groups-rings-fields

[stone fulcrum]

As we describe field extensions as a vector space over the original field

Sniped D:

[upper pivot]

lol

similar af wording

[bleak abyss]

I mean it depends a bit with group theory, you can define a group and do a lot of counting arguments and talk about a bunch of examples (S_n, A_n, kinda D_n, Z/n, etc) with 0 linear algebra

But then matrix groups are arguably the most important examples. And the more you wanna do the more LA you'll want

[gentle pendant]

It's also a good playground to learn about the basic ideas/principles of abstract algebra, where the answers are easy and the theory can be neatly packaged.

[marble bolt]

Could someone explain what N(a) is supposed to be measuring?
a is just an element, so shouldn’t N(a)=1?

[bleak abyss]

Eh N is just any function which does the trick

Like for example

If R = k[x]

Then you can let N be the degree map

[marble bolt]

So, we are checking the degree of a mapping of an element

[bleak abyss]

Wait what

Hold on so

We have a map N:k[x] -> \mathbb{Z}^+ \cup {0}

Which is N(f) = deg(f)

This satisfies the requirement to be a Euclidean domain

[marble bolt]

Ohh

[bleak abyss]

For a general ring R, N is some function such that blah blah blah

[marble bolt]

Ok, thanks!

[valid bridge]

i literally don't get quotient groups

does anybody know what this is isomorphic to

im guessing its something easy like Z/2Z x Z but idk

is it Z/2Z x Z/2Z ?

[lethal cloak]

dont guess

here's a hint, do it in steps

first quotient by (1,1)

then by (1,-1)

[tribal pasture]

Isnt M = Z² since (1,-1) + (1,1) = (2,0) and (1,-1) - (1,1) = (0,-2) and thus it has the same basis as Z²?

[lethal cloak]

yes M is Z^2

[valid bridge]

wait so Z^2/M = Z^2/Z^2 which is just the trivial group?

[lethal cloak]

no

[valid bridge]

wtf

[lethal cloak]

the notation is misleading

there's many different subgroups isomorphic to Z^2 inside Z^2

[valid bridge]

fuck

[lethal cloak]

some, like M, are proper subgroups

[tribal pasture]

Ah just like 2Z is isomorphic to Z but Z/2Z ≠ Z/Z

[lethal cloak]

you should draw a picture

it might help

of the lattice points Z^2

and what the cosets are

[valid bridge]

the picture i drew is just like

two lines in the xy plane which make a cross

[lethal cloak]

the full subgroup should be a crosshatch

not just 2 lines

but that's the idea

[ocean magnet]

is the answer Z/2

[lethal cloak]

yes

[valid bridge]

so the two cosets are all the diagonal lines with +1 slope and then all the diagonal lines with -1 slope?

[lethal cloak]

nope

so let's see the definition

what's a coset

it's a subset of the form x + M

the first example is 0 + M = M

so M, a crosshatch made by diagonal lines, is a coset by itself

another coset is (1,0) + M

draw both cosets and see what you can deduce

[valid bridge]

@lethal cloak thank you

nice pfp btw

[harsh kettle]

isnt coset xM and not x+M ?

[mild laurel]

its really the same

multiplicative vs additive notation

[tribal moat]

a=Curve(16sin^(3)(t),13cos(t)-5cos(2t)-2cos(3t)-cos(4t),t,-15.71,15.71) this is the heart equation guys

[steel owl]

You mean a cardioid?

[solemn rain]

Let G be a group and let K be the kernel of some homomorphism from G to another group. Then the set whose elements are the left cosets of K in G with operation defined by uK o vK = (uv)K

i think if some1 proves with me this theorm

i think im oging to udnerstand alot of

understand*

can some1 help

i would understand that simply G/K means cosets of K in G

and i would understand why from the proof

[hot lake]

what theorem ?

[solemn rain]

lmfaaao sorry

i got it anyways b

the theorem was that this set defined with this operation

is G/K

is the same as the set of fibers

[tribal moat]

@steel owl yes bro

But I don't like using "cardioid" name because it doesn't look like a real heart

[slate brook]

does my answer make any sense..?

[slate brook]

<@&681260374879633482>

[chilly ocean]

I'm trying to prove $x^x$ has no elementary antiderivative. I probably need to use the corollary to Liouville's theorem that says a function $f(x)e^{g(x)}$ has an elementary antiderivative if and only if there exists some rational function r with $f=r'+rg.$ I've rewritten $x^x$ as $e^{\ln{x^x}}$ to get it into a usable form and tried some equation manipulation, but I can't see anything yet and don't know what to look for really. Something about the transcendentality of $e$ maybe?

[cloud walrus]

Botnuke:

[gentle pendant]

Isn't it g' not g in that criterion? In any case, f=1, g=xlogx here. Assuming such an r exists, we take the equation f=r'+rg', plug in the functions f,g and you are able to solve this equation to get log(x)={some rational function}. This is absurd for many reasons, take your pick.

[solemn rain]

let G be the dihedral group of order 16 and G bar = G/<r^4>

( notice <r^4> commutes with every element in the group hence a<r^4> = <r^4>a for all a hence normal )

show that G bar is of order 8

help XD

i tried thinking if maybe G bar is cyclic

and then saying r has order 8

but idk how woulod i show that or even if thats correct

[stone fulcrum]

@solemn rain
What's the order of <r⁴>? How many cosets would it generate?

[solemn rain]

fuck yea

|r^4| = |r|/gcd(|r|,4)

= 8/gcd(8,4)

2?

@stone fulcrum

[gentle pendant]

Yep so the order of the subgp is 2, and so there are 8 cosets.

[stone fulcrum]

@solemn rain
Pretty much the order of G/H is |G|/|H|

Because that's how the cosets fit

[worthy geode]

Indeed

That's the Lagrange theorem for finite groups

[solemn rain]

tysm every one

i get it now

ur amazing all

ty

sorry for the late hope-so-not-rude replies

@stone fulcrum im sorry but how do you know this

that |G/H| = |G|/|H| ?

[gentle pendant]

Because each element of G/H is a coset, which is an equivalence class of elements of G containing |H| elements. So you have |G/H| disjoint collections of |H| elements whose union is all of G...

[solemn rain]

fuck

yea

wow thats cool

they partition the group

[gentle pendant]

yeah it's pretty cool how much you can get from such a simple fact.

[solemn rain]

yea man awesome

the fucking textbook didnt say that

in that section

i was assumed to do this without it

tysm anyways

[gentle pendant]

np

[leaden finch]

can someone check my work for this one

for c

[latent anvil]

How did you get that final answer?

