice proof

[grizzled spindle]

What’s standard notation for normal closure

My prof uses «R»

[rocky cloak]

Wikipedia suggests five different notations
https://en.m.wikipedia.org/wiki/Normal_closure_(group_theory)

[white oxide]

Where do we use unique factorization? I know that if gh in (f), then f divides gh, but don't we have irreducible <=> prime in a PID (and k[x_1, \dots, x_n] is UFD but not a PID)

[rocky cloak]

I'm not sure I fully understand your question, but irreducibles being prime is one of the defining features of being a UFD.

It is true that PIDs are UFDs though if that's your question

[white oxide]

Oh really

Huh I wonder why it was only stated for PIDs

I just assumed it wasn't necessarily true for UFDs

That irreducible => prime

[rocky cloak]

UFD is just every element is a product of irreducibles + every irreducible prime

[white oxide]

Hm okay

Thanks

[white oxide]

Why in considering infinite sums do we often require all but finitely many terms are zero

[vocal pebble]

Im not sure what this is in reference to, but say in an abelian free group you want to consider only finite sums because the infinite sums with infinitely many nonzero terms may not "converge" to anything in the group

[white oxide]

I see

It was for infinite sums of ideals

[rocky cloak]

Also a sum of ideals is really the smallest ideal containing the others.

An ideal is only defined to be closed under binary sums (because that's usually the only type of sum that is defined). Repeated binary sums can become arbitrary finite sums, but not infinite

[tardy hedge]

Thanks jagr for always being a great help 👍

[next obsidian]

Thanks jagr

[white oxide]

Ty jagr

[potent condor]

np

[woeful sage]

Lie Groups

[woeful sage]

is that not doing stuff with groups?

I think classification is a major part of mathematics

People try to classify groups, manifolds, lots of stuff

[chilly ocean]

thanks @rocky cloak because of you i solved a huge 20 point question in end sem

[tardy hedge]

Its studying groups for the sake of groups but its not doing stuff with groups

[woeful sage]

That's still group theory

[woeful sage]

But I suppose they were asking about applications of group theory within mathematics

[thorn jay]

Is there a name for the subdirect product G \rtimes G where G acts on itself by conjugation?

[cinder fox]

Is this statement correct?

If R and S are isomorphic rings, let $f:R\to S$ be an isomorphism. If $T$, a subring of $R$, is a field, then so is $f(T)$.

[cloud walrus]

struct ∅ {};

[thorn jay]

yes

(assuming you use the correct convention that rings are unital) that's not even needed here

[tardy hedge]

why what were you thinking of initially?

[thorn jay]

I have no clue

I don't even remember writing that, frankly

[tardy hedge]

How can a field not have identity?

[thorn jay]

The rings

A ring can be a subrng of a rng without identity

[tardy hedge]

ohh ok

[tardy hedge]

tensor product of abelian groups is like taking tensor product over Z with Z-modules right

[thorn jay]

it's literally that, yes

[crystal vale]

I am thinking about the group which is finitely generated but there is a subgroup of that group which is not finitely generated

[rocky cloak]

The free group would be such an example

(on more than one generator)

[thorn jay]

many such examples, what a world it would be if every f.g. group was Noetherian

😔

[chilly ocean]

i was trying to find a subgroup with same property inside SL_2(Z)

[crystal vale]

I don't know much about free groups, but I think it is a group which has a generator without any relation?

So can I say Z × Z is a free group with generators (1,0) and (0,1)?

[rocky cloak]

It's free abelian, but not a free group

Commutativity is a relation

[chilly ocean]

it needed a word relation

[crystal vale]

Actually I thought about it

So what is an example of free group with two generators?

[rocky cloak]

The free group generated by {x, y} is the set of all words with letters x, y, x^-1, y^-1 where the operation is concatenation of words

And x x^-1 etc is the empty word

[crystal vale]

I see

[tardy hedge]

i always forget about examples of fg R-module that has non fg submodule. Is the typical example that of considering polynomial ring in infinite many variables as a module over itself?

[rocky cloak]

Sure

Noncommutative polynomial ring in more than one variable also works

[tardy hedge]

is this also an example of a free module with finite rank that has a free submodule with not finite rank?

[rocky cloak]

The noncommutative polynomial ring is

[tardy hedge]

the ideal generated by all variables is free but not finite rank no?

[rocky cloak]

It's not free

[tardy hedge]

oh ok yea

[chilly ocean]

i love module over pid

[tardy hedge]

its free as R-mod not R[x1,....]-mod

[rocky cloak]

Yes

x1 x2 - x2 x1 is a relation

[tardy hedge]

me too i want to review why the pid condition makes things work so nicely

[rocky cloak]

In the commutative case the rank of a submodule is always smaller

[tardy hedge]

rank here meaning size of a basis or size of biggest set of independent elements?

[chilly ocean]

pids are close to fields

[rocky cloak]

Either

[tardy hedge]

in the commutative case of free module finite rank, youre saying if a submodule is free then it has finite rank that is smaller?

[rocky cloak]

Yup

Or equal of course

[chilly ocean]

which definition of rank you are using

[tiny jolt]

I proved both of these facts without using the fact that pi was irreducible, should i be suspicious?

[chilly ocean]

send your proofs

[tardy hedge]

Is this like an obvious fact or. Tbh, the nuances with free modules ive been weary of for a while

[tiny jolt]

in the first part, "is an ideal" means "is closed under multiplication by anything in ..."

[tardy hedge]

Hard to remember them

[rocky cloak]

Not obvious no, though the rank 1 case should be, since
x*y - y*x is a relation between x and y

[chilly ocean]

what is R

[tardy hedge]

That relation is ring elements?

[tiny jolt]

oops, Z[i]

[tardy hedge]

Oh ok so it cant be more than rank 1 because you always have a relation that contradicts freeness

[rocky cloak]

Like if x and y are two elements in R, they are not linearly independent

[tardy hedge]

So that shows in comm case a rank 1 free module with a submodule that is free cant have rank more than 1. But how do we know its for sure fg?

Maybe silly questions but i havent thought too much about free modules that arent fg

A free R module that is not fg is still just a direct sum of copies of R right

[rocky cloak]

If it's not finitely generated then it's basis would be infinite

But the basis can't have size more than 1

[tardy hedge]

Im a bit confused by this still. A free submodule cant have rank more than 1 because ..

[rocky cloak]

So the rank is the size of the basis. A basis consist of linearly independent elements. No two elements are linearly independent

So there cannot be two different elements in the basis

[tardy hedge]

Ok that makes sense but why were you saying that relation with just ring elements? Isnt that relation always true

[rocky cloak]

What do you mean?

[tardy hedge]

Im thinking since the module is rank 1 there cant be more than 1 module elements that are lin ind, so thats why a free submodule must have rank 1. Im not sure where the xy - yx thing comes in

I guess im confused cause im thinking of equation using module elements and ring elements

Not just ring elements

[rocky cloak]

Well apriori it's not obvious that the size of a basis and the maximal number of linearly independent elements is the same

(and indeed it's only true in the commutative case)

[tardy hedge]

Aghh yea thats the kind of stuff that makes me go 😩

[chilly ocean]

i have a problem on modules

[tardy hedge]

Omg im going crazy lol, im like {2,3} is lin independent in Z over Z right? But no because -32+23 = 0. I thought that contradicted bezouts lemma but i guess 0 is still a multiple of gcd of 2,3

I thought it was independent initially cuz i interpreted bezout lemma wrong

I thought it was like since gcd is 1 any combo is 1 or a multiple of 1 but i forgot 0 is a multiple of 1. And obviously no integer pair would be independent but im just speaking out loud

Anyway

[tardy hedge]

Post it

[chilly ocean]

We give the C-vector space $C[x] $ a $C[t]$-module structure by setting $(t.f) = x^2f, f\in C[x]$ Show that $C[x]$ is a free module over $C[t]$ and compute its rank

[chilly ocean]

Posting isn't the problem,

[cloud walrus]

Akhi Mishra(Riemman's Cat)

[next obsidian]

Is ts from an exam bruh

[chilly ocean]

yeah

[next obsidian]

…

…….

[thorn jay]

The humble loading bar

[tardy hedge]

Wdym

[chilly ocean]

wtf

[tardy hedge]

Is guy currently doing an exam and asking for help on it?

[chilly ocean]

no

[tardy hedge]

ts 🥀

[thorn jay]

French interval notation??

