groups-rings-fields

[next obsidian]

Regular means that ab = ac => b = c

[hidden haven]

is this a ligma joke

[next obsidian]

And same with a on the right

[hidden haven]

ok nice

[next obsidian]

Magma is a set with a binary operation

Associative is… well you know

[hidden haven]

ye that one's fine

[next obsidian]

Finite means finite as a set, not something weird

[hidden haven]

I hate the finite module thing so much

[next obsidian]

Eh

I almost never have to deal with a finite as a set module so it’s whatever

[hidden haven]

Finite CW complex I am good with tho

[next obsidian]

It also has weird implication for your ring

[sharp sonnet]

left cancellative magma

[next obsidian]

Like it has to have a finite as a set ring quotient

[hidden haven]

What's weird about that? That is part of the definition of a ring

Right next to all rings are commutative

[foggy merlin]

what are your guyses strategies to keep track of lots of different steps and implications during a longer proof. I'm currently trying to read through a proof of Krull's principal ideal theorem, more precisely the height part of the theorem (which for some reason is a lot longer than the one given on wikipedia)

[terse crystal]

I guess you just read it… no short path

But the version of proof I read is only one page I don’t see that how that’s a long proof anyway 😂

[foggy merlin]

idk how my prof turned it into a long proof either

[winter solstice]

Probably the same way I turned the proof of n^2 even iff n is even into a whole page a long time ago.

[hidden haven]

Break the proof into steps and then try to explain it to someone in your head

[rustic crown]

(who's that someone >.<)

[hidden haven]

[lethal dune]

moldi got many

[hidden haven]

Many imaginary students

[lethal dune]

[foggy merlin]

I think the breaking into steps part was kind of exactly what made my profs proof long lol.

[delicate orchid]

Better to go into too much detail than not enough

[hidden haven]

Making the proof modular has a lot of advantages, but yeah if the proof is already short then it will make it longer

[next obsidian]

TBF the Principal Ideal Theorem is just a rally long proof. There's a lot of stuff that goes into it

[pastel cliff]

what's an example of a non-commutative algebra

[next obsidian]

matrices over a field

[pastel cliff]

wdym over a field

[chilly ocean]

Chmatrices

[pastel cliff]

still very new to this lol

[tribal moss]

Just n×n matrices with real entries, then.

[pastel cliff]

ohhh that makes more sense

im assuming you mean matrix multiplication then right

[tribal moss]

Yes, with addition of matrices.

[pastel cliff]

but addition of matrices is commutative

am i doing a dumb

[hidden haven]

Non commutative algebra just means that multiplication is non commutative

[pastel cliff]

oh

[hidden haven]

If addition is not commutative in any structure you don't even call it an addition 🙈

[pastel cliff]

im studying from lecture notes rn, that was written no where lol

[pastel cliff]

probably why i was having a hard time coming up with non-commutative addition lol

[pastel cliff]

for context im not being terribly dumb, the lecture is next week, prof just uploaded notes early

so he'll probably actually say that in the lecture lol

[hidden haven]

Why are you studying it a week early

[pastel cliff]

because.

[hidden haven]

diligentNitezba

[pastel cliff]

idk bc i can lol

im gonna have an ~interesting~ semester ahead

five classes, one of them is independent study, so any headstart helps

also bc he dropped the hw along the notes

[hidden haven]

[pastel cliff]

is being closed under addition implied?

[hidden haven]

What being closed under addition

Wait algebra (S,*)

[pastel cliff]

the algebra

[hidden haven]

What kind of algebra is being talked about here

[foggy merlin]

algebruh

[hidden haven]

Can you send definition

[hidden haven]

Bruh moment

This is algebra in the algebraic structure sense

[pastel cliff]

something being closed under multiplication but not addition feels weird so im just gonna assume it's for definition's sake

[hidden haven]

Though I guess the earlier answers still make sense

You don't even have an addition in the first place

So your question makes no sense

[pastel cliff]

oh

that's so cool

that you can just define an algebra and say fuck you, you no add

[hidden haven]

1984 algebras

[delicate orchid]

Formal sumz

[pastel cliff]

[chilly ocean]

Does anyone know what this means? If I could get an example it will help greatly

[hidden haven]

Try to apply the definition to find the order of every element in Z/6Z

[delicate orchid]

^

[hidden haven]

[delicate orchid]

That’s a good one

I was going to use -1^2 = 1 as an example

[hidden haven]

Also good

[chilly ocean]

Z/6Z?

with multiplication?

[hidden haven]

Addition

x^n should then be read as nx

[chilly ocean]

what would the elements in the set look like

yeah

{0,1,2,3,4,5} ?

[hidden haven]

Yep

[hidden haven]

And 1 as 0

[chilly ocean]

so 0 isnt in the set

[delicate orchid]

No 0 is in the set, it’s the additive identity

[chilly ocean]

orders:
1: 6
2: 3
3: 2
4: 3
5: 6

right?

[delicate orchid]

Looks right to me

[chilly ocean]

i know how to find these

[chilly ocean]

but what does the last sentence of this means

these are the orders of indvidiual elements of the set

how is it "the same as..."

[delicate orchid]

Consider the subgroup of Z/6Z generated by <2>

You get a subgroup containing every multiple of 2 mod 6

But you’ve just shown that 3 times 2 = 0 times 2 so there’s only 3 elements in that subgroup, which is the order of 2

[chilly ocean]

<2> looks like {e,2,4} right?

