groups-rings-fields

[misty meadow]

And as x - 3 | 2 as well

So x - 3 | 5 & x - 3 | 2

Therefore x - 3 | 10

x mod 10 = 3

Or just in quick terms

Does that make sense?

I went very quick

Can explain better if you're still unsure

[hollow shore]

the line is mod?

[misty meadow]

so ≡₂ means is congruent mod 2 I believe

[hollow shore]

yes

[misty meadow]

I'm pretty sure it is although I've not seen much of that notation

[hollow shore]

i havent seen the one u used earlier

[misty meadow]

do you want me to break down how I did it?

Ohhh

[hollow shore]

what does it mean by

[misty meadow]

You mean x | y ?

[hollow shore]

this

[misty meadow]

Okay

[hollow shore]

yea

[misty meadow]

So x | y means x divides y

Like x is a factor of y

So

[hollow shore]

why u divide by the mod?

[misty meadow]

so do you agree x ≡₂ 1 means x is congruent to 1 (mod 2)

[hollow shore]

yes

[misty meadow]

^ I'll explain for you

Okay

[hollow shore]

okie

[misty meadow]

So do you agree x - 1 ≡₂ 0 ?

[hollow shore]

yep

although im not sure if we are allowed to do that

cuz i havent been taught that

[misty meadow]

So that is the same as x - 1 / 2 = n remainder 0

[hollow shore]

right

[misty meadow]

Idk if you can however It's basic stuff if that makes sense ? You might just use different notation

You don't have to use | if you don't like

But like you know when I divide modulo n there will be a remainder from 0 to n-1 right

[hollow shore]

yes

[misty meadow]

So this is basically saying we can divide by 2

As there's no remainder

Then minusing 2 again is also going to be remainder 0

as we have (x - 1) - 2

x - 1 is divisible by 2 in this system so so is - 2 right

so x - 3 is divisible by 2

we know x ≡₅ 3

So we also know x - 3 ≡₅ 0

Thus x - 3 is divisible by 5

hcf(2,5) = 1 (as they are coprime) (and also prime)

So x - 3 must also be divisible by 10

Thus x - 3 ≡ ₁₀ 0

[hollow shore]

hmmm i understand this but applying this in the exam is another matter

[misty meadow]

x ≡ ₁₀ 3

[hollow shore]

so for this question

i got x is congruence to 13 mod 40

is that right or am i wrong

but i worked that out using Euclid's Algorithm

[misty meadow]

ohhh gotcha

Sorry I kinda skimmed the explanation he gave initially

(it's 330am for me!)

[hollow shore]

all g

[misty meadow]

Euclid's algorithm kinda guarantees stuff

[hollow shore]

can u double check if i did that right?

[misty meadow]

I think so?

I'm bit sleepy

Can I ask a quick question which might help

Are you doing group theory or ring / field theory

[hollow shore]

ring n field

i got a question, if i got 2x is congruence to 2 (mod 4)

can i divide both sides by 2 and get x is congruence to 1 mod 4?

[empty rose]

"dividing by 2" isn't really well-defined mod 4

[misty meadow]

Yeah it's not really

I try to avoid division imo

[empty rose]

like 2 divided by 2 mod 4 could also be 3

[hollow shore]

the other option is x4

[misty meadow]

Yeah as ℤ/4ℤ isn't even an integral domain

[hollow shore]

to find the inverse

wait no not 4

aye how would u guys work that out o.o

[empty rose]

i think you can divide by anything coprime to the thing you're modding by
so for instance it's valid to divide by 4 modulo 9, and it's the same as multiplying by 7 because 4*7 = 1

[hollow shore]

so bee did i do this right

[empty rose]

(i'm not a guy)
well i think if 2x is congruent to 2 mod 4, then x is congruent to 1 mod 2

[misty meadow]

Idk as can't we can't get generators?

Sorry sleepy

[hollow shore]

implying i gotta also divide the modulo by 2

[empty rose]

no, look at 33

[misty meadow]

Good revision though

I got my algebra exam coming up in like just under two weeks

[empty rose]

that's congruent to 3 mod 10, and to 1 mod 4, but not to 13 mod 40

knowing a number mod 10 and mod 4 is only enough to determine it mod lcm(10, 4) = 20

[hollow shore]

this is all over the place

can you pls show me how to work it out using Euclid's Algorithm

[empty rose]

...doesn't euclid's algorithm compute a greatest common divisor? i'm not really sure how that would be useful here

[hollow shore]

yea but like im trying to avoid CRT

😱

[hollow shore]

how do we work this out guys

and or girls o.o

[summer path]

#elementary-number-theory

[hollow shore]

anyone good with proof?

[sonic coral]

rings are abelian under addition

[empty rose]

i'm not sure what level of formalism they're expecting here
are you "allowed" to just do algebra on the elements of a ring as if they're something more familiar like integers, or do they want you to explicitly cite axioms of rings?

[hollow shore]

ring

yes

[long nebula]

Seems like the latter

Based on the quesiton

Cephy do you know the axioms of a ring

[hollow shore]

not expert at it but yes

cant i just say

(a+b)(a-b) = .... expanded it out

[long nebula]

what axiom are you using to do that?

[hollow shore]

use the axioms of a ring to cancel out the middle term

commutative ofc

ab = ba

[long nebula]

okay take it step by step

[sonic coral]

that’s not true in all rings

[long nebula]

what axiom are you using to justify that (a+b)(a-b) = a^2 - ab + ba - b^2?

[hollow shore]

addition is commutative in a ring

if my memory doesnt betray me

[sonic coral]

.

[long nebula]

No you need to use another axiom

[hollow shore]

oh

that?

[empty rose]

what's "that"

[long nebula]

No, that's the same as what you just said

[hollow shore]

rings are under abelian under addition

[sonic coral]

why can you cancel out the terms

[long nebula]

that's true, but it doesn't justify why (a+b)(a-b) = a^2 - ab + ba - b^2

I mean, the entire statement is false, but understanding why this step is true will probably be helpful

[hollow shore]

why is the statement false?

im confused

[long nebula]

because ab isn't necessarily equal to ba

[long nebula]

but this equation is true

[hollow shore]

dont we just substitute ab = ba

[long nebula]

why is ab = ba? can you come up with a ring where this isn't true?

[hollow shore]

nope

[long nebula]

what about the ring of 2×2 real-valued matrices

is AB always equal to BA for A, B being matrices?

[hollow shore]

nope

i see

[long nebula]

So can you give a counterexample for 13i)

[hollow shore]

two matrices A and B

[long nebula]

what are A and B equal to

[hollow shore]

um i need to work that out lol

[long nebula]

you need to give a specific example

oh okay lol

[hollow shore]

not good with matrices

[long nebula]

that's fine

in the meantime do you want to talk about ii)?

