groups-rings-fields

[shy fossil]

lemme see

[void cosmos]

@oblique thorn u are probably using christof par's textbook

there is a youtube playlist that covers it and does many examples ( the hwole course + EEA )

even if ur not using it

check out this video

https://www.youtube.com/watch?v=fq6SXByItUI&list=PL6N5qY2nvvJE8X75VkXglSrVhLv1tVcfy&index=11&ab_channel=IntroductiontoCryptographybyChristofPaar

[shy fossil]

Here's my notes that I had with me rn

don't worry, i am taking cs courses for reasons

[oblique thorn]

Here are my notes

It’s 4 pages

And they want us to do a = 51 and Fp = 2^17 - 1

[shy fossil]

i see, we did it in 2 pages lmao

[oblique thorn]

Do you get what the pseudo code is?

[shy fossil]

who writes pseudocode like that

[oblique thorn]

I see

[oblique thorn]

I have no idea

It’s baffling

[shy fossil]

that's an iterative algortihm, that's why it's so convoluted

[rocky cloak]

Cryptographers

[shy fossil]

the recursive algorithm i sent you is much cleaner

[oblique thorn]

Can’t read that lol

[shy fossil]

I have taken crypto classes, and they don't so it like this lmao

[shy fossil]

really? need a doctor

[oblique thorn]

Yeah our ta admitted he doesn’t really know what he’s doing

[oblique thorn]

Psh. Well I’m on my iPhone mini that’s prob y

[shy fossil]

maybe

[oblique thorn]

But my university is notoriously bad at teaching and it’s in Florida

They are known to have the worst schools

But anyways how would we go about a = 51 and b/p= 2^17 - 1

[shy fossil]

just set up a code

and plug

[oblique thorn]

So a would be 51

And b would be 2^17 - 1?

But change it to p cause it’s prime I think he said

Yea I think i figured it out. I just want to show you how confusing he made it

[long obsidian]

Is Q[x]/<x^2-2> isomorphic as fields to Q(√2)?

[formal ermine]

yeah

prove it

[long obsidian]

Okay I guess this is just the evaluation map thank you

[white oxide]

is this map from M to aM? if that's the case then obviously it's image is contained in (v1,...,vn) since aM is a subset of (v1,...,vn). and since aM is isomorphic to M, we get that M is free since a submodule of a free module is free? just wanted to make sure I understood this correctly

[elder wave]

It's a map M->M but the image is exactly aM

[white oxide]

ye the image is aM which is contained in (v1, ..., vn) which is obviously a free module

[elder wave]

yes

7.1 is that submodules of free modules over a PID are free i suppose?

[white oxide]

yup

[white oxide]

here, by isomorphic to the product of cyclic modules they mean $E \simeq E_1 \times E_2 \times \dots \times E_s$ where $E_i \simeq R/(p^{r_i})$?

[cloud walrus]

okeyokay

[agile burrow]

Yes

[white oxide]

ok thansk

[white oxide]

how does this argument show that the sum is direct?

or how is it equivalent to stating that their intersection is trivial for that matter

[coral spindle]

That's the definition of a direct sum

[white oxide]

oh

[coral spindle]

A module is the direct sum of two submodules if (1) they sum to the whole module and (2) their intersection is trivial.

[white oxide]

ye sorry i meant like how is that argument equivalent to the second condition you just posted

[delicate orchid]

This looks more like the whole “can be uniquely expressed as a sum” definition

Which is equivalent

[coral spindle]

Right I misunderstood the question yes

^ what wew says

[white oxide]

right i could prove it but i'm too lazy so i just asked the server lmfao

like ig that's my question

why are they equivalent

[steel light]

may I interrupt with a question

[white oxide]

sure go for it

[steel light]

Does the proof that if x is nilpotent then 1+x is a unit involve the fact that x^n is a zero divisor?

[coral spindle]

Not really

[solar vessel]

x being nilpotent exactly means that that thing is eventually 0

[steel light]

Unluck

[coral spindle]

It does in the sense that x^n is a zero divisor because x is nilpotent

but in no place do you need to say "and since x^n is a zero divisor there exists..."

[solar vessel]

and ofc it depends on the fact x is nilpotent because otherwise that wouldn't be there

[steel light]

Well since x^n itself is 0 it’s true for every element of the ring no?

Or well

[delicate orchid]

If X = Y o plus Z then we can write each element as y+z - if the intersection of Y and Z is trivial then we can’t have y’+z’ = y+z as this would imply that y’-y = z’-z is in both Y and Z

[steel light]

Ah ic what you’re saying

[delicate orchid]

I forgot how to do it lol sorry for the delay

[steel light]

Oh wait I’m goofy

mfw when I just showed 1 + x^n = 1

[night onyx]

Not really, for a hint look at the proof of why a + ax + ax^2 + ..... converges to a / (1 - x) when |x| < 1. They're surprisingly related

[steel light]

https://tenor.com/view/think-about-it-use-your-brain-use-the-brain-think-brain-gif-7914082

[steel light]

What the

[delicate orchid]

This is why I don’t like R[[x]] it’s SILLY

[white oxide]

ah and y' - y and z' - z are nonzero

[delicate orchid]

Yur

[white oxide]

ok cool thanks

[night onyx]

lol like just the normal geometric series from calculus, I know it seems crazy but trust, the proofs are similar

[steel light]

I don’t want to look

[night onyx]

lol okay well for another hint, consider (1 - x)(1 + x + x^2 + .... + x^n)

if n is big

[solar vessel]

don't want to look at what

[steel light]

hints 😭

[delicate orchid]

I simply cannot bring myself to peep

[solar vessel]

oh have you not done

[steel light]

It’s too late so I’ll try it out and see what I get

But it feels like if I can’t come up with it myself it’s kinda pointless to do the exercise, unless it’s some super arcane trick or whatever

[solar vessel]

then keep trying

[steel light]

You lose the magic of discovery and the learning that comes with the struggle

Yeah

[solar vessel]

I agree you need to feel the pain more

[delicate orchid]

false.

[steel light]

No shade to you kryojyn

I appreciate you trying to help

[night onyx]

yeah no worries, no offense taken!

[steel light]

hehe..

[solar vessel]

stop giggling and go solve the problem

[steel light]

I’m making tea

I’m sure you understand if I put my teatime first

brit.