[delicate orchid]

Yes

[chilly ocean]

woah

so the order of the group

is 3

adn the order of the indvidual element is 3 too

[delicate orchid]

Yup

The way you’d define a general cyclic subgroup of order n from the free group is <a such that a^n = 1>

The connection there is clear

[hidden haven]

Also 0: 1

[chilly ocean]

so 0 is part of the set

1 isnt

[hidden haven]

Yes

[delicate orchid]

Yes a group needs the identity

[hidden haven]

No

1 is too

[delicate orchid]

No 1 is as well

[chilly ocean]

fuck

this class is so aids man

[delicate orchid]

It’s really not that bad

[chilly ocean]

[delicate orchid]

Why wouldn’t 1 be in the set

[hidden haven]

This is the set

Under addition that "loops around"

[delicate orchid]

I think you saying 1 is 0 really threw them off moldi ngl

[hidden haven]

It is the definition of Z/6Z

[hidden haven]

oof perhaps

[chilly ocean]

i get it

[pastel cliff]

isnt this modular stuff

[chilly ocean]

0 is the identity for additive right

[delicate orchid]

yeah

[hidden haven]

Yes

[chilly ocean]

so the identiy to the first power get u the identity

i get it

[hidden haven]

Yes

[pastel cliff]

an operation in an algebra can be anything right

[hidden haven]

Yes

[chilly ocean]

algebruh

[delicate orchid]

well aktually

[hidden haven]

It is one of the definitions of Z/6Z

[delicate orchid]

[hidden haven]

Here have your other definition

[delicate orchid]

[pastel cliff]

set of all humans is closed under the operation of sex

[delicate orchid]

sex literally makes new humans how can it be closed

[hidden haven]

That doesn't make sense

[pastel cliff]

aw

sad

[delicate orchid]

I'm going to cry

[hidden haven]

What you said would mean

That sex is a fixed arity operator

And also

[pastel cliff]

all humans already exist

[hidden haven]

You can sex literally any given pair of people

[pastel cliff]

[gritty sparrow]

Bit of a stupid question, but can you find such a H and a map which is zero on 6Z but doesn’t reference Z/6Z?

[pastel cliff]

tfw humans arent an algebra

[hidden haven]

Bro 0 map is zero on 6Z

[delicate orchid]

x -> x mod 3

x -> x mod 2

god I love being annoying

[gritty sparrow]

Ker=6Z is what i mean

[pastel cliff]

algebra feels like such a meme to study for some reason

[hidden haven]

I was smartasser than you before you smartassed

[delicate orchid]

then no if you want a homomorphism

[pastel cliff]

precisely because people can be annoying

[delicate orchid]

by the first iso theorem

[hidden haven]

Consider the symmetries of a hexagon

[gritty sparrow]

Not really, Z/6Z can be a subring, that will work just fine for me

[delicate orchid]

ohhhhh you want RINGS

[gritty sparrow]

Sorry group

[delicate orchid]

ok but you want ker = 6Z right?

[hidden haven]

Poor guy was so hyped

[gritty sparrow]

Yeah

I want ker=6Z

[delicate orchid]

you can't handle the ker = 6Z

[hidden haven]

So done

[delicate orchid]

yeah moldi's kinda right

[hidden haven]

[delicate orchid]

map all the rotations to 0 and keep reflections

i.e. the image is C_2

[pastel cliff]

wtf is that first lecture gonna be about lmao

oh is that just closure

[gritty sparrow]

Point is, H can be some group that contains Z/6Z, and still be defined without mentioning it

[hidden haven]

I was thinking map 1 ∈ ℤ to the 60° rotation

[hidden haven]

D_6 is defined without reference to Z/6Z

[delicate orchid]

*D_12

[gritty sparrow]

Yep exactly what i wanted

[hidden haven]

a+b=c

It's a good number theory problem

It's like fermat's last theorem except there is a solution

Pretty cool

[delicate orchid]

I could find a solution

[pastel cliff]

no way

get this man a nobel prize

[delicate orchid]

nobel prize
maths

[pastel cliff]

that's the joke

[delicate orchid]

what joke

[pastel cliff]

YOUR MOM

[delicate orchid]

take it back... now

NOW

[pastel cliff]

https://tenor.com/view/cope-gif-23689217

[delicate orchid]

#abstract-chill

anyway what were we talking about

something to do with a^n = 1

[hidden haven]

a+b=c

[delicate orchid]

we looking for integer solutions or something

[hidden haven]

What's this class about btw

[pastel cliff]

it's just called "algebra"

[hidden haven]

[delicate orchid]

run

run as fast as you can

[pastel cliff]

full syllabus

class starts monday

[delicate orchid]

wtaf is this

[hidden haven]

Group theory + sneak peek ring theory

[delicate orchid]

ok around week 6 it starts making some semblance of sense

god reading this makes me nostalgic

I'm so OLD moldi

I am withering away

[pastel cliff]

i love my math department

[delicate orchid]

yeah that looks alright

[pastel cliff]

you are 12

look at your pfp

clearly 12

[delicate orchid]

mfw

,av

[cloud walrus]

Wew Lads Tbh#0864's Avatar

Click here to view the image.

[delicate orchid]

anyway!

[hidden haven]

Wew did this stuff in grade school

[delicate orchid]

course looks good

first few weeks are bizzare

but the actual meat is good

[hidden haven]

Doesn't seem too bad to me

Except maybe boring initially

[delicate orchid]

they're introducing polynomial algberas before groups

[pastel cliff]

yummy

book is A First Course in Abstract Algebra with Applications, 3rd Edition by Joseph J. Rotman

[hidden haven]

But not as rings though

[delicate orchid]

idk about you but I didn't see my first algebra algebra until my 4th year research

[hidden haven]

Or as algebras over rings

[delicate orchid]

yes that's what makes me concerned

[hidden haven]

Here algebra just means algebraic structure

[delicate orchid]

ohhhhhhh

[hidden haven]

Which is set with operations

[delicate orchid]

right ok

bit less daunting

[pastel cliff]

does it mean something else???

[hidden haven]

Ye

[delicate orchid]

yes

[pastel cliff]

[wicked zephyr]

2+3=5 give me my fields

[delicate orchid]

imagine you have a vector space but you can also multiply the vectors

that's an algebra to my understanding

[hidden haven]

omg

[delicate orchid]

I've only seen group-rings though and weird FG-modules so there ARE more wacky ones out there, like equipping R^3 with the cross product that's a classic

[hidden haven]

An R-algebra is a ring homomorphism R → S

With image in the center of S

[pastel cliff]

god these WORDS

[delicate orchid]

I simply do not believe it

where's my rep theory book

[pastel cliff]

https://www.youtube.com/watch?v=JDGPnLfH1Io

really starting to feel this one

[hidden haven]

What you said earlier can be generalised to modules

An R-algebra is an R-module with a compatible ring multiplication

But since it itself is a ring S

[pastel cliff]

ok i have a few days to self study some stuff before this class starts

[delicate orchid]

oh I see now

[hidden haven]

You can say that it is a ring S with a compatible R module structure

[pastel cliff]

anything yall recommend i look at ahead of time

[delicate orchid]

ok I'll let it slide... for now...