[next obsidian]

French write ]x,y[ for (x,y)

[tardy hedge]

Thats crazy

[thorn jay]

Jumping off a bridge

[next obsidian]

Or at least used to, and you’ll still see it occasionally

[chilly ocean]

the problem is given, because it is related to your discussion

the problem is already solved

[thorn jay]

I mean I hope so if it was on an exam

Lol

[chilly ocean]

yes it was

[thorn jay]

Not getting my joke qwq

[chilly ocean]

how one can ask it on live exam

[thorn jay]

This is an insane bingo

I love it

[chilly ocean]

yeah but not in exam

[keen badge]

Why $\frac{\mathbb{Z}\times \mathbb{Z}}{\langle( 1,-1) \rangle}\cong \mathbb{Z}\times \mathbb{Z}_2$?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[wraith cargo]

Last time I did an NSFW shit post was like last September
I gotta step up my game

[wraith cargo]

(a,b) |-> (a+b, a+b mod 2)

Actually no

That isn't surjective lol

Actually that isomorphism isn't even true I think

It should be Z

Not Z x Z/2Z

[lusty marlin]

Yes

[next obsidian]

(1,-1) is the kernel of the addition map

[thorn jay]

This just in, Z has torsion

[keen badge]

This is what I thought, but I know that the first homology group of the Klein bottle is Z X Z_2 and when I tried to solve it via Mayer Vietoris I got that H_1(K) is this group, so they have to be isomorphic. I just dont see how...
(Im sorry for being off topic with homology groups, but this is the context for the question)

[next obsidian]

Z’s balls:

https://tenor.com/view/torsion-gif-16742133732546221725

[wraith cargo]

https://tenor.com/view/how-do-we-tell-him-gif-8272342975911687520

[thorn jay]

Then you probably made a mistake somewhere, because these two are not isomorphic

[keen badge]

I don't think I had a mistake tho

[thorn jay]

I'd reckon you making a small mistake is more likely than two objects both not being isomorphic and being isomorphic

[next obsidian]

Lmfao

You should now

[keen badge]

I should now what

[next obsidian]

You should now think you made a mistake

[keen badge]

https://tenor.com/view/no-gif-23548728

[next obsidian]

Z and Z x Z_2 are not isomorphic

[keen badge]

I just used this exact sequence

[next obsidian]

And your work would have you conclude they are

So you made a mistake

[rocky cloak]

Klein bottle is two Möbius strips glued along their boundary right?

In which case MV should give you Z^2 /(2, -2)

[next obsidian]

This is what I imagined happened cuz then you do get Z x Z_2

[keen badge]

Of course they’re not. It’s just a bit frustrating that I can’t find my mistake...
Oh well — to hell with those 5 points.

[keen badge]

yeah, but I did not look at the bottle like this, rather as two cylinders glued along S^1

[next obsidian]

How do you do that

[keen badge]

I am not sure.

Maybe

[wraith cargo]

You can do that but it's tricky

You have to do an extra twist

And you get an extra 2 factor in there

[next obsidian]

Isn’t the standard way to do this computation to just use a polygonal description anyway

[wraith cargo]

Yeah

[keen badge]

The point is to do it using Mayer-Vietoris

[next obsidian]

Tough

[wraith cargo]

Ruff

[stone elbow]

Hi all, does there exist a free resolution of $\mathbb{C}[z, \overline{z}]$ as a complex algebra?

[cloud walrus]

a-square

[next obsidian]

There’s a free resolution of everything

But what do you mean a resolution as an algebra?

This is a real thing, but I don’t think it’s what you want

Also I guess wtf is C[z,z-bar]

I guess what I am now realizing is maybe “complex algebra” means something different than what I was interpreting it as

[stone elbow]

I’m trying to represent trigonometric polynomials as restrictions of complex polynomials to the unit circle, but the conjugation is ruining my day 🙂

[next obsidian]

That doesn’t answer anything for me lol

[stone elbow]

So I’m trying to come up with the “correct” ring, and currently I’m stuck trying to express the ring of polynomials in z, z-bar with generators and relations

[next obsidian]

What is z-bar

[stone elbow]

Complex conjugate of z

[next obsidian]

What is z

[stone elbow]

z is a variable

[thorn jay]

😭

[next obsidian]

What the hell is the conjugate of a variable

[stone elbow]

that’s what I’m trying to understand lol

[next obsidian]

…

[thorn jay]

is this perhaps some algebra equipped with an involution?

linear involution*

[next obsidian]

I mean I guess you can say you have a

Uhhh let’s say it’s called a

Qolynomial

Which is a function inputting a single complex number z

[stone elbow]

Probably, from the analysis point of view this is just real-smooth function x + iy to x - iy

[next obsidian]

Which is formed by sums, products, and taking complex conjugates

Where z-bar represents taking the conjugate

So you can then say the ring of qolynomials is denoted C[z,z-bar] as a subset of all real-smooth maps C -> C

But this is (I think) different than a formal polynomial ring

But let’s say this is the ring you’re dealing with

Then just as a ring, you have a free C-resolution both as a vector space and as an algebra

[stone elbow]

This tracks

[next obsidian]

For the existence of a free resolution as a vector space just any resolution is free

For a free resolution as an algebra (which I truly don’t believe is what you want) there’s a canonical resolution for any A-algebra, A a commutative ring

[thorn jay]

you get a polynomial-ish "ring" over an extended language, which essentially behaves the same as complex conjugation is a ring hom

[next obsidian]

If the algebra is R, you take the (hella infinite) polynomial ring A[R] where all elements of R are now new variables

And the map A[R] -> R is the map sending the variables r to the element r

And extending A-algebra-ically

[stone elbow]

Freaking monads 😄

[next obsidian]

And then you just repeat this process

This gets you a simplicial resolution by free A-algebras

Or whatever you want to call this, it’s a simplicial A-algebra with an augmentation to R

But I seriously doubt this is what you want

I did see tteg typing tho

She probably has smart things to say

[boreal inlet]

Oh right, I completely forgot about the divisibility condition, yeah, my bad

[glad plover]

What is the best way to learn to understand Abel-Ruffini Theorem?

[boreal inlet]

I'm not sure if I should ask this here or alg-geo but I'm trying to understand the structure sheaf of Spec(A) where A is a commutative unital ring.

How do I show that stalk at x \in Spec(A) is isomorphic to the localisation of A by the complement of the prime ideal p_x containing x?

I have defined the map, and also have proved its surjective

But I'm not sure how to show its injective.

The map comes from the direct limit cone that defines the stalk, and that also shows why it's surjective

[chilly ocean]

alg geo of course

[boreal inlet]

Alright then

[fading skiff]

The map comes from the direct limit cone that defines the stalk, and that also shows why it's surjective

[boreal inlet]

Uh.. i mean yes

How do I show its injective?

[fading skiff]

equality in the big localization, “comes from” equality in some smaller localization. Basically ensuring injectivity of the map from the direct limit.

[boreal inlet]

I don't think I get why it's useful

I was thinking more on the line of showing that kernel is zero here

Let's say there is one element in the stalk which gives zero when mapped via the map we defined via the direct limit cone

[fading skiff]

If an element dies after localizing at p_x, it must already be zero somewhere “close enough” in the directed system of localizations.

so I Guess yeah

[boreal inlet]

Okay I think I am somewhere

[chilly ocean]

on the moon

[boreal inlet]

😭

[rocky cloak]

If your goal is just to understand Abel--Ruffini, then there is a fairly elementary proof due to Arnold.

This video goes through that proof
https://youtu.be/BSHv9Elk1MU?si=X1Un0GtJU9yJUdQ2

Otherwise Abel--Ruffini is usually proven as a corollary if Galois theory, and then you're also able to say something more quantative about which polynomials can be solved.

For this I guess just pick up a book in Galois theory, might require you to know a little about groups rings and fields first.

[stone elbow]

Can you check if I got it right, please?

[stone elbow]

Nvm found a mistake

[boreal inlet]

This is probably a very stupid question but let's assume S and T to be two multiplicatively closed subsets of a commutative unital ring A such that S \subset T.

Then do we have an inclusion of S^-1A into T^-1A?

[rocky cloak]

There's a natural map, but it's not necessarily an inclusion

[boreal inlet]

The natural map is sending a/s to itself right

[rocky cloak]

Yes

[boreal inlet]

I mean the classes, they might be different in each licalizayon

Localization*

oof typo

[rocky cloak]

So you can reduce this to the case S = {1}

[boreal inlet]

Oh, A to T^-1 A

[rocky cloak]

You have a map A -> T^-1 A, but it's not necessarily an inclusion

[boreal inlet]

Oh wait actually I need an example because I'm not sure I see it for this particular case

because a goes to a/1 in T^-1A no- oh

Wiat my bad

I got it

Sorry

[thorn jay]

Any ring A with any multiplicative system containing a zero divisor will not yield an embedding

[boreal inlet]

Precisely yeah

I remembered the example I did while solving Atiyah

The reason I asked this because I was trying to actually make sense of the restriction maps in the sheaf structure of an affine spectrum

They need not be inclusions, as long as they are ring homomorphisms it would be fine

[thorn jay]

It would be rather uncharacteristic for them to be inclusions

Because two nonequal continuous functions on U can have the same restriction to U' ⊂ U, and that is in fact almost always the case for some pair of functions

[boreal inlet]

Okay, good, thanks a lot actually

But this now makes me ask this question:

Let's say we take the presheaf of rings of the affine spectrum of A, and we have an element b/s in one O(U). Let's say in a restriction from O(U) to O(V), b/s becomes zero. Can I comclude b/s is zero? And Why?

[thorn jay]

U and V are distinguished open sets here?

[boreal inlet]

[If one needs to define O(U) explicitly, first we define ∆(U) which is the complement of the union of prime ideals in U, and then localize A wrt ∆(U).]

[rocky cloak]

I guess you can think about the analogy again.

If a function is 0 on an open set V must it be 0 everywhere?