[south patrol]

tea time to me means like 3rd meal of the day lol

[steel light]

I usually have it in the morning and evenings but I pulled an all nighter so I woke up in the afternoon

[white oxide]

i'm confused, why don't 1 and 2 just immediately follow from the hypothesis? bc f is injective it the sequence is exact so that there exists a left inverse right, and g is surjective so that there exists a right inverse

like how does 1 directly imply 2 if the hypothesis that the sequence is exact already implies those two

[agile burrow]

The sequence being exact doesn't imply those two though. Just because f has a set-theoretic left-inverse doesn't mean the left-inverse is a homomorphism

[white oxide]

ah yea good point

sigh

this will involve commutative diagrams huh

[hushed hamlet]

not really

[formal ermine]

hi pseudo

[hushed hamlet]

the usual approach is constructing maps

when i proved that thing once, i did it via finding a lot of isomorphisms, because i couldnt figure out the map ¯_(ツ)_/¯

[hushed hamlet]

hello

[white oxide]

ok, i'll attempt it thanks

[formal ermine]

welcome to the server

[hushed hamlet]

thx

[formal ermine]

are you a phd student?

[hushed hamlet]

not yet

but im looking for positions at the moment

[formal ermine]

what area

[hushed hamlet]

i found a potential advisor, but i need to get through the university screening to get a scholarship

[hushed hamlet]

supergeometry generally

close cousin of complex geometry

so its a field where you can do use differential geometry and algebraic geometry

[formal ermine]

oh geometry

[hushed hamlet]

where as algebraic geometry is my strong suit

[formal ermine]

I'm terrible at geometry

[hushed hamlet]

well i like to say, that these things are not really geometry

i never once used geometric intuition to solve a problem in algebraic geometry

[formal ermine]

opens wikipedia page
sees "sheaf"
it's geometry

[formal ermine]

Real

[hushed hamlet]

more so differential geometry

[hushed hamlet]

true

[hushed hamlet]

why did you ask btw

[formal ermine]

because of your message in adv lounge

[hushed hamlet]

ah right

but yeah, looking for a phd position is kind of rough

especially since i dont want to go to america for a phd where its relatively straightforward

[topaz solar]

Z/2 moment

[hushed hamlet]

exactly

[topaz solar]

Ah so that variant of super

[hushed hamlet]

is there another variant of super?

[topaz solar]

Big

Why not US

[hushed hamlet]

im working on formalizing string theory, so that stuff popped up

so glad i took algebraic geometry courses

[hushed hamlet]

because its a 5 year program, with non dodgeable coursework

and also qualification exams

i might add, that i usually do horrible on exams

[topaz solar]

Well idk about the 5 year thing always being that long but

[hushed hamlet]

most phd programs are straight research everywhere but US and canada probably

[topaz solar]

Yeah ok fair

[hushed hamlet]

like the first two years of US phd is basically a masters coursework

and i dont want to do that agian

[topaz solar]

I mean US PhD starting with a masters is kinda literal tbh

[crystal turtle]

In some places, if you can pass quals from the beginning wouldn't you get out of the coursework?

Idk entirely how it works

[hushed hamlet]

a friend of mine informed himself about taht stuff, according to him you can dodge at most 1 year of coursework

[hushed hamlet]

and then again im usually so bad at taking exam

[topaz solar]

I think you can do something similar some places, or if you already have a master’s-ish

[summer path]

I think this is true of most places in the US

[hushed hamlet]

not taking the chance of potentially wasting 2 years

[formal ermine]

wtf

what am I reading

[topaz solar]

I wouldn’t recommend coming here in particular, but you can definitely just skip things here

As in, my school

[hushed hamlet]

well im hopefully going to australia

but if that doesnt work out i prolly stay in germany

[steel light]

is (-1)^n = 1 for n even and -1 for n odd in an arbitrary communital ring? I want to say yes but I’m having trouble proving it :pepelaugh: and I can’t find anything on the web

[hushed hamlet]

and just try to get some money by teaching

luckily i have existing parents that support me

[formal ermine]

yes?

[topaz solar]

-1 * -1 = ?

[crystal turtle]

It suffices by induction to just prove that (-1)^2 = 1

[steel light]

I don’t know

[crystal turtle]

If that helps

[steel light]

I think I did this

Okay

That does help

I will go by induction

[crystal turtle]

Then you're done lol, simple induction

[topaz solar]

Hope this helps 👍

[coral shale]

2?

[steel light]

don’t monke me u fkn losers I’ve been forced to abandon all intuition bc of this damn subject

[formal ermine]

it doesn't help, feather is looking for a simple and intuitive answer

[crystal turtle]

Feather it's -1 * -1

[topaz solar]

I basically suggested the same thing as Ryx

Except he said induction

[formal ermine]

but is it simple and intuitive?

[topaz solar]

Yes?

[steel light]

illumi you’re going to get murdered in your sleep

watch out german

[coral shale]

i wouldnt wait for him to sleep to act...

[formal ermine]

shuri look behind you

[steel light]

I think I’d just have to give him a linear algebra problem

he’ll probably die from thinking too hard

[formal ermine]

bro

[steel light]

IM SORRY

[formal ermine]

you didn't know what -1 * - 1 was

[steel light]

[coral shale]

bro u didnt know Z dense whatevertheshit

[formal ermine]

THAT WAS HALF A YEAR AGO BRUH

[topaz solar]

“Z is complete because it’s dense-in-itself”

[formal ermine]

I started learning math a year ago

[steel light]

new flex for my little sister taking algebra 1

[coral shale]

u wont hear the end of it in a decade

[topaz solar]

I unironically don’t know where dense in itself came from

[crystal turtle]

Real (me too)

[formal ermine]

stfu ryx

[topaz solar]

Bro skipped right to p-adics

[formal ermine]

my anniversary is coming up soon

[crystal turtle]

Perhaps he was thinking of closed? ==> complete

[coral shale]

dont mess up the date then

[formal ermine]

I explained

I asked in discussion what dense linear order is

like what that property is called

(I now know it)

[topaz solar]

Density

[formal ermine]

someone said "dense-in-itself"

[topaz solar]

In the name

[formal ermine]

so I thought "idk any topo so I'll just believe them"

[topaz solar]

Z is very much so not a DLO

discrete linear order

[formal ermine]

Yeah

so every cauchy sequence is eventually constant

that was my thought process

here

this is when I started learning math

[crystal turtle]

Discrete topology moment

[topaz solar]

Yeah and still having troubles with power series 😔

[formal ermine]

real

[crystal turtle]

