groups-rings-fields

[cloud walrus]

Liria ^(;,;)^:

[hot lake]

https://cdn.discordapp.com/attachments/496784958430380033/723289142263414890/unknown.png

yes

uh

[shy bluff]

So $\chi_S(g^{-1} \cdot x)$ will be 1 if $g^{-1}x \in S$ and 0 if $g^{-1}x \not \in S$, is this correct?

[cloud walrus]

Liria ^(;,;)^:

[hot lake]

does it say somewhere the action of G on {0;1} ?

[shy bluff]

Nope

that's all of it

Yea that's what I'm caught on

I don't know what the action of g on {0, 1} is?

[hot lake]

it says it's the trivial action on {0;1}

[shy bluff]

Trivial action = maps to itself?

[hot lake]

that means g.z = z forall g in G and z in {0;1}

[shy bluff]

Oh ok

I see

[shy bluff]

Ah and then if $x \in gS, then g^{-1}x \in S$ and vice versa yea?

[cloud walrus]

Liria ^(;,;)^:

[hot lake]

well, you have to use (once again) the definition of g.S to show that

it's all about unpacking all the definitions until you get a tiny tiny logical proof to do at the bottom of it

[vapid dust]

Can someone explain to me how polygon symmetries and cayley tables work?

[stone fulcrum]

That's a multiplication table. If you pick out 6 on the left, and 4 on the top, you can easily "battleship" to find that 6×4 = 24

Now, you can construct a similar table for any group

Here's a table for mod 5 addition that I pulled off Google lol

@vapid dust

[kindred rivet]

cayley tables

basically are

[vapid dust]

So what is the difference between a reflection and a rotation?

[kindred rivet]

a table of the elements

of a group

and the group action between any two elements of the group

and their output

[chilly ocean]

@vapid dust Reflection is flipping an object across a line without changing its size or shape. Rotation is rotating an object about a fixed point without changing its size or shape.

[kindred rivet]

...

[stone fulcrum]

It's worth googling "dihedral group" to see this. It's pretty hard to describe as it's very geometrical

[kindred rivet]

i presume he means in terms of group action or somethin

[vapid dust]

I have to be able to describe the symmetries of various polygons, prove or disprove that the symmetries form a group, then do cayley tables.

So, I am being told that a Rhombus has 4 rotations, whereas a rectangle has 4 reflections and 2 diagonals. I am trying to process what that means.

[shy bluff]

If b is the standard inner product on R^n, what is the "Usual action on functions" of GL_n(R) on it?

Is that just matrix multiplication?

[solemn rain]

yea

matrix multiplication / function composition

[elder valley]

In classical algebraic geometry, what's the motivation for working in projective space? Like the most basic problem in the field is solving systems of polynomial equations over affine space, but how does projective space come into play in trying to solve those problems?

I know you can embed the variety into projective space by taking its projective closure and homogenizing the equations, but why would you even want to do that

[next obsidian]

You need projective space to make the whole intersections work like they “should”

Like two lines might not intersect, but they do in projective space

And Bezout’s theorem holds in projective space when you count intersections correctly

That’s as much as I know

[elder valley]

But if they don't intersect then there's no solution to the system. Why is there a need to go into this expanded setting where solutions do exist

[next obsidian]

Idk, it gets rid of the anomalies

In almost all cases two lines intersect

And in almost all cases a parabola and a line intersect twice

In this context that always holds

[elder valley]

Hmm. Sort of like working in the closure of a field when solving a single polynomial equation

[next obsidian]

I think projective space is also of geometric significiane just, inherently

Sort of, perhaps

Like projective geometry was used for like renaissance art or some shit to understand perspective

[mild laurel]

If you've seen complex analysis things, the riemann sphere is just the projectivization of the complex plane

[next obsidian]

And how things appear to humans when things go away in the distance

[mild laurel]

And the riemann sphere is used when sometimes you need a "point at infinity"

[next obsidian]

That’s a good point. You can’t really understand Möbius transforms properly unless you go to the riemann sphere, but they’re important even just in the plane

And more generally for groups of automorphisms and stuff like that.

[mild laurel]

Yeah, and another related reason is that in arithmetic dynamics (also with mobius transforms), you'd like to divide by zero in some sensible way, and projectivization allows you to do this

[elder valley]

I don't see what any of this has to do with algebraic geometry. Or at least with solving polynomial equations

[next obsidian]

A lot of classical algebraic Geometry was really just multivariable complex analysis

And it’s to illustrate the point that sometimes adding in extra stuff provides the more natural environment to work in even if what you’re interested in is stuff in the original context

Like in real analysis you’re curious about some function, but the best proof is to extend the function to C, use the sweet structure there, then come back to R

Like computing an integral via a path integral in C and using residues

[mild laurel]

Also, I really don't think you should think of algebraic geometry as solving polynomial equations. The way I look at it, its more like studying the geometry of sets of polynomial solutions

[elder valley]

It's always seemed like a lot more algebra than geometry to me

[latent anvil]

Mathemagician has told me he just ignores it when people talk about geometry in AG

[next obsidian]

Yeah pretty much

But that’s because I come from a not-geometrically oriented point of view

[mild laurel]

Yeah idk, I just don't feel like AG helps you solve polynomial equations at all

[next obsidian]

Me neither

There’s some like algorithmic stuff that does I believe

[elder valley]

Isn't that backwards though? Shouldn't the subject be about using algebra to answer questions about geometry

[next obsidian]

There’s some books on it, but the modern way of learning AG doesn’t cover that stuff

Eh, the two are so related you can go both ways

Me and shamrock (actually shamrock) figured out a proof of something purely algebraic using the geometry of schemes

About localization and stuff

[latent anvil]

I think that's backwards from what auvera is asking

They're asking why AG is so algebraic if the point is to study geometry using algebra

(i might have misunderstood)

[elder valley]

Well sort of. Obviously algebra will be used a lot but it just seems every book I read completely ignores the geometry part of the subject

[latent anvil]

I think Hartshorne does actually give a lot of geometric motivation in the first chapter iirc

Specifically when it comes to singularities and blowing up

That felt very geometric to me

[next obsidian]

It also depends on what angle you come at it from

Hartshorne’s stuff on schemes has like no geometry until like the 4th chapter when you start to talk about curves

I think there’s a trend where people try and set up the huge tools that have been developed for AG without any sort of motivation for the geometry behind it

It’s something I struggle with too deeply and I’m interested in the subject

[elder valley]

I read his first chapter long ago but didn't get farther than that. I don't remember relating it to geometry but my memory is hazy

[next obsidian]

Personally I just pretend it’s all algebra atm and have faith at some point I’ll get it

