groups-rings-fields

[oblique river]

figure out what Z[i]/(2 - i) is

meaning: Z[i]/(2-i) is isomorphic to Z/nZ for some n. Which n?

Hint: Don't forget that Z[i] = Z[x]/(x^2 + 1)

[ripe crest]

Oooh - okay, I'll work on it and let you know what I get! Thanks!

[oblique river]

I'll be back in a bit

and can give you some tips

[ripe crest]

A preliminary question... in Z[X]/(X^2+1), X^2 (from Z[X]) gets sent to 1, and X^2 + 1 gets sent to 0. So in the quotient, X^2 = 1, and X^2 +.1 = 0. So 1 + 1 = 0? What am I doing wrong

[mild laurel]

X^2 = -1 in the quotient

[ripe crest]

So X^2 does not get sent to 1 in the quotient?

[mild laurel]

no

[ripe crest]

Ah, that makes sense - I was doing the division incorrectly

Ahh C[X]/(X-a), sends X to a by euclidean division .. for some reason, for a few moments I thought it got sent to -a xD

[oblique river]

that negative sign will get you haha

in the quotient Z[x] --> Z[i]

x gets sent to i

and so x^2 gets sent to i^2 = -1

[ripe crest]

right

So far, I've been looking at Z[X] and figuring out it's maximal ideals

I found that (p, X) is maximal when p is prime

and, the maximal ideals of Z[X]/(X^2 + 1) are of the form:

The image of (X^2+1, X, p) where p is prime. Which would be (X, p) in the quotient

Idk if those are all of them, but they are maximal in the Z[i]

[oblique river]

careful. (x^2 + 1, x, p) is just the unit ideal

[ripe crest]

Well, I guess (p, i) in Z[i]

[oblique river]

i is a unit in Z[i]

because i^4 = 1

if an ideal contains i, it will contain 1 and be hte whole ring

[ripe crest]

ah fk

right, can't have units

in ideals

[oblique river]

it's ok :) as it turns out, it's actually kind of a subtle question what hte maximal ideals of Z[i] are

the trick for trying to understand Z[i]/(2-i) is to use the [whatever-th] isomorphism theorem and realize that Z[i] = Z[x]/(x^2 + 1) with x --> i in the quotient, and therefore

Z[i]/(2-i)
= (Z[x]/(x^2 + 1))/(2-i)
(remember i is the image of x, so (2-i) is the image of the ideal (2-x))
= Z[x]/(x^2 + 1, 2-x)

and now just quotient in the other order first

Quotient by 2-x. Where does that send x to? Where does that send x^2 + 1?

[ripe crest]

wait

Z[i]/(2-i) = Z[X]/(X^2+1, 2-X) ?

Sorry - I sent that before you typed that up (slow internet)

oooh that's a nice approach

lemme see

Z[X]/(2-X) is isomorphic to Z

sends X to 2

sends X^2+1 to 5

[oblique river]

Yes!

[ripe crest]

so we have Z/(6)

ie, Z/6Z

[oblique river]

6? 😳

[ripe crest]

uhhh wait

5

lmaooo

I already added 1

[oblique river]

hahaha yep

I figured tha'ts what happened

but yeah -- that's it!

Z[x]/(2-i) is isomorphic to Z/5Z

[ripe crest]

haha xD

That's actually super cool

[oblique river]

I agree

I remember doing that problem in my first class on rings

and it really stuck with me

so now I keep it in my back pocket

[mild laurel]

I remember you were asking me about algebraic NT yesterday

[ripe crest]

I'm going to go through it one more time to make sure I really get it

Yup!

[mild laurel]

one of the things the theory allows you to do, is to just note that you can calculate (2 - i)(2 + i) = 5 and that's the cardinality of the quotient

[ripe crest]

Is it true though that the (p, X) are maximal ideals of Z[X]?

[oblique river]

those are maximal ideals of Z[x], but there are lots more

[ripe crest]

huh. that is quite surprising (the cardinality calculation)

[oblique river]

For example, Z[x]/(p, x+1)

[ripe crest]

ah yes

[oblique river]

and Z[x]/(3, x^2 + 1) will end up being a field (although that takes a little more work and knowledge of polynomials over Z/pZ)

[cloud walrus]

All functions are continuous:

[oblique river]

just to add on to what Zoph said, it is in fact true that the cardinality of Z[x]/(a+bi) will be equal to a^2 + b^2 which you can also write as (a+bi)(a-bi)

@tribal pasture why don't you try to prove that? Show that it is a homomorphism and that it's both injective and surjective

[ripe crest]

and Z[x]/(3, x^2 + 1) will end up being a field (although that takes a little more work and knowledge of polynomials over Z/pZ)
@oblique river I actually did a question about polynomials over Z/pZ and found that there are p(p-1)/2 irreducible monic polynomials of degree 2 in F_p and that there exists a field with cardinality p^2. That was a good problem

[oblique river]

Oh cool

yeah the relevant part that you need for the thing I wrote is knowing: "when is x^2 + 1 irreducible over F_p"

[ripe crest]

We have F_3[X]/(X^2+1) and X^2+1 is irreducible

yeah

[oblique river]

yes it is in fact irreducible over F_3 (but not over all F_p!)

and so the quotient of a poly ring over a field by an irreducible poly is a [blank]

[ripe crest]

I learned in k[X], k a field, P irreducible, iff (P) prime, iff (P) maximal... but i never really learned the proof - I should probably go over that

field!

[oblique river]

yes -- so (x^2 + 1) is max'l in F_3[x] and therefore (3, x^2 + 1) is max'l in Z[x]

[tribal pasture]

@tribal pasture why don't you try to prove that? Show that it is a homomorphism and that it's both injective and surjective
@oblique river I am trying but I am struck at injectivity, in trying to show f o i1 = f o i2 = 0 implies f = 0

[oblique river]

great. if f o i1 = 0, that means that (f o i1)(a) = 0 for all a in A

i.e. f(a,0) = 0 for all a in A

similarly, f(0,b) = 0 for all b in B

[tribal pasture]

Wait so I can do f(x,y) = f(x,0) +f(0,y)?

