groups-rings-fields

[chilly ocean]

yes, except needing cube instead of square roots

[cloud walrus]

tree3:
Compile Error! Click the reaction for details. (You may edit your message)

[stone fulcrum]

You've got the unfortunate problem that ρ is in both sets of roots, so there's a weird way that they do mix

[steep hull]

Yeah

[chilly ocean]

oh crap

It's not S3 x S3

[stone fulcrum]

Oh boy it's been a while let me think

Might be best to extend by ρ, then √2, then √3? Does one need all those?

[chilly ocean]

To make the splitting field K? Yes

[stone fulcrum]

Or, sorry, cube roots

[steep hull]

It’s indeed degree 18

[chilly ocean]

@steep hull how do you prove that?

ah wait there's a theorem in Dummit and Foote that may be helpful

[steep hull]

The extension over Q is Galois so you can argue by looking at the Galois group. There is a very general theory behind all this, however.

[chilly ocean]

hmmm

Proposition $14.21$ of D&F: Let $K_1$ and $K_2$ be Galois extensions of $F$. Then $(1)$: The intersection $K_1 \cap K_2$ is Galois over $F$ and $(2)$: The composite $K_1 K_2$ is Galois over F. The Galois group is isomorphic to the subgroup $H = {(\sigma,\tau)|\sigma|{K_1\cap K_2} = \tau|{K_1\cap K_2}}$

[cloud walrus]

Azu:
Compile Error! Click the reaction for details. (You may edit your message)

[chilly ocean]

Let $K_1$ be the splitting field of $x^3-2$ and $K_2$ that of $x^3-3$

[cloud walrus]

Azu:

[chilly ocean]

Then in this context, $K_1K_2$ is the splitting field $K$

[cloud walrus]

Azu:

[chilly ocean]

It seems obvious that $K_1 \cap K_2 = \mathbb{Q}(\rho)$

[cloud walrus]

Azu:

[chilly ocean]

@stone fulcrum this is right so far, right?

[leaden finch]

can someone help a sistah out for this problem

[sharp sonnet]

the hint in 8. literally tells you what to do

in the other one just assume that there are 2 identities/inverses and play around a bit with the group axioms

for the cancellation law consider that each element has an inverse

[tidal pier]

@leaden finch did you manage to solve it?

If not, ping me when you are back and I’ll assist you through it

[obsidian zealot]

How do we prove S4 is a group, S4 is non-abelian

[woven delta]

How do you prove the set of permutations of 4 elements is a group?

Is there an identity permutation?

If you compose 2 permutations do you get another?

Are permutations invertible?

[obsidian zealot]

are those 4 ways to prove that s4 is a group

[chilly ocean]

That is quite poetic if you ask me

[obsidian zealot]

true

[woven delta]

Lol

Anyway if the answer to those 3 questions is yes, then S_4 is a group

[obsidian zealot]

got it, this helps

thanks for help

[toxic zephyr]

Is there a connection between vector fields and fields from abstract algebra?

[woven delta]

Not really

[gilded trellis]

erm.. I'm kinda stuck here... can someone please elaborate more on that implication ?

how does he go from gcd(ord(H),ord(G')) = 1 to phi(H) = {1}

[woven delta]

Lol

Take an element in H

[gilded trellis]

yeah

[woven delta]

The order of the element is a factor of the order of H

[leaden finch]

@tidal pier i havent solved it yet since i am confused D:

[gilded trellis]

oh

[solemn rain]

|Phi(H)| must divide both |H| and |G'|

[woven delta]

Now the order of the image of that element is a factor of the order of the element

[gilded trellis]

yes

[woven delta]

So the image must be 0

But yeah what mo2men said works as well

[solemn rain]

@leaden finch come another channela nd ping me there

if you want help

i can try

[gilded trellis]

hmm, alright ! thanks :3

I appreciate the help !!

[topaz sundial]

If I have z= nth root of unity in C, and y=z+1/z
How do I show the minimal polynomial of y over Q, splits in Q(y)?

[mild laurel]

Q(y) is a subfield of Q(z), you know the structure of the galois group for Q(z), try to use that to say something about Q(y)

@topaz sundial

[topaz sundial]

So Gal(Q(z)/Q) is isomorphic to Z/nZ*
|Gal(Q(z)/Q(y))|=2 (I believe it has the identity and phi(Z+1/Z)=Z-1/z )

So then |Gal(Q(y)/Q)|=Euler phi function(n)/2

But god, here I'm bad at applying what this tells me.

[mild laurel]

Figure out what the phi(n)/2 conjugates of y must be, and show that they're all in Q(y)

[topaz sundial]

Ok. Thanks.

[gilded trellis]

what would the dependency chart of Artin's algebra be ?

erm

I worked through 70% of the group stuff (until homomorphism restriction) and temporarily jumped to the linear algebra stuff (real vector spaces) hoping they'll be a good example for what I studied in groups

(that chart is for Dummit and Foote's btw)

[solemn rain]

whats your question

[gilded trellis]

I jumped back and finished the first chapter on group theory

what would the dependency chart of Artin's algebra be ?

what things can I skip and come back to later

[solemn rain]

i think you can skip the group theory chapter in artin

[gilded trellis]

I knew it was okay to temporarily jump ahead and comeback to that material from harvard's course

[solemn rain]

just go matrices ---> vector spaces ---> linear transformations

[gilded trellis]

ah

[solemn rain]

bilinear forms etc

then after knowing some group theory

you can learn about 'linear groups'

[gilded trellis]

the linear algebra stuff is necessary for rings ?

[solemn rain]

chap 8

uhh no

in dummit and foote rings are b4 vector spaces

[gilded trellis]

in Artin

oh

sokka

ah I see

[solemn rain]

i really dont know tbh but i dont thinnk so

[gilded trellis]

What do I need for rings ?

because I really need to get a quick grasp on the basic stuff

[solemn rain]

nothing ig

u can really start with it

but most of ppl do groups first

[gilded trellis]

I did groups

[solemn rain]

then yea you can do rings ofc

[gilded trellis]

alright, I'll try to check out the ring theory part

[solemn rain]

thats how its done in dummit

[gilded trellis]

but Group actions ?

