groups-rings-fields

[chilly ocean]

and that breaks homomorphism stuff

[next obsidian]

Yup

Well

More so I’d just say like

The expression nk just doesn’t make sense

Ya know or something like that idk

[chilly ocean]

like f(ab)=f(ba)

[next obsidian]

Right

[chilly ocean]

so there would be a contradiction

[next obsidian]

Anyway

[chilly ocean]

ok i see why it doesnt work for non abelian

[next obsidian]

So for the Hom(Z^n,G) thing

If the image of ei is ai

ei being like

1 in the i-th spot

0 everywhere else

Then you’re forced to have

f(m1,…,mn) = Sum mi ai

Cuz like

Break up by linearity

The LHS

[chilly ocean]

ye

[next obsidian]

Yeah okay yee

And then like

Because Z^n is free

As in like, there’s no relation among the basis vectors if you wanna call it that

It just pops out that

Defining f like that will always be a homomorphism

Like you can show it’s linear and stuff

This isn’t the case when you have say

[chilly ocean]

yeah lol

[next obsidian]

Z/4Z

It’s generated by 1, yes

But you have to send 1 to some x which has 4x = 0

[chilly ocean]

0=4

[next obsidian]

Otherwise you won’t be a homomorphism

Yeh

[chilly ocean]

oh

so thats where gcd comes in right

[next obsidian]

Uh

Idk does it?

Kekw

[chilly ocean]

nvm im thinking of maps between cyclic groups

[next obsidian]

Oh yes

!

This is a nice point

Also I think I lied

I think the Hom(Z,G) = G

Is always true even if G isn’t abelian

[chilly ocean]

yeah thats true

[next obsidian]

I was thinking about it

Anyway

By that fact

Combined with univ prop of kernel

Maps from (Z/nZ, G)

Are in bijection with elements of G such that the order divides n

You define the map as

f(m) = x^m

And then you can show this just is a homomorphism

[chilly ocean]

oh lol

so order of elements in G need to divide n

[next obsidian]

Not all of them

[chilly ocean]

but the ones that do have maps

[next obsidian]

Just those are the ones which give you a map from Z/nZ

Yee

[chilly ocean]

so Hom(Zn,G) uniquely determined by number of elements of order k such that k|n

for G abelian again probably

or maybe not abelian too

[next obsidian]

Nah

This works for any group

But no it’s not uniquely determined I think

[chilly ocean]

uh

[next obsidian]

Cuz the size

Does not uniquely determine a group

Also what I can say

[chilly ocean]

oh yeah i meant something else

[next obsidian]

Is Hom(Zn,G) is not an abelian group

If G isn’t abelian

But maybe it’s just also a group

Lol

Idk

[chilly ocean]

ilol

[next obsidian]

I think not

[chilly ocean]

this reminds me

[next obsidian]

Probably no inverses

[chilly ocean]

i was reading about closed categories

[next obsidian]

Is that when they have a Hom object which lives in the category?

[chilly ocean]

pretty much from what i learned, a category is closed if its Hom sets are objects in the category

yeah

[next obsidian]

Yeah

Categories fibered in the category

[chilly ocean]

i think some types of abelian categories are closed but idk how to prove or anything

i just read on nlab

ngl nlab is bad

[next obsidian]

Nlab good*

Where * means

[chilly ocean]

if you have knowledge of everything

[next obsidian]

If you’re a certain type of person at a certain point in ur career

[chilly ocean]

wikipedia better tbh

ye

[next obsidian]

I use nlab at times now

[chilly ocean]

slowclap*

also uh

remember or convo yesterday about ideals of rings of functions?

maybe we didnt have conversations

but rn im trying to get better understanding of ideals of a ring R by looking at Hom(R,S) or Hom(T,R)

[next obsidian]

Prime ideals are all given by maps into a field

Consider the map R -> R/P -> Frac(R/P)

Maximal ideals by surjective maps onto fields

This is all I know

[chilly ocean]

woah

what lol

if P is prime then R/P is integral domain?

i think right

but idr proof

ab in P implies a or b in P

so if we suppose ab=0 in R/P

we have that a in P or b in P

which corresponds to a 0 or b 0

oh lol

when can you stop finding prime ideals?

so say P0 is prime in R and P1 prime in R/P0,…

does this end

im only thinking of this because im thinking of solvability for whatever reason and wondering what an analogue would be

[next obsidian]

It might jot end

You’re starting to think about krull dimension

Which is the length of chains of primes inside of A

[chilly ocean]

woah

chains of primes?

like their inclusions?

[next obsidian]

Yeah

That’s what P1 a prime of R/P0 is

It’s a prime containing P0

[chilly ocean]

oh cool

so many moving parts in theory of ideals and it seems super widely applicable to everything because it pops up so often, you said algebras show up everywhere but im not matured enough to see it

maybe i need to take up something like algebraic geometry so that i can have and learn more examples maybe

also if u dont mind checking imma try and compute Tor of abelian groups

[shell brook]

G \ H doesn't mean set minus right

i dont know what it could mean but no chance its set minus

I've never seen it defined

[chilly ocean]

nah just think of it as sending every element in H to identity

but G/H={gH|g in G}

set of left cosets of H

[shell brook]

I know what G/H means

[chilly ocean]

oh bruh wtf

[shell brook]

lol

[chilly ocean]

im dyslexic

they meant G/H

right?

[shell brook]

no

cause look

[chilly ocean]

not set minus right

[shell brook]

the entire question uses it

the mistake would've been noticed by now

[chilly ocean]

oh it is set minus

[next obsidian]

It definitely does

[unreal portal]

yeah that's setminus

[shell brook]

alr

okay

[chilly ocean]

uh

order of G/H is p

[next obsidian]

Yeah but this actually means setminus

[chilly ocean]

lol this is dumb

why wouldnt they use minus sign

[shell brook]

its probably easy but

i hope

[chilly ocean]

it is yea

[shell brook]

yeah kinda weird notation choice cuz we've never seen anything like this lol

[next obsidian]

I always use $\setminus$

Tf?

