groups-rings-fields

[bleak crystal]

a•a =0 means either a is zero or zero div

obviously

i see that

anyway b

how do you expand that

[ebon osprey]

(a+b)^n = a^n + a^{n-1} b a +...

Stuff like this I guess

[bleak crystal]

(a+b)^n+m

[old hollow]

Have you done (a) @bleak crystal

[ebon osprey]

Whatever

[bleak crystal]

he mentions it in the hints

idk why

[ebon osprey]

Be it n or n+m

[bleak crystal]

yes

[old hollow]

Oh

Ok cool

[ebon osprey]

It's because then you expand and in each term you'll get like a^n or b^m

[bleak crystal]

how do you expand that

[ebon osprey]

Something like that

Well

You write (a+b)(a+b)...(a+b)

And you suffer

[bleak crystal]

i don't understand

[ebon osprey]

Wait

[old hollow]

Here

I think he’s saying

Suppose n is the integer where a^n=0

And m is the integer where b^m=0

[ebon osprey]

Yes

It's the nilpotence index

[old hollow]

If you expand (a+b)^(n+m) you will get a^n and b^m terms in there

[bleak crystal]

why does that matter

[old hollow]

Try it

Wdym

[bleak crystal]

i don't know how to expand it

[old hollow]

Binomial theorem?

Remember

[bleak crystal]

oh

[ebon osprey]

Bruh

[bleak crystal]

what is n+m choose 1

fuck

[ebon osprey]

Dont care about that

[bleak crystal]

I'm so fucking tired

what do i do

how do i expand this

i don't know how to do this

[ebon osprey]

Just write it's equal to $\sum {n+m \choose k} a^kb^{n+m-k}$

[cloud walrus]

Jean-Jérôme

[old hollow]

^

Does that make sense

That’s literally the formula for binomial theorem

Plugged in with (n + m) for the exponent

[bleak crystal]

wouldn't there be middle terms...?

so it's not nilpotent

[ebon osprey]

And think about "is k bigger than n? Is n+m-k bigger than m?"

[bleak crystal]

what

i don't follow at all

god i fucking hate this class, i still don't understand why I'm taking it

[ebon osprey]

If k<n then n+m-k > m and b^(n+m-k) = 0

[bleak crystal]

such a waste of time

[old hollow]

Is this abstract algebra

[bleak crystal]

yes

[old hollow]

Nah dog it’s lit

[bleak crystal]

it's such a waste of time

[ebon osprey]

If k> n then a^k = 0

Pussy

[old hollow]

It’s annoying because it’s tedious in the beginning and a bit lame

[ebon osprey]

Haha

[bleak crystal]

it's been tedious all semester

i have yet to find a single reason high school teachers take this class

[ebon osprey]

No complaints you're a math warrior

[old hollow]

It’s the same level of lame as proving derivative rules in calculus

Wait what

[bleak crystal]

it makes me want to fucking scream

[old hollow]

This is a high school class?

[bleak crystal]

this is a math ed course

for high school teachers

[scarlet estuary]

you dont understand why they'd want math teachers to know basic algebra??

[bleak crystal]

yes, abstract algebra is basic algebra

[old hollow]

Wait what? You’re required to take this to become a teacher?

[bleak crystal]

mhmm yup sure

[bleak crystal]

yes

[old hollow]

Huh I never knew that

[scarlet estuary]

"algebra" in a mathematical sense yes

[old hollow]

Wel that’s pretty cool

[bleak crystal]

not really

i have 4 other classes to worry about

[old hollow]

Oh

[bleak crystal]

I'm shooting for a C-

[old hollow]

Hmm

[bleak crystal]

as long as i pass this class I don't give a fuck

i will literally NEVER use this in teaching

"hey kids, today we're gonna learn about zero divisors!"

[ebon osprey]

Dude I gave you the answer

[old hollow]

Well obviously you wouldn’t use this in teaching, unless you’re literally teaching this course

[bleak crystal]

exactly

sorry I'm just losing my mind at this class

[scarlet estuary]

the point is for the teacher to know more than their students

[old hollow]

But there’s a lot of stuff you learn that you don’t teach lol

[bleak crystal]

it's really driving me nuts

[bleak crystal]

calc 3 is literally 2 years beyond the scope of high school

[old hollow]

None of my math teachers know this stuff and it’s depressing, so I feel kinda happy bout it

[bleak crystal]

linear algebra is beyond high school

[scarlet estuary]

plenty of high schools teach calc 2

at least in the US

[bleak crystal]

must be a rich school

my school didn't even have ap calc

[old hollow]

Maybe

[ebon osprey]

Dude you have to know more stuff than they do to be a teacher

[bleak crystal]

i transferred schools just to take calc

they literally would have had to make a special course for like 3 kids in my class

[old hollow]

Their point is that this feels artificial, like why would they learn it for the sake of knowing more

Right? Idk

[bleak crystal]

yay, I'm smarter than a bunch of 16 year olds!

what a surprise!

okay I'm done ranting. just needed to vent a little

[scarlet estuary]

well understanding the basics of how algebra actually works is pretty important if you want to teach a botched version of it (which is what HS algebra is)

imo

[ebon osprey]

It's not about being smarter

[old hollow]

Hahaha HS algebra is botched abstract algebra

[ebon osprey]

It's about having a deeper understanding of the field

[scarlet estuary]

also FWIW the binomial theorem is like

standard content in high schools

at least here in canada

[ebon osprey]

A tad deeper

[bleak crystal]

i appreciate math less because of this course

[scarlet estuary]

so you should probably know it

[bleak crystal]

modular arithmetic is cool

[old hollow]

Dude I’m guessing it’s just really boring rn

[bleak crystal]

my school had a TERRIBLE math department

[old hollow]

Yes modular arithmetic is super cool

[ebon osprey]

Man the exercises you're solving,I learned them in first year of uni

[bleak crystal]

i just learned how to complete the square THIS SEMESTER

i literally never learned it

[ebon osprey]

So it's not that far

[worthy haven]

The best teachers I had in high school had graduate degrees in mathematics

[bleak crystal]

oh i also learned the rational roots theorem because of abstract algebra

you had teachers with grad degrees in math?