[oblique river]

now you want to conclude that f = 0, i.e. that f(a,b) = 0 for all a in A and b in B

what do you mean "you can do"?

that is a true fact (why is it true? what adjective describes f which makes that a true fact?)

and you are allowed to use true facts in your proof

[tribal pasture]

Ah I am just confused a bit on how mod hom works on product of modules

[oblique river]

it's not relevant that it's a product

it's a hom so f(x + y) = f(x) + f(y)

(a,b) = (a,0) + (0,b) -- that's just how modules work (no homs here)

therefore f(a,b) = f(a,0) + f(0,b)

[tribal pasture]

Ah right true true, sorry for a silly quesiton

[oblique river]

ur fine

[tribal pasture]

And surjectivity is true by defining f to be a+b where a in Hom(A,M) and b in Hom (B,M) right?

[oblique river]

yes

[ripe crest]

man this stuff is so cool

let's see if I can do the question I couldn't do on my midterm...

[oblique river]

agreed

[ripe crest]

Showing that F_p[X]/(X^2+1) is an integral domain iff p != 1 (mod 4)

where p is odd

[clear obsidian]

[ripe crest]

okay, I had to look at the solution.. still couldnt get it lol

endomorphism as groups I'm guessing

Well, what does it mean for x to be in the kernel of f?

Right, what are some cases when that happens?

right

When does that 3rd case never happen?

You'd have to show that, but yes I think so

So, if m and n are coprime, what is kernal and image of f_n?

yup!

Okay, so when m and n aren't coprime, what things are in the kernel?

Hmm, wait a second

I'm not sure if that's super easy to answer

Clearly if n is a multiple of m, then the image is 0 and the kernel is everything

brb, but think about using the gcd

well idk if ill be back soon 😛

not sure if the gcd is the right approach, but it seems like a worthy approach

[ripe crest]

take p prime, p > 2, why does F_p[X]/(X^2+1) not being an integral domain imply that X^2+1 has a root in F_p[X].

[oblique river]

it comes down to the theorem you mentioned earlier

in k[x], f(x) irred. <==> (f) prime <==> (f) maximal

remember that R/p integral domain iff p prime

so the question is about x^2 + 1 being irreducible

[ripe crest]

Right, so X^2+1 is not irreducible, I thought about that - but I couldnt see why that implied it has a root

[oblique river]

and this is a cool fact:
if f(x) has degree \leq 3, then f(x) irred if and only if f(x) has a root

[ripe crest]

has degree not equal to 3?

[oblique river]

sorry, I meant less than or euqal to

but then I panicked but then I fixed it

[ripe crest]

ah

That is really cool - huh.

[oblique river]

basically:

[ripe crest]

is this over a general field?

[oblique river]

yep!

if f(x) reducible, factor f(x) = g(x)h(x)

both g and h have positive degree

deg(f) = deg(g) + deg(h)

so if deg(f) is at most 3

either deg(g) or deg(h) = 1

therefore one of them has a root

therefore f has a root

[ripe crest]

g and h have degrees 1 and 2 or 2 and 1

[oblique river]

yep

so one of them must look like (ax - b) which has root b/a

[ripe crest]

a degree 1 polynomial always has a root.... hmmm

[oblique river]

so f(x) has a root iff f(x) irreducible

yep :)

[ripe crest]

yeah - I can see that

[oblique river]

for x^2 + 1, the reason is a little "simpler"

[ripe crest]

sorry, im sitting far away from my router - so I get your messages a little late 😛

[oblique river]

if it's reducible, it factors as (ax+b)(cx+d)

and so it has roots -b/a and -c/a

haha sorry :)

[ripe crest]

So, in Z[X], (aX+b) is prime iff b is a multiple of a?

Hmm, no. (2X + 0) isn't prime

Wanted to look at Z[X] since Z isnt a field

[oblique river]

0 is a multiple of 2

It's still true in Z[x]

what isn't true in Z[x] is that an ideal is prime iff maximal

irreducible iff prime is still true

[ripe crest]

Gotcha

Is the best way to understand Q[X,Y]/(X^2-Y) by euclidean division? I dont know the maximal ideals of Q[X,Y]

[oblique river]

in that case

quotienting by x^2 - y means "y = x^2"

(rearrange y - x^2 = 0)

so you can just replace y = x^2

[ripe crest]

right - in the quotient, Y = X^2

[oblique river]

which is to say that Q[x,y] / (y = x^2) is canonically isomorphic to Q[x]

via the map Q[x,y] -- Q[x] where x --> x and y --> x^2

[ripe crest]

" Q[x,y] / (y = x^2) " actually makes so much sense wtf. I can see this as looking at each parabola as a single point.

[oblique river]

:D

this is kinda what I wanted to communicate earlier, like instead of thinking "k[x]/(x-a)", think "k[x]/(x = a)"

[ripe crest]

not sure if that image is totally right though... hmmm

this is kinda what I wanted to communicate earlier, like instead of thinking "k[x]/(x-a)", think "k[x]/(x = a)"
Right!

[oblique river]

that's great btw :) that was one of the single most useful realizations I had about quotient rings

if not the most useful

so it's great that you're into it haha

[ripe crest]

Is the image right though? Of parabolas that are vertical translations of y = x^2?

[oblique river]

I'm not sure if parabolas are the best way to think abou tit

what "map" do you ahve in mind?

that you're talking about teh image of?

[ripe crest]

Oh, I meant a mental picture. Like for R/Z we get a circle, R^2/R as vector spaces give us R because we have a bunch of horizontal lines that we treat as points

be back in a bit

[oblique river]

kk

[ripe crest]

yeah my picture is incorrect

Then for Q[X.Y]/(Y^2-X^2), we have Y = X or Y = -X. Is this not also isomorphic to Q[X] because we can send x -> x, and y -> x.

Hmm, but that doesn't take care of the Y = -X part..