You should learn topology

[formal ermine]

I should

school gives me no time

[topaz solar]

Should learn real analysis

[formal ermine]

6 am to to 4 pm every day

is a pain in the ass

I either get no sleep or no time to do anything

[formal ermine]

that's what I'm planning on doing once I graduate high school

I will take real analysis 1, 2, and 3

and linear algebra 1 and 2

at uni

[crystal turtle]

Swag

[coral shale]

smh. Or let epsilon = 0.5 in the defn of cauchy

[solar vessel]

bro is lost

[topaz solar]

Actually learning determinants

From Axler

[formal ermine]

I just find it really annoying that people keep on bullying me for the same stupid things I said

like trash talking me for it once is ok

but not every single day

[steel light]

yeah

I’m sorry

[coral shale]

ok maybe i should stop counterclowning

[summer path]

Iwwumi

[crystal turtle]

I will stop clowning iwwum okay

[steel light]

I think the fact that it’s math makes it easy to do that

[solar vessel]

what did you guys do

[coral shale]

-1 * -1 = 1 tho feather ok

[steel light]

bc once someone learns something it sticks with them forever and it’s “easy”

[topaz solar]

I’ll still find it odd to jump right to p-adic analysis

[coral shale]

now prove it

[steel light]

uh

[coral shale]

nuh uh uh

no uh-ing

only pwoving

[steel light]

-1 + 1 = 0
(-1)(-1) + (-1)(1) = 0
(-1)^1 = 1

[crystal turtle]

People would be clowned less if they didn't skip straight past prerequisites

[formal ermine]

it's also hilarious that ultra went back 20k messages and half a year worth of messages just to trash talk me

[steel light]

I don’t

[topaz solar]

Bro had a vendetta

[formal ermine]

because I want to take algebraic number theory 2

[steel light]

yeah

[formal ermine]

took algebraic number theory 1 this semester

the proof is my favorite one

[crystal turtle]

Because then they would have better intuition for stuff

[steel light]

people want to study the stuff that’s interesting

[formal ermine]

and he said I should take the seminar on p-adics that he hosts

because it will help (and is a prereq) for alg nt 2

[topaz solar]

I think you’ll run into issues but

[steel light]

obviously better to build the prereqs properly

[topaz solar]

Not my problem

[steel light]

but who cares it’s their life

[coral shale]

im genuinely struggling to follow this

[steel light]

kekw

[coral shale]

stop making me verify shit in my head

all of u

[steel light]

now you see why I was doubting myself? 😂

[formal ermine]

it's also more fun to skip to harder stuff, tho it's probably not fun for the people that try to help me when I ask on here

[coral shale]

apply one axiom per line

[steel light]

-1 * -1 = ??? fuck if I know

[steel light]

ya I feel this hard

Oh I

Typo’d

[summer path]

It's probably fine

[steel light]

At the end

[topaz solar]

feather

[steel light]

It should say (-1)^2

lmao

[steel light]

Yes

[topaz solar]

(-1)*a = ?

[steel light]

-a

[topaz solar]

Yeah

[crystal turtle]

I get wanting to go straight to the advanced stuff

[white oxide]

can somebody please give me a hint for 1 --> 2, idk where to start

[topaz solar]

-(-1)

[steel light]

= 1

oh

[topaz solar]

Yeah that’s it

[steel light]

lmfao

[topaz solar]

I’m clowning you for this one

[steel light]

deserved 😭

[summer path]

But it's more of a you say you're studying p-adic analysis, but then you ask some very very fundamental questions that make it seem like you don't understand the basics necessary to study what you say you're studying

[coral shale]

-1 + 1 = 0
-1 . (-1 + 1) = -1 . 0
-1 . -1 + -1 . 1 = 0
-1 . -1 + -1 = 0
-1 . -1 = 1

[crystal turtle]

This is the problem every time people skip basics

[coral shale]

ok happy now

[topaz solar]

The power series thing really felt like this

[steel light]

LMFAO shuri yes

[crystal turtle]

Like I get it you can do it

[formal ermine]

I wouldn't have studied p adics if my prof wouldn't have recommended me to do so

[crystal turtle]

But you're missing baby fundamentals

[formal ermine]

I just wanted to take alg nt 2

[crystal turtle]

P-adics are cool tho I get it

[steel light]

yeah don’t let anyone discourage you illumi :)

[crystal turtle]

Like I was thrown into some more advanced stuff by my prof

[formal ermine]

which ones

[crystal turtle]

But we also started with covering the backgrounds

[void cosmos]

this should be the splitting lemma

u prove it by the five-lemma

[white oxide]

yea i'm tryna prove it without

cuz like lang doesn't have the splitting lemma

unless like

our prof wants us to prove/find the splitting lemma

[crystal turtle]

Not pointed at you specifically, but being comfortable with basic real analysis and metric space analysis/topology is very helpful for understanding p-adics

[white oxide]

💀

[formal ermine]

eg?

[crystal turtle]

Again, in part for building intuition, but also since terms form them will often just be thrown around since they are considered "basics",

[agile burrow]

Let $m \in M$. Show that $m - \varphi \circ g(m)$ is in $\ker g = \mathrm{im} f$. That might hint at how to define the map $\psi$

[cloud walrus]

walter

[agile burrow]

oops i forgot to reply to okeyokay

[white oxide]

huh okay thanks, i'll solve the other parts of the question first then i suppose

[void cosmos]

yea

[crystal turtle]

Idk, like compactness, connectedness, completeness, ... Stuff like that shows up a lot of.

[void cosmos]

btw this is like the goat of theorems okeyokay

the fact that u have these maps imply that ,

M is a direct sum of M' and M''

[summer path]

I think what you mean is like typical first course in analysis material

[hushed hamlet]

what does algebraic number theory 2 cover

[summer path]

And topology I guess

[formal ermine]

global-local

neukirch chap 2

[agile burrow]

That seems like cool stuff, idk anything about local fields

[white oxide]

yea i remember reading about this in hungy

[void cosmos]

yeah

what walter said is the proof in hung

[crystal turtle]

Something like that

[white oxide]

lmao so this lang is exercise is prove this theorem essentially

F

[void cosmos]

yeah

its not like terribly hard ig

or idk

i never tried it on my own

just say by the splitting lemma this is the diect sum and then use projections to find ur split

and run

u only get 10% of the problem pooints so worth

[formal ermine]

idek man

math is the only fun thing I have

[white oxide]

idk if i can use the splitting lemma tho so maybe i have to prove that too

here's the actual exercise

[void cosmos]

yea i was joking

use walters hint

[white oxide]

bro's doing crazy math in hs

i barely got a 3 on the ap calc test senior y ear

[void cosmos]

if the proofs already been done in 3.2 but ur just verifying details then yeah should be ez

[hushed hamlet]

maybe you are more comfortable with this statement

[summer path]

I thought neukrich didn't have that high prereqs, like just a year of group/rings/modules/galois is enough

[white oxide]

nah i've never seen short exact sequences of groups lmfao

[hushed hamlet]

well modules are also an abelian category

basically the same thing

trust me

[white oxide]

nah he showed M = Imf dsum Kerpsi

true

all abelian groups are Z modules!