[worthy haven]

people who have a deep, real understanding of math beyond the high school level will be able to teach everything up to the high school level better, will be able to make everything feel more motivated, and will be able to generate better examples

[bleak crystal]

holy fuck i grew up poorer than i thought

i think my school had ONE dude with a master's

[worthy haven]

one had a masters, one had a PhD and used to be a professor, but retired to spend more time with her kids + to help high schoolers

I went to a very wealthy high school that was like 2 hours by bus from my house lol

[bleak crystal]

yeah

wow

what an experience you guys must have had lol

[old hollow]

My high school is def not poor but we don’t have any teachers that know math beyond calc 3

[worthy haven]

but anyways that aside, the teachers who really really knew math got me into math

[bleak crystal]

I'm lucky i got to go to a magnet school for the last 2 years of HS lol

[worthy haven]

taught their stuff way better

and had way more interesting assignments and tests

[bleak crystal]

man I'm gonna be teaching calc 2 at MOST

[old hollow]

Dude I would hate teaching calc

At least, AP calc

[bleak crystal]

because of how lackluster my high school teachers were, i decided to become a teacher

[old hollow]

It’s so like, algorithmic

It sucks

[bleak crystal]

yeah i love calc

it's just

beautiful

[old hollow]

Calc is great

[bleak crystal]

it's beyond fascinating, it hasn't ever lost its charm for me

every time i tutor it i appreciate it more

holy shit

[old hollow]

Jesus Christ

[worthy haven]

holy shit I don't think that's legal here

every high school needs to offer calculus, and lots of university programs straight up require it for admission

[bleak crystal]

what country are you guys in?

[worthy haven]

I'm Canadian

[bleak crystal]

CALC IS REQUIRED FOR ADMISSION

NO FUCKING WAY

dude my university offers algebra 2 equivalents

[worthy haven]

most science (especially physics)/engineering/math programs require you take calculus in high school

[bleak crystal]

people like you are the reason i want to teach

[worthy haven]

importantly, here the high school calculus course covers no integrals

[bleak crystal]

completely fucked by the system

totally unfair

[old hollow]

lol

[scarlet estuary]

alberta high school calc courses did cover single-variable integration

fwiw

[old hollow]

well the only reasonable, far-reaching solution to the problem is

online

[worthy haven]

I should've said ontario, you're right

IB and AP classes here do too ofc, but like normal math courses

[scarlet estuary]

though they didnt do much beyond derivative formulas / integration techniques / basic word problems, extrema, curve sketching sstuff

it was a fairly light course

[bleak crystal]

I've never seen a school with IB

[scarlet estuary]

but it was a high school course so

makes sense

[worthy haven]

also high school teachers here are paid decently (most of my HS teachers made around 80k-100k), which means more qualified teachers in general

[bleak crystal]

my first high school had like 1 AP course, comp/lit

[old hollow]

100k??? oisdfoisodijfs

[bleak crystal]

eEIWKSIWB

EIGHTU TJISAND DOLLARS

NO WAY

[old hollow]

that is insane

[scarlet estuary]

thats canadian BTW

[bleak crystal]

wanna guess the minimum in my state?

[worthy haven]

we also pay COL in canadian

[bleak crystal]

do ya?

[scarlet estuary]

take off about 30% to get USD values

[bleak crystal]

do ya?

go ahead guess

[worthy haven]

but we also pay for things in CAD so should you convert?

[bleak crystal]

first person within 5k gets a high five

[worthy haven]

35k

[old hollow]

US employs people they pull off the street

[scarlet estuary]

yeah since CAD prices are higher

[bleak crystal]

THIRTY TWO

[scarlet estuary]

ever looked at the prices on a book label

[old hollow]

you got a high five!

[bleak crystal]

AND YOU GOTTA PAY HEALTH INSURANCE

AND TAXES

[old hollow]

yeah

[scarlet estuary]

canadians still pay for health insurance btw

well more accurately, its deducted from their paycheque by their employer

[old hollow]

Cassius you said you liked modular arithmetic

[scarlet estuary]

like it is in the US

its generally cheaper than american insurance though

[worthy haven]

but this isn't for general medical care, it's only for like dental/vision/mental health/physiotherapy/etc.

[scarlet estuary]

and focuses more on dental/eyecare/cosmetic surgery and stuff

[bleak crystal]

yeah but it doesn't make sense to make a salary so low AND not include insurance

[scarlet estuary]

stuff that isnt covered by the national plan

[bleak crystal]

32 fucking thousand, can you believe that?

[worthy haven]

yeah a lot of stuff is covered by the national plan

[bleak crystal]

I'm moving to wyoming either way, indiana sucks so bad

[scarlet estuary]

fair, teachers in many regions are certainly super underpaid

[worthy haven]

https://www.ontario.ca/page/public-sector-salary-disclosure-2020-all-sectors-and-seconded-employees here's a list of ontario public servants who earn > 100k, that's how I knew about my hs teacher who made over 100k

[bleak crystal]

insurance that covers mental health? damn

[worthy haven]

there are 160 000 teachers in my province, and 27 000 of them make over 100k

[bleak crystal]

wow

highest paid in my state is 96

and that's with a LOT of experience

[worthy haven]

highest paid a teacher can be here is like 120k ish? 110?

there's a standardized progression

[bleak crystal]

this pains me to hear

i read that hs math teachers in Wyoming make like 60

also land is VERY cheap there because low population density

luckily i don't have to stay here forever basically

anyway, math

[old hollow]

yes

ok I was saying

[bleak crystal]

$\sum {n+m \choose k} a^kb^{n+m-k}$

[cloud walrus]

Cassius

[old hollow]

ah that

mess

[bleak crystal]

what do i know about this summation

[old hollow]

if you can figure out a way to prove that the sum is zero, you're done

[bleak crystal]

i see

[old hollow]

well since a^n = 0 and b^m = 0 (what we defined), you could try to prove that k is always bigger than n, and n+m-k is always bigger than m

do you see why that would make every term 0?

