groups-rings-fields

[formal ermine]

immediately follows from a pid being a ufd

[white oxide]

ah too bad that's on the next page 😭

[wraith cargo]

ok well

you know that <d>=I

[formal ermine]

fun fact a ufd is characterized by that theorem

[wraith cargo]

and you have that <ra> = <d>

but since r is a unit it has an inverse

[white oxide]

wait you mean <a> = <dr> right

[wraith cargo]

so since <rd> is an ideal you can multiply it by the inverse of r

[wraith cargo]

yes sorry

[white oxide]

oh right

then it contains d

[wraith cargo]

and get that <rd> = <d>

[white oxide]

so they're the same

[wraith cargo]

yeah I meant trivially to see it you have that every element of <rd> is of the form rdx for x in R
But since r^-1 <rd> = <rd> we can see that r^-1<rd> = <d> hence <d> = <rd>

[white oxide]

ah ok got it thx!!

[white oxide]

this might be a stupid idea, but could we possibly consider G/H U G/K where H is the normal subgroup in G of order 5 and K is the normal subgroup in G of order 3? if we can show that they're intersection is trivial then we can perhaps consider a canonical homomorphism of some sort, don't know if this goes anywhere though (will work on it in the meantime)(

[south patrol]

Not really sure what that union would mean when you're quotienting out by different things

[white oxide]

yeah and i suppose the union is not always a subgroup as well

i guess i can just consider HK lmfao and just show that their intersection is trivial

don't know why the hint is asking me to quotient shit out

[chilly radish]

In fact union of subgroups is a subgroup iff one contains the other

[chilly radish]

This would work, but note that it only works because H,K are normal, therefore HK is a subgroup (it's a sufficient condition that one of the subgroups be normal)

[south patrol]

Hm what happens if we allow bigger unions

I mean tbf it is possible then I think

[white oxide]

yes but any subgroup of order 30 contains a normal subgroup of order 5 and a normal subgroup of order 3

[south patrol]

Lol

[chilly radish]

But uh

[south patrol]

Just union over all proper subgroups of a finite group

[chilly radish]

You're not guaranteed oth a subgroup of order 3 and order 5

[chilly radish]

Or

Not and

Reread

[hollow mica]

wait that’s so messed up

[chilly radish]

This is for 2 subgroups ofc

It makes sense, unions are set theoretic, they don't preserve any structure

[chilly radish]

If you allow for 3 subgroups there's nontrivial examples where the union is again a subgroup

Which imo is more fucked up

[south patrol]

Ah interesting

[hollow mica]

Oh that puts me at peace then

[chilly radish]

There's actually a classification for it

[south patrol]

Yeah first thing that came to mind was Z/2 z Z/2Z and the three subgroups of order 2

But that is trivial enough lol

[coral spindle]

you're joking

what is it then?

[chilly radish]

A group is the union of 3 proper subgroups iff it has klein group as a quotient

[coral spindle]

Oh lmao

[south patrol]

Nice

[coral spindle]

that's hilarious

the klein group, like?

[chilly radish]

Yea

Z/2xZ/2

[coral spindle]

that's actually amazing

[south patrol]

I mean I gave an example just now

Lol

I wonder

How hard this is

[chilly radish]

Which makes it the prototypical example in a sense

[coral spindle]

hahaha good observation potato

[south patrol]

Hm

[chilly radish]

https://math.stackexchange.com/questions/959886/can-a-group-be-a-union-of-three-subgroups

There's further results for 4,5,6 subgroups

Where the klein 4 group is replaced by some finite set of subgroups

This is so funny

[south patrol]

Oh wait

[hollow mica]

seems like something appropriate for the american math monthly

[south patrol]

I think I know how to do

[chilly radish]

It breaks again at 7 subgroups

So sad

[south patrol]

Maybe

Oh wait no

[chilly radish]

No group is the union of 7 proper subgroups

[hollow mica]

how much do you wanna bet

[coral spindle]

Do you have a paper that worked this out, Shin?

[hollow mica]

that I don’t find one

[coral spindle]

I'd love to see it

[chilly radish]

It's all in the mse thread

[coral spindle]

nice

could u link?

[white oxide]

nvm this is not true in general but we certainly do know that G either has a normal subgroup of order 5 or a normal subgroup of order 3

[chilly radish]

Boop @coral spindle

[white oxide]

so still works

[chilly radish]

They provided references and an overview that was published in american math monthly

[chilly radish]

But you're not guaranteed both a subgroup of order 3 and a subgroup of order 5

So you can't always generate one of order 15 by multiplying them

[white oxide]

nah you are by the first sylow theorem, if p is a prime and p divides 30 then there's a subgroup of order p

unless im reading smt wrong

[coral spindle]

ty

I just didn't see it

huge blind

[chilly radish]

Sorry, I forgot the word normal silly me, but you're right, as long as one of them is normal it still works

I made an oppsie

[white oxide]

yea it's all good

thanks for helping tho

[chilly radish]

No problem

Ftr I think they wanted you to use correspondence theorem

But that's just a roundabout way of saying the same thing

[white oxide]

correspondence theorem?

well i looked in the answers and my intuition was somewhat right, they just used some canonical homomorphisms i believe

[chilly radish]

The correspondence between subgroups of the quotient group and subgroups and subgroups containing the normal subgroups

[delicate orchid]

weird how they teach that for rings so often but not groups

[chilly radish]

They should teach it

It's important

[coral spindle]

Huh, I learned that in my first classes on groups and rings

[south patrol]

We did it in rings and used it in groups without properly acknowledging it lol

I think it must be taught in some form for Jordan-Hölder right?

Or composition series more generally

[formal ermine]

we did neither

[chilly radish]

In my gt course (very same one I am TAing rn) we teach it explicitly

We don't do jordan hoelder tho, but we do mention it

[coral spindle]

Good to hear

[elder wave]

Because your prof skipped a bunch of stuff to squeeze in Galois theory

[frigid lark]

Wait, Galois theory, without Jordan Holder?

[hidden haven]

Why do you need Jordan Holder to do Galois theory?

[south patrol]

I guess it'd be helpful to understand solvability of groups a bit better

[hidden haven]

Yeah I guess that'd be it

[south patrol]

Like to show that any composition series is enough to check solvability

I suppose

[hidden haven]

Well solvability is the existence of one series anyway

[south patrol]

Yeah indeed

But I mean to check smth isn't solvable right

[hidden haven]

Ah true

[south patrol]

Hm

Still not necessary tho

Cause we know subgroup of solvable group is solvable etc

[hidden haven]

Ye just classify all subgroups bro

[south patrol]

So as soon as you reach an obstruction cresting a tower or smth we know it ain't solvable

[hidden haven]

True

[distant chasm]

Any help here?

[ivory trail]

think of examples of finite cardinality Z_2[X]-modules and see what they have in common

if i give more of a hint it pretty much gives it away

[distant chasm]

I know it is something like the number of elements of a Z_2[x]-module needs to have 2^n elements because of the characteristic but I don't know how to explain why

Or is it what i wrote correct?

[ivory trail]

you can formalize this by saying Z_2[x]-modules are automatically Z_2-modules, which are vector spaces and therefore easy

this is an example of "restriction of scalars"

[distant chasm]

Ok, thank you !

[ivory trail]

no problem. you can actually think of k[x]-modules as vector spaces over k with a linear operator given by the action of x

[winter yew]

Hi. Any idea how to show that factor-ring of a finite ring with unity is finite?

[wraith cargo]

R -> R/N is surjective

[uncut cloud]

Because you have a canonical morphism R -> R/N which is surjective

[winter yew]

Thanks!

[uncut cloud]

This morphism is probably the most important tool to study factor rings

[solemn garden]

how is that possible

[solemn garden]

Why is that discussion’s greatest moment

[fossil shore]

Hey, won't any subring containing F be the whole ring?

[next obsidian]

No

Consider F

[fossil shore]

Oh wait... the subring doesn't have to be an ideal right...