[topaz solar]

Modules & Galois stuff is the trick there

[summer path]

Maybe

But I think that's still a second semester kind of thing

[hushed hamlet]

given that statement the map you should construct should be relatively clear tho

[white oxide]

i give up i can't think

i'll come back

[coral shale]

finitely generated only right

[delicate orchid]

:breakfast2:

[topaz solar]

Second semester algebra here starts at rings

[delicate orchid]

no, all abelian groups

[topaz solar]

[coral shale]

what have i managed to confuse... lets see

[delicate orchid]

here's a hint nerd... all modules are abelian groups... and all rings are Z...

[coral shale]

oh nvm

[topaz solar]

No Sylow, no iso theorems beyond the first iirc, etc

[hushed hamlet]

here first course algebra is groups to galois theory

second course algebra is atiyah macdonald

until like integral closure

[topaz solar]

Yeah it’s not very intensive here

[summer path]

All rings are Z

The uni I'm at doesn't even require algebra to graduate with a degree in math

[topaz solar]

Real

[delicate orchid]

what the fuck

anyway yeah obviously they're not Z but Z is inital in Ring

[topaz solar]

Mfs from that uni gonna be as funded as Wew

[delicate orchid]

so they basically are

[summer path]

You can literally take computational calc, some analysis then some random stats and cs classes and be done

I don't know lmfao

[delicate orchid]

sad

algebra as a whole seems very under-represented in undergraduate courses

but it could just be that all of pure is

[summer path]

I think my uni is only cares for engineering or something

[topaz solar]

Same

Bro secretly goes to my uni

[summer path]

,ti 218876477146398720

[cloud walrus]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[summer path]

Is sharp in eu

[topaz solar]

[crystal turtle]

It should only be algebra anyways /s

[hushed hamlet]

only mandatory algebra courses here are linear algebra 1 and 2

but thats not really algebra

[summer path]

Just make the entire math major a survey of algebra

[elfin summit]

I've been given an equation to solve over the quaternions. Since I don't know much about it, I moved the problem to a matrix equation using a very well known ring isomorphism and got 2 non trivial solutions. My issue is that I'm a bit confused about how to get the corresponding quaternions to those solutions.

that's the isomorphism I'm using

[delicate orchid]

go back through the isomorphism? that's kind of the whole point of an isomorphism

[elfin summit]

yeah, but there's an issue I didn't expect

every quaternion is expressed a+bi+cj+dk, where a,b,c,d are real numbers

when I find the linear combination of the solution with respect to the matrices I J K and the identity I get complex coefficients

what's going on? I don't get it...

[delicate orchid]

yes this is an isomorphism with a C-algebra

if you want it as an R-algebra you will need a 4x4 matrix

[elfin summit]

hmm, good point

well, I already spend the whole afternoon doing calculations by hand 😆

better to continue with this approach

[reef trench]

Is ths question even possible?

I was actually very stuck w this, so i looked this question up, and the only answers I see are for the finite abelian groups

[crystal turtle]

It must mean abelian

[reef trench]

The proof for the abelian case is rather straightforward

[crystal turtle]

The order of the product isn't specified

so that would not really be a well-defined product if not abelian

[reef trench]

Oh hell nah

How was i supposed to know that :(

[delicate orchid]

you weren't

[reef trench]

This has to be a error on the textbook

[delicate orchid]

I agree that's some bullllshitttt

[crystal turtle]

perhaps it's true and the order doesn't matter? But I'd be a bit surprised if that's true

[delicate orchid]

what if it's not dependant on the order of the

yeah

let us think

[crystal turtle]

done thinking

I don't think so

[delicate orchid]

C_2 x any 2'-group LMFAOOO

[topaz solar]

The council meditating on the correctness of the book

[reef trench]

For anyone curious, the book is algebra chapter 0 by paulo aluffi

I chose it bcs it introduces abstract algebra from a category theoretic view

[agile burrow]

https://www.math.fsu.edu/~aluffi/algebraerrata.2009/Errata.html

It's a nice book, but you should be cautious. There's a link to errata for the book

[reef trench]

Thank you!!

[void cosmos]

Why is a finite simple ring isomorphic to a matrix ring over a field?

[elder wave]

Not sure what you're asking

for a proof of artin wedderburn?

[void cosmos]

i only can prove some basic stuff on semisimple rings

Yeah thats whats missing from what i need to know

ig

[elder wave]

pretty standard you can find this by just looking up the name

even wikipedia sketches a proof iirc

[void cosmos]

wouldnt this be a coroally tho?

Cuz the ring is finite?

[elder wave]

oh FINITE simple ring

i read semisimple

my bad

[void cosmos]

Np

[verbal tulip]

I was reading a book on group theory and came across a spectacular theorem that didn't seem to get much attention in the book. It says that for every vector space whose mappings form a representation of a group, every member of the vector space can be written as a sum of functions which form partners of the different irreducible representations of the group. This seems to be an amazing way to find basis of arbitrary (finite or infinite) vector spaces. Is this a famous theorem with its own name?

[wooden ember]

Unless I’m misunderstanding what you’re saying you’re basically describing a variant of Maschke’s theorem?

[delicate orchid]

yeah this is Maschke's, and the functions you're describing are characters

every complex vector space though

has to be complex

[verbal tulip]

Any good resources on it?

[summer path]

fulton & harris is standard text for rep theory i think

(which i also plan on starting to read soon hopefully)

[void cosmos]

anyone?

[coral shale]

can you define those 2 objects

finite ring - finite elements?

matrix ring over a field - matrices with entries of that field?