[bleak crystal]

yes

[old hollow]

(idk if this method would work, I'm just shooting different stuff)

ok

[bleak crystal]

a^n is zero

so if k>n, a^k also zero

[old hollow]

yes

[bleak crystal]

a^n•a^k-n

anyway

[old hollow]

hmm

[bleak crystal]

how could i argue k>n

[old hollow]

$$\sum_{k=0}^{n+m} {{n + m}\choose {k}} a^kb^{n+m-k}$$

[cloud walrus]

abs_0

[old hollow]

is this the sum

yeah I think it is

ok

well immediately that method won't work because k starts at 0, and n>0

[bleak crystal]

right

[old hollow]

but maybe we could try something else

[bleak crystal]

n is at least 1

a+b is nilpotent

such a hard proof

[old hollow]

oh wait

bruh

[bleak crystal]

yes?

[old hollow]

\begin{align*}
\sum_{k=0}^{n+m} {{n + m}\choose {k}} a^kb^{n+m-k} &= \sum_{k=0}^{n+m} {{n + m}\choose {k}} a^k b^{n-k} \cdot b^m\\
&= \sum_{k=0}^{n+m} {{n + m}\choose {k}} a^k b^{n-k} \cdot 0\\
&= 0?
\end{align*}

[cloud walrus]

abs_0

[old hollow]

wait but

[bleak crystal]

oh shit

nice

[old hollow]

lol

but wait

[bleak crystal]

there's more?

[old hollow]

do we know for sure that $0 \cdot n = n \cdot 0 = 0$ for all $n$ in this ring?

[cloud walrus]

abs_0

[old hollow]

i don't remember

is that a property of rings

[bleak crystal]

uhhhh

not necessarily

hang on

[old hollow]

frik

[bleak crystal]

yes R is communtative in this problem

with identity

and 1≠0

[old hollow]

yeah but do we know that anything multiplied by 0 is 0

is that an axiom of a ring

[bleak crystal]

that's what 0 is

[old hollow]

I feel like i'm being dumb rn

[bleak crystal]

the 0 identity in this case

[old hollow]

well in the normal sense of thinking about it

[bleak crystal]

but yes

it's the additive neutral

[old hollow]

ok cool

[bleak crystal]

that's the standard defintion

yeah so since communtativity the sum is zero

right

[ebon osprey]



I kinda advanced a bit

[bleak crystal]

oh you're trying it too huh

[ebon osprey]

I kinda finished it although last question needs to be more formal

[old hollow]

oh ez shmoney

o

[bleak crystal]

okay so it's all zero

[old hollow]

yeth

[ebon osprey]

And bonus question is “im too lazy”

[bleak crystal]

0+0 n+m times

nice

so a+b is nilpotent

alright 1/3 done

with this question

then one more

[bleak crystal]

yeah

pain

[old hollow]

well $(ra)^n = r^n \cdot a^n = r^n \cdot 0 = 0$

[cloud walrus]

abs_0

[bleak crystal]

yeah

[old hollow]

that was significantly easier lol

[bleak crystal]

i actually was thinking that believe it or not

yay im not completely stupid!

1+a is a unit

interesting

[old hollow]

haha

[bleak crystal]

(1+a) ^ n

does that work

[old hollow]

yeah thats the logical choice here

[worthy haven]

you can prove this very quickly
0 * x = (0 + 0) * x
= 0 * x + 0 * x
adding the additive inverse of 0 * x to both sides gives 0 = 0 * x, and an identical argument works for x * 0

[bleak crystal]

wait no

nice

[old hollow]

ah

that's clever

[worthy haven]

rings have distributive property + additive inverse + if 0 is around you're assuming it's the additive identity

so you're guaranteed it

[old hollow]

\begin{align*}
(1 + a)^n = 1 + {n \choose 1}a + {n \choose 2}a^2 + \cdots + {n \choose n}a^n
\end{align*}

[cloud walrus]

abs_0

[old hollow]

hmmm

oh I'm dumb

we're proving it's a unit

not nilpotent

[bleak crystal]

so the rest are 0

they'd have to be

[ebon osprey]



[old hollow]

we're proving $1 + a$ has a multiplicative inverse

[cloud walrus]

abs_0

[ebon osprey]

There with more details

[bleak crystal]

wow

i wish i followed that better than i do

let me go over again

lines 2 and 3

how does that work

like the transition between

wait no i see

[old hollow]

he distributed the sum to the 1 and a

[bleak crystal]

duh

yes

[old hollow]

I don't get what's after that tho

also seems very random that you would choose that sum to multiply it

it's like, okay cool it works, but how did you come up with that particular sum lol

[bleak crystal]

alright what about 3 to 4

i also don't see why its zero

[old hollow]

well $$(-1)^k(\text{something}) + (-1)^{k+1}(\text{something}) = 0$$ because they're being subtracted

[cloud walrus]

abs_0

[bleak crystal]

ah i see

that makes sense

[old hollow]

ye

[bleak crystal]

how do you make that transition between 3 and 4 though

[old hollow]

combine it into one sum

[cloud walrus]

abs_0

[ebon osprey]

I take this sum because it’s equal to 1-a^n

[old hollow]

ah

[ebon osprey]

It’s the formula

[old hollow]

ohhhh

I get it

[ebon osprey]

^^

[bleak crystal]

that's a neat proof

i like that one

[ebon osprey]

That’s how I got there

[old hollow]

yea

ok next

[ebon osprey]

I’m struggling with bonus

[bleak crystal]

i'm struggling with life

[ebon osprey]

Let’s talk about it after yeah ?