[solemn rain]

groups ---> rings ---> modules and vector spaces

what about group actions

?

[gilded trellis]

Do I need that stuff ?

[solemn rain]

yes

[gilded trellis]

so can I skip to it ?

[solemn rain]

yea you can after the first chap on groups ofc

[gilded trellis]

my goal is just to grasp basic ring theory (because we need it)

and then I'll come back and read everything

[solemn rain]

then yea the fastest way is just df way

[gilded trellis]

yea you can after the first chap on groups ofc
@solemn rain Ah ! I see !! Great :3

Thanks !

[solemn rain]

groups ---> rings

thats it

[gilded trellis]

Thanks !!

[solemn rain]

np

[gilded trellis]

heh

I was worried because Artin's ... is kind of weird

[solemn rain]

idk i havent tried it

[gilded trellis]

and tbh at times It gets irritating

[solemn rain]

they say its good for learning both lin algebra and algebra

[gilded trellis]

to understand

bad choice for me ..

self-learning

[solemn rain]

dk why it misses semidirect products tho

[gilded trellis]

he focuses more on lin algebra

instead of symmetry and permutations

they're left till chapter 5

[solemn rain]

tbh i would say try to learn lin algebra

most of ppl never learn any abs alg b4 lin algebra even

and probn thats how it should be ^

[gilded trellis]

well... in our curriculum it's number theory -> group theory -> matrices -> ring theory -> linear algebra

though it's not really in-depth .. the group and ring theory part

[solemn rain]

yea thats cool too

[chilly ocean]

sounds like any part isnt indepth if its in that order no offense

[gilded trellis]

high school mate :"""3

[chilly ocean]

oh

[gilded trellis]

lol

[solemn rain]

i mean i really dk what ur missing tbh

if u covered first and sec chapter

of gt in artin

other than like

smidirect products which u can covr easily

on a youtube video

[chilly ocean]

u dont need much group theory for rings tbh

[gilded trellis]

what really surprised me tbh

[solemn rain]

yea just get your feet wet id say

[gilded trellis]

cuz I used artin

is that we start with : binary algebraic structures -> semigroups -> monoids -> groups

so when I tried to do some exercises

from our textbook

I was like : "wth is this !?"

(because artin only covered groups..)

[solemn rain]

yea some texts do this

df for example doesnt say anything about monoids magmas whatsoever

some do

[gilded trellis]

welp ! imma come back to lin algebra later

I've speeded through dummit and foote

the group theory stuff

and it's more lucid

[solemn rain]

well id say try to learn lin alg as early as possible but like

[gilded trellis]

erm

[solemn rain]

if ur having fun with df gt or rings

then fuck it lmfao

[gilded trellis]

more understandable than artin's

[solemn rain]

i dont think my failure in df was bec of lack of knowledge in lin alg XD

[chilly ocean]

I liked df more as well

its for dumber people

[gilded trellis]

hah

[solemn rain]

XD

[gilded trellis]

I didn't have to spend 5 mins rereading a paragraph

just to understand what the heck he's talking about

[solemn rain]

tbh im getting really bored tho from df

[gilded trellis]

lol

[solemn rain]

like he talks too much

[gilded trellis]

yeah

was going to point that out

[chilly ocean]

but its cool to do df then artin

once you know whats going on

[gilded trellis]

artin's makes you put your all into it

[solemn rain]

i really wanted to do these more 'terse' books mostly grad texts

but that wouldve been bad for me ig cuz its my first exposure

[gilded trellis]

so you feel a sense of achievement

i really wanted to do these more 'terse' books mostly grad texts
thought about that too

but I was worried about losing motivation

idk

[solemn rain]

yea well about that df has a reputation for being 'dry'

[gilded trellis]

the difficulty just has to be perfect

it is dry

totally agree with them on that

[chilly ocean]

I dont think its dry

why do you think so

[gilded trellis]

erm

[solemn rain]

its jusut boring

[gilded trellis]

the style of writing

[solemn rain]

XD well it doesnt matter

anyways boys

if you have anything just ask

gl hf

[gilded trellis]

kay ! :3

thanks mate !

by the by

Allan Clark's :"3

Elements of abstract algebra

checked that out too ...

[chilly ocean]

never heard of it

[gilded trellis]

it's ... weird

erm

it basically defines something

and then gives you exercise after exercise

so you do all the work by yourself

[solemn rain]

^ isnt that good? XD

[chilly ocean]

im so fucked araragikun its unbelievalbe

[gilded trellis]

no correction and the exercises can be hard af

im so fucked araragikun its unbelievalbe
why ???

[chilly ocean]

I dont know shit im supposed to know

[gilded trellis]

you don't have time ?

[chilly ocean]

I have a shit ton

[gilded trellis]

to revise the material

[chilly ocean]

I mean

I have a lot of free time during a day

[gilded trellis]

yeah ?

[chilly ocean]

but I gavent learned anything past week

fuck

[gilded trellis]

that's alright

treat it like a short break

you know

if you're well rested you'd have more energy to tackle learning

you don't have a deadline do you ?

[chilly ocean]

not yet

but there will be

[gilded trellis]

well then, I see no problem with that !

start at it afresh tomorrow morning (or today.. depending on your time)

and trust me you'll be able to pull it off

what have you been doing for the past week ?

[chilly ocean]

Im trying to, not sure if I should even atten my zoom class tomorrow since I wont understand shit

[gilded trellis]

record it !

[chilly ocean]

ok I have an idea, I will tell my friend to record the calss

yeyeye

[gilded trellis]

yep

and then study on your own to catch up

and rewatch the class

tbh our maths teacher has just disappeared on us lol

[chilly ocean]

yeah I will wake up and during the clss instead attending it I will just try to learn as much as I can

[gilded trellis]

we had 2 zoom sessions with him

and then he sent us some exam in a pdf

and just poof

[chilly ocean]

dude yeah thats why im also fucked, Ive had only 1 class since the start of march of 1 course

[gilded trellis]

idk where he went, we sent messages but nothing 0.0

[solemn rain]

would u mind showing me the exam

just for like

[chilly ocean]

yeyey recently I got mail that he is back and we need to speed the fuck up

[solemn rain]

fun x

[gilded trellis]

xD it's easy

aight

gimme a sec

(it's in french btw)

(there is an arabic version of it too)

[solemn rain]

yea send the arabic version if yo ucan

[chilly ocean]

arabic numbers?