[hollow shore]

so that mean i is not always true

[long nebula]

Yes i is not always true

[hollow shore]

for ii give me a sec

i ned to read it again

ii should be true tho right

from a(b+2) = ab + 2a

compare that to the rhs

[sonic coral]

why is ab + 2a equal to 2a + ab?

[slow cloak]

a(b + 2) = ab + a2

(it is the case that a2 = 2a, but you need to show that)

[hollow shore]

o my

a2 = 2a

[slow cloak]

use 2 = 1 + 1

[gilded gull]

Maybe it would help if you noted each property you use when simplifying?

Just make sure you arent skipping steps :p

[hollow shore]

a(1+1) = (1+1)a

[gilded gull]

Ok but multiplication isn't necessarily commutative...

R isnt necessarily a commutative ring

[hollow shore]

can u give a counter example?

[gilded gull]

No you're right in that 2a = a2 in this case

[hollow shore]

should be true yes?>

[gilded gull]

But this step is incorrect logic

[hollow shore]

oh okay

[gilded gull]

Use the distribute property

On the lhs

[gilded gull]

^^ and make sure you are using the axioms of rings in each one

[hollow shore]

ty

[gilded gull]

Yw

If you want to type everything out here it shouldnt take too long and I can check

[long nebula]

ayyyy Rishi

🥺

[hollow shore]

i wrote it out

i think

what axiom would i say for

[gilded gull]

Hola

[hollow shore]

2a = a2

[gilded gull]

Ok so

The hint before was 1+1 = 2

Then distribute

[hollow shore]

so

(1+1)a

applying the distributive property

[white oxide]

What do they mean by $F(\alpha_1)$ is algebraic over $F$? Do they mean that it's an algebraic extension of F?

[cloud walrus]

okeyokay

[white oxide]

wouldn't that imply that F is algebraically closed as well since $F \leq F(a)$

[cloud walrus]

okeyokay

[long nebula]

every element of F(alpha_1) is algebraic over F

[long nebula]

No, how so?

[white oxide]

every element of F(alpha_1) is algebraic over F right

and F lives in F(alpha_1)

therefore every element of F is algebraic over F

and my book defines a field F as being algebraically closed if every nonconstant polynomial in F[x] has a zero in F

[long nebula]

every element of F is already algebraic over F (1 for example is a root to x-1)

Your definitions are right but I think you're confusing the logic

[white oxide]

oh i think i see what you mean

[long nebula]

how does every element of F being algebraic over F (which is true for every field F) imply that it's then algebraically closed?

for example take F to be Q

[white oxide]

here's the definition: A field $F$ is algebraically closed if every nonconstant polynomial in $F[x]$ has a zero in $F$.

[long nebula]

Q is algebraic over Q, Q(sqrt(2)) is algebraic over Q, but neither of those fields are algebraically closed (for example they don't have a cube root of 2, so x^3 - 2 has no root)

[cloud walrus]

okeyokay

[long nebula]

yes this is correct

[white oxide]

doesn't being algebraic over F imply that it's the zero of some polynomial with coefficients in F?

oh wait

[long nebula]

Yes, but just because each element is the solution to some nonzero polynomial doesn't mean that all nonzero polynomials have roots in F

[white oxide]

I'm thinking about the converse

that every single element of F is a zero of some nonconstant polynomial in F, but I can just take x - c for c in F

duh

[long nebula]

Yeah, in F(alpha_1) => root of a nonzero polynomial in F[x] but not the other way around

[white oxide]

ahhh ok got it that makes so much sense

thanks!!

[long nebula]

yup, F is algebraic over F for that reason

no problem! :)

[white oxide]

just to clear things up, the sentence: Let $\overline{\mathbb{Z}_p}$ be an algebraic closure of $\mathbb{Z}_p$. is saying that $\overline{\mathbb{Z}_p}$ is a maximal algebraic extension of $\mathbb{Z}_p$? I know an equivalent definition is that it contains all zeros of $\overline{\mathbb{Z}_p}[x]$

[cloud walrus]

okeyokay

[lethal dune]

yes equivalent

[hollow shore]

halpp

[rustic crown]

whut u try so far

[hollow shore]

prove the commutativity + associativity + distributivity + identity element

what is the inverse?

for part a, what i tried so far

[rustic crown]

yea so you need tell the additive inverse of (n, a).

[hollow shore]

I worked it out

I think

Additive inverse is just (-n, -a)

[rustic crown]

yep!

[hollow shore]

And multiplicative inverse should be

[rustic crown]

so you've shown it's a commutative ring

[hollow shore]

(1, a) and (-1, a)

[rustic crown]

[zenith hollow]

[summer path]

[hollow shore]

hey guys

we can reduce this by mod 5 right

to check whether or not this poly is reducible

[lethal dune]

no it's not conclusive, also what coefficient

[hollow shore]

i worked it out

[frigid lark]

Did you show that this polynomial is irred mod 5?

Or that it has no roots

[lusty marlin]

0 is a root of this polynomial mod 5

[hollow shore]

reduce mod 5

prove that the reduced poly has no real rooot

[frigid lark]

Oh lol

It is

[hollow shore]

then by definition the main poly is also irreducible

ya 0 is a root for the reduced poly

otherwise irreducible

[frigid lark]

As Knight watch said, the reduced polynomial, f, can be written as X^2(X^2+2)

Well, reducing mod 5 doesn't do anything for you

[hollow shore]

hm thats true

should i reduce mod 3 2 or 7 next smh

[frigid lark]

Also to tell a degree 4 and above polynomial, f in k[x] is irreducible, takes more work than just saying f has no roots in k

For example (x^2+1)^2 in Q[X]

[hollow shore]

Ugh so lazy ima stop here for tonight

Enough maths for the day :3

[frigid lark]

Ok, reducing mod 2 gives you the answer

But remember to show that your reduced polynomial cannot be split into quadratic factors

[hollow shore]

oh i did that

but for mod 5 i forgot that 0 is a root

lol

[hollow shore]

sweet thanks

[tribal niche]

Can I get a hint for the following problem?

If S contains the identity then S is a subgroup so we're done, but I don't know how to approach the general case

Maybe you can biject S to a subgroup in some way?

[coral spindle]

A set S with this property may not be a subgroup of G in general.
Hint: what other kind of subset of a group has the property that |S| divides |G|?

Your idea is a good one

[tribal niche]

Oh, I think I've got it - thanks!

[slim kayak]

Why does this tell me that $M \otimes_{R} A$ is an A module?

[cloud walrus]

Kerr

[coral spindle]

It's a right A-module because we can define the multiplication on simple tensors as (m x a) . b = m x ab

using ascii notation here, forgive me

the isomorphism has nothing to do with it

[slim kayak]

How does one check if it distributes over sums of simple tensors?