[hidden haven]

And then r•1 where • is scalar multiplication

Is the image of r ∈ R under this structure homomorphism

[delicate orchid]

basically just shove a ring over a ring rather than a group over a ring right?

[pastel cliff]

i know basic definitions but dont understand them annd their significance yet but that's just a matter of time

[delicate orchid]

like which ones

[hidden haven]

So this homomorphism R → S is precisely all the data you need

[pastel cliff]

but yall scaring me lol

[delicate orchid]

we're talking about fairly advanced nonsense

[pastel cliff]

ring, group, "algebra", field, isomorphism

[delicate orchid]

emphasis on nonsense

you will see all of those but algebra

[hidden haven]

And another interesting definition you'll see in terms of tensor products

[delicate orchid]

I'm stopping you there

[hidden haven]

[delicate orchid]

I need to go get my tensor time hat on first

[pastel cliff]

nerd

[hidden haven]

Need a generic universal property sticker

[delicate orchid]

you just posted one

[hidden haven]

That's just the first isomorphism theorem

[delicate orchid]

it's tensor time chat

[pastel cliff]

pls 👉 👈

oh not that

asking for things to self study

or just prep in general for this inevitable shitstorm

[pastel cliff]

.

[delicate orchid]

look bro, you've just walked into a fluid dynamics lecture before taking calc 1 - don't let us spook you

[hidden haven]

Generic one would be like

x → Fu u
\ | :
↓ ↓ ↓
Fy y

[delicate orchid]

moldi you did not just type that

please

[hidden haven]

I regret it

[delicate orchid]

I regret it as well

Moldi I want to go back and struggle to understand quotient groups again it was so simple (get it chat) back then

[hidden haven]

Just stare at the universal property enough

And you'll start struggling again

Better yet see the Yoneda lemma proof of the isomorphism theorem (G/H)/(N/H) ≈ G/N

[pastel cliff]

[hidden haven]

Then you'll struggle with the other proof

[delicate orchid]

I still have fuckin no clue what the universal property of the tensor product actually means
I know it has something to do with guaranteeing some bijective map from Hom(M, N) to Bil(M, N)

[pastel cliff]

im gonna go play minecraft now

[hidden haven]

Based

[pastel cliff]

too much lingo for a day

im gonna have so much fun with this class

literally bursting at the seams with joy

[wicked zephyr]

Yoneda Lemmas is just spicy Cayley's Theorem, change my mind

[hidden haven]

Bijection between Bil(M ⊗ P, N) and Hom(M × P, N)

[delicate orchid]

that's the one

[wicked zephyr]

i'm literally playing right now, hi brother

[sharp sonnet]

you understand linear maps, but dont understand multilinear maps
wouldnt it be cool to transfer your understanding of linear maps to multinear maps

[hidden haven]

It turns bilinear maps into linear maps, that is all

[delicate orchid]

oh yeah I get the motivation

[next obsidian]

except Yoneda's lemma is useful

[wicked zephyr]

Univeesal properties are the best

[delicate orchid]

I just don't know what a "universal property" actually is

[next obsidian]

pls name me 3 cases you actually have used Cayley's theorem

[wicked zephyr]

Is a terminal object in a suitable category

[next obsidian]

That's why I find this analogy kind of weak like

[hidden haven]

Spicy = more useful, chmonkey

[delicate orchid]

[next obsidian]

Idk I don't even view it as a generalization of it or something

[delicate orchid]

this might as well be in Slovakian

[next obsidian]

it's just so much better than Cayley's

it's like saying a ring is a spicy Z

I dunno

[wicked zephyr]

A group is a groupoid with one object

[next obsidian]

I know this lol

I'm not saying it won't even prove it

[wicked zephyr]

yeah I mean

[hidden haven]

Actually true

[wicked zephyr]

it was kind of a joke

[hidden haven]

That's the technical term

Spicy ℤ

[next obsidian]

I'm just saying people say "oh have you seen Cayley's theorem? Yoneda is like that but souped up"

except now you set it up to be some theorem that you learn and never use

[weak oriole]

At this point I won't be surprised if there's an actual category theory formulation of this "spicy"

[wicked zephyr]

It is interesting

[next obsidian]

or is there to just provide some sort of like philosophical meaning like "oh woah everything embeds into a functor category woooooooooooooooooooaaaaaaa"

[delicate orchid]

#category-theory PLEASE

[wicked zephyr]

It is philosophical

[next obsidian]

Someone drank too much of the cat theory kool-aid

[hidden haven]

Just today I was reading about integrals and fubini's theorem in Category theory

[next obsidian]

I don't really see the philosophical stuff in this

[wicked zephyr]

Was it interesting?

[hidden haven]

Yes

[wicked zephyr]

just motivation, it is useful when explaining, stuff like that

[hidden haven]

And I don't mean integrals done categorically or something

[weak oriole]

May god have mercy on u bro

How do u even do that with just objects and arrows

[next obsidian]

was it a coend?

[hidden haven]

I mean literally integrating a functor

Over a categroy

[hidden haven]

Yes

[next obsidian]

knew it

[wicked zephyr]

Yeah, once I read about coend calculus

[hidden haven]

Ends and coends lmao

[wicked zephyr]

but dont understand them

[next obsidian]

I almost learned about coends

to try to understand how a tensor product is a coproduct

[weak oriole]

Oh you weren't joking

[wicked zephyr]

I think yoneda lemma can be stated in terms of coends and stuff like that

[hidden haven]

Oh ye I just learned about that today

[wicked zephyr]

share it! sounds interesting

[next obsidian]

I mean so like

the underlying abelian group

ohw ait

No I know that it's a coproduct

fml

[hidden haven]

So once you have tensor of abelian groups

[next obsidian]

I mean a cokernel

[hidden haven]

You can write tensor product of any 2 R-modules over R

As a coend of their tensor over Z

And this case be used to define a more general notion of tensors over stuff in general monoidal categories

Have to read further to know why this is useful though

Mac lane just says yeah this is a thing you can do

ok

[next obsidian]

https://math.stackexchange.com/questions/1037864/tensor-product-and-commutation-category-theoretical-argument?rq=1

I was trying to figure out how to understand this

The thing I didn't understand was like since we're doing tensor products over different rings in which category this is a coequalizer