[boreal inlet]

Yes, they are open in Spec(A)

[thorn jay]

Distinguished - are they the sets of prime ideals not containing some distinguished element

Or just general open sets

[thorn jay]

But this

[boreal inlet]

Not necessarily right? Because just take the floor function and check on (0,1) and (1,2)

[boreal inlet]

Yes, the example I'm taking have both of them to be distinguished. Sorry.

U = D(s) and V = D(st)

[rocky cloak]

Your example isn't continuous, but anyway. You shouldnt expect that, so you shouldn't expect it here either

[thorn jay]

They're continuous wrt to the discrete topology

Lol

[boreal inlet]

Thank you, I was thinking of that exactly. Okay, now - which leads me to the actual question - I'm actually trying to prove that stalks of the structure sheaf on Spec(A) are isomorphic to A_p_x.

I have defined the map (using the direct limit cone) and also have shown it is surjective. How do I show it is injective?

Someone here suggested that showing restricting the elements to a smaller set and showing its zero should work here, and that's why I asked the original question

I was trying to prove that the kernel is zero basically

[rocky cloak]

Well are you aware of the descriptions of elements in a direct limit and when they are 0?

[boreal inlet]

I am not really sure, is it somewhat like: if a restriction map sends one element to 0, at some index, then it's actually zero in the limit?

If it's true, I haven't proved this yet

[rocky cloak]

That is indeed true

[boreal inlet]

Oh. That explains why it's true here...

[rocky cloak]

Okay, but if you don't know what the elements of the direct limit is, how can you hope to explain which of the elements are in the kernel?

[boreal inlet]

Yeah... i'll try to look it up then. I do know the definition of how the direct limit is constructed when the directed system is in place, but if that doesn't help I'll try something else

[rocky cloak]

I guess a different approach could be showing that A_p satisfies the universal property of the direct limit

[thick tapir]

doesn't entering 1 into this algorithm output multiple values (either 2 or 3), since 1 is in multiple different cycles? why does 1 => 3 if the cycle (1 2) exists as well

[cinder bluff]

It looks like your book follows the convention of reading cycles from the right to the left

So to get the output for 1, first perform (2513), then (13425), and last (12)

[thick tapir]

ah ok, thanks

[ivory ore]

if e is an idempotent element

[cloud walrus]

longboard kayak

[ivory ore]

is also a subring of R

also

[cloud walrus]

longboard kayak

[next obsidian]

I disagree

[ivory ore]

well i have done the proof but is there any use that i can get

[next obsidian]

Because my rings got 1

A subring has to have the same identity element for my definition, and eR, Re, eRe won’t have the same identity

I think eRe is a subring of eR though…???

Cuz e is the identity on both…?

[ivory ore]

yeah

right, ok then it is at least a ring on it's own

[next obsidian]

Anyway, idk about the eRe thing but

Knowing that eR is a ring in its own right is good

[ivory ore]

alr

[next obsidian]

Let’s say we are dealing with commutative unital rings so I know what I’m saying is def true

Maybe it’s true more generally

But if e is an idempotent so is 1 - e

And you get that R ≈ eR x (1-e)R

And likewise if R ≈ S X T then (1,0) and (0,1) are orthogonal idempotent

Meaning idempotent where xy = 0 and x + y = 1

[ivory ore]

sounds like direct sum type shit

[next obsidian]

So splitting of rings into products are exactly the same as having non trivial idempotents

And this is also the same as having a disconnected Spec

[next obsidian]

Yeah

Splitting a ring into a direct sum of (not quite sub)rings are idempotents

[ivory ore]

https://planetmath.org/cornerofaring

corner of what

A classic example is the Peirce corner eRe, where 'e' is an idempotent in R (e² = e). This corner is not a subring of R unless e is the identity element.

what you mentioned as a caution

[next obsidian]

An okay so I think what I said is the simplification when you’re commutative

Otherwise I guess you need to do eRe to be closed or something

Yeh cuz multiplication bad

If not commutative

And if you are commutative then obviously eRe = eR = Re

I’m the chmoat

[ivory ore]

in anyways ex = xe

[next obsidian]

Sure if e is in the center

Is technically a stronger statement

[ivory ore]

so idempotent always lies inside?

that's a nice thing i learnt

I found a lemma being proved in m.SE

[cloud walrus]

longboard kayak

[hot goblet]

Does anyone have any tips for computing splitting fields of polynomials with horrendous roots? Is the best strategy to compute the extension E as E=F/<irr> and try to simplify that field down?

[ivory ore]

something i found which seems relevant to the prior chat

[south patrol]

Wow your rings are rings

[chilly ocean]

galois radical guy

[next obsidian]

Potato self react, they wanna be nG soooooo bad

[south patrol]

I mean this is true

[chilly ocean]

does anyone know about arnold cat map

[hot goblet]

im stupid the answer is use the galois group

that tells you the order of the splitting field extension

[chilly ocean]

what is horror roots

[hot goblet]

its only a cubic but the roots look like this

[chilly ocean]

what u want to do with this

[hot goblet]

compute the order of the splitting field

i think i know how to do it

ill get back to you if I cant figure it out

[chilly ocean]

yes galois theory is needed, so you found a minimal cubic irreducibe polynomial right?

[hot goblet]

yeah no dont worry I got it, figured it all out

[chilly ocean]

whats the ans

[hot goblet]

6

deg 3 extension to add the real root and deg 2 extension to add the two complex roots

[hot goblet]

Is it true all cubics with roots outside Q have field extension degree 6 over Q

[chilly ocean]

you mean irreducible cubic

[rocky cloak]

No, there are irreducible cubics with Galois group C3

[glad osprey]

Do you mean if f(x) is an irreducible cubic in Q[x], does the splitting field always have degree 6? Then no, the splitting field is Q(a, sqrt(D)) where a is a root of f(x) and D is the discriminant. I think x^3 + 2*x^2 - x - 1 is a counter example, it's irreducible with discriminant 49, and its splitting field has degree 3

but if f(x) has complex roots, then the splitting field has degree 6

[hot goblet]

Ah thank you everyone who responded

[hot goblet]

ah I see

[hot goblet]

Im assuming this counterexample is the Z/3Z case @rocky cloak mentioned, if so where does this cubic come from/how was it constructed

[rocky cloak]

So there is a formula for the discriminant in terms of the coefficients.

So you can basically play with it a bit until you find something where the discriminant is a square

[hot goblet]

ahh I see

[glad osprey]

I constructed it by tedious experimentation in sage (by spamming discriminant on various polynomials) 😅 there's probably easier ways to find it

[hot goblet]

you did mention discriminant here so yeah ig it is relevant

this is a really good counterexample ill keep this one in my back pocket

[hot goblet]

Sorry I was thinking further on this and what is the concept of discriminant in this cubic? Is it something to do with the resultant quadratics when factored linearly?

I havent really looked closely at cubic behaviour since I was 16 and those were on very nicely behaved cubics

[rocky cloak]

In general the discriminant of the polynomial is the product of the differences of its roots

[hot goblet]

so in (x-1)(x-6) = x^2-7x+6 the discriminant should be 25

how does this relate to the roots?

[rocky cloak]

Well, the relationship to the roots is pretty explicit I guess.

For a quadratic it is irreducible iff the discriminant is not a square (the discriminant appears in the quadratic formula).

In the degree 3 case the splitting field of the polynomial is given by adjoining a root and a square root of the discriminant, so it tells you something about the degree of the splitting field.

In general the discriminant generates the field fixed by the even permutations, when viewing the Galois group as permuting the roots, so whether the discriminant is a square tells you whether the Galois group is contained in the alternating group

[hot goblet]

ahhh this is fascinating

thank you very much

galois theory truly is amazing

[amber burrow]

Is it important to know how to prove Sylow theorems?

Everytime I look at the two page proof in my book and i'm like, i dont have the mental capacity to actually read and understand this

[south patrol]

In applications just the numbers are important

But sometimes doing small variants of the group actions is ok

[chilly ocean]

there is a simple few lines proof available

using embedding

[chilly ocean]

for sylow first theorem it is mentioned in excersises of foote

[frigid shard]

i wrote up some notes on sylow for a course a while back

might be of use to u

[final trellis]

what are some methods to find inverse of a permutation polynomial?

[stone elbow]

It’s pretty cool that the complex form of trigonometric polynomials can be viewed as the embedding $\mathbb{R}[\sin x, \cos x] \to \mathbb{C}[z, z^{-1}]$!

[cloud walrus]

a-square

[chilly ocean]

so both of them are pid

[stone elbow]

No, $(1 - \cos x)(1 + \cos x) = (\sin x)^2$, so the former isn’t even a UFD

[cloud walrus]

a-square

[thorn jay]

What's the embedding?