[olive mirage]

algebraic geometry is that duckling that's so ugly it is beautiful or something

I think, actually, Eisenbud's Commutative Algebra is an algebra text that really emphasizes the geometry, so that kind of gets at it the other way around

"The Geometry of Schemes" is a must too

[latent anvil]

lol

I liked Reid's undergraduate commutative algebra because it puts CA in the context of geometry

but it doesn't cover a lot compared to atiyah macdonald or something and it's not an AG book

[next obsidian]

The first picture in there doesn’t even make sense to me tho, with the cone

[latent anvil]

Yeah me neither lol

[mild laurel]

Also, maybe a slightly more "solving polynomial equations" reason to care about projective space is that for elliptic curves, equations of the form y^2 = x^3 + ax + b, it turns out that you can put a group structure on the points of this curve over some field K, but it turns out that the identity of this group is a "point at infinity"

[next obsidian]

The rest is good but I spent like an hour trying to understand that and gave up

[mild laurel]

@latent anvil I took a class on that book, it was pretty good

[latent anvil]

We used it last quarter

[next obsidian]

Different related question, is the fact that x^2 = y^3 has a cusp related to the fact that it’s an elliptic curve with the degenerate a = 0, b = 0?

[latent anvil]

Mathemagician and I

[olive mirage]

I'm obviously biased, but I think elliptic curves are a great place to focus as you start to think about all these entwining subjects

[elder valley]

Elliptic curves are what I mainly work with and yeah the group structure you get is certainly nice. But this seems like more of an extra feature on top of just looking for solutions and doesn't really seem intuitive

[olive mirage]

well, at first sight, the group law is coming from Bezout's theorem

but once you dig a few layers deeper, it's coming from the Picard group

and I think understanding that makes all this machinery feel worth it

[elder valley]

I read about that in Silverman's book but he goes very quickly through it

Doesn't seem to do it justice

[olive mirage]

Hindry/Silverman does a good job going over it at a high level

every curve has a Jacobian Variety which has a group structure, and the curve embeds into its Jacobian

it's just that in an elliptic curve that variety is itself

[next obsidian]

Damn

Zeta are you an algebraic Geometry?

Geometer*

[olive mirage]

I'd say an arithmetic geometer

but yeah

[next obsidian]

Mkay

[elder valley]

On a related note is there an easy way to compute the genus of a curve? Or in particular to know if it's genus 1

[olive mirage]

although I do bounce around a good bit, but I'd say my heart lay with arithmetic geometry

[next obsidian]

That last comment just sounds super cool about Jacobian varieties and embedding it and shiiit

[golden pasture]

genus-degree formula:)

[olive mirage]

but combinatorics/finite group theory stuf I think about a lot too.

[next obsidian]

Riemann-Roch?

[olive mirage]

and lately I'm focused on understanding analytic number theory, becuase I do a lot of undergradaute research projects, and arithmetic geometry is not a good topic for undergraduate research haha

[elder valley]

I thought Riemann roch only gave existence

[olive mirage]

The degree theorem only works if its smooth, or if you accept a (pathological) definition of genus

[next obsidian]

It can let you compute it if you understand the cohomology of it IIRC

But that part of my class was sketchy

For me at least

[olive mirage]

but Riemann Roch is the "standard" way to calculate genus

but there are all sorts of problems floating around with exactly how your curve is given to you and so on

but every genus one curve can be put in Weierstrass form, so it is traditional to use that

[elder valley]

Say polynomial equation in 2 variables?

[olive mirage]

and there is a nice formula for the degree of a hyperelliptic curve (one of the form y^2 = f(x) )

[next obsidian]

Oooooohhh we covered that. It has to do with P^n right?

[mild laurel]

@next obsidian Chapter 5 of Griffith's book talks about the Jacobian variety

At least, that's where I learned all that I know

[olive mirage]

proving the existence of the Jacobian variety is ugly

[elder valley]

In cryptography I encounter all sorts of equations and it'd be nice to be able to tell at a glance if it's elliptic or not

[olive mirage]

but no one makes you, you can just pretend 😛

[next obsidian]

I should really learn some stuff about varieties. I really hated them on my first pass but liked schemes a ton.

So I learned nothing about them since I disliked them

[olive mirage]

well, you can turn it to "group scheme" in your head instead

[next obsidian]

I think that’s a highly controversial opinion

Haha

[olive mirage]

I think optimal pedagogy is both at the same time

[latent anvil]

schemes are hard

that's my hot take

[next obsidian]

Schemes r cool

I couldn’t sleep and was thinking about how to glue affines together

It was bad

[elder valley]

Challenge question: intuitive definition of a scheme?

[latent anvil]

no

[next obsidian]

Locally looks like the spectra of a ring

[latent anvil]

locally the opposite of a ring

[mild laurel]

@ buncho bananas talks about it here https://www.reddit.com/r/math/comments/a0dd6j/mathematicians_what_is_your_favourite_algebraic/eah33w3/

[latent anvil]

:^)

[olive mirage]

Like, I think you have to bounce back and forth and talk about how you are algebrizing geometry, and then being like "hey, polynomial rings are the only rings!"

[latent anvil]

oh that's a mood zeta

[olive mirage]

haha I even meant to say aren't haha

[latent anvil]

Also zoph I didn't realize that was buncho

[next obsidian]

I mean they are. All rings are polynomial rings over themselves in 0-variables

:^)

[latent anvil]

I've said dumb things and gotten called out for not knowing number theory by the user lol

[olive mirage]

number theorists are the worst

[next obsidian]

Take that Thomas

Both of you

[mild laurel]

: (

[latent anvil]

?

zoph it's okay, I accept your life choice

[olive mirage]

fortunately there are a lot of younger number theorists that are more, shall we say, human

[latent anvil]

To study number theory

[olive mirage]

Like Ravi and Jordan, for example

[next obsidian]

Who is Ravi?