I think the answer is as abelian groups

and then like, I think the module structure comes in abstractly because an R-module structure is just a map R (x)_Z M -> M

and I think you can naturally induce this via the direct limits or some shit, idfk

[hidden haven]

This seems different from the coend thing though

Too sleepy to think

[next obsidian]

I read parts of "tensor categorical foundations for algebraic geometry" or something

well like, to show it is a coequalizer when I was looking into it involves writing it as a coend

[hidden haven]

But then again the coend thing comes down to coequalising the obvious tensors

[next obsidian]

or something, idk I went hard on this last year then gave up because I decided this was dumb as fuck

[hidden haven]

Ye right

[next obsidian]

and I had better things to do

[hidden haven]

Lmao

[next obsidian]

Like after reading someone's PhD thesis I decided I was done with this bs

since I can just write a direct proof of the fact

and it's even in Bourbaki lol

[hidden haven]

Having read >10 chapters of mac lane, I haven't even used anything past the first 5 outside of cat theory itself

[next obsidian]

yeah

[hidden haven]

I'll be using the 6th one in homological algebra though 🙈

[next obsidian]

I read a bit of it to figure out the diagonal being cofinal thing

This actually mattered for me before

since I was doing weird direct limits over doubly indexed stuff

[hidden haven]

Right

[next obsidian]

I still don't really understand how that stuff works

I just try to avoid situations that this becomes important

lol

[hidden haven]

It's just diagram chasing

[next obsidian]

Yeah I mean like I'm confused if I'm doing a direct limit followed by another

or if I'm like doing a single direct limit over I x J

or something

so I just try to avoid having to ever think about what's going on lol

by coming up with different arguments

[hidden haven]

I think those should be the same lmao

[next obsidian]

¯_(ツ)_/¯

[hidden haven]

Assuming all existence

[next obsidian]

It popped up when I was reading Serre's local algebra

to define the like, completed tensor product

piece of shit

[hidden haven]

[gritty sparrow]

were you reading martin brandenburg's thesis?

[next obsidian]

I did for a second

then went "okay why am I doing this"

[delicate orchid]

would you read my PhD thesis Prof. Chmonkey

[gritty sparrow]

I was planning on reading it sometime, it seemed pretty cool (or atleast the idea was cool)

[next obsidian]

If you make it on commutative algebra or AG, maybe

yeah idk, I don't really at the moment care about these big categorical generalizations

I just cared about how to frame things in a more categorical way for the specific thing I cared about at that time, and I found the thesis a source that explained the things I needed

[gritty sparrow]

I see

[waxen hedge]

Everything is a presheaf

[chilly ocean]

would someone be able to enlighten me as to what is meant by a transformation-group groupoid

[chilly radish]

I'm probably missing something, but I don't see why the hypothesis of IBN is needed for 1, and also maybe for 2, since R^n is always finitely.generated by definition while R^infty is not

[gritty sparrow]

For 1 i'm pretty sure it isn't necessary but for 2 it probably is. The point being that R^infty is not finitely generated by the particular basis that you used to define it but there is a chance that there is some other finite basis for R^infty, without IBN I don't see how you can immediately conclude that such a basis doesn't exist.

[chilly radish]

I see, that's fair. I'm still not sure how you'd reach a contradiction even with IBN, if such a basis did exist, then you'd want to construct some other finite basis with a different cardinality for a contradiction, but i'm not sure how you'd do that since you don't have much to go off of

[next obsidian]

Wait what

[gritty sparrow]

Direct sum with R on both sides and use the isomorphism in 1

[next obsidian]

Produce an infinite basis for R^infinity

Produce a size n basis for R^n

Then IBN handles the rest

[chilly radish]

Ohhhh right

[chilly radish]

IBN only concerns finitely generated modules

[gritty sparrow]

IBN is for two finite direct sums

[next obsidian]

The fuq

I always used it as R^omega is defined just by |omega|

[chilly radish]

Well this exercise shows that you can conclude that from IBN

[next obsidian]

Oh

Lol

[chilly radish]

part 3 specificlaly

[next obsidian]

I proved this for commutative rings

Then just took IBN to be that property

Lol

[chilly radish]

Commutative rings always have IBN so it doesn't matter

[next obsidian]

Yeah you have to prove that is my point

And I proved it directly for all cardinalities

IIRC

Or maybe at some point I just extended the finite case without knowing I can

[chilly radish]

For some reason I didn't think of using 1 at all, but it's immediate from that

[chilly radish]

Assuming choice

Or not?

The proof I saw used zorn to assert that existence of a maximal ideal but looking it up it seems to hold in constructive settings too

[next obsidian]

I did it using maximal ideals

And you reduce to vector spaces

[gritty sparrow]

You can do it via constructing a theory of determinants/adjugate matrices for free modules over commutative rings

[next obsidian]

[chilly radish]

Yea with commutative rings you can do this directly to conclude for cardinalities without using IBN

[lavish nexus]

is the smallest transcendental extension of F always isomorphic to F(x)?

it contains an injective map from F[x] so if we extend the map to be from F(x) it should be still injective right?

[gritty sparrow]

Yes the map is still injective (because any map from a field is injective)

[prisma shuttle]

yes this should be true

[lavish nexus]

right there are no nontrivial proper ideals in a field

thank you!

[lavish nexus]

I don't understand the circled part

v_i's are cosets of E

how can I sum them and get an element in K

nvm this is not a basis for the quotient K/E but for K as a v.s. over E

[oblique river]

I think you should review the notation

K/E is not the quotient of K by E

oh sorry lmao you figured that out on your own

yeah the notation is somewhat odd, but it's standard when talking about field extensions to write K/E to denote that K is a field extension of E

the reason is that in this kind of field theory, the "fundamental object" you're studying is not a field by itself, but the extension of fields, as a pair

and so we want to keep that in the notation

and "K/E" emphasizes that

[prisma shuttle]

hey guys can someone explain how to do this

how can we find all the ideals of $\mathbb{Z}/12\mathbb{Z}$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

I know the definition of an ideal but i am not sure how to find all of them

[gritty sparrow]

well if you know all the ideals of Z, you can see that ideals of this ring are in correspondence with ideals of Z containing 12Z. If you haven't seen that before you can find all the ideals by hand by first noticing that all ideals are just principal ideals and then seeing what ideal each element generates by hand

[prisma shuttle]

oh i see

so is it just $\mathbb{Z}/n\mathbb{Z}$ which $n|12$?