[bleak crystal]

unfortunately it's pretty basic

[ebon osprey]

I mean bonus, not life haha

[bleak crystal]

covid, lockdown, etc

[ebon osprey]

Hahahaha

[bleak crystal]

and this fuckin class man

it physically pains me

i can't say i follow part e

the handwriting too

do you wanna see something funny

e and f are not easy either

[old hollow]

what is "check this out"

[bleak crystal]

man this assignment is tough, it makes me feel as though i'm in way over my head

it's a hint he offered on the problem

[old hollow]

yeah that's the exact sum @ebon osprey used

but he wrote it in expanded form

[bleak crystal]

right

[old hollow]

wait I'm stupid you obviously alreayd knew that lol

ok finna uhh

[bleak crystal]

yeah i really don't follow the handwriting on e lol

[old hollow]

me neither

ok let's see

we're proving $u + a$ has a multiplicative inverse

[bleak crystal]

u+a is a unit

[cloud walrus]

abs_0

[bleak crystal]

right

this is similar to the last one

[old hollow]

yeah

[ebon osprey]

Ok handwriting bad got it

[old hollow]

whip naenae

[bleak crystal]

Write u+a= u (1 + u^−1 *a) and use Parts (c) and (d).

is the hint

[old hollow]

this feels so annoying

write $u + a = u(1 + u^{-1}a)$?

[cloud walrus]

abs_0

[bleak crystal]

yes

[old hollow]

ah

well now it's trivial!

wait is it

[bleak crystal]

uh

[old hollow]

what was I thinking

[bleak crystal]

u^-1 * a on the inside

so it's not just 1+a

uh

i don't know

[thorn delta]

u^-1a is nilpotent, so....

[bleak crystal]

wait why is that

[thorn delta]

a previous part you proved, part c

[old hollow]

$a$ is nilpotent, and by (c) we have $u^{-1}a$ is nilpotent, and by (d) we have $1 + \text{nilpotent}$ is a unit, and a unit times a unit is a unit? lol

[ebon osprey]



[cloud walrus]

abs_0

[thorn delta]

yea, that's right abs

[old hollow]

wait but is a unit*unit a unit

[ebon osprey]

What abiut handwriting now ??

[thorn delta]

yes

(ab)^-1 = b^-1 a^-1 in general

[old hollow]

haha it's lit now

[bleak crystal]

yes better

[ebon osprey]

So in bonus I manage to prove a0 is unit but no clue about the others being nilpotent

[old hollow]

@bleak crystal do you see this

wow these hints are coming in clutch

[bleak crystal]

yes that makes sense

wow this is a killer problem

[old hollow]

lol

[bleak crystal]

imagine coming up with this

[old hollow]

it's not actually that bad, but without the hints it would be super ridiculously cringe

[ebon osprey]

I like algebra ^^

Basically because of this

Sometimes you can just parachute a solution

[bleak crystal]

so f is very obvious right

like we said a unit + nilpotent = unit

[ebon osprey]

Yep !

[bleak crystal]

that's exactly what they're telling us

nice

now, is f(x) itself nilpotent

it is right

[ebon osprey]

Ahem

Not nilpotent

Unit

[bleak crystal]

oh sorry

the a1,a2... sequence is nilpotent so f(x)-a0 is nilpotent

so f(x) is a unit

and that extends to the polynomial ring

is there any reason it shouldn't?

or couldn't?

damn, what a problem. riveting

here's his #11 walkthrough

[old hollow]

christ

[ebon osprey]

Nice !!!

I like it

[sour warren]

What are you guys working on btw

[cloud walrus]

mirzathecutiepie

[ebon osprey]

What is $(U\cap v)$ ?

[cloud walrus]

Jean-Jérôme

[bleak crystal]

that bad?

[ebon osprey]

Is it the smallest subgroup containing U cap v ?

[old hollow]

nah it's longer than I expected

[thorn delta]

v is also a normal subgroup

[bleak crystal]

i see

[thorn delta]

its just funky notation

[ebon osprey]

Hmmm I’m not too familiar with those layers of group theory, is it obvious that U cap v is a group ? I mean v is normal in V

Sorry if dumb question, im curious )

[proud bear]

pretty sure yeah, since the intersection of two subgroups is another subgroup

[ebon osprey]

Oooooooh

Hum

Is it ?

But i got that we dont need it to be a group

Yeah ok it is a group

[cloud walrus]

mirzathecutiepie

[ebon osprey]

I agree with everything

Every answer to my questions haha

[bleak crystal]

I'm not smart enough to comment

spectating

[ebon osprey]

I am not familiar with this kind of diagram

But I think you should explicit some stuff

“An easily seen fact” and “normality bs”

Ok

[proud bear]

idk but i thought normality wasn't transitive so i'm not sure if the last step holds

[ebon osprey]

I agree

With the explanation

[proud bear]

,ti --at karachi

[cloud walrus]

The time in Asia/Karachi is 09:01 AM (PKT) on Thu, 25/03/2021.
Stain is 12 hours behind, at 09:01 PM (PDT) on Wed, 24/03/2021.

[proud bear]

nice lol

[ebon osprey]

Same

[proud bear]

truly

[ebon osprey]

Where are u

[lavish pike]

what is the weird p lookin thingy regarding homomorphism like p:R[x]->R

[ebon osprey]

What are u talking about

[lavish pike]

lmfao

one sec

that weird thing whats that called

what is it in word any clue

[proud bear]

[ebon osprey]

It’s a greek letter

[sturdy marsh]

is it fai or fee

[old hollow]

Fee

[ebon osprey]

It’s used as an f in modern greek and russian

[proud bear]

no they're asking how to type it in the word typesetting program

[lavish pike]

its late relaxxxxxx

i gotcha lol

[ebon osprey]

Fai

[old hollow]

You heathens

[cloud walrus]

mirzathecutiepie

[ebon osprey]

Oh idk about word

[scarlet estuary]

its fee in OG greek

but fie in english

(actually it's pee in really old greek)

even "pee" isnt quite accurate

it's more like

"peh-eh" where the first "eh" is high pitch

and the second "eh" is middle pitch

ancient greek was somewhat tonal

in modern greek its just "fee" though

nothing fancy

since modern greek isnt a tonal language

[cloud walrus]

mirzathecutiepie

[ebon osprey]



Doesnt this work ?