@leaden finch

[leaden finch]

ohh i think i mess up

umm since we have ( 1) x^7

we would subtract 2 to 1?

[fading wagon]

2 is not a multiple of 4

[stone fulcrum]

@leaden finch
You also have the + 1 at the end there. A possible final answer is x^7 + 1

It happens to be true in Z2[x] that x² = x, so you can go further to say that this is x + 1. But that is not likely expected of you.

[wind parrot]

Well as polynomials in Z_2[x], x^2 is definitely not equal to x. They induce the same function Z_2 -> Z_2, but they are still different polynomials

[stone fulcrum]

Yeah that's fair. Ignore that last part

[vital night]

Not sure how to do this problem

[thorn tree]

basically you want a matrix M such that M * u = v

with det(M) = 1

[vital night]

k so how do I figure out what M is supposed to be?

[lethal cloak]

write a general element of SO(2)

draw a picture

and stare at it for a bit

[vital night]

what is SO(2)

[lethal cloak]

orthogonal 2x2 matrices with determinant 1

[thorn tree]

well you don't need to calculate the entries of M

the question just asks you to prove one exists

giving an explicit M in terms of u and v is one way to do it, but you could also use some basic linear algebra arguments

[vital night]

How would I go about finding M

[leaden finch]

can someone help me with part b

[mild laurel]

what have you tried?

[leaden finch]

i havent tried b though

im having a hard time understanding it

i did a though

[mild laurel]

then what are you having trouble understanding

[leaden finch]

im not sure what to do for part b

[bleak abyss]

If you don't have an immediate plan of attack write out the definitions fully

In this case write out the definition of a subring and that'll show you what you need to verify about the stuff (or falsify)

[merry pollen]

possibly dumb question: in an integral domain why can't polynomials of degree greater than 1 be units

[mild laurel]

deg(fg) = deg(f) + deg(g)

[merry pollen]

hmm

could you expand on that

[mild laurel]

take g= f^{-1}

[merry pollen]

so 0 = deg(f) + deg(g) right?

[mild laurel]

yes

[merry pollen]

but why does that imply the polys of degree > 1 cant be units

[mild laurel]

what are possible degree of polynomials

[merry pollen]

sorry connection failed for a sec

but thanks for the help I see now

[upper pivot]

also you should try proving deg(f+g)=deg(f)+deg(g) for IDs. do you see why this isnt true for non-IDs

[merry pollen]

dammit internet keeps disconnecting

why what isn't true for non-IDs?

[upper pivot]

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

[mild laurel]

non-ID polynomial rings

[upper pivot]

or yeah, that

[merry pollen]

for non-IDs theres a zero divisor right, so then one of them could be zero so deg(f + g) could be less?

[gentle pendant]

I think you mean deg(fg)=deg(f)+deg(g) @upper pivot 🐶

[upper pivot]

or wait yeah sorry

oof i typoed

and copy pasted the typo

@merry pollen sorry thats what i meant

[merry pollen]

np

[leaden finch]

can someone pleasee help meeee

i'll give a generous donation if someone helps mee

[gentle pendant]

<@&684286608136208429>

[kindred rivet]

@gentle pendant dont ping me needlessly; my powers are scantly utilized due to their shear impact

[quiet cave]

sheer*

[mild laurel]

Uh, I'm not sure this is the right channel

[halcyon siren]

Why is this true?

The n(H, v) being taken to be dim(V) part

[stone fulcrum]

@leaden finch
Help with waht

[spare arrow]

Hey,

Find out how many different groups you can form with 144 elements. (except for isomorphism, of course.)

How would you answer this question? Is there any efficient method/smart ways to answer the question?

Thanks in advance

🤔

[upper pivot]

try the usuals

namely the sylow theorems

[mild laurel]

That's where you'd start, but I'm not sure this question has any easy method, there are a lot of them

[upper pivot]

yeah so apperantly theres

197 groups of order 144

i dont think sylow would be too useful to find all of these now that i think of it lol

[gentle pendant]

Just manually check group axioms for each possible binary operation on a set with cardinality 144. 🐱

[hot lake]

then manually check every possible isomorphism between them

[scarlet estuary]

This is how I use mathematica

I feel personally attacked

[mild laurel]

Yeah imagine using Mathematica

[oblique river]

@gentle pendant you joke but I had a student do that to verify that Z/4Z was a group when I TAed for a group theory class a couple years ago. in order to check associativity they wrote out all 64 triples by hand

(0+0) + 0 = 0 + (0+0)
(0+0) + 1 = 0 + (0+1)
etc

[thorn tree]

👀

it's just addition lamao

[oblique river]

yes

that's why it was a waste of time

[gentle pendant]

hahaha that is amazing

[upper pivot]

lmfao

[delicate bloom]

why work smarter when you can work harder?

[scarlet estuary]

if the shoe fits

[valid bridge]

Is the answer just Z/3Z x Z/3Z x Z/12Z x Z

wanna make sure i didn't fuck up the operations

[hot tusk]

can someone help me with notation

what does the x on top of the Z_8 mean

also can someone help me figure out wtf a coset is

[mild laurel]

For your first question

It means the set of natural numbers less than 8 and coprime to 8

[hot tusk]

nice

got it

ty

[mild laurel]

maybe you should start by telling me how you understand cosets

[hot tusk]

im looking at this example and it seems that g is just increasing by 1 every time and getting added onto {0,3}

[mild laurel]

Notationally at least, that's what $1 + H$ means

[cloud walrus]

Zopherus:

[hot tusk]

but then this example just some wtf

[mild laurel]

$1 + H = {1 + h \mid h \in H}$

[cloud walrus]

Zopherus:

[mild laurel]

I mean, its the same thing in that example

$(12)H = {(12)h \mid h \in H}$

[cloud walrus]

Zopherus:

[mild laurel]

It's not addition anymore, but its the group product of S_3

[hot tusk]

dam this hurts to ask but whats s_3 again

[mild laurel]

The permutation group on 3 elements

you should probably go back and figure that out

[hot tusk]

yea brb

[leaden finch]

can someone explain too me what they are trying to say here

[ocean magnet]

that's x*y=0 iff x=0 or y=0

Z6 breaks that because 2*3=6=0

@leaden finch

[leaden finch]

oh okay i seee it now

we can create a multiplication table for it

right?