[chilly ocean]

are they irreducibles

[stone elbow]

Yes, because they have the trigonometric degree of 1

[stone elbow]

Note that on the unit circle, $z^{-1} = \overline{z}$

[cloud walrus]

a-square

[chilly ocean]

so you mean the embedding isnt surjective

[thorn jay]

Usually embeddings aren't surjective

Lol

[stone elbow]

Correct, the image consists of real-valued functions

[chilly ocean]

Replace R by C

THEN what happens

[thorn jay]

I don't see it

As in, the embedding

[chilly ocean]

yes same

your map isnt clear

For R

[thorn jay]

Neither for C

[stone elbow]

$\cos x \mapsto (z + z^{-1})/2$

[cloud walrus]

a-square

[thorn jay]

Ah, I dee

See

And that satisfies the necessary relation

[chilly ocean]

FOR C it is straight forward

[thorn jay]

Not really, you still need to say where cos and sin go

Lol

Cuz if there's an embedding from C[sin x, cos x] then there is also one from R[sin x, cos x]

[stone elbow]

I haven’t checked, in this case we might get an isomorphism

[chilly ocean]

yes u will get it

[thorn jay]

I guess the image of cos(x) + sin(x) is z, naturally

[stone elbow]

$\cos x + i \sin x$

[cloud walrus]

a-square

[thorn jay]

Close wnough

[stone elbow]

This embedding is the best way to solve exercise 11 in Lang:

BTW Springer’s print-on-demand is the best way to get a hardcover, Amazon’s paperbacks are quite flimsy

[stone elbow]

In the ambient ring, $2iz \sin x = (z - 1)(z + 1)$, which is also pretty sweet in and of itself

[cloud walrus]

a-square

[stone elbow]

$z \pm 1$ are irreducible because they are prime (the localisation only kills one maximal ideal)

[cloud walrus]

a-square

[stone elbow]

It’s a bit ironic that functions on one conic describe functions of a different kind on another conic

[chilly ocean]

Without embedding how one can do it

[stone elbow]

We did it in undergrad analysis somehow, I don’t remember how (probably by the orthogonality of the Fourier basis)

How do you teach it to undergrads?

[white oxide]

Here r(a) is the radical of a. Why is (p_1, \dots, p_r) contained in r(a)? We don't necessarily have p_i^k \in (m) for any k in Z, and I don't know why an average element a_1p_1 + \dots + a_rp_r is in r(a). Is it because we can raise it to some integral power so that we can factor out m?

[barren sierra]

It's not $(p_1, \ldots, p_k)$ it's $(p_1 \times \cdots \times p_k)$

[cloud walrus]

Spamakin🎷

[white oxide]

Oh lol

That makes a lot more sense thanks

Didn't notice that there was no comma 😂

[sturdy spear]

Lang mentioned

[tardy hedge]

I feel like the vibes here have been more quiet recently

Maybe im imagining

Where the algebros at

[tough raven]

Here, something for them to do.

[crystal vale]

so in my recent interview, they asked me about the quotient group {Q\{0} } \ H, where H = { x^2 | x in Q\{0} }. so first I found all the possible cosets, and i know every non-identity element has order 2, what can i conclude about that?

[chilly ocean]

what is the exact question

[crystal vale]

they said, tell us more about this quotient group

[chilly ocean]

can you find non trivial proper subgroup of this group

[crystal vale]

yes

every non-identity element has order 2, so we can

[chilly ocean]

of infinite order

[crystal vale]

and i think it is related to vector space

[crystal vale]

okay

but what is the motivation to find this non trivial proper subgroup?

[chilly ocean]

you want to tell more about this group

maybe there is no finite index subgroup

[crystal vale]

i think they want to show that it is isomorphic to product of infinites copies of Z/2Z

[chilly ocean]

whats your vector sp structute a.x=x^a ? a \in Z/2Z

[crystal vale]

maybe thats works

we have to verify that

[chilly ocean]

you can verify all the properties

[crystal vale]

okay

[chilly ocean]

addition is group operation and multiplication is just given above and in Z/2Z it works

[ivory ore]

Rings and Modules: i'm lost

[thorn jay]

Hiding in my thread

[ivory ore]

so aut(M) are the only units in End(M)?

[thorn jay]

By definition essentially

An invertible endomorphism is an automorphism

[chilly ocean]

M is gen module or ring?

[ivory ore]

M ia abelian group

[velvet hull]

so module

[chilly ocean]

wait what

[velvet hull]

Z-modules are exactly abelian groups

[thorn jay]

Who in god's name denoted a ring with M 😦 I just would like to talk to them

[rocky cloak]

They study mings and rodules

[thorn jay]

Rroup gepresentations in M-rodules

[chilly ocean]

not everybody went to the same school as you

[thorn jay]

I am self taught

[chilly ocean]

for module we need R also to write End_R(M) that is why i asked

[thorn jay]

Often it's implicit from context

[chilly ocean]

was there a context

[thorn jay]

"unit" only makes sense when End(M) is a ring, which is not the case for rings

[chilly ocean]

oh i m not as clever as you

so we have a M-Zodule

[tawny dune]

Wait why not what fails

[thorn jay]

Pointwise addition of endomorphisms of a ring may not be endomorphisms

[tawny dune]

Oh sorry I thought M was a module

[thorn jay]

If this were the case in general, then the ring operations would intercommute

[thorn jay]

Exactly

[tribal moss]

Not just "may not" -- the pointwise sum of two ring homomorphisms is not a homomorphism, because it maps 1 to 2.

(Unless the codomain is the zero ring).

[thorn jay]

A universal algebraist always takes into account the edge case of the trivial algebra :P

That fucker satisfies every (nonnegated) thing

[hard hearth]

how would one go about doing this? i tried the following:
if there is an isomorphism, then the image of x + (x^2) must be nonzero in F[x]/(x^2 - 1). further, the image p of x + (x^2) must be such that p^2 = 0 in F[x]/(x^2 - 1). so we need a polynomial ax+b a,b\in F (a or b\neq 0) such that (ax+b)^2 is divisible by x^2 - 1. so a^2x^2 + 2abx + b^2 must be divisible by x^2 - 1. by the factor theorem, x=+- 1 must be a root of this which implies a^2 + 2ab + b^2 = 0 and a^2 - 2ab + b^2 = 0 which implies 2(a^2 + b^2) = 0 which implies ???

yeah so im not sure, what to do from here

i want to conclude that a^2 + b^2 is not 0 so F must be such that 2 = 0

but a^2 + b^2 may be 0, even if a,b \neq 0

[tardy hedge]

why if there is an isomorphism, the image of x+(x^2) must be nonzero in F[x]/(x^2-1)?

[vocal pebble]

because x + (x^2) =/= (x^2) and isomorphism is injective

[crystal vale]

If char F = 2 then it will be isomorphic

[hard hearth]

that is clear

[hard hearth]

thats why i am trying to prove this

so we can say its isomorphic iff char F = 2

[hard hearth]

this is easy to verify

there i am trying to show F[x]/(x^2-1) has no nilpotents for char F =/= 2 instead

[thorn jay]

This method would work too:
a^2 + 2ab + b^2 = 0
a^2 - 2ab + b^2 = 0
Subtract
4ab = 0
If char F ≠ 2 then 4≠0
=> ab = 0

[thorn jay]

Yeah that's what I meant, used idempotent instead of nilpotent

[hard hearth]

oh haha i added instead 😄

but that does not imply a and b = 0, so maybe still there is a nonzero nilpotent??

[thorn jay]

Either a or b is zero

[thorn jay]

Per case you get from the first equations that the other must be zero too

[hard hearth]

yeah

thanks for helping 😄

[thorn jay]

Np!

[crystal vale]

Yss

[crystal vale]

Say char F ≠ 2, then x-1 and x+1 are co-maximal elements, now use Chinese Remainder Theorem

So F[x]/ < x^2 - 1 > isomorphic to F × F

So this ring has 4 ideals, but F[x]/< x^2 > has only 3 ideals

[chilly ocean]

of course if char(F) is not 2 then they are not ISO use chinise remainder on second ring

[crystal vale]

What?

[keen badge]

f(x) is irreducible over some field F if and only if f(x+α) is irreducible for all α in F?

[glad osprey]

Yep, it's easier to see if you substitute irreducible for reducible tho: if f(x) = g(x)h(x) then f(x + a) = g(x + a)h(x + a)

[keen badge]

Oh thx.

[hard hearth]

any element a of R[x]/(tx-1) can be expressed as r(t)*(x+(tx-1))^k for some k\in N with r(t) \in R.

Let k_a be the smallest such k and r_a be the r occuring with k_a.

so can i map a to the Laurent polynomial t^{-k_a}*r_a(t) to identify the two rings?

the idea here is that x + (tx - 1) is an inverse for t in Rx, so i map it to t^{-1}

this map is invertible because if we have a laurent polynomial l(t), then you can take out the smallest power of t from it to express l(t) as t^{-k}*f(t) where f is some polynomial, and then map this to f(t)*(x+(tx-1))^{k} in R[x]/(tx-1)

i hope my solution makes sense 😄

[hard hearth]

i have not learnt chinese remainder theorem, but as you can see above it was easily solvable without it

[tough raven]

An alternative at the step after a^2 x^2 + 2 ab x + b^2 = 0 (mod x^2 - 1) is to do long-division of the polynomial by x^2 - 1 (because it is monic, you can do this without worrying about the characteristic). You get the remainder 2 ab x + (a^2 + b^2). Since x^2 - 1 cannot divide a non-zero polynomial of degree < 2, a^2 + b^2 = 0 and 2ab = 0 in F. From here, you can show that 2a = 2b = 0, from which it's clear how things go in characteristic other than 2 and characteristic 2.