[gilded trellis]

no no

lol

[chilly ocean]

araragi-kunUWUWUWUWU

[gilded trellis]

hai ?? :3333

what are you studying rn ?

I'm guessing graduate ?

:3

[chilly ocean]

no

anal,probability and compuational math

[gilded trellis]

0.0

anal,probability and compuational math

anal,pro....

anal

[chilly ocean]

uwu

[gilded trellis]

:"""3

[chilly ocean]

its fun you should try it

[gilded trellis]

we have probability

but I know NOTHING about it

except counting

[chilly ocean]

same

I cant even count tbh

only up to 73

[gilded trellis]

xD

damn boi, you can count to 73 omg

I can only do 69 :3

alright we're going off topic here...

[chilly ocean]

OwO

[gilded trellis]

(sorry)

but I think you can pull it off Godel ! I mean, from what I've seen you're good ..

[chilly ocean]

im good in looking good

thanks for motivation though I will study tomorrow I promise

[gilded trellis]

0.0

still, what did you "waste" your time on ?

[chilly ocean]

playing games, watching movies

[gilded trellis]

Yare yare

shouganai ne..

was going to suggest you organize your time but that never works out for me xD

[solemn rain]

how do i normally prove a subgroup is characteristic in G

i have to prove Frattini subgroup of G is characteristic in G

[mild laurel]

Show that automorphisms on G preserve maximal subgroups

In other words, if H is a maximal subgroup of G and phi is an automorphism of G, then phi(H) is also a maximal subgroup of G

[solemn rain]

assume not

assume H is a maximal subgroup of G but phi(H) is not maximal

then there exists a subgroup of G , call it C , where phi(H) < C < G

but since C is bigger than phi(H) , phi^-1(C) = {x in G | phi(x) is in C} would be bigger than H , a contradiction

?

is that right

[mild laurel]

yes

[solemn rain]

cool

ty

[mild laurel]

now think about why this helps

[solemn rain]

umm

knowing its characteriistic means its normal?

and means its the unique subgroup of its order?

idk

[solemn rain]

H is normal subgroup , K is any subgroup , H intersects K = {1} , G = HK

then HK = H x K = H x_phi K where phi is the identity element

H x_phi K is semidirect product

the semidirect product is the set of (h,k) wherre h in H k in K

where the operation is defined as

(h1,k1)(h2,k2) =(h1 k.h2,k1k2)

where k.h2 is the action we get from the homomorphism phi: K ---> Aut(H) ( which is always jusut phi(k) = khk^-1 for h in H , conjugation ) ?

am i laying out definitions out right

we use this theorem 'then HK = H x K = H x_phi K where phi is the identity element' to classify stuff

so for certain orders of n we just find those 'ingredients' and construct new groups

am ir ight folks

[strong stone]

The positive integers are not a group or WHAT.

[mild laurel]

Phi is not always that

In this case, it is that conjugation action

[strong stone]

Did someone delete me

[mild laurel]

@strong stone this channel is for on topic discussion. If you want to make jokes please do it in #chill

[strong stone]

That wasn’t a joke.

But anyway

May I join discussion.

[solemn rain]

what case

i did not say any groups it was all arbitairy

[mild laurel]

In the case of inner semidirect products

[solemn rain]

okay so for groups of order say n

isomoprhic copies vary as the homomorphism vary?

like if i change hte ephi

i get new group?

the ph*

the phi*

[mild laurel]

Uh what

[strong stone]

Yes

Wait..

[solemn rain]

H x_phi K is = G for some phi

[strong stone]

It depends though

[solemn rain]

if i change phi

i get new group

?

[mild laurel]

I'm not sure what you're trying to say

[topaz sundial]

Ok.

So I'm trying to figure out what exactly Gal(Q(z)/Q(y)) looks like (where z is the nth root of unity, and y=z +1/z )
I know it should have two elements, so one is the identity automorphism, and the other is maps y to whatever the other root is. But I'm not sure how to figure out what the other root is 😑

[solemn rain]

you said in this case its conjugation

[mild laurel]

I mean, yes?

This is the point of outer semidirect products

[solemn rain]

i really dk what you mean by outer inner

[mild laurel]

Inner is when they're both subgroups of a larger group

And in this case, the action is always the conjugation action

[solemn rain]

okay

outer is just when i get 2 groups and multiply them out

and get a new 1?

@strong stone yo bro i am trying to get help plz no troll 😦

[strong stone]

Im not im just trying to learn too man

[mild laurel]

@strong stone please stop saying things you have no idea about

[strong stone]

Guys im sorry

[mild laurel]

Yes momen

[strong stone]

I just want to learn

[solemn rain]

okay cool

so thats it about semidirect products?

[mild laurel]

This isn't where you should start learning

Uh

[solemn rain]

i classify groups of order n

[mild laurel]

I guess

[solemn rain]

with this theorem

and i also like

[strong stone]

Please direct me to where do I go to learn.

[solemn rain]

generlize direct products

phi = identity

then i get a normal direct product

[mild laurel]

@topaz sundial think about the automorphisms of Q(z), which ones fix Q(y)?

[solemn rain]

okay thank you very much @mild laurel

[strong stone]

Wait zopus please direct me to where you think I should learn.

[mild laurel]

You should learn linear algebra and calculus first

[strong stone]

I basically know calculus

Are you sure you need to know those maths for this?

I though abstract algebra was well...

Abstract

[mild laurel]

What's your point

[topaz sundial]

Oh. Since phi(z+1/z)=phi(z)+phi(1/z)=z+1/z, then phi(z)=z or phi(z)=1/Z

[strong stone]

Listen im sorry im not trying to argue or anything i just want to learn math

[topaz sundial]

Linear algebra is pedagogically taught before abstract algebra. While technically you don't "need" it, most of the most prominent examples of groups come from linear algebra, and by the time you get to field theory without linear algebra you aren't getting far

[strong stone]

I understand

Thank for explanation

Time for liner algebra

#linear-algebra

[topaz sundial]

Basically my point is, if you really want. Yes you can learn some abstract algebra first, but linear algebra is often more graspable to a first time learner compares to abstract algebra.

