groups-rings-fields

[solar glacier]

so youre considering a restriction of a map from G to G/N to a map from H to G/N

[tribal moss]

Yes, in the part where I conclude it divides |H|.

[solar glacier]

im unable to see the second black thing you sent lol

[tribal moss]

That's you quoting the first one!

[next obsidian]

I think that’s just your reply

[solar glacier]

i take it youve seen this problem before lol cause DAMN lol

[tribal moss]

Not that I remember -- but as you alluded to yourself, considering the quotient and the projection map into it is pretty much the automatic thing to do.

[solar glacier]

i got stuck after that part though on the exam and was running out of time

[solar glacier]

still unsure how he duduced that |H|/|ker\psi|=1

[next obsidian]

This counts the size of the image

This divides |H| and divides [G:N]

So it’s 1

[solar glacier]

ohhhhhh

it also divided [G:N] thats what i was missing

and theyre corpime

thus it must be 1 if it divides both

very slick argument

[ruby sundial]

are there any resources on why generating functions work to give helpful solutions to recurrence relations

because it seems like a miracle that they work

[carmine fossil]

Isn't it more like you create a generating function based on the problem to be solved?

Like ordinary generating functions are pretty easy to motivate

I guess a general generating functions text is probably what you want
https://www2.math.upenn.edu/~wilf/DownldGF.html
@ruby sundial

[ruby sundial]

no i know about this lol

the books pretty much goes over handbook methods of finding generating functions

but im asking from a more algebraic side why do these methods work

[tender wharf]

i love normal subgroups

aH Ha

[delicate bloom]

algebraic in what sense, like x^m * x^n = x^{m+n} sort of explains it all to me I think

[terse wadi]

I have a question regarding proposition 33 on quotients of graded rings. What type of multiplication do we use on the ring to the right? If I use componentwise multiplication I get that $(S_{i}/I_{i})(S_{j}/I_{j})=0$ when $i\neq j$ which can't be correct

[cloud walrus]

GUNILLA62

[next obsidian]

You multiply it like in S

If you have the class of a homogeneous element s_i in S_i and s_j in S_j you send it to the class of s_is_j in S_i+j

The fact that this like, works, depends on I being a homogenous ideal

[terse wadi]

Okay, so we aren't using the usual (external) direct product multiplication?

[next obsidian]

Nah

That actually doesn’t even make sense

Because each S_i/I_i is only an abelian group!

🤯

Graded stuff are whacky

[terse wadi]

I didn't realise this at all

Thank you!

[next obsidian]

[cloud walrus]

Mecejide

[pallid oracle]

The Lie algebras of double covers are the same as the original Lie algebras for Lie groups over C, so I wouldn't expect it to magically happen in characteristic 2, but it might idk.

[delicate bloom]

for ideals of a commutative ring is $(I+J)(I \cap J)=IJ$? I'm not asking for a proof since I'd like to prove it myself later; I'm just asking since it seems like this would be well known but possibly has some caveats, since it's essentially $\gcd(a,b)lcm(a,b)=ab$ in ideal form.

[cloud walrus]

mOwOsity

[white grotto]

Can someone help wiht some intuition on lie brackets
what properties does C have when AB-BA=C
i have dozens of questions on lie brackets in my textbook and they all seem very difficult cause i have no idea what it tells me
like if C=A then A is nilpotent, if C is diagonalizable and commutes with A and B then C is the zero matrix, if A is diagonalizable and C commutes with A and B then C is also the zero matrix, if C= aA+bB(a,b scalars) then A and B are simultaneously trigonalizable

[rotund aurora]

Its false in general

https://math.stackexchange.com/questions/263027/for-i-j-ideals-show-that-ij-subseteq-i-cap-jij

[delicate bloom]

cool thanks

[rotund aurora]

it is interesting to ask when does that hold

[delicate bloom]

it's true in Dedekind domains, that's really all I care about

at least, if I trust that random person saying so in the answer haha

[delicate bloom]

but I'd be interested to know to what extent it holds outside of Dedekind domains too while I'm here thinking about it, good point

[rotund aurora]

As a warmup: Do there exist matrices A and B such that AB-BA=I where I is the identity matrix?

[white grotto]

no

tr(AB-BA)=0 while tr(I)!=0

[rotund aurora]

nice

[white grotto]

so tr(C) always =0

and ive seen the property that if tr(M)=0 for any matrix M, then there exist A,B such that M=AB-BA

with complicated proof. Not sure that would help

[formal ermine]

how do I show that Z2 actually is a subgroup of S3 without checking all of the elements and stuff? do I use cayley or something?

[rustic crown]

nu

find an element in S3 of order 2

that gives a map Z --> S3 sending 1 to that element of order 2, and kernel is exactly 2Z so this embeds Z/2Z --> S3

[tender wharf]

alternatively cayley's theorem

[rustic crown]

.<

[devout crow]

it feels like this is the same problem as what they asked

[rustic crown]

oh so i didn't understand the question >.<

yee then it makes sense what DerpZ is saying >.<

Z/2Z --> S2 by cayley
and S2 --> S3 by permutations fixing 3 or something

[formal ermine]

can't I use cayley for Z3 too?

[rustic crown]

yea you can

Z/3Z also embeds into S3

[formal ermine]

Z/3Z is iso to S2 no?

[rustic crown]

S2 = Z/2Z

[formal ermine]

OH

LMAO

[rustic crown]

permutations on 2 elements is id and swap

[formal ermine]

that was a brain fart

isn't S_n always a subgroup of S_k for a k ≥ n

[rustic crown]

yep

[formal ermine]

yeah

[rustic crown]

and in multiple ways if k > n

unless your definition of S_n is exactly the permutations on 1,2,...,n

[formal ermine]

why?

[rustic crown]

because then there is a canonical way to put S_n --> S_k

a permutation f is sent to the permutation which is f on {1, 2, ..., n} and identity on the rest of the elements {n+1, ..., k}

[rustic crown]

here i meant that there are multiple injective maps S_n --> S_k with different (but isomrophic) images

[formal ermine]

I thought with "unless..." you meant it's no longer possible to embed Sn lol

[devout crow]

S_n has a bunch of subgroups isomorphic to isomorphic to S_k, but only one of them is the symmetry group of the set {1,...k} < {1,...,n}

but this is kind of an artificial thing to think about

[white grotto]

yo, was there some follow up to the warm up

[formal ermine]

yeah ok

thanks!

[rotund aurora]

the questions of your textbook haha

they seem pretty nice tbh

[rustic crown]

what if char is p dividing size of the matrices?

[rotund aurora]

[delicate bloom]

I was thinking about that earlier, I think you can then use the trick [d/dx, x]=1 with a basis of p monomials to construct examples explicitly since as functions x^p=x lets you do it finitely

[white grotto]

wdym?

[delicate bloom]

Tr(I)=0 mod n

if I is nxn

just checking by hand earlier I found $A=\mqty{0&1\1&0}$ and $B=\mqty{0&1\0&0}$ works for $\mathbb{F}_2$ to make $AB-BA=I$

[cloud walrus]

mOwOsity

[white grotto]

yeah we're in R or C

tho nice

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

ive been thinking about this since morning

I am dying

[white grotto]

i have the solutions BUT i dont want to look xd

cause it's useless if i have no strings to follow

if C commutes with A^2 and A diagonalizable

then does C commute with A?

what does commuting with a nilpotent matrix tell us?

[delicate bloom]

only intuition I have about lie brackets is that the jacobian identity is written more intuitively like as if it's a product rule, think of the first term "distributing" to the other two terms:
[A,[B,C]] = [[A,B],C] + [B,[A,C]]

makes it feel more like a derivative

[rustic crown]

wait help me with that last part, how to do it finitely?

oh do you mean x^p = 1?