[cloud walrus]

The Chmonkey

[chilly ocean]

lol wtf

btw i think i have misunderstanding

order of a group is size of set, but it also corresponds to order of each element

[next obsidian]

No those are different

I mean they’re related

[chilly ocean]

wait

[next obsidian]

But “corresponds” isn’t the right word

[chilly ocean]

order of each element divides

[next obsidian]

Yes

That’s (basically) Lagranges theorem

[chilly ocean]

yea

is a lot to remember at times );

[robust pollen]

More generally, if e is an idempotent in a ring R, then eRe is a ring with unit e. Note that the set-inclusion eRe -> R is a rng morphisms, but can only be a ring morphism if e = 1 to begin with

It's like the familiar situation for algebras, where you have something called "idempotent subalgebras". These are linear subspaces which are closed under multiplication from the full space, but their unit is not the unit of the full algebra, but rather an idempotent.
For example, look at matrix rings, e.g. M_2(k) and embed k into it by sending a |-> {{a,0}, {0,0}}

(actually, it's not "like the familiar situation for algebras", it is exactly that situation, since rings are just algebras in Z-mod)

[next obsidian]

When they’re commutative

[robust pollen]

what do you mean?

[next obsidian]

Oh huh

I guess the image of Z -> R does always end up in the center

Let’s gooooooooo

[robust pollen]

wha

[next obsidian]

I was thinking when you aren’t dealing with a commutative ring S, and R algebra is a map R -> S landing in the center of S

And so I was worried about if you had a noncommutative ring it might not be a Z algebra

But this is just dummy

[robust pollen]

Oh ok 😄

btw, I don't understand the statement that R-algebra structures on an abelian group S are the same as ring morphisms R -> S. How does this work?

[final pasture]

S needs to be a ring for this to make sense, except if I'm missing something

What does a ring morphism going from a ring to a group means ?

If you have S a (commutative) ring, then any ring morphism f: R -> S defined an R-algebra structure on S, given by r . s = f(r) x s

[robust pollen]

I mean that having a ring morphism in particular implies that S is a ring

[final pasture]

Doesn't really make sense stating it that way, but yeah okay

[robust pollen]

So if you start with an R-algebra structure on an abelian group S, the fact that s is a ring is already baked into it

It makes total sense actually 😬

[final pasture]

Well then you're not working over an abelian group S, you're working over a commutative ring S

You don't say that a group G has dimension 28 and expect people to deduce from this that G also has a vector space structure

[next obsidian]

So like

The point is that given f: R -> S

You define r•s by f(r)s

And given an algebra structure in the “usual” sense

You can recover f by looking at what r•1 is

But for this to work as you’d want you need f to land in the center of S

Else you have this weird shit with left multiplication and right and it’s just whack

Or something

Idk you can probably define a left algebra

And a right algebra

[final pasture]

(Yeah you can but it's ugly )

[next obsidian]

But anyway if your rings aren’t commutative you already want to die so why make it harder on yourself

[robust pollen]

(R is commutative I guess, S doesn't have to be)

[next obsidian]

Yeah

The issue is like

What if you want to do

s(r•s’)

Uh oh

[final pasture]

S does need to be commutative if you want an algebra in the usual sense, I think

[next obsidian]

Nah

If im f < center of S

[robust pollen]

What is an algebra over a ring for you guys? For me it's a monoid in the category of R-bimodules

[final pasture]

Oh

[next obsidian]

You can make it workout basically the same

[final pasture]

yeah that's enough, then, yeah

[next obsidian]

A ring map R -> S landing in the center of S

Kekw

But actually

[final pasture]

You can also define it the explicit and annoying way

[next obsidian]

In my world everything is commutative

So it’s just a ring map

😩

[final pasture]

Like R is a ring, A is an R-module and then you add a multiplication by an R-scalar law that satisfies compatibility conditions

like distributivity over addition

and mixed-commutativity with multiplication

Explicitly, r.(x+y) = rx + ry, same thing with (x+y).r and (r . x) * y = x * (r . y) = r . (x * y)

(I probably forgot half of the compatibility conditions, the list is long and it's intuitive anyway )

[robust pollen]

You mean A is an R-bimodule, though, right? Otherwise (x+y).r doesn't make sense. Or is R commutative?

Also you probably want multiplication in A to be R-balanced

and then it should just give you the statement that A is an R- algebra iff A is a monoid in R-bimod

actually, this is related to the most recent question in #category-theory lmao

[next obsidian]

Is anyone familiar with André-Quillen homology? If so, how did you learn it?

[winter thorn]

Is (G, *) notation for "G is a group"?

I need notation badly

[paper flint]

I'm not aware of any "notation" that would tell you right away that the given algebraic structure is a group

(G,*) could also represent a monoid, or semigroup, or a poset, none of which is necessarily a group

Good practice is to just say "A group G.."

[winter thorn]

bummer\

[rustic crown]

G in Obj(Grp)

but yea "G group" is just better to say lol

[chilly ocean]

you should actually get used to just writting the words

its much easier to read

[winter thorn]

unfortunately, yeah

[chilly ocean]

i think once you get used to this you will think "fortunately"

like this reads like trash

$g,h\in G \implies g\cdot h \in G$

[cloud walrus]

lime_soup

[winter thorn]

I don't know what you are talking about now

This is beautiful

[chilly ocean]

just write,
for any two elements $g$ and $h$ in $G$, their product $gh$ is also in $G$

[cloud walrus]

lime_soup

[winter thorn]

ehhh, wastes too much time imo

maybe I don't want to write much.

or at all

[chilly ocean]

for what its worth

when you are TAing

[chilly ocean]

if you see someoen writting assignments like this

it always seems to end up being a poor asignment

[winter thorn]

hmm

I don't want to write for anyone else's sake

just some personal things

[rustic crown]

write for your own sake

[winter thorn]

I still wanted to get it right if I could

[chilly ocean]

The biggest tip i ever got for math writting was to make sure that you say the type of object a thing is

like

Let $f$ be a polynomial in the ring $\mathbb{C}[x,y]$

reads so much nicer than

Let $f\in \mathbb{C}[x,y]$

modulo getting the latex correct

[viscid pewter]

incredible

[rustic crown]

lol

[hidden haven]

Yeah the ldotsx_n is confusing

[cloud walrus]

lime_soup

lime_soup

[hidden haven]

🥳

[chilly ocean]

y=ldotsx_n

its funny to imagine

just how poor you could make some notation

Consider the polynomial ring in one variable. Denote this single variable by $x_1,\ldots,x_n$

[cloud walrus]

lime_soup

[winter thorn]

I need to show that in an Abelian group G, H being the set of all elements whose order is odd forms a subgroup.