[cloud walrus]

JustKeepRunning

[prisma shuttle]

thats the set of all ideals?

[gritty sparrow]

no

[prisma shuttle]

wait i look at the answer key and it says it is?

this is in d&f

[gritty sparrow]

it will be nZ/12Z where n divides 12

maybe d&f are talking about the corresponding quotient rings

[prisma shuttle]

oh shoot i misread sorry

[gritty sparrow]

np

[prisma shuttle]

wait why isn't it just nZ why is it nZ/12Z

because doesn't nZ contain 12Z?

and it also satisfies ideal definition?

[gritty sparrow]

nZ isn't even a subset of the ring

[lavish nexus]

If [E:F] = m and [E’:F] < inf, prove [EE’:E’] <= m

can I suppose every element in EE’ is a sum of a*b where a is in E and b is in E’?

And if so I feel like my argument is flimsy

Suppose E/F has a basis {e_1, … e_m}
Then writing a’s as linear combinations of the basis elements we have that everything in EE’ is a linear combination of them and we cannot have more than m linearly independent elements

Because I’m concerned about infinite sums

[terse crystal]

There is no infinite sum…

[lavish nexus]

umm why

[terse crystal]

You didn’t define infinite sum…😂

[dreamy jewel]

If a, b are algebaric over F, and $F(a) \cap F(b) = F$, does this imply that $[F(a, b): F] = [F(a): F][F(b): F]$?

[cloud walrus]

CelesteCrow

[terse crystal]

EF is defined to be E(F) , extension of E containing elements of F, so any element in EF has the form of finite sum of ab where a is from E and b is from F

[lavish nexus]

<=

ok I guess it’s the same thing as the smallest field containing E and F
which is the definition I got

[terse crystal]

Yeah, equivalent

[lavish nexus]

Then my argument works?👀

[terse crystal]

Yeah

[lavish nexus]

excellent ty

[dreamy jewel]

Oh shoot thanks i just came up with a counter example

[terse crystal]

What is the counterexample?

[dreamy jewel]

well i wanted to prove or disprove it for a general n algebaric number, so that their pairwise intersection is F, and i thought about i, \sqrt(2), and i\sqrt(2) for Q

[terse crystal]

? [Q(sqrt(2),i):Q]=4 I think

Since it equals Q(sqrt(2)+i)

[dreamy jewel]

well so lemme rephrase my question

Suppose we have a_1, ..., a_n algebraic over F, if
[F(a_i) \cap F(a_j) = F]
for i \neq j, then does this mean that
[F(a_1, ..., a_n): F] = product of [F(a_i): F]

That's the counter example i was talking about ^

[terse crystal]

So why your counterexample is a counterexample?

That’s 4=2*2

[dreamy jewel]

okay i might be tripping lol

well that's 4 \neq 2^3

[frank lake]

is the relationship that if we had a group S, this would satisfy as the subgroup of S containing all orders? And we can prove this by proving that this subgroup is closed and has an inverse

idfk

I'm dumb

because like |a| -> a^n = e, and |b| -> b^n = e

[dreamy jewel]

Ahh if gcd(|a|, |b|) = 1, then |ab| = |a||b|?

[frank lake]

so |ab| -> (ab)^n = e

this is order

not absolute value

[dreamy jewel]

yeah

[frank lake]

the order of an element in a group

[dreamy jewel]

i know

[frank lake]

if |a| and |b| are relatively prime you say?

[dreamy jewel]

yeah

If the order of a and b are relatively prime, and i think the group has to be abelian

[frank lake]

yeah it doesn't specify if the group is abelian or not

idk

[dreamy jewel]

lollll

do you have the full question?

[frank lake]

it just asks to find a relationship

this is the whole question

[dreamy jewel]

bruh

[frank lake]

OHHHH

NVM

I FOUND IT

I was reading the wrong part

[frank lake]

I've never heard of that property until now

[dreamy jewel]

yeah it's a cool exercise if you wanna try it (it only works in abelian groups tho)

Also I think for 13 you can just think of the dihedral group

[frank lake]

dihedral groups 😰

that's the case for dihedral groups tho?

a^n = e when n = 2?

[dreamy jewel]

well so in a dihedral groups, 2 flips = 1 rotation right

[frank lake]

I forgot what the identity of a dihedral group is

[frank lake]

yeah

[terse crystal]

Where is 3?

[dreamy jewel]

So I have a_1, a_2, a_3 where a_1 = i , a_2 = \sqrt(2), a_3 = i\sqrt(2)

[lavish nexus]

D_2n groups

You have flips and rotations

and rotation*flip is another flip

[frank lake]

so the identity element is the rotation feature then

[dreamy jewel]

yeah the identity of dihedral group is to literally rotate by 0 degrees

[terse crystal]

Oh I see

[frank lake]

I see

[dreamy jewel]

which is the same as rotating by 360 degrees

[frank lake]

yeah

now how do I do a. b. and c.

like we have a group

but what do we do with it

[lavish nexus]

you have r and s

r is rotation s is flip

rs is another flip

try for example D_8

[frank lake]

so squares

[lavish nexus]

now what is the order of flips?

[dreamy jewel]

well if you pick 2 different flips, you can also compose them to get a just a standard rotation

[lavish nexus]

yes rs*s=r

[frank lake]

2

[lavish nexus]

ok so s^2= 1

[frank lake]

since it takes 2 flips for 1 rotation

[frank lake]

mhm

[lavish nexus]

you see what you want for a and b now?

[frank lake]

so |ab| = |s^2*s^2| right?

or just |ss|

[lavish nexus]

you want two elements both are order 2

so both are flips

[frank lake]

would this be right?

[lavish nexus]

but when multiplied gives something of a different order

[frank lake]

oh

so |ab| is not equal to |a||b| for non-abelian groups

[lavish nexus]

that gives you order 1
you want 345

no it’s not

|ab| can be whatever you want

[frank lake]

|ab| = 3 -> (ab)^3 = e

[lavish nexus]

yes so what should n be

[frank lake]

n is 3

[lavish nexus]

yes

[frank lake]

is there anything more I can do with this?