[delicate bloom]

no, x^{p-1}=1 is fermat's little theorem but the problem is this is false for x=0. x^p = x for all x in F_p though

[white grotto]

is that helpful in any way that is clear to you in the case of matrices?

[rustic crown]

ah okie, i was thinking like we're considering the group-algebra k[Z/pZ] and derivative makes sense on this

[delicate bloom]

I dunno, I didn't work it out myself, I just figured taking the basis {1,x,...,x^{p-1}} and looking at what d/dx and x* as matrices would just work out since as functions x^p=x

there could be problems haha

thinking about it, seems like the derivatives will break, so maybe not

[rustic crown]

yea

[rustic crown]

but i think this still works

d/dx (x^n) = n * x^(n-1) and since we're in char p, value of n only matters mod p

so n = m (mod p) would mean the derivative of x^n and x^m are same

[delicate bloom]

ah yeah I see, that sounds good

[rustic crown]

here x^(p+1) and x^2 are same as functions, but their derivatives are different

.<

[delicate bloom]

hmm I wonder if we could fix it using p(p-1) dimensional matrices

3 scary 5 me

[white grotto]

I just need this and everything else is easy, if uv-vu= some linear combination of the 2, then they're simultaneously trigonalizable

[delicate bloom]

it would be consistent with my p=2 example I found since 2*(2-1)= 2 hah

[white grotto]

If theres any hope of finding intuition for this ill pay a million dollars

some day. when i have a net worth of 150 billion

[delicate bloom]

my guess is most of this stuff is motivated by people doing things in physics

[warm wyvern]

What can we say, in general, about the automorphism group of a group?

[delicate bloom]

you can split it into the inner and outer parts

[warm wyvern]

Lol

Are outermorphisms just non-inner automorphisms?

[delicate bloom]

yeah, the outer automorphism group is the quotient of the automorphism group by the inner automorphism subgroup

idk, probably much more can be said, I'm no expert

[white grotto]

Okk tell me if i am correct, after seperation of this equality we get u(a Id+v)=v(bid+u), which is a kind of commutativity that preserves the fact they are simultaneously trigonalizable right?????

because..... i am not sure

[rotund aurora]

Can someone recall me why we wrote F-automorphisms of the rational funciton field F(x) as a 2x2 matrix in GL_2(F) ?

I mean, the fact that all automorphisms are of the form x --> (ax+b)/(cx+d) where ad-bc not zero is clear, but idk what notational advantages does looking at these automorphisms as matrices have

[warm wyvern]

More specifically, why do we care about actions of a group on itself other than a source of examples

[white grotto]

I GOT IT

And i am done

thanks discord

[coral spindle]

Conjugacy is a very fruitful way to analyse groups.

[rotund aurora]

Uh I think the point is that its a homomorphism

[coral spindle]

It's actually in PGL_2(F)

You're right, it's just notational really.

It should be said that these things (Moebius transformations) occur naturally a lot

[rotund aurora]

Why PGL? Like the determinant neednt be 1, just not zero, right?

Uhm wait I think I said nonsense

Ill check that later

[chilly radish]

Any element of the galois group fixed Delta iff it has even sign. Recall that the fixed field of G is exactly the base field, so Delta is in the fixed field iff all the elements in G are even

i.e. G is in An

[glossy crag]

What's this from, i.e. is this a set of notes available online?

[white grotto]

if a nilpotent matrix commutes with a diagonalizable matrix that is not lambda id, is that nilpotent matrix the zero matrix

[rotund aurora]

the zero matrix is nilpotent and commutes with everything

uhm wait this is definitions, idk if zero is considered nilpotent xD

[white grotto]

no i mean is it necessarily the 0 matrix

[rotund aurora]

I think actually no, oops

oh wait sorry I read that wrong

[rustic crown]

0 is definitely nilpotent >.<

[white grotto]

0 is nil

i am asking if a matrix is nilp, commutes with a diag that is different than lambda I, is it necessarily 0

and how to prove that

[white grotto]

this is part of my unending journey of trying to understand what commutes with nilpotent matrices

i mean is it that hard to answer this :(((

[rustic crown]

ig the better question is "why care?" :p

[white grotto]

lol. just a crucial insight in most exercises involving some kind of commuting matrix

[knotty frigate]

Depends but it does it really matter?

[glossy crag]

Not at all, by Jordan decomposition every matrix (over an algebraically closed field or one whose minimal polynomial splits) decomposes into the sum of commuting diagonalizable and nilpotent parts. The diagonalizable part doesn't have to be the scaled identity and neither does the nilpotent part have to be 0.

[white grotto]

ughhh. yeah i see. okay

thanks

[rotund aurora]

I was gonna give an example but wolframalpha is broken ahha

[white grotto]

looool

[rotund aurora]

that product should be

and you can see they commute

[white grotto]

yeah

:(

[rotund aurora]

why sad?

[rustic crown]

https://tenor.com/view/mala-mishra-jha-pat-head-cute-sparkle-penguin-gif-16770818

[white grotto]

cause i am nowhere near understanding what commutes with a nilpotent matrix

and ive been at it since morning

[knotty frigate]

Lol

Sad

[rustic crown]

i think ocean man gave a really nice class of examples

[white grotto]

when 2 nilpotent matrices commute?

and when a diagonalizable matrix commute with a nilpotent one

how to find the matrices that commute, diagonal and nilpotent one in the jordan decomposition

where it's D+N

[rustic crown]

convert to jordan canonical form, decompose there and change basis again?

[white grotto]

i dont think that gives us the D and N that commute. at least most of the time it doesnt

[rustic crown]

it should tho

if A is any matrix, then you can find P such that PAP^-1 = J, now J = D + N where D and N commute and so A = P^-1DP + P^-1NP

clearly P^-1DP is diagonalizable, and P^-1NP is nilpotent

because changing the basis can't possibly do anything to some intrinsic property of an operator

iirc if x is any operator you can actually write the semi-simple part and nilpotent part as polynomials in x (with no constant term)

[chilly radish]

No problem!

[ruby sundial]

Ok so you start off with a recurrence relation which is an equality formula that generates a sequence in the integers for example. How does it work to turn this object into an equality about power series

[delicate bloom]

@ruby sundial that sorta stuff is covered in here

once you put the coefficients on a power series, then things like multiplying by x shifts the sequence on the power series, then you can start to interpret other things like multiplication of the series turns into convolution on the coefficients for instance.

[celest cairn]

Hi, I wanted to find the splitting field of $x^{8} + 11x^{4} + 24.$
I found that the splitting fields for the individual factors are $\mathbb{Q}(\sqrt[4]{-3}, i)$, and $\mathbb{Q}(\sqrt[4]{-8}, i)$.
I don’t know what to do from here.

[cloud walrus]

Sapphire Gaming

[delicate bloom]

not sure what you're doing but your polynomial is a quadratic in y=x^4, which means you should be able to get the roots pretty directly

[rotund aurora]

how is this possible

The minimal poly of 1+cbrt 2+cbrt 4 is x^3-3x^2-3x-1, and it is not divisible by x^3-2

and like they are coprime

[gilded gulch]

To be short, let a^4 = (-3) and b^4 = -8.
The splitting field must contain Q(a, b, i).
However, Q(a, b, i) splits already, because

1. Q(a, i) is a splitting field of x^4 + 3 = 0 over Q.
2. Q(a, b, i) is a splitting field of x^4 + 8 = 0 over Q(a, i)

[delicate bloom]

maybe the wording just means "do these as two separate exercises", idk

[rotund aurora]

[delicate bloom]

The splitting field must contain Q(a, b, i).
how do you know it contains i?