[void cosmos]

yes

yes

[coral shale]

well firstly Id hope finite simple rings have to have prime characteristic?

mmm then maybe u think about frobenius kinda things but ive only seen those kindof results for groups

[coral shale]

bezout

[white oxide]

bro saw it immediately

[coral shale]

(b, c) is gcd

right?

[white oxide]

god i suck at number theory

like i'm so bad

yeah

[coral shale]

or is that ideal

ok whatever

[crystal turtle]

it's ideal since it's (1) not 1

[topaz solar]

either way it’s essentially bezout

[crystal turtle]

but yeah w/e same thing

[white oxide]

oh

[coral shale]

is that why they do that silly notation for gcd

[topaz solar]

there’s something to get 1 as a sum

[coral shale]

(S) = (gcd S) in a good enuf ring

[crystal turtle]

[coral shale]

probably needs euclidean domain

mm no, i think bezout doesnt need that

bezout domain

[crystal turtle]

:o

[coral shale]

but theres no gcd unless ur in ed ig

[white oxide]

wait why is Eb \cup Ec = 0

[crystal turtle]

gcd domain

[white oxide]

E_b is the kernel of x --> xb btw

[crystal turtle]

\oplus

not \cup

[white oxide]

what

[coral shale]

cap?

n is cap

[white oxide]

no i was asking about where he said finally, E_b \cap E_c = 0

cap

[crystal turtle]

oh

[white oxide]

(as one sees immediately)

[crystal turtle]

oop

[white oxide]

can you see it immediately Ryx?

apparently not

sigh

[coral shale]

their gcd is 1

[crystal turtle]

(he also said E_b \oplus E_c so uhhh)

[white oxide]

tell me what that means, number theorist

nah jk

i know what that means

[coral shale]

ok no, I can guess the idea but im sure i dont see it

[white oxide]

yo i'll venmo somebody 50 bucks if they can tell me why E_b \cap E_c = 0

[crystal turtle]

i don't see it

[topaz solar]

Ok

What’s E_b

[white oxide]

this some number theory shit

number theory's for losers

[topaz solar]

whuh

[white oxide]

the kernel of x --> bx

bro i'm aboutta post this shit on r/explainlikeimfive

[coral shale]

can u gib example of finitely generated torsion module

[crystal turtle]

so if cv = bv = 0, what can you say about v?

[white oxide]

o

our right

[coral shale]

nvm i defo do not know any of these

[white oxide]

ur right

[crystal turtle]

perchance, it has something to do with v = xbv + ycv

[white oxide]

what i thought it was just x = 0

i give up

[crystal turtle]

take this

[topaz solar]

I’m not sullying that but I will your self sully

[coral shale]

[crystal turtle]

you know what cv and bv are by assumption of v in the intersection of the kernals

what does this tell you about v

[white oxide]

v = 0!

[crystal turtle]

🤯

[white oxide]

bro i'm too good at this shit

i'm like

half asleep

and i solved that question

[topaz solar]

This seems odd

[white oxide]

shittttt

[crystal turtle]

go sleep my man

[white oxide]

nah

i have like 4 problems left due on tuesday im fugged

[topaz solar]

I’d have to look further back in it to see how some of these things are actually defined

[white oxide]

Let m in R, m nonzero. We denote by E_m the kernel of the map x \mapsto mx.

[crystal turtle]

nah just wing it

that's what I always do

[topaz solar]

So the things annihilated by m?

[white oxide]

ye or having exponent m if you wanna get fancy

Why did the finitely generated torsion module get an award?

Because it had a finite amount of "torsion" for acceptance speeches!

[crystal turtle]

bruh that's what "exponent a" means

[topaz solar]

So

[white oxide]

nah

[topaz solar]

cxbv = xav

[white oxide]

W or L joke by chat gpt btw

[topaz solar]

Is this commutative or smth

[crystal turtle]

if we're taking gcds

probably a domain

[topaz solar]

I would hope so but I’ve known moamen’s book to be screwy

bc=a and E has exponent a

So if cxb = xbc = a, then that’ll hit you to 0 no?

[white oxide]

yo i've collected 11 s in the last 20 minutes

that's some good shit

[topaz solar]

At least half are your own get outta here

[rotund dragon]

you gotta beat the person in #book-recommendations

[crystal turtle]

think i know which one you mean but you gotta be specific

there's at least one in the past month that has like 15 sullies

[topaz solar]

If we had v in E_b and E_c, then bv = cv = 0

[white oxide]

ayyy man what can I say I'm my #1 supporter ya feel me 😭

[topaz solar]

(c-b)v=0

And for any multiple of c, b too

So $(xc-yb)v=0$ for all x, y

[cloud walrus]

Dragonslayer Sharp

[topaz solar]

That’s why it’s obvious

Namely, we can choose x, y to get 1

[rotund dragon]

good point

[topaz solar]

1v = 0

[rotund dragon]

pedestrian mathematicians

[topaz solar]

v = 0

@white oxide

[white oxide]

Nah yea I got it but tyy

[topaz solar]

[white oxide]

did you see it immediately?

[crystal turtle]

oh

[topaz solar]

Once I saw how the things were defined

[white oxide]

Damn bro sharp being seeing everything immediately

[topaz solar]

I still have no idea what E(p) means

[white oxide]

Bros on addy or some shit

[topaz solar]

Why would addy help

That’s a focus thing

[white oxide]

Lmao thats hilarious

[topaz solar]

Not a sight thing

[white oxide]

oh ya true

[steel light]

What are some cool polynomial identities that hold in commutative rings?

My motivation for this question was the binomial theorem

[snow condor]

don't see why k_2 + mz is in K in that case, anyone explain?

also sorry the typesetting in this book is really awful, if a symbol looks random just replace it with one that makes sense

[topaz solar]

if m is a multiple of p part?

For that one, look above and see pz=k in K

[snow condor]

yes just that sentence

[topaz solar]

so mz = npz = nk for some n

if p|m

therefore mz is in K and that's a subgroup

[snow condor]

ah of course

thx

[topaz solar]

divisible group moment

[snow condor]

[topaz solar]

divided

[snow condor]

divide my... wait wrong channel

[topaz solar]

[coral spindle]

Divide and conquer

[coral shale]

waffles

[rocky cloak]

I guess pretty much any polynomial identity that doesn't involve dividing should hold no matter the ring. So binomial theorem. Geometric series:

1 - x^n+1 = (1 - x)(1 + x + x^2 +...+ x^n)

Some less familiar ones, a polynomial is nilpotent if all coefficients are. A polynomial is invertible if the constant coefficient is invertible and all other coefficients are nilpotent.