[mild laurel]

what is this table supposed to represent

[leaden finch]

i think the mod table

or i think im doing it wrong

[mild laurel]

I mean that's right? But I don't see why its relevant

[leaden finch]

he said that " Z6 breaks that because 2*3=6=0" but i think thats because the 6 lands in 0 ?

[mild laurel]

okay yeah that's true

[lavish gale]

can u give me link to this pdf @leaden finch

[leaden finch]

for which one ?

can someone help me with this one

heres my work

[neat patio]

ur working looks correct, quotient is 1/3x^2 -1/3x +1/3 and remainder is 2, whta do u need help with

i fink this q belongs elementary num theory as well idk

[scarlet estuary]

@neat patio its from Z_5[x]

so thats not the final answer

anyway, do you know what 1/3 is in Z_5? @leaden finch

[neat patio]

ohh

is Z_5 integers base 5?

not seen that notation before

[scarlet estuary]

why are you answering questions in #groups-rings-fields if you havent seen baby's first ring

anyway, Z_5 is the integers modulo 5

with standard addition/multiplication

[neat patio]

ah oke, my baddddd

[clear obsidian]

So is it canonical to think of quotient groups like a literal division of real numbers? Because it definitely works out in a very majestic way

[scarlet estuary]

that's where the term comes from.

[clear obsidian]

makes sense

really nice how quotient lattice is the literal reverse of a subgroup lattice

[eager willow]

I've proven that the nilradical N of a commutative ring with identity R is an ideal. Does anyone know much about the quotient R/N, besides that it has no nilpotent elements?

[mild laurel]

I don't think there's much more you can say in general

[leaden finch]

im getting confused with Z 5

ik its from Z5 : { 0,1,2,3,4}
.

[upper pivot]

spec(R/N) is homeomorphic to spec(R) is a nice general property

but yeah not much you can say thats general

[eager willow]

interesting, thanks! I'm new to algebraic geometry, the topology for spec is the Zariski topology?

[oblique river]

yes

[leaden finch]

can someone help me understand this one

how did they get 6?

why is it 2?

[stone fulcrum]

Where did they get the 6 @leaden finch?

[leaden finch]

where it says 3( 4^-1) = 3(2)

[stone fulcrum]

In Z7, the multiplicative inverse of 4 is 2

[leaden finch]

=6

hmm but how

[stone fulcrum]

In any ring (field), the multiplicative inverse of x is the unique number y such that xy = 1

4×2 = 1 in Z7
So 4 and 2 are multiplicative inverses

[leaden finch]

hmm but i still dont get it how they ended up with 6

[stone fulcrum]

They're taking the inverse of 4 (which is 2)
And they're multiplying it by 3

I'm not sure why they're doing this, but the algebra works out

[leaden finch]

okay, so ik that in Z 7:{ 0,1,2,3,4,5,6}

which we can create our table

shouldnt it be 3 ( 4^-1)= 1?

[mild laurel]

why do you think that?

[leaden finch]

isnt that the deintion for mult inverse?

im still confused

[mild laurel]

3 and 4^(-1) are not multiplicative inverses

3 and 3^(-1) are multiplicative inverses

[leaden finch]

mmm okay so then

[stone fulcrum]

2×4 = 1
Correct?

So that means 2 and 4 are multiplicative inverses.

When they write 4^(-1), that's just another way to write 2

[leaden finch]

hmm yes

so for ths one i am lost

for part b

ik Z5: { 0,1,2,3,4,}

so then i think we would do it like this 1( 3^-1)

can someone please help mee

i have homework due and i have no idea how to do it

[delicate bloom]

show your work

try to do it just like they did in that example you put a picture of earlier

[upbeat burrow]

@leaden finch were you successful?

[oblique river]

aphrodite and I walked through part b last night

they have their work written out

[leaden finch]

yes, i got it already since i was super confused last night lol

[vestal snow]

I had a question about taking graduate algebra

I'm gonna be taking it next semester as a sophomore and I wanted to know how difficult it is compared to undergraduate Algebra

More specifically, is it gonna be very difficult for me to get an A?

I am doing my UG Algebra sequence and got an A the first semester and will most likely get an A the second semester too

[mild laurel]

you'll probably be fine

grad classes will be harder, but grade very leniently

[vestal snow]

Are they usually curved?

[leaden finch]

can someone help me out with b

[mild laurel]

In some sense of curved, yeah

Usually it's just heavily weighted on homeworks

And take home tests or no tests

Or like writing a paper or giving a presentation

[vestal snow]

Thanks!

@leaden finch do you know the algorithm for integers?

Of finding gcd

It's essentially the same idea

[leaden finch]

im struggling since Z 5 confuses me

ik Z5 is { 0,1,2,3,4}

[vestal snow]

Well here's something that night help

In Z_5, 1/2 = 3; 1/4 = 4

By 1/n I mean n's inverse

So your goal when dividing is to multiply g by some cx^k such that the leading term of g times c gives you the leading term of what you're dividing at each step

Of course multiplication is happening in mod 5 here

So here's an explicit example

f = 2x^2 + 3x + 1 and g = 4x-1

So first you try to find something to multiply 4 by so that you get two

And the way you do that is by first multiplying by something to make 4 a 1

In this case, the number is 4 itself

Now you wanna multiply 1 by 2 to make that into 2

So in total, you multiplied by 8

Which is 3 mod 5

And now you multiply by x to make the x in 4x an x^2

Does that make sense?

[leaden finch]

hmm im confuses how did you en dup with 1/2 in the beginign

[vestal snow]

My bad, that was bad notation

It would be better to say that 2^{-1} = 3 in Z_5

Because 2*3=6=1 mod 5

[leaden finch]

i think im getting lost with this inverse stuff

[vestal snow]

Let's see

Alright try this

Write the following on your margin

1^{-1} = 1, 2^{-1} = 3, 3^{-1} = 2, 4^{-1} = 4

Now for a second, pretend you're dividing the polynomial in Q

What would your first step be

Wait

f = 2x^2 + 3x + 1 and g = 4x-1

Make a long division table and all that

Pretend you're in Q

What would you do?

[leaden finch]

we would divide 4x/ 2x^2 which is 1/(2x)

[vestal snow]

You're dividing f by g not the other way around

Does this look familiar?

It's exactly what the first step would be in Q

Maybe you need a refresher in polynomial division in Q?

[leaden finch]

oh yes

[vestal snow]

Wait did that mean you understood it or you need a refresher?