[hard hearth]

actually thats the most natural thing that i should have done, especially given that i said

...a^2x^2 + 2abx + b^2 must be divisible by x^2 - 1.
in my message. i dont know why i didnt do that xD

[white oxide]

Maybe I'm thinking too literally here, but why must there be a bijection from X to X'? Is it something like we can assume without loss of generality that they have the same cardinality otherwise we can remove/append elements to X' lol

[velvet hull]

I think it's just telling you to start with a collection of letters X, and then assume that the inverse of each letter is unique & distinct, and assume that the identity is also unique & distinct

it's just formalism

[white oxide]

I see so we just assume that they exist

[velvet hull]

well you can construct X' from X if you really wanted to

[white oxide]

Ye

[coral spindle]

We are assuming a bijection exists and calling it ^-1

[tribal moss]

Strictly speaking it is a bit sloppy and they should have said "pick some set X' that is disjoint from X and has the same cardinality as X".

Such sets always exist -- e.g. { {x,X} | x in X } -- but it would be more confusing to insist on a particular construction for the purpose of this proof.

[amber burrow]

is the classification of finitely generated abelian groups and semi-direct products just completely absent in artin?

i dont see it in contents

and is chapter 6 about isometries actually useful, or just motivation for group actions?

ive read up to 4.5 of dummit and foote and was getting bored so i wanted to try artin

[velvet hull]

can't find it, but it does give the classification theorem for modules over PID polynomial rings, which is a more general statement

but you can prove the structure theorem for fg abelian groups yourself with the sylow theorems

[velvet hull]

the first half is about isometries, the rest is about group actions

honestly imo isometries is a skip

[amber burrow]

alr nice

i think ill just start from ch 7

[velvet hull]

group actions are important

they cannot be skipped

[amber burrow]

i did them in dummit and foote

[velvet hull]

okay then

[round zodiac]

so a representation of a group G is just a k[G]-module? (k a field)

[velvet hull]

that's one way to say it 😂

it's a group acting on a vector space which is a field action so I guess it is a module on a group action

[swift current]

How to prove that $3$ is irreducible in $\mathbb{Z}[\sqrt{5}]$ but not prime

[cloud walrus]

rando

[velvet hull]

do you mean Z[sqrt-5]? because it seems like it is prime in Z[sqrt5] lol

but anyways Im not aware of a better method for proving it's not prime than just brute force searching for zero divisors mod 3

you have at most 9 cases to check here anyways

[jade mason]

You could argue with norms

[velvet hull]

for irreducibility sure

[rocky cloak]

In general it is useful to check something is prime by showing that the quotient is an integral domain.

It is then useful to use that
Z[sqrt d]/(3) = Z[x]/(3, x^2 - d).

To show irreducibility you just need to show that it can't be factored. It's useful to use that the norm N(a+bsqrtd) = a^2 - db^2 is multiplicative

[velvet hull]

for primality? to prove it's not prime you need existence no?

[jade mason]

Oh right

[velvet hull]

oh shit that's good goddamnit

[harsh scarab]

I have arrived to nod assuredly while i pretend to understand what I'm looking at

[boreal inlet]

Let S be a multiplicatively closed set. Then, is it true that saturation of S is equal to the contraction of extension of S wrt the canonical homomorphism from A to S^-1A?

I am not sure about anywhere to start from, and I don't know if I need to prove or disprove it.

Only thing that I'm aware is that if S is saturated, the complement is union of prime ideals, and for the saturation, the complement is the union of prime ideals which do not intersect S

[next obsidian]

What’s the extension of S?

[boreal inlet]

Oh wait, that's going to be the whole ring

Because f(S) has units

[next obsidian]

Yah

[boreal inlet]

Oh okay this does nothing then

I thought it could be similar to the case for sets

[rocky cloak]

The saturation of S should be the preimage of the units in S^-1 A

[next obsidian]

This is not hard to prove

So u can give it a shot

[boreal inlet]

That is true, but all units of S^-1A not necessarily need to be of the form f(s) where s \in S right?

all elements of f(S) are units

[rocky cloak]

Correct, for example 1/s you wouldn't expect to be in the image of f

[boreal inlet]

Okay if I can show like the complement in A of the preimage of the units in S^-1A is union of prime ideals in A, not intersecting S we are done

[keen badge]

Say I=(a,b) and J=(c,d) [ideals in a ring]
Then is I*J=(ac,ad,bc,bd)?

[keen badge]

The problem Im having with it, is that I dont understand how I(or J) is a subset of IJ.
Say.. how can a in IJ?

[coral spindle]

I and J are typically not subsets of IJ

E.g. I = J = (2) in Z

[coral spindle]

Yes

At least if R is commutative

[storm kiln]

You can even prove IJ lies in the intersection of I and J

[stone elbow]

It’s the other way around: IJ is a subset of IR, which is just I

[white oxide]

Can somebody please explain the line "if for some i we have x_i \notin p_i, we are through"? We are assuming that for each 1 <= i <= n - 1, there exists x_i \notin p_j whenever j \neq i, however 1 <= j <= n - 1. We still haven't included the case where j = n, so how is this applying induction?

[keen badge]

Is it not the intersection of I and J lies in I*J?

[wraith cargo]

this is the case j=n

if such an x_i exists there's nothing to prove, the theorem is true
If such an x_i doesn't exist then you get a contradiction

so the theorem has to be true

the idea is that you have some index 1<=j<=n and you exclude it from your first look

then for all other indices except j you know such an x_i exists

[white oxide]

Ah that makes sense

thank you

[storm kiln]

If a in I and b in J, then ab is in I since I closed under right multiplication by elements from R (and thus from J as well). Similarly ab is in J since J is closed under left multiplication by elements from R. So sums of the form \sigma a_i * b_i are both in I and J.

[keen badge]

It's funny, because "logically" it makes no sense that the multiplication of sets(in this case, ideals) lies in their intersection

[storm kiln]

Yeah, you'd think multiplication makes things bigger

[tribal moss]

Except in this case the sets are ideals, which are pretty well defined by you cannot get out of them by multiplying by anything.

(Oh, what Jelle said, really).

[keen badge]

Same thing goes for radical I guess..
(Because it is denoted by sqrt(I) one might think(me) that it makes the ideal smaller)

[tribal moss]

Perhaps think more in terms of "multiplication" ~ "intersection" from boolean algebras.

[keen badge]

IDK what boolean algebra is 🙂

[tribal moss]

Basically, an axiomatic abstraction of how "union", "intersection", and "complemet" from elementary set algebra work.

It also turns out to be equivalent to rings satisfying xx=x for all x. (These rings all have characteristic 2).
You get such a ring from a set algebra by declaring a+b = symmetric difference of a and b; a·b = a intersect b.
On the other hand, from a ring with xx=x, you get a boolean algebra by defining a OR b = a+b-ab; a AND b = a·b; NOT a = a+1.

[stone elbow]

I’ve seen the proof of the prime avoidance lemma for like 100 times and yet ever time it’s baffling 😄

[ebon gyro]

the geometric statement of prime avoidance is more lucid to me

if you have finitely many points contained in an open subset of an affine scheme you can always find a smaller distinguished affine subset containing those points

[rocky cloak]

Can you expand about how this gives a geometric interpretation of prime avoidence?

I'm not able to connect the dots

Okay, I see what you're saying.

You have finitely many primes in the open set V(I)^c, then there is an f in I such that the primes are contained in D(f).

Doesn't necessarily feel that geometric to me though. It's kinda particular to the distinguished open sets...

Idk

Feel like just, if an ideal is contained in a finite union of prime ideals it's contained in one of them is lucid enough.

[stone elbow]

Can you check my proof, please? I have a feeling that I messed up somewhere because I haven’t fully used the hint

[white oxide]

Here is the ith entry in the direct sum of b_i equal to the 0 ring

Otherwise b_i doesn't live in A (set theoretically)

[next obsidian]

Yah

[rocky cloak]

There's no set theoretic issues with taking the sum of ideals

The sum of ideals aj will always be an ideal of A, and in this case the sum is direct.

[south patrol]

I'm confused about what the question is

Okay sure I mean I guess yeah the point is that these are internal direct sums so there aren't even any abuses, but more generally there is just the standard "abuse" that V is a submodule of V (+) W etc

I assume that is what the confusion is

[grizzled crow]

Do we tend to consider the empty product to be a product, when someone makes a statement like: "Prove everything in set A can be expressed as a product of stuff in set X"

[tough raven]

I know I do.

[drifting mauve]

Could someone pls check my logic for this question? I think my proof seems a bit dodgy but I can't put my finger on what I've missed. Let $R$ be an integral domain such that $R[x]$ is a PID. Prove that $R$ is a field.

[cloud walrus]

theaveragejoe6029

[drifting mauve]

take the ideal in $R[x]$, $(x)$. Then we have that $(x)$ is a principal ideal since $R[x]$ is a PID. Thus, we have that $R[x]/(x)$ is a field. and since $R[x]/(x)$ is isomorphic to $R$, we then have that R is a field.

[cloud walrus]

theaveragejoe6029

[velvet hull]

well (x) is a principial ideal by construction

not because R[x] is a PID

[velvet hull]

and the quotient of princial ideals is not guaranteed to be a field, you're mixing it up with maximal ideals

because otherwise what you have just proved is that all rings are fields

because R is always isomorphic to R[x]/<x>

[drifting mauve]

oh yeah, true lol. I misread my notes.

fr thought I was being slick haha

[velvet hull]

oh, there is a slick proof for this theorem, keep searching for it!