[next obsidian]

Indeed

[fossil shore]

brain ain't working

thx

[next obsidian]

🧠 ⚙️

[glossy crag]

I'm having a brain fart, how does $U_1\oplus\cdots\oplus U_n=V_1\oplus\cdots\oplus V_n$ and $U_i\subset V_i$ imply $U_i=V_i$?

[cloud walrus]

leave_no_norm

[chilly radish]

See the discussion below. The klein 4-group is an example

[void cosmos]

just think of the natural map

[glossy crag]

Does this work or am I wildly overcomplicating things:
Take $U'i=\bigoplus{j\neq i}U_i$, likewise with $V'_i$, let $M$ be the full module, then $U'_i\subset V'_i$ and the surjective map $M\to M/V'_i\cong V_i$ factors through to the surjective $U_i\cong M/U'_i\to V_i$, so there is a submodule $W_i\subset U_i$ such that $U_i/W_i\cong V_i$ and $V_i/U_i\cong\frac{V_i/W_i}{U_i/W_i}\cong0$, so $V_i=U_i$.

[void cosmos]

what does full module mean

[glossy crag]

the full direct sum, U_i and V_i are submodules of some fixed module.

[cloud walrus]

leave_no_norm

[void cosmos]

yea

i think so

[south patrol]

Sorry misread lol

[hidden haven]

This is an internal direct sum

[south patrol]

Oh lol

[hidden haven]

You have a direct sum and you make any of its terms strictly larger, what happens?

[south patrol]

What are the U_i

Or am I just late to the party

[hidden haven]

I think any submodules

[glossy crag]

just some submodules of a fixed module M

[south patrol]

Just left R-modules for any ring?

Okay

[hidden haven]

I inferred that it's internal because of the inclusion U_i ⊂ V_i

[glossy crag]

yeah, i guess i am wildly overcomplicating things

[hidden haven]

Otherwise idk what that would mean

[south patrol]

I think you are like

Take v in V_i

[glossy crag]

you are correct, yes

[south patrol]

Write it as a unique R-linear combination of elements of the U_i

[hidden haven]

[void cosmos]

^ yea

like

it can be done manually

also

easily

right?

[south patrol]

And then you'll see

Well actually better than that lol like just straight up intersect both sides with V_i right

[hidden haven]

You're basically proving this

Ok no

[south patrol]

I initially was worried about like

[hidden haven]

Answering this

[south patrol]

Infinite dimensional vector spaces

But no

[hidden haven]

Ye still works the same

Direct sum very delicate

[south patrol]

Hug

[hidden haven]

[south patrol]

Where do I learn about Witt vectors

[void cosmos]

what are those

[glossy crag]

Lorenz's Algebra II

he has a chapter on those

[south patrol]

Oh that was a serious question as well lol

[glossy crag]

but i haven't read it

[south patrol]

Inch resting

[next obsidian]

Bourbaki Algebra chapter 8

Only in French

[south patrol]

[next obsidian]

Also local fields by Serre

Has a little section on em

[south patrol]

I will check these refs out thank you

I imagined could probs learn some stuff like

[next obsidian]

The Bourbaki ref is mainly a meme

[south patrol]

In the process of learning about some applications

Lmao

[next obsidian]

But it is the one I used

I think it’s chapter 8. Maybe it was 9

[south patrol]

Lol thank you

Is it actually only in the french lol

[next obsidian]

Yea

Only up to chapter 7 is in English

Or was it in comm alg

I think it’s in comm alg

Chapter IX?

I forgot lmfao

But either way, it isn’t in one that’s been translated

[south patrol]

Yes you're correct

[next obsidian]

I’m a genius

[south patrol]

Found a lecture series on them lol

https://maths-people.anu.edu.au/~borger/classes/copenhagen-2016/

[next obsidian]

I’m a genius

[south patrol]

Oh lmao

This page also links it, hot

Actually how advanced are the prereqs yikes I might need to backtrack more lol but I will see how stuff goes

Thank you CHMONKEY

[next obsidian]

Glomed

[karmic moat]

borger 🍔

[void cosmos]

lmfaao

[glossy crag]

what's that about, i'm unfamiliar with server lore

[wraith cargo]

Eisenbud ch. 7 and Serre - Local fields

[south patrol]

Thank

Been reading Bourbaki which is nice

[teal vessel]

question because I'm dumb: how would I prove that function composition associates? I know it intuitively, it's because the embedding of one function into another can basically act as a placeholder to be expanded on at any time, but idk what the formalization would look like

OH I can just do a function map diagram, right?

[next obsidian]

Just show they’re equal

((f•g)•h)(x) = (f•g)(h(x)) = f(g(h(x)))

All equalities by definition

Do the same for f•(g•h) and you end up with f(g(h(x)))

So (f•g)•h = f•(g•h) by definition of equality for functions

[open sluice]

two functions are equal if they take on the same value everywhere on their domain

[teal vessel]

I think I figured it out, both have the same domains and codomains, and I can demonstrate a mapping from x to f(g(h(x))) using a mapping diagram showing how both end in the same place.

[maiden ocean]

I remember helping you translate this i think

[next obsidian]

A little haha

[south patrol]

Lol

Hm

[hollow fjord]

Hi guys! The set of nilpotent elements of a commutative ring is an ideal, is this real for non-commutative rings?

[solar glacier]

I don’t think so

Think of 2x2 matrices over Z

[hollow fjord]

i was trying to see if a could find a necessary condition and then prove that not every non-commutative ring satisfies it.

sorry for my english

uwu

[solar glacier]

With x being 0,1
0,0 y being 0,0
1,0 these are nilpotent but their sum isn’t

[hollow fjord]

Thank you!

[solar glacier]

You’re very welcome!

[south patrol]

Let $K = \mathbb F_p(t)$. Any hints on how to show the Artin-Schreier polynomial $f:= X^p - X - t\in K[x]$ isn't solvable by radicals? I know that the splitting field of $K$ is generated just by any root $\alpha$, since the roots are $\alpha,\alpha+1,\dots,\alpha+ (p-1)$, and that $\mathrm{Gal}(K(\alpha)/K) \simeq \mathbb Z/p$ correspondingly, but I can't find any obvious obstructions to $\alpha$ somehow lying in some bigger radical extension

[cloud walrus]

potato

[south patrol]

Ah okay, here's an attempt:

Eh okay didn'tq uite work gr

[wooden ember]

isnt going through the dual rep unnecessary here to show representations are completely reducible? We only have to show this for $r=1$ since it's easy to split $V$ into representations of each matrix algebra, and in this case, we have an intertwiner $\phi:A\oplus ... \oplus A \to V$ given by $\phi(a_1,...,a_n)=a_1v_1+...+a_nv_n$ for ${v_1,...,v_n}$ a basis of $V$, and so its kernel is a subrepresentation, giving us $V\cong A^{\oplus n}/ \ker{\phi}$ as an $A$-representation, and this is $W^{\oplus mn}/W^{\oplus r} \cong W^{\oplus mn-r}$ where $W$ is the unique irreducible representation of $A$

seems like it's about the same idea as what they do but without having to worry about the dual representation

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

this is basically just repeating the argument of "free modules over semisimple rings are semisimple, and thus all modules being quotients of free modules they are also semisimple"

oh i guess you can even make things easier by explicitly decomposing representations of A as direct sums of the actions of elementary matrices

[glossy crag]

Funny, I was about to post a question on a related topic.

[distant chasm]

what is the splitting field of the polynomial x^7 + 1 over F_2 ?
x^7+1 = (x+1)(x^6+x^5+x^4+x^3+x^2+x+1) where (x^6+x^5+x^4+x^3+x^2+x+1) has no roots but it can be written as (x^3+x+1)(x^3+x^2+1), when i calculate the splitting field, do i need to take in consideration the (x+1) as well? Would it be F_8 or F_64?