Haha

I won !

Hum let me see

Because its in v

Aaaah

Ok

Good remark

Yes you are right

Exactly

Yes

Gg

Lets go

I actually have to teach this on friday

So thanks

I havent really read the course yet

Whats N

[sour warren]

Normality bs is actually a technical term

[cloud walrus]

mirzathecutiepie

mirzathecutiepie

[ebon osprey]

Hey

Actually i feel you could do everything at once by using the $xH = Hx$ characterisation of normality

[cloud walrus]

Jean-Jérôme

[proud bear]

shouldn't you be considering gCAg^-1?

[cloud walrus]

mirzathecutiepie

[proud bear]

yeah i think this works

idk i'm still learning this stuff. i'm pretty smol brain

[cloud walrus]

mirzathecutiepie

mirzathecutiepie

[proud bear]

i don't really know about this stuff

[cloud walrus]

mirzathecutiepie

[proud bear]

what's the difference between this and a lattice of subgroups

[lavish pike]

uhhh is there like a general way to prove something is surjective?

like how do i prove that every b has some a

[prisma ibex]

You just do it

There’s not much you can do other than like

Applying the definition

And it really depends on understanding the objects and the morphism involved

[lavish pike]

relax with that homie lmfao

magic words magic person

so i legit just say $\varphi(c_0+c_1x+...)=a=h$

[cloud walrus]

cRaZyNiChOlAs12

[ebon osprey]

Just find an a for each b yes

Haha

[lavish pike]

[carmine fossil]

Well, You can say that's the same as constructing a right inverse

[ebon osprey]

Hey mirz can you post the theorem again ?

Thanks

So isnt there something like an isomorphism theorem that could kick ass in the last part ?

Oh ok

Isomorphism theorem mvp

Since Lagrange

[proud bear]

wait mirza. how old are you

2004?

[latent anvil]

subtraction goes in #prealg-and-algebra

[proud bear]

[ebon osprey]

Whats Nb

Ok

[cloud walrus]

mirzathecutiepie

[proud bear]

are you going to sleep soon or no

[cloud walrus]

mirzathecutiepie

mirzathecutiepie

mirzathecutiepie

[proud bear]

👏

[cloud walrus]

mirzathecutiepie

mirzathecutiepie

[proud bear]

is \unlhd a custom command

$\unlhd$

[proud bear]

oh damn

only knew about $\trianglelefteq$

[proud bear]

$\normsub!!!!!!!!\unlhd$

i guess they look the same

[cloud walrus]

mirzathecutiepie

[proud bear]

lol

was seeing how close they matched up. i'll go to #bots

yeah they're the same

[terse orchid]

isnt the squaring the circle problem fairly easy? i dont see where field extensions come in

[cursive temple]

why is it easy

[terse orchid]

maybe not easy

but just not enough stuff to write a 4000 word paper about

[scarlet estuary]

then write a shorter paper

but i mean youre right

in that all you really need to prove is the transcendentality of pi

which isnt hard

and then a basic field theory argument applies

[vestal snow]

If $F/k$ is galois, then that means that for any two embeddings $f_1,f_2$ of $F$ into $\bar{k}$ such that they restrict to the identity on $k$, $f_1(F)=f_2(F)$.

[scarlet estuary]

(gelfond-schneider)

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Is this statement true?

I'm pretty sure this follows from normal

and F being a splitting field

[terse orchid]

i cant namington, the paper has to be 4000

[scarlet estuary]

well perhaps you can pad it appropriately with historical context and whatnot

or introduce more complicated galois theoretic constructions or whatever

[chilly ocean]

oh, my guess is that this is true regardless that F/k is galois. but i am not that smart at field theory

[vestal snow]

You definitely need galois because there are 3 ways to embed Q(cube root of 2) into the algebraic closure

[chilly ocean]

ah hm

at the least, a result like this does sound familiar from galois theory class

[vestal snow]

Is the second part of this trivially easy?

Since there is only one way to embed K into an algebraic closure if K/k is galois (by that I mean all embeddings have the same image), they are equal

But then I don't see why we need F to be galois

[rustic crown]

well how do you define composite of 2 fields?

that depends on the chosen embedding of F in k bar, and all have same images when the extension is galois

[vestal snow]

Haven't we fixed an embedding though?

When we say $F \subseteq \bar{k}$, we're fixing some isomorphic copy of $F$ to become the canonical copy of $F$ in $\bar{k}$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

This seems like a badly worded problem

[rustic crown]

if you're fixing... then i don't see any problem

[vestal snow]

Thanks

[lavish pike]

so i have to prove that T is closed under subtraction and multiplication right?

[golden pasture]

essentially

yes

[lavish pike]

Could i say something along the lines of $x=(b_{0}+b_{2}x^{2}+...+b_{2k}x^{2k}) and y=(n_{0}+n_{2}x^{2}+...+m_{2k}x^{2k})$ where x and y are both in T

[cloud walrus]

cRaZyNiChOlAs12

[rigid cave]

Yeah so x+y will belong to T and so will x*y since an even number times an even number is in an even number. Also, R is a ring so the new coefficients after the multiplication will belong to R. However, do not take my word for it since I am not that good at this

Do you guys have any "medium" exercises? Not too hard but not too easy

[lavish pike]

na that makes sense, but what do u mean by the new coefficients

like the $(a_{2}b_{2})x^{2})$ ?