[vocal patrol]

In the integers mod 4 under addition, is 3 the inverse of 1?

[coral spindle]

No. 3 is the inverse of 3.

Oh, addition

My bad

[gilded gulch]

Oh because he got Q(a, i) already.
In detail, whenever a is a solution of x^4 + 3 = 0, ai is a solution too, so we get ai / a = i in the field

[coral spindle]

You are right: it is 3

[vocal patrol]

Thank you

[coral spindle]

Under multiplication it is different

[vocal patrol]

Yes

[coral spindle]

My mistake for misreading

[vocal patrol]

No problem :)

[light iris]

the inverse in that group is the number which you add to the element to get something equivalent to 0 mod 4, and 3+1 is equivalent to 0 mod 4

[gilded gulch]

Agree. I think "the monic polynomial" in the text is misleading.

[delicate bloom]

oh ok, I thought you were just reasoning immediately from the first line you wrote not what they put

[rotund aurora]

Given an arbitrary polynomial p over a field F, how do you denote its galois group?

Something like Gal(p(x)/F) or simply Gal(p/F) ?

[glossy crag]

I don't think there's a consensus, I've seen John Milne denote it by G_p.

[gilded gulch]

I think your Gal(p/F) notation is short and nice, but I've never seen that being used.

[glossy crag]

I found the proof a little too compact, did I fill in the details correctly?

If J is an ideal containing pR_p, then p is contained in J\cap R, but the only way the latter doesn't intersect R-p (from which would follow J=R) is if p=J\cap R and hence pR_p=J, i.e. pR_p is maximal. If M is any other maximal ideal, then M\cap R has to be contained in p (otherwise M wouldn't be proper), hence M=pR_p.

[gilded gulch]

I see no issue in your proof.
BTW I spent a little time to understand p = J cap R -> p R_p = J,
which follows from the p R_p <= J and J <= p R_p, where the formal comes from the assumption.

[glossy crag]

pR_p=J because p=J\cap R and contraction/extension are inverses on (all) ideals of R_p, i.e. pR_p=(J\cap R)R_p=J.

[gilded gulch]

That's more clear to me now. Thanks

[white grotto]

yooooo question

hope youre still there

does this explain the fact that if fg-gf=f then f^ng-gf^n=nf^n

cause the right term looks suspiciously like a derivative

[formal ermine]

An is a subgroup Sn because its the kernel of the sgn homomorphism?

[rustic crown]

yep

[formal ermine]

oke

[gilded gulch]

Nice reasoning. And therefore An is a normal subgroup, additionaly

[formal ermine]

yea

[rustic crown]

and you get an iso, Sn/An = Z/2Z

[formal ermine]

also because An has index 2

how do I compute $\bL^H$

[cloud walrus]

rectangle cube

[formal ermine]

like with $H = S_2$ and $\bL = \bQ(\alpha)$ where $\alpha = \sqrt[3]{7}\sqrt{-3}$

[cloud walrus]

rectangle cube

[formal ermine]

(that's the splitting field of $x^3 - 7$ btw)

[cloud walrus]

rectangle cube

[formal ermine]

so the definition is $\bL^H = \Set{x \in \bL | \sigma(x) = x ~ \forall \sigma \in H}$

[cloud walrus]

rectangle cube

[formal ermine]

but like

wait

isn't it just everything in L except something that contains the roots?

so it's just the base field?

because that's what the automorphisms are supposed to fix?

[rustic crown]

which S2 is that?

the galois group is S3, so you need to tell me the exact copy of S2 you want

[formal ermine]

wdym

[rustic crown]

S3 has 3 subgroups isomorphic to S2

and the exact fixed field will depend on the this subgroup

[formal ermine]

S2, Z2

what's the other one

[rustic crown]

umm

i mean

the subgroups generated by (12), (23) and (31) in S3 are all isomorphic to S2

[formal ermine]

won't the resulting fixed fields be isomorphic too

[formal ermine]

ah wait

I forgor about things like (12)

[rustic crown]

i don't think that's true

but i may be wrong

[formal ermine]

omw asking on mse to get downvoted into oblivion

[rustic crown]

but in any case, if you're doing galois theory you're interested in the actual lattice you get and not just intermediate fields upto isomorphism

[solar glacier]

MSE is an ego fest lol

[rustic crown]

for example, one copy of S2 in the galois group correponds to the complex conjguation

and the fix field is then just the intersection of your field with R

which is Q(cbrt(7))

[chilly ocean]

there are certainly people like that there, but there's also a lot of people willing to help and give good answers

[solar glacier]

true

[formal ermine]

actual lattice?

[rustic crown]

yea the inclusion reversing bijection and stuff

[formal ermine]

what's a lattice

[rustic crown]

a place where it makes sense to take max and min

of two elements

[chilly ocean]

sup and inf

to be precise

this doesn't have to be max/min and often will not be

[rustic crown]

was it sup and inf?

dun remember

yee makes sense

[chilly ocean]

do you want me to give you an easy way to remember

[rustic crown]

yee

[chilly ocean]

real values functions form a lattice

[chilly radish]

I mean a lattice doesn't have to have infinite joins and meets

[chilly ocean]

but you don't always have that sup of two functions is a maximum

[chilly radish]

Just every two elements have a max and min

There seems to be a mixup here

[chilly ocean]

.

[formal ermine]

my homework wants me to find the intermediate fields tho

[chilly radish]

The max of two functions is just their pointwise sup

But it's called a maximum in the lattice

Or join and meet whatever

[rustic crown]

yea exactly, you don't wanna just find it up to iso

[chilly ocean]

source?

[formal ermine]

why

[rustic crown]

this is the lattice of intermediate fields

[formal ermine]

ah so like a diagram

[rustic crown]

and that's the subgroups thingy

[chilly radish]

I guess max and min are a bit confusing in this context ok

I did mean a least upper bound and a greatest lower boind

[rustic crown]

yea that makes more sense

[chilly ocean]

I mean it makes sense to say max(f, g) where f, g are functions because we don't mean the literal maximum in the lattice of functions, but their pointwise maximum

but yeah it's a little confusing

[rustic crown]

yee

[chilly ocean]

do you like my explanation

det

[rustic crown]

yea it's super nice

[formal ermine]

it's not true

I went through an example

[rustic crown]

so something like a galois extension with group Z/2Z x Z/4Z would give a counterexample

if you use Z/2Z x 0 then fixed field will have galois group the quotient Z/4Z

[formal ermine]

I just checked the isomorphism class of S2 in S3 lol

[rustic crown]

and if you use 0 x 2Z/4Z then the galois group of the fixed field will be Z/2Z x Z/2Z

[formal ermine]

{id, 12} and {id, 23} give different results

[rustic crown]

.<

we wanted non-iso

i think they'll be iso in your example

right?

[formal ermine]

Q(cbrt(7)zeta_3^2) and Q(cbrt(7))

[rustic crown]

yea both are iso to Q[x]/(x^3-7)

[formal ermine]

oh

[formal ermine]

@rustic crown conjugate subgroups yield isomorphic fixed fields

[rustic crown]

yee

if F is the fixed field of H, and f is an automorphism, then fHf^-1 fixes f(F)

[vocal patrol]

Why is Z mod p with p being prime always a field

[chilly ocean]

because elements invertible mod n are precisely the elements relatively prime to n

if p is a prime, then every non-zero element is relatively prime to p i.e. invertible

[warm wyvern]

how is this true?

shouldn't it be $S_\frac{|G|}{p}$?