[drifting mauve]

actually. since we have that every prime ideal in a PID is maximal we have that (x) is maximal anways right?

*every non-zero prime ideal

[velvet hull]

that proof works, the one I had in mind was to take any nonzero r in R. consider the ideal <r,x>

[tough raven]

Once you show that (x) is prime, yes!

[drifting mauve]

Yeah, but that's pretty easy.

[drifting mauve]

Do you mean <r,x>? Where do you go from there?

Sorry for the late reply btw. I had to run off to a lecture 🏃‍♂️💨.

[velvet hull]

right, typo.
well, it's a principal ideal by assumption, so let it be generated by f
then f has to divide r and x at the same time, what do you conclude about f?

[drifting mauve]

That r is invertible?

[velvet hull]

well that's the goal, but you're skipping a lot of steps there

[vivid anvil]

In mathematics, a group is a set with a binary operation that satisfies the following constraints: the operation is associative, it has an identity element, and every element of the set has an inverse element.

based on this Wikipedia definition of group, is the boolean set of values {0,1} a group with the AND operation?

the identity element would be 1, since 1 & 1 = 1 and 0 & 1 = 0

[thorn jay]

No, 0 has no inverse

[vivid anvil]

you're right, i didn't realize that. i was focusing on 1. Thanks

[thorn jay]

There is another logic gate/operation that does make it into a group though, where 0 is instead the identity

[vivid anvil]

could u tell what it is

[thorn jay]

XOR

[vivid anvil]

ah

[vivid anvil]

but 1 XOR 1 = 0

wait

[thorn jay]

.

[vivid anvil]

yea

i see it now

yea it satisfies all the condition

[thorn jay]

C2 my beloved

[mighty kiln]

XNOR works too, with identity 1

In particular [(x == y) == z] == [x == (y == z)]

[rocky cloak]

Gonna start using = as my symbol for group operation

[thorn jay]

Well that's just the induced group from the map x -> x' (where ' denotes negation)

I mean of course it is they're both cyclic and that's the only nontrivial bijection on { 0, 1 }

[vivid anvil]

So if I define a set {A, B, O} which is commutative and associative, and O is the identity element, and A*B=O, it's a group

and A and B would be inverses of each other

[next obsidian]

Arki red fox

[vivid anvil]

A commutative group is called an abelian group

[velvet hull]

it's also called a commutative group!

[rocky cloak]

Don't stomp on my boy Niels Henrik

[vivid anvil]

i see

[chilly ocean]

all homomorphism on circle are of the form z goes to z^n

right?

[tribal moss]

Perhaps all continuous homs, but I think there will be wildly discontinuous ones too.

[chilly ocean]

one discontinuous exampls which isnt obvious

[tribal moss]

I also expect the discontinuous examples to require the axiom of choice, and not be able to be written down explicitly...

[chilly ocean]

so how we can write it

[tribal moss]

We can't -- it would be a nonconstructive existence proof.

[chilly ocean]

please give the existeence proof

[tribal moss]

I don't have all the details ready -- but my gut feeling is that one can use Zorn's lemma to show that the circle group is isomorphic to the direct sum of Q/Z and continuum many copies of Q, and if we have such an isomorphism we can use it to interchange two of the copies of Q, giving a nontrivial automorphism that preserves all the rational fractions of a whole turn, which would be discontinuous.

Wikipedia agrees with my gut, FWIW: https://en.wikipedia.org/wiki/Circle_group#Group_structure

[south patrol]

Yeah worth mentioning that C can have 2 autos without choice but tons with choice

[tribal moss]

(But, to avoid doubt, those wild automorphisms of C won't generally preserve the unit circle).

[keen badge]

What about radical ideals?
If I,J are radical ideals, does it mean that their intersection is equivalent to their multiplication?

[chilly ocean]

comaximal

[tribal moss]

Not necessarily - for example in Z, (10) and (15) are radical ideals. Their product is (150); their intersection is (30).

[keen badge]

what is comaximal

[rocky cloak]

I+J = R

[rocky cloak]

The radical of their product is their intersection though

[keen badge]

oh ok

[keen badge]

oh wait it's not good..
I need the radical of their intersection..

More specifically,
(In algebraic geometry) If I want to describe the radical ideal that vanishes at (x0,y0) and (x1,y1)

Than for (x0,y0) I take (x-x0,y-y0) and for (x-x1,y-y1)

so now I need to look at Rad((x-x0,y-y0) intersection (x-x1,y-y1))

[tribal moss]

The intersection of radicals is already itself radical, if I'm not mistaken.

[keen badge]

yes it is.

[keen badge]

I just know that the answer for this question is their multiplication

So I am trying to make sense of it

[tribal moss]

Hmm consider the straight line containing the two points. I don't think its equation is in the product of your ideals (though I cannot rattle off an airtight argument for this), but it is in their intersection.

[keen badge]

Ok, so say I want I(x-x0,y-y0) intersection I(x-x1,y-y1)=X

How can I find f1,dots,fn with I(f1,dots,fn)=X

[rocky cloak]

The ideals are comaximal, so the product should be their intersection

[thorn jay]

That itself doesn't prove much, the stronger property is that the radical of the intersection is the intersection of their radicals though

[rocky cloak]

(assuming the two points are not identical anyway)

[tribal moss]

Hmm, to be concrete let's consider (x,y) and (x-1,y). I claimed that y would not be in their product, but I was wrong: xy-y(x-1)=y.

[rocky cloak]

So is the question just what the generators for the ideal that vanishes at two points is?

[tribal moss]

But a set of generators for a product ideal could just be the cartesian product of the generators for each factor. It might not be a minimal set of generators, though.

[rocky cloak]

I guess exemplified in how
(x, y) * (x-1, y) = (x(x-1), y)

[keen badge]

Why these two ideals are comaximal?

Ohh I think I got it

(x0,y0)!=(x1,y1)
wlog x0!=x1 then we take
f= (x-x0)/(x0-x1) in I=(x-x0,y-y0)
g=(x-x1)/(x1-x0) in J=(x-x1,y-y1)
this is well-defined because again x0!=x1 and we have
f+g=1

[tribal moss]

Yeah.

[keen badge]

🙂

[tribal moss]

Another way to phrase the same argument (which at least to me feels less at risk of sign errors) is that I+J contains (x-x0)-(x-x1)=x1-x0, and if that constant is nonzero we can scale it afterwards to get 1.

[alpine island]

Are there any other rings that are Morita equivalent to Z? (other than integer matrices)

[tribal moss]

That sounds like an #advanced-algebra question.

[rocky cloak]

No, any Morita equivalent ring is equal to End(P) for a progenerator P, and the only progenerators are Z^n (which you can see from the classification of fg abelian groups for example)

[hard hearth]

what is the idea behind the yellow underlined sentence? (hint) i feel like we need to take some nonzero noninvertible element b and consider polynomials like bx and x and try to contradict the existence of a division algorithm

[rocky cloak]

The easiest would probably be to show that its not a PID, by showing for example that (r, x) is not principal when r is not a unit.

But I guess you can have a look in section 2 to see how they prove it

[chilly ocean]

So basically,
The yellow line means that R[x] can't be an Euclidean domain unless like R is a field because the division algorithm for polynomials only works if you can always divide by any nonzero coefficient, which isn’t possible if
R has noninvertible elements, which you thought correctly by the way.
And as said by @rocky cloak Look more into section 2 for the proof and stuff.

And before anyone says that I am using ChatGPT like the last time they did, I was just being a bit formal and do note that answers from ChatGPT are pretty long.

[hard hearth]

actually its much more simpler 😄 R[x] is Euclidean domain => R is PID => R is integral domain. (x) is a prime ideal, so (x) is maximal (R[x] is PID) but R[x]/(x) = R so R is a field

so actually this also holds for when R[x] is just a PID

this argument is by dummit and foote, not by me

[chilly ocean]

Ahh that textbook

Abstract Algebra

[hard hearth]

do you know how to use the noninvertibility, since you said that idea works?

[chilly ocean]

Yes

[glad osprey]

well? we're waiting

[chilly ocean]

Oh u want the answer ig

Well how do I type out the symbols though

If R is not a field, then there exists a nonzero element b in R that is not invertible (i.e., there is no b⁻¹ in R).
Now consider the polynomial ring R[x]. The division algorithm says we should find polynomials q(x), r(x) in R[x] such that bx = x * q(x) + r(x),
where deg(r(x)) < deg(x), so r(x) is a constant.
We can let q(x) = b and r(x) = 0, since:
x * b + 0 = bx.
Try reversing the process: divide x by bx.
And we would want:
x = bx * q(x) + r(x)
Again, deg(r(x)) < deg(bx) = 1, so r(x) must be a constant.
However, to make the leading terms match, we’d need q(x) = b⁻¹.
But since b is not invertible in R, b⁻¹ does not exist in R, and therefore not in R[x].
So the division algorithm fails, which means R[x] is not a Euclidean domain unless R is a field.

There we go, it was kind of hard to type it out since I never actually used Latex before which would make my job easier

[chilly ocean]

Anyways I have a question for you, how did you make your role blue, I have the he/him role but it isn't blue for me for some reason

[rocky cloak]

he/him is the blue diamond next to your name.