I've shown that for some a in H, a^-1 is in H.

How do I even show that for some a,b in H, a*b is in H?

[viscid pewter]

what's the order of ab

[winter thorn]

It's the LCM of the order of a and the order of b?

[viscid pewter]

there we go

well strictly it divides the LCM of the order of a and the order of b

but same difference

[winter thorn]

Hmm, I'm still clueless how to incorporate this fact into 'a*b is in H'.

[viscid pewter]

can you say anything about the odd/evenness of: the LCM of: the order of a and the order of b?

[chilly ocean]

i think you might be over looking this

to show that ab is in H, its enough to show that the order of ab is odd

[winter thorn]

OH

[chilly ocean]

and moverover

[winter thorn]

Order of ab is odd

[viscid pewter]

yeah

[chilly ocean]

there is this subgroup criterion

to check that H is a subgroup you need only check that a^{-1}b is in H

[winter thorn]

I'm stupid, it mentioned in the subgroup test that you need to use the specific property of the set to prove a subgroup.

[viscid pewter]

ehhhh

[chilly ocean]

this automatically gives you the closed under inverses

[viscid pewter]

i don't like that test

[chilly ocean]

but

[viscid pewter]

clearer to just do each criterion separately

[chilly ocean]

it often can be easier to just check the two seperately

[viscid pewter]

yeah

[winter thorn]

Gallian did mention that, it's the One Step Test.

[chilly ocean]

i still think its good to know

[winter thorn]

I'm stuck on a question concerning Dihedral groups.

I need to show that if H is a subgroup of D_n, then either H is all rotations or half of H is made up of rotations.

[chilly ocean]

thats a good exercise

pick a small example of a dihedral group and look at its subgroups

[winter thorn]

okay

[viscid pewter]

either it has s or it doesn't

[molten silo]

Can anyone find a transversal for this action on R^2?

[viscid pewter]

is it not just {(0, b)}?

[molten silo]

So any real number right?

nevermind

@viscid pewter Can you explain why?

Would {(1,b)} work?

[potent briar]

"Prove that if y is a linear functional on an n-dimensional vector space V, then the set of all those vectors x for which [x, y] = 0 is a subspace of V; what is the dimension of that subspace?"

I think the dimension of that subspace is 1?

[molten silo]

how do i explain this?

I think of it as a set of One element representative for each orbit

I know there is only one element in the transversal, since there is only one orbit

ok

[potent briar]

i don't have that theorem yet

[molten silo]

So they are vertical lines?

[potent briar]

Rank and Nullity is chapter 50. this question is in chapter 17

ehehe

[molten silo]

i see thank you

[potent briar]

this book studies spaces themselves a lot before doing any transformations

yeah that's the question

with y = 0

then the set of all those vectors x for which [x, y] = 0 is a subspace of V

then the set of all those vectors x for which [x, y] = 0 is a subspace of V

oh halmos uses stupid bracket notation for some reason

[x, y] = y(x)

the set of all vectors

such that y is 0 at those vectors

i haven't had anything on transformations yet so idk what kernel is

but no y is not 0

it's any functional y : V -> F

[robust pollen]

Do you know Gram-Schmidt, or the fact that you can extend a set of m =< n linearly independent vectors to a basis?

[potent briar]

yes I proved that you can extend a non-basis independent set of vectors to a basis

well actually, only kind of proved it

because i haven't proved that all spaces have a basis

but i can believe it on faith for now becuase i was told it was difficult to prove xd

[robust pollen]

the fact that vector spaces have a basis is an axiom

[potent briar]

hmm i don't think that's independent

in this textbook it's an exercise to prove it

[robust pollen]

Yeah, assuming the axiom of choice.

[potent briar]

it says "Prove that every vector space has a basis. (The proof of this fact is out of reach for those not acquantied with some transfinite trickery, such as well-ordering or Zorn's lemma.)

[robust pollen]

But conversely, if you assume that every vector space has a basis, you can prove choice. So in fact, one might as well view the existence of bases as an axiom 🤷‍♂️

[potent briar]

i see

anyway

my intuiton was that if y(x) = 0

[robust pollen]

Anyway, show that there is some x such that y(x) is not zero, and try some extending to basis stuff

[potent briar]

then y(a*x) = 0

so the set of all vectors that make y 0 is a subspace of dim 1

because i already proved that if {x1, ..., xn} is a basis and {a1, ..., an} a set of scalars
then there is only one functional such that [xi, y] = ai

[robust pollen]

Given y, can you find x such that y(x) is non zero? Can you then find x' which is linearly independent from y and such that y(x') = 0?

[robust pollen]

I don't understand (1.1). Nowhere is it stated explicitly what the $(R,R)$-bimodule structure on $C \otimes_S C$ is, and then in (1.1) it is sort of implied that it's given by $r \cdot (c \otimes_S c') \cdot r'= c r' \otimes_S r c'$.
But this does not seem to be the most natural R-bimodule structure, I would have expected $r \cdot (c \otimes_S c') \cdot r'= r c \otimes_S c' r'$. so what gives?

[cloud walrus]

expectTheUnexpected

[robust pollen]

And actually, also the centralizer thing, equation (1.2), implies that the action is the one I write. lolwut. Am I dupid?

[potent briar]

ok @robust pollen i don't understand

[robust pollen]

ok, so first of all, is there a vector v in V such that y(v) is not zero?

[potent briar]

yes

becuase of the annhilator of the whole space is only y = 0

[robust pollen]

well, if and only if y is not the zero functional, to be precise

(in your original question, y was any functional, so to solve the exercise completely you'll have to distinguish between y = 0 and y =/= 0)

[robust pollen]

ok

[robust pollen]

so, you also said here that you know you can extend any set of linearly independent vectors to a basis.

[potent briar]

sure

[robust pollen]

the set {v} is trivially a set of linearly independent vectors

[potent briar]

keep appending indepedent vectors until you're out

[robust pollen]

Ok, so, since Vwas n-dimensional, we can find vectors v_1 = v, v_2, ..., v_n that form a basis, right?