[lavish nexus]

so ab is a rotation

[frank lake]

ohhh

WAIt

SO A AND B WOULD BE FLIPS

right?

[lavish nexus]

yes

[frank lake]

😄

TYSM

wait

so |a| and |b| would be flips or just a and b

[lavish nexus]

Umm the first two They are orders

[frank lake]

yeah so just a and b

[lavish nexus]

yeah those are elements

[frank lake]

so since ab is a rotation and a and b separately are flips, |ab| = 2, and |a| and |b| are confusing for me

what is |a| and |b|?

like a and b are flips, but you need to find how many times some element repeats to get 1 flip

[lavish nexus]

This is wrong

[frank lake]

what's wrong about it?

[lavish nexus]

why is the order of a rotation 2

[frank lake]

wait

[lavish nexus]

unless you’re in D_4 but that’s abelian

[frank lake]

oh wait I think I messed something up

wouldn't the order be 0?

[lavish nexus]

What is order

[frank lake]

the order of ab

since ab itself is a rotation

it doesn't even need to move to reach itself

[lavish nexus]

yeah it needs to be nonzero

the smallest nonzero integer

[frank lake]

?

why can't it be zero?

[lavish nexus]

because it’s the definition

otherwise the order of everything is 0

[frank lake]

I mean the rotation is the identity element, if it was a different element the order would be different

[lavish nexus]

Rotation is not id

[frank lake]

what does id mean?

[lavish nexus]

id is just id

identity

[frank lake]

oh

[lavish nexus]

means you do nothing

[frank lake]

ohh

that cleared up some stuff

so it would be 1

it takes 1 rotation to get to the identity element

so |ab| = 1

[lavish nexus]

no

[frank lake]

|ab| = 4? ;-;

[lavish nexus]

|ab| can be arbitrary

[frank lake]

doesn't it depend on the dihedral group you're dealing with?

[lavish nexus]

yes

[frank lake]

man

I hate dihedral groups

[lavish nexus]

I mean I can rotate by 180 degree

or I can rotate by 1 degree

or 0.0001 degree

or infinitesimal degree

[frank lake]

yeah

[lavish nexus]

In D_2n the order of rotation is n

[frank lake]

I mean

[lavish nexus]

because r in D_2n is rotation by 360/n degree

rotate n times you get id

[frank lake]

my problem is just to find specific groups such that |a| = |b| = 2, |ab| = 3, |ab| = 4, |ab| = 5

[lavish nexus]

Yes and I already told you what you can take to be a and b

[frank lake]

you used the dihedral group as an example

[lavish nexus]

here

[frank lake]

and since there is 2 flips in a rotation, the first property is satisfied

I have to find groups that satisfy all these properties listed

[lavish nexus]

I can find a group that simultaneously satisfies all three

[frank lake]

what group is it?

[lavish nexus]

D_120

Just taking n = lcm(3,4,5)

[frank lake]

;-;

mhm

[lavish nexus]

you can then find powers of rotation with order 3 4 or 5

[terse crystal]

Or <a,b,c,d,e,f:a^2=b^2=c^2=d^2=e^2=f^2=(ab)^3=(cd)^4=(ef)^5=1>😂

[frank lake]

🧠

[dreamy jewel]

that's some spicy group presentation right there

[frank lake]

it's alright

[dreamy jewel]

you should try a tietze transformation to simplify it down

[frank lake]

I'll do a different problem, but thanks for all the help

@lavish nexus @dreamy jewel

[dreamy jewel]

No problem, hope I was helpful

[frank lake]

what subgroup would that be of?

it has the subgroup brackets

<>

😂

[lavish nexus]

so in D_inf for any n positive there exists corresponding a and b

[dreamy jewel]

it's this thing called a group presentation, you usually should learn what a "free" group is first and it takes a while to get to that

[terse crystal]

The quotient group of free group generated by {a,b,…} over normal subgroup generated by {a^2,b^2,…}

[lavish nexus]

That’s generator relationships

[frank lake]

damn

[frank lake]

wow

[fickle brook]

oh there's a thing going on here isn't there

[dreamy jewel]

That's some High-Key way to do this when you learn more algebra

[frank lake]

I'm sooo fucking nooby at this stuff

[dreamy jewel]

it's okay, you will get there soon enough

everyone's been in your place before

[frank lake]

lemme do a simpler problem and I'm gonna come back to check

[dreamy jewel]

Let F(T) be the field of fractions, and consider the map T -> T^2, is there anyway to show this is a homomorphism without like doing a lot of messy algebra

[thorn delta]

what is T?

in the F(T)

[gritty sparrow]

well, here is a fun thing you can say: first construct the map F[T]->F(T) which sends T to T^2, such a map exists as that is pretty much the defining property of the polynomial ring (read: universal property). Then notice that all nonzero elements map to invertible elements (since they map to nonzero elements in F(T) all of which are invertible) hence by the defining property of the field of fractions (read: universal property) this factors as a map from F(T)->F(T)

[dreamy jewel]

sorry let me clarify F(T) is the field of rational functions (aka. the field fraction of F[x])

[thorn delta]

ah okay

[dreamy jewel]

Thank you so much!

[thorn delta]

I thought you were trying to say T was some ring, and F(T) was its field of fractions

[dreamy jewel]

So just making sure I am getting this correctly.

Actually I am not getting this correctly, my power level in category theory is way too low for this

[gritty sparrow]

what part do you want me to clarify

[chilly ocean]

How do I get good at this subject

[chilly ocean]

This is literally the hardest math course I’ve ever taken

[hidden haven]

It will get easier

probably

[chilly ocean]

Idk why

Like this shit is so abstract

I can’t wrap my head around it

[hidden haven]

Ye you get used to the abstraction

Just gotta accept the fact the fact that it is abstract and everything suddenly works out 🙏

[chilly ocean]

Alrighty man

Just came here to vent lol

[hidden haven]

Understandable

[dreamy jewel]

So if I am understanding this correctly, you are saying that by the Universal Property of Polynomial Rings, for any ring homomorphism between F and F(T), there exists a unique homomorphism \phi from F[T] to F(T) such that \phi(T) = T^2 (in fact you can send it to any element of F(T))

Then by the Universal Property of the Field of Fractions, for any injective ring homomorphism from F[T] to F(T) (which the map from T to T^2 is), there exists a unique field homomorphism \psi: F(T) to F(T) such that \psi \circ i = \phi, where i is the inclusion of F[T] into F(T)