[strong stone]

Wait a minute

Isnt linar algebra just y=mx+b type stuff?

[topaz sundial]

No.

[strong stone]

..

[topaz sundial]

It deals with an abstract concept called a vector space, which mimics many of the properties of vectors over R^n

[strong stone]

uhhh

interesting

[topaz sundial]

If you don't know what those are, id suggest learning R^n vector algebra, and build from there.

[strong stone]

wait so

i have to learn vector algebra to even learn linear alhebra

[mild laurel]

There's a nice way to describe all the automorphisms of Q(z)

[topaz sundial]

Yes phi(z)=z^k, where gcd(k,n)=1

[strong stone]

Wow your right ig this is complicate type stuff

Imma head ocer to #linear-algebra to see what their up to

[topaz sundial]

So if phi an automorphism of Q(z) fixing Q, and f is one fixing Q(y) and g is one of Q(y) fixing Q
Then phi(z)=z^k=f(g(z))
So then g(z)=z^-k, since f(z)=z^-1

Wait that's if f(z)=z^-1 but that's not gautanteed 😤

Wait if g(z)=z^k, then depending on choice of f, that will depict both phi(z)=z^k and phi(z)=z^-k

Which would have the property that the set of all g is half of the set of all phi, like I need.

But since I'm working from Q(y) to Q on g, it's fixing Q and sending y to its other roots. So g(y)=z^k+z^-k
Or in other words g(2cos(2pi/n))=2cos(2kpi/n)

So then the minimal polynomial of y over Q should just be F_y= product_{gcd(k,n)=1} (x-2cos(2kpi/n)

This feels wrong tho, cz then Fy has degree of Euler phi function (n),
Which it should have half that, Ugh.
If I put the limitation that k≤n/2 is that enough?

[topaz sundial]

Pretty positive with that limitation we are good.

If anyone wants to confirm please tag me so I see.

[leaden finch]

can someone help me with a

i need help and i dont get it . iF someone can explain it to me in steps it would be great !

[chilly ocean]

I think its LCM of all elements

so order of 1 is 2, same with 4. Order of 2,3 and 5 is 3 so the entire thing has order 6 as you can check manually

[leaden finch]

how do i check it ?

[chilly ocean]

Do you understand what that thing ina) means?

and definition of order?

[leaden finch]

its an infinite order right ?

[chilly ocean]

no

Do you know how this element squared looks like?

[leaden finch]

hmm no D:

[chilly ocean]

thats the problem, you dont understand what it says

think of this element as a function described f(1)=4, f(2)=3, ... f(5)=2. If you 'square' it by doing f(f(x)) you will get a different element,

You want to find the number of times you need to do f(f(f.... such that each element from 1 to 5 stays on the same place

not sure if that explains anything, probably should let others help since im a bit scuffed today

[sharp sonnet]

you should probably just re-read the definition of what S_5 actually is, how an element in it looks and what it does

i.e. this thing here is a bijective function from {1, 2, 3, 4, 5} into itself

and the question is asking how often you have to compose this function with itself to get the identity function

[chilly ocean]

Quick question:\

Let $a,b \in \mathbb{Z}^+$ and let $d = \text{gcd}(a,b)$ and $m = \text{lcm}(a,b).$ In the group $G = \mathbb{Z}_a \times \mathbb{Z}_b,$ let $M$ be the subgroup generated by $(1,1).$ Find a cyclic subgroup $H$ of $G$ with $G = M\oplus H.$

[cloud walrus]

wstayman:

[woven delta]

@chilly ocean what is the order of M?

[chilly ocean]

@woven delta |M| = m

[woven delta]

Okay so what do we know about a relationship between a, b, m, and d?

[chilly ocean]

We know that $\mathbb{Z}_a \times \mathbb{Z}_b \simeq \mathbb{Z}_m \times \mathbb{Z}_d.$

[cloud walrus]

wstayman:

[woven delta]

Lol no we don't

[chilly ocean]

We

[woven delta]

Idk actually maybe we do

We know what ab=md, sure

[chilly ocean]

We do, we've shown before that this is true (in class)

[woven delta]

Oh okay

Well if you've shown that before you shouldn't have too much trouble

[chilly ocean]

I know we need |H| = d. And I think the subgroup generated by (1,0) is the answer. My problem is in proving it

[woven delta]

Okay so you're actually wrong

[chilly ocean]

yeah I'm lost

[woven delta]

(1,0) does not have order d

It has order a

[chilly ocean]

what I meant to say was the group ${(1,0),(2,1),(3,2),....}.$ Not sure if it's even a subgroup, and I am unsure of how to denote it

[cloud walrus]

wstayman:

[woven delta]

It's not

It's a coset

It is the coset (1,0)M

[chilly ocean]

Right, I see that now.

[woven delta]

And it has the same order as M

But anyway what if you divide out the gcd so to speak

Like consider (a/gcd(a,b), b/gcd(a,b))

What is the order of that

[chilly ocean]

The order would be their least common multiple... so d?

[woven delta]

No

Try again

Oh actually you were right

You just used the wrong words

[chilly ocean]

when d = gcd(a,b) right?

[woven delta]

Yeah

Well the order is a divisor of the gcd

And you can prove that it has to be the gcd

Then you have to prove that the subgroup generated by that element does not intersect M

But once you show that you are done

@chilly ocean

[chilly ocean]

Okay thanks I understand how the proof will follow now. How did you come up (a/gcd(a,b),b/gcd(a,b)) as the generator for the subgroup?

[woven delta]

Just asking myself what element will have order gcd(a,b)

[chilly ocean]

Oh true, I was confused about that part as I was still trying to use a coset. Thanks for the help!

[ornate arch]

How can I use this proof (part b) https://faculty.math.illinois.edu/~hildebr/347.summer19/induction2sol.pdf to prove Cassini's identity?