[coral shale]

galois ones

[wet zodiac]

how should i go about disproving that if $R$ is a comm ring and ${I_\alpha}{\alpha\in A}$ are ideals such that $I\alpha+I_\beta=R$ for all $\alpha,\beta\in A$ we have that $R/(\bigcap_{\alpha\in A} I_\alpha$) is isomorphic to $\prod_{\alpha\in A} R/I_\alpha$

[cloud walrus]

most likely to get denied G+

[wet zodiac]

[formal ermine]

2Z + Z = Z but Z/2Z isn't isomorphic to Z/2Z x { 0 }

[wet zodiac]

yes it is

[chilly ocean]

[formal ermine]

it is?

[wet zodiac]

dude

phi(r)=rx0

[crystal turtle]

[formal ermine]

Wait fuck { 0 } has size 1

I thought it's size 2

lmaoo

[crystal turtle]

[wet zodiac]

ryi

illum

if that was true then CRT for n=2 would be false

[crystal turtle]

Timo is dying in the background

[wet zodiac]

ok back to this nasty problem

[coral shale]

this is what u get for writing oplus as cartesian x

[agile burrow]

Well let's take the basic example of R = Z. Then for any two distinct non-zero prime ideals (p) and (q), we have (p) + (q) = Z. But what's the intersection of all these non-zero prime ideals?

[crystal turtle]

(i think there's a simply counterexample in ||the ring of integers||)

[agile burrow]

good work team

[crystal turtle]

oh oop walter had the same idea

[rocky cloak]

Its true in the finite case, so a counterexmaple would have to be infinite

[wet zodiac]

(pq.....)Z

[agile burrow]

Is that ideal non-zero?

[wet zodiac]

i think its 0

[agile burrow]

I guess I should clarify that I'm looking at the family of ideals {(p) : p is prime}

Yeah, the intersection is zero

[wet zodiac]

so now we have that Z/0 is isomorphic to Z/2ZxZ/3Z...

but theres only 1 element in Z/0

[agile burrow]

One way to see that is to note that an integer n is in the intersection of all of these iff p divides n for all primes p. But every non-zero integer has finitely many prime divisors

[wet zodiac]

and this is obviously not surjective

[agile burrow]

Hmm Z/0 is isomorphic to Z

[wet zodiac]

qhar

[agile burrow]

If I quotient out by the zero ideal, I'm not making any identifications so I just get my original ring back

[wet zodiac]

oops

[agile burrow]

Z/Z would be the zero ring

[crystal turtle]

a cardinality argument still works tho

since the infinite product is uncountable

[wet zodiac]

ok yeah

nevermind this problem wasnt that bad at all

[agile burrow]

It's also the case that one side has zero-divisors and the other doesn't

[wet zodiac]

thanks guys

i thought it was gonna be some wacky eldritch horror problem

[agile burrow]

Happy to help

[crystal turtle]

oh true

[wet zodiac]

because my proof for finite ideals was

pretty nasty

[stone depot]

Is it necessary for a element of a group to be in a some subgroup?

[delicate orchid]

consider the subgroup generated by that element

[stone depot]

you mean cyclic?

[peak root]

or consider the additive group G with only 2 elements 0 and 1 with 1+1=0. The only subgroup is the trivial group {0} and G itself.

[delicate orchid]

there are groups without order 2 elements? not sure why you brought C_2 up

[delicate orchid]

if it's generated by a single element it's automatically cyclic

by definition

[stone depot]

You mean like for any element say 'a' belongs to finite group G so
a^order(G) = e (identity element)
am I right?

[delicate orchid]

no

if a is an element of G

then <a> = {1, a, ..., a^whatever} is a subgroup of G

the statement you made is correct but I don't see the direct relevance

[coral spindle]

I think Valentino was trying to point out, in a roundabout way, that G is a subgroup of G, and in some cases you can do no less than that

[delicate orchid]

oh I see

le cyclic p-group has arrived

[stone depot]

actually I want to prove that group of order 9 is abelian,
so subgroup of only 3 is possible which is cyclic and abelian too, so I am wondering about other 6 elements, I thought maybe those can form subgroups of 3 or is it something else? please if possible give hint only

[delicate orchid]

do you know anything about like, centers of groups

or the sylow theorems

[stone depot]

Yes, was proving sylow second only but then decided to do some problems

should i try by class equation

[delicate orchid]

could be a good idea

I'm not actually sure how to prove this without citing results about p-groups and G/Z(G) so this should be fun

actually I just need the result about G/Z(G)

[rocky cloak]

Similar to what Wew said, I think you'll need to involve some properties about group actions of p-groups. Specifically noting that G acts on itself by conjugation and the fixed points of the action is exactly the center of G.

[delicate orchid]

ah that sucks there's no nice way to do it specifically for p = 3

write out all 9x9 cayley tables

[rocky cloak]

Well, I mean there might be

We just have to be smarter

[delicate orchid]

could invoke group extenstions perhaps lol

but that's going the WRONG way

[coral shale]

Z/0 triggers me

[delicate orchid]

like it's pretty clear that there are no non-direct split extenstions of C_3 by C_3

[stone depot]

I don't think it involves that much its in herstein just after homomorphism section is finished

[delicate orchid]

assume non-abelian, and take two elements a,b, if either is order 9 then the whole thing is cyclic so both must be order 3

so we know that everything in this group is order 3

no that is absolutely not true

D_8

Q_8

[stone depot]

sorry quanterian was there

[delicate orchid]

upper triangular matrices with det 1 over F_3 ;)

gah if p = 2 this would be immediate

and there are exponent p groups which are non-abelian so we have to somehow use the fact it is both exp p and order p^2

this group has to be 2-generated (minimally), as it's order 9 and the generators must have order 3 each (and it can't be cyclic)

[rocky cloak]

Maybe there is a brute force solution you know that the weekend are
1, a, a^2, b, ab, a^2b, b^2, ab^2, a^2b^2

Then for each of these see ba equal to them and derive a contradiction

[delicate orchid]

I was trying to show that all commutators are trivial

yeah that's basically the same

[coral spindle]

It is true for prime square order groups that aren’t cyclic

[delicate orchid]

let a, b be the two generators, obviously we cannot have ab = b or ab = a or ab = a^2b or ab = ab^2 as nothing involved is equal to the identity
ab = b^2 implies that a = b, likewise ab = a^2, and ab = 1 implies a = b^2 which isn't possible