[coral spindle]

We simply define it as such

You then need only check that it respects the defining relations of the tensor product, i.e. bilinearity.

[slim kayak]

Check it for simple tensors?

[coral spindle]

No

I'm stating that you need to check well-definedness

I don't know if you've seen the definition of the tensor product as a quotient of the free module

if you've seen that, then I'm asking that you show that this definition stabilises the module that we quotient by

[slim kayak]

What do you mean by stabilizes? Like in the group action sense?

[coral spindle]

Let M be an R-module and let N be a submodule of N. Let phi : M -> M be any map. When I say here that "phi stabilises N", I mean that phi(N) is a subset of N

The point is that we need zero to be sent to zero in the tensor product for well-definedness, and that's the same as the module that we quotient by being stabilised

[slim kayak]

So in this case verifying that for all b and elements (n,m) in N, that (n,mb) lies in N?

[solar glacier]

cani ask a question regarding a particular group theory proof i have a solution to i wanna know if im on right track

let G be group of order n >2 show there does not exists a subgrou H with order n-1 (without use of LaGrane). Here is what I have thus far:

Suppose there exists such a subgrou and G has order at least 3 and H has order at least 2 but order 1 less and let g \in G \ H and let h \in H \ {e} which exists as H is of order at least 2

then if gh \in H then ghh^-1 \in H which implies g \in H so gh \in G \ H and thus gh = g forcing h = e which is a contradiction by our choice of h done. does this argument hold

[rustic crown]

yee that works

[solar glacier]

cool thanks det 🙂

[true drum]

Is this a question from ch. 1 of dummit and foote?

[south patrol]

feels like it's basically the same length just to prove lagrange's theorem lol

[south patrol]

hm are there any nice tricks to check whether elements of k[x_1,...,x_n] are irreducible lol

[coral spindle]

Well I mean

[south patrol]

atm the main thing i know is just to consider it as an element of e.g. k[x_1,...,x_{n-1}][x_n]

lol

[coral spindle]

if k = frac(D) for some UFD D, we have tricks like Eisenstein

[south patrol]

so like y^2 - x^2 - x^3 is irreducible in k[x,y] = k[x][y] since no square root of x^2 + x^3 lol

[coral spindle]

Oh in multivariate stuff

wow idk

[south patrol]

But mroe generally seems a bit of a pain

I will try Whitneying and inducting, thanks Tterra

[chilly ocean]

hope that helps 👍

[south patrol]

👍

Can anyone give a simple and intuitive explanation of Whitney and Induct

[chilly ocean]

yes but only if you're aware of the fact that 0 is not linear

[south patrol]

What

[chilly ocean]

WHAT!

[coral shale]

[coral spindle]

How did this simple and intuitive thing come about

I was away from this server for a while

[chilly ocean]

some guy who kept asking questions with words such as "simple" and "intuitive" bolded. the questions were often neither simple nor intuitive

on the occasion that there was a simple and intuitive answer it usually involved the bare minimum of thinking

[coral spindle]

Oh yes, that guy

[slim kayak]

Well, you see... You take your element and just draw the higher-dimensional algebraic set produced by that particular element and see if it is a variety or not /s

[formal ermine]

no he kept posting the same question for like 7 days straight

in the abstract algebra channel

and it had like nothing to do with algebra lol

[elder wave]

this isn't when it started

[formal ermine]

is it not?

he was the first person to use it

[elder wave]

yeah but that wasn't the first time/question

[coral spindle]

It had been happening for a while

[elder wave]

simple and intuitive lore

[formal ermine]

it was the first time that I had seen that so I thought that would be the beginning lol

[elder wave]

there was this one AG question that made chmonkey mald

that's when i drew this

[lethal dune]

Lmao

[elder wave]

yes this is the one

[coral spindle]

why is this guy

[lethal dune]

Was it resolved at the end lol?

[slim kayak]

within a few minutes of the same time each day, bro had a schedule

[sage lodge]

"simple and intuitive" is what you say to ChatGPT

[formal ermine]

yes

by tteg

[lethal dune]

Lol

[lethal dune]

Should have sent a chatgpt solution

[slim kayak]

Jesus, first instance of them posting the same question was in may

[formal ermine]

hi det

[slim kayak]

4 months of simple and intuitve problem statements

[rustic crown]

det sleepy

illu so quic

[formal ermine]

how are you besides sleepy

did you have a day off today too?

[rustic crown]

sat is usually off

[formal ermine]

oh

I thought it's friday

[slim kayak]

"the determinant is very tired, he is eppy"

[rustic crown]

eppy?

[formal ermine]

for the past 4 days my days have been:

• wake up at 3-4 pm
• play the new zelda till like 8 or 9 pm
• do math
• sleep

oh wow fancy discord formatting

having a little school break is nice

I fell off a bike today though lol

[coral spindle]

hey don't do that

[formal ermine]

I only have a few injuries but my phone is damaged

[slim kayak]

The two ways of life: gaming and math addiction

[warped fable]

Could be interesting!

[formal ermine]

it's these stupid train grooves' fault

[south patrol]

how do you do that fancy formatting

[formal ermine]

- a

[south patrol]

-cringe
-based
-bluepiled

[formal ermine]

• a

[south patrol]

Oh

• cringe

Nice!

[coral spindle]

A friend of mine was very seriously injured by clipping his bike wheel in tram tracks. They're a serious danger

[rustic crown]

• 
• 
• 

[formal ermine]

ah tram tracks

that's what they're called

yeah I have a big wound on my leg but I'm still sad about my phone

there are so many splitters

[warped fable]

oh damn

[coral spindle]

Honestly I think you came off lucky

[formal ermine]

the display is half broken

[coral spindle]

I'm glad you're not injured

[formal ermine]

would have rather gone to the hospital but with an intact phone 👍

[coral spindle]

My friend had broken bones lol

[formal ermine]

anyway time for some algebra

my prof wrote:

$$\bC[x_{i, j}, y : 1 \leq i, j \leq n]/(\det(x_{i, j})\cdot y - 1) \iso \bC[x_{i, j}, \operatorname{det}^{-1} : 1 \leq i, j \leq n]$$

what does the det^-1 here mean?

bonus points if your explanation is simple and intuitive

[elder wave]

inverse

[elder wave]

👍

[formal ermine]

inverse in \bC(x)?

[elder wave]

tbh i don't get your notation

[formal ermine]

don't blame me

[rustic crown]

localization

[elder wave]

no it is that

[chilly ocean]

det is a polynomial in C[x_{i,j}] so it has an inverse in C(x_{i,j})

timo........

[formal ermine]

yeah

that's what I meant

is it just that?

ok epic

thanks

[summer path]

Inverse of is ?