Can you change the vectors v_2, ..., v_n in such a way that y applied to each of them is zero?

[potent briar]

what?

[robust pollen]

do you agree with the fact that we have a basis v_1, v_2, ..., v_n?

[potent briar]

yes

[robust pollen]

Ok. So, if for example y(v_2) is not zero, can you somehow build another vector, say w_2, for which y(w_2) = 0?

hint: the definition of w_2 will include v_2, v, and y (more precisely y(v) and y(v_2))

If you can do that, you can do it for all v_2, ..., v_n. Then you show that the corresponding w_is, for which y(w_i) = 0 holds, are linearly independent. So you get a lower bound on the dimension of the space you're looking for

[potent briar]

hmmm

ok let me think of a concrete example

in R^2 with the canonical base

let's say y = 2x - 7y

that's a linear functional no?

[robust pollen]

uh, no

[potent briar]

what??

[robust pollen]

how is y = 2x - 7y a linear functional?

first of all, there are ys on both sides, and secondly, this expression does not live in R, but in R^2

[potent briar]

... fine, let the vectors of R^2 be of the form (x, y)
and let f(v) = 2x - 7y

[robust pollen]

Ok. f(x,y) = 2x -7y then

but examples really don't help too much here I think. And certainly not R^2, R^3 would be much better

[potent briar]

ok right R^3 then

f(v) = 2x - 7y + z

(0, 0, 0) makes it 0
(7, 2, 0) and its multiples make it 0

i dont know what vector with three non-zero coords makes it 0 hehe

(0, 1, 7) and its multiples, (1, 0, -2) an its multiples

[robust pollen]

a bit more systematical:
Standard basis e_1 = (1,0,0), e_2 = (0,1,0), e_3 = (0,0,1), then f(e_3) = 1

But f(e_1) and f(e_2)? They are non-zero.

What about f(e_1 - 2 e_3) and f(e_2 + 7 e_3)?

Are e_1 - 2 e_3 and e_2 + 7 e_3 linearly independent?

if they are linearly independent, you know that the kernel of f (fancy name for those vectors that get sent to 0) has dimension at least 2. Now, can it be 3-dimensional?

[winter thorn]

Need to prove that C(a) = C(a^3) whenever a^5 = e.

I showed that ax = xa <======> a^3x = xa^3

is this enough?

[robust pollen]

C(a) = centralizer of a?

[winter thorn]

yes

[potent briar]

they are linearly independent

[winter thorn]

???

[potent briar]

i guessing it can't be 3-dim

[winter thorn]

bruh at least highlight the message you are replying to

[robust pollen]

other discussion 😄
And regarding your question, well the <===> means that x is in the centralizer of a if and only if it is in the centralizer of a^3

[robust pollen]

well, yeah: if it we're 3-dimensional, f would just be the zero map

[robust pollen]

also, if a^(2n-1) = 1, then C(a) = C(a^n).

[winter thorn]

Haven't gotten that far, and I can prove that now

but

whenever ax = xa, xa^3 = a^3x

does this imply that C(a) = C(a^3)

[potent briar]

right

[robust pollen]

so, try to generalize and then prove that if you have a non-zero linear functinoal, its kernel has dimension n-1

[robust pollen]

Well, what does C(a) = C(a^3) mean? Write it out.

[winter thorn]

The centralizer of a is equal to the centralizer of a^3

[robust pollen]

Write it out in terms of sets

[winter thorn]

C(a) = {x in G: ax = xa, a in G}

C(a^3) = {x in G: a^3x = xa^3, a in G}

[robust pollen]

So what does your "xa = ax <=> xa^3 = a^3x" show, in terms of these sets and elements?

[winter thorn]

If x is in C(a), x is in C(a^3)

and by extensionality, C(a) = C(a^3)??

[robust pollen]

Not quite. Your "if and only if" shows that x is in C(a) if and only if x is in C(a^3). And that is the definition of equality of sets.

[winter thorn]

okay

I think I got it

[final pasture]

you're not taking one out

You're quotienting

[winter mesa]

so 6?

[next obsidian]

the size of H is 4

[winter mesa]

wait what elements would be in H? im confused

[proud bear]

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

[long obsidian]

Is the space of nxn matrices isomorphic to the space of nxn matrices without 0

[winter mesa]

so there would be 2 left cosets and they would <j> and <k> right

[viscid pewter]

without 0 in what sense

also what's the group operation

[long obsidian]

My bad I should be more specific. Let M be the matrix ring with the usual matrix addition and multiplication over a division ring. I read somewhere that $M\cong M/{0}$. But I'm not sure if this is a mistake

[cloud walrus]

fajitas

[teal mantle]

hey guys is this correct?

[rustic crown]

it's not wrong. but depends on the prof how much details you should put in.

[teal mantle]

alright thanks, this isnt hw, im just prepping for an exam so I didnt go super into details

[next obsidian]

I think you can more succinctly prove the latter one by noting if Z_16 -> Z_2 (+) Z_2 is onto then the latter is a quotient of Z_16 which is cyclic, but clearly the latter isn't cyclic

It's basically the same sort of idea as what you did

[barren sierra]

is this not trivial using Lagrange's theorem?

right?

[next obsidian]

it is trivial, yes

oh well

no yeah

Maybe they want you to show that if K < H < G and K,H are subgroups of G, then K is a subgroup of H?

[barren sierra]

I think the fact that H and K are subgroups is given

that's how I'm reading it

[next obsidian]

It said that K is a subgroup of G

you "need to show" that it's a subgroup of H as well

this is like totally fucking trivial

but you need it to apply lagrange to get a statement about [H:K]

[barren sierra]

gg

ok

[robust pollen]

even easier, just apply Nichols-Zoeller theorem to the group (Hopf) algebras of K,H,G over C

[teal mantle]

If I have a finite group $G$ and two normal subgroups $H$ and $K$, what can we say about the statement [G/H \cong G/K \Leftrightarrow H \cong K.] Is either direction true?