[ripe crest]

Ah, the solution says to see that Y^2-X^2 = (Y-X)(Y+X) = 0, but neither of (Y-X) or (Y+X) are in (Y^2-X^2), so Q[X,Y]/(Y^2-X^2) is not an integral domain, while Q[X,Y]/(X^2-Y) is an integral domain. So they are not isomorphic

[oblique river]

yes that's correct

[ripe crest]

I think that's enough for one day

Thanks for all the help today! I learned a lot!

Finally getting a handle on things

[oblique river]

good to hear!

gn

[solemn rain]

prove fermats little theorem by using lagranges theorem :

in the multiplicative group (Z/pZ)

the order of the group is p-1

for any a , a^(p-1) = 1 ( lagrange )

---> a^p=a

in mod p

is that right?

uh oh

2 typing alot

prob wrong X

[smoky cypress]

Lol a^(p-1) = 1 is not lagrange, but you can deduce that from lagrange

[solemn rain]

D

okay

[upper pivot]

this is almost right

[solemn rain]

fuck

[upper pivot]

no the correction is literally trivial

[smoky cypress]

Yeah

[solemn rain]

for any a

?

[upper pivot]

so basically you showed this is true for all a in Z/pZ* right

[solemn rain]

yea

[upper pivot]

is 0 in the multiplicative group

[solemn rain]

no

i ahve to show

for 0

yea

fuck

is it hard?

[upper pivot]

yeah lmao

[solemn rain]

nvm

[upper pivot]

idk is 0^p=0 hard to prove

[solemn rain]

XD

yea

i gotchu

tysm

[smoky cypress]

ok so um

[solemn rain]

idk any NT

is it hard to prove this without algebra?

would be cool if it is

[upper pivot]

theres a nice proof of this

[mild laurel]

its harder than what you just did

[smoky cypress]

Suppose the subgroup generated by a (i.e. 1, a^1, a^2, ....) has order b

[upper pivot]

that is elementaryt

[smoky cypress]

Then

b|p-1

[mild laurel]

congratulations

[solemn rain]

yea i got that

[upper pivot]

an equivalent statement of lagrange is that x^|G| = 1 for x in G

[smoky cypress]

Is that equivalent?

Interesting

[solemn rain]

yea i meant that

well it my book

its stated as

corrrolarly

coralrarly

corrolarly

[smoky cypress]

[solemn rain]

corollary

[smoky cypress]

Corollary

[solemn rain]

i think a grader would like whoever's answer more

tysm

all

[smoky cypress]

Lol

Imagine doing math and thinking about the graders

[upper pivot]

i can write out a proof of FLT using elementary stuff

[solemn rain]

yea i even suck

at grades

[upper pivot]

if u want

[solemn rain]

i dont know any elementary stuff tbh

and i am abit slow

[woven delta]

Why would you jds?

[upper pivot]

why not

its nice looking

[woven delta]

The proof using Lagranges theorem is super clean

[solemn rain]

i mean the fact that proving this is easeir

is blowing my mind

literally

[upper pivot]

yeah obviously

[solemn rain]

is this what algebraic nt like?

using algebra to do nt?

[upper pivot]

but idk the classic proof is nice too and its nice to know both

[woven delta]

No

[solemn rain]

whats difference

[woven delta]

Lol

[solemn rain]

@upper pivot do it

i want to see it

im slow tho

so no flame please

[woven delta]

Algebraic number theory is the study of algebraic numbers

[solemn rain]

haha

oh

my bad

lmao

[woven delta]

Not number theory using algebra

[solemn rain]

okayu

[mild laurel]

thats a nice way to put it

[woven delta]

Lmao

[upper pivot]

first $(a+b)^p\equiv a^p+b^p \pmod{p}$, then basically this: $(a+\underset{1+\dots+1}{p \text{times}})^p\equiv a^p+\underset{1^p+\dots+1^p}{p \text{times}} \pmod{p}$ FLT follows

ful

i hate latex

brb

[solemn rain]

(a+b)^p = a^p + b^p

i know why but i dont think i can prove that

using binomial theorem ig

[upper pivot]

yep

try it, its easy enough

[solemn rain]

yea

okay so (a+b)^p = a^p + b^p\

what now

[upper pivot]

too lazy to latex

[solemn rain]

yea jusut write it out

lmao who uses latex

xd

[upper pivot]

so do that on

(a+1+..+1 (p times) )^p

[solemn rain]

a^p + (1+1+1 p times)^p ?

[upper pivot]

keep doing it inductively

[solemn rain]

whjat

[mild laurel]

you don't even need to lmao, the second term is already 0

[upper pivot]

$(\sum a_i)^p = \sum a_i^p \pmod{p}$

[cloud walrus]

JohnDoeSmith:

[upper pivot]

oh right

[mild laurel]

p^p is 0

[upper pivot]

lmao

fair nuff

[solemn rain]

yea ayea fuck

👀

anyways

[upper pivot]

yeah ok (a+p)^p = a^p+p^p (p) is better looking

[solemn rain]

yea

okay boyss

im going to attempt 1 more and ask u guys if the solution is right

its not given a name

but its predictable

[upper pivot]

right the other proof i was thinking of was

$(\sum 1)^p \equiv \sum 1^p \pmod{p}$

[cloud walrus]

JohnDoeSmith:

[upper pivot]

i mixed the two up in my head oof

[solemn rain]

yeaa

ty anyways

ugh tbh i wanted it

to be like a super hard proof lmao

just so ic an feel lucky being born in the 200s

2000s and having groups

lmao

and like having someone like me to be able to prove it

If |G| = pq --->Z(G) = 1 or G is abelian

proof:

suppose |G| = pq

|Z(G)| = p or q or pq or 1

if |Z(G)| = p

or q or pq then G is abelian

qed?