[latent anvil]

I would call buncho a human

[next obsidian]

Uhhh

To me@he isn’t younger

[olive mirage]

I assume buncho is much too nice to visit math overflow

[next obsidian]

But I forget you’re like... a prof

[olive mirage]

oh when I say younger, I mean like, 45ish probably

[next obsidian]

And married and so our definitions might be different

Oh

I thought you meant like

Me, aged

Or like grad student

Haha

[latent anvil]

on this server younger usually means high schooler imo

[next obsidian]

Fair

[olive mirage]

older than me, but on he younger side of people who have the power to influence the culture surrounding the discipline

[latent anvil]

As in "wow you're pretty young to be reading Rudin and atiyah macdonald at the same time"

lol

[next obsidian]

I got naynay’d on earlier by a bunch of hsers

[olive mirage]

like, number theory has historically been dominated by people who seemingly want to make you feel dumb and inadequate, and was really toxic

[next obsidian]

Hmm

[olive mirage]

who knows, maybe by now mathoverflow isn't a cess pit anymore 😛

[next obsidian]

I guess I wouldn’t know since the only number theorists I know of are like the super famous ones

Like Gauss

And Euler and shit

Lol

[latent anvil]

So like, that's kinda of my perception of math in general

I wasn't aware it was worse for number theory

[olive mirage]

That is definitely my experience, anyway. And it was pervasive enough that it kind of rubbed off on even the nicer number theorists. Like I've definitely caught myself saying things and been like "I really do not have to say it in a way that makes me look like a prick"

I think I've gotten better 😛

[latent anvil]

that kind of thing is something I need to work on

I can be very, like, showy

I need to use the word obvious less lol

[next obsidian]

Sometimes you said some stuff to 398 kids I was like

Uhhhhh... uhhhhh

[latent anvil]

Yeah...

[olive mirage]

Yeah, "it is clear after some reflection that..." is a good phrase

[next obsidian]

Honestly I think I only use phrases like “one can see that”

When I don’t want to take the time to do some argument

It doesn’t make any valuation statement on the difficulty

It just says you can see, but I will not show

[olive mirage]

or a Zagier-ism "If you sit down by yourself with a sheet of paper and attempt to prove this, I can all but guarantee success"

[next obsidian]

That’s really wordy tho haha

[latent anvil]

I use obvious when
• it's obvious
• I've been thinking about it for several hours and now it's obvious

[olive mirage]

it's for giving talks

[latent anvil]

Nice and unambiguous

[olive mirage]

The worst is when a book says it is obvious, and I stare at it and can't figure it out and think about it for an hour, and then just it is obvious, and I try to figure out what I could have said to make it easier, and I can't think of anything

[latent anvil]

yuuuup

I'm starting to write our paper in my reu

[next obsidian]

I don’t know, I saw someone say that when a teacher says “this is easy” or “it’s an easy exercise to...” that their brain turns it into “don’t try to do it because it’s easy and if you can’t you’re dumb”

[latent anvil]

and this is very difficult

[next obsidian]

But I don’t know how you can like avoid that or work around that

[latent anvil]

How to communicate obvious things

[olive mirage]

Math writing is really really hard

[solemn rain]

is this taken boys?

can i ask something stupid here?

[latent anvil]

one thing that's tough is that if you tell students something will be hard they'll think it's hard

Oh yeah go ahead

[solemn rain]

cool

[next obsidian]

If it’s algebra related yeah

[solemn rain]

ty sorry for interrupting

yea its an algebra proble,m

Let G be a finite group. Prove that for any prime p, there exists a unique normal subgroup N such that G/N is a p-group and any homomorphism is trivial on N

i missread the problem at first and thought G was a p-group

and so i thought immediately on Z(G)

[next obsidian]

I don’t get what that means for any homomorphism is trivial on N

[solemn rain]

phi(N) = 1

for homo phi

[next obsidian]

For any homo? To where?

[solemn rain]

oh fuck yea missed hthis one

[next obsidian]

Unless you specified a target it’s impossible

Since you can take identity

[olive mirage]

I feel like there are words missing here

[solemn rain]

the homo f is of G into any p-group

[next obsidian]

Even then I think you need a specific map

[solemn rain]

is trivial on N

[olive mirage]

Gotcha

[solemn rain]

yea sorry

[olive mirage]

so my very dumb proof would be "Take the intersection of all of them"

[solemn rain]

fuck

thats so nice

[latent anvil]

you only need to consider finite groups

[solemn rain]

fuck wow

[latent anvil]

And there's countably many isomorphism classes finite groups

so this is well defined

Like you don't need to take a class sized intersection

[solemn rain]

zetamath very nice tysm

[next obsidian]

Wait what

[latent anvil]

Intersect all kernels

Of all maps into p groups

[next obsidian]

Okay

Ohhh

[latent anvil]

I was arguing that this is well defined

[next obsidian]

I see why you need the countable iso classes

[olive mirage]

I mean, that is a subset of the powerset of G

[latent anvil]

?

[olive mirage]

so you don't need to worry about anything ugly

[next obsidian]

But how do you show that G/N is a p-group then

[latent anvil]

wym zeta?

Oh I see

Yeah you're right

[olive mirage]

G/N definitely won't be a p group

[latent anvil]

The actual collection you get is a set

[next obsidian]

Yeah sham that’s why I was sort of confused

[olive mirage]

well, maybe it will be, but not usually

[latent anvil]

Even if it's indexed by a proper class

[kindred rivet]

Hello guys

[solemn rain]

how do i find the p group

[latent anvil]

sorry mb

[kindred rivet]

mo2men

[solemn rain]

yo

[latent anvil]

okay yeah so the quotient won't be a p group probably

[kindred rivet]

Do u mean the p sylow group

[latent anvil]

But N will be contained in this intersection

[next obsidian]

You’d have to find some subgroup of N then right?

[solemn rain]

no no sylow

read problem

[next obsidian]

Err

Yeah what sham said

[olive mirage]

I always tell my algebra students that if they ever think they have to worry about whether something is a set or not they don't

so I hope I don't find an exception :-{P

[next obsidian]

It actually hurt me once

I did Zorn’s on a proper class

[solemn rain]

quicky

if G is a p-group

then the psubgroup would be Z(g)

Z(G)8

and thats trivial rright?

[next obsidian]

No

[solemn rain]

why

isnt Z(G) a p-subgroup

[next obsidian]

Z(G) is never trivial for a p group

[olive mirage]

if G is a p-group, it's the identity

[solemn rain]

yea

what

yea mathe

and G is a p group

so the quotient would be a p?

[latent anvil]

yeah zeta I think the problem was to show there's a minimal essential extension of a module? In order to construct an injective hull

and Mathemagician zorned on the class of all essential extensions

which does not work

[next obsidian]

Yeah, but I thought you said Z(G) is trivial?

[solemn rain]

no no i meant

the problem

xd sry

[next obsidian]

Ohhhh

I see

Lol

[solemn rain]

so would it be correct

[next obsidian]

Errrmmm. Maybe? I don’t see why Z(G) will be acted trivially on by all morphisms into p-groups tbh

[latent anvil]

yeah that sounds fake

[olive mirage]

but if our group is A_100 and p=7, what's N?