[formal ermine]

what order does have

[summer path]

Zanzan in aa

[hollow shore]

morning guys n girls :3

similar question

part d

is c just 1?

[coral spindle]

But it's 11:53pm D:

What you need to find is those values of c for which x^2 - x + c has no roots

Oh wait part d, not part c

[hollow shore]

yes i found c

how do we work out d?

[empty rose]

i'm not either of those things and also it's not morning here

[coral spindle]

Well we know x^2 - x = - c, so note that x(x - 1) = -c

So maybe you can think about what the value of x^-1 is in terms of c.

[empty rose]

(what was your answer to (c)?)

[hollow shore]

1, 5, 8, 9, 10, 11

c is an element of {1,5,8,9,10,11}

[hollow shore]

i dont get this

why is x^-1 in term of c?

[coral spindle]

x(x - 1) = -c, so what is x^-1?

[hollow shore]

1/x = 1/-c?

is that what u meant?

[coral spindle]

Is that what you get from that equation? I don't like to see guesses

[formal ermine]

wrong, it's 1 am

[empty rose]

actually it's midnight tomorrow

[hollow shore]

where and how u got x^-1

did u just want me to sub x^-1 in

[formal ermine]

you live in the wrong timezone

[formal ermine]

what is the definition of x^-1?

[hollow shore]

inverse

[formal ermine]

yes but what is the definition of an inverse

[hollow shore]

reciprocal?

[formal ermine]

no...

if a is the inverse of b then...

[hollow shore]

a is opposite of b?

unless you're implying the identity element

[coral spindle]

Gonna give you a HUGE hint here

let's assume that c is nonzero

then c has an inverse in Z_13

so x(x-1)/(-c) = 1

Think now about what x^-1 is

[hollow shore]

-1

okay waitttt

[coral spindle]

I'm not going to respond to random guesses

[south patrol]

Okay you may want to look up what inverses are

[hollow shore]

isnt it just x-1?

[formal ermine]

nu

[coral spindle]

I suggest you stop fishing for answers and instead describe your thought process

If you don't have a reason why, then your answer is incomplete anyway.

We cannot help you with any misunderstandings if you just guess.

[tired horizon]

To prove that the polynomial X^4+1 is irreducible over Q, can we use Eisenstein's criteria even tho we only have 2 coeffs?

[hollow shore]

is this not right

[coral spindle]

I'm a bit insulted that you'd ask chatGPT instead

But no, it's evidently wrong.

[coral spindle]

Well Eisenstein will not be helpful here, since 1 is not divisible by any prime.

But yes, the number of nonzero coefficients does not really matter (it matters a bit, but hey ho)

[hollow shore]

i checked the working for x^-1

[coral spindle]

You may find it helpful, instead of showing that f(x) = x^4 + 1 is irreducible, to show that f(x+1) is :)

just a guess tho

[hollow shore]

why r u insulted smh

[tired horizon]

Why?

[coral spindle]

Try it and see

[fossil shuttle]

Lmao

[tired horizon]

Is there a theorem or smth

[south patrol]

This is a very well known polynomial

Well, it is one of a family of polynomials known as the cyclotomic polynomials, all of which are irreducible

[tired horizon]

Oh yeah I've seen that on the lecture, we still didn't do it but I get it yes

[south patrol]

You can also just see there are no roots and try a factorisation as a product of two quadratics

[tired horizon]

Its where we use the roots of unity right?

[south patrol]

Or factor it over R lol

[tired horizon]

So I just start by supposing that it can be reducible then find a contradiction

[south patrol]

Yes, given $n \ge 1$ we define $\Phi(x) = \prod_{\mathrm{\omega} \text { a primitive nth root of unity}} (x-\omega)$

[cloud walrus]

potato

[tired horizon]

Like it has a factorisation over R but not Q

[south patrol]

Yes

Basically it's like

Each irreducible factor (in Q[x]) is a product of irreducible factors in R[x]

[tired horizon]

Alrighty

Thanks

[south patrol]

owo

[tired horizon]

Is this considered a proof too?

I'll say that it doesn't have a root in Q

[long nebula]

Please don't ask ChatGPT for problems like these; you'll likely get misinformation. Especially don't post screenshots from ChatGPT here because you'll likely spread misinformation.

[tired horizon]

Thus not irreducible in Q

[coral spindle]

Not having a root in Q is not sufficient

it could split as a product of two degree-2 polynomials with no root!

You must rule out that case

[tired horizon]

So I have to always check that it does not have a factorisation

Too

[coral spindle]

Well there might be a shortcut!

[long nebula]

I second this suggestion ^

[coral spindle]

Wow what a cool and hot guy suggesting this

he must be epic

[long nebula]

Use Eisenstein's criterion on (x+1)^4 + 1 and then argue that if that polynomial is irreducible, x^4 + 1 must be too

[tired horizon]

Ok ill be a hot and cool girl and try it then

[coral spindle]

Yaaas queen

(sorry)

[tribal moss]

(For example, x^4+4 has no rational root, but is still reducible in Q[x]).

[tired horizon]

Poopy

Nvm

So now tell me why if we have f(x+1) irreducible means f(x) is

[coral spindle]

Well if f(x) = g(x) * h(x), then the same happens for f(x+1) right?

[tribal moss]

if f(x+1) = g(x)h(x) then f(x) = g(x-1)h(x-1).

[coral spindle]

yeah, what tropo said

[tired horizon]

Yes that was so obv somehow

[coral spindle]

the only thing left to do is to rule out the possibility of the latter factorisation being trivial

but this isn't hard

[lavish spoke]

ChatGPT and its consequences have been a disaster for the human race

[tribal moss]

Hmm, no, exactly what I said proves the wrong direction ...

[celest furnace]

It can only get better from here

[coral spindle]

haha good point

[tired horizon]

Fun story

This is my last undergrad semester

We had a small research thesis

Just to practice for grad

I had a classmate who used it to make his thesis

He was too lazy

Like WAY TOO lazy , even to check

[coral spindle]

"as a large language model"

[tired horizon]

In his acknowledgment part

[next obsidian]

Thesis paragraph 3:

“As I am a large language model”

[next obsidian]

FUCK

[coral spindle]

that was in his thesis, wasn't it?

suck shit monke

how does it feel

sit down.... oh wait YOU ARE THE CHAIR

[tired horizon]

He thanked a person working in a place

In another country

As if he was ruled in that internship

In china

[sonic coral]

doesn’t the mod 2 irreducibility test work on x^4 + 1?

[celest furnace]

What about eisenstein on (x+1)^4 + 1

I hope

[coral spindle]

Yeah we did eisenstein

[tired horizon]

Yeah i asked about that but it gives me the heebie jeebies cause it doesnt have other coeffs in between 0 and 4

[coral spindle]

oooh spooky zeroes

[tribal moss]

Eisenstein is slick and fast here, but I think "x^4+1 factors in R[x] and neither factor is in Q[x]" feels more instructive.