(G/Z(G) would be cyclic )

[upper pivot]

prolly add in a line to say |Z|= p or q leads to contradiction

but yeah seems fine

[solemn rain]

what

why

[upper pivot]

if G is abelian then center is the whole group

[solemn rain]

i didnt say G is abelian/

right/

?*

lmao

XD

XD

yea yea got you

sorry

[upper pivot]

alright yeah, but proof works otherwise

[solemn rain]

tysm

are these proofs uspposedto be hard?

or are they trivial

[upper pivot]

not really

how do you feel about them

[solemn rain]

idk

like it took me time to think of thinking of the quotient group

when i asumed |Z| prime

and someitmes i just google them

but some i do alone

which is cool ig lmao

there are some things that i dont think i can ever prove alone tho

like this one

if p divides |G|

then G has element of order p

im going to google this 1 soon

and therea re some that i need u guys to correct me cuz sometimes i dont think that my proofs are correct

like this 1 , prove that |G:H| = |G:K|*|K:H|

like lmao i just said |G:H| = |G|/|H|

and just canceleld stuff

dk if taht supposed to be that easay

[upper pivot]

thats correct

[solemn rain]

cool

im going to attempt this 1 which looks hard af

Suppose H and K are subgroups of finite index in the (possibly infinite) group G with IG : HI = m and IG : Kl = n. Prove that Lc.m.(m, n) ::: IG : H n Kl ::: mn. Deduce that if m and n are relatively prime then IG : H n Kl = IG : HI·|G:K|

[upper pivot]

oh yeah nvm on what i said

thats not exactly right

[solemn rain]

XD

[upper pivot]

cause it assumes your group is finite oops

[solemn rain]

XDDD

literally

you are a boss

cuz it literally says

( do not assume G is finite )

XD

[upper pivot]

rip

look at the cosets

basically

cosets of K in G and then cosets of H in K

[solemn rain]

okaay

ugh idk

[upper pivot]

ok how many cosets of K are in G

[solemn rain]

|G|/|K| ?

[upper pivot]

no remember G is infinite

so we dont wanna use that notation

[solemn rain]

yea okay

[upper pivot]

whats the defination of [G:K]

[solemn rain]

|G/K| ?

XDD

umm

[upper pivot]

i mean yeah

[solemn rain]

yea but bad here

[upper pivot]

and whats G/K

[solemn rain]

the set of cosets of K in G

[upper pivot]

mhm

so how many cosets of K in G

(and then label em all like a_1 K, a_2K....)

[solemn rain]

idk how many cosets of K in G

[upper pivot]

well its just [G:K]

[solemn rain]

yea

okay

[upper pivot]

now do the same for cosets of H in K

[solemn rain]

well its just [H:K]

[upper pivot]

[K:H]

[solemn rain]

sorry

yea

[upper pivot]

but yeah

now as i said, label them right

[thorn flint]

Hi, I have a question regarding group actions:
@thorn flint

[solemn rain]

i think this is like

[thorn flint]

Oops

[solemn rain]

combinatorial stuff

lmao

[upper pivot]

group actions are related to combi

[solemn rain]

i meant

this

okay so i labelled them

well

[upper pivot]

now can u write K as cosets of H and then make the cosets of K cosets of H in G

[solemn rain]

okay so for example

cosets of K in G

g_1K , g_2K...

cosets of K in H

k_1H,k2_H...

k_1 = g_nK for some n?

[upper pivot]

(also i assume we cannot use 3rd iso theorem here btw lol, cause that would trivialize this)

more like we can say that g_i k_j H are the cosets of H in G

[solemn rain]

mmmmmmmmmmmm

okayyy

hard

oh

g_i *k_J is in G

lmao

that was messing me u

p

[upper pivot]

if u dont like this, if ur allowed to do 3rd iso u can do that

[solemn rain]

idk any iso theorems yet

next section

[thorn flint]

Hi @solemn rain

[solemn rain]

well okay ig

@thorn flint heyy

im going o attempt this hard 1

fail and not know and feel stupid and miserbale

then ask again or maybe google to save you from the trouble

Suppose H and K are subgroups of finite index in the (possibly infinite) group G with IG : HI = m and IG : Kl = n. Prove that Lc.m.(m, n) ::: IG : H n Kl ::: mn. Deduce that if m and n are relatively prime then IG : H n Kl = IG : HI·|G:K|

[upper pivot]

whats :::

[solemn rain]

10

[upper pivot]

oh <= lol

[solemn rain]

yea im dumb i copied

and pasted lol

[upper pivot]

wait looking at that problem set

do u know

[smoky cypress]

Does every finite group of composite order has nontrivial normal subgroup?

[upper pivot]

3rd and 2nd iso theorem

[solemn rain]

no i only know first just from reading

why?

[upper pivot]

no whoever smh

A_5

[smoky cypress]

I kinda expected it to be no xD

ok

alternating group?

[upper pivot]

3rd iso would make that coset problem

trivial

so there are groups with no non-trivial subgroups

called simple

[solemn rain]

u talking to me/

[upper pivot]

classifying them was one of the big problems of the last century

google it to get more groups that are simple

nah sorry was talking to whoever

[solemn rain]

holder program

athe sectiona fter isomorphism theorems

[smoky cypress]

ok

welp

idk what simple means yet

[solemn rain]

anyways

no normal subgroups

[smoky cypress]

but i do know that's a big problem last century

[solemn rain]

except itself and 1

[smoky cypress]

oh ok

[solemn rain]

like prime groups

haha

[upper pivot]

i know a lot of quotient groups of linear groups and the A_n for n>= 5 and Z_p are simple

[thorn flint]

Do you mind helping with the question I had yesterday? I know that I have to show it is reflexive, symmetric, transitive but not sure if I can just say that if we let g be x, then $x \cdot x$ thus reflexive

[cloud walrus]

Mac:

[upper pivot]

i know the group in zophs name (monster group) is simple

but thats it lel

[solemn rain]

yea sure @thorn flint

we wish to show xRx for all x

so we wish to show x=g.x for some g for all x

[smoky cypress]

I know there are tits group

[solemn rain]

do u know some g

that satisfies this? @thorn flint

[thorn flint]

identity?

[solemn rain]

yes

1

x=1.x for all x

so xRx

for all x

thats reflextivity

[thorn flint]

Okay so for symmetric, I have to show that x = g.y -> y=g.x?