[delicate orchid]

do you think I wasn't completely aware

[rocky cloak]

Oops I may have accidentally assumed b and b^2 have different cosets here

[delicate orchid]

ok now for the hard ones - why can't it be a^2b^2

[coral spindle]

No but I was saying the argument would’ve worked, it seemed like you were saying otherwise

[white oxide]

please help

what is a R-canonical form

[delicate orchid]

my question is what's the structure theorem for modules over R[x]
is it literally just "these are the modules" lol

[white oxide]

i have no clue about that either

[delicate orchid]

oh wait it's the REAL numbers so it's just f.g modules over a pid

got it

[white oxide]

yo could I get a hint for showing that if there exists a submodule $N$ of $M$ such that $M = \text{Im}f \oplus N$, then there exists $\varphi: M'' \to M$ such that $g \circ \varphi = 1_{M''}$, I've already shown the existence of $\psi: M \to M'$

[cloud walrus]

okeyokay

[white oxide]

so I know that $g \circ f = 0$, so I suppose this question reduces down to showing that there exists a bijection between $M''$ and $N$?

[cloud walrus]

okeyokay

[hushed hamlet]

Do you know why they write 0 on the left and on the right?

[white oxide]

yea, f is injective and g is surjective right

[hushed hamlet]

Also which direction are you proving, since in the case of assuming it splits

You get way more than just the direct sum

[white oxide]

i'm trying to show that if there exists a submodule N such that blah blah blah, then the sequence splits. i've already shown the existence of psi such that psi(f) = 0

[hushed hamlet]

Refer to this me thinks

I would use the other equivalent notions of being split exact

That you recently proved

[white oxide]

wdym

i didn't prove any other conditions

[hushed hamlet]

Wait is there not a third part, that says this splits

[white oxide]

yea there is a third part, i didn't include everything

and i gave up on trying to prove the equivalence of the conditions lol

[hushed hamlet]

This exercise is basically the same thing, but its more self contained, so maybe you can try to solve that

[white oxide]

hm ok i'll try this then

thanks

yea

this is really fucking helpful lol

[hushed hamlet]

Lmao

Did you manage to do the exercise?

[white oxide]

yea

[hushed hamlet]

[white oxide]

So $\phi: M \to M' \oplus M''$ given by $m \mapsto (\psi(m), g(m))$ should be an isomorphism

[cloud walrus]

okeyokay

[white oxide]

also new sharp pfp

nvm

wait yea

[hushed hamlet]

Ya

[white oxide]

wait hold up how does injectivity follow

cuz psi and g are surjective but not necessarily injective

[hushed hamlet]

At some point you need to actually use exactness

Im semi sure that to get surjextive or injective you need to use that

[white oxide]

hm ok

[white oxide]

is it the case that exactness goes the other way? as in, Im$\varphi$ = Ker$\psi$

[cloud walrus]

okeyokay

[spare totem]

Can we just say this is true always since in case these x_i*y_j sum could not be all broken down into some ideal representative and the other constant parts, we could just argue that in such case we always have identity element to represent the ideal through?

oh wait

no

I need to think a bit more

[elder wave]

i'm not sure what you're saying here

[spare totem]

Actually let me think a bit more really

I will rephrase if needed then

[delicate orchid]

just show it's closed under addition and multiplication in R, I have no idea what you're talking about with the representative stuff

[spare totem]

yeah ignore it

It was me messing up assuming this forms some sort of equivalence classes

but I recalled

[white oxide]

if you still remember how to can you give me a hint on how to use exactness to show injectivity 💀

[spare totem]

commutative rings don't need that to be true

[white oxide]

bc like $\phi(m) = \phi(n) \implies \psi(m) = \psi(n)$, but i can't really express $m$ and $n$ as outputs of $f$ if that makes sense, nor take inverses and what not, not rlly sure how to use exactness to show m = n

[cloud walrus]

okeyokay

[spare totem]

do you think it's really an ideal? If I assume J1 and J2 are both ideals of some commutative ring R, I can't really seem to be sure of fact that the J1 J2 will be a subgroup of R under addition

[delicate orchid]

it's definitely an ideal

remember that J_1J_2 isn't the product of all elements in J_1, J_2 - it's all sums of products of that form

[spare totem]

but if I take two elements from it, and add them, it's not closed imo, let's say I have x1y1 + x2y2 and x1y1, and add them to get x1y1 + x1y1 + x2y2

this is not in that set

[night onyx]

yeah it is, it's a sum of products of things in J1 and J2 so by definition its in the set

[spare totem]

what value of k in that expressoin will lead to x1y1 + x1y1 + x2y2?

[delicate orchid]

3

[spare totem]

but that would be x1y1 + x2y2 + x3y3

[delicate orchid]

there are three terms

ok? and set x_1 = x_2 = a_1 and x_3 = a_2 likewise for b

[night onyx]

it doesn't say that x_1 and x_3 can't be equal

[delicate orchid]

then a_1b_1+a_1b_1+a_2b_2 is in that set

I'd suggest not overloading notation

[spare totem]

I think it should not be equal

[delicate orchid]

cause it can lead to things like this

[spare totem]

otherwise it does not make sense to me at all

[delicate orchid]

you're just taking all finite linear combinations of ab with a in J_1 b in J_2

[spare totem]

I am really weirded out

[delicate orchid]

you're really overthinking it

[night onyx]

I mean just think of span{v_1, v_2} in a vector space. It's all linear combinations of v_1 and v_2, including v_1 + v_1 or v_2 + v_2 + v_2, it's just all possible linear combinations

[delicate orchid]

it's exactly like that

[spare totem]

then why it didn't write it as sum with index i and then sum with index j instead?

[delicate orchid]

what

because that's not what it is

[night onyx]

all it says is that its a_1 + ... + a_k where each a_i is x_iy_i for some x_i in J1 and y_1 in J2

that's all the definition says, so if you add two such sums, you just have a bigger sum

[delicate orchid]

It’s just the abelian group generated by the set {xy : x in J_1, y in J_2}

I don’t know how to make this simpler sorry

[spare totem]

both x and y use same index there

so I am not really able to make sense of it

no matter how hard I try to think of it

[delicate orchid]

What’s the problem with them both using the same index

You have some sequence of products x_iy_i

[spare totem]

yeap

[delicate orchid]

And then you add them up

[spare totem]

and also every x_i is unique

is all I can think of there

[delicate orchid]

Well no because 2xy is also in the ideal

As you’ve shown already I presume

[spare totem]

a part I missed from it

[summer path]

This is the same image as before

[night onyx]

don't think this lol, they specifically don't say unique

[spare totem]

He said 2xy should also be in it

by assuming it's already an ideal

[night onyx]

uniqueness of each x_i isn't part of it

[spare totem]

so I think he missed the part that it's True/False problem

[spare totem]

Why should I not?