[latent anvil]

Let H be the big intersection

[next obsidian]

I feel like if anything it would have to be {1}

Like take Z/pZ

[kindred rivet]

Zeta

[next obsidian]

Or Z/p^2Z

[olive mirage]

(If G is a pgroup, the identity map G to G has kernel the identith)

[kindred rivet]

Ty

[next obsidian]

Yeah

That’s a good proof of that haha

[solemn rain]

i think i have to use this theorem

that saays if p is the smallest divisor of order G such that [G:H] = p then H is normal

thats just waht comes to my stupid brain

tbh

[latent anvil]

I don't think that will be useful here

You don't have much control over p

It needs to work for all p, not just the smallest prime dividing the order of G

[solemn rain]

oh yea

yea yea

lmfao thats strong no?

[next obsidian]

I mean

[solemn rain]

'for any given prime p' there is a normal such that the quotient is psub

lmfao

[next obsidian]

You’re the one who gave the problem dude

I’m the one thinking that seems hella strong lol

[latent anvil]

I have a dumb idea that doesn't work

[solemn rain]

haha help me guys

[latent anvil]

take all the p groups

And all maps out of G

[olive mirage]

I think you'll see it isn't so strong once you see what is happening in the A_100 case.

[latent anvil]

Take the product to get a map from G to that big product

[next obsidian]

A_100...

[latent anvil]

The kernel of that will be the H we talked about above

[next obsidian]

Yeah

[latent anvil]

and the image is a finite p group I tho

*think

By looking at the orders of the elements of the image?

I think?

[next obsidian]

Image of G?

Hmmmmm

[solemn rain]

wait

[next obsidian]

If it is then that’s the answer

I don’t see how to show uniqueness

[solemn rain]

A_100 is simple

right?

[latent anvil]

Consider the collection of maps φ_α : G -> P_α for P a p group

[solemn rain]

then doesnt this like

make this wrong

[next obsidian]

Would it be... all of A_100 then

Is the trivial group a p-group?

[latent anvil]

Let P be the product of all P_α (we can choose a countable subset to make this work)

[olive mirage]

Yeah, all of A_100 is correct

[latent anvil]

We have a map φ : G -> P

[olive mirage]

because there are only two normal subgroups of A_100

[solemn rain]

lmfaaao

lmfaaaaao

yea

[olive mirage]

and we know it isn't the trivial one

[latent anvil]

the kernel of this is the thing we care about

[next obsidian]

Yeah

[latent anvil]

the subgroup we talked about above

[next obsidian]

How do you show the image is a p-group tho

[latent anvil]

I think the image will be a p group

so if it's not, there's an element in the image with order q ≠ p

[olive mirage]

so this "very strong" result isn't saying anything deeply profound

[next obsidian]

And even if you do that, how would you go about showing uniqueness

[latent anvil]

By cauchy

I think uniqueness will be easy

I haven't thought about it though

[solemn rain]

okay boys i cant seem to follow u

[next obsidian]

Zeta are you saying that it isn’t profound because often times it’s just gonna be everything?

[solemn rain]

so a recap if u guys finish please ? 😄

[next obsidian]

Take all the P_groups

Like one rep in each isomorphism class

Err

You need to go by maps

[solemn rain]

whats rep

[next obsidian]

Representative

[solemn rain]

ok

[next obsidian]

I’ll let sham explain lol

[latent anvil]

Magician if h = φ(g) and h^k = 1, then φ_α(g)^k = 1 for all α

Does that make sense?

[next obsidian]

Oh yeah

[latent anvil]

this is impossible unless k is a power of p because φ_α(g) lives in a p group

[next obsidian]

Yeah

Nice

[latent anvil]

So the image is a p group by cauchy

[next obsidian]

So every element has order some power of p

[olive mirage]

(You are really taking the intersection across all normal subgroups H such that G/H is a p-group)

[latent anvil]

that's existence

Oh that's a much better way to put it zeta

[solemn rain]

how can i prove those

exist

zetamath

[latent anvil]

we're really taking the minimal such subgroup

which is an easier way to phrase it

[next obsidian]

I mean

You don’t have to

[latent anvil]

Mo2men, the whole group is such an H

[next obsidian]

Because you only need one

And H = G works

[solemn rain]

lmfao yea

[latent anvil]

I think I can phrase it better now

[solemn rain]

yea yea

[next obsidian]

Then you can take the intersection of all

[solemn rain]

yea

[latent anvil]

first show the intersection of two of them has quotient a p group

There are finitely many so the intersection of all of them has the same property

this satisfies has quotient a p group

[next obsidian]

Ohhh then it acts trivially

By the first iso

[latent anvil]

Yup

Because it's contained in all the kernels

[next obsidian]

Since the image is a p-group since it’s contained in a p-group

[latent anvil]

So uniqueness is really saying that the second condition forces it to be minimal

I think

[next obsidian]

Yeah that sounds right

[olive mirage]

this is a universal property

[next obsidian]

How do you show the intersection has that property tho

[olive mirage]

those come with uniqueness for free

[solemn rain]

p^0 is power of a prime p right?

quicklie

[next obsidian]

Like third iso??

Yeah

[solemn rain]

okay yea i get u

[next obsidian]

Second?

[latent anvil]

Fuck I keep doxxing you lmao

[next obsidian]

Idc

Really

[latent anvil]

Just do the thing we did above

[next obsidian]

Which thing

[latent anvil]

Map into the product

the kernel of that is the intersection

the image sits inside a product of p groups

And so is a p group

[solemn rain]

okay so for any prime p , the intersection of all subgroups H that make G/H p-group is such H?

dumb it down for me boys

hahaha

[latent anvil]

yes

[solemn rain]

okay i got it

and atleast

G is one of those

( the original gropu )_

[latent anvil]

yup

[solemn rain]

okay

now why the homo part

again?

[next obsidian]

But justifying that is a bit whacky. Do you have to do that thing over all homs?

[solemn rain]

any homo

that sends G into a p-group is trivial in N

[next obsidian]

Yeah

[solemn rain]

ig i have show that N is a subset of ker(f)?

[latent anvil]

yes exactly

[solemn rain]

does not have to be proper

right?

ye

[latent anvil]

no

[next obsidian]

This is just automatic tho right

[solemn rain]

why

mathematgicain

[latent anvil]

You're doing the thing

[next obsidian]

If f goes to a p-group

[latent anvil]

that we were just talking about

[next obsidian]

Fair sham

Then the image is a p-group right?