[void cosmos]

its the smallest field containing the roots of x^7+1 😛

extns

[wraith cargo]

x^7 + 1 doesn't factor like that

it's x^7+1 = (x+1)(1-x+x^2-x^3+x^4-x^5+x^6)

[glossy crag]

What's the difference lol, it's over F_2

[wraith cargo]

oh lmao

[distant chasm]

Yep, its F_2

[wraith cargo]

ok I didn't catch that

[wraith cargo]

(you don't consider x+1 since it has a root in F_2)

I think those two polynomials generated the same splitting field so it's F_8 but lemme think abt it

I remember seeing it in D&F but I'll look into it

[glossy crag]

Actually nvm, what am I talking about.

You're right.

Adjoining a root of either produces a primitive 7th root of unity and so x^7-1 splits

[wraith cargo]

ye that's the idea iirc

[glossy crag]

If you know what a primitive root of unity is, the right answer is F_8.

[distant chasm]

Ok thank you

I was thinking about F_8 as well

[glossy crag]

If you adjoin a root of either cubic, you get an element \zeta\neq1 that satisfies x^7-1, therefore its order in the multiplicative group is 7, so you get 7 distinct roots 1,\zeta,\zeta^2,\dots,\zeta^6 of x^7+1 and x^7+1 splits.

[glossy crag]

In case you need the argumentation.

[wraith cargo]

why must field theory be so fucked sometimes

[glossy crag]

Au contraire, I think it's very neat and logical.

[formal ermine]

hey this is beautiful to my eyes

category looks fucked to me

[delicate orchid]

yeah because the category of fields is fucked

[formal ermine]

🤝

[distant chasm]

I don't need the argumentation thank you, i'm allowed to use something that says F_p^n * F_n^m is F_n^s where s = lcm[m,n]

but i wasn't sure about the (x+1)

Thank you again

[wraith cargo]

noooooo

it's neat and logical

[cloud walrus]

leave_no_norm

[glossy crag]

If I want to reduce the proof of $M_{n_1}(D_1)\times\cdots\times M_{n_k}(D_k)\cong M_{m_1}(E_1)\times\cdots\times M_{m_l}(E_l)\implies n_i=m_i,D_i\cong E_i$ to the case of 1 factor, can I argue like this: the factors are the unique (non-zero two-sided) ideals of either ring, so the isomorphism induces a bijective additive and multiplicative map between them (perhaps after a permutation), which is unital, since the identity of either ring is the sum of identities of the factors, so we get $M_{n_i}(D_i)\cong M_{m_i}(E_i)$.

[cloud walrus]

leave_no_norm

[glossy crag]

How do I delete this, the react isn't doing anything

[chilly ocean]

you don't, you need to get a mod to do it

honorables used to be able to do this

[south patrol]

I am in pain

[static yew]

disclaimer: everything I know of abstract algebra is 100% self-taught (mostly from Wikipedia), limited primarily to finite groups and sets (primarily GF(2^n)) , stemming from their relationships to my work as a software engineer in error detection/correction, random number generation, and public/private key cryptography

(snipped: my end goal. let me know if you want to hear that)

My current journey of discovery has me trying to understand what the algebraic closure of a (prime) finite field looks like.

The only concrete thing I can find is a line on Wikipedia that says "For a finite field of prime power order q, the algebraic closure is a countably infinite field that contains a copy of the field of order q^n for each positive integer n (and is in fact the union of these copies)."

What's the effective difference betweeen the algebraic closure of, say, F_2, versus the positive integers? Where can I find something that will help me understand the finer details here?

[dreamy fiber]

Well so the positive integers don’t form a group (or even a monoid) under addition

whereas the algebraic closure of F_2 is a field

[static yew]

sorry, nonnegative integers

[dreamy fiber]

Equipped with an abelian group under addition and its multiplicative structure gives it a field structure

Well even the non/negative integers

They form a monoid under addition

Another example is the characteristics here

[static yew]

okay, not a group b/c no additive inverse

[dreamy fiber]

For any element in the algebraic closure of F_2

If you add the same element to itself, it is 0

[formal ermine]

do you know what an algebraic field extension is?

[dreamy fiber]

This is not true in

the non-negative integers

[static yew]

I do not understand what an algebraic field extension is. That's one of the things I'm having lots of trouble with

[formal ermine]

a field extension is just like

you have one field contained in another

[glossy crag]

As sets there is nothing to distinguish them, since they're both of "same size", but it's not just about that, is it.

[formal ermine]

we write that as L/K

K is contained in L

and both of them are fields

an algebraic field extension is a field extension L/K such that every element in L is the root of some polynomial with coefficients in K

example: C/R where C are the complex numbers and R are the real numbers is an algebraic field extension

[static yew]

so that means that all non-constant polynomials with coefficients in R [and integer exponents?] have roots in C, right?

[formal ermine]

not quite

[static yew]

oh

[formal ermine]

every element in C is a root of some polynomial in R

[static yew]

OH but not all polynomials necessarily have all of their roots in C?

[delicate orchid]

they do in this case

[formal ermine]

yes, not necessarily, but it turns out that's also true

we can now define what an algebraic closure of a field is

[delicate orchid]

but no in general

[static yew]

and if they do then C is an algebraic closure?

[formal ermine]

Yes

[delicate orchid]

yes!

[static yew]

okay, that part I can grasp

What I'm having trouble with is "extending" that concept, if you will, to finite fields

[delicate orchid]

an algebraic closure of a finite field is a field that contains every possible root to every possible polynomial with coefficients in that finite field

[formal ermine]

we call a field L algebraically closed if it fulfills one of the following, equivalent properties:
a) every non constant polynomial in L has a root in L
b) there are no algebraic extensions of L

[delicate orchid]

replace "finite field" with "R" if you want:
"an algebraic closure of R is a field that contains every possible root to every possible polynomial with coefficients in R"

[formal ermine]

the algebraic closure (n.b. simplifying here) of a field K is the smallest field above K, let's call it L, such that the extension (L/K) is algebraic and the L is algebraically closed

[static yew]

I'm almost there. If a polynomial p of degree k is irreducible in GF(2^k), then it doesn't have roots, which (if k>1) means that GF(2^k) is not algebraically closed, right?

[delicate orchid]

if the degree is > 1 then yes I think so

[formal ermine]

yes, finite fields are never algebraically closed

[static yew]

I'm really good at GF(2^k) b/c of LFSRs and CRCs

[formal ermine]

this can be proven easily:

,, f(x) = \prod_{a \in \on{GF}(2^k)} a + 1

[formal ermine]

replace GF(2^k) with whatever finite field you wish

[static yew]

but the union of all fields of prime powers of 2: F_2, F_2^2, F_2^3, ... is an algebraic closure?

[formal ermine]

as you can see, this polynomial has no root in GF(2^k)

=> GF(2^k) is not algebraically closed

[formal ermine]

the union of two fields is almost never a field

I can't even think of a way to meaningfully define the union of two fields without a bigger field containing both

[static yew]

yeah but this is the union of an infinite number of fields

okay, sounds like you're just as confused by this definition I found as I am

[next obsidian]

You take the “union”

You can make sense of the Union inside the algebraic closure

But when you don’t a priori have it, you can define it as a colimit

[delicate orchid]

told you the category of fields was fucked

[next obsidian]

Or you can do some annoying trick with adjoining variables for each irr poly and mod out

Do this over all n, take a colimit

[static yew]

you guys are going a little too fast for me here

[next obsidian]

This is an algebraically closed field containing your thing

[delicate orchid]

chmonkey my friend we're talking to a software engineer

[next obsidian]

And then take the sub field of algebraic elements

Oh

[delicate orchid]

no offense, stevie

[next obsidian]

I’m describing how you form algebraic closures in general

[formal ermine]

oop I forgor

[next obsidian]

You can define the algebraic closure of finite fields “easily”

[static yew]

like I said: I'm self-taught, and my focus has been on stuff I needed to understand to do my job, so I'm missing a lot of seemingly-irrelevant (to me) supporting knowledge

this all got messy when I got into non-binary prime fields

[next obsidian]

Then assume the field exists

[static yew]

I don't even have a mental model of what F_4 looks like
prime fields are easy cuz I can just do regular arithmetic and take the modulus

[next obsidian]

Just assume it exists for some abstract reason

Once you do this you can think of it as the Union of all finite fields

[static yew]

ah, but I'm an engineer and I'm trying to do engineering, so I need stuff that not only exists, but is constructible

[delicate orchid]

ok here's a fun fact, F_4 is GF(4)

[next obsidian]

I don’t know why you need F_p-bar at all then, and I have no idea then

[delicate orchid]

since you probably don't care too much about proofs (like me) I'll just let you know that all finite fields of the same size are isomorphic

[next obsidian]

I think of finite fields as just formally adjoining a root of a funny polynomial to F_p

[static yew]

right, but what is GF(4)?