[celest furnace]

How do you factor x^4 +1 in R[x]?

[coral spindle]

(x^2 + sqrt(2)x + 1)(x^2 - sqrt(2)x + 1) if I'm not mistaken

no, not quite

oh no that does actually work

[celest furnace]

Totally forgot you could do that

[sonic coral]

that was my approach when i did this problem for the first time

just ruling out all the cases

[tribal moss]

Every real polynomial splits into factors of degree at most 2 over R -- just combine every pair of complex conjugate roots into one irreducible quadratic.

[tired horizon]

Can you explain to me the mod method?

Idk it

[tired horizon]

.

[coral spindle]

if f(x) is reducible (in Z), then it will still be reducible modulo 2

[solar glacier]

Work In mod 2 on the poly

[coral spindle]

therefore, if it is irreducible mod 2, then it is irreducible (in Z)

[tired horizon]

Why mod 2 specifically

[coral spindle]

it was a lucky guess

could be mod any prime

[solar glacier]

Easy to check only

2 elements

[tired horizon]

Oh okay

So we just check with primes

[coral spindle]

It's worth noting that a polynomial is irreducible in Z[x] iff it is irreducible in Q[x], which is known as Gauss' lemma

[solar glacier]

Mod any prime yes then it checks out over Z

Gausss lemma 👌

[solar glacier]

z mod 2 would be the easiest to check since there’s only 2 elements to verify your irreducibility

[wraith cargo]

is mod 2 irreducability an application of Hensel's lemma?

[tribal moss]

Hmm? There are fairly few candidate factors to check modulo 2, but why not at least 3?
x+1, x²+1, and x²+x+1

[slim kayak]

Laziness

[chilly ocean]

you don't want to worry about orientations

oops wrong channel

[tired horizon]

This is the theorem right?

[sonic coral]

yup

[tired horizon]

You guys saved me idk why the teacher never talked about it

[slim kayak]

Wasn't there a criterion involving the formal derivative of f? Kinda weird to use f' here as notation

[tired horizon]

They used it as a second polynomial

Instead of f becoming g or h after the mod operation

[celest furnace]

I think it’s more standard to put a bar over f

[slim kayak]

Hats are also popular in fashion, or tildes

[solar glacier]

$D_xf(x)$ is what’s used in algebra for@derivative

[cloud walrus]

MyMathYourMath

[coral spindle]

f'(x) is also common

so is \partial, so is just D.

[coral shale]

its just notation, anything sensible will do after defining it

[next obsidian]

I love D

For derivative

Don’t take this out of context

[sonic coral]

i’m sorry what did you say?

[solemn garden]

(x + 1)(x^3 + x^2 + x + 1) = x^4 + 1 in Z2, does this mean it’s reducible mod 2?

[sonic coral]

well normally you’d check for roots in Z2, unless i’m misunderstanding your question

[solar glacier]

There’s only 2 roots to check for in z2 to show its irreducible

[solemn garden]

1^4 + 1 = 1 + 1 = 0, so it seems like 1 is a root?

[boreal inlet]

how do I show this last statement?

Because if this holds, strong induction lets us do the rest

[solar glacier]

You’re saying the product of the first n-1 elements plus the n element have only the trivial ideal in common

[boreal inlet]

I mean they are said to be pairwise coprime for any i not equal to j

and if this holds then

[solar glacier]

By induction

You’re done

[boreal inlet]

$\mathfrak{\alpha}n \cap \mathfrak{\beta} = \displaystyle \prod{i = 1}^{n} \mathfrak{\alpha}i = \bigcap{i = 1}^{n} \mathfrak{\alpha}_i$

[cloud walrus]

rikusp2002

[boreal inlet]

but how do I show that

I basically have to show that there exists elements in alpha_n and beta such that x + y = 1 when x belongs in alpha_n and y belongs in beta

problem is

y is of form $\displaystyle \prod_{i = 1}^{n - 1}x_i$ where $x_i \in \mathfrak{\alpha}_i, i = 1, 2,3... $

[cloud walrus]

rikusp2002

[boreal inlet]

how is this equivalent to 1 modulo alpha_n

[torn warren]

For a finite group G, all elements with odd orders forms a subgroup, can I get a counter-example?

this is true for abelian group

[boreal inlet]

try S3 maybe

oh nah it does form a group my bad

A3

[boreal inlet]

anyone? 🥲

[torn warren]

no, A3 is cyclic

[boreal inlet]

yeah

I meant A3 was the group

from S3, if you take the odd order elements

Tbh I don't know about counter example here, because I remember showing even some group with order 2^n * k is not simple if the Sylow 2-subgroup is cyclic

The resulting normal subgroup only has elements of odd order

so I guess you can try breaking those conditions and try to find something?

In case of any group of order 10, exactly 4 elements have order 5

D_10 also has a cyclic sylow 5 subgroup...

okay so not these either

wait

what about D_12

@torn warren

only elements of odd order is e, r^2, r^ 4

fuck this also forms a subgroup

😭

@torn warren S5

yes

45 elements of odd order (calculate the conjugacy classes of 3 cycles and 5 cycles, every other representattive has even order, which gives us 20 + 24. get the identity. 45.)

group order is 120

45 does not divide 120

no subgroups exist of order 45

[torn warren]

right, thank you so much!

For 3cycle, I did $\binom{5}{3}\cdot 3! /3=20$, is this correct? I mean is there a general way to compute this?

[cloud walrus]

Witness

[boreal inlet]

Yes

I mean it's like you're filling places so

So that's how you do it I think

In case of 5 cycles

5C5 * 5! / 5 = 120/5 = 24

[torn warren]

yes, I did it in this way, but I don't know the name, conjugacy classes

[boreal inlet]

Oh it's just like the set of elements that we can get from the element a

Let's say we have (123)

now (12)(123)(12)^-1 = (132)

So (132) belongs in the conjucacy class of (123)

Like this you'll see, for S_n, you'll always have the same cycle type for all elements in one such set

The action is called Conjugation, (this bab^-1 thing) and it defines an equivalence class on the set of the group

So the group gets divided into disjoint sets according to those, each of which are named conjugacy classes, and each element of that class can be generated through one representative element

Like in case of 3-cycles, (123) is the representative

For n cycles, it's (12....n)

For cycles like (ab)(cd), (12)(34) is the one

So like that

S5 has uh... 7 such classes

Identity (generated by identity itself, only one element here, all elements in this class has odd order, 1)

Transpositions (generated by (12), all elements have order 2)

3 cycles (generated by (123), all elements have order 3)

4 cycles (generated by (1234), all elements of order 4)

2-2 cycles (generated by (12)(34), all elements have order 2)

3-2 cycles (generated by (123)(45), all elements have order 6)

5 cycles (generated by (12345), all elements have order 5)

Fun fact, 7 is the number of unordered positive integer partitions of the integer 5, also called partition number of 5. π(5)

It has a recurrence relation for generating it, which I frankly forgot 💀

[boreal inlet]

RIGHT

i now remember

Index of Centralizer of an element a, equals the cardinality of the conjucacy class represented by a.