[sharp sonnet]

apply determinant

[ornate arch]

@sharp sonnet wdu mean

i think i got it

[sharp sonnet]

if you apply determinant to both matrices you instantly get cassini's identity

[ornate arch]

yeah just realized

thanks 😄

one follow up q, how do i go from there to show that F_mF_n + F_m−1 F_n−1 = F_m+n−1 is true?

[obsidian zealot]

how do if this is the multiplication table of a group

[mild laurel]

Check the axioms, is there an identity element? Do all elements have an inverse?

[obsidian zealot]

i think a is the identity element since a * a gives me back a

[mild laurel]

Is that all an identity element must satisfy?

[ornate arch]

How can I prove this? Lambda is the golden ratio for fibonacci

[bleak abyss]

Most of this is just a computation

Like if you just use known formulas for matrix multiplication and inverses, you'll automatically get

[cloud walrus]

Daminark:

[bleak abyss]

Then the trick is recognizing that the top left entry is 1, this is where you need the input that gamma is the golden ratio.

Also tbh this would be better suited for #linear-algebra

[ornate arch]

so i can just 'prove' it computationally then

and my bad, ill post on there next time

[strong stone]

What is the notation for subgroup.

[golden pasture]

i usually see $H\subset G$ or $H<G$ for any subgroup

[cloud walrus]

Ariana:

[golden pasture]

$H\triangleleft G$ for normal subgrp

[cloud walrus]

Ariana:

[topaz solar]

(with a line underneath occasionally as well)

[golden pasture]

shhh that doesnt exist

2<2

[somber bramble]

you propagate bad notation

I have very strong opinions on this matter

[chilly ocean]

Let $f(x) \in F[x]$ with splitting field $K$ over $F.$ Define the multiplicity of the root $u \in K$ to be the integer $n$ such that $f(x) = (x-u)^ng(x)$ where $(x-u) \nmid g(x)$ in $K[x].$ Prove that if $f$ is irreducible over $F,$ then every root $u \in K$ of $f$ has the same multiplicity.

[cloud walrus]

wstayman:

[mild laurel]

A root has multiplicity n if it is also the root of the (n-1)th derivative

[chilly ocean]

I hadn't thought of that. I see how it's true, but how would I implement that into the proof? Should I start by supposing f has two roots, a,b in K with different multiplicities? Or is there a more direct way to prove this?

[golden pasture]

consider (f,f')

if theres repeated root this gives you something

[chilly ocean]

does (f,f') denote gcd(f,f')?

[golden pasture]

oh uea

more bad notation from me

[mild laurel]

it's good notation

[golden pasture]

ily

[chilly ocean]

Off topic, but if (a,b) is used for gcd(a,b), what is the shorthand notation lcm(a,b)

[golden pasture]

[a,b]

[mild laurel]

(a) \cap (b)

[chilly ocean]

$\cap$

[cloud walrus]

tet:

[mild laurel]

Which mirrors the main reason why we use (a,b)

[leaden finch]

can someone help me with this one

[sharp sonnet]

what is the problem?

[leaden finch]

i dont know how to start it

[sharp sonnet]

then look up the definitions

this is a straightforward computation

[simple agate]

what would be the best way to approach this?

by showing that (x^2 - 1) is maximal in k[x] hence the quotient is a field? or by showing that quotient is isomorphic to k

[mild laurel]

The first thing is the right idea

[simple agate]

ty

oh wait I was operating on the assumption that was a field. is it not?

[oblique river]

(that's part of what you need to figure out)

[simple agate]

since k field => k[x] PID => (x^2 - 1) is maximal iff it's irred

but it's reducible as (x+1)(x-1) and there's another iff about maximal ideal <=> k[x]/(x^2 - 1) field

sorry I don't really have any lecture notes for this class and books are so dense and difficult to reference for problems like this

[oblique river]

seems like you've figured it out

[simple agate]

thanks

[oblique river]

👍

[chilly ocean]

Hi guys, I've been struggling with this problem for several hours now and I was wondering if I could ask for a hint?

Let $K/F$ be the splitting field for a separable poynomial $f(x) \in F[x]$, let $\theta \in K$ be a root of $f(x)$, and assume $Gal(K/F) \cong S_n$ where $n \geq 5$ where $n$ is the degree of $f(x)$.

[cloud walrus]

Azu:

[chilly ocean]

Prove that there does not exist a radical extension $E/F$ such that $\theta \in E$

[cloud walrus]

Azu:

[chilly ocean]

Also, $ch(F) = 0$

[cloud walrus]

Azu:

[chilly ocean]

Further, I already proved, as an earlier part of the problem, the following two sub-problems:

1. $F(\theta)/F$ is not Galois

[cloud walrus]

Azu:

[chilly ocean]

2. For $L/F$ a Galois extension, if $\theta \in L$, then $K$ is isomorphic to a subfield of $L$

[cloud walrus]

Azu:

[chilly ocean]

Anybody? 😦

[wind steeple]

mh

If theta can be expressed with radicals, then every of its conjugates does

if f is irreductible it's clear that there's a problem by Galois correspondance

f would be soluble by radicales and Gal(K/F) would be soluble

it is not true for Sn, n >= 5

Maybe you can show that f is in fact irreductible to conclude

Idk if it's true but I think it is

yeah it's true, you can have information on the cardinal of Gal(K/F) if f is reductible

[lime skiff]

I don't really get how to do the following problem: Let G be a simple group. If there is a subgroup H with [G:H] \geq 2, then G is isomorphic to a subgroup of Sym(G/H).

[wind steeple]

look a the translation action of G on G/H

[lime skiff]

so just like Cayley's theorem?

[wind steeple]

Idk what do you mean by cayley theorem

[lime skiff]

every group is isomorphic to a subgroup of its symmetric group

[wind steeple]

for me cayley's theorem is that you can inject G in S(G)

[lime skiff]

yeah

that

[wind steeple]

yes, it is not exactly the Cayley theorem

[lime skiff]

got it

thanks

[junior meadow]

Why in each ordered field 1 > 0 and a^2 > 0 for each a?