GF(2) is {0,1}
GF(3) is {0,1,2}
GF(2) doesn't seem like {0,1,2,3} cuz 2*2=0

[next obsidian]

And it satisfies some relation like a^p^n = a

[delicate orchid]

you're right, it's not Z/4Z

it's a different structure

I'd prefer to think of it as {0, 1, x, 1+x} where 2 times anything is 1

[static yew]

I'm trying to understand Schoof's algorithm, which seems to involve finding the roots of x^3 - y^2 + ax + b for some constants a,b and x,y in F_p

which involves the Frobenius transform of the algebraic closure of F_p

[next obsidian]

I have nothing helpful to say because this isn’t the sort of thing I ever deal with so I’ll just have to tap out, sorry

I’ve got no idea how to constructively describe these things ¯_(ツ)_/¯

[static yew]

if that was supposed to say "2 times anything is 0" then I 100% get it, cuz I thought GF(2^k) has characteristic 2

[delicate orchid]

that's not what characteristic 2 means

replace 0 with 1 and you're correct

[static yew]

characteristic 2 means that 1+1=0 right?

[delicate orchid]

if anything non-zero multiples to get zero you definitely do not have a field

fuck nm you're right

[next obsidian]

Wew yeah, what the ruck

[delicate orchid]

or are you

[next obsidian]

Char 2 means 2•x = 0

Because well, 2 itself is 0

[static yew]

I don't know if I'm right that GF(p^k) has characteristic p

[delicate orchid]

you are

[delicate orchid]

I got confused swapping between the two definitions give me a break

[static yew]

hah! the score is now Stevie-O 1, people who know what they are talking about |Z|

[delicate orchid]

I do NOT know what I'm talking about that is a grave error to make

[static yew]

anyway if you tell me that GF(2^k) is the polynomials of degree k-1 or less
or at least is isomorphic the set of those polynomials
then I think I get it

[delicate orchid]

they're isomorphic as sets sure, even as F_2 vector spaces

do you know what a quotient is

[static yew]

I don't understand vector spaces (they've never been relevant to my work)
and I only know "quotient" as the value q such that a = qb + r

[south patrol]

pls learn about vector spaces before doing fields stuffs if u wanna make more prorgess

[next obsidian]

Yeah

If you’re doing anything remotely needing math, especially if it’s applied, linear algebra will be very helpful

[south patrol]

Pls someone help me with this solvaility by radicals q bleak

[next obsidian]

It’s by far the most useful branch of math

[south patrol]

Idk if it's like actually easy or smth

[ivory trail]

linear algebra is the only math with applications

[south patrol]

what lol

[next obsidian]

This might be a based opinion

All math which has applications ultimately reduce to linear algebra

That’s certainly provocative

[south patrol]

calculus

Tbf

just linearisation

[next obsidian]

calculus is trying to turn analysis into linear algebra

[south patrol]

[ivory trail]

i'm mostly joking

[next obsidian]

No

I think you’re onto something

[ivory trail]

can you express dynamic programming problems in linear algebra

[glossy crag]

Doesn't that make your polynomial solvable by radicals?

(in this definition)

[south patrol]

If you extend the definition lmao

not in the normal definition

[formal ermine]

"assuming the necessary assumptions ..."

[glossy crag]

Idk man, maybe that's the appropriate definition in char p, wouldn't know it.

[south patrol]

It is defined like that to get rid of situations like this extension I think aha

[formal ermine]

I am sullying you for the first sentence, not the content of the second sentence

[delicate orchid]

likewise

[formal ermine]

yeah it's senseful but I don't think it helps

[coral spindle]

I'm sullying because other people did it and I want to feel involved

[distant chasm]

can someone give me two examples of Z[i]-modules with 20 elements?

[coral spindle]

Have you tried looking for one on your own?

[distant chasm]

I don't actually know how to find one

[coral spindle]

OK, do you know any finite Z[i]-modules?

[ivory trail]

what's the easiest example of a Z[i]-module

[coral spindle]

Hint: maybe take quotients by ideals

[south patrol]

This theorem is mostly for characteristic 0

That's the point of this exercise

To show how it breaks in char p

[spice whale]

yes field of rational functions

[formal ermine]

wait non char 0 fields exist? Lmao

very unbased

[static yew]

based on my current reading,

1. modules are like vector spaces
2. vector spaces are basically groups that have an operator that is associative+commutative , an identity, an inverse for all elements, and the concept of (from my perspective) basically exponentiation

am I reading this right?

[dreamy fiber]

You can always show that the quotiehnt of Z[i] for any non-zero ideal is finite

[south patrol]

not exponentiation at all

[dreamy fiber]

You can get a good estimate on the cardinality as well

[formal ermine]

yes modules are like vector spaces but over a ring, meaning that you can't define by scalars

[south patrol]

X is like Y except with lots of other things lol

[coral spindle]

I wouldn't agree with 2., but 1. is fine.

[spice whale]

so Q(x) is the field of rational rational functions

[formal ermine]

if I had one penny for every time I used this in alg nt I would be rich

[static yew]

note that when I envision groups I envision multiplicative group of integers modulo n

that's not what other people seem to envision, but it's how I see it

[coral spindle]

Oh well that's not a good way of envisioning groups

[spice whale]

if you gave all the elements sapience it would become the field of rational rational rational functions

[delicate orchid]

those are the most basic groups possible so that's really not a good mental model

[coral spindle]

Not even a good way of envisioning Abelian groups

[static yew]

It's a very useful way of envisioning if most of your math is done modulo 2^(8k)

[spice whale]

that's not even necessarily a group

[coral spindle]

Riiiiiiiiiiiiight

[delicate orchid]

it's a useful way to visualise... the multiplicative group of Z/nZ... lol

[spice whale]

zero divisors..... shudder

[coral spindle]

Anyway, the most helpful way I know of thinking about vector spaces is frankly just imagining R^3 as a space and then saying the dimension/field 50 times over.

Eventually you don't have to think about problems re: dimension or field change because you know the formalism well.

It's not generally a great idea to get hung up on how to conceptualise a structure; rather start working with them and you will build intuition after reflection.

[spice whale]

i visualise groups as these funny collections of people with social connections

[static yew]

also the Galois group of polynomials with coefficients in F_2 modulo an irreducible polynomial of some degree k

[spice whale]

oh wait wrong group

[delicate orchid]

that's a graph

[static yew]

computing the the transitive closure of those groups is a pain

[spice whale]

kevin bacon

[coral spindle]

gevin pacon 😔

[spice whale]

it was a joke you missed the joke

[delicate orchid]

some might say a perfect graph

[static yew]

Kevin Bacon \in S

[coral spindle]

not as perfect as the epic people reading this

[delicate orchid]

https://tenor.com/view/reddit-gold-reddit-gold-slot-machine-gamble-gif-27586860

[coral spindle]

We did it reddit

[spice whale]

when the voicing moves 😳😳😳😳

thank you kind stranger

[coral spindle]

You won the internet today

[spice whale]

i had a cupcake today does this mean it's my cake day

[coral spindle]

yes

also why didn't you share

I'd like some cake

[spice whale]

I don't know where you are

[static yew]

wait
so if a module is a generalization of a vector space where the scalars come from a ring rather than a field

and all fields are rings
then all vector spaces are modules, right?