[cloud walrus]

OmnipresentCoffee

[rustic crown]

nope

both are false

work with abelian groups to get rid of checking normality

G/H =~ G/K could be because these groups are too small

G =  Z/2Z +  Z/4Z
H =  Z/2Z + 2Z/4Z
K =   0   +  Z/4Z

and if H =~ K we don't really know how they live inside G

G = Z/2Z +  Z/4Z
H = Z/2Z +   0
K =   0  + 2Z/4Z

[chilly ocean]

Let M be a module over R

We can localize M by S, where S is a multiplicatively closed set in R

If S contains 0, then S^-1M is 0

Is this also true if S contains a zero divisor

I can see if there is and s*m=0 then m/s=0 in S^-1M

But will all of S^-1M be zero?

[hot tinsel]

In Lang's algebra pg 206 he states that if $A$ is a complete local ring then the power series ring $A[[X]]$ is also complete local.

[cloud walrus]

K零ꓘ

[hot tinsel]

I can see the local part here, can someone help me about the complete part?

[chilly ocean]

no

localize Z/6Z at 2

Ah thank you

So basically what’s going on is

Z/6Z has 2 torsion and 3 torsion

So to kill it we would need to inver 2 and 3

But then let’s say if we had Z \oplus Z/6

We could only ever kill everything if we invert 0

[thorn flint]

Hi, for this question a

Do I just need to mention that g(0) is not equal to 0 in R and R is an integral domain so there are no zero divisors and the only way to get a zero divisor is if f(0) =0. So R* is defined as R with f(0) not equal to 0?

[lethal dune]

g(0) neq 0 is already given

[chilly ocean]

would anyone be able to spend 1 hour ish helping me understand ext and tor and uct later at around 11est

[potent briar]

i have an exercise like prove that if you have m linear functionals on an n-dim space with m < n there is a vector that makes them all 0

my intuition is that it's true because theres n - 1 base vectors that make a dual base vector 0

so given some functionals you can choose a vector thats a combination of these other basis vectors

[stark sigil]

Hint; what is the zero set of a linear functional?

[potent briar]

im guessing its the set of vectors that make it evaluate to 0

what lol

[stark sigil]

Why’s that a guess

[potent briar]

i havent been introduced to it formally lol

[hidden haven]

Ye that is the definition

But you can say more about it

What properties can a vector space or a subspace have?

[stark sigil]

Cocatthink lol

[potent briar]

if you have a functional which is a linear comb of the ith dual basis vector

it evaluates to 0 on every other vector

no?

[stark sigil]

What’s the definition of a linear functional?

[hidden haven]

All other basis vectors yes

[potent briar]

what

yes

so if the evaluated vector is a combination of any of these vectors

[stark sigil]

This method will lead to a brick wall because none of the linear functionals have to evaluate to 0 on any basis vector

[potent briar]

i.e. not the ith basis vector

[stark sigil]

So, what’s the definition of a linear functional?

[potent briar]

wdym

[stark sigil]

What is the definition of definition?

“definition of a linear functional” is like the clearest math question ever

[potent briar]

i know what they are

i meant why wont it prove it

[stark sigil]

I said why after “because” in that post

[potent briar]

any linear functional is a combination of the dual basis

[stark sigil]

Is that the definition of linear functional?

[potent briar]

no

[stark sigil]

What is it?

[potent briar]

im not going to type it

just believe me i know it

[stark sigil]

Is the sum of all dual basis vectors a linear functional, even though it evaluates to 0 on none of the basis vectors?

[potent briar]

thats why the question is

m functionals

with m < the dim of the space

[hidden haven]

It won't help to think in terms of basis elements vik

[stark sigil]

The object i asked about is one linear functional

[potent briar]

ok

ooh i see

[stark sigil]

So what is the definition of a linear functional?

[potent briar]

youre wasting time i already didnt come up with the obvious cobsequence of the definition

[stark sigil]

Oof didn’t expect this in an abstract algebra student

[hidden haven]

[potent briar]

i just find it an annoying question

i do know the definition by heart

[hidden haven]

Can you say anything about dimension of the kernel of a functional?

[stark sigil]

You need to write it down to proceed with the problem

That is the way mathematics works

[potent briar]

the book studies spaces thoroughly before transformations

idk what a kernel is

[hidden haven]

Zero set

[stark sigil]

It also helps to see what definitions your book uses

[potent briar]

scalar valued function on the space with linearity

[stark sigil]

Excellent

Now I know what I’m dealing with

[potent briar]

what

were you

expecting

does another definition exist jesus

[stark sigil]

Ya when a subject has multiple orders of presentation

Different things can have different definitions to start with

Determinant is a prime example

[potent briar]

and in those books the words linear functional mean something different

[stark sigil]

Could

[potent briar]

right

[stark sigil]

It never means something different

It will eventually be proven equivalent to other definitions

But depending on your book, you know a certain definition and have to use it to prove things so as to not use circular reasoning

[potent briar]

i mean might as well recite the whole book

the def of space could be different

[stark sigil]

Gotta go for an hour

[potent briar]

anyway

[stark sigil]

Good luck!

[barren sierra]

Not sure where else to ask this

is the book "Naive Lie Theory" by Stillwell any good?

Just got it for free from my library

[chilly ocean]

I wouldn't trust it.

[wise igloo]

yeah it might try dealing drugs to you

[untold cloud]

Hi, guys. What does it mean for pi(g)|w? if W is a subspace of V, then the size of pi|w(g) will be smaller than the size of pi(g)? So it means there are some entries will be removed in pi(g) in order to have p(g)|w?

[robust pollen]

That is some terrible notation

[untold cloud]

agree, but my lecture uses the same notation😅 , just do not understand what it means

[robust pollen]

What do you mean by "size of pi|w(g)"?

[untold cloud]

for example, 2 by 2 matrix if dimW＝2

[stark sigil]

That’s standard notation for restriction of a function to a smaller domain

It also applies here

[robust pollen]

Nah, it's note, pi : G -> End(V)

[stark sigil]

And pi(g) is an element of End(V)

So pi(g)|_W is the restriction of that endomorphism to W

[robust pollen]

Well, pi(g)|W is standard notation, but pi|W(g) is not standard (but it's defined in the statement, so all is good - except for the fact that it's terrible notation imo)

[stark sigil]

A better notation eludes me at the moment

The second best after that is just using pi again

[robust pollen]

I like that one better 👍

[untold cloud]

sry, i am confused, doesn't a subrepsentation is like (pi|W,W), pi|W: G-->GL(W), but if dimW=n, then GL(W) is isomorphic to GL(n,F), where F is the field for W\subset V. Then for example dimV=n+1, then the representation(pi,V), pi:V-->GL(V), GL(V) is isomorphic to GL(n+1,F)? how can image of a restriction of a function not in the codomain anymore?