[cloud walrus]

DarQ

[rustic crown]

nope

[next obsidian]

Index p

Means there’s p-cosets

[warm wyvern]

OH

mb

[barren sierra]

p = index of H = |G / H|

[warm wyvern]

I confused index with order

[vocal patrol]

I get the proof until the step of the last implication

How does the gcd for every a being 1 with p imply that there exists this multiplicative inverse x in the field?

[formal ermine]

fermat's little theorem

[vocal patrol]

Thank you I’ll look that up

[formal ermine]

if a and p are coprime then a^(p - 1) = 1 (mod p)

[vocal patrol]

I got how it works now

Sorry if I’m slow, but how does this give me the assurance that x is in the field

[chilly ocean]

huh

gcd(a, b) = 1 implies there exist constants x, y such that xa+yb = 1

here xa+yp = 1, or modulo p, xa = 1 (mod p)

that is, a has a multiplicative inverse

[vocal patrol]

Makes sense thanks a lot

[chilly ocean]

I mean, Fermat's little theorem works, but imo you prove Fermat's little theorem using that set of non-zero elements is a group

and Lagrange theorem

so it might be a little loop here

[formal ermine]

the proof I have in mind is using euler's theorem

[warm wyvern]

x = a^(p-2)

[chilly ocean]

Euler's theorem is proven the same way

[warm wyvern]

isn't euler's theorem a generalization of flt?

lel

[chilly ocean]

you guys are using a result that you prove using what you want to prove

is what I'm getting at

[formal ermine]

isn't that just looking at the cyclic subgroup generated by the element and lagrange?

[next obsidian]

You should use bezout’s lemma about integer combinations like blitz did

[chilly ocean]

and how do you know it's a group?

that's the point

[next obsidian]

This is the lowest-tech way

[chilly radish]

There definitely are non group theoretic proofs of both Theorems

[next obsidian]

And you’ll need to use that like a billion more times during the course of a group theory course

[chilly ocean]

Someone would have to provide me with one to convince me tbh.

[next obsidian]

Broooo plsssss

Fermat existed before group theory

😭

[chilly ocean]

I mean that if we want to do it step by step, then you'd have to also provide your proof of Fermat's little theorem that doesn't use group theory

but at that point we might as well go more basic

and use Bezout lemma

[formal ermine]

let $\bQ(\cbrt{7},\zeta_3)/\bQ$ be the splitting extension of $x^3 - 7$. I want to know the fixed field generated by $\Set{\operatorname{Id}, (12)}$. my basis is $\Set{1, \cbrt{7}, \cbrt{7}\zeta_3, \cbrt{7}\zeta_3^2, \zeta_3, \zeta_3^2}$. how would I know where $\sigma \coloneqq (12)$ sends $\cbrt{7}\zeta_3^2$ to? like if I number my roots: 1 = $\cbrt{7}$, 2 = $\cbrt{7}\zeta_3$, 3 = $\cbrt{7}\zeta_3^2$. then $\sigma(\cbrt{7}\zeta_3^2) = \sigma(\cbrt{7}\zeta_3)\sigma(\zeta_3) = \cbrt{7}\sigma(\zeta_3)$ but where does it send the $\zeta_3$ to?

[cloud walrus]

rectangle cube

[warm wyvern]

op should've just used Wedderburn’s little theorem tbh

[formal ermine]

ah wait I think I got it

sigma(zeta_3) = 1/zeta_3

because sigma(cbrt(7)zeta_3) = sigma(cbrt(7))sigma(zeta_3) <=> sigma(zeta_3) = ... = 1/zeta_3

no this doesn't feel right

because then the fixed field would just be Q

1 -> 1

cbrt(7) -> cbrt(7)zeta_3

cbrt(7)zeta_3 -> cbrt(7)

cbrt(7)zeta_3^2 -> cbrt(7)/zeta_3

zeta_3 -> 1/zeta_3

zeta_3^2 -> 1/zeta_3^2

where's my mistake

[tribal moss]

The most basic would be to bypass Bezout and start all the way back from Euclid's lemma.
Fix a with 0 < a < p. When we take the remainder of dividing na/p for each n with 0 < n < p, we will never get 0 due to Euclid's lemma. But that means we never get the same remainder twice, because if na/p and ma/p have the same remainder, then |n-m|a would give a remainder of 0. By the pigeonhole principle, then, we must have gotten 1 as a remainder in one of the cases. Thus a, which was arbitrary, has a multiplicative inverse.

[glossy crag]

Can anybody explain this? The definition of DD this uses is "noetherian ring whose localisation at every prime is a DVR". First line is clear to me, that's a foregoing lemma. The proposition it refers to is how every ideal in a DVR is of the form (r^k) where r is the unique prime. An ideal of R_p/p^aR_p is an ideal of R_p containing p^aR_p and all ideals of R_p look like (r^k), but how does he get p/p^a from this?

[tribal moss]

What is "DD"?

[gilded gulch]

Putting k=1, we know r^1 is the maximal ideal of R_p,
But since R_p is a local ring with the unique maximal ideal p R_p,
the unique prime ideal r you said is p R_p.

Therefore each ideal in the DVR R_p is p^k R_p for some k
ideal of R/p^a <---> ideal of R_p /p^a R_p <---> ideal of R_p which contains p^a R_p
<---> ideal with k >= a case

and for k >= a cases,
(R / p^a) / (R / p^k) = (p^k / p^a) = (p / p^a)^k
where the step in the middle is obtained by thinking R, p^k, p^a as R-algebras.

[glossy crag]

Dedekind domain, I forgot this book calls them "Dedekind rings".

[gilded gulch]

My mistake- k <= a cases, I mean.

[glossy crag]

1. Just to double-check, I assume p^kR_p=(pR_p)^k holds (for all localisations, not just with respect to a prime, i.e. for JR_S)?
2. You're right, all ideals of R_p/p^aR_p are of the form p^kR_p/p^aR_p with k<=a, but how does this translate back to R/p^a?

[gilded gulch]

1. in the set-theoretic level,
   p^k R_p = {(a_1 * a_2 * ... * a_k) / s : each a_i in p and s is in (R - p)} = (p R_p)^k

2. The translation I used:
   ideal of R/p^a <---> ideal of R_p /p^a R_p <---> ideal of R_p which contains p^a R_p
   the latter is "ideal of R / I <---> ideal of R containing I" in general, which is already clear.

The former you just asked is, thinking R, p^a and R_p as R-algebras,
(R_p) / (p^a R_p) = (R tensorprod R_p) / (p^a tensorprod R_p) = (R / p^a) tensorprod R_p
where the second equality holds since tensoring commutes with cokernel.

ideals of the left-hand side <---> R_p-subalgebra of the left-hand side
<---> R_p-subalgebra of the right-hand side
<---> R-subalgebra of the R / p^a (?!?!?!)
<---> ideal of R / p^a

The step (?!?!?!) in the middle is mysterious because,
tensoring does not preserve injectivity in general.

Perhaps we have to show this step is valid for our R_p,
but I guess lemma 3.11 in your capture already deal with it.
Is that right?

[formal ermine]

how do I show that every irreducible polynomial $f \in \bQ[x]$ has at max one root in $\bQ(\cbrt{2})$

[cloud walrus]

rectangle cube

[formal ermine]

I have like no idea how to begin

[warm wyvern]

can I have help with this?

I am so lost on what I should do

I'm not even convinced this theorem is true lel

[formal ermine]

which part?

[warm wyvern]

there's only one question lel

[formal ermine]

I swear I proved a similar result for field extensions lol

hmm

maybe look at G -> Sym(left cosets of H), phi(x)(aH) = (xa)H?

[warm wyvern]

this is 9.11 btw

[glossy crag]

Lemma 3.11 shows that R/p^a and R_p/p^aR_p are isomorphic.