[coral spindle]

Yes

[spice whale]

if you send your current place of residence, mother's maiden name, and the name of your first pet, I'll share next time

[coral spindle]

would u also like my credit card numbers

[static yew]

you also need the name of the street they grew up on

[spice whale]

all of that yes

and the 3 numbers on the back

[static yew]

My credit card number is 5 (mod 10)

[spice whale]

expiration month and year

[formal ermine]

hey @coral spindle we are friends right? c-can you gwive me ywour s-social sewurity number uwu~?

[coral spindle]

OK so if you're a troll please don't look. Thanks everyone. My current address is 10 Downing Street, London SW1A 2AA.

[spice whale]

thank you

I'm breaking in as we speak

[delicate orchid]

https://media.discordapp.net/attachments/937528941248393226/1078355280456323203/kotyaka_blekmyaka.gif

[formal ermine]

I am sending you a postcard with further instructions

[coral spindle]

I hope y'all know what's at 10 downing street

otherwise the joke kinda falls flat

[delicate orchid]

no it's not a flat it's a house

[formal ermine]

I hope I get mail back

[spice whale]

wew and i are literally English

[formal ermine]

my deepest condolences

[spice whale]

well wew pretends to be english

[coral spindle]

Oh nice I always assume other people here are yanks

But I am indeed british

Anyway, I guess I can say something about modules

[static yew]

okay, kids!, we're gonna have some fun today!

1. take your credit card number and treat it as an integer
2. multiply it by 100 and add the expiration month
3. multiply by 100 and add the expiration year mod 100
4. multiply by 1000 and add the three digits on the back
5. take it to the 65537th power
6. divide by 1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139 and take the remainder
7. DM me the answer

[delicate orchid]

comically loud gulp

[formal ermine]

hey if you wanna say something about modules then you should definitely answer my question #advanced-number-theory message

[coral spindle]

D:

You can broadly think of modules as abelian groups that have some structure imposed upon them by the ring

This is very broad though

[spice whale]

i know wew is a fed in disguise

[coral spindle]

THe ring can have very diverse structure

[delicate orchid]

p-adic modules or as I like to call them, graded modules

this is incorrect

but I like to call them that

[coral spindle]

Bruh idk

WIsh I knew more about the p-adics that would be helpful

[spice whale]

in the case of the integers that's no restrictions

[coral spindle]

Yar

[spice whale]

har har

[static yew]

I, personally, think of modules as a highly complex abstract concept that I won't need to know or care about unless and until someone invents an error-detection/correction algorithm, random number generator, or public-private key algorithm based on them

[delicate orchid]

I'd unironically just think about modules over the p-adics like, categorically

there's no point in actually thinking about them

so just throw diagrams at it

[coral spindle]

You will literally never believe what I'm gonna say next

Linear codes.

They're vector spaces. I.e., modules.

[formal ermine]

I know of this one result that like uh

[delicate orchid]

ur entire PC operates on a finite ring nerd

[formal ermine]

every abelian group has a unique Z_p structure

[static yew]

I'm pretty sure I said something earlier about everything being 2^(8k)

[dreamy fiber]

I mean to study Elliptic curves, you need to be good at modules

[coral spindle]

Not sure about that.

[static yew]

yes, that was the RSA-100 challenge. it was factored in 1991

[formal ermine]

trust

I read it on MSE

[delicate orchid]

hmmmmm

[formal ermine]

so it must be true

[coral spindle]

In the first place it needs things like 3-divisibility in Z_2 for example

[formal ermine]

the user has 17.1k rep

[coral spindle]

And you probably want continuity conditions

[spice whale]

ok what's the structure of C_2 over Z_3

[coral spindle]

But I can believe after you've added that that there is at most one Z_p-structure

[delicate orchid]

consider (2,0,0,...)

what does "structure over" even mean here

are we taking p-adic reps or what

[coral spindle]

C_2 as a Z_3 module

[spice whale]

it is a module over

i guess

[delicate orchid]

I refuse to think about this

[coral spindle]

protip it ISN'T, BICHES. SUCK IT

[dreamy fiber]

C2 as in the cyclice group of order 2?

[delicate orchid]

it's too hot to think about it

[formal ermine]

https://math.stackexchange.com/a/4260665/943349 fwiw lol

[coral spindle]

Z_3 contains 1/2 so C_2 cannot be a Z_3-module

:(

[delicate orchid]

how does that follow

[coral spindle]

U forgor 💀 to say

[formal ermine]

I forgor

[delicate orchid]

.... LOL?

[dreamy fiber]

I mean any finite dimensional Z_3 vector space

has to have cardinality divisible by 3 right

[coral spindle]

Well let C_2 = <g>. What should 1/2 . g be? It cannot be g^0 or g, since g^2 = g^0

[dreamy fiber]

(except for 0)

[coral spindle]

Z_3 is not a field

[delicate orchid]

thanks for doing my thinking for me boytjie it's much appreciated

[coral spindle]

Sowwy wew

[dreamy fiber]

Oh I thought you meant Z/3Z

lmaooooo

[formal ermine]

the 3-adics

[coral spindle]

if you want me to shut my whore mouth I can do so

[delicate orchid]

no, I was being sincere

[dreamy fiber]

the notation was confusing lol

[spice whale]

finitely generated moderatorule

[coral spindle]

Oh right, sorry, yeah the notation sucks

[delicate orchid]

it was basically exactly what I thought it would happen

[formal ermine]

yeah we should just write $\mathcal O_{\bQ_p}$ for less confusion

[delicate orchid]

oh my golly gosh what happens if you consider le whoilesome divisible group as a p-adic module oh my goodness gracious

[dreamy fiber]

$\bQ_3$

[coral spindle]

Lol

[cloud walrus]

ScarletScorch

[coral spindle]

maybe $\hat \bZ_p$

[cloud walrus]

Boytjie (never-to-be-glomed)

[coral spindle]

I think that's decent notation

[delicate orchid]

order of a ring

I said no thinking

[formal ermine]

then profinite completion of the integers would be $\hat \bZ_\infty$?

[static yew]

wait if Z_3 isn't Z/3Z what is it

[coral spindle]

Ye

[spice whale]

to find out more, watch the video from veritussy, YouTube's most trusted mathematics commentator

[coral spindle]

The 3-adic integers

[spice whale]

p-adic integers

[delicate orchid]

ok tbh seeing that the p-adics are a lattice in Q_p isn't really thinking

[static yew]

that sounds like some pohlig-hellman stuff

[formal ermine]

what constructions of R do you know?

[spice whale]

every abelian group is a zoinkers module

[coral spindle]

uh no?