[solemn rain]

umm

yea

[next obsidian]

Then by first iso

[solemn rain]

cuz its asubgroup

[next obsidian]

G/ker f iso to p-group

So ker f is in the huge intersection you formed N by

[solemn rain]

lmfao

cool af

tysm

all

i got it

i would have never figured tha tout

🧠

[olive mirage]

you could also do something really dumb and define the subset {g \in G | g is in the kernel of every homomorphism from G to a p-group}

prove that set is a subgroup

[solemn rain]

i like dumb things

[olive mirage]

and it is automatically a characteristic subgroup

and hence normal

[solemn rain]

lmfao i like these solutions

by not solving the problem

[olive mirage]

characterstic ==> normal is low key one of the best ways to not do work

[solemn rain]

yes

i dont know how to prove that tho

[olive mirage]

though this is a much longer to wrtie up solution than the one discussed above

[solemn rain]

yea

cuz id have to prove this rihgtr?

i have a qauestion in proving thing sgenerally

you could also do something really dumb and define the subset {g \in G | g is in the kernel of every homomorphism from G to a p-group}
@olive mirage

when you do this ^

do you have to show existence ?

[next obsidian]

Zetamath how do you show it’s characteristic?

[solemn rain]

that such thing is there?

[next obsidian]

1

The identity is in there

[solemn rain]

oh my god

yea

so u do

[next obsidian]

So you don’t need to worry

[solemn rain]

yea yea

but in general u need to show

( just like we did with the intersection )

that such 'thing' exists

atleast for a trivial case

[next obsidian]

Not really

The set will always exist by the axioms of set theory

You just would have to show it’s a normal subgroup

I.e it’s normal

and non-empty

[latent anvil]

it's minimal among normal subgroups whose quotient is a p group

automorphisms will preserve normality and having quotient a p-group

and they preserve the lattice structure enough to make that work

probably not what zeta was thinking though lol

since it's just the proof we talked about above

[next obsidian]

Oh

I was thinking like

The Thomas proof

That you hate

[latent anvil]

oh you can just do it explicitly lol

[next obsidian]

“Proof”

[latent anvil]

composing with an automorphism of G will still give a map into a p group

[olive mirage]

yeah, morally speaking, any time you describe a subgroup by this kind of property that doesn't mention specific elements it will be charactersitic

you get a feel for it

[latent anvil]

you intuited the proof magician was referring to lol

[next obsidian]

YES

[cloud walrus]

shamrock:

[next obsidian]

Sham haaatea that proof

I like it

🙂

[latent anvil]

yeah because you pretend logic doesn't exist

[next obsidian]

But it’s sooooooo goooood

[olive mirage]

but like, the subgroup generated by the squares is automatically charactersitic, so automatically normal

and so the squares in A_100 generate A_100

you get to do all kinds of nonsense like this

[latent anvil]

oh that's so sick

[next obsidian]

Squares...?

[latent anvil]

g^2

is a square

[next obsidian]

Ohhhh

I see I see

[solemn rain]

u smart boys

i dont know what u talking about gl

lmfao

ty agian

i didint care muchj about char subgroups

ddidnt know u guys think they are so cool 😄

[olive mirage]

I was mostly mentioning that because someone the other day in here was like " Idon't think charactersitic subgroups are ever useful"

[latent anvil]

oh yeah characteristic subgroups are gr8

big fan here

[olive mirage]

and they're incredibly useful for things you define with this kind of proeprty

though they also have a tendency to produce proofs that don't tell you anything haha

[solemn rain]

yea i thinnk i saw very weird stuff, so char ssubgroups are subgroups that are trivial for every auto on G

[latent anvil]

the best kind of proof

[solemn rain]

very weird things are char subgroups

[latent anvil]

all computation and magic

no insight

[solemn rain]

like a subgroup that is unique of its order is char

lmfao

[next obsidian]

Yeah

[solemn rain]

very werid shit

and their proofs were just too scary for me

without even seeing them

[next obsidian]

Our TA is like “if there’s a subgroup you define by ‘the unique subgroup such that...’ it’s characteristic”

[solemn rain]

yea

[next obsidian]

And if that property is preserved under automorphisms then it is

But just morally that sort of thing seems to be true

[latent anvil]

"invariantly defined"

i think that was the wording?

[olive mirage]

if I laid out all the elements of this group on the table could you tell me which was in the subgroup? What if you turned your back, I swapped them around, and then you looked back? IF so, it's charactersitic

[next obsidian]

Yeah so sadly

[olive mirage]

that may not be useful, but that is the intenal vision I have

[next obsidian]

The unique subgroup made up of these specific elements isn’t

[latent anvil]

?

[solemn rain]

yea i get it

the swap being the auto ig

[next obsidian]

Like “the unique subgroup of S_6 which is ‘list of the elements’”

That’s just every subgroup lol

[latent anvil]

hmmm i wonder what the field fixed by a characteristic subgroup of Gal(K/k) is

sounds weird

like for normal we have a nice characterization

galois extensions of k

[next obsidian]

Zeta you said rhombuses tiling the plane is some deep elliptic curve stuff

[latent anvil]

ig Aut(Gal(K/k)) doesn't really seem meaningful

[olive mirage]

yeah, every ellipic curve over C is isomorphic to C mod one of those lattices

[next obsidian]

Is that sort of thing at all penetratable for someone who’s finished a grad course in algebra and knows a bit of AG?

Emphasis on like

[latent anvil]

oh I think thomas carr tried to teach me this

[next obsidian]

Bit

[olive mirage]

So, the proof actually runs through differential equations and so called "doubly periodic functions"

[next obsidian]

Ohhhhh

[olive mirage]

and it connects in with some of the motivation for modular forms

[next obsidian]

We did something with those last year sham

[latent anvil]

in complex?

[olive mirage]

the best book that gets at all of this, in my mind, is McKeen and Moll

[next obsidian]

Doubly periodic

Yeah

[latent anvil]

yeah I think I remember you talking about those

[next obsidian]

No like 336

[latent anvil]

oh

[next obsidian]

We had a problem on them

[latent anvil]

do not remember

i was so checked out lol

[next obsidian]

My thing was about like

Groups of auts without fixed points

[olive mirage]

McKean, sorry

[next obsidian]

And it was a similar flavor

[latent anvil]

honestly I don't think I remember any complex analysis

at all

that's a lie but not really

[next obsidian]

Shame

It’s cool stuff

[latent anvil]

no, sham

[next obsidian]

Like the universal entire function

Fake

I proved it exists, but fake

[latent anvil]

every so often I try to remember the proof of liouville

isolated zeroes are relevant maybe?? idk

currently I'm blanking

[next obsidian]

Iirc it’s just Cauchy estimates

[olive mirage]

Complex is like the one course I've really wanted to teach but haven't gotten to

[latent anvil]

I would like to relearn it

but there's so much math

it's relatively low on the list

[next obsidian]

You can show the first derivative is 0 with the Cauchy estimates

[olive mirage]

so therefore I will just have to make my entire number theory course into complex analysis

[latent anvil]

lollll

[next obsidian]

Take any point, draw a big ass circle around it

By boundedness as you increase the radius, derivative at that point is arbitrarily small, so 0