[robust pollen]

Subrepresentation just means that if you act with pi(g) on the subspace W, you will again land in W.

Also, pi doesn't have domain V, but G

[untold cloud]

oh yeah, that domain is a type. But dimW<dimV, then how can we mutiply p(g) with w\in W

lol, type is a typo

[robust pollen]

W is a subspace. In other words, "Pad with zeroes" until you get a vector of length dim V, act on that

[untold cloud]

ok, so it is like (w_1,w_2,...,0,..,0)? something like this?

[hot lake]

if W is a subspace of V

and pi(g) is a map from V to V

then you can apply pi(g) to an element of W

because elements of W are also elements of V

that would give you an element of V

but if W is a subrepresentation

that means all the pi(g)(w) are actually in W

[untold cloud]

oh,yes,lol, i see i see, thanks

[hot lake]

and so pi(g)|W can be a linear map from W to W

[untold cloud]

But i saw someone is asking online. he\she uses the definition that a subrepsentaion is pi:G-->GL(W). is this uncorrect?

[hot lake]

this is all correct

well subrepresentation can mean either the subspace

or the map into GL(the subspace)

I'm not sure which one is more used

[chilly ocean]

im stuck on question 5 im not sure of the formula to find the pint parallel to equation given

[barren sierra]

#multivariable-calculus will be of more help

[hot lake]

I don't think you're in the right channel

[untold cloud]

but how can GL(W) is an n by n matrix, for example, dimW=n=dimV-1, then GL(W) v, where v =(v_1,...,v_{n+1})

[hot lake]

what

GL(W) isn't a group of matrices until you choose a basis for W

[untold cloud]

oh, I see! Thanks! it is set of linear bijection from W to W?

[hot lake]

yes

[untold cloud]

Thanks! really helpful!

[chilly ocean]

ill just join a vc at est11 and hope for a savior

[chilly ocean]

Est11 is like in 5 hours or sth right?

[agile shore]

Hey guys! I just started my Introduction to Algorithms class for my masters and have hit a bit of a brick wall. It's been since high school that I've had to take a math class, and so I've been trying my best to hit the books and catch myself up to speed. I'm running into an issue with an exercise in my textbook though (not part of an assignment), and I've really been trying to figure it out. The question is specifically finding which values insertion sort beats merge sort for a given implementation, where insertion sort is 8n^2 and merge sort is 64n * lg n (I'm not sure if this notation is standard, so lg == log_2). I wrote this down as 8n^2 < 64n * lg n.

I tried my best to simplify it, dusting off some of my old online resources to relearn logarithms, but I simply couldn't figure it out. I decided to put it into Wolfram Alpha to see the answer and work backwards from there, but it completely blind-sided me with something I've never seen before, being the Lambert W function. I won't write it here simply because I don't know how to feed LaTeX to the bot, or if that's possible, but the link is this: https://www.wolframalpha.com/input/?i=8n^2<64n*log_2(n)+solve+for+n

I'm mainly wondering, what is the Lambert W function, how do I solve it, how do I use it, and how was it used in this context? And, if possible, the steps taken to solve for the expression would be extremely helpful. I'm not the sharpest tool in the shed for this kind of mathematics, so I might need a bit more than others to follow along 😓. I've looked at some online resources and figured out that it's the inverse of f(x) = x * e^x, but it's been so, so long since I've done this sort of math that I'm struggling to understand it. I've really been trying to learn it on my own, and this was the last resource I had, as I didn't want to trouble you guys with something I could learn on my own. If you take your time to help me understand this, I thank you greatly.

If I'm in the wrong channel, let me know and I'd be happy to move. I'm not quite sure what Lambert W function falls under

[scarlet estuary]

this isnt really #groups-rings-fields but in fairness there isnt a great place for this question on the server, and no one else is using the channel rn, so its fine here

anyway, the problem is that one cant solve this in exact values through elementary functions

in general, we cant solve things of the form f(x) e^x in an elementary way for "most" f(x)

which is what this becomes after some rearranging

the lambert W solves specifically equations involving xe^x, which turns out to be useful here

i could show the full derivation but it seems tedious and not particularly insightful

since i dont think you care about exact values

after all, n will be an integer

[agile shore]

Yeah I just need approximations, then round up/down for the final answer

[scarlet estuary]

then click this button?

[agile shore]

That does solve my question, but doesn't help me to understand how

[scarlet estuary]

you plug in values lmao

as i said, theres no elementary way to express this solution

[viscid pewter]

you're asking how a computer approximates values?

[scarlet estuary]

thats why the lambert W had to get involved

[agile shore]

Which is what I want to mainly tackle, as logarithmic algorithms are relatively common among the most popular ones

[scarlet estuary]

i mean, im not even sure what your question is exactly

you cant "solve a function"

you can evaluate it at a point, but you'd never do that for the lambert W by hand

you'd either do a computer approximation or, lacking access to a computer, you'd try random numbers

[agile shore]

I see, so for these kinds of problems it's best to leave it to the computer?

[viscid pewter]

yes

[agile shore]

Unfortunate...

Thank you for the clarification though!