[cloud walrus]

cRaZyNiChOlAs12

[rigid cave]

Yes exactly. Those coefficients will belong to R since it is a ring and multiplication is a binary operation. This means that the polynomial will belong to T

[lavish pike]

so am i able to just say that we know a even integer minus a even integer is an even integer or do i have to actually like write out the proof of that

[cursive temple]

the proof is like half a line

[mint seal]

In ring theory?

[rigid cave]

Doesn't really matter

[mint seal]

Hmm

I don't know what you know or don't know

Have you showed that Z/pZ is a field if p is prime?

[rigid cave]

Yes that I have done

[mint seal]

And have you showed R/I is a field iff I is a maximal ideal ?

[rigid cave]

No, that I have not done I think. Let's try it out!

[mint seal]

that's a nice generalization of the earlier fact

as pZ is a maximal ideal of Z

[rigid cave]

Oh, I did not notice that!

[mint seal]

I should have mentioned R is a ring with identity, for that to hold

I'm used to rings always having identities

[rigid cave]

Oh, okay!

[lethal cipher]

Quick question. When looking at whether $(A_v,v)$ is a subring, does closure under addition imply $v(x+y)\in A_v$ or $v(x)+v(y)\in A_v$?

[cloud walrus]

dackid

[lethal cipher]

I'm pretty sure it's the first, but I want to make sure

[next obsidian]

v(x) shouldn’t ever be in A_v

v(x) takes values in some ordered group

Closure under addition would mean that if x,y satisfy v(x)>= 0 and v(y) >= 0 then v(x + y) does which follows by the condition valuations must satisfy in addition

(Greater than or equal to the min)

[lethal cipher]

Oh my apologies. A_v is the set and v(x)>=0 is the restriction of the operation v(x).

I understand now, thank you

[next obsidian]

It’s not a restriction on the operation

It’s a restriction on what belongs to the subset A_v

[lethal cipher]

Ah okay. So v(x) can do what it wants, but we are only taking a subset for what x values follow the rule v(x)>=0

[next obsidian]

Yes

Well the function v doesn’t exactly do what it wants

We’re assuming it already exists for A

Since A was a valuation ring

But it just stays doing what it was doing

[lethal cipher]

It does have some restrictive properties

[next obsidian]

Yes, in order to be a valuation

You need those properties in order to show A_v is closed under addition and multiplication

[lethal cipher]

Oh, so is (A, v) the ring?

[next obsidian]

Yes

[lethal cipher]

It never explicitly mentioned that

Got it!

[next obsidian]

A is the ring, and v is some valuation on the ring

Oh sorry, I slightly misspoke

[lethal cipher]

Lastly, is a field of quotients, K, the field where the inverse of A exists?

[next obsidian]

A never existed, my bad

v was a valuation on K

And you’re finding a subring of K

Which consists of all non-negatively valued elements

[lethal cipher]

But if K is a field, what's ring?

[next obsidian]

A_v

Is the subring of K

[lethal cipher]

Sorry, I meant ring

[next obsidian]

What’s a ring?

[lethal cipher]

Know, what is the ring in this scenario?

[next obsidian]

A_v is the subring

Of K

Oh

[lethal cipher]

But a subring is a subset of a ring right?

[next obsidian]

Right

Well every field is a ring

So if you need to think of A_v as a subset of a ring then consider K a ring

[lethal cipher]

Ah in that case, what exactly is a field?

[next obsidian]

A field is a ring where every no -zero element has a multiplicative inverse

Example Q

What isn’t a field is Z

[lethal cipher]

So it is a field of quotients?

[next obsidian]

Because 2^-1 doesn’t exist

I mean... sure

But that’s an odd way to think of them

[lethal cipher]

Well it says K is a field of quotients.

So I'm trying to understand what that means exactly

[next obsidian]

It looks like it says K is just a field

A field of quotients is something you can associate to certain rings

You just add in all fractions

[lethal cipher]

"which has K as a field of quotients."

[next obsidian]

So everything gains an inverse

Yeah

So if you have a ring A where the only zero divisors are 0

[lethal cipher]

So K is a field, but in relation to A_v, it is a field of quotients.

[next obsidian]

Then you can make a field of quotients

Yes

Well

It is A_v’s field of quotients

A field of quotients is something you do to certain rings

[lethal cipher]

True, I need to be more specific

[next obsidian]

Any field of quotients is a field

Field is the type of algebraic structure it is

Where as field of quotients is like...

Something associated to certain rings

[lethal cipher]

So a field of quotients contains the multiplicative inverses of A_v

[next obsidian]

In a sense it’s the smallest field containing A_v

All you do is add in the inverses of everything

And then the ratios

[lethal cipher]

So every element that is not in A_v must be the inverse

[next obsidian]

Not exactly

Take Z for example

Its field of quotients is Q

[lethal cipher]

I agree, but, I see what you're getting at

[next obsidian]

So you take all non-zero integers and add in 1/x for them

But also you have say

3/2

[lethal cipher]

Why is that the smallest field then?

[next obsidian]

This isn’t the inverse of an integer

[lethal cipher]

Why not just have 1/x's?

[next obsidian]

It’s not closed under multiplication

Try doing 2* 1/3

This has to exist but 2/3 isn’t a 1/x

So you’re adding in as little as possible in order to make a field

[lethal cipher]

Oh shoot you're right! Do rings need to be closed under multiplication?

[next obsidian]

Yes, by definition

Multiplication is an operation so you need to have the multiplication of any two elements in your set be in that set

[lethal cipher]

Oh duh, multiplication is a binary operation

Quick question: what exactly is a unit element?

[viscid pewter]

an element which is a unit

[lethal cipher]

But what do we mean by unit here?

[viscid pewter]

an element which is invertible multiplicatively

ew

no go away

[lethal cipher]

So the identity?