[gritty sparrow]

yeah pretty much, except that the universal property of fraction fields talks about nonzero elements mapping to invertible elements, not non zero elements mapping to non zero elements (in case of F(T) it doesn't matter since all non zero elements are invertible)

[hidden haven]

That depends

What upendra gave isn't actually a universal property

formally

This is

if you make sure that the codomain is a field

You are talking about a universal to the forgetful functor from Field to Dom_injective

[dreamy jewel]

okay there's just 1 more question (well i mean there's a ton of questions running through my mind rn)

How do I know that the \psi I found is like the original homomorphism I described that sends T to T^2 in F(T), for all I know \psi only agrees on T to T^2 for the isomorphic copy of F[T]

I might be making a very trivial mistake here

[hidden haven]

The original homomorphism you described was "identity on F, T maps to T^2"

[dreamy jewel]

yes

[hidden haven]

This does those 2 things

By construction

[dreamy jewel]

and it's unique

[hidden haven]

Yep

[dreamy jewel]

bc of universal property

[hidden haven]

yes

[dreamy jewel]

well okay i guess my problem is that

it sends T to T^2, but like what if I have like \psi(1/T)

[hidden haven]

You can figure out what it will be from the homomorphism properties

[dreamy jewel]

Like what does \psi do to elements outside of the isomorphic copy of F[T]

i just realized i asked a pretty dumb question

[hidden haven]

That's not a problem

Universal properties are magic so always good to ask dumb questions

[dreamy jewel]

So for like any 1/f(T), \psi(1/f(T)) has to be (1/f(T^2))

bc of the homomorphism property stuff

[hidden haven]

yep

[dreamy jewel]

okay

Thank you so much! @hidden haven @gritty sparrow

[gritty sparrow]

Np

[dreamy jewel]

How do I get good at Algebra

[hidden haven]

(I don't have any valuable input)

[dreamy fiber]

That’s 1st isomorphism

I can tell you that much? 😂

[hidden haven]

Universal properties

[dreamy jewel]

what exactly are universal properties

[hidden haven]

They are ways to define things without referring to their elements

[lethal dune]

oh boy here we go

[dreamy fiber]

(￣▽￣)

[hidden haven]

The point is, that most of the time (all of the time) you will be working with homomorphisms in and out of things rather than those things themselves

And in that case, it becomes a lot more convenient to have a definition that tells you exactly how those maps behave rather than giving a messy element level definition and proving this as a theorem

Or you could of course just treat the universal property as a theorem and use it

But there are philosophical benefits too

For example you could define the polynomial ring to have the underlying set of finite formal sums

[dreamy jewel]

So like you kind of get a more dynamic view of algebraic structures rather than the static interpretations?

[gritty sparrow]

Philosophical benefits: virginity protection

[hidden haven]

or you could define it to have the underlying set of infinite tuples of coefficients, only finitely many of which are non zero

But then you would want to call both of these the polynomial ring

So you would do a definition by elements, and then say anything that is isomorphic is also a polynomial ring

which is dumb

[dreamy jewel]

That's some direct sum flummery right there

[hidden haven]

[hidden haven]

Yeah you could say that

[dreamy jewel]

I feel like you had to stop yourself from talking about modules

which admittedly i know nothing about

[hidden haven]

not really

[dreamy jewel]

the infinite direct sum gives me unpleasant memories lol

[hidden haven]

It is slightly different since you also have a ring multiplication here

You can multiply tuples and this is not done coordinatewise

[lethal dune]

but ngl definitions through universal property is hard to comprehend sometimes

[dreamy jewel]

One might say part of algebra is that you relearn the same thing 10 different times

[hidden haven]

Only the first 3 times

[lethal dune]

[hidden haven]

Not with universal properties there is only 1 product

[dreamy jewel]

ah

[lethal dune]

I still don't understand tensor products through UP

[dreamy jewel]

please don't tell me there's a thing called universal product

[hidden haven]

Try to understand simpler universal properties first

[lethal dune]

[hidden haven]

There is a universal property of products

[dreamy jewel]

well tarnation me

[hidden haven]

and this universal property doesn't change whether you talk about rings, groups, modules, topologies, etc

Any category you could imagine

Products in that category are defined with the same universal property

Which is another advantage of universal properties

[dreamy jewel]

bc ... they are universal?

[hidden haven]

Products are not always Cartesian products, and in the long run it is harmful to think that product thing = cartesian product of underlying sets with thing structure

Because products are not that

[hidden haven]

lol

I guess

[hidden haven]

This is easier to see with coproducts

[dreamy jewel]

That's not even a pun, that's just reality

[hidden haven]

The coproduct of 2 sets is their disjoint union. The coproduct of 2 modules is their direct sum. Of 2 rings, their tensor product. Of 2 groups, their free product

None of these look like any other

At all

But they behave very similarly

Because they have the same universal property

[dreamy jewel]

wait hold on so does this universal property here have a version where you amalgamate something?

[hidden haven]

Coproduct of A and B is an object C along with maps i_A: A → C, i_B: B → C, such that whenever you have another object D with maps j_A and j_B, there is a unique f: C → D such that j_A = f i_A and j_B = f i_B (what we formally call "everything commutes ")

The i_A and i_B are called "inclusions" because that's what they often are in the above examples

You can try to think of why coprod of 2 sets is their disjoint union with this definition

[dreamy jewel]

So like

So just to be clearly,

Okay i admit ican't wrap my head around this

[hidden haven]

Ye usually it takes a while to see why this property is satisfied by the usual disjoint union

Have to work through the proof once in your life

[dreamy jewel]

okay i will take your word for it

[hidden haven]

Nice

And then there is a general fact about universal properties

[lethal dune]

don't, moldi is lying, verify it yrself

[hidden haven]

That the objects satisfying the universal property

are always unique up to isomorphism

(and a nice isomorphism at that)

This is something that can be checked completely formally once you know that formal definition of a universal property

And then you can start proving facts like this about universal properties in general

[dreamy jewel]

Okay so real quick, on a scale of 1 to 10

I am taking Commutative Algebra next semester, how fucked am I, with 1 being absolutely fucked

[hidden haven]

uh

damn that is pessimistic

[dreamy fiber]