[leaden finch]

i need a refresher on this stuff

[vestal snow]

Alrighty

[leaden finch]

i know how to do polynommial stuff

hmm i think i just get confused when i see Z5

[vestal snow]

Give me like 10 minutes

Also, I think it would be much easier to do this over voice

[leaden finch]

yes

hmmm i have headphones

[vestal snow]

Ready whenever you are

[tribal pasture]

Can somebody verify this?

[halcyon siren]

I'm going mad, or am I right in thinking non-zero C-linear endomorphisms necessarily have a non-zero eigenvectors

[hot lake]

you are going mad

[chilly ocean]

👀

[halcyon siren]

Whys that

It's an invertible C linear map

So it can be a non-zero invertible matrix wrt some basis right

So it has a non-zero eigenvalue 🤔🤔🤔

What say ye

[hot lake]

you can have nonzero endomorphisms that are not invertible

[halcyon siren]

@hot lake eee

Isn't that the definition of an endomorphism of a vector space

[hot lake]

no

if it was, you wouldn't need to say nonzero because the zero map is not invertible

[halcyon siren]

Oh wait I might be thinking of automorphisms

[hot lake]

and if it was, you wouldn't need a new word for automorphisms

[halcyon siren]

Give me a mo'

I'm confused

Just in the basis case

How do we know pi of the generator of g even has an eigenvector

[hot lake]

the dim(g) = 1 case ?

[halcyon siren]

Ye

Don't we still need that pi(generator) to have an eigenvector

[hot lake]

well 1-dimensional vector spaces are a pretty restricted kind of vector spaces

it wouldn't be a representation

I guess

[halcyon siren]

Yeah

It wouldn't

[hot lake]

oh wait it could

we are talking lie algebras here

[halcyon siren]

Nah it has to be a homomorphism

[hot lake]

not lie groups

you can perfectly have pi(g) = 0 for all g

and still have a common nonzero eigenvector for all the pi(g)

by picking any nonzero element of V

in fact you can pick any nonzero element of V and that gives you a common nonzero eigenvector for all the pi(g)

whoever the pi(g) are in End(V)

because dim(V) = 1

[halcyon siren]

Is dim V = 1?

The lie algebra is 1 dim, not the rep

So I just need that for some g in the lie algebra the corresponding endomorphism has a eigenvector.

[hot lake]

hmm oh

yes

K is algebraically closed might help there I guess

[halcyon siren]

Tbh I'm only interested in the K = C case

So it's a bit simpler

[reef axle]

every matrix has eigenvectors

put it in jordan form

[halcyon siren]

That's what I originally thought 😅

But yeah cheers everyone

Mental lapse

[reef axle]

yeah idk why you were told you were going mad

[hot lake]

oh wait

xD

I'm sorry xD

I thought it said eigenvalues ?

so you are not mad

[vestal snow]

I think the claim is incorrect. Could someone verify?

f=x²-1, g=1, h=x-1 is a counterexample

[mild laurel]

think it should say that h(x) and g(x) are relatively prime

[vestal snow]

Yeah

That makes sense

Thanks

[halcyon siren]

How do we write a general element of the inverse quotient map?
Say q^{-1} (w) where w in V/U

Like err I want to write w plus some random thing in U

Is it w + U?

I don't remember

[woven delta]

You write w+U in the quotient space

[somber bramble]

w+U represents

1. the element in the quotient space represented by w and
2. the subset of V of all elements in the same quotient class as w (i.e. q^{-1}(w+U) with w+U in the first sense)

an arbitrary element of 2 would be written perhaps as "w+u for a u \in U"

[tribal pasture]

What does Z/(n,m)Z means?

[stone fulcrum]

I'm guessing (n,m) is the gcd? Do you have any context?

Or it's the ideal generated by n and m, perhaps

[tribal pasture]

I am guessing gcd too as well

[mild laurel]

Those are the same thing

the ideal generated by n and m is the same as the ideal generated by gcd(m,n)

[stone fulcrum]

Yus. Just wondering which interpretation I should be thinking in

[mild laurel]

doesn't really matter

[cloud walrus]

creamy shits:

[wind steeple]

I1 n ... , In subset of p then Ii subset p = I1 n ... n In subset of Ii

[cloud walrus]

creamy shits:

[wind steeple]

then p = Ii

you don't need any induction

n = $\cap$

[cloud walrus]

Zak:

[wind steeple]

$I_1 \cap \dots \cap I_n \subset p \Rightarrow I_i \subset p$
$I_i \subset p = I_1 \cap \dots \cap I_n \subset I_i \Rightarrow p = I_i$

[cloud walrus]

Zak:

[hot lake]

my god wtf are you guys doing

[wind steeple]

you haven't done i yet ?

[cloud walrus]

creamy shits:

[hot lake]

no you want to show that I1 subset p or I2 subset p

because I1 inter I2 subset p does not imply that I1 subset p

[cloud walrus]

creamy shits:

[hot lake]

yes

you can try to prove that if I1 is not a subset of p then I2 is a subset of p

"A or B" is logically equivalent to "(not A) implies B"

and if you want to go full contradiction you assume both not A and not B and try to get a contradiction from that

[cloud walrus]

creamy shits:

[hot lake]

yes ... ?

[lavish gale]

how do i approach this problem?

[hot lake]

you can show a set has even size if you can pair its elements together (aka if it has an involution without fixed points)

[lavish gale]

involution without fixed points?

[unique valve]

Is e the identity

[errant cave]

probably

[unique valve]

Just pair up elements with inverses

For the second one

And for the first one

Pair elements with their squares

[lavish gale]

hmm

[ripe crest]

Here's something I don't get. Suppose R is a unitary, commutative ring and I an ideal of R. Consider pi:R -> R/I. Now, J is an ideal of R containing I iff pi(J) is an ideal of R/I. Great.

Now, look at pi1:Z[X] -> Z[X]/X and pi2:Z -> Z/2Z.
Now, (2) is an ideal of Z and hence Z[X]/X, but the preimage of (2) under pi1 does not fully contain (X). Where am I messing things up?

[mild laurel]

the preimage of (2) under pi1 does contain (X), since 0 is an element of (2) and the preimage of 0 under pi1 is exactly (X)

And this is exactly how you prove the theorem that you stated

[ripe crest]

But, X is not in (2)

[mild laurel]

so?

[ripe crest]

Oh.

[mild laurel]

the preimage of (2) is not (2)

[ripe crest]

The preimage of (2) is larger than (2). My mistake.

Yeah.

Thanks!

[mild laurel]

anytime

[ripe crest]

I have a hard time working with polynomial rings with more than one variable. For example, C[X, Y]. Any tips?