[scarlet estuary]

1 > 0 is usually a matter of definition

you could define 1 < 0, and then you'd have the same structure except with all inequalities reversed

so we make 1 > 0 by convention

as for a^2 > 0, try and prove it!

[junior meadow]

That's what I am trying to do

[scarlet estuary]

[though note that its actually >=]

well, idk how you're defining an ordered field

theres a bunch of different but equivalent ways to define it

[junior meadow]

I know only total order definition

1. $if a < b then a + c < b + c$
2. $if 0 < a and 0 < b then 0 < a⋅b$

[cloud walrus]

Dark Archon:

[junior meadow]

Oops.

Okay, for positive a it's going from definition

[hot lake]

isn't 1 > 0 a consequence of a² > 0 for each nonzero a ?

[junior meadow]

For negative a I think it can be said from the fact that $-a * -a = (-1) * a * (-1) * a = 1 * a *a$

[cloud walrus]

Dark Archon:

[lime skiff]

depending on how much you know, it boils down to showing (-1)^2 = 1

[junior meadow]

But yeah, this

[lime skiff]

if that's something you can safely assume, then good

pretty sure I had to show this in my algebra class somewhere

[hot lake]

I think you just need to prove the other usual rules from your two axioms

[lime skiff]

you can do so via the distributive property

or some other axiom somewhere

[junior meadow]

Okay. If -1 is inverse to 1, that means that 1 is inverse to -1, via commutativity of addition. That means that -(-1) = 1. But how to connect that to fact that (-1) * (-1) = 1 ?

[scarlet estuary]

why prove that its equal to 1? you just have to prove that its positive

oh wait, people were taking the proof in that direction

nvm

i guess that works

[stone fulcrum]

First instead prove that anything times 0 is 0

(-1)x = -x follows from that

[scarlet estuary]

^

this is what i had in mind

[junior meadow]

Huh? How those two are connected?

[scarlet estuary]

or perhaps a more intuitive approach

you want to show that x + (-1)x = 0, right?

[junior meadow]

I want to show that (-1) * (-1) = 0 from field axioms

[scarlet estuary]

well yes

wait

what

do you mean >

[junior meadow]

(-a) * (-a) = (-1) * (a) * (-1) * (a). Because a*a for a>0 is positive by definition, showing that (-1) * (-1) = 1 proves statement automatically

[scarlet estuary]

so you mean >

okay

[junior meadow]

Well yeah, > 0 works too

[scarlet estuary]

im referring to this

anyway

you want to show, basically, that (-1)(-1) + (-1) = 0, right?

since [by uniqueness of additive inverses] that implies (-1)(-1) = 1

so apply distributivity et voila

[you do need that -1 \neq 0, but this should be fairly obvious]

[it follows from 1 \neq 0]

[junior meadow]

Yeah, I've found proof that a * 0 = 0 for all a. From that one can prove that (-1) * (-1) = 1, and then formula for squares is given. And then I can clearly see that complex field cannot be made ordered field.

[lime skiff]

err this might seem obvious but I'm stuck on something simple

let I be a principal ideal in a commutative unital ring R

let I = (a)

then I is prime iff a is prime in R

I know that if bc is in (a), then bc = a^k = aa^k-1

[chilly ocean]

nooo

[lime skiff]

but how does it follow that either b or c is in (a)

[chilly ocean]

bc is in (a) iff bc = ax for some x

[lime skiff]

oh dang it

you're right

[chilly ocean]

i.e. a divides bc

if a is prime that means a divides b or c

[lime skiff]

okay that's much better

I was getting confused because of my prof's notation <a>

which made me think of cyclic groups

wew

thanks

[chilly ocean]

also you don't need to assume the ring is unital here

[ornate arch]

how can i show this not computationally

[chilly ocean]

two quadratic polynomials are equal if they agree at 3 distinct points

plug in a,b,c

[ornate arch]

is there a way to not plug in random numbrs and show they're equal

[lime skiff]

unity came from earlier

[chilly ocean]

yes, I just told you how

[ornate arch]

the problem says theres a non computational proof

[lime skiff]

this is an eight part exercise

that was part 5

[chilly ocean]

I just gave you the proof

[ornate arch]

how do we just know that

[chilly ocean]

know what?

[ornate arch]

"two quadratic polynomials are equal if they agree at 3 distinct points"

[chilly ocean]

polynomial division

[lime skiff]

what's the degree of f-g?

[chilly ocean]

n'th degree polynomial has at most n roots

[lime skiff]

agree means f-g has a zero there

[chilly ocean]

Does there exist software to compute the splitting field of polynomials of degree >=5 and determine if Galois (separable) or not?

cocalc maybe?

[mild laurel]

Sage can do this

[lime skiff]

if f-g has more zeros than its degree in this case, it means f=g

[mild laurel]

sage = cocalc

[chilly ocean]

superset no?

[mild laurel]

cocalc is just a web server version of sage

[chilly ocean]

oh ok

Excellent

Can it handle polynomials over $F_p$ rather than $\mathbb{Q}$?

[cloud walrus]

Azu:

[chilly ocean]

probably

it can handle a lot

[mild laurel]

yeah it can

[chilly ocean]

how steep is the learning curve?

[mild laurel]

its basically a python package

[chilly ocean]

ok

How much work would it be to write the code you need to factor a polynomial into irreducibles in $F_p[x]$?

[cloud walrus]

Azu:

[chilly ocean]

very little

just look it up

[mild laurel]

sage: f = (x^3 -1)^2 - (x^2 - 1)^2
sage: f
x^6 + 96*x^4 + 95*x^3 + 2*x^2
sage: f.factor()
(x + 23) * (x + 76) * x^2 * (x + 96)^2

its this easy

GF(97) is how sage does F_97

[chilly ocean]

wow

what's F_97?

[mild laurel]

Uh

F_p

when p = 97 lmao

[chilly ocean]

oh

duhhh

my bad thx

[mild laurel]

I forgot whether 97 was prime or not too its fine

[chilly ocean]

so those are the irreducible factors of f(x)?