[latent anvil]

ahh okay yeha ii remember now

Idk if i'll take 534 in the spring

[next obsidian]

Oh but if you do the Don Marshall

[latent anvil]

I only need 2 more classes to graduate

[next obsidian]

Power series first

[latent anvil]

you too i assume

[next obsidian]

It’s a result of isolated zeroes or something

And you get it very fast

Or no

[latent anvil]

I went to advising and they made everything official, 50x and 54x are counting towards my degree

[next obsidian]

We get identity theorem very fast

Oh I should do that

I’m so bummed I can’t take the prelim

[latent anvil]

we should move out of algebra

[next obsidian]

I’ll be annoying and email again next year

[latent anvil]

this is not algebra lol

[next obsidian]

Fair

[thorn delta]

My book says that a subgroup $H \subset G$ is normal if $gH =Hg$ for any $g \in G$. i.e. $gHg^{-1} = H$. I always thought the definition was more like $gHg^{-1} \subset H$, i.e. $H$ is closed under conjugation. Is going from $gHg^{-1} \subset H$ to $gHg^{-1} = H$ trivial or something? I don't quite see it.

[cloud walrus]

kxrider:

[latent anvil]

it is

but you can't do it for just one fixed g

[oblique river]

replace g with g^(-1) and multiply over

[latent anvil]

if you know it for all g, you get the implication

^

[thorn delta]

ah i see, cool.

[woeful mist]

Any good places to learn abstract algebra? Could apply that to topology and differential geometry

[solemn rain]

check pinned in #book-recommendations

[woeful mist]

What about websites?

[solemn rain]

idk about websites

u could try ictp youtube playlist

on algebra

there is also a diff geometry playlist too

and topology

[woeful mist]

Thank you

[mild laurel]

@woeful mist It would be wise to learn linear algebra and some sort of intro to proofs before trying to study abstract algebra

[woeful mist]

Okay

Khan academy has linear algebra

[eager willow]

is x^5-5a^4x+a irreducible over Q for all integers a > 0 ?

I know for some a by eisenstein's criteria, and that there are always 3 real and 2 complex roots

[solemn rain]

induct

[chilly ocean]

do a few chapters of velleman and then artin @woeful mist

[woeful mist]

I don't think spending hundreds of AED would be worth studying abstract algebra. I'll just have to wait for uni then.

[chilly ocean]

what is AED

if that's referring to money use libgen

[latent anvil]

@eager willow so if it factors, the complex roots stick together, so it's gotta be 2+3 or 1+4 or 1+2+2 or 1+1+1+2. So either it has a rational root or factors as an irreducible quadratic (with two complex roots) times an irreducible cubic (with three real roots)

this seems to simplify a little

[sour plume]

@chilly ocean apparently it's UAE-money! https://en.wikipedia.org/wiki/United_Arab_Emirates_dirham

the more you know

[latent anvil]

If you have a rational root, a is 0 mod x, so a = bx. Then x^5 - 5(bx)^4 x + (bx) = 0, and canceling x we get (1 - 5b^4)x^4 + b = 0

[knotty mason]

I have one of those coins

[latent anvil]

then 5 b^4 - 1 divides b, whcih seems sus

like in particular 5 b^4 - 1 <= |b|, so b^4 <= |b| + 1, so b is ±1

and then it's false that 5b^4 - 1 <= |b| automatically

yeah what am I talking about. The point is that that polynomial has no rational roots

and so if it factors it factors as an irreducible cubic with real roots times an irreducible quadratic with complex conjugate pair roots

you can write that down explicitly and get some messy relations

[eager willow]

Why not an irreducible cubic with 1 real, 2 complex roots and a quadratic with real roots?

[latent anvil]

hmm good point

not sure why I thought that couldn't happen, sorry

It's probably too late for me to do math tonight

Still much simpler

you can definitely write x^5 - 5 a^4 x + a = (x^2 + Ax + B)(x^3 + Cx^2 + Dx + E) if it's reducible

I think

lol

[eager willow]

Yeah for rationals, not necessarily integers though

[latent anvil]

no for integers

by gauss's lemma

the stuff I did for the root only worked for integer roots anyways

wolfram alpha gives
-5 a^4 x + a + x^5 = A C x^3 + A D x^2 + A E x + A x^4 + B C x^2 + B D x + B E + B x^3 + C x^4 + D x^3 + E x^2 + x^5

not great

[eager willow]

I was trying to separate them out with symmetric operations on the roots as well but that wasn't going anywhere

[latent anvil]

Equivalent to
-5 a^4 x + a = (A + C)x^4 + (AC + B + D)x^3 + (AD + BC + E)x^2 + (AE + BD)x + BE

I'm not sure what you mean

[eager willow]

Using relations on the roots like the trace being 0

[latent anvil]

sorry why would the trace of a root be zero?

[eager willow]

Because the coefficient on x^4 is 0

[latent anvil]

In the case where the complex roots pair up to a quadratic that seems false

the trace of a root will be twice the real part, right?

It's Galois conjugates will just be itself and its conjugate

[eager willow]

The sum of the 5 roots is 0

[latent anvil]

Oh, I agree

But that won't be the trace if the quintic isn't irreducible

[eager willow]

Oh ok I thought trace applied to polynomials

Not just irreducibles

[latent anvil]

so to me trace is defined for any Galois extension K/k

so in particular if you have an element algebraic over the rationals

You can take its trace to get a rational number

this will be the sum of all the roots of its minimal polynomial

But the polynomial isn't really important

it's better to define it without the polynomial

ahhh

sorry

I have literally no idea how I just did that

okay so

those relations are good

But I'm not really sure how to get more info about the roots

or like, what knowing about the roots tells us

We get A = -C, B + D - A^2 = 0, A(D - B) + E = 0, AE + BD = -5 a^4, BE = a

Whcih is ugly

I mean it tells us things like B, E divide a and A divides E

If we multiply the second equation by A we get 2AB + E = AB + A(B + (D - B)) = A^3

or E = A(A^2 - 2B)

Idk I don't really want to think about this anymore because it got ugly

Maybe you can do something with that

[eager willow]

The way the question was phrased makes it seem like the galois group is the same for all a, not that I'm actually taking that as a given, but I'll look for a contradiction with those

[latent anvil]

what was the question exactly?

[eager willow]

For an integer a > 0, compute the galois group of the splitting field for x^5-5a^4x+a

Of course it's irreducible for some a, so for some a it's just S_5

[latent anvil]

How does that follow?

You can be irreducible with Galois group smaller than S5

[kindred rivet]

Ya

[eager willow]

Over Q?

[latent anvil]

Yes

[kindred rivet]

Granted statistically S_5 is quite likely

If it’s irreducible

[latent anvil]

Do you know about the discriminant?