I guess it might be a good idea to give a problem about that

Let's say showing that the ideal (X-a, Y-b) is maximal in C[X, Y]

If this were something like R[X], I know immediately what modding out by an ideal means because it's a Euclidean domain so I just look at the remainders to see what the quotient space looks like

C[X] also isn't too bad, but I'm not as comfortable with it as R[X], and then when we get to C[X, Y], I lose all intuition

It's not a Euclidean domain, it's not even a PID and it's not as easy to work with (for me) as Z[X], even though Z[X] isn't those things either

To tackle that problem, I'd probably want to show that C[X,Y]/(X-a, Y-b) is a field

[mild laurel]

Yeah, I mean, its harder for those reasons and you should expect that

That's the right idea though

[ripe crest]

Do you have any advice for how to gain intuition with C[X, Y]?

Just do problems?

[mild laurel]

learn all of algebraic geometry

yeah, just do problems

[ripe crest]

lmao

[mild laurel]

I'm not sure what type of intuition you're looking for really, things are going to be harder in 2 dimensions and this just limits the things you can do

[ripe crest]

Yeah - we had a problem on our last homework which our prof said was a weaker version of Hilbert's Nullstellensatz

Showing that the maximal ideals of C[X,Y] are of the form (X-a, Y-b)

I just skipped it because I didnt budget time properly, but now I'm trying to go back and do it

Hmmm - fair enough..

I guess it's something like modding out by ideals generated by multiple elements

At first I had no idea how to approach that - but if I had known I could quotient out by the ideal generated by each element term-wise, that would've helped so much

That one thing makes it so much easier to work with stuff like this R/(r1, r2, r3, ... , rn)

[mild laurel]

Yeah that's true

[ripe crest]

Do you enjoy algebraic geometry?

[mild laurel]

Yeah it's pretty cool

I mostly do number theory things though

[ripe crest]

algebraic number theory?

[mild laurel]

Uh, more modular form and elliptic curve stuff

but alg nt is cool too

[ripe crest]

Gotcha - do you need algebraic number theory for that?

Would you recommend a class on alg nt if I am enjoying algebra?

[mild laurel]

not really for the basics, all of these fields definitely intersect though

[ripe crest]

gotcha

[mild laurel]

if you like number theory yeah go for it

you'd need to be pretty comfortable with galois theory though to learn alg nt

[mild laurel]

@ripe crest

[ripe crest]

yeah - it'd be after the galois theory course

I did not enjoy my elementary number theory course lol

[mild laurel]

yeah idk

algebraic number theory is pretty different so

[ripe crest]

Is the stuff you're studying - modular forms and elliptic curves - fun?

[mild laurel]

haha, I'd study something else if I didn't find this fun or interesting

[ripe crest]

Elliptic curves sound cool - heard they have applications in cryptography

Can you describe to someone who knows basic ring theory and modules what sorts of things you're interested in when studying elliptic curves?

[mild laurel]

Yeah, there are quite a few cryptographic protocols that use elliptic curves

there are a lot of different things that we care about

maybe like, one example is finding integer solutions to elliptic curves

If you think of elliptic curves as just equations of the form y^2 = x^3 + ax + b, then you could be interested in the integer solutions

This is kind of because 2 variable equations of degree 2 are easy to figure out, and so this is kind of the next step

[ripe crest]

So, is that related to looking for ideals of Y^2 - X^3 - aX - b? ignore this lmao

[mild laurel]

uh, not really

[ripe crest]

Wait sorry

Finding solutions over the rationals?

[mild laurel]

no over the integers, which is what we care about when solving diophantine equations

And yeah, I guess its hard because the integers aren't a field so things aren't as nice

[ripe crest]

Ah okay

maybe like, one example is finding integer solutions to elliptic curves
oops missed tht

[mild laurel]

Maybe the more important part is that elliptic curves over fields form groups

abelian groups

[ripe crest]

ooh that sounds much more fun lol

over finite what?

[mild laurel]

typo sorry

[ripe crest]

What is the operation? Addition that is done pointwise?

[mild laurel]

no its complicated lmao

well

[ripe crest]

ah

[mild laurel]

over C, its just the addition of points in C

[ripe crest]

But I guess that complicated operatoin is used because it means something?

[mild laurel]

well because its a group

you know how nice groups are

[ripe crest]

sure

all those symmetries

[mild laurel]

and the fact that you can take the set of solutions to a polynomial and put a group structure on them means that we can use the group structure to tell us things about the set of solutions

[ripe crest]

But working with GL_n(k) with multiplication is different from working with M_n(k) with addition

Interesting - i never thought about just making a set into a group - not really caring about the operation - and then using group theory to tell us more about the underlying set

Does that involve representation theory?

[mild laurel]

oh boy

oh boy oh boy oh boy

The idea of using representation theory to study arithmetic objects, which include elliptic curves

is called the Langlands program

which you may have heard of

[ripe crest]

I've heard about it, yeah

[mild laurel]

or is part of Langlands I guess

[ripe crest]

dont understand it at all 😛

[mild laurel]

neither do I lmao

[ripe crest]

what are arithmetic objects?

[mild laurel]

elliptic curves, modular forms, l-functions mostly

[ripe crest]

p-adics?

[mild laurel]

well, automorphic forms I guess, which are a generalization of modular forms

not really, we can define all of these objects over the p-adics and that happens often but

the p-adics themselves are well understood

[ripe crest]

My algebra prof recommended the first couple of chapters in "A Course in Arithemtic" to learn about p-adics

p-adics arent arithmetic objects?

[mild laurel]

Arithmetic object isn't really a term thats used

[ripe crest]

gotcha

[mild laurel]

I mean, I guess you could call Z/pZ an arithmetic object but eh

its like we understand Z/pZ well and its used to do other things

not really to study Z/pZ itself

and its the same for p-adics

[ripe crest]

Makes sense

Similar to linear algebra i guess

[mild laurel]

yeah pretty much

we do things over the p-adics, but the p-adics aren't something we study

[ripe crest]

I see

Do you need analysis in what you do?

Beyond just undergrad Rudin stuff

[mild laurel]

sometimes

I studied some analytical number theory last semester

and there I needed a lot of analysis, mostly complex analysis

[ripe crest]

haha, well it is "analytic" number theory

I mean more with elliptic curves and modular forms

[mild laurel]

modular forms are really a complex analytical object though, so analysis things come up pretty often

[ripe crest]

ooh

[mild laurel]

and you can study the analytical side of modular forms

growth rates and such

[ripe crest]

analytic side seems kinda gross

[mild laurel]

not my favorite either

[ripe crest]

ew

no measure theory or functional analysis though?