[glossy crag]

Don't get 2) yet, probably have to stew on it a little.

[gilded gulch]

Sorry, I'm not a native English user.
What is the meaning of "stew on it"?

My dictionary doesn't know the phrase...

[warm wyvern]

probably "think about it"

[gilded gulch]

thanks!

[warm wyvern]

np :)

[formal ermine]

@warm wyvern does this help?

[gilded gulch]

btw this one looks simple but tricky...

[formal ermine]

yeah

[gilded gulch]

Perhaps you already derived a fact I did:
if dim ((splitting field of f) cap Q(cubic root of 2)) >= 2, then it is 3, so the latter field is a subfield

But I don't have any progress beyond it

[formal ermine]

why are we looking at the intersection?

[warm wyvern]

I'll think about it tomorrow

I'm currently spent

[glossy crag]

I just meant I need to think about it some more and absorb it, thanks for the answer.

[gilded gulch]

My attempt is: (assuming the number of root >= 2)

1. thinking of intersection, the splitting field of f contains Q(cubic root of 2) as its subfield
2. therefore it contains the splitting field of of Q(cubic root of 2) as its subfield as well.
3. hence it contains sqrt(3) * i
4. x^2 + 3 = 0 is factorized in the splitting field of f
5. ... somehow we get a contradiction

[gilded gulch]

OK

[gilded gulch]

I think... this question is too difficult to me.
Sorry, I quit.

[formal ermine]

lol dw

[formal ermine]

I think looking at the splitting field is a good idea though

so we gotta show that the splitting field is never Q(cbrt(2))

for an irreducible polynomial f of degree > 1

[gilded gulch]

I see.

[formal ermine]

I think I got it?

ok so I think I have it for deg > 3

assume deg(f) > 3. because f is irreducible it must be separable because we're working over a char 0 field, therefore the galois group of the splitting field over Q has at least degree deg(f) (as we're permutating deg(f) roots). this implies that the degree of the extension is also at least deg(f) (because our extension is galois). however the degree of Q(cbrt(2))/Q is 3. therefore Q(cbrt(2)) cannot be the splitting field as that has at least degree 4 over Q

[coral shale]

still on this?

[formal ermine]

yes

[formal ermine]

and for deg(f) = 2, deg(f) = 3 we can just check by calculation

multiply out etc

[coral shale]

Did u try assuming 2 roots and constructing a factor that divides f

[formal ermine]

no

it's 2:18 am and I need sleep lol

[coral shale]

just an idea

[formal ermine]

does this look correct tho?

I am showing that for deg(f) > 3 the splitting field cannot be Q(cbrt(2))

[coral shale]

😓

[formal ermine]

wait

I forgor my reasoning why that should show that it has at max one root in there

[coral shale]

Q(cbrt2) is simply not splitting because it lives in R

[formal ermine]

wdym

[coral shale]

K = Q(cbrt2) subset R

[K : Q] = 3
[split(xxx-2) : Q] = 6

xxx-2 has only 1 root in K. Adjoining either of the other 2 roots splits it.

Ok this works:
Try some algebraic manipulation with
f(a+bx+cxx) = f(d+ex+gxx) = 0
(x denoting cbrt 2)

[rotund aurora]

Any hints on showing that any finite subgroup of PGL(2,Q) occurs as a Galois group over Q?

[small bramble]

I really don't get this highlighted line from Dummit and Foote. It defines neither a bilinear binary operation (I think) nor a ring homomorphism f:R -> A mapping 1_R to 1_A such that the subring f(R) of A is contained in the center of A

[wooden ember]

The algebra structure mapping simply sends r in R to the map multiplication by R in Hom(M,M)

The line you highlighted is just defining the module operation on the right by the one already seen on the left if you will since algebras should have a module operation that can be used both on the left and right

[formal ermine]

Ah my basis is wrong

[analog zephyr]

Hi guys

[rustic crown]

hewwo

hi ryu

[lethal dune]

hello

[chilly ocean]

where's the algebra

[lethal dune]

encrypted

ok here's one, $A$ is finitely generated module with $A \cong A\oplus A$, then show $A=0$

[lethal dune]

I do have a solution but I would like a different take

[rustic crown]

you could do it with some tiny ag language

think of A as a quasi-coherent sheaf on Spec R

if A is non-zero then it will be non-zero at some stalk (say p in Spec R)

so this basically reduces the problem to the case where your ring is now local

now if you go further to residue fields, you get that A_p/pA_p must be 0 as it's a k(p)-vector space isomorphic to its square

and by nakayama this means A_p = 0

i could remove the AG language if you wish

[chilly radish]

Finite dimensional vector space. This is really the only place you use the finite generation, right?

[rustic crown]

and nakayama

[chilly radish]

Ah right

It follows from the jacobson radical version right, cuz the radical is just the maximal ideal

[rustic crown]

yep

maybe i should also say that if R^n --> A is surjctive then at stalks, (R_p)^n --> A_p is still surjective

(can be seen by tensoring with the flat R-module R_p)

[chilly radish]

I think you could do this without reducing to the local case first? For any maximal ideal m you get that mA=A by the same reasoning when you quotient, then in particular you get the same result on the radical and apply nakayama

[rustic crown]

i was thinking that earlier but then wasn't sure how to get the jacobson radical working nicely

[chilly radish]

Hmm yea ok maybe you're right this isn't immediate

[rustic crown]

ig it was clear if there were finitely many maximals? but idk

[chilly radish]

Yea you need to do some finagling here to get it to work

Checking locally is always good practice anyways

Makes life easier

[warm wyvern]

can someone help me with this?

[tender wharf]

how far have you gotten

[warm wyvern]

I'm pretty sure the proof looks like this sketch from 9.11

[warm wyvern]

here ker's index would've been either p or 1

if it's 1 then H=ker and if it's p then it's trivial then G iso S_p then G=H (in both cases H is normal)

I'm not sure how to generalize the result tho

[warm wyvern]

I suspect I wasn't supposed to use this reasoning I shouldn't have had to consider the index=p case maybe?

I'm not sure how I'd do that tho

[lethal dune]

that's the proof I have

except the AG terminology

lol

[rustic crown]

so the left action of G on G/H gives you a homomorphism
G --> S_n

set K to be the kernel of this and verify it gives you what you're looking for

[rotund aurora]

For which pairs of positive integers (n,m) does there exist an irreducible polynomial over Q with n real roots and m complex roots?

https://math.stackexchange.com/questions/130047/the-existence-of-a-polynomial-of-degree-n-with-m-real-roots-when-m-equiv-n

mmh I guess this is it

[rustic crown]

there was a nice exercise in aluffi similar to this

but now i wanna see an algebraic proof >.<

i think the idea in aluffi was this... say m = n + 2k
define f(x) = x(x-2)...(x-2(n-1))(x^2+2)(x^2+4)...(x^2+2k)

so f is in Q[x] and has n rational roots at 0, 2, 4, ..., 2(n-1)

and it has 2k complex roots which are not real

now the reason we made everything even is that f = x^n (mod 2)

so if q is any odd integer, then q * f(x) + 2 by eisenstein is irreducible

after that you use analysis to say that if q is HUGE then roots of f(x) and f(x) + 2/q are really close (since all the roots of f have multiplicity 1, this should be easy to show)
so for all large odd q, it will have n real roots and a total of m complex roots

[rotund aurora]

nice

Will this embedding be compatible with the ordering?

that question doesnt make sense

[rustic crown]

it should because squares map to squares?

yea right, but not every positive thing is a square

[rotund aurora]

Also, if we have a subfield of R cap A, then its orddering is not necessarily unique

[rustic crown]

right

[rotund aurora]

but R cap A is realclosed, therefore the ordering is unique. So I guess the order should be respected if F is iso with R cap A

[rustic crown]

we can give an order on Q(sqrt(2)) using the embedding Q(sqrt(2)) --> R sqrt(2) --> -sqrt(2)
and everything would be weird

how do you define realclosed?