[formal ermine]

every non aphromic group can be realized as an l adic prüfer rep

[static yew]

I am agnostic to the axiom choice and Zorn's lemma and thus don't trust that there is a valid construction of R

[formal ermine]

[coral spindle]

That's nice do I care tho

[formal ermine]

constructing R doesn't require the axiom of choice

[static yew]

dedekind then

[formal ermine]

you chose the wrong one

[delicate orchid]

take the algebraic closure of Q and then examine the largest proper subfield

I think

[static yew]

look any axiom that results in banach-tarski is amongus-levels of sus in my books

[next obsidian]

I am not very knowledgeable on this stuff

[formal ermine]

🫂

that's what I feel like all the time

[next obsidian]

But Stevie, I think you are maybe not concerned with constructive math, but with computable math

[static yew]

this is true

I'm a software engineer. the only fields we get to work with are finite ones

actually I mostly work with using polynomials with coefficients in F_2, modulo an irreducible

[delicate orchid]

oh so you do know what a quotient ring is

[static yew]

if I'm doing anything outside basic arithmetic mod 2^32 or 2^64

I might know what a quotient ring is, but not by that name

[delicate orchid]

you're quotienting F_2[x] here

"modulo" <-> "quotient"

which is how you formally define a field extenstion

[proud zenith]

F_2 modulo an irreducible is a finite field extension, e.g. if poly has deg d, extension of F_2 of deg d

[static yew]

yeah I do that a lot

I use the irreducible polys as seeds for PRNGs to produce long sequences of unique values

[delicate orchid]

I do not like wireless technology

[static yew]

maybe? I just wanted sequences with linearity less than the absolute trivial of 1,2,3,4,5

a common one is degree 33
I use x^37 as a generator, and while I'm sure the NSA could reverse it, a casual observer would not be able to infer much from nearby outputs

[delicate orchid]

wahh whah whahwahhwah WAHHHH

[static yew]

hold on, I'm processing that, eggman

[delicate orchid]

yeah, it's even true for an arbitrary field

[static yew]

multiplication of vectors
So wait, what's the difference (beyond homomorphism or isomorphism or remorphism or whatever) between the vectors in this d-dimensional vector space over F_2, and the polynomials in the (polymial ring? what's the name for that set of polynomials?)

[delicate orchid]

none

[static yew]

but you can multiply polynomials

[delicate orchid]

ok?

doesn't mean we have to

[static yew]

in fact I do. that's how I do all my stuff with them.

[delicate orchid]

if we include that operation it's an algebra, which is a vector space where you can multiply stuff in a nice way

but we don't have to

[distant chasm]

I still don't know the answer, i think i will say something very stupid but is Z[i]/(4+2i) good?

[static yew]

to generate the k'th poly, I take the polynomial x and multiply it by itself 37k times

[next obsidian]

I think you really have to understand what a vector space is to understand what they said

[static yew]

is it that, if you're defining it as a vector space, you don't get to do poly multiplication?

like, it's possible to define a second operator * with uv = vu and (uv)w = u(vw) and an identity for that operator

but you're just not using that?

[coral spindle]

Well that's certainly a finite Z[i]-module. I can't tell at a glance if it has 20 elements or not, but if you can write all 20 down then sure

[delicate orchid]

if you want me to be specific: the SET F[x] together with scalar multiplication from F and addition is a F-vector space

the same SET but with just polynomial addition is a group

[next obsidian]

Precisely

[static yew]

okay, so it is what I was understanding

you're taking the (ring? is it a ring? I don't think it's a field) of polynomials of degree less than d and coefficients of F_2

and you can define + as the "usual" polynomial addition, except everything is mod 2
and you can define * as the "usual" polynomial multiplication, except everything is mod (some irreducible poly)

but you don't define * and you just leave it with + and say "ok this is my vector space"

[delicate orchid]

well, yeah but we've only talked about modulo irreducibles so far

[distant chasm]

Can you give me another example or another hint? Is there a way to count them without writing them?

[delicate orchid]

we're taking the set, but yes

and you still need multiplication by elements in F

[coral spindle]

Well perhaps you have seen this if you are doing this question as part of algebraic number theory: the cardinality of Z[i]/(4+2i) is |N(4+2i)| = |(4+2i)(4-2i)| = 20.

Someone can correct me here if I'm getting this wrong

Oh wait I calculated wrong

[delicate orchid]

here lemme do something fancy for you stevie

[coral spindle]

There, fixed

[distant chasm]

Yes, i was thinking about that when i came up with the example

[coral spindle]

Well that's an excellent approach.

[distant chasm]

But i don't know how to find another one

[static yew]

okay here's a question, purely for my own edification:

you can take the set of polynomials of degree d and coefficients in F_2
and add the mod-2 '+' operator
and you get a vector space

what if you take the set
and add the mod-a-polynomial '*' operator instead
is that a vector space?

• associative: a(bc) = (ab)c
• commutative: ab = ba
• identity: "0" is the constant polynomial 1
• I don't know if everything has a multiplicative inverse though
• scalar multiplication = exponentiation
• wait is it a module

[coral spindle]

No

[static yew]

damn

[coral spindle]

It's not even a group

[static yew]

I thought I was onto something

what? yes it is

[coral spindle]

No, 0 is in there.

Groups can't have an absorptive element unless they are trivial

[static yew]

oh right
drop out the constant polynomial 0

then it's definitely a group

[coral spindle]

How might you get something with the same norm as 4+2i, maybe that's not just 4-2i or an associate

There's a really easy one to find

[coral spindle]

Still no

[delicate orchid]

no inverses

[coral spindle]

Has almost no inverses

[distant chasm]

If i change 4+2i with 2+4i, are they isomorphic?

[static yew]

groups don't have to have inverses do they?

[coral spindle]

I don't actually know. You might have to check

[static yew]

those are fields

[distant chasm]

Ok i will, thank you

[static yew]

less sure about rings

[delicate orchid]

Let $\mathbb{F}[x]$ be the set of polynomials with coefficients in $\mathbb{F}$. And let $+$ be polynomial addition, $\times_{\mathbb{F}}$ be scalar multiplication by elements in $\mathbb{F}$ and $\times$ be polynomial multiplication
then:
\begin{enumerate}
\item{$(\mathbb{F}[x], +)$ is a \emph{group}}
\item{$(\mathbb{F}[x], +, \times)$ is a \emph{ring}}
\item{$(\mathbb{F}[x], +, \times_{\mathbb{F}})$ is a vector space}
\end{enumerate}
and with all 3 it's an algebra

[coral spindle]

Yes, they do. They must have inverses. That is part of the definition of a group

[cloud walrus]

wew lads or something

[delicate orchid]

this shows a fundemental gap in your knowledge

[coral spindle]

Bro idk just read the wikipedia article on groups, it's not hard to find this out

[delicate orchid]

fields have two operations

groups have one

[static yew]

don't they literally call it the multiplicative group of integers modulo n

[coral spindle]

They literally do

[delicate orchid]

yes... because it only has one operation... multiplication

[static yew]

but it's not a group?

[coral spindle]

It is a group.

[delicate orchid]

but it is

the multiplicative group of integesr mod n is NOT on the set {1, ..., n}

it is on the set of integers coprime to n

(Z/6Z)^{\times} = ({1,5}, *)

(Z/8Z)^{\times} = ({1,3,5,7}, *)

and so on

[coral spindle]

If you have any ring R, you can look for the elements that have multiplicative inverses, which form the set R^\times. This is then called the group of units of R.

[spice whale]

yes they do...
that's part of the definition of a group

[static yew]

right
and if you have the set of nonzero polynomials of degree less than d with coefficients in F_p
and you have an operator * which is "usual" polynomial multiplication, modulo an irreducible polynomial

isn't that a group?

[coral spindle]

The multiplicative group of integers mod n is (Z/nZ)^\times.

[coral spindle]

Yes, modulo an irreducible polynomial, excluding 0.

That is in fact the group of units of a field.