Blue name is the very active role, whitch you get by being here often

[chilly ocean]

Oh ok thanks

[white oxide]

Does this work for showing that if $x$ is nilpotent, then $1 + x$ is a unit? Here $N(A)$ is the nilradical of $A$. If $1 + x$ is not a unit in $A$, then $1 + x \in \mathfrak{m}$ for some maximal ideal $\mathfrak{m}$. Since $\mathfrak{m}$ is prime and $N(A)$ is the intersection of all prime ideals in $A$, $N(A) \subset \mathfrak{m}$. But $x\in N(A)$, so $1 + x - x = 1 \in \mathfrak{m}$. This contradicts the fact that $\mathfrak{m}$ is maximal.

[cloud walrus]

okeyokay

[coral spindle]

Yeah I think that works and that's pretty slick too

[white oxide]

Okay because i looked up the answer and people used geometric expansion or smt

[coral spindle]

Ofc you know the elementary proof right? By considering the power series of 1/(1+x)?

Yeah the geometric expansion is a neat proof

[south patrol]

Yes. In fact you have shown that if x is in all maximal ideals (I.e. in the Jacobson radical) then 1 + rx is a unit for all r, and the converse is also true

[white oxide]

Damn I'm fucking cracked /s

thanks

[tardy hedge]

I didn’t know blue name means v active

I didnt even know the colour of name means anything

[stone elbow]

Hi all, if $A$ is a commutative ring, and we embed $A\textrm{-Mod}^{\textrm{op}}$ into some $B \textrm{-Mod}$ by the Mitchell theorem, is B necessarily commutative?

[cloud walrus]

a-square

[south patrol]

The answer is no for somewhat trivial reasons: BMod is equivalent to M_n(B)Mod, and M_n(B) is never commutative (except trivial cases)

Perhaps a better question is: is it always possible to embed AMod^op into BMod for a commutative ring B?

And then I am unsure to be honest, but suspect the answer to be "no"

[next obsidian]

I believe the answer is no

If so you could just embed any (barring size conditions) abelian category into the category of modules over a commutative ring which I am 99.99% sure is not possible

Since that’s like the first thing people tell you to be careful about with the embedding theorem

[peak root]

How can I prove that in $\mathbb{Z}[x]$, the ideal $(4,x)\cap (2,x^2)=(4,2x,x^2)$? $\supseteq$ part is not hard, but how can I do $\subseteq$ part? I want to show for any $r\in (4,x)\cap (2,x^2)$, $r$ is in $(4,2x,x^2)$. And I only know that $r$ can be written as $r=4f+xg=2m+nx^2$. How can I proceed from here?

[cloud walrus]

Dong_Valentino

[chilly ocean]

that is also easy

[peak root]

Is there any hint?

[chilly ocean]

one side is clear as IJ is in I intersection J

FOR opposite part focus on generators

[peak root]

I figured it out. $r=4f+xg\in (2,x^2)$ implies $xg\in (2,x^2)$. So either $2|g$ or $x|g$. In both cases $xg\in (4,2x,x^2)$.

[cloud walrus]

Dong_Valentino

[peak root]

Thanks.

[white oxide]

What do they mean by annihilated by m, hence naturally a A/m module? Do they mean that the action of A/m on M/mM is well-defined since it's annihilated by m?

[chilly ocean]

yes

[chilly ocean]

can you share 2.6 i want to read that

[rocky cloak]

So A-Mod^op is not a small abelian category, so the Mitchell embedding theorem doesn't apply, and I believe there are arguments for why A-Mod^op cannot embedd fully faithfully into B-Mod.

If A is Noetherian you could use the category of fg A-modules though. In which case I think you should be able to take B = End(E) for an injective cogenerator.

Now, should one expect this to be commutative...

[mighty kiln]

How does this work

Say for A = Z

[stone elbow]

Is E the tensor unit?

[rocky cloak]

Actually, here's an argument for why it can't be commutative for A=Z.

So we just consider the category of fg abelian groups Z-mod.

Then we have a fully faithful exact contravariant embedding
F: Z-mod -> B-Mod
Now F(Z) is a B--Z-bimodule and using exactness and finite projective presentations you can show that F = Hom(-, F(Z)).

Let's call F(Z) for E. Since F is exact E is injective, and since F is an embedding E is a cogenerator. So E contains Q/Z as a direct summand.

Let's say E contains some extra summands of Z[1/p]/Z. Then
F(Z/p) = (Z/p)^I. Since F is fully faithful this must have endomorphism ring Z/p as a B-module. Since over a commutative ring multiplication by an element is a homomorphism, the action of B on (Z/p)^I is just multiplication by elements in Z/p, but then the endomorphism ring is much larger, contradiction.

So E must be Q/Z direct sum some Q-vector space. So E = Q/Z (+) V.

Now again multiplication by an element in B should be a homomorphism, so since this is fully faithful B simply acts by multiplication by integers. And then the endomorphism ring is much to large.

I think you can embed Z-mod into B-Mod for B = End(Q (+) Q/Z) though

[stone elbow]

Thanks, it will probably take me a while to understand this argument though

[chilly ocean]

Jagr are the application for like Postgraduate role closed because I applied for it but I have the undergraduate role for some reason.

[rocky cloak]

I think the mods are just slow to aprove the roles that require manual aproval, but I don't know the inner workings

[stone elbow]

I need to go through the modules exercises section in Lang first at least 🙂

[chilly ocean]

Oh ok 👍

[rocky cloak]

The main idea is really that for a commutative ring, multiplication by an element is a homomorphism.

So for modules with few endomorphisms you are very limited in how B can act.

[rocky cloak]

So refuting this conjecture then I guess.

For A=Z, E=Q/Z is a cogenerator, but the embedding into End(E) = profinite integers cannot be full. In particular Hom(Z, E) will have the profinite integers as endomorphism ring, which is of course larger than Z.

[chilly ocean]

every irrational number has unique continued fraction ?

[chilly ocean]

Yes they do

unique infinite continued simple fraction expansion to be even more precise

[chilly ocean]

how did u show that

[chilly ocean]

How do I write the fractions here tho

write sequence

Because like I really don't know how to use latex

[rocky cloak]

I guess
[a; b, c, d, ...]
is a notation people use for continued fractions

[chilly ocean]

mhm

[chilly ocean]

So like it sucks to not have any notation but anyways here is a sentence explanation for you @chilly ocean
Every irrational number has its own unique infinite continued fraction because when you break it down using the continued fraction algorithm (taking the integer part then flipping the decimal part over and repeating), the steps always follow a consistent pattern. Since irrational numbers are infinite and do not repeat like rational numbers, the process never ends at all which gives each irrational number a unique continued fraction.

[next obsidian]

Nice circular argument gpt

[chilly ocean]

I am just being formal, if you think being formal is being an AI then I don't really know what to say.

[next obsidian]

Your argument is literally “because irrational numbers are infinite, they have infinite continued fractions”

And you don’t address uniqueness at all

[chilly ocean]

Well, I can't type out fractions here and can't really give the notational proof and stuff, so I have to type out what I can in my own words in sentences.

[chilly ocean]

how did you detect that

[coral spindle]

I'm not convinced this is a ChatGPT answer but it is certainly a very low-quality one

Please let me know if this comes up again

[topaz solar]

well, they've had similar answers in, say, AG channel

[coral spindle]

I have been looking around

Concerning. Looking into it. Etc

[topaz solar]

W

I sure think its suspect

[next obsidian]

The other one also had an em-dash

Just saying

[tribal moss]

I think I would prove by induction on n that:

1. For fixed a1, a2, ..., an, all positive integers, the function x |-> [0;a1,a2,...,an,x] maps (1,infty) monotonically to an open interval of (0,1).
2. When n is fixed, different choices of a1, a2, ..., an lead to disjoint ranges which together cover all of (0,1) except some rational endpoints.
   which together with appropriate handwaving about limits prove that
   a) different simple continued fractions must have different values.
   b) an arbitrary irrational is in the range of one of the functions at every n, which fit together to produce an infinite continued fraction.

[topaz solar]

well, ios devices do that automatically

[next obsidian]

Wat

[topaz solar]

if you type -- on ios it will spit out one of the dashes

[next obsidian]

Yeah but I mean that very few people use them, but chatgpt loves them

Not that it’s a difficult character to produce lol

[stone elbow]

To play devil's advocate you can easily type "—" by pressing Shift+Option+- (but I don't know many people who actually bother to do it)

[next obsidian]

I AM NOT DAYING ITS HARD TO PRODUCE

RAHHHHHHH

[long swan]

to be fair, you can just hit alt+0151 to type an em dash...

[next obsidian]

I am going to fill my mouth with popits and bite down hard

[coral spindle]

I have an autohotkey script to put this in windows lol

- – —

[coral spindle]

wait hold on this isn't just circular, it's factually wrong

There are plenty of irrational numbers with a repeating continued fraction expansion

However this appears to claim the opposite

In the first place, sqrt(2) = [1; 2, 2, 2, 2, ...]

[rocky cloak]

I think their just claiming that process doesn't terminate, which is clear from a finite fraction just being a rational number.