[viscid pewter]

just say 'multiplicative identity'

no

usually it means an element with a multiplicative inverse

but sometimes ppl use it to mean the identities

confusingly

[lethal cipher]

So when you say it has a multiplicative inverse, do you mean the inverse is contained in the group? or it just doesn't have a unique inverse at all?

[viscid pewter]

yeah

so like 2 wouldn't have an inverse in Z4

[lethal cipher]

What is Z4?

[viscid pewter]

like

mod 4

there is nothing that multiplies 2 to give 1

[lethal cipher]

Oh, Z\4Z is how my book writes that

[viscid pewter]

yeah

probs less confusing that way

[lethal cipher]

That makes sense too, the congruence class of 1mod4 only has odd numbers

I'm not sure if my explanation is right, but I do get the idea

[viscid pewter]

uhhh

i don't think it's about congruence classes from like Z or whatever

it's just that it doesn't have a multiplicative inverse

[lethal cipher]

What I mean to say is 4k+1 is odd for any k

[viscid pewter]

yeah, i'm not sure that's related

[lethal cipher]

And so multiplying anything by 2 will never give an odd number

[viscid pewter]

that's more like it

[lethal cipher]

That's the logic I was trying to get at

[viscid pewter]

but 4k+1 being odd doesn't have much to do with it? or

oh well

ok i see it now

i don't even think about conjugacy classes now when doing modulo

[lethal cipher]

So what exactly about this claim makes this a maximal ideal?

[small bison]

anything bigger contains a unit

[lethal cipher]

Is that enough to conclude something is a maximal ideal?

[small bison]

yeah any ideal that contains a unit is the whole ring

[lethal cipher]

Ohhh, interesting

[small bison]

it's cause 1 is in the ideal in that case

and the ideal generated by 1 is the whole ring

[lethal cipher]

Thank you all so much for the help. My research has lad me to a grad textbook where I need some understanding of abstract algebra, which I do not have much. I was able to get the research I needed done today thanks to you all. 😀

Nope!

Yes indeed. We learned that the best way to think about matrix multiplication for our intended goal is with rings.

Not in the way you're thinking tho 😁

If you are interested, I can share after I finish eating my food. It's definitely something I'd rather draw out than do on LaTeX

[lethal cipher]

Actually nevermind. I may mention it tomorrow once I gather a better understanding.

[uncut girder]

What's this book abt

What are you reading it for

I know a little about valuations

[lethal cipher]

Well, for the moment I am reading it to get an idea of why {x\ in A|v(x)>=0} is a subring of A and what the maximal ideal is for this ring

Once I actually understood it, it's honestly pretty straightforward. However, I haven't taken an abstract algebra class or learned too much about it, so that was a big jump for me

Then the goal is to try and see if two matrices are diagonally dominant w. r. t. P-adic absolute value, then the Product of the two is also diagonally dominant.

That's the end goal

A similar thing goes for the inverse

[jagged dune]

Hi.

I have a question about rings.

I'm struggling to see why ri (r is any real, i is the square root of 1) fails to form a ring.

[cloud walrus]

slimvesus

[jagged dune]

yes

it needs to be an abelian group, be associative and distributive under multiplication

for some reason I can't think of a scenario where it doesn't work

I'm pretty sure the reals are.

Well the reals are under complex, so you'd still be closed, no?

Actually that's where I have a question

so this is a complex set

if I have a number not including i, I guess where r = 0, am I still closed?

ok

what if I have just 5 or something

so maybe 5i * i

that's -5

so this isn't a ring

it's no longer in the set

[uncut girder]

What is diagonally dominant?

[jagged dune]

Thank you.

I don't know how I didn't think of that lol.

[lethal cipher]

It's a little different for normal absolute value, but we are dealing with an ultrametric distance, so the respective inequality is considered instead.
So the absolute value of the diagonals of each row is strictly greater than the maximum of all the other entries in the row.

[uncut girder]

Ok

[delicate bloom]

you're only restricting it to be greater than the entries in the row not the column too?

according to your definition these are two diagonally dominant matrices which multiply to make a non diagonally dominant matrix: $$\begin{pmatrix} 1 & -p \ p & 1 \end{pmatrix} \begin{pmatrix} p^2 & p^3 \ p & 1 \end{pmatrix} = \begin{pmatrix} 0 & p^3-p \ p^3+p & p^4-1 \end{pmatrix}$$

[cloud walrus]

Merosity

[chilly ocean]

could I ask a question?

seems like the question's done so I'll go ahead

anyone know how to do this? I'm not sure what a semidirect product is but I've proved homomorphism

this is what I have so far

[mild laurel]

Uh, I feel like you should go look up or learn what a semidirect product is before trying this problem?

[chilly ocean]

I haven't really been able to find anything concrete googling

[sturdy marsh]

https://en.wikipedia.org/wiki/Semidirect_product#Inner_and_outer_semidirect_products

[mild laurel]

If this is an exercise for a class, then have you guys not covered semidirect products? Or if it's an exercise from a book, then it should be in the corresponding section?

[chilly ocean]

I'm just looking at practice problems at the end of the chapter, I guess it's assumed that I knew what a semidirect product was beforehand

it's ok tho I'll look at the wikipedia, appreciate it

[lethal cipher]

That is definitely something we may have to do.

[delicate bloom]

yeah considering I gave a counter example to what you're trying to prove heh 😛

[uncut girder]

Whats the purpose of considering these matrices

[delicate bloom]

there's something called the Gershgorin circle theorem

basically tells you stuff like if it's invertible, where the eigenvalues are, and helps with computing eigenvectors/eigenvalue algorithms

[lethal cipher]

Yea, we proved that a while ago. Now we are just trying to see if there is any variance with certain matrix operations and ultrametric diagonal dominance.