[mild laurel]

yeah

[chilly ocean]

@mild laurel you're a hero

[mild laurel]

Sage is super nice for all this stuff

(I'm biased because I've developed a bit of sage), but its genuinely good I think

from there, you can just do

f.splitting_field('y') to get the splitting field

Over F_p, splitting fields are a bit boring, it'll just tell you the order of the splitting field

since all fields of order p^n will be isomorphic

[chilly ocean]

right

[mild laurel]

but this works over Q too, except I think you just write f.splitting_field()

http://doc.sagemath.org/html/en/reference/number_fields/sage/rings/number_field/splitting_field.html

you should probably just read this

[chilly ocean]

@mild laurel you're the best, thanks so much

[mild laurel]

anytime

it'll probably be easiest for you just to use cocalc.com too

installing sage takes a bit

[chilly ocean]

got it

[mild laurel]

cocalc.com runs pretty slow though, so if you need to do a big computation, it'll be better to download sage

[chilly ocean]

it's just a degree 5 poly in this case so hopefully not too bad

[mild laurel]

yeah, sage is pretty efficient overall even though it runs python

[chilly ocean]

uwu

@mild laurel does it use a numerical algorithm to find the splitting fields for poly of degreee >=5?

[mild laurel]

https://github.com/sagemath/sage/blob/develop/src/sage/rings/number_field/splitting_field.py

I gotta go eat, but the method is in here

Looks long, so I'll look at it later

[ornate arch]

If K is a field, and we consider the space W ofd polynomials of degree less than n over K. We let x1,x2..., xn be distinct elements of K. How can I find a polynomial Pi for all 1<=i<=n which takes value 1 at xi and 0 at xj for j=/i

[chilly ocean]

[lime skiff]

might as well ask here as well

can anyone help me show that Heis(p) is solvable?

the heisenberg group on F_p

uppertriang matrices with diag elements = 1

[chilly ocean]

@ornate arch look at the last question you asked

try to see the structure int here

@lime skiff each iteration of [,] eliminates a diagonal

because the elements directly above the diagonal add

when you multiply matrices

[lime skiff]

err what's [,] exactly

[chilly ocean]

commutator

[a,b]=ab-ba

wait no

lmao

[lime skiff]

oh

aba^-1b^-1 you mean

[chilly ocean]

(ab)(ab)^-1

yeah

same argument works i think

[lime skiff]

so I look at the series Heis(p) < Heis(p)' < Heis(p)''

?

etc

[chilly ocean]

exactly

[lime skiff]

I see!

that makes a lot of sense

[chilly ocean]

each iteration eliminates one diagonal

[lime skiff]

OH

that's what you meant by

ok

I proved earlier that it's solvable iff the derived series ends

at some G_n = {1}

[chilly ocean]

yeah yeah

[lime skiff]

okay I see now

it all makes sense

[chilly ocean]

whats the definition of solvable again

exists composition series where all the quotients are abelian?

[lime skiff]

yep

[chilly ocean]

ok yeah

[ornate arch]

Lagrange Interpolation?

[chilly ocean]

yeah

[lime skiff]

also I was wondering

how do you see directly that quotienting will just remove a diagonal?

do you have to actually compute [a,b] for a,b in Heis(p)

that's kinda tedious but easy I guess

or is there another way you see this immediately?

[latent anvil]

Let M be a fg module over k[x, y] (or any polynomial ring over a field/nice ring). Is it true that M_p being free for all p implies M is free? How about projective?

[ornate arch]

@chilly ocean i have one more follow up q about the same problem, how can I use that Pi to prove that the n polynomials Pi from the previous problem form a basis of W?

[ornate arch]

im assuming i show it is linearly independenet and that Pi spans w

[lime skiff]

Okay here's a stupid question

What's Z[x]/(x-2)

It's Z[2] = Z?

Okay it is, good

[mild laurel]

Theres a homomorphism from Z[x] -> Z that sends f to f(2)

the kernel of this homomorphism is exactly (x-2)

[obsidian zealot]

What is a good example where H is a subgroup of G, and |H| = 3, and G is not abelian

[solemn rain]

G = S_3

H = <(123)>

@obsidian zealot

[obsidian zealot]

that seems to work

How could i find |G/H| = 4

[solemn rain]

u want |H| = 3

|G/H| = |G|/|H|

|G|/3 = 4

idk there exists such group

whata

no

u want a group of order 12

which is non abelian

D_24

[obsidian zealot]

is D_24 non abelian

[solemn rain]

yes

[obsidian zealot]

and does the condition |G/H| = 4 satisfy

[solemn rain]

i define the dihedral group as D_2n of order n

the order is 12

oh wait

[obsidian zealot]

this is the example i have to prove

[solemn rain]

okaay well

i suppose u know not alot of nonaeblian groups

can u give me an example

we work through it together

[obsidian zealot]

a non abelian means its not symmetric across the diagonal

[solemn rain]

u basically want a nonabelian group that is of order 12

and has a subgroup of order 3

3 is a prime that divides 12 so it has an element of that order

so u can generate the subgroup of that order easily

we just need to find the element

so for example D_24

try to find an element of order 3 in that group

then the subgroup generated by it is your 'H'

and it does satisify |G/H| = 4

as 12/3 =4

got it?

[obsidian zealot]

oh i got it

[solemn rain]

i think the only noabelian groups ik are like

the matrix groups

symmetric groups

ddihedrals

and Q_8

cant think of something else lmao

[obsidian zealot]

dihedrals are the ones i can think of

[solemn rain]

okay

can you find an element of order 3

?

[obsidian zealot]

yes im working on it in a paper

[solemn rain]

good luck

[obsidian zealot]

thank you

[solemn rain]

u still stuck?

[obsidian zealot]

Yes, but I also know that alternating group A4 is non abelian

[solemn rain]

yea

[obsidian zealot]

and that it has order 4

order 12

sorry not 4

[solemn rain]

yea

can u find an element of order 3

[obsidian zealot]

for A4 im not sure but I can try to see

[solemn rain]

how do the lements in A4 look like

[obsidian zealot]

(1), (1, 2,3), (1,3, 2), (1, 2,4), (1,4,2), (1,3,4), (1,4,3), (2,3,4), (2,4,3), (1, 2)(3,4), (1,3)(2,4), (1,4)(2,3)

[solemn rain]

damn

[obsidian zealot]

those are the elements of A4

[solemn rain]

okay]

can u find an elmeent of order 3

[obsidian zealot]

(1,2,3)

is that order 3

[solemn rain]

yes

the order of a permutation is the length fo the cycle

of*

[obsidian zealot]

oh i didnt know

[solemn rain]

can u construct a group

a subgropu*

?