[coral spindle]

I was focusing really on this line:

Since irrational numbers are infinite and do not repeat [...]

[rocky cloak]

Well, that's not wrong right, just kinda irrelevant

As for uniqueness, I think the claim is simply that if you have a continued fraction and use the standard algorithm for computing them you recover the same coefficients.

I think this follows just from continued fractions always covering to some positive real (which I guess actually might depend on how you define convergence of continued fractions)

But for the usual way it should be clear

[coral spindle]

No it is wrong, because sqrt(2) = [1; 2, 2, 2, 2, ...] is an irrational number which does repeat, if indeed we interpret that line in what I think is the only reasonable sense

[wraith cargo]

Tbh idk what you mean by reasonable because I have no idea what they meant

I feel like what they wrote isn't a complete sentence

Idk if they were ever referring to the coefficients repeating?

Oh wait maybe they meant like the decimal expansion doesn't repeat lmao

Tho idk how it's relevant

[rocky cloak]

Typically when people talk about numbers repeating this way they mean the decimal expansion. And with this interpretation the statement is actually correct, so I think it is a more reasonable interpretation

Not that it would be relevant in either case

[coral spindle]

Fair enough, but I think this more indicates at best a confusion about continued fraction and decimal expansions

[chilly ocean]

I meant a pattern like rational fractions or stuff so sorry for that

[coral spindle]

Oh you're online Cyber

Can I ask you a quick question?

[chilly ocean]

yes

[coral spindle]

How would you define what an integral domain is?

[woeful sage]

just ask chatgpt if the answers sound like chatgpt

[coral spindle]

It's interesting that this answer is taking so long. You yourself answered a question not too long ago about Euclidean domains, Cyber, so I would have thought you'd know integral domains - a much simpler concept - like the back of your hand!

[woeful sage]

it's just a domain that has integrals in it, simple.

[chilly ocean]

Bro I am giving explanations and examples

[coral spindle]

You haven't answered my question though

[woeful sage]

he just asked for the definition though

[coral spindle]

It's been three minutes. Any student could tell you the definition in about 10 seconds

[chilly ocean]

I know but I need examples and stuff

https://tenor.com/view/me-atrapaste-es-cine-its-cinema-cinema-esto-es-cine-gif-17729711691959966457

[coral spindle]

You need me to give you examples?

[chilly ocean]

No

[coral spindle]

OK maybe just name a single example of an integral domain

[chilly ocean]

I will give

[coral spindle]

no need to justify

just name a single example

[chilly ocean]

then just say Z, easy example

[coral spindle]

Spoilers!

Maybe except Z then

[chilly ocean]

Q

[coral spindle]

Hey well done!

[chilly ocean]

R

[woeful sage]

what about integral domains that aren't a field

[chilly ocean]

what about non integral domain?

[coral spindle]

I've seen enough fwiw

[chilly ocean]

That's Z6 for example

I don't know how to write it using the latex and stuff

[woeful sage]

There are helpful resources in pinned messages in #latex-help

[glad osprey]

just curious: since you're a postgraduate, you've obviously had to write some papers or something. How did you do it without knowing latex?

[woeful sage]

word

[chilly ocean]

I don't know in Discord

[thorn jay]

Bro has the undergraduate role though

[glad osprey]

he said earlier he applied for postgrad

[chilly ocean]

yes I applied for it

the postgrad role

[thorn jay]

Lol

Doesn't say postgraduate pending, meaning youve been rejected

Probably

[glad osprey]

hmm, weird, why would they reject such a promising candidate?

[kind temple]

this is my favorite tv show

[chilly ocean]

they also rejected me

[chilly ocean]

absolute ads

[thorn jay]

Aye beats me man

[glad osprey]

you mean you don't know how to use latex in discord specifically? But you know how to use it otherwise?

[coral spindle]

Just gonna throw it out here that the mod team is aware of this whole situation and we're working on it; there's no need to continue investigating

But ofc do as you please

[glad osprey]

Okay, sorry

[coral spindle]

It's fine I was just letting you know

[glad osprey]

I'm pretty convinced for my own sake atleast

[thorn jay]

The council will decide your fate

[chilly ocean]

No I usually write in Word

[sly crescent]

What situation?

[woeful sage]

Me learning algebra

[rocky cloak]

Someone accusing Cyber of copying answers from gpt as replies to questions here I guess

[chilly ocean]

i coudnt follow the induction

[tribal moss]

I didn't give details of the induction step; they're left for the reader to reconstruct.

[chilly ocean]

your map is injection right

[tribal moss]

Monotonic maps between subsets of R totally ordered sets are always injective.

[chilly ocean]

so we have to use induction to show every term of sequence is unique for irrational

[tribal moss]

I don't know whether we "have to", but that was the way I could imagine taking when I wrote my suggestion.

[south patrol]

Lmao

[hard hearth]

hey! sorry for not replying yesterday, but doesnt this just disprove that R[x] is not a euclidean domain with norm(f) = degree(f)? we need to show that it is not euclidean domain for any norm function

i actually already knew this but was wondering what to do for any general norm function 😄

[hard hearth]

hey! i did think about this. so what i got is that if (r,x) = (k), then because x \in (x,r), we must have x = kp(x) for some polynomial p => k is a unit => (r,x) = R[x]. but we know that the map R -> R given by x -> rx is not surjective, so there will be some constants missing from (r,x) showing that (r,x) =/= R[x] (contradiction)

i like this proof much better because its direct, thanks for the hint! (the book uses a more abstract argument. see #groups-rings-fields message)

[tardy hedge]

Lolll just read all that above

Funny

[keen badge]

The intersection of co-maximal ideals is their multiplication, but what if J,K,L are ideals in R where J,K are co-maximal and K,L are co-maximal, but J,L aren't co-maximal.
the intersection of J,K,L is still their multiplication?

What if only J,K are co-maximal but K,L and J,K aren't?

[rocky cloak]

Consider for example the case where J=L

[keen badge]

J=L=(2) and K=(3) then JKL=(12) and J inter K inter L=(6)... that's too bad 😦

[hard hearth]

does anyone know a source of interesting ring theory problems?

[lusty marlin]

Dummit and Foote has some nice problems

Although I guess any standard algebra textbook should work

[dull kite]

*not

[crisp roost]

can you write (C_2)^3 to represent the group (C_2)x(C_2)x(C_2)? (where C_2 is the cyclic group of order two)

if thats more clear

[thorn jay]

Yes

[white oxide]

In free abelian groups is ma = a + a + ... + a m times or nah

[coral spindle]

In latex we would write this as follows, just so you don't feel the need to go to word every time:

write $C_2 \times C_2 \times C_2$ as $(C_2)^3$

[cloud walrus]

Boyt(ji=-k)e

[coral spindle]

LaTeX is actually super friendly, don't fear it :)

[sonic coral]

@chilly ocean

[swift current]

I am trying to prove that $3$ is irreducible but not prime in $\mathbb{Z}[\sqrt{-5}]$, was hoping someone could read through my proof and tell me if I have made any errors:

Let (R = \mathbb{Z}[\sqrt{-5}]). Recall that the norm (N\colon R\to\mathbb{Z}_{\ge0}) is given by
[
N(a+b\sqrt{-5}) ;=; a^2 + 5b^2.
]
It has the key multiplicativity property
[
N(xy) = N(x),N(y),
\quad
\forall,x,y \in R.
]

\medskip

(i) (3) is irreducible in (R).\

Suppose, for sake of contradiction, that
[
3 = \alpha,\beta
\quad\text{with}\quad
\alpha,\beta\in R
\quad\text{non-units.}
]
Then
[
9 ;=; N(3) ;=; N(\alpha),N(\beta).
]
Since (\alpha,\beta) are non-units, (N(\alpha)) and (N(\beta)) must both exceed (1). The only factorization of (9) into two integers (>1) is
[
9 = 3\cdot 3.
]
Hence
[
N(\alpha)=N(\beta)=3.
]
But no element of (R) has norm (3), because
[
a^2 + 5b^2 = 3
]
has no integer solutions because (a^2 \equiv 3\pmod 5) has no solutions. Thus the factorization (3=\alpha\beta) with both (\alpha,\beta) nonunits is impossible. Therefore (3) is irreducible in (R).

\medskip

(ii) (3) is not prime in (R).\

We exhibit a product of two elements that (3) divides in (R), but does not divide either factor. Observe:
[
2\cdot (1+\sqrt{-5}) ;=; 2 + 2\sqrt{-5},
]
and
[
2 + 2\sqrt{-5}
= (1+\sqrt{-5}), (1-\sqrt{-5})
= 1 - (-5) = 6.
]
Hence
[
2,(1+\sqrt{-5}) = 6 = 3\cdot 2.
]
So (3 \mid 2,(1+\sqrt{-5})) in (R). If (3) were prime, it would have to divide one of the factors (2) or (1+\sqrt{-5}). But
[
N(2)=4,\quad N(1+\sqrt{-5})=1^2+5\cdot1^2=6,
]
and neither (4) nor (6) is a multiple of (9=N(3)). Hence neither (2) nor (1+\sqrt{-5}) can be divisible by (3). This contradiction shows (3) is not prime in (R).\

Q2. Let (0\neq a\in R). We will show that (a) is a unit in (R). Consider the ideal