[scarlet estuary]

the problem is that there is no elementary way to express solutions to this

by "elementary" i mean using like

polynomials, radicals, fractions, exponentials/logarithms, trig

[agile shore]

Yeah

[scarlet estuary]

you know, the "common" functions

so we're forced to "make something new up" to extend our ability to solve these problems

in this case, that is the lambert W

but its very hard to compute

(hence why you dont learn about it with the other functions)

[agile shore]

I see... it does feel a bit unsatisfying to learn something is far outside your reach to solve though

[scarlet estuary]

but you dont need exact values for this problem anyway

[agile shore]

well, more than just a bit

[scarlet estuary]

since n is a natural number

you just need approximations

you CAN do calculus to maybe speed up the process of finding an approximation a bit

(see any numerical methods class)

[agile shore]

Yeah, but the pursuit of knowledge was the main driving force

I see something that I don't understand it and I want to try to understand it

[viscid pewter]

we understand it

it's just that it's literally impossible to get from here to there

where here is the nice functions we use all the time, and there is the lambert w

i mean there's taylor series ig

[scarlet estuary]

if you want a pseudo-closed form, the lambert W on this part of its domain bears the following taylor series

[viscid pewter]

when will that even converge tho

[scarlet estuary]

r = 1/e

[viscid pewter]

lmao

[scarlet estuary]

but you can uniquely extend it to a holomorphic function

so

¯_(ツ)_/¯

again, not really practical to compute

but uh, it exists

[agile shore]

https://tenor.com/view/math-what-i-dont-understand-confused-lost-gif-9121307

[scarlet estuary]

fyi this is far from unique to these sorts of things

[agile shore]

I sadly am just not knowledgeable enough to understand this information

[scarlet estuary]

hell, the polynomial equation x⁵ - x - 1 = 0 has no elementary solution either

[viscid pewter]

ok do you know what taylor series are

[agile shore]

I do not

[viscid pewter]

ok well

[scarlet estuary]

okay then ignore what i just said lmao

[agile shore]

🤣

[viscid pewter]

they're like

[scarlet estuary]

except for:

hell, the polynomial equation x⁵ - x - 1 = 0 has no elementary solution either

this is a true fact

its not that mathematicians havent figured it out

its that we KNOW theres NO POSSIBLE WAY

to express it without inventing "new" functions

(in the case of degree 5 polynomials, one possible "new" function is the bring radical: https://en.wikipedia.org/wiki/Bring_radical)

the lambert W is the same principle

its a bit more theoretically hefty because of domain/convergence issues and whatnot

but thats not really relevant unless youre taking a complex analysis course

so whatever

[agile shore]

Man, I wish I understood this stuff

Let me just steal your knowledge

[viscid pewter]

basically the taylor series of a function is an infinite sum of polynomial terms, derived by taking repeated derivatives of the function, that is usually equal to the function at a certain point

[agile shore]

https://tenor.com/view/run-running-fast-basketball-court-sprint-gif-16667898

[scarlet estuary]

anyway, i will confess that this is a bit unsatisfying at first

[viscid pewter]

to approximate a function, you can then sometimes just take the sum of the first few terms in that series

[scarlet estuary]

fortunately, computer techniques to approximate this stuff have gotten very very good

[viscid pewter]

and sometimes it'll be a good approximation

[agile shore]

I never learned derivations

[viscid pewter]

yeah but the actual series is like

known

ppl have worked it all out

[agile shore]

ahh ok

[viscid pewter]

so you can just copy the formula from somewhere, and approximate it that way if you want...

[scarlet estuary]

it only converges on a small interval though so youre still SOL with this question lmao

[viscid pewter]

so this one approximates it for x close to 0

[scarlet estuary]

this is more practical

[final pasture]

I don't really know any Galois Theory but do we know that we still can't do it if we're allowed to use, say exponentials and logarithms ?

[scarlet estuary]

and a programmer might actually know what o(1) means!

[viscid pewter]

oh, sick

[agile shore]

o(1)? I don't know what little o is, only big O 🤪

[viscid pewter]

when is that one a good approximation

[scarlet estuary]

probably a corollary of liouvilles theorem somehow

[agile shore]

So basically, to solve this question, I either need to plug in random values or let a computer solve it. Unsatisfying, but it works I guess

[final pasture]

The one about integrals that can't be expressed using elementary functions ?

[scarlet estuary]

yeah

[final pasture]

Okay, thanks

[scarlet estuary]

you can translate a lot of results from galois theory into results on exponentials using it

idk exactly how it works though

been a while

[final pasture]

Any reference I could check to learn more about that ?

(I should probably actually learn galois theory before that but I'm just being curious )

[scarlet estuary]

+o(1) means the "relative error" approaches 0 as your inputs get larger

(informal definition)

[scarlet estuary]

dont have any recs but the broad name is "differentiable galois theory"

you quickly end up doing algebraic geometry though

so uh

[agile shore]

Ahhh I see!

[scarlet estuary]

not for the faint of heart

[agile shore]

Makes sense

[final pasture]

Alright thanks

[vestal snow]

There should be another equivalence here right?

$(v+v')\otimes w = v\otimes w + v'\otimes w$

[cloud walrus]

Finitely Many Bananas

[vocal wolf]

its already there

[vestal snow]

This only has the property about scalar multiplication

We would also need addition, right?

[scarlet estuary]

einstein convention

[vocal wolf]

einstein convension is used so its really $(\alpha_1 v_1 + ... \alpha_n v_n ...$

[cloud walrus]

Brian485

[scarlet estuary]

those are all sums

[vocal wolf]

yeah

[vestal snow]

I see

Thanks

Is this correct?

In terms of Einstein notation

[vocal wolf]

yes

[vestal snow]

Thanks

[vocal wolf]

what are your doubts?

[vestal snow]

Just the convention

I try to avoid Einstein notation so I thought I'd make sure I was using it correctly

[chilly ocean]

geometers hate him! click here to learn his trick now

[scarlet estuary]

and yet another new notation for differential geometry is invented.

i swear literally every student develops a unique one

[robust pollen]

Einstein convention is an absolutely blessing

[chilly ocean]

ok so I know if $B$ is a ring containing $A$, then $B\otimes_A A[x_1,\dots,x_n]\cong B[x_1,\dots,x_n]$. also if $M$ is any $A$-module, then $M\otimes_A A/I\cong M/IM$. I can't see how to combine these rules to get a rule for determining the tensor product $B\otimes_A A[x_1,\dots,x_n]/I$

[cloud walrus]

wren

[chilly ocean]

I is just an ideal in the corresponding ring both times it appears

I feel like it would be nice if $B\otimes_A A[x_1,\dots,x_n]/I\cong B[x_1,\dots,x_n]/IB$, but I am getting confused

[cloud walrus]

wren

[rustic crown]

I is an ideal of A[x1, ..., xn] right?

what is IB?

[chilly ocean]

ye

yes

IB is the product of I by all elements of B

[robust pollen]

Is B tensor I an ideal in B tensor A[x...]?