[static yew]

okay so that's where we diverged

[coral spindle]

OK

[static yew]

I said "mod a poly" and I meant an irreducible polynomial

not just any polynomial

[delicate orchid]

wait is that where we diverged? I have forgotten lol

[proud zenith]

Find a book in peace and quiet and get the definition of group, ring, field, then algebra, then possibly module if you don't mind a bit of nuance

[spice whale]

do you understand what a group is

now

[static yew]

I don't mind the nuance, but I do have trouble keeping all the nuances straight

[delicate orchid]

module before algebra, imo

still don't know what a module really is

it's just like a dude

[proud zenith]

I like algebra because you can say "an algebra is a ring which is simultaneously a vector space"

[coral spindle]

"this guy" points to letter M

[proud zenith]

oh btw I should have added vector space in the list

[spice whale]

clearly just an algebra over a monad of the form $R\otimes -$ in CRing

[cloud walrus]

ally ❤ (semaluhtounuyulohowwah)

[delicate orchid]

you need therapy

[spice whale]

yes but not for that reason

[delicate orchid]

there can be multiple reasons

[spice whale]

how will a therapist help me

[delicate orchid]

no more category theory

[spice whale]

with the perceived problem beforehand

[spice whale]

i will teach my therapist category theory

[lyric epoch]

Guys how can I find second factorisation of 40

[delicate orchid]

ok I think I've gotten all the funny squares in my head to actually believe this now

[spice whale]

btw @static yew ignore this it's a joke

[delicate orchid]

I still don't really see it

[coral spindle]

40 = 4 * 10

Or maybe you'd prefer 8 * 5

[static yew]

40 = 2*2*2*5

[coral spindle]

The point is, don't forget about integers

[delicate orchid]

someone ain't reading the paper

[static yew]

if I'm testing out a pohlig attack I care that the prime factors are 2 and 5

[spice whale]

sorry not in CRing it's in R-Mod

[coral spindle]

You know what I can't deny that

[delicate orchid]

uhh it's 3 mod 4 that are irreducible in Z[i]

[delicate orchid]

because they literally wrote the prime factorisation at the top mr smarty pants

[coral spindle]

Yeah I assumed they'd already found a nasty ass one

[static yew]

okay what the heck is Z[sqrt(-13)]

[delicate orchid]

Z adjoin sqrt (-13)

lemme think about this one

[spice whale]

read a book

[lyric epoch]

Oh sry

[delicate orchid]

,w (a+sqrt(-13)bi)(a-sqrt(-13)bi)

[cloud walrus]

Results provided by WolframAlpha

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

Requested by wewladstbh#0

[static yew]

so, the set of all integers, plus the root of the polynomial x^2 = -13 ?

[delicate orchid]

thanks wolfram VERY cool

yes

[spice whale]

where did you get the extra i from

[static yew]

how and why would that ever be useful to anyone

[delicate orchid]

forgot

[delicate orchid]

I thought you did cryptography lol

you mfs care about weird and wacky rings

[static yew]

even I have limits, man

[delicate orchid]

colimits perhaps

[spice whale]

number theory is literally the core of cryptography

what are you talking about

[coral spindle]

Yet here you are, complaining about it

[spice whale]

you will be joining me in therapy

[static yew]

it's bad enough that I have a subset of 2-D vectors in Z/(large prime)Z plus an element named "the point at infinity" that is mathematically "zero"

[delicate orchid]

oh I've got it

[formal ermine]

wait what's 13 mod 4

[delicate orchid]

wait lemme double check

[static yew]

anyway, so you want a prime factorization of 40 that is not 2*2*2*5 ?

lemme see if I can do this

[coral spindle]

One of the things listed in 🔢

I can't see one immediately

[delicate orchid]

fuck I was doing Z[sqrt(13)] nooo the wholesome chungus noo

[coral spindle]

It's probably not going to be a nice difference of squares

At least, I can't see any that are the difference of two squares

[delicate orchid]

yeah sqrt(13) is BIG

[lyric epoch]

It's is but I have to prove that they are not unique factorisation domain so I want do two different factorisation of same elements into irreducible elements

[coral spindle]

Well I mean that alone isn't hard, right? You can do (1+sqrt(-13))(1 - sqrt(-13)) = 14 = 7 * 2 and 2 is definitely irreducible in Z[sqrt(-13)]

[lyric epoch]

It's not sqr13 its sqr -13

[coral spindle]

I know

[delicate orchid]

yeah there are way easier ones lol

[delicate orchid]

jsut do this

[coral spindle]

But the question asks for 40 specifically

[lyric epoch]

@coral spindle it's not 14 it's 40

[coral spindle]

honestly, I don't see how.

[delicate orchid]

grrrrrrrrrrrrrr

[coral spindle]

YES I KNOW

That's what I'm SAYING

[delicate orchid]

but that doesn't matter in finding if it's a UFD or not

stupid question I'm going to go sulk

[vagrant zinc]

Guys can someone explain me what is the order of root of 3 in R ?

[coral spindle]

Let's say x = sqrt(3). Can you tell me 2x, 3x, 4x — just the first few?

[formal ermine]

R not R*

[coral spindle]

Oh right. I did indeed jump the gun.

Fortunately fixable.

Saw (c) not (b).

Could still be a typo, who knows... I love it when people ask questions and then jump off the earth for a few minutes.

[ivory trail]

Z[sqrt(-13)] is the set of all numbers a + b sqrt(-13) where a,b are integers. it is understood to have a ring structure, i.e. addition/multiplication and units for those (0 and 1), additive inverses, distributivity, associativity, commutative addition. (in the case the multiplication is also commutative, so it's called a commutative ring)

[delicate orchid]

so yeah what stevie said

[ivory trail]

well not if you don't understand what a generating set is

we want our thing to be closed under the above operations

[delicate orchid]

we're talking to a software engineer

[ivory trail]

yes, so i'm saying exactly what the elements are

it's not $\mathbb{Z} \cup {\sqrt{-13}}$

[cloud walrus]

196883

[static yew]

yeah I think I see the difference

I'm not sure I see the difference

[formal ermine]

it's $\set{f(\sqrt{-13}) | f(x) \in \bZ[x]}$

[delicate orchid]

this does not have 2sqrt(-13) in it

[delicate orchid]

this does

perhaps I should stop assuming things

nahhhhh

[static yew]

technically $\mathbb{Z} \cup {\sqrt{-13}}$ doesn't have, say, 2 sqrt(-13), or 3 sqrt(-13), or 3187503 sqrt(-13) in it, but Z[sqrt(-13)] does

[cloud walrus]

Stevie-O

[coral spindle]

Um actually it's the image of the unique map $\phi \colon \bZ[x] \to \bC$ with the property that $\phi(x) = \sqrt{-13}$ nerd underscore face

[cloud walrus]

Boytjie (never-to-be-glomed)

[ivory trail]

yeah exactly

[static yew]

Um actually it's a Frobenius anthropomorphism with an Ableian Group Project using a modular finite ring polynomial vector space to math at things until you convince someone to give you a grant

[delicate orchid]

Nobody gave me a grant

[static yew]

obviously you didn't use enough buzzwords

[delicate orchid]

I am in crippling student loan debt

[static yew]

wait

you guys said Z[sqrt(-13)] is a ring, right?

[delicate orchid]

Yes sir

[ivory trail]

yes, a commutative ring in fact

[coral spindle]

It is a subring of C

[static yew]

yeah
so if it has Z and sqrt(-13) in it
it has to have all integer multiples of sqrt(-13) or it wouldn't be closed under addition

[coral spindle]

Yes ofc

[delicate orchid]

Exactly!

[formal ermine]

should've mentioned its connection to (p, q)-adic analysis and thus to the solution of the collatz conjecture...

[wraith cargo]

It's also a dedekind domain!

[static yew]

look this stuff may be obvious to you but it's not obvious to me

[coral spindle]

And indeed it does contain all such multiples

[delicate orchid]

Here come the NERDS

[formal ermine]

wait what was 13 mod 4 again

[delicate orchid]

1 you muppet

[static yew]

dedekind domains aren't real
they're like birds

[ivory trail]

same thing for multiplication, but sqrt(-13)^2 = -13 which is already an integer so you don't have to worry about it. if it was the cube root of -13, then this would be important

[delicate orchid]

Are you doing the indicator thing

[delicate orchid]

Because if you’re doing the indicator thing -13 is 3 mod 4

Which is what you need

[static yew]

what's this about 3 mod 4?

[delicate orchid]

Dw about it it’s nerd shit

[wraith cargo]

NT shit

[static yew]

I'm into nerd shit!