[upper pivot]

no

[solemn rain]

if xRy then yRx

thats symmetryh

okay i leave

let johndoe help you he is much better

tysm johndoe

im done feeling stupid i cant do any problem i go watch elite

ty agaain

[upper pivot]

you want to show that if there is some $g\in G$ such that $x=gy$ then there is some $g'\in G$ such that $y=g'x$

[cloud walrus]

JohnDoeSmith:

[upper pivot]

np, and dw ur doing much better now

then a month ago for example

[solemn rain]

how was i like a month ago

i dont rememebr

[upper pivot]

not as good

[solemn rain]

but what i did new now is i started to do exercises

i never did exercises

and jsut watched utube videos

[upper pivot]

oh that was a bad idea

definately do exercises

[solemn rain]

yea trhats what im doing

but they are hard tbh

[upper pivot]

thats the meat of learning a new topic

[solemn rain]

like most of them are hard for me

[upper pivot]

reading the ch/watching videos w/e are like 10% of the actual learning, less

[solemn rain]

yeaa

[upper pivot]

(also dw most problem sets in books are intended to take several hours)

[solemn rain]

yea

all of them tho

i dont do all of them tbh

but id o like alot

the thing is i dont manage to get them alone

most of the time

i do get some tho

the easy ones

like i just proved eulers theorem

the nt 1

[upper pivot]

takes practice

[solemn rain]

that was easay

bad quesrtion:

problems get harder as you move on?

[upper pivot]

i would argue the opposite

they get easier

cause u get more familiar

[solemn rain]

yea

dark souls philosophy

cool

[upper pivot]

also oof sorry @thorn flint we kinda flooded ur question up

how far did u get

[solemn rain]

yea ayea

my bad

bye guys

[upper pivot]

see ya

[thorn flint]

No worries, I'm still on the symmetric part

@upper pivot Its not possible to use inverses here right?

[upper pivot]

it is

inverse is how you prove symmetry

try it

[thorn flint]

Okay

let me try

Hmm furthest I can get is when it becomes both inverses of the original

But not y=g'x

[upper pivot]

if x=gy, then we want y=g'x for some g' right. how should we isolate the y

[thorn flint]

If we take the inverse of g

we just get xg^-1=y

[upper pivot]

why is g^-1 on the right of x

[thorn flint]

Sorry g^-1 x = y

[upper pivot]

yeah

so is y=g' x for some g' in G

[thorn flint]

Oh right

So g's is just this inverse element

[upper pivot]

yep

[thorn flint]

And for transitivity, we show that x=gy and y = g'z implies x=g''z?

[upper pivot]

yep

[thorn flint]

If I multiply x = gy and y = g'z, then I get xy = gg' yz, if I let gg' = g'', then xy = g'' yz. Then taking y^-1, x = g''z

Does that work? @upper pivot

[upper pivot]

you cannot do y^{-1}

its not a member of the group

[thorn flint]

How do I know if it is the member of the group?

[upper pivot]

ok so we are saying $G$ is acting on $S$. and when we say $gx$, it means the action of an element $g\in G$ on $x\in S$

[cloud walrus]

JohnDoeSmith:

[upper pivot]

anyhow for your proof, x=gy and y=g'z right

x=gy replace the y with something

[thorn flint]

Oh its that easy

x = gg'z

Then gg' = g''

So x = g''z

In general if G is acting on S and when gx, then we can only make inverses of g?

[thorn flint]

Hi can someone help for (b)? I'm not sure how to start

[mild laurel]

what are you confused about?

[thorn flint]

How do I determine A?

We have the mapping that G = D6 and also a =e.
So

[mild laurel]

Use the definition of A

[thorn flint]

A = { g.e with g in G}
Then the map f:G ->A maps h to h.e?

[mild laurel]

yes

But it wants you to be more explicit than that

What exactly are the elements of A

[thorn flint]

if a=e then there is only the identity element?

[mild laurel]

why?

[thorn flint]

Oh wait no getting confused

elements of A is just g in G

isnt it?

[mild laurel]

why?

[thorn flint]

Because A = {g.e with g in G}?

[mild laurel]

but what does . mean?

hint: it's not the group multiplication

[thorn flint]

Its the group action?

[mild laurel]

yes

[thorn flint]

So every g in G is acts as a permutation with e?

[mild laurel]

What do you mean

[thorn flint]

Is A just all the elements of D6?

[mild laurel]

why?

[thorn flint]

Not sure

Any other hints?

[mild laurel]

I am not quite sure what you're confused about

[thorn flint]

How do I know the elements of A?

[mild laurel]

By the definition

A = {g.e with g in G}

just calculate g.e for every g in G

[thorn flint]

How do I do that? Can you show me an example?

[mild laurel]

What's e.e?

[thorn flint]

identity

[mild laurel]

okay, and how'd you get that?

[thorn flint]

Any element acting with the identity is the identity of that element

[mild laurel]

okay sure, well, how do you calculate g . x in general?

Maybe reread your question and see what the group action is

[thorn flint]

Isn't it just taking the ordered pairs from each element in G and each element in X?

so the elements of A are {e, (r,e),(r^2,e),(s,e)...etc}? But then it would just be {e,r,r^2,... etc} in the end?>

[mild laurel]

No

Read the problem again

It says that g . x = gxg^(-1)

[thorn flint]

Hmm

I'm not sure why A is not = G

Since the elements basically aren't changing

[mild laurel]

What does that even mean

What does the element r get sent to

Like what's r.e

[thorn flint]

rer^-1?

[mild laurel]

And what does that simplify to

[thorn flint]

just e

Oh wait

Just e

[mild laurel]

Yes

[thorn flint]

So A = {e}

[mild laurel]

Yes

[thorn flint]

Okay thanks!

[trail hamlet]

Let's denote the hadamard/pointwise product by (.*).