[mild laurel]

measure theory or in particular harmonic analysis comes up every now and then

haar measure of p-adics has come up

This comes up more in arithmetic geometry stuff I think though

I don't think I've ever seen functional analysis

although I wouldn't really be that surprised

[lavish gale]

why we can write $ a^{i-j}=e $?

[cloud walrus]

Curiosity Zero:

[golden pasture]

cuz its a group and a^i=a^j

[lavish gale]

$ a^i=a^j$ and we multiply left side $ a^{-j} $ to get $a^{i-j}=a^{j-j}$ but we dont know yet $a^{-1} \in H$ and $ a^{-j}= a^{-1} \cdots a^{-1} $ j times

[cloud walrus]

Curiosity Zero:

[lavish gale]

or im over thinkin 😂

[golden pasture]

a^k for all k is in H

j-j=0

[lavish gale]

so cant we just say

$a^{-1} \in H $ cuz $a^n \in H$ for all n

[cloud walrus]

Curiosity Zero:

[fickle brook]

no

[golden pasture]

that isnt true if H is infinite tho but in this case yes

[fickle brook]

they say a^i = a^j with i>j

so i-j > 0

and it's fine

[lavish gale]

i get that part

[fickle brook]

a^(i-j) is part of our sequence

[lavish gale]

can we muntiply by something which inst part of our sequence?

[fickle brook]

wdym

a^i = a^j

this is just an equation

[cloud walrus]

Curiosity Zero:

[fickle brook]

no not necessarily

we can multiply both sides by a^-j just fine bc we're working in G

a^-j is still in G

[lavish gale]

yes

oo

tq tq

[somber bramble]

and importantly, a^{i-j} is in H because i-j>0

[lavish gale]

part I had confusion was miltipication by a^{-j}

[tepid thunder]

http://prntscr.com/rgp49v

Hi so I understand what to do if I am asked to show that C determines a G-map from sig and psi but not sure how to do it if it is from |R^2 to |R^3

[fickle brook]

what's a G-map

[tepid thunder]

Think it is just Group maps

[hot lake]

a morphism of G-modules

[tepid thunder]

<@&286206848099549185>

[golden pasture]

just define the map explicitly and show it is a map

[thorn flint]

Hi, I have a question regarding group actions:

I'm not sure I understand group actions completely but I'm confused how $y=g \cdot x$ relates to this

[cloud walrus]

Mac:

[thorn flint]

So far I know that the group action needs to satisfy two properties:

1. $g_1 \cdot (g_2 \cdot x) = (g_1 g_2) \cdot x$ for all $g_1, g_2 \in G, x \in X$ and

2)$1 \cdot x = x$ for all $x \in X$ Where 1 is the identity

[cloud walrus]

Mac:

[solemn rain]

@thorn flint okay just do the usual

@thorn flint show reflexitivty , transitivity , and symmetry

what are u having troubles with

[thorn flint]

Ah right! @solemn rain I was just confused exactly what the mapping signifies

[solemn rain]

xRy iff there exists g in G such that y=g.x

thats it

what are u confused about

[thorn flint]

Okay, i guess i'm just trying to get a visual understanding but the way you phrased it makes senes

thanks

[solemn rain]

i justwrote what was written haha but okayy

maybe you can think of it as like

xRy iff

x can be moved to y

by the group action

thats why u would call this a transitive group action

if X is in the relation

.

[thorn flint]

So I should show that:

• reflexive: $x \tilde x$ for all $x \in X$
• symmetric: $x \tilde y \implies y\tilde x$ for all $x,y \in X$
• transitive: $x \tilde y$ and $y\tilde z$ $\implies x \tilde z$ for all $x,y,z \in X$

[cloud walrus]

Mac:

[solemn rain]

xRx for all x

xRy ---> yRx

xRy and yRz --> xRz

yea y

[thorn flint]

Yea

[ripe crest]

In C[X,Y], does the polynomial X go to -a in the quotient C[X,Y]/(X-a)? I'm having a hard time seeing what happens in the quotient because I can't just do Euclidean division because C[X,Y] is not a PID so it's not a Euclidean domain

[wind steeple]

C[X,Y]/(X-a) is C[Y]

[ripe crest]

So I have C[X,Y]/(X-a) is isomorphic to (C[X]/(X-a)) [Y]

Now, C[X]/(X-a) is isomorphic to C

So we get C[Y]

nice!

thanks!

[wind steeple]

ye

[ripe crest]

Guess I just needed to read Dummit and Foote lol

[solemn rain]

Assume both H and K are normal subgroups of G with H intersects K = 1. Prove that xy=yx for all x in H and y in K

help ^ thats from dummit and foote lmao

[mild laurel]

Rearrange that equation so you can use normality

[solemn rain]

i tried to say that

axa^-1 = x , aya^-1=y for all a in G

xy = axa^-1aya^_1 = axya^-1

yx = ayxa^-1

dk how these are equal now

[mild laurel]

That's not what normality says

[solemn rain]

yea then i got them mixed up always

normality says aHa^-1=H for all a in G

no?

[ripe crest]

Yeah, but axa^-1 doesn't have to equal x

[solemn rain]

? why

[ripe crest]

Because that would imply ax = xa for all a in G and x in H. Meaning that H is in Z(G), but clearly normal subgroups dont have to be in Z(G), right?

[solemn rain]

YEA

lmao

okay

i gotta review ig

ty

[ripe crest]

np

[solemn rain]

okay this 1 is supposedly easy

prove that intersection of normal subgroups is normal

proof: let H and K be normal in G

---> ghg^-1 is in H , gkg^-1 is in K for all g in G

( h in H k in K )

let x be in H intersects K

we wish to show gxg^- is in H intersects K

x is in H , x is in K

gxg^-1 is in H , gxg^-1 is in K ( normality )

gxg^-1 is in H intersects K

done?

[ripe crest]

Yeah

[solemn rain]

cool

tysm

now im going to try the first one

Assume both H and K are normal subgroups of G with H intersects K = 1. Prove that xy=yx for all x in H and y in K

i was giveen a hint that i can just show that xyx^-1y^-1 is in H intersects K

but whatever

let g be in G

gxg^-1 is in H

gyg^-1 is in K

x is in G

xyx^-1 is in K

xyx^-1y^-1 is in K closure

same with H

so both in H interects K

right?