[delicate bloom]

if we define 'totally dominant' to mean a matrix that is both row and column dominant, then I was able to prove the product of two totally dominant matrices is totally dominant

and then realized we can loosen this to if A is row dominant and B is column dominant then AB is totally dominant

so a little less restrictive, I guess a simple corollary is if X is column dominant then X^T X is totally dominant, which is kind of interesting

[lethal cipher]

I haven't proved it yet, but I've definitely made that deduction.
On another note, how'd you jump to X^TX being totally dominant?

Ohhh, I see.

Because X^T is row dominant

Since you're already 5 steps ahead of me with this stuff, I think you can also see if A is totally dominant, then so is A^-1

[delicate bloom]

interesting, I see how A^-1 is guaranteed to exist now if A is totally dominant, but I guess my next question to ask myself is, when can we say like, A, C are totally dominant and we have A*B=C does this imply B is totally dominant? Does it imply it's column dominant? Or none at all?

obviously A*A^-1 = I as a special case of that

ty ty

[lethal cipher]

Well, I think we can answer that question quite nicely.

Since both A and C are totally dominant, and inverse exists and is also totally dominant.
So $A^{-1}AB=A^{-1}C \Rightarrow B=A^{-1}C$

[cloud walrus]

dackid

[lethal cipher]

And we have already established that product is totally dominant

Well, you have. I still have yet to prove it. :P

[delicate bloom]

I never proved the inverse was totally dominant yet, I was hoping to prove it by this but

I haven't thought about it yet either lol just a bit occupied with something else atm

[ebon osprey]

Man can you give me the ref of your book, I think it's pretty nice

Is a refinement just a longer tower ?

Ty

So is it j=s ?

Okay

Lang loves algebra at any level and in any form

Wow it's a bourbaki member

[lethal cipher]

In regards to this, it says in the next page that K/v=A_v/P_v. Why is this exactly?

Here, this may be needed

[golden pasture]

the residue field of k is defined as A/P

[visual hare]

Can someone explain to me how these function compositions make sense? Topics in Algebra 2nd edition

[mild laurel]

iirc Herstein writes his composition in the opposite orders, so that phi dot psi means do phi first and then do psi

[elfin furnace]

the composition of permutations?

[visual hare]

oh my god...

why, just why

[mild laurel]

iirc he also writes function application in the opposite direction too?

So he writes (x)f instead of f(x)

[visual hare]

I escaped Fraleigh's book because of this notation

now Herstein does it too

[elfin furnace]

hernsteins abstract algebra doesnt have that wacky notation

iirc

[mild laurel]

oh maybe i was messing them up

[visual hare]

he does the function notation in the opposite way

[mild laurel]

oh yeah

[visual hare]

are you sure? are there other editions?

[elfin furnace]

im sure

its the normal notation

[visual hare]

which edition is this?

[elfin furnace]

wdym

its just "Abstract Algebra"

oh yeah

third edition

yeah composition of permutations is normal in this book too

[mild laurel]

one is called topics in algebra and the other is abstract algebra so they're different books I think

[elfin furnace]

yeah they are different books

same author using different notation

[vestal snow]

Is there a slick way to do this or do I have to suffer?

[mild laurel]

maybe it'd help to note that the irreducible factors of x^n - 1 over Q are exactly the dth cyclotomic polynomials for d | n

i don't know if this is really any faster than just calculating though

[vestal snow]

Yeah that's what I had in mind

Wait hold on

[vestal snow]

This method is MUCH faster than direct computations right?

FOILing stuff out is gonna be a pain

[mild laurel]

yeah probably

oh, you can also use mobius inversion on that fact I mentioned earlier to solve for the cyclotomic polynomial instead

$$\Phi_n(x) = \prod_{d \mid n}\left(1 - x^{n/d}\right)^{\mu(d)}$$

[cloud walrus]

Zopherus

[vestal snow]

I used wikipedia

[mild laurel]

oh I didn't realize this was on the wikipedia page

[vestal snow]

Can we classify problems like these as cruel and unusual punishment?

[rustic crown]

(i remember doing this by hand 😶 )

(only to get sad that cyclotomic polynomials can have coefficients other than 0,1, -1)

[cursive temple]

I hope you didnt do them by hand untill you found that out

[golden pasture]

lmaooooo

[unique juniper]

whats the best way to do this?

firstly, how do i efficiently get initial sylow 2 and 3 subgroups?

[chilly ocean]

Is there a result like every p subgroup can be placed in a sylow p subgroup? Then maybe starting with a 4 cycle like (1 2 3 4) can be a starting point

[unique juniper]

i see

[chilly ocean]

(Finding a sylow 3 subgroup is trivial)

But I'm not sure if that statement is true, I just pulled it out of my ass

[unique juniper]

it is

what i did was

consider the subgroup generated by (1 2 3 4) and (1 2 ) and got it that way, but that just a guess

[chilly ocean]

Er, I don't think that generates a sylow 2 subgroup

[unique juniper]

maybe i did something wrong lol

[chilly ocean]

(1 2 3 4)(1 2) = (1 3 4) is a 3 cycle

[unique juniper]

er

let me check

$\langle (1 2 3 4), (1 2) \rangle = {1, 1234, (13)(24),(1432),(12),(134),(234),2413}$

[cloud walrus]

Yes

[unique juniper]

maybe i did something wrong?

[chilly ocean]

I didn't check it, but it looks like you agree that there are 3 cycles

But sylow 2 subgroup of S4 is order 8

[unique juniper]

hmmmmm

ye

[chilly ocean]

So there should be no elements of order 3 right?

[unique juniper]

im not sure

[chilly ocean]

This follows from lagrange's theorem

[unique juniper]

how

[chilly ocean]

If x has order 3, then the cyclic group generates by x has size 3. but 3 does not divide 8

[unique juniper]

yes

but

<(1234),(12) > why are we saying that this is 3 cycle

[chilly ocean]

But you said so yourself, that subgroup contains (1 3 4)

[unique juniper]

sure

i see

i see

my bad

sorry lol

so if we start with a subgroup of order 4

it would be contained in a sylow 2 subgroup

then what?

experiment ?