C(a) is a subgroup of G if a in G, so that's covered

If you want I can try to derive the general formula, I haven't done it in a long time so I gotta think

[torn warren]

I see, so for $S_n$, for a single k-cycle case, there are $\binom{n}{k}(k-1)!$, right?

For $k, n-k$ cylce, where $n\neq 2k$, we have $\binom{n}{k}(k-1)!(n-k-1)!$
For $n/2, n/2$ cycle, we have $\frac{1}2\binom{n}{n/2}(\frac{n}2-1)!^2$

[cloud walrus]

Witness

[tired horizon]

If we have a degree n irreducible polynomial and we have a root B of that polynomial and they ask u to find the other roots, we always have to find n roots?

[boreal inlet]

Yeah kinda like this

[torn warren]

I see, so for $S_n$, for a single k-cycle case, there are $\binom{n}{k}(k-1)!$

For $k, n-k$ cylce, where $n\neq 2k$, we have $\binom{n}{k}(k-1)!(n-k-1)!$

For $n/2, n/2$ cycle, we have $\frac{1}2\binom{n}{n/2}(\frac{n}2-1)!^2$
is this right counting? @boreal inlet

[boreal inlet]

I mean that's probably what they are asking

[cloud walrus]

Witness

[boreal inlet]

I would do it like

Uh lemme write this might take time

[tired horizon]

I only found 4

Lemme show u the polynomial

[boreal inlet]

$\frac{n!}{\displaystyle \prod_{j}(j)^{a_j} (a_j !)}$

[cloud walrus]

rikusp2002

[boreal inlet]

Here a_j is the type of cycle in the cycle type L

[boreal inlet]

Let's say you have a k cycle

It's actually an k-1-1-1....-1 (n-k) times partition.

For that your a_1 = k, and rest of your a_j = 1

Spread the profuct

[torn warren]

I don't get it, what is $a_j$, I think the most easy mistake is when like the 2-2 cycle, and we need to divide a factor 2

[cloud walrus]

Witness

[torn warren]

in this case, it should be $\binom{n}{k}(k-1)!$ ?

[cloud walrus]

Witness

[boreal inlet]

I'll be providing examples in a bit

@torn warren this

Yours was missing the (n - k)!

2-2-1 cycle in n = 5 is same as 2-2 cycle

And so on

[torn warren]

wait, it shouldn't have this factor (n-k)!

if you check the 3-cycle in S_5

$\frac{5!}{1^22! 3^1 1!}=\binom{5}{3}\frac{3!}{3}=\binom{5}{3}(3-1)!$

[cloud walrus]

Witness
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[boreal inlet]

How does the 3 cycle look like in S5?

It's of form (123)(4)(5)

So it would be of form

Oh yeah my bad

Yeah (n-k)! Is already in the denominator which cancels the next (n-k)! Part, so it doesn't need to be there @torn warren

My bad, calculation erros

[torn warren]

I derive it in this way @boreal inlet

[boreal inlet]

yee

okay

[torn warren]

but I think this way is too complicated, is there a shortcut to get this?

[sly rain]

The concept of localization got introduced on a practice sheet.

Ultimately I want to show that $ Z/pZ \cong Z_{(p)} / pZ_{(p)}$. Where $Z_{(p)}$ is the localization at p.

I showed that $pZ_{(p)} = Z_{(p)} \setminus (Z_{(p)})^{\times}$ is the unique maximum ideal. All elements $(a/b) \in Z_{(p)}$ with $ p \vert a $ are non-units.

E.g. for Z/3Z, I get the structure and which elements are equal. What are the elements in $Z_{(3)} / 3Z_{(3)}$?

-3,0,3,6,9...

3/2, 6/2, 9/2...

etc.

These should get mapped to zero.

Am I approaching this the wrong way and/or how should I think about this conceptually?

As for the proof I think consider the map $ Z \rightarrow Z_{(p)} \rightarrow Z_{(p)} / pZ_{(p)} $ and since pZ is the kernel and the map is surjective (this I have to show), it follows from the homomorphism theorem that $Z/pZ \cong Z_{(p)} / pZ_{(p)}$

[cloud walrus]

aabb

[wraith cargo]

I'd think abt it through the first isomorphism theorem as you mentioned at the bottom

[tribal moss]

If all you're after is "odd-order elements do not necessarily form a subgroup", then just observe (123)(124) = (13)(24).

[wraith cargo]

oh or are you asking how should you think abt the quotient Z_(p)/pZ_(p)

[slim kayak]

the ideal generated by p or the point p?

[wraith cargo]

Well the ring is local so the latter

but aren't they the same thing

you usually say you localize at some prime ideal

[sly rain]

Yep, mostly this.

[slim kayak]

With one you are adding everything not in the ideal (p) as denominators i.e. with the complement of (p), when localizing at a point p your elements look like x/p^n

[wraith cargo]

ahh then it's the former

[sly rain]

I think so too

[wraith cargo]

The elements are 0+3Z_(3), 1/b+3Z_(3) and 2/b + 3Z_(3) but you can show that in the quotient both of those latter two are equivalent to 1/1 and 2/1, so your elements are 0,1,2

(at least when you're talking abt Z_(3)/3Z_(3))

[tribal moss]

Suppose you have 1/a in Z_(p) where a is not a multiple of p. Then a is invertible modulo p, so there is b in Z such that ba=kp+1.
But then 1/a - b = (1-ab)/a = -kp/a which is in pZ_(p), so 1/a equals b in the quotient ring, and then it's easy to see that the quotient is at most Z/pZ.

But the quotient can't be trivial either because 1 is not in pZ_(p).

[sly rain]

Thank you

[sly rain]

Thank you too

[twilit notch]

My professor is saying that in Q[sqrt(-1)], 3 is a prime. But the norm of 3 is 9 right? Which is not a rational prime? So how is 3 a quadratic prime

(assuming that if N(x) is not a rational prime, then x is not a quadratic prime. which could be wrong, but that is what my professor was using. i thought textbook only said that if N(x) is a prime, then x is a prime, and not the other way around, so i was confused about that too.)

[next obsidian]

Idk what a quadratic or rational prime is, but the only direction you have is N(x) prime => x prime, not the other way around

[twilit notch]

i know, i was confused why my professor was using it backwards

[next obsidian]

The definition of prime I’m thinking of involves taking the ring of integers and asking if it’s an actual prime there. In this case you look at Z[i], and some extra work gets you that a prime p in Z becomes not prime iff it is equal to 1 mod 4

[twilit notch]

wait why does it become not prime?