You can use that to cook up an example where the galois group is A5

Or embeds in A5

it's still very important to know when it's irreducible ofc

https://math.stackexchange.com/q/286944/408287

here's an example

[kindred rivet]

I wonder statistically

On average polynomials have a Galois group of S_n for some polynomials of degree n

[eager willow]

huh... i guess I hadn't seen that.

[latent anvil]

yeah it's worth keeping in mind

Have you tried adjoining a root α and factoring in Q(α)?

[eager willow]

Would that do much good without already knowing the minimal polynomial of alpha?

[latent anvil]

Yeah, I think it'd be useful

It could even tell you things about that minimal polynomial

[eager willow]

So there is a theorem that applies here. An irreducible polynomial of prime degree p with p-2 real roots has a galois group of S_p. But I didn't know that when I made the assumption at first

[eager willow]

since for some root r there is a subgroup corresponding to Q(r) which is order p. Then the subgroup is cyclic order p so it's generated by a p cycle in the S_p embedding. Complex conjugation restricted to the splitting field also must be in the galois group and since only 2 roots are non-real this is a transposition in the S_p embedding. Finally, if g is a transposition and h is a p-cycle, then {g,h} generates S_p

[knotty mason]

I think they do have S_n on average

cant remember where I read it

[half nebula]

I feel kinda dumb for asking this but what does |G|=23 mean when G is a group? Does that mean it has 23 elements/ the order is 23?

[knotty mason]

that's correct

[half nebula]

Okay cool.

[delicate mauve]

it also means G is cyclic

[knotty mason]

since 23 is prime it implies G is cyclic

[half nebula]

I got this question here that's asking me if G is isomorphic to Z_23 if |G|=23. I was just making sure I was right on that part.

[delicate mauve]

is it? didn't notice it

[half nebula]

Also I feel like this one is a trick question. $D_4 is isomorphic to Z_2xZ_2xZ_2$. Like it's asking if that is true or not.

[cloud walrus]

Soulgiver831:

[half nebula]

God I really need to learn Latex XD

[delicate mauve]

no

the order of D_4 is 4

[olive mirage]

haha how can that be a trick question?

assuredly the person who wrote this question thinks the order of D_4 is 8

[delicate mauve]

then also no

[half nebula]

Like that's what I was thinking. We just haven't ever had one that's false before so I wasn't sure.

[delicate mauve]

think of the orders of elements

isomorphic groups have the same number of elements of a given order

[olive mirage]

I think orders of elements is even one step harder than you need for this one (-:

[half nebula]

Like this is the first homework problem all semester where I think I can just say "it isn't true."

Well I should probably give the reason too.

[delicate mauve]

@olive mirage are you thinking of commutativity?

[olive mirage]

Yeah, I think that's easier

[delicate mauve]

no

[half nebula]

It's okay I just used order.

Thank you two though ^^

[delicate mauve]

np

[half nebula]

God these questions are getting me nervous though. It's review for our final on monday and it's kinda freaking me out ngl. Like these seem like some pretty tough questions.

Like I'm doing them fine now but I don't do good with a time limit.

[half nebula]

So with congruences, I'm not sure if I'm doing this right. I have the congruence 7x is congruent to 1(mod 17). I looked around online (because our book doesn't explain diddly when it comes to this) and I found this.

So, I followed it and that would mean that 7p+17q=1. The only way for that to happen would be for p and q to both equal 0 right?

[next obsidian]

7p + 17q = 1 has a solution since 7 and 17 are relatively prime

This is Bezout’s lemma

[half nebula]

I can't say I've heard of that before. How do I find the solution?

[next obsidian]

Also is p and q are both 0 then there isn’t an equality

Idk, start plugging shit in haha

I mean

[half nebula]

Oh wait I forgot. p and q can be negative.

[next obsidian]

Okay let’s actually try and solve it haha

Yes this is crucial

[half nebula]

I'm guessing I want to keep p positive though?

[next obsidian]

Probably

[half nebula]

since that's what's used for the thing.

[next obsidian]

That’s a good way to go about it

[half nebula]

Okay.

[next obsidian]

A solution is (5,-2)

I just did it in my head tbh

[half nebula]

I'm not all that good at that ><

But yeah that works.

So it's x is congruent to 5(mod 17)

[next obsidian]

Ummm

That definitely does work

But I’m struggling to see if it’s unique

I think it is

Oh it totally is yes

That’s the answer, because inverses in a group are unique

Like if you saw 4 <= x <= 4 then clearly x = 4

This just holds in the limit is all it says

[latent anvil]

@eager willow where you able to show irreducibility?

also yeah the p-2 real roots thing is clutch, I can't believe I forgot it

[shy bluff]

If I have a group G that acts on X, then the stabilizer of $g \cdot x$ is all $g \in G$ such that $g \cdot (g \cdot x) = g \cdot x$? Is this correct?

[cloud walrus]

Liria ^(;,;)^:

[delicate mauve]

don't use the same letter g

all h in G such that h * (g * x) = g* x

[shy bluff]

Ok

Thank you

And g is fixed?

[stone fulcrum]

No hol up.

Take an element of the set S. There's going to be some elements G such that gs = s, that is, g "stabilizes" s

If g doesn't change x, then g is in x's stabilizer.

[shy bluff]

Oh

Oh ok I see

G_x is actually a subgroup of elements of g that don't change x

For any x in X

[stone fulcrum]

For a given x

[shy bluff]

Oh for a ggiven x?

[stone fulcrum]

For any given x, haha

Any x you choose generates a new stabilizer subgroup of G

[shy bluff]

Yea, so x is fixed

And the stabilizer is the set of all g's that stabilize your fixed x

[stone fulcrum]

And, like you said, this set is a subgroup of G

[shy bluff]

So then the stabilizer of $g \cdot x$ is the set of all $h \in G$ such that $h \cdot (g \cdot x) = g \cdot x$ right?

[cloud walrus]

Liria ^(;,;)^:

[stone fulcrum]

Oh haha yes I suppose that's true

Oh, then the above was pretty true too

Mb you have a better grasp on this than I realized

[shy bluff]

No my grasp is uh pretty loose

I'm having a lot of trouble in this class tbh 😦

[stone fulcrum]

Well you've got the stabilizer pretty down

[shy bluff]

🙂

Oh and then $gG_xg^{-1}$ is just $gjg^{-1}$ for all $j \in G_x$ right?

[cloud walrus]

Liria ^(;,;)^:

[stone fulcrum]

Yeah that's the "lazy shorthand" that appears often in group theory

We write the group into the equation to mean "any element of the group can go here"

[shy bluff]

Yea I'm beginning to see that lol, it confused me for quite a while :x

And this

Basically means show tahth they're both subsets of each other?