How to derive the general identities diag(u)*(A.*B) = (diag(u)*A).*B = A.*(diag(u)*B) and diag(u.*v) = diag(u)*diag(v) as described in https://chat.stackexchange.com/rooms/105551/discussion-between-joel-sjogren-and-kevin ?

[chilly ocean]

Let's use $\circ$ for hadamard

[cloud walrus]

kawaii-snore:

[chilly ocean]

$u \circ v = [ u_1 \cdot v_1, \dots, u_n \cdot v_n ]$

[cloud walrus]

kawaii-snore:

[chilly ocean]

too lazy to latex out the diag() of that but it's just that vector along the diagonal of a n x n matrix

then you can easily see diag(u) * diag(v) (this is regular matrix multiplication) is the same

As for the first identity, multiplying a matrix by a diagonal matrix $diag([ u_1, \dots, u_n ])$ means that the ith row is scaled by $u_i$,

[cloud walrus]

kawaii-snore:

[marble bolt]

So, if we let M be an R-module and I be an ideal of R with
IM={r1m1+...+rkmk}
Whats an example where I is not equal to the annihilator(M/IM)

[oblique river]

If M = 0 then M/IM = 0 so its annihilator is all of R, regardless of what I is

[marble bolt]

Ok, thanks

[ripe crest]

Hey bananas, why was it obvious that the map Q[X,Y] -> Q[T] that sends X -> T and Y -> T^2 is a homomorphism?

[wind steeple]

by universal property

Q[X,Y] is the free algebra of 2 generators over Q

btw it's only the fact that you can extend by morphism your application from {X,Y} to Q[T]

[topaz solar]

It’s not particularly hard to see even without invoking universal properties (if you haven’t seen them that is)

[ripe crest]

Hmm I don't know what a universal property is. I don't know what a free algebra is 😛

[topaz solar]

Universal properties are v nice and u should like them

[solemn rain]

what are those

[topaz solar]

Basically they just deal with stuff like “unique object satisfying these things up to unique isomorphism”

[ripe crest]

How would one see it without using universal properties?

I know Aluffi covers it in his Algebra text, but we didnt go over them in class

[topaz solar]

The product is universal in how X->A and X->B factors through X->AxB

[mild laurel]

Your intuition should be that you can choose X and Y to be sent anywhere and that will give you a homomorphism

Like, do you know anything about generators of groups?

[topaz solar]

And because it’s the unique object (up to unique isomorphism) that does this splitting

[ripe crest]

Group theory was 2 years ago... So I remember a bit, but not too much

[topaz solar]

X and Y generate Q[X, Y] as a Q algebra

If you send them somewhere, the rest can follow

[ripe crest]

And Q too right

[topaz solar]

They don’t generate Q

Z[X] isn’t generated by X except as a Z algebra

[ripe crest]

No, as in X, Y and Q generate Q[X,Y]

[topaz solar]

Oh

No

Q isn’t a generator

[ripe crest]

how do you get the polynomial 3 using just X and Y?

[woven delta]

Lol Z algebra

[solemn rain]

wtf is an algebra

[woven delta]

You mean a ring?

[ripe crest]

right, Z is not an algebra

[topaz solar]

Yes I know I’m rarted liquid, I’m relating it to the Q[X, Y] thing

[ripe crest]

no field at its centre

[topaz solar]

Anyway, my mental retardation aside, Q doesn’t generate Q[X, Y]

[ripe crest]

how do you get the polynomial 3 using just X and Y?

[topaz solar]

3 isn’t made from combining X and Y, for example, but Q[X, Y] has a unit, yes?

[ripe crest]

Yeah, 1

and -1

actually a lot

all of Q

besides 0 is a unit

[topaz solar]

3(as a scalar) • 1 (unit of Q[X,Y])

Whoops turns out doing this gets you Q “inside” Q[X, Y]

[ripe crest]

Right, but you're saying we can generate Q[X,Y] using X and Y - so we should be able to get to 3 using X and Y. If we define X^0 = 1, then I see how we can do it, I don't see any other way. Which does make sense to me because Q[X,Y] is a vector space and 3 is the infinite vector (3, 0, 0, 0, .....)

[topaz solar]

Alright lemme stop you for a moment

All your fields have an identity

Good ol 1

[ripe crest]

yeah

[topaz solar]

How do you define Q[X, Y]

You get just 3 as an element of you define it as sums of elements (like polynomial ring)

That’s an obvious thing of course, so that’s no crazy realization

[ripe crest]

Q[X,Y] = Q[X][Y]

[topaz solar]

Your problem seems to be figuring how you get 3 when you treat Q[X,Y] as this free algebra thing

[woven delta]

When you say the subring generated by X and Y, that must by necessity include 0 and 1

You don't have to say garbage like X^0 = 1

[topaz solar]

You’re not generating it as a group my G

It’s a Q algebra

[woven delta]

Yeah so it must include 0 and 1, be closed under multiplication by elements of Q, and be closed under multiplication

In particular

[ripe crest]

okay - this isn't really helping me xD - I think it's fine

[topaz solar]

Since it contains the unity 1, you can just do scalar multiplication

So the constant 3 polynomial is just 3•1

This is annoying tho so you can just write 3 since it’s obvious

(Like how you just use naturals in rings to denote 1+1+....+1)

[ripe crest]

yeah - makes sense

[topaz solar]

I’m a bit of a retard so the way I’ve (attempted to) explain this is almost certainly subpar, but hopefully gets you to where you can better follow whatever it is you’re doing

[ripe crest]

thanks 🙂

In C[X,Y]/(X-a, Y-b), it is clear that X gets sent to a and Y gets sent to b in the quotient

But for something like R[X,Y]/(X^2Y^2 +1), there are many different values that satisfy (XY)^2 = -1. Such as X = i, Y = -i, or X = -i, Y = i. Or X =1, Y = i. Or X = i, Y = 1.

So what happens here?

[solemn rain]

is the second isomorphism theorem important

[ripe crest]

probably - I never used it though xD

first is the most important

[solemn rain]

yea

i get first

and i get third

2nd and fourth ugh

lmao

[upper pivot]

fourth is very important

2nd not so much, i think it can be used to prove CRT but thats about it

[solemn rain]

fourth?

the lattice shit?

dude this looks fucking trash ugh

XD

[upper pivot]

lattice??