[chilly ocean]

yes, works perfectly

[solemn rain]

cool

tysm

[ripe crest]

What is K closure?

[solemn rain]

thats just me saying

why

its in K

[chilly ocean]

wait, both in H intersect K? both what?

[solemn rain]

y is in G

yxy^-1 is in K normality

ugh i mixedu p

variables

now im confused

[ripe crest]

Write it out on paper first

Then type it out when you think you have a solid proof

[solemn rain]

okay

[ripe crest]

Might help

[chilly ocean]

you already noticed all that was needed about xyx'y'

[solemn rain]

1 more

11 more problem

idk if its supposed to bethat easy

let A and B be groups. show that T = {(a,1) | a is in A} is normal in AxB

proof: we wish to show (x,y)(a,1)(x^-1,y^-1) is in T

(x,y)(a,1) = (ax,y)

(ax,y)(x^-1,y^-1)=(a,1) which is in T

qed?

[chilly ocean]

yes

[solemn rain]

okay

[chilly ocean]

uhh no

ax vs xa but other than that yes

[solemn rain]

what

?

i dont understand

anyways the other part

of the problem

show that AxB/T is isomorphic to B

im not so sure about this 1 cuz im not that good with quotients but heres my problem

solution*

AxB/T = { (x,y)T | (x,y) is in AxB }

(x,y)(a,1) =(ax,y) where a is in A

any element in AxB/T is in the form (ax,y) for a in A x in A Y in B

define the isomorphism f:AxB/T --> B, (ax,y) --> y

is that right?

[chilly ocean]

ye

[solemn rain]

col

[chilly ocean]

I think

[solemn rain]

ty

[chilly ocean]

I read only last two lines though

[solemn rain]

what i would not be so sure about

is htis 'any element in AxB/T is in the form (ax,y) for a in A x in A Y in B'

[ripe crest]

Read this to gain more intuition about quotients: https://www.math3ma.com/blog/whats-a-quotient-group-really-part-1

https://www.math3ma.com/blog/whats-a-quotient-group-really-part-2

[solemn rain]

okaay

tysm

[ripe crest]

"any element in AxB/T is in the form (ax,y) for a in A x in A Y in B" That's not true. The elements of the quotient are sets

We work with representatives of those sets

[solemn rain]

(x,y)T

oh

yea

yea yea i got you

any representative

right?

[ripe crest]

yup

Question for you, what is the identity element in the quotient?

[solemn rain]

the

divided group

1N

idk how to say it

the dividend XD

[ripe crest]

It's T

[solemn rain]

yea for that

example

eya

aT * T =aT

io think its so much easier

to think of quotient groups

as just the set of equivalence classes

of the relation a = b mod H

idk whats so fucking cool with defining them as fibers

in df

[ripe crest]

lol

[solemn rain]

but hey im ignorant XD

i htink im going to take a break then go to nextr section ig

lagranges theorem

[ripe crest]

(Y)

[solemn rain]

tysjmn

for the help

all;

[chilly ocean]

Does this proof look solid?

[ripe crest]

idk whats so fucking cool with defining them as fibers
@solemn rain I mean, one of the reasons that's used is because we want to say stuff like: if f:G -> H is a homomorphism, then G/ker(f) is isomorphic to f(G)

Which, imo, is much more obvious if we take the fiber viewpoint

[solemn rain]

uh oh

spoiler alert

thats an important theorem

u spoiled it for me

[ripe crest]

oh no

[solemn rain]

thats like 2 sections ahead..

ruining my fun over there

[ripe crest]

sorry :X

Should I not mention stuff you haven't learned for next time?

[solemn rain]

im joking

hahaah no

its cool

[ripe crest]

ah ok - hard to detect sarcasm over the interwebs

[solemn rain]

ok boomer

[ripe crest]

heh

[mild laurel]

yes tet

that's the standard proof

[chilly ocean]

hooray!

[ripe crest]

The maximal ideals of C[X] are just of the form (X-a) right?

[chilly ocean]

yes

[ripe crest]

Uh

In one of the solutions for a homework question, they preform euclidean division in C[X][Y].... i thought you cant do that since C[X,Y] isn't a PID

[chilly ocean]

yeah that seems wrong

[mild laurel]

did they write C(X)[Y]?

[ripe crest]

No.. It's C[X][Y]... I'm gonna ask on Piazza

[mild laurel]

Technically, C[X][Y] is a single variable polynomial ring where the coefficients are in C[X]

[ripe crest]

But, C[X][Y] is still not a PID though, right?

[mild laurel]

yeah its not

because C[X] is not a field

[ripe crest]

Right - I thought it would be enough it was a PID or even a euclidean domain, but Z[X] is a counterexample

[mild laurel]

yeah it has to be a field

[ripe crest]

Well, I've posted - now to wait

[oblique river]

you can still perform euclidean division in Z[x] as long as the thing you're dividing by is monic (or more generally as long as the thing you're dividing by has a unit for its leading coefficient)

The same will be true in C[x][y]

[ripe crest]

Oh - I remember that from Z[X]

I did not know the same applied in C[X][Y]

Is there a theorem for this?

[oblique river]

it's just the euclidean division algorithm

you can just see that the algorithm works

whatever proof you like for C[x], the same will work here

[solemn rain]

what are those

sorry just curious

[ripe crest]

Polynomial rings

A ring is an abelian group with an additional binary operation

that satisfies associativity (and distributivity)

And some (most?) people want their rings to contain an identity wrt that new operation

[solemn rain]

yea ring theory is up next_

for me

hope its fun

[ripe crest]

I really like it

[solemn rain]

yea

[ripe crest]

Favourite math class i've taken in a while

[solemn rain]

i really dont know if i should try learning NT or not

cuz i hear

ring theory is basically made for NT

so idk maybe i should try them

like is ring theory useless to learn if ur not interested int

in nt?

[ripe crest]

All i rmbr from number theory is congruences and prime numbers and that every natural number is the product of primes

If you know the basics, I think you'll be fine

[gentle pendant]

no, ring theory is just a core part of higher algebra. It does a lot in service to algebraic number theory, but also galois theory, algebraic geometry etc.

[ripe crest]

And I don't think it's useless to learn if you're not interested in nt. I'm enjoying ring theory even though I didn't like (elementary) nt

[gentle pendant]

although elementary number theory is a good thing to keep in mind when learning the basic theory of rings, as a many of the theorems/defns are just the natural generalisations of theorems/defns in elementary number theory.