[coral spindle]

Stevie, you need to get a good book on AA

[ivory trail]

my proof of the mod 4 thing was like a page lmao

[wraith cargo]

Z[sqrt(-13)] is the ring of integers of the number field Q[sqrt(-13)]

[static yew]

is there a name for the 3 mod 4 thing

[delicate orchid]

I don’t think I’ve ever seen the proof nor do I want to

[ivory trail]

well it is kind of weird a priori so you should see it at least once

[delicate orchid]

Yeah, the indicator tells you if it’s like this or if you need to do something weird

[ivory trail]

wait i think you have it backwards

https://en.wikipedia.org/wiki/Ring_of_integers#Quadratic_extensions

oh no i have it backwards

13 isn't -13

idiot

[delicate orchid]

It’s ok I thought the same too

[ivory trail]

i blame you

i am free of error

[coral spindle]

you have literally done nothing wrong ever and you're perfec,t rho :thum bsup

[delicate orchid]

I have done nothing ever

[wraith cargo]

Generally if your number field is Q[sqrt{D}] where D is is a squarefree integer, then the ring of integers of this ring is Z[1/2 + sqrt{D}/2] if D = 1 mod 4 and Z[sqrt{D}] if D = 2,3 mod 4

[rustic crown]

buii

[delicate orchid]

Yeah that’s the one

Hate it

Stupid

[wraith cargo]

NOOOO
It's cool

[delicate orchid]

Generalise it to nth roots then I might care

[wraith cargo]

That's actually a good question lol

[coral spindle]

It's so gross

So so gross

[wraith cargo]

NT is coooool

[coral spindle]

There's really no justice on this bitch of a number theory......

[proud zenith]

I know what you are talking about regarding first impression of "the point and infinity", it can be rigorously formulated but you have to move into projective space. That is not as bad as it sounds. You take y^2 = x^3 + Ax + B, and you homogenize it, now we have Y^2Z = X^3 + AXZ^2 + BZ^3. You allow any points (X:Y:Z) except (0:0:0). In this setup, the point at infinity is the much more tangible (0:1:0)

[delicate orchid]

Oh wait I missed this message

This is just the group on an elliptic curve lol

Perfectly normal

[ivory trail]

every non-annoying planar cubic with a point is an elliptic curve right

[delicate orchid]

Yeah something like that

[proud zenith]

you can just move back to F_p x F_p by taking (X:Y:Z) to (X/Z, Y/Z)

[delicate orchid]

Scary!!!!!!!!!

Oh nvm not that scary

[glossy crag]

In the book I'm using the Wedderburn theorem for simple rings is stated for left-Artinian rings, i.e. a left-Artinian ring is simple if and only if R\cong M_n(D). I assume this holds more generally (two-sided) Artinian rings?

[ivory trail]

the homogeneous coordinates (X:Y:Z) should be looked at as ratios, so (X:Y:Z) is the same point as (aX:aY:aZ) for nonzero a

[delicate orchid]

Yes

[static yew]

yeah I've read about it but ngl that sounds like a band-aid

"you want to add a point "at infinity"? On a plane? did you fail geometry? In a CAVE? WITH A BOX OF SCRAPS?"

"uhh noo you see it's actually a 2-d simplification of a 3-d projection..., if you look at this"

[proud zenith]

yes and there's an equivalence relation

[ivory trail]

projective geometry is actually magic and you should be grateful

[glossy crag]

Why not state it like that then, lol. It's not like the book shies away in other aspects.

[delicate orchid]

I’m adding a point at infinity because I CAN and if you want to CRY then leave

[delicate orchid]

I’ve got no clue, do they prove left artinian + simple <=> right artinian?

[static yew]

ngl "point at infinity" doesn't seem any more or less nonsense than Z[i] or Z(t)

actually Z(t) makes my head hurt

[ivory trail]

imagine saying this: "every distinct pair of lines in the plane intersects at exactly one point, except when they are 'parallel'"

[static yew]

why would you do that

[ivory trail]

why would you have an exception

very annoying

[delicate orchid]

Z(t) is quite natural

[proud zenith]

look at infinite train tracks! they meet at infinity on the horizon

[delicate orchid]

You should see some of the shit I deal with

[static yew]

I see you one parallel postulate and raise you the surface of a sphere

[glossy crag]

Hm... I haven't seen that yet, but I suppose they might, seeing how that follows trivially from the theorem.

[coral spindle]

a few minutes has become 23, btw

[ivory trail]

well a sphere isn't really a plane

[ivory trail]

it has curvature: any non-degenerate triangle has angles adding up to more than 180 degrees

[proud zenith]

best advice Stevie-O is put a few weeks of evenings into a book on abstract algebra, and force yourself to not move completely past too many exercises

[ivory trail]

there's also a nice axiomatic basis for projective geometry, and you get something cool called projective duality

[delicate orchid]

Yeah it’s not aerodynamic enough

[static yew]

Exam question: "A PE teacher lines the boys up at one end of the gym and and the girls at the other end
every minute, he blows the whistle, and the boys walk half the distance remaining across the room. When do they meet?"

• the physicist says "they never do"
• the mathematician says "in an infinite amount of time"
• the engineer says "technically they never do, but pretty soon they'll be close enough for all practical purposes"

[ivory trail]

so if you take your plane and add these "limit points" where the train tracks meet on the horizon

you get a line at infinity

boom, projective plane

[delicate orchid]

The physicist spouts some gibberish about SU4(C)
The mathematician says “trivial”
And the engineer wonders why they’re walking so inefficiently

[glossy crag]

Is this meant to be some kind of invariant ring of D_8?

[coral spindle]

ur a king, wew

[delicate orchid]

Kind of, it’s the ring of S_4-stable representations of D_8

[coral spindle]

[ivory trail]

my point is that the projective plane isn't so contrived, so neither is the point at infinity on an elliptic curve

[coral spindle]

Oh is this the grothendieck ring?

nice

cool

neat

[delicate orchid]

There’s an even worse one in 4 variables later on but I cba to get the paper

[formal ermine]

grothendieck ring?

[coral spindle]

get all yer reps

there's addition like (+) and multiplication like (x)

add some inverses

you've got a ring

[delicate orchid]

False

[coral spindle]

that's the grothendieck ring SHUT

[formal ermine]

that clears things up 👍

[delicate orchid]

I’m going to give the exact sequence definition in an abelian category because I hate illuminator so much

[coral spindle]

Wew where did I go wrongggggggggggg

OK fine

nerd

[delicate orchid]

It just so happens to be what you claimed because every C[G] module is projective

[formal ermine]

hey that's fine I know some cat theory now

[coral spindle]

do u know Abelian categories

[formal ermine]

maybe

[delicate orchid]

Tbh abelian categories are just exact categories in my head

[coral spindle]

MAYBE AIN'T GREAT

[delicate orchid]

I simply don’t care for them in any other context

[agile burrow]

who up braiding they monoidal categories

[formal ermine]

YES UNLESS YOU ASK ME TO PROVE SOME WEIRD SHIT

hey I heard monoidal subalgebra recently

[static yew]

yeah I came in here in the first place cuz I'm trying to understand Schoof's algorithm for counting the number of points in an elliptic curve over a prime field

and I'm getting hit with algebraic closures and Frobenius endomorphisms and mathematical notation that makes no sense to me

like:

"the frobenius endomorphism given by $\phi$ satisfies the characteristic equation $\phi^2 - t\phi + q = 0$ except if phi is a mapping how do you square a freaking mapping

[cloud walrus]

Stevie-O
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[coral spindle]

$\phi^2$ is defined as $\phi \circ \phi$

[cloud walrus]

Boytjie (never-to-be-glomed)

[coral spindle]

The composition of phi with itself

It's an endomorphism, so you can do so.

[vagrant zinc]

I'm sorry it's just group theory, and I went to take a bath since I've been reading and exercising all day, I don't think I know anything about what they are talking about, I know they ask me the order of each element, so I wouldn't understand well with advanced topics.

[static yew]

wait so if it's (x, y) -> (x^q, y^q) then the "square" is ( (x^q)^q, (y^q)^q) ?