[next obsidian]

Idk, some math

You get to write it as

(a + bi)(a - bi)

For some a,b

[twilit notch]

yeah

[next obsidian]

I don’t get this proof outline though using though

[twilit notch]

but if a prime p is in Z, b = 0 right?

[next obsidian]

No

5 = (2+i)(2-i)

So 5 does not stay prime in Z[i]

[twilit notch]

i meant p defined in Q[sqrt(-1)] itself

[next obsidian]

What does that mean

I meant a prime number in Z

You can say exactly when it becomes not prime

Adjoining i

In this case p = 3 which is not 1 mod 4

[twilit notch]

oh i see, yeah that makes sense

but it is not exactly what i was referring to

[next obsidian]

But so even ignoring that this uses a not true statement

Idk what this proof is supposed to get

It seems like you’re asking why 3 is a quadratic prime or whatever, but the conclusion of this is that an x isn’t a quadratic prime?

[twilit notch]

well.. yes, if the theorem was only used based on the correct definition there isn't an issue because N(x) not being prime would not imply that x is not prime

the thing is i'm confused whether the theorem is an iff or an if statement, because i saw it being used both ways

but if that is wrong then it makes sense

[next obsidian]

It is only true that
N(x) prime => x prime

This is equivalent to x not prime => N(x) not prime

[twilit notch]

yes, ok

[next obsidian]

Cool

[twilit notch]

thank you!

[next obsidian]

[summer path]

i always thought chmonkey only lives in AG channel >.<

[next obsidian]

No

Kekw

I have more messages here I think

Yeah like 5x more

AG channel didn’t use to exist

[summer path]

[next obsidian]

Drink water and eat food

[summer path]

thanks for reminder

[formal ermine]

I'm not sure if this already counts as geometry but

if $$f(x, y) = \sum_{i + \mu j = v} c_{i, j} x^i y^j$$

and if f is irreducible

[warped fable]

if you have to write a formula it probably isnt geometry

[formal ermine]

then how do we know that x^v and y^(v/mu) appear in f

[formal ermine]

well it is connected to the newton polygon of f

[warped fable]

give me a good reason i should care about the newton polygon

[formal ermine]

solving algebraic curves

[warped fable]

how do i solve a curve

[formal ermine]

that's what I still have to learn

[warped fable]

a curve is a wiggle on the plane

what does solving a curve even mean

[agile burrow]

v + \mu * 0 = v

[formal ermine]

yeah but why shouldn't c_v,0 be 0

[agile burrow]

oh i mean i guess it depends on what the constants are

but idk the context so

[warped fable]

probably irreducibility

[agile burrow]

you get a summand of x^v with some coefficient

[formal ermine]

if the support of f lies on a straight line x + mu y = v

then we call f quasi homogeneous wrt mu and v

we somehow gotta use irreducibility though

[agile burrow]

I mean

If coefficient of x^v is 0 then every summand has a y

[formal ermine]

how so?

oh

[agile burrow]

bruh

[formal ermine]

wait I think I see

wait

[agile burrow]

If j = 0 then i = v, so the only summand with no y factor is x^v

[slim kayak]

You can't have the sum of two squares ever equalling 3, so there simply is no number of norm 3.

[formal ermine]

🤦

yeah ok lmao

and y^v/mu has to appear for the same reason right

[agile burrow]

yes

[twilit notch]

not norm 3, norm of 3

[agile burrow]

also is your reference using a v or a nu

[formal ermine]

,, \nu

[formal ermine]

I can't tell the difference

[agile burrow]

ok

[slim kayak]

No, norm 3. If 3 wasn't prime you need two numbers a and b, neither of which a unit, so that ab=3. But then a and b must have norm 3.

[twilit notch]

oh i understand now, yeah that makes sense

[warped fable]

another german masquerading as an algebraist

am i surprised? no

[formal ermine]

fuck

newton polygons are just numerical analysis in disguise

[warped fable]

😂

[formal ermine]

my prof's lecture notes are just "this works" without explaining why it works

[warped fable]

german superior mathematics

its funny how germans became the french, math culturewise

every german i know does algebra and cant understand a single piece of visual reasoning

[slim kayak]

spill the tea

[formal ermine]

lol

ahahaha

omg he finally explains wtf we are actually doing and how it works

[warped fable]

chad

[south patrol]

Lol

[formal ermine]

hi potato

[formal ermine]

nvm I still don't know what's going on

the quasi homogeneous case I understand

but if it's not he splits it up into multiple quasi hom parts and then does an approximation????

[south patrol]

Hiii

[formal ermine]

ah wait

so

we do approximation

infinitely many often

(the chad way)

then write it as a series

and get a puiseux series or whatever that's supposed to be?

[thorn monolith]

How to prove that S_6 has no subgroups of order 40?

[agile burrow]

GOOD question

If you have a subgroup H of order 40, then it has a 5-Sylow subgroup isomorphic to Z/5 which is necessarily characteristic in H, hence normal in S_6

But S_6 definitely has distinct 5-Sylow subgroups

Actually yeah, I quite like that

[thorn monolith]

didn't quite get it

[agile burrow]

Oh

Well like, the groups generated by (1 2 3 4 5) and (2 3 4 5 6) are distinct 5-Sylow subgroups

But all 5-Sylow subgroups are conjugate, so none of them can be normal

[thorn monolith]

oh okay

[warped fable]

6 doesnt divide 40 so does the problem reduce to proving S5 has no subgroup of order 40, also

um no lol

[agile burrow]

That too

Or wait

[warped fable]

yeah this approach doesnt work immediately

[agile burrow]

Maybe you can say something like if you had a subgroup of order 40 then it would be conjugate to something contained in a copy of S_5 inside S_6

Either way, I would use this argument for the same question about S_5 lol

[warped fable]

I like your argument because it would take me ages to come up with that, if ever

Very smart

[agile burrow]

I liek group theory

[formal ermine]

i fucking hate newton polygons

they are making me go insane

[summer path]

I like det, very cute

[thorn monolith]

i have a few more questions about group theory ahaha

[rustic crown]

tubu kawaii

[agile burrow]

If you post them here, someone will usually get around to answering it

[lethal dune]

Why being char subgroup of H should imply it's normal in S_6?