I would give a solid 6.9

[dreamy jewel]

I have realized i have retained absolutely nothing from my last semester of algebra

after this enlightening conversation here

[hidden haven]

You are welcome 😌

[dreamy jewel]

Thank you, I am grateful

[rustic crown]

i had a call with uppu, and i realized i started to forget raq already >.<

[hidden haven]

You probably don't need to know all the universal property stuff beforehand though

[hidden haven]

death

[dreamy jewel]

We touched like a very little smidget of 🙀 theory at the end of last semester

[hidden haven]

🙀 theory turns into 😺 theory very quickly

Don't worry too much about it

[dreamy fiber]

Cats are the living incarnation of evil

They are cursed by nature

[dreamy jewel]

Thanks, I hope I know my flummery before going into CA then

[hidden haven]

Do you play town of salem

[dreamy jewel]

lol yes

[hidden haven]

Seeing how you are censoring shit to flummery

[dreamy jewel]

I have an atuomatic filter on my cursed words in an academic discord server

[hidden haven]

Nice

[dreamy jewel]

Tarnation Me

[hidden haven]

doodle

[dreamy jewel]

Thank you so much for your patience

[south temple]

Hey guys, I'm trying to prove the following but it just feels like I'm hitting a brick wall:
Let $\nu : R \to R/I$ be the natural homomorphism. If $K$ is an ideal in $R$, then $K = \nu^{-1}(\nu(K))$.

[cloud walrus]

vov&sons

[hidden haven]

That is not true

[south temple]

oh it's not?

[hidden haven]

You need K to contain I for this

[south temple]

how come?

[rustic crown]

try K = 0

[hidden haven]

The correct equation is ||K+I = nu inverse (nu(K))||

[south temple]

ah right

but how does having K contain I make this easier

or like

why is that key

to my equation

[hidden haven]

If K doesn't contain I, then the preimage of its image contains I (because of what det said) so certainly can't be equal

If K contains I, then this is true because that's what you have to figure out

[south temple]

yeah but I've tried doing this with K containing I and it still felt like I was missing something

we know that nu is surjective

and the preimage is also an ideal that contains K

[hidden haven]

Try the first isomorphism theorem

[rustic crown]

moldi let the people use elements for once >.<

[hidden haven]

Yeah probably not good advice

I mean you would chase elements even with the first isomorphism theorem

[cloud walrus]

det

[south temple]

yeah that's where I've been hitting the brick wall haha

I just don't see how to incorporate I

but I'll keep trying

I've been doing it by proving one is a subset of the other

or trying to

[rustic crown]

right, you're almost there.

[cloud walrus]

det

[rustic crown]

we need to prove that a is also inside K

[late thicket]

I started group subgroup concepts

can anyone help me?

should I read galian?

[hidden haven]

Probably read whatever book your course is following unless it is bad

[rustic crown]

unless it's artin >.<

(jk)

[hidden haven]

Throwback to when I actually read Artin

[rustic crown]

F

never looked at artin ever since clare recommended it for group theory >.<

[hidden haven]

I read it for both alg2 and alg3 because Clare followed it religiously Some of the chapters were so unreadable I gave up on the course and still got AB because she didn't ask anything from the post midsem syllabus in the endsem 🥴

[south temple]

I think I got it

!!!

[rustic crown]

[south temple]

we have nu(a) = nu(b) for some b in K, so nu(a) - nu(b) = I. But that means that a - b is in I...

which implies that a - b is in K, and so a is in K + b

but b is in K

so a is in K

[rustic crown]

nu(a)-nu(b) = 0 >.<

[south temple]

right that makes more sense

thank you, it just wasn't getting through to me that there must be an element b that is also mapped to nu(a)

[hidden haven]

Here is the solution using the first isomorphism theorem You have 2 quotients, R/I and R/K. The natural map to R/K contains the kernel of the natural map to R/I, so there is an induced map f: R/I → R/K. Now the kernel of the natural map to R/K is exactly K, but it is also the preimage of the ker f under the natural map to R/I. This is exactly the equation you have to prove

[south temple]

we start with R, and we draw two arrows branching out of R: one to R/I and one to R/K yes?

[hidden haven]

ye

[south temple]

I'm just trying to do this visually

[hidden haven]

You should

[south temple]

what is the induced map?

explicitly?

[hidden haven]

Have you seen the first isomorphism theorem?

[hidden haven]

this thing

It is
a+I maps to a+K

There are a lot of issues like well definedness etc

but first isomorphism theorem says that this all works out when K contains I

Also you don't need to read this solution lol

Just read it if you are interested

[south temple]

I want to understand it

[hidden haven]

It is more abstract and no one would expect you to do this kinda tricks early on, but it could be helpful sometimes

[south temple]

I'm taking an alg. geometry course and I'm just derusting with these first few assignments

[hidden haven]

ah I see

[south temple]

it's been a long while since I took an algebra course 😢

[south temple]

but even with this, you have to prove that the preimage of ker f is K

so you still have to do things element-wise

[hidden haven]

Here is a general thing

ker(fg) = g inverse (ker f)

Works for any maps f and g

Or a more symmetric phrasing

(fg)^-1(0) = g^-1(f^-1(0))

[south temple]

hmm I'm gonna be honest this isn't really resonating with me rn. I think I need to take a break for a bit haha

nevertheless, I really appreciate the help

[hidden haven]

Ye understandable

[prisma shuttle]

hey guys can someone check this solutoin for this problem cuz it seems a bit too simple

Can't we just say that becaluse all the elements in $p\mathbb{Z}/p^m\mathbb{Z}$ are multiples of $m,$ raising any of the elements to the $m$th power results in the zero ideal in $\mathbb{Z}/p^m\mathbb{Z}$ so we are done?

[cloud walrus]

JustKeepRunning

[lethal dune]

multiples of m?

you mean p?

[prisma shuttle]

yea multiples of $p$ sorry

[cloud walrus]

JustKeepRunning

[next obsidian]

Yeh

You can show (a)^n = (a^n) for any element a, in any ring

[prisma shuttle]

ok thx

bro some of these problem in d&f are worded so weirdly

like for this problem

they say "the ideal" but like ima pretty sure ideals aren't unique right

so they should say "an ideal" I supppose?

[next obsidian]

No

They say what the idea I is afterwards

It’s “the” because I is a specific ideal