4th is just corrospondence theorem

[solemn rain]

its called the lattice isomorphism theorem in df

wait

it says that

if N is normal in G

there is a bijection between A and A/N

where A is the set of subsets that has N

[upper pivot]

yeah thats about right

but theres more

mainly that the bijection is order preserving

and more importantly normal subgroups corrospond to normal subgroups

[solemn rain]

i dont understand

wdym

corrospond

[upper pivot]

well in the bijection

if H is a normal subgroup of G containing N, then H/N is a normal subgroup of G/N

[solemn rain]

let G bar donate G/N

H bar < G bar?

[upper pivot]

< meaning normal subgroup? then yes

[solemn rain]

yea yea

i have to prove these

in exercises 2

exercise 2*

loks trash

[upper pivot]

oh yeah those are all the important facts

like (1) is order preserving

and (5) is the usual statement of the 4th iso theorem

[solemn rain]

i can try to prove

5

lmao thats so cool lmfaao

5 is easy

so in summary

fourth isomorphism theorem says

if A is a subgroup of G and N is normal in G and din A

if A is normal in G , A/N is normal in G/N

?

i proved that

the iff

[upper pivot]

well heres how i would state it

There is an order preserving bijection between the normal subgroups of G containing N and the normal subgroups of G/N given by $H\mapsto H/N$.

[cloud walrus]

JohnDoeSmith:

[solemn rain]

okay

gotchu

ty

what the fuck

lmao

i randomally scrolled

in the book and i literally found something totally weird

[upper pivot]

this is really important when u look at quotient groups for instance. although i have mostly used this in context of rings

[solemn rain]

why is there an isomorphism theorem in rings?!

its literally the same

how

is ideal here analogous to

[woven delta]

What do you mean?

[solemn rain]

normal

[woven delta]

Yes

Lol

[solemn rain]

lmaaao

[upper pivot]

yeah

[woven delta]

Smh

[solemn rain]

how are those things similar

[woven delta]

[solemn rain]

like

i thought rings are completely different structures

why are there like

these similarties

like isomo theorem

[woven delta]

Why are you so shocked by this

Smh

[solemn rain]

idk

[woven delta]

Like calm the fuck down

[solemn rain]

there is literally lagranges theorem

lmfao

yea there is definitely a god

[woven delta]

No

Fuck off

[solemn rain]

lmao u fuck off

[upper pivot]

lmfaooo

[solemn rain]

there is a problem from last section that i couldnt do

any help?

Prove that S_4 has no normal subgroups of order 8 or normal subgroups of order 3

@upper pivot i see what you mean now by the coset problem using third isomorph theorem

i wonder why i didint discover it when i saw the problem

hehe xd

[ripe crest]

Reasking my question in case someone can help

In C[X,Y]/(X-a, Y-b), it is clear that X gets sent to a and Y gets sent to b in the quotient
But for something like R[X,Y]/(X^2Y^2 +1), there are many different values that satisfy (XY)^2 = -1. Such as X = i, Y = -i, or X = -i, Y = i. Or X =1, Y = i. Or X = i, Y = 1.
So what happens here?

[upper pivot]

x^2y^2 gets sent to -1

[ripe crest]

that's true

but what about X and Y by themselves?

They must get sent to a unique value, no?

it is a function

[upper pivot]

no

take an easier example first

R[x]/(x^2+1) is C right

basically x^2=-1 and all elements are like a+bx

[ripe crest]

Yup

X -> i

[upper pivot]

but you dont really send X to anything

well why not -i

its the same ring yeah if it gets to -i

[ripe crest]

hm - you're right

[upper pivot]

yeah, so basically you dont really send X,Y to anything

[ripe crest]

But, we have a quotient map

[upper pivot]

certain expression are just 0 in the quotient rings

[ripe crest]

it does send X to something

[upper pivot]

its better to think of quotient maps as cosets/modulo classes rather than "sending X to something"

like Z/pZ for instance

[ripe crest]

Sure, in Z/pZ, every number gets sent to its remainder mod p

[upper pivot]

one could say Z/(p) is sending p=0, but prolly better to say that right

similarly R[x]/(x^2+1) is taking the remainder of all R[x] elements mod x^2+1

[ripe crest]

Sure - but still what does X get sent to?

It must go somewhere

[upper pivot]

nothing, x is x

[ripe crest]

Ah right

thanks I had a brainfart

or a series of brainfarts

[upper pivot]

its fine, but as i said, just think about it as the remainders or cosets

ofcourse since x^2+1 has remainder 0 mod x^2+1, we can say x^2=-1 in a informal way

[ripe crest]

right, thanks!

What would be the remainder of X^2 when modded out by X^2 Y^2 + 1? If it was just X^2 + 1, the remainder would be -1. But I'm not sure how to proceed with the Y^2

[upper pivot]

x^2 would just remain x^2

[ripe crest]

How do you see that?

[upper pivot]

well whats the remainder of x^2 mod x^2y^2+1

[ripe crest]

ah right - no y's in there.

Okay, then y^2 would also remain y^2 then?

[upper pivot]

mhm

[ripe crest]

it's interesting tho, because if we had X^2 mod 3X^2, that would be 0 in k[X]...

ah it's because we don't have a field

thanks again!

[upper pivot]

yeah np

[ripe crest]

If I'm trying to learn more about C[X,Y]/(Y-X^2-X-1), how should I approach it?