[rotund aurora]

R cap A is formally real (orderable) and it has no proper algebraic extensions that are orderable

R cap A = P cup {0} cup -P where P is the set of squares in R cap A

unless Im delusional, no?

Like if its iso is clear, because that will happen only when F is realclosed, so its order is unique and is determined by the set of squares

but if F is not realclosed and its just an embedding R cap A, I guess the question doesnt make much sense

[rustic crown]

right

[warm wyvern]

I mean

That's what I was trying to do lel

[rustic crown]

lol

[warm wyvern]

But how do I know it's not just the trivial group or smth?

[rustic crown]

K could be trivial ig

[tribal moss]

0 is a square, so you don't need a separate term for it.

[rustic crown]

but you just wanna find K such that [G:K] divides |G| and n!

so if |G| does divide n! then ig just pick K = 0

[warm wyvern]

How do you know such a subgroup exists?

[rustic crown]

that's why we gave a weird construction of it

start with the action of G on G/H

to get the homomorphism G --> S_n

and kernel of homomorphisms are normal, so at least one requirement is fulfilled

usually you would use it for small values of n, so that |G| doesn't divide n!, therefore this map won't be injective in that case and K would be a nice non-trivial normal subgroup

[warm wyvern]

IG i can see why a non trivial kernal might exist

But how tf do I prove its index can always be GCD(|G|, n!)?

[rustic crown]

is that even true?

that exercise only asks you to show it divides the gcd

[warm wyvern]

Wait wat

omfg

[rustic crown]

lmao

[rotund aurora]

Isaacs talks about positive squraes, thats why I wrote it like that (I should of said positive square, yes, but I omitted it because it wasnt important). Talking about nonzero postiive elements is probably better, because you can then say that P and -P are disjoint

[warm wyvern]

That's such a dumb result

[rustic crown]

lol

it's not that dumb :p

you can use in some cases to say if a group is not simple

(okie saying that is not that useful :p)

[warm wyvern]

Lmao

Yeah, i know simple groups are super important

I just don't know why yet

[rustic crown]

i don't know much either lol

[rotund aurora]

What signature is he talking about? I thought signatures of bilinear forms were pairs or triples of numbers

mmh I guess its the trace of the (a) matrix of the bilinear form?

mmh but like the trace can be negative I think xD. Idk if its then the positive entries of the diagonal

Apparently, some authors call the trace the signature

The proof relies on hyperbolic planes, and idk what they are. This is the proof

I guess its not that important anyway

[small bramble]

hmmm I feel like dummit and foote is doing a terrible job at explaining R-algebras since it never explains that. So just defining an left and right R-module operation is enough to have an R-algebra also?

this is all it states (and I did exercise 22 too)

oops

wrong snippet

or rather not enough of the right snippet

[warm wyvern]

[elder wave]

the more you know the more you become aware of things that you don't know

it's a never ending cycle

so i guess everyone feels like that

[rustic crown]

i mean i just know jordan holder theorem, so every finite group is has finite simple groups as its building blocks, but beyond this idk anything :p

[formal ermine]

building blocks as in?

[rustic crown]

composition factors

you try to filter a group as much you can, i.e. find a series of subgroups 0 = G_0 < G_1 < ... < G_n = G
such that G_i is normal in G_{i+1} and the filteration can't be refined further

then composition factors are quotients G_{i+1}/G_i

by jordan holder theorem, any two filterations will give the same composition factors up to a permutation

since you can't refine the filteration further, the quotient G_{i+1}/G_i doesn't have any non-trivial normal subgroups, so it's simple

[agile burrow]

What if simple group, but infinite

[glossy crag]

What's this from?

[glossy crag]

JH holds for infinite groups as well, it's the finite length assumption that is pertinent.

[rustic crown]

yee ik but idk if people really care about infinite simple groups lol

[agile burrow]

They are frightening

[formal ermine]

someone once sent this cool diagram for algebraic structure

like a flow diagram

found it

[agile burrow]

Ngl I have no idea how to read this

[formal ermine]

start from a magma

do you want an identity?

go to unital magma

do you want inveritibility?

go to loop

[agile burrow]

Oh wait

Maybe this is dark mode discrimination

It is

[chilly ocean]

pretty sure this is from wikipedia

[lavish spoke]

Are these words used much? I've heard semigroup and monoid but not the other middle ones

[chilly ocean]

tfw inverse semigroup with identity is a group

[chilly ocean]

unital magma, no

quasi-groups and loops exist more in combinatorics

[lavish spoke]

ah ic

[chilly ocean]

but inverse semigroups, yeah
they're just in theory of semigroups so you might have not heard of the term

it's an attempt to make some kind of theory

and they correspond to partial functions

[lavish spoke]

Oh I think I've only come across monoid in a different sense of the word (I assume a category theory monoid is a different thing)

[chilly ocean]

just how semigroups correspond to functions, and groups to bijections

[lavish spoke]

Algebraic theory of semigroups or something that'd pop up in analysis?

[chilly ocean]

you identify the two

because if you have a monoid then you can cook up a category in which it's monoid in that sense of the word

and conversely

it's very straightforward

[chilly ocean]

or?

oh

algebraic theory of semigroups, yeah

I don't think they pop up in analysis

[rotund aurora]

Clark's field theory notes http://alpha.math.uga.edu/~pete/expositions2012.html you can find the pdf there

[chilly ocean]

it's kind of related to this thing called Penrose inverse of a matrix

[narrow marsh]

you should try classifying them, its a nice and easy exercise

[coral spindle]

This is on wikipedia btw, on this article https://en.wikipedia.org/wiki/Semigroup

[warm wyvern]

magmas: every algebraist's punching bag
semigroups, monoids: somewhat useful somewhat pervasive
groups: what everyone actually cares about
everything else:

[coral spindle]

True lmao

[coral spindle]

They're not, btw. An inverse semigroup has a different notion of 'inverse' to the one in a group – in fact the diagram is misleading in that way. There are inverse monoids (i.e. inverse semigroups with an identity) which are not groups, e.g. the bicyclic monoid.

[chilly ocean]

I know

I was being sarcastic

[coral spindle]

Mb I just read the rest

[chilly ocean]

Idk, Moufang loops seem to appear in some obscure physics

just because quasigroups lack associativity doesn't mean they are not useful - it means they're painful to work with

particularly in octonions

https://en.wikipedia.org/wiki/Octonion#Properties

for me it seems that it's semigroups and monoids that seem to lack applications

like... what do you apply it to. Automata theory?

if someone wants to prove me wrong, feel free too, that'd be nice

[tired horizon]

Im trying to solve this equation in Z^2

I did all this work

I looked for a particular solution

Then went to lok for the general one

I feel stupid for not getting this

But after I looked at the teacher's solution ...why do we have the same k here

Is it because their gcd=1

Idk why I'm not getting it

[rustic crown]

this is probably more appropriate for #elementary-number-theory

but i'll just answer it since you're already here >.<

so you saw that (-2, 4) is a particular solution

[tired horizon]

Oh I didnt even notice it lol

[rustic crown]

i.e. you have 11(x+2) + 6(y-4) = 0

[tired horizon]

Yep

[rustic crown]

now 11 divides 6(y-4) as it divides the first term and 0

since 11 and 6 are coprime you get that it has to divide y-4

[tired horizon]

It doesn't devide 6

[rustic crown]

so y-4 = 11t

where t is some integer

plug this back into your original equation

11(x+2) + 6*11t = 0

[tired horizon]

Oooooh

Damn

[rustic crown]

and so x = -2 - 6t uwu

[tired horizon]

I get it now

Thanks :3

I shouldn't be studying at 3 am lol

[rustic crown]

yee getting a good sleep is important :p

(i'm sleeping at 5am since the last few days >.<)

[tired horizon]

Yeah I think I gotta stop now XD

Thanks tho

[rustic crown]

maybe since you're here i could also tell you how to solve these in higher dimensions or if you have more equations

say you're solving Ax = b where A is an m x n matrix over Z and b is a m x 1 vector

the idea is to use the smith normal form of A

write A = PDQ where P and Q are invertible matrices over Z of the appropirate dimensions and D is the smith normal form (which is a diagonal matrix)

so it reduces to solving D(Qx) = (P^-1)b

since D is diagonal, this isn't hard at all

[tired horizon]

What's does smith normal form mean?