[obsidian zealot]

Not really sure but let me try

[solemn rain]

just generate the cyclic subgroup

[obsidian zealot]

like the permutation of the cyclic

[solemn rain]

<(123)>

|<x>| = |x|

<x> is alwys a subgroup of a group G for any xi n G

if u dont know

[obsidian zealot]

oh I see

[solemn rain]

<x> = {x^n | n is an integer}

[obsidian zealot]

so <(123)> is a H in my example since its a subgroup of G

[solemn rain]

yea

|G/H| = |G|/|H| = 12/3 = 4

[obsidian zealot]

so |H| = 3 the order of <(123)>

[solemn rain]

yea

[obsidian zealot]

oh i get it

Now i just need to find the left cosets in G/H

[solemn rain]

yea

any problems with that?

[obsidian zealot]

I kind of know I can try working them out here

[solemn rain]

sure

[obsidian zealot]

let me look at my notes

gH = {gh: h in H}
G = A4, H = {(1), (1, 2,3), (1,3, 2), (1, 2,4), (1,4,2), (1,3,4), (1,4,3), (2,3,4), (2,4,3), (1, 2)(3,4), (1,3)(2,4), (1,4)(2,3)}

12H =

[solemn rain]

how

H must have order 3

how did u get these

H =<(123)>

[obsidian zealot]

oh yea, i forgot its not the elements of A4

[solemn rain]

gH = {gh: h in H } g in G

[obsidian zealot]

i know that the first one if the identity

and (123) is the other one

is it H = {e, (123), 132)}

H(123) = { e, (123)(123), (123)(132)}

[solemn rain]

gH = H iff g is in H

by just closure

[obsidian zealot]

oh

[obsidian zealot]

Suppose we have H space subset of G, and for all h, h' in H, we have h to the power of minus 1 end exponent h apostrophe space element of H. Then H is a subgroup of G. is this true or false

[outer meadow]

huh?

[steep hull]

I’m assuming H is a subset of G. Then set h=h’ to see that e is contained in H, so H contains the identity. But then set h’=e to see that given some r, r^(-1) is also in H, so inverses exist in H. Finally, given r,s in H, (r^(-1))^(-1)s=rs is in H, so it has closure. (Associativity is obvious since H is a subset of a group G).

[leaden finch]

which of yall know group theory? LOL

[solemn rain]

idk about me

but if u have ap roblem just aks

a problem* or a question just ask

[fading wagon]

@leaden finch juat ask

[solemn hollow]

juts ask

[shrewd wasp]

what does the ring $\mathbb Z [i]$ mean?

[cloud walrus]

Shipreck:

[chilly ocean]

Gaussian integers

Complex numbers of the form a+bi, where a and b are integers

[shrewd wasp]

okay cool

thank u

[solemn hollow]

I realised you can think of it as Z[x] but with i instead of x. not sure this is the meaning in general tho

[buoyant relic]

ring of polynomials in the indeterminate x over the ring Z

[chilly ocean]

Z[x]/(x^2+ 1)

[buoyant relic]

wait whats that

[chilly ocean]

the ring Z[x] quotiented by the ideal (x^2+1)

[random crag]

its basically like mods

[solemn hollow]

which is isomorphic to Z[i] right

[chilly ocean]

personally I prefer Z[x]/(x^2 - 1)

[random crag]

multiples of x^2+1 are treated as 0

theyre isomorphc

doesnt really matter

[chilly ocean]

lol sonja

[wind steeple]

Z[x]/x²-1 is Z

[random crag]

oh well not -1

[chilly ocean]

i don't think so

Z[x]/x²-1 is Z

wrong

it would be some split-complex thing

[buoyant relic]

what r quotient groups even

[chilly ocean]

rings*

[random crag]

theyre a way to construct new rins from old ones

[wind steeple]

Z²

[buoyant relic]

DID I STUTTER @chilly ocean ???

[chilly ocean]

you identify all numbers that differ by the thing you're quotienting by

[buoyant relic]

i said group

[chilly ocean]

yeah it may be ZxZ

[wind steeple]

it doesn't have the algebraic properties of Z[i]

[chilly ocean]

ZxZ is isomorphic to Z (set-theoretically speaking)

no

i don't think Z[x]/(x^2-1) is isomorphic to Z^2 either

if you ignore multiplication, they're isomorphic as abelian groups

not as rings

[wind steeple]

ax+b -> (a,b)

[chilly ocean]

that's not a ring homomorphism

not surjective

it is a set-theoretic isomorphism though

(a.k.a. a bijection)

its not calle set theoretic isomoprhism

lol

[wind steeple]

but it not Z[i]

[chilly ocean]

yeah it isn't

it would be Z[j] i guess

whats j

j^2 = 1

but j != 1

k

lol

try to wrap your head around that!

:^)

wrapped

how do i unwrap it now

[wind steeple]

(a,b)(c,d) = (ac+bd,ad+bc)

[chilly ocean]

I mean wouldnt it be isomoprhic to Zx/x-1 x Zx/x+1

that's just ZxZ

so no

lol

yeah but why wouldnt it be iso to ZxZ

more elements that square to 1 \🙂

are there tho?

since that's a sentence you can formulate in ring-theoretic language, the rings are not isomorphic

sure

how many are in z[x]/x^2-1

4

oh wait

you're right

smh

sorry

I think CRT might say it is iso to ZxZ

my brain has been hijacked by a low IQ goblin for a moment

excuse me

(x-1) and (x+1) are not relatively prime though

arent they tho?

can you express 1 in terms of them?

no

why

[wind steeple]

j² = 1 isn't it ?

[chilly ocean]

yes

yes

what are the elements x such that x^2 = x in Z[j]?