I know how to find its maximal ideals since I know the maximal ideals of C[X,Y]

[oblique river]

@ripe crest the best way to think about C[x,y]/(y - x^2 - x - 1) is as C[x,y] / (y = x^2 + x + 1)

it's canonically isomorphic to C[x]

via the map x --> x and y --> x^2 + x + 1

howdy

yeah

same tbh

rip :(

[bleak abyss]

Sorry my mom was talking to me for a while

Okay so

Some results about algebraic curves

The local ring of a curve at a smooth point is a DVR

A "uniformizer" is an element whose valuation is 1, aka a generator of the maximal ideal. Okay so the proof I just went through is that if K is perfect, C/K is a curve, P is a smooth point, and t\in K(C) is a uniformizer at P

Then K(C) is a finite separable extension of K(t)

That should be it

Like okay if char(K) = 0 then everything in sight is separable. If char(K) = p then you start having to play some games

And the nice relevant fact is that if you have a char p perfect field, then every element is a pth power

So it's obvious that K(C) is a finite (hence algebraic) extension of K(t), yeah? Transcendence degree 1 and shit

So this is one step in a bunch that eventually leads to an equivalence between the category of smooth curves over K with morphisms being non-constant rational maps (equivalently surjective regular maps) and the category of finitely generated field extensions L of K with transcendence degree 1 such that K = \overline{K}\cap L

Which may or may not be old news to you but

I kinda like it, it's pretty dope in itself and it kinda gives a nice bridge. Like the business about primes in Z being ramified when you pass to rings of integers in number fields should reasonably transplant to the case of function fields

And I think in the above, letting K = C says that this should have a lot to do with ramified covers of Riemann surfaces

(Silverman later starts talking about ramification in the setting of algebraic curves)

So, for affine varieties a regular map is just, restrict a polynomial on A^n to the variety

Now for locally closed people you can talk about rational functions

Turns out, a map on an affine variety is regular iff each point has neighborhood on which the map is rational

Good because my parents are back and they're gonna ask for things lol

[ripe crest]

@ripe crest the best way to think about C[x,y]/(y - x^2 - x - 1) is as C[x,y] / (y = x^2 + x + 1)
@oblique river so basically, if we can rewrite the polynomial in terms of either X or Y, it will be isomorphic to C[T]

[oblique river]

yeah, if you can get it to y = f(x) or x = f(y) form

[ripe crest]

Gotcha

got my midterm tmr... gonna try to learn as much as I can .. kinda cramming lol

[oblique river]

you'll do great :)

[ripe crest]

midterm at home - let's see how this goes

So the maximal and prime ideals of C[x,y]/(y - x^2 - x - 1) will be of the form (X-i)?

[oblique river]

what is i?

[ripe crest]

i^2 = -1

[oblique river]

(x-i) is only one ideal

the maximal ideals will be of the form (x-a, y-b) where b = a^2 + a + 1

[ripe crest]

ah right - lemme write down my thought process

hmm

but

[oblique river]

alternatively we can write (x-a, y-a^2-a-1)

for a in C

[ripe crest]

the quotient is isomorphic to C[T] and its maximal ideals are of the form (T-a)

[oblique river]

mhm

[ripe crest]

I guess I was just confused because we had a single variable polynomial ring being isomorphic to a multivariable one

But, yes you're reasoning makes sense

[oblique river]

*isomorphic to a quotient of a multivariable one

you could also write (x-a)

[ripe crest]

yup, but even the quotient is a multivariable polynomial ring, no?

[oblique river]

usually "polynomial ring" is something of the form "C[x1, ..., xn]"

I wouldn't call C[x,y]/(x^2 + y^2 - 1) a polynomial ring

it's a quotient of a polynomial ring

[ripe crest]

makes sense - even though there are X and Y in the quotient, they're not the same as the ones in the polynomial ring

equivalence classes and all that

[oblique river]

yep

[ripe crest]

i know in a PID, irreducible <=> prime <=> maximal

and maximal => prime always

in a UFD, we dont necessarily have prime <=> maximal

do we have irreducible <=> prime?

[oblique river]

I think that's true in a UFD

but not in general

[ripe crest]

gotcha - so the prime ideals of C[X,Y] are the irreducible polynomials

[upper pivot]

also prime => irreducable always iirc, but the other way is only for UFDs

[ripe crest]

maximal => prime => irreducible always

[oblique river]

no, be careful

[ripe crest]

goot to know

[oblique river]

if f(x,y) is irreducible, then (f) is a prime ideal

and if (f) is a prime ideal then f(x,y) is irreducible

but there are ideals of C[x,y] that aren't generated by one element

and those ideals can also be prime

[ripe crest]

errr, but prime <=> irreducible in C[X,Y] since it's a UFD, no?

[oblique river]

yes

well, I think you're being a bit too slick here

you can't say "prime <=> irreducible" if one side is about ideals and the other is about elements

[upper pivot]

(btw my comment was on elements not ideals so dont let that confuse you sorry)

[oblique river]

like I said, irreducible polynomials generate prime ideals

and if a prime ideal has a single generator, that generator is irreducible

[ripe crest]

Ah I see what you're saying

[oblique river]

but there are non-principal prime ideals

[ripe crest]

(X-a, Y-b) is prime, but we don't really have irreducibility here

[oblique river]

yes, there is no single generator that you can call irreducible

[ripe crest]

hmm okay

If I want to find the prime ideals of C[x,y]/(x^2 + y^2 - 1), can i just look at the prime ideals of C[T]?

[oblique river]

no, that ring isn't isomorphic to C[t]

[ripe crest]

I know 0 will be one because that quotient is isomorphic to an integral domain

sorry, wrong one

C[x,y]/(y - x^2 - x - 1)

I know in C[T], the prime ideals are of the form (T-a), because maximal <=> prime because C[T] is a PID

[oblique river]

yes

[ripe crest]

So, I need some condition on a such that (T-a) is isomorphic to a prime ideal of C[x,y]/(y - x^2 - x - 1)

Oh, if I can find the inverse map, then it should be easy I think

[oblique river]

the map is x --> t

so the inverse of (t-a) is (x-a)

you might ask -- what about y? well, in the quotient C[x,y]/(y = x^2 + x + 1), we have y = x^2 + x + 1

if you think about it from the other direction, namely, "maximal ideals of C[x,y] containing y - x^2 - x - 1", well that's going to be ideals of the form (x-a, y-(a^2 + a + 1)) as said before