That's kind of true of math in general though, it is fruitful to keep examples you know well (in this case, the ring Z) in the forefront of your mind when learning more general and abstract theory.

[solemn rain]

ty

[ripe crest]

Let A = C[X,Y]/(X^2 - Y^3). We denote by x and y the projections of X and Y in A.

What do they mean by projections of X and Y?

[wind steeple]

the classes

modulo X^2-Y^3

[ripe crest]

Ahh ty

[clear obsidian]

So I think this should be correct

[solemn rain]

i thin kthis is

isomorphism theorem

[ripe crest]

yeah, that's the third or second one. He's trying to prove it

[solemn rain]

lmao i thlought

he was saking if the proposition itself

was true

not the proof

asking*

[ripe crest]

We have A= C[X,Y]/(X^2 - Y^3).

Remark that the maximal ideals of A are in one-to-one correspondence with the complex points of the planar curve X^2=Y^3.

The maximal ideals of A are of the form M/(X^2−Y^3) where M is a maximal ideal of C[X,Y] containing (X^2−Y^3). So M is of the form (X−a,Y−b) where a^2=b^3.

I get everything except the last bit. Why does a^2 = b^3?

[chilly ocean]

you can think y = b and x = a since the ideal is quotiented out

[ripe crest]

So I need (X^2 - Y^3) \subset (X-a, Y-b).
That means that X^2 - Y^3 \in (X-a, Y-b).
That means that X^2 - Y^3 = (X-a)P + (Y-b)Q.
Which means, X^2 - Y^3 = XP - aP + YQ - bQ.
That implies P = X, Q = -Y^2.
So, we have X^2 - y^3 = X^2 -Y^3 -aX + bY^2.
Now, we need -aX + bY^2 = 0...

[oblique river]

x^2 - y^3 is in (x-a, y-b) if and only if x^2 - y^3 = 0 in C[x,y]/(x-a, y-b)

C[x,y]/(x-a,y-b) is iso to C

and the quotient map C[x,y] --> C[x,y]/(x-a,y-b) = C is given by f(x,y) --> f(a,b)

so f(x,y) = x^2 - y^3 is in the kernel of this map if and only if f(a,b) = 0, i.e. if and only if a^2 = b^3

[ripe crest]

hmm, lemme read that..

holy that's so smart

i'll need to spend some time digesting it - but I think I get it

Why does my method not work?

[oblique river]

Which means, X^2 - Y^3 = XP - aP + YQ - bQ.
That implies P = X, Q = -Y^2.

that's just not true

trying to do specific polynomial manipulations is just really not feasible in multiple variables with more complicated expressions

[ripe crest]

Oh yes, I see

[oblique river]

You can do the same logic as I did without referring to quotients though

[ripe crest]

because P and Q can have X and Y in them

[oblique river]

x^2 - y^3 = (x-a)P + (y-b)q. The right hand side is 0 once you plug in x = a and y = b. Therefore the left hand side must also be 0 when you plug in x = a and y = b

(yep -- that's the issue indeed)

[ripe crest]

do you mean (y-b)?

[oblique river]

yep sorry

You could also do a division algorithm type thing. Divide x^2 - y^3 by x-a and you'll get
x^2 - y^3 = q(x,y)(x-a) - r(y)
you know the remainder only has y terms because it's degree in x must be less than 1 (by the euclidean algorithm). then you can try to argue that r(y) must be divisible by (y-b)

but I'm not exactly sure how

seems hard

[ripe crest]

Ahhh that's good. We look at the evaluation map that sends x->a and y -> b and see what happens

[oblique river]

that's the best way to do this kind of arithmetic with ideals imo

in any ring R: r is an element of I if and only if r = 0 in R/I

going back to C[x,y], it tells you immediately that the element x^7 - 2xy + y is in the ideal (x-1, y-1) because if you plug in x = 1, y = 1 you get 0

[ripe crest]

Huh right.

I always thought I needed euclidean divsion to check where things got sent, but I can just plug it in even when I dont have euclidean division and see what I get

[oblique river]

yep

[ripe crest]

Okay, for something like C[X,Y]/(X^2 + XY + 1). The maximal ideals of that are maximal ideals in C[X,Y] containing (X^2+XY+1) modded out by (X^2+XY+1). Since maximal ideals of that ring are of the form (X-a, Y-b), I get that a^2 + ab + 1 = 0 in the quotient. So all maximal ideals that satisfy that relation are maximal ideals of the quotient.

[oblique river]

yes

the maximal ideals of the quotient are of the form (x-a,y-b) where a and b satisfy a^2 + ab + 1 = 0

[ripe crest]

Uh wait... here x and y are what, complex numbers?

No - they can't be - (X^2 + XY + 1) isn't maximal

[oblique river]

no, x and y are elements of the ring C[x,y]

for example, (x-1,y+2) is a maximal ideal of that quotient

because 1^2 + 1(-2) + 1 = 0

[ripe crest]

I thought: the maximal ideals of the quotient are of the form (x-a,y-b) where a and b satisfy a^2 + ab + 1 = 0, modded out by (x^2 + xy +1)

Do you not need to mod out the maximal ideals in C[X,Y] containing (X^2 + XY + 1) by (X^2 + XY + 1) to get the maximal ideals of C[X,Y]/(X^2 +XY + 1)?

[oblique river]

sure, you can also say that maximal ideals of the quotient are of the form (x' - a, y' - b) where x' and y' are the images of x and y in the quotient

[ripe crest]

ah okay..

[oblique river]

I mean, yes, technically (x-a,y-b) is an ideal of C[x,y] not C[x,y]/(x^2 + xy + 1)

but in a case like this there is no risk of confusion. If I'm talking about (x-a,y-b) as an ideal of C[x,y]/(x^2 + xy + 1) then I clearly mean the image of if

[ripe crest]

Ah gotcha. The image of an ideal is an ideal because we have surjective map

[oblique river]

indeed

[ripe crest]

Okay, and the image of (X+i, Y) would also be a maximal ideal of the quotient, right?

[oblique river]

yep

as would (x-i, y)

and (x-a, y-b) for any a and b with a^2 + ab + 1 = 0

:)

[ripe crest]

phew

finally seems like I get these questions

[oblique river]

I have to run cuz the store closes in 30 minutes and I need food

great!

[ripe crest]

Thank you!

No problem!

are you getting bananas

[oblique river]

of course :)

[ripe crest]

xD

[oblique river]

here is a fun exercise to think about that puts a lot of these things together