I am really puzzled

[night onyx]

because you're adding things to the question that aren't in it and getting stuck on it

[spare totem]

isn't it much saner to assume the general case that they're just some elements

and prove general case

assuming uniqueness seems weirder

[summer path]

Why would adding assumptions be more "general"

[night onyx]

imo the saner thing is to prove what the question asks

[spare totem]

It asks true/false question

is it an ideal?

[spare totem]

adding assumption is not general

that's what I am saying

[delicate orchid]

then don't add an assumption that the x_is have to be unique lol

have you never seen a constant sequence? x_i = 423 kinda deal

[spare totem]

yeap

but we still prove stuff assuming every x_i is just some element

[delicate orchid]

some element in J_1 yeah

[spare totem]

not like "oh yeah for every x_i there is j so x_i = x_j"

I really can't...

ugh

this is weird af

[summer path]

Tbh I still can't really see what you're getting stuck on

[spare totem]

maybe I am just missing some common knowledge

but I can't really point to what either

[summer path]

Do you know what the definition of an ideal is?

[spare totem]

subset (say I) of elements of commutative ring (say R) such that it forms subgroup under addition and for every a in R if i is in I, ai is also in I?

[delicate orchid]

of course you can

[spare totem]

what happens if J1 and J2 are not finite?

[delicate orchid]

is there a mean looking guy with really big bushy eyebrows glaring at you from across the room with a sign that says "x_i \neq x_j" or something

[delicate orchid]

again I ask, have you never seen a constant sequence in like, the real numbers

[spare totem]

I have but..

[night onyx]

https://tenor.com/view/wink-johnny-rose-johnny-eugene-levy-schitts-creek-gif-20808979

[delicate orchid]

there's just no issue whatsoever here

[summer path]

You have J1J2 defined by finite sums...

[spare totem]

this goes against all the Math I have done so far

[summer path]

What

[spare totem]

tho I agree that I am a dumbo CS student

well thanks anyways

I will just think a bit more maybe and ask later

It's 1 AM too for me

[delicate orchid]

it really really doesn't I'm sorry but this just isn't correct

take the sequence a_1 = 0, a_2 = 1, a_3 = 1

and another sequence b_1 = b_2 = b_3 = 1

[spare totem]

I never denied existence of these sequences

[delicate orchid]

then the corrisponding sum would be 0*1+1*1+1*1 = 2

[spare totem]

Come on

[delicate orchid]

like

you certainly seem to be!

brb playing civ 6

[spare totem]

I am saying assuming these to hold true for that set is absolutely wrong way to do it

[topaz solar]

You’re denying it because you seem confused about 2xy = xy + xy being in it

[spare totem]

or I believe so so far

but I could be wrong

[topaz solar]

No

[spare totem]

yes

[topaz solar]

It’s for every finite sequence yes

[spare totem]

please elaborate

[topaz solar]

That is a finite sequence

That’s it

There is nothing deeper

[summer path]

Your definition of J1J2 is over a finite sum of j1j2

[spare totem]

yeap

[summer path]

So if you believe xy is in it, then xy+xy is also in it

[spare totem]

how come?

J1 J2 is not proved to be ideal yet

we have to prove or disprove

[topaz solar]

bro it’s a finite sum of j1j2

[spare totem]

yeap

[topaz solar]

Yes that’s it

[spare totem]

wait

Wait

[topaz solar]

[spare totem]

okay maybeI am getting it

I think I am attaching too much importance to indices

nothing more

[summer path]

You are trying to show that the set defined as finite sums of j1j2 is an ideal

If you don't want to call it anything yet that's fine

[spare totem]

I think my confusion was caused by indices

[topaz solar]

Yes

[spare totem]

but I kinda get it now

yeah

fk

I am sorry

Thanks again everyone

Now I get it

problem was this

I thought x1, x2, etc to not be just any element of X, but be specific elements

this caused whole fiasco

[hollow mica]

Now you get it 👍

[spare totem]

I think it was a huge mistake on my part on misreading the set notation

but oh well

[summer path]

It's okay, you figured it out in the end

[coral shale]

til the smallest finite group in which an element of the commutator subgroup is not a commutator, has order 96

[white oxide]

can somebody tell me if I'm on the right track for $1 \implies 2$? so I know that every $x \in \text{Im} f$ can be written as $m - \varphi(g(m))$ since $x \in \text{Ker} g = \text{Im} f$. therefore I want $\psi$ to somehow transform $m - \varphi(g(m))$ back into $m'$, where $f(m') = m - \varphi(g(m))$ if that makes any sense. at first I thought about applying $g$ then adding $m'$ but obviously that's not well-defined

[cloud walrus]

okeyokay

[void cosmos]

u are

[white oxide]

ok cool

yeah unfortunately i can't use isomorphisms or the splitting lemma or anything right

bc that uses the fact that the two conditions hold

[void cosmos]

yea i would guess so

[white oxide]

or that the sequence splits

yea

[void cosmos]

its the point of the exercise

unfortunantely

splitting lemma is the goat

[white oxide]

oh wait i can take the left inverse of f

i think

wait no

i'm trippin

[white oxide]

oh wait no

i can use the isomorphism

namely M is isomorphic to the direct sum of M' and M''

since the proof of that only uses one of the conditions

nvm

nvm

[hushed hamlet]

the splitting lemma also holds in every abelian category

which si nice

[white oxide]

oh wait i think the easiest thing is to define an isomorphism from M' direct sum M'' to M using the homomorphism phi/the sum

then i can do some shit with projections and inclusions and i think that should work

please please please

[hushed hamlet]

yes

usually you just see the right homomorphism define it and check its an iso

[white oxide]

oh god i should've known holy

[hushed hamlet]

but as youve probably seen in my solution, i just didnt see it

[white oxide]

i was trying to define something from M straight to M'

no offense but like

[hushed hamlet]

so i did it another way, but in the end after all those identifications i finally saw it

[white oxide]

when i looked at ur solution and saw first iso i was like damn that's a lot of effort

[white oxide]

ye