A diagonal matrix form?

English isn't my 1st language lemme google it

[rustic crown]

yee sort of, if you know some linear algebra there is a notion of diagonalizing a square matrix over a field

there you can write a matrix A as PDP^-1

here we're working over Z which isn't a field, so it makes it much harder

[tired horizon]

Ah yeah now I get it

And since its not a field how do u diagonalize it

[rustic crown]

Z is still a euclidean domain, so you could carry out euclid's division algorithm as long as possible with all rows and all columns

for example the smith normal form of the matrix
[a b] would be [gcd(a, b) 0]

no body would want to compute smith normal form by hand lol

okie maybe i could give you an example

by doing the above problem with this theory

but i'll remove all the computational steps

since usually you'd just use a computational software and ask it to find the smith normal form for you

we had

[11  6][x] = [2]
       [y]

we have

[11 6] = [1] * [1 0] * [11  6]
                       [-2 -1]

therefore one would be solving,

[1 0](something) = [2]

and solution to this can be just read of

the first variable is forced to be 2, and the second can be arbitrary

so the something is,

[11  6][x] = [2]
[-2 -1][y]   [k]

and the matrix on the left is invertible

so you get,

[x] = [-1 -6][2]
[y]   [2  11][k]

which is

x = -2 -6k
y = 4 + 11k

this might be a good candidate of one of the most useless examples i've ever done uwu

killing a cute rabbit with a nuclear weapon >.<

oh

hewwo walter

[agile burrow]

hi det

i <3 smith normal form

[rustic crown]

[tired horizon]

Damn

Thats a big pain

[rustic crown]

but that's exactly why it's super useful

with you can prove some nice theorems really quickly

jordan canonical form, rational canonical form, structure theorem for fg modules over pid etc

(i write etc because i can't recall more :3)

[tired horizon]

Nice

I'll take screenshots and write it down and go brag to my teacher lol

[rustic crown]

[tired horizon]

When are we gonna learn that :"))

[rustic crown]

so ig the usual route is, you study groups, rings, modules and while doing some module theory you wanna understand structure of modules over a pid which is where it's usually proven

in our undergrad it was a course offered in the 4th sem

[tired horizon]

Im still studying groups rings and fields

[light iris]

structuuuuure

[tired horizon]

Im taking 3 algebra classes

Goups and rings / arithmetics / fields and polynomials

Along with other 3 classes mesure theory / ordinary differential equations/ calc 5

[rustic crown]

woah

[light iris]

good luck

[tired horizon]

Thanks

[rustic crown]

[solar glacier]

whats calc 5

[tired horizon]

Functional analysis

[solar glacier]

ah

[tired horizon]

Idk its internationnally like that

[solar glacier]

youre grad right not undergrad

[tired horizon]

Nope

Undergrad

[solar glacier]

ugrad>>

??

woah nice taking measure theory

and functional

[rustic crown]

i thought calc 5 was calc on manifolds with some de rahm chomology stuff :p

[tired horizon]

Its not up to me to choose :"))

We are forced to take 6 classes for one semester

[rustic crown]

wait so how long is your undergrad? is it 3 year of 4 year?

[tired horizon]

3

[rustic crown]

teaching functional analysis to kids is scary

unless they choose it on their own

[tired horizon]

Speed learning math lol

[rustic crown]

you live in a scary place >.<

[tired horizon]

We struggled so much with it omg

Cause we didn't finish our topology class

We didn't take normed vectorial spaces

[rustic crown]

rip

[tired horizon]

So we had to study them on our own with the teacher speed teaching functional analysis

I spent like 10h a day studying on weekends for like a month to get a hold of everything :"))

[solar glacier]

your schools in session rn??

[tired horizon]

What does in session mean

[solar glacier]

being so close to xmas/new years

like classes are being taught

and exams and homeowkrs are being given

[tired horizon]

Yeah its the winter holidays

We get like 2 free weeks before exams

[solar glacier]

woah

[rustic crown]

reminds me of my algebraic geometry 1 course in undergrad >.<

[solar glacier]

what country is this in lol

[tired horizon]

North africa

[solar glacier]

oh wow

[tired horizon]

Its a weird country

[tired horizon]

I've always wanted to take that class

[solar glacier]

so it must be super late rn for u

[solar glacier]

same

[tired horizon]

I think I'm gonna take it on grad

[solar glacier]

my prof menrioned hes teaching my alg part II

[tired horizon]

3h18

[solar glacier]

in alg geo point of view

am?

[tired horizon]

Yeah

[rustic crown]

it felt like the prof did 2 semester's worth of stuff in half a semester 🙈

i remember going to a new class and learning 5 new definitions each time

[solar glacier]

i took a riemann sufrace course taught using algebraic curves and riemann surfaces text which is as close to alg geo as ive taken

[tired horizon]

My teacher did that with normes spaces

[solar glacier]

like zero sets of polynomials n stuff and compact riemann surfaces

[tired horizon]

She just read through a pdf

56 pages

In 3h

[rustic crown]

[tired horizon]

And she was like... yeaaah

Thats pretty much it

[light iris]

How much more abstract algebra do i have to study before i “get it” and not just memorizing theorems

[tired horizon]

We only learned the riemann integrals

And sums in series

[tired horizon]

At this point I think its just memorizing

[solar glacier]

yeah we learned about zero sets of homogeneous polynomials

which form RIemann surfaces apparently. that course was tough for me.

[rustic crown]

when you do harder abstract algebra, the stuff you thought you didn't get are crushed by the new stuff you don't get, so it feels you finally understand the old stuff

[light iris]

I feel like some people i talk to are really understanding it

[tired horizon]

Dude why math stuff in english sound so scary

[light iris]

Loll

So youre saying

[tired horizon]

Well it takes time tho

I am getting the grasp of it

[light iris]

You encounter even more crazy stuff that makes you realize you understand it evne less than the previous topics

[tired horizon]

Cause I have to its my field for now

I do algebra and cryptography so I had to learn how to undertsad stuff

[light iris]

Wow

[tired horizon]

Personally mindmaps helped me a lot

[rustic crown]

no more like you're not scared of the old stuff, and once you're not scared you can look at the theory much more easily

[light iris]

Mindmap?

[light iris]

The new stuff puts the old stuff into context?