[stone fulcrum]

That's a good way to do it! Yep

Gx is the stabilizer of x?

[shy bluff]

I assume so

Like this is the whole question

Like I'm so confused so as to where g is fixed and where g is not fixed

[stone fulcrum]

I'll not overuse g.
Gx is the set such that hx = x, for any h ∈ G

Note that x is fixed, in this sense

[shy bluff]

Yes

[stone fulcrum]

G(gx) is the set such that hgx = gx, for a fixed gx

[shy bluff]

Ye

[stone fulcrum]

Well, I shouldn't say that. g is free here

It so happens that this is true for any g you choose, but you have to choose that g for Ggx to make sense, haha

Am I making sense?

[shy bluff]

Yes

I think that it's the gG_xg^[-1} bit that's messing me up

[stone fulcrum]

Oh that's just gjg^(-1) for some j in Gx

Since j is in Gx, it obeys hj = j

[shy bluff]

yea

Wait hj?

Uh like this is what I have so far

hrm

Oh wait no I think I can do this directly from $(kg) \cdot x = g \cdot x$ by multiplying both sides by $g^{-1}$

[cloud walrus]

Liria ^(;,;)^:

[next obsidian]

@olive mirage do you mind helping me with an AG thing if you’re free?

[olive mirage]

haha I can try!

to assume that I can do a random Hartshorn exercise is... ambitious haha

[next obsidian]

So I’m trying to do exercise II.3.7 in Hartshorne about generically finite morphism that’s dominant

Does that at all sound familiar?

Basically it suggests you show some field is a finite extension of another, so it suffices to show its finite as an algebra

But there’s some whack stuff where you have to pass to stalks and then take fields of fractions and I have nooooooo idea why finite generation as an algebra would be preserved under that

[olive mirage]

haha my AG professors were adamant I not touch Hartshorn

[next obsidian]

Oh lmaoooo

No worries then

I mean honestly, maybe that’s for the better

But I already spent some good time doing everything in the first two sections so... ¯_(ツ)_/¯

[olive mirage]

Hmmm... This sounds like a commutative algebra thing

But, in all honesty, I love and do a lot of algebra, but when I'm doing algebraic geometry I don't think about the algebra of the underlying rings at all.

(I think a lot about the Galois theory of the field extensions, and the group structure of the Jacobian and that kind of thing)

But I think mostly about smooth things mostly of dimension <= 2, so a lot of the more technical stuff I get to narrowly avoid.

[latent anvil]

What's the problem magician?

I might be able to help with the algebra at least

[half nebula]

Hey is this chat being used?

[chilly ocean]

no

[half nebula]

Dang sorry hold on. Family stuff.

[shy bluff]

I have a quick question

In part b over here,

How do we know that each decomposition on the left is isomorphic to the one on the rigHT/

[next obsidian]

Shamrock it’s II.3.7

But I don’t think it’s quite commutative algebra yet

I don’t see how to use the generic finiteness

[latent anvil]

yeah I pulled it up

[next obsidian]

@shy bluff it doesn’t matter the order of the product on the left right?

[shy bluff]

I don't know

[next obsidian]

Well it doesn’t

G x H is iso to H x G

[shy bluff]

Like they went and wrote it as 4 seperate disticnct abelian groups?

[next obsidian]

Is all I mean

[shy bluff]

Oh no I mean like

[next obsidian]

Ohh sorry I’m doing the first example

[shy bluff]

They have like $Z_4 \times Z_9 \times Z_5 = Z_180$

[cloud walrus]

Liria ^(;,;)^:

[next obsidian]

Yeah that’s because of coprimality

Z/nZ x Z/mZ is iso to Z/mnZ when m and n are coprime

You can do it by working with elements

I can outline a proof really quick if you’d like

[shy bluff]

Like how do you know that $Z_2 \times Z_2 \times Z_9 \times Z_5 \cong Z_{90} \times Z_2$ rather than $Z_{60} \times Z_3$?

[cloud walrus]

Liria ^(;,;)^:

[next obsidian]

You don’t have a 3 in any of the stuff on the left

Z_60 x Z_3 has an element of order 3

The left doesn’t

Err

That isn’t quite right Z9 has an element of order 3

What’s the easiest way to see this.

Z_60 decomposes like Z_5 x Z_3 x Z_4 by coprimality stuff

[shy bluff]

Oh I see

[next obsidian]

And that isn’t iso to Z_2 x Z_2 x Z_9 x Z_5

[shy bluff]

I see

[next obsidian]

I recommend you try and go back over the classification of finite abelian groups

I don’t think I’m doing a great job communicating it haha

[shy bluff]

No I tihnk I understand

[next obsidian]

It basically comes down to coprimality stuff

[shy bluff]

So Z_4 x Z_9 x Z_5 = Z_180 because all 3 of them are coprime

[next obsidian]

And the fact that Z/nZ x Z/nZ is never Z/n^2Z (unless n = 1 lol)

Yes

[shy bluff]

So you basically just multiply all the coprime ones together

Ok I see

[next obsidian]

Yeah

And G x H is iso to H x G

[shy bluff]

Yea

Tha tbit's fine

[next obsidian]

So you can group them up into “blocks” as you see fit

[shy bluff]

Twas just the coprime bit that I did'nt get

[next obsidian]

The like first one is an example where you have to do that

[shy bluff]

The explanations in this course are uh aimed at a higgher level than my understanding :p

[next obsidian]

Ahhh sure sure. Try and see if you can’t prove it, you can just find an element of the correct order to imply Z/nZ x Z/mZ is iso to Z/mnZ when m and n are coprime

As in, you can find an element of order mn

And since the group has size mn it must be cyclic

Ping me if you have more group theory questions, I really like finite groups

[shy bluff]

o will do

I am having trouble with my group theory homework

[olive mirage]

There are canonical forms for finitely generated abelien groups that let you check for isomorphism in the obvious way, but I don't know if your text/professor went over them

[next obsidian]

Sham, do you have any ideas about the Hartshorne thing btw?

[latent anvil]

okay so first I want to understand why/how K(X) extends K(Y). This would hold if the generic point of X pushes forward to the generic point of Y, but that's not true in general. However f is dense! Let η be the generic point of Y and ξ the generic point of X. Then cl(f(ξ)) contains f(cl(ξ)) by continuity. The closure of ξ is X by definition, so cl(f(ξ)) contains f(X), and it's closed so it's all of Y by density. Thus f(ξ) = η by uniqueness of generic points

[next obsidian]

It does

[latent anvil]

I have gotten to the point where I understand the hint lol

[next obsidian]

Yeah