[agile burrow]

Oh wait

[warped fable]

oh H needs to be normal

[agile burrow]

Yes you're right

[lethal dune]

yeah

[south patrol]

Good question

[agile burrow]

I'd have to look at it

[south patrol]

I may have a think lol

[agile burrow]

Or maybe you could look at like normal core or something

[south patrol]

This is too big a group lol

[warped fable]

This works for S_5

The index is sufficiently small, the normal core has order 40 or 20

[agile burrow]

GOOD point

So then can we use your earlier point of reducing this to a problem about S_5

[warped fable]

6-cycles arent there, 3+3 cycles arent there…

[summer path]

Is normal core something one usually learns in group theory class?

[lethal dune]

no you just google

[summer path]

Yeah I just did

[warped fable]

H definitely has a 5-cycle, conjugate so that it contains (12345). It’ll also contain a 2-cycle, but can that 2-cycle contain 6?

[summer path]

I'm just wondering since I'm trying to gauge how much group theory I am expected to know when I study for quals

[warped fable]

Seems like you’ll get all the 5-cycles of S_6 if it does lol

[agile burrow]

That would be a lot

More than 40, perhaps

[chilly ocean]

Does anyone have an example of a scalar triple product or vector triple product for complex vectors?

I can't find anything

I'm assuming this operation is well defined for complex vectors, but not useful?

[south patrol]

well both are kinda geometrically motivated things right so makes more sense for R I suppose

and the scalar triple product is basically just the determinant

[chilly ocean]

Oh lmao

I completely forgot about this, I can use that for testing

Thanks!

[chilly ocean]

Uhm...

Is... there something equivalent to the vector triple product too?

[thorn monolith]

well i have one more question about S_6... how to figure out how many exist subgroups of H such that H is a subgroup of order 36 of group S_6, and for the action on the set {1, 2, . . . , 6} H has more than one orbit?

[south patrol]

Just extend C-linearly but yeah I'd just call it the complex vector triple product

[glossy crag]

What's a good reference for learning the basics of semisimplicity and Wedderburn stuff (enough to understand CSAs, crossed products, and Brauer groups). This is one of those topics I never learned properly that's a stumbling block for me.

[pliant raptor]

In $\mathbb{Z}[t]$, how can I show that $t \not\in (4,2t,t^2)$?

[cloud walrus]

ImHackingXD

[coral spindle]

find a nice description (or at least a nice property) of the elements of that ideal, and show that t doesn't have that property

Maybe you can give a good guess at a property that might be helpful

Hint: look at the coefficient of t in elements of the ideal

[south patrol]

This is the kernel of a specific map you can find, too, which gives another nice description

||Z[t] -> Z/4Z; t |-> 2||

[pliant raptor]

Oh I see, the coefficient of t must be even, thanks!

[broken stirrup]

I'm supposed to find a 2-generator group containing an isomorphic copy of every countable abelian group. I know that we can embed a countable abelian group into a divisible group which is direct sum of isomorphic copies of Q and prufer groups of type p^/infty. Then if i take direct sum of infinitely many copies of Q and prufer groups for every p, it contains isomorphic copy of every countable abelian group. Then as it's countable, we can embed it to a group generated by two elements of infinite order

Am i making sense?

[white oxide]

is it usual for algebra books to list lagrange and the first isomorphism theorem as corollaries? i suppose technically they are corollaries if you establish some results about counting before lagrange and other things before the first isomorphism theorem, but they seem to be too powerful to be referred to as corollaries

iirc fraleigh really placed emphasis on them whereas hungerford lists them as corollaries doesn't even give proofs in some parts, which is justified by the previous theorem but idk seems pretty casual to me (maybe because it's a graduate level text?)

[coral spindle]

Yeah

https://ncatlab.org/nlab/show/The+Rising+Sea

[white oxide]

whoaaa this is so cool!

thanks for posting

[coral spindle]

LA MER QUI MONTE!!!!!!

[white oxide]

french

[warped fable]

nlab

[lethal dune]

thik e ache

[coral spindle]

grothendieck j'aime trop ta métaphore mais sauve-moi, je vais me noyer

[formal ermine]

je deteste la langue francais

[warped fable]

https://tenor.com/view/cedric-villani-danse-dance-math-silly-gif-17541486

[delicate orchid]

🚨⚠️ HIGH LEVELS OF FRANCE-ACTIVITY DETECTED ⚠️ 🚨

[formal ermine]

ja wir sollten auf deutsch wechseln.

[warped fable]

this is not an improvement

[formal ermine]

du bist kein improvement

[chilly ocean]

#groups-rings-fields

[formal ermine]

tut mir leid herr terra

[agile burrow]

PSL(2, 7) is simple

[delicate orchid]

I find it rather complicated actually

[warped fable]

just the orbifold fundamental group of H^2 modded out by the (2, 3, 7) von Dyke group

easy

[formal ermine]

I swear newton polygons aren't meant to make sense

I don't know a single person that understands them

[warped fable]

You know maybe youre right

[formal ermine]

istg at this point I question whether my prof even understands them well

[agile burrow]

Lol I'm trying to get into some of this stuff actually

Discrete subgroups of modular group

[warped fable]

yeah thats good stuff but i know very little

[agile burrow]

Fair enough, seems like a pretty big area

[warped fable]

yep

lots of geometry

but alas, little topology

[agile burrow]

Yeah idk, I really liked the topology and group theory/rep theory interactions I saw

There are a lot of directions to branch into, it's just hard to commit to any one thing lol

[warped fable]

i get that a lot

[agile burrow]

You should do a write up on newton polygons when you become the first person you know who understands them

[warped fable]

ideally one should learn one topic every five years

then move into something entirely different

[formal ermine]

I swear I'm already planning on doing that

it will be the first entry on my blog

[warped fable]

id read that

non ironic

[formal ermine]

do you not understand/know them either?

[warped fable]

i have come across the concept before in relation to pusieaux series but never looked deeper.

however one spells pusieaux

puiseaux?

i have been looking at newton polytopes of junk a lot lately

which is related but different i think

[formal ermine]

YES that thing

[formal ermine]

yeah we will cover those too later I think

this is a course on newton okounkov theory

[warped fable]

ah

based russians once again

[formal ermine]

so far we've only done algebraic curves, newton polygons, and right now we're doing classical ag

as in varieties over C

[warped fable]

sounds cool, id like to hear/read bite sized info on what the main ideas are

newton polytopes appear in mirror symmetry which is why im encountering them, but so far not in a serious way lol

[formal ermine]

you're not the first

[warped fable]

(Landau Ginzburg A branes) toric variety -> fan polytope <- polynomials (Landau Ginzburg B model)

whatever all of this means

[formal ermine]

I remember hearing something about toric varieties

[warped fable]

F

[formal ermine]

anyway

time to learn for my german exam on wednesday

[warped fable]

lmao

what does gedankenzuge mean?

“train of thought?”

[formal ermine]

train of thought

yes

gedankenzuge is a declination of gedankenzug

[warped fable]

gotcha

theres a black metal band with that name whose screams are ridiculous which is why i remembered