[tired horizon]

Let's say u don't understand rings

[rustic crown]

sometimes

[tired horizon]

And the theoremes

[rustic crown]

like i used to think that modules and tensor product and stuff are kinda hard about a year ago

[tired horizon]

U try to draw what u know

[solar glacier]

you cant really draw out ring concepts lol

[tired horizon]

Lemme show u an example of a mindmap I made after I was revizing for a test

[rustic crown]

now i have seen their sheafified versions with modules over sheaf of rings, so they seem super trivial 😂

[solar glacier]

you can draw in topology by not much in alg

[light iris]

Ah isee

[tired horizon]

Not literally lol

[solar glacier]

oh lol

[light iris]

So mindmap but actually making one, not just in your head

[tired horizon]

Wait lemme show u an example

[solar glacier]

i was like how do i draw an ideal lol

[light iris]

Ohh

[tired horizon]

This is an example for a topology test

[light iris]

This is cool

[solar glacier]

i like this

[tired horizon]

I started doing it like every lecture we have

[solar glacier]

wish it was in english lol

[tired horizon]

Until I managed to make this one for one concept

[light iris]

I expect its very useful for showing how things are connecred because this kind of stuff is all about showing the structure and showing how things have similar or different structur

[solar glacier]

but i think i can manage to translate

[rustic crown]

i can't imagine doing math in a language which isn't english >.<

[tired horizon]

Thats how I knew i understood that

[tired horizon]

I do it in 3 languages lol

[rustic crown]

[tired horizon]

You're american?

[rustic crown]

nu

[solar glacier]

im american lol

california/Texas

[tired horizon]

We speak 2 languages

[rustic crown]

aren't these like super scary place?

[tired horizon]

And I learned english too cause everyone has to lol

[rustic crown]

i can't even imagine doing math in my native language lol

[solar glacier]

LOL^

same w me

i tried learning some of the words once

[tired horizon]

U study in another language in univ?

[solar glacier]

me? no english im in america

[rustic crown]

english is not native for americans?!

[tired horizon]

I was talking to u

Cause ik in the US they study in english

[rustic crown]

nope, all my studies were in english

[tired horizon]

Ah okay

[rustic crown]

i did my ug in india, and doing masters in germany

[tired horizon]

We actually study in arabic here

Then in uni

We switch to french

[rustic crown]

scary

[tired horizon]

Yeah but we basically speak french too on a daily basis

So its not that hard getting used to it

Doing math in french is cute

English math is scary to me

Everything sounds hard

[rustic crown]

i had to read some math article and my prof gave me it in french because she didnt' know any other nicer place to find it >.<

[tired horizon]

How is math in germany?

[rustic crown]

i had to guess so many things

[rustic crown]

mostly in german, idk if they even do the bachelors in english at all

[tired horizon]

I'm planning to go to either france or germany in 2 years

[tired horizon]

Aw man so I have to learn it

[rustic crown]

but since masters also get a lot of international students they're shifting to engilsh

[glossy crag]

I have a very basic question I just can't get past: if p is a maximal ideal of R, consider the localisation R_p and the ring homomorphism R/p^a->R_p/p^aR_p, r+p^a goes to r+p^aR_p. Why is this injective? If r is in p^aR_p, then it is of the form b/c with b in p^a and c not in p, so you get rc is in p^a and hence in p and r is in p. But how do I get that r is in p^a?

[rustic crown]

you wanna learn a 4th language?

[tired horizon]

I already speak 4

Lol

German is hard tho

Its more complicated than french

[rustic crown]

ig you can induct

say r is not in p^i but it is in p^(i-1)

so in the R/p vector space p^(i-1)/p^i, the vector r is non-zero

but when you multiply it with the scalar c in R/p, it becomes 0, as rc lives in p^a which is a subset of p^i assuming a >= i

so c is 0 in R/p

which gives c in p

which is sad

so r lies in all p^i for i = 1, 2, ...,a

:3

[glossy crag]

Hmm that works. I expected the answer to be more trivial, something incredibly basic I'm missing. Thanks.

[rustic crown]

idk, i was just trying to continue from where you left of :p

[rustic crown]

btw a cautionary comment, you don't directly get that b = rc

the definitions only say there exists a z such that bz = rcz

but the thing is, you could have just changed the fraction b/c to bz/cz

and this new c works now

.<

[glossy crag]

Wdym, R imbeds in R_p as r/1, so if r/1=b/c you get b=rc.

[rustic crown]

yea but the definition of equivalence isn't that a/b = c/d iff ad = bc

[glossy crag]

Are you talking about the definition of localisation for general rings?

For domains it's enough to consider ad=bc.

[rustic crown]

is R a domain? >.<

[glossy crag]

For domains it is, afaik.

[rustic crown]

yee for domains yep

[glossy crag]

I mean... The proposition doesn't outright assume it iirc, but the context of the book is all domains and localisation in the text is defined as ad=bc.

[rustic crown]

oh okie

[glossy crag]

Thanks for the pointer regardless.

[rustic crown]

i haven't ever worked with localizations just for domains lol

but it can be fun with non-domains

[glossy crag]

Yeah I've heard it comes up in AG.

What German uni you at btw, I'm thinking about future prospects (currently finishing bachelor in Germany).

[rustic crown]

for example if you have two rings and you take their product, R x S, then inverting the element (1, 0) basically kills off the component S and you get R back :3

[glossy crag]

That's neat.

[rustic crown]

i was at muenster >.<

[rustic crown]

yee and one nice thing you get is that R is a flat RxS-module

so tensoring will automatically be exact

[glossy crag]

Oooh, I've seen a poster for your department on our bulletin board. It had a purple-ish logo, right?

[rustic crown]

yea lol

i came because of that poster

😂

[glossy crag]

Kek. How was it?

I've seen one of your guys on YouTube talking about Hochschild homology.

Big dude with glasses and long hair, iirc.

[rustic crown]

it's been great... i just wish there were more people taking the course ag3 with me >.<

[glossy crag]

Are you at a different German uni now?

[rustic crown]

oh i haven't seen their youtube or twitter

[rustic crown]

nah, muenster was the only place in germany i found which had a scholarship for master students

i might visit bonn sometime as it's not that far

[glossy crag]

How was your overall impression of the faculty and course offerings, particularly algebraic ones?

[rustic crown]

i would be happy if there were even more courses lol

but ig they have a good bunch

and facultys have been pretty good so far

(that wasn't always the case in india)

so it was already an upgrade for me, so i'm pretty happy :p

[glossy crag]

Do you have a module handbook or a list of offered courses on hand?

[rustic crown]

okie the website of course catalogue is currently down

[glossy crag]

Thanks for trying anyway.

But in words, what have you taken.

And what are the mandatory ones.

[rustic crown]

for masters there aren't anything mandatory ig... only thing is that you have to do at least one applied math course

i'm doing these
algebraic geometry 3
differential geometry 1
etale cohomology seminar
topology 1 (just attending)
topology 3 (doing a thing called type 2, where you don't need to write exam)

[high cypress]

Wow

Topology is so difficult that’s crazy

[rustic crown]

topology 1 is the easiest among all the courses i'm taking lol

which is why i'm just attending it

don't have the energy to do weekly hw for it

[maiden ocean]

whats in topology 3

and AG 3

[rustic crown]

ag3 did some flatness, smoothness, then some homological alg and should end with the proof of riemann roch

[maiden ocean]

[rustic crown]

[gilded gulch]

It must be a "top 3" hardest course haha

[rustic crown]

top3 is homotpy theory, and later half was more unstable stuff

[maiden ocean]

neat

[high cypress]

No way

Ain’t no way

I’m reading through topology without tears and I’m still crying

[rustic crown]

.<

https://tenor.com/view/kanna-kamui-pat-head-pat-gif-12018819

[maiden ocean]

I mean det is a masters student and of the things he listed only topology 1 is standard undergrad material

[rustic crown]

yea

dg1 is also sort of undergrad stuff

[high cypress]

Oh come on

If ur masters degree then it’s obviously easy for you

[maiden ocean]

what are you guys doing in it? lee type stuff?

[high cypress]

This has a filter damn