[chilly ocean]

i think so

yes

[robust pollen]

Can you find a surjection B tens A[x ..] -> B tens A[x...]/I?

[chilly ocean]

yes, just take the quotient of the second component

oh....

teehee

the kernel is B tensor I

yeah ok I see

tyty

I just got a bit confused

[next obsidian]

You need IB[x_1,\dots,x_n] in the bottom

but you can prove this by doing

[chilly ocean]

well the only way for IB to make sense is interpreting it as B tensor I

and B tensor I works on the bottom

[next obsidian]

$B\otimes_A A[X]/I = B\otimes_A A[X] \otimes_{A[X]} A[X]/I = B[X]\otimes_{A[X]} A[X]/I = B[X]/IB[X]$

[cloud walrus]

The Chmonkey

[next obsidian]

Where $X$ just means $x_1,\dots,x_n$

[cloud walrus]

The Chmonkey

[chilly ocean]

yeah

you just did some very nice trickery

[next obsidian]

This is really common, once you get some practice you'll see it immediately

Sneaking in a A (x)_A cuz it does nothing

[chilly ocean]

yeahhhh

well I assume it's common cause quotients of A[X] are just general rings that contain A

[next obsidian]

well

they might not contain A

the map might not be injective

[chilly ocean]

oh yeah

[next obsidian]

quotients of A[X] are finitely generated A-algebras

[chilly ocean]

yes

[next obsidian]

which you are correct are very common

if A is a field then yup they contain A :)

[chilly ocean]

hahahaha of course

[chilly ocean]

man I'm just staring at this

it just works so well

but either IB[X] or B tensor I are both perfectly fine ways of saying almost the same thing right

[next obsidian]

Frankly, I'm not totally certain

Because you can only tensor B with I over A

not over A[X]

Since B isn't a module over A[X]

I think IB[X] is just... nicer

Like if I is something like $x^2 - 1$ or $x^2$ basically generated by polynomials in $\mathbb{Z}[X]$ then $IB[X]$ will be the same ideal

[cloud walrus]

The Chmonkey

[next obsidian]

As in like it's generated by those same polynomials, just where we now multiply it by all polynomials in B[X]

[chilly ocean]

yeah

[next obsidian]

And like

[chilly ocean]

it is nicer

[next obsidian]

B (x) I

doesn't "exist" inside B[X]

it would have to be the image

[chilly ocean]

yeah

[next obsidian]

and I think the map won't be injective

you'd need some sorts of flatness

[chilly ocean]

but B (x) I is what naturally pops up when you do this

[next obsidian]

well...

no

the image of it is

Because you lose exactness on the left

B (x) I doesn't embed into B (x) A[X]

[chilly ocean]

oh yeah I remember that

you only get right exactness

[next obsidian]

yeah

[robust pollen]

Right, this works over fields but not over general rings :(

[next obsidian]

The image will end up being IB[X] tho

after you associate B (x) A[X] with B[X]

that's why I don't like modding out by B (x) I because it kinda doesn't make sense

[chilly ocean]

that's kind of mean of the modules to do that to me

[next obsidian]

tensor product is not terribly well-behaved but you get used to it

and learn how to become buddies with it

I like it a lot

[chilly ocean]

yeah I'm not yet buddies with it 😔

[next obsidian]

because it's useful and has that goldilocks level of badly behaved

it makes it interesting without just being trash

Like

B (x)_A A[[x]]

is not B[[x]]

that's weird to me

I think if B is finitely presented it's true

this has to do with the fact tensoring doesn't commute with infinite products (but it's true when the module you're tensoring the product by is finitely presented)

the fact taht B (x)_A A[x] = B[x] is because it commutes with infinite direct sums

[chilly ocean]

oh yeah

[robust pollen]

Can I envision A[[x]] just as a product of A over Z then?

[chilly ocean]

but isn't this just sums of modules

[next obsidian]

It is

Yeah

like A[x] is a countable direct sum of A

so that's why it becomes B[x]

[chilly ocean]

you lose the multiplicative strucutre

[next obsidian]

a countable direct sum of B

Oh well like the isomorphism respects the ring structure of B

this isn't immedaite but like

if you know what the map does

you can verify it

[chilly ocean]

yeah but it feels a bit forgetful

[next obsidian]

Yeah view it over N tho, it makes more sense. the i-th coordinate gives you the coefficient on x^i, you need product not direct sum cuz a power series can have infinitely many non-zero coefficients

[robust pollen]

Sorry, was thinking Laurent

[next obsidian]

Laurent series are weird

[chilly ocean]

why finitely generated?

you can have large polynomial rings in lots of variables

[next obsidian]

because you require that only finitely many non-zero negative coefficients

[next obsidian]

well if you want finitely many variables

[chilly ocean]

oh

[robust pollen]

Finitely generated as algebra

[chilly ocean]

ok

[next obsidian]

any A-algebra is a quotient of a "polynomial ring" in infinitely many variables

but if you want to be able to say A[x_1,\dots,x_n]

you need f.g.

[chilly ocean]

yeah

what resources are good to learn about tensor products in more detail

[next obsidian]

Umm

a commutative algebra book

there's notes by like

Keith Conrad online

Those are good too

establishes good base properties to build off of

besides that it's just experience

you use it and you get used to it

it acts unexpectedly and you remember how it can go wrong and learn to avoid that

[hidden haven]

chgonkey

[chilly ocean]

omg my brain slipped

[next obsidian]

Also to really get good with tensor product stuff you need to learn what Tor is

since it controls stuff about the failure of left-exactness

[chilly ocean]

yeah I don't know what a tor is

[robust pollen]

What is an ideal in an algebra?

[scarlet estuary]

https://en.wikipedia.org/wiki/Algebra_over_a_field#Subalgebras_and_ideals

[robust pollen]

I mean, conceptually

;rotate

One should also be able to define "polynomial algebras" in general monoidal categories with biproducts, right? Since a polynomial algebra seems to be nothing more than a special case of a "graded algebra object"

[median pawn]

Hey. Let $G$ and $H$ be groups, where $H$ is abelian. Also, let $\tilde G = G/[G,G]$ denote the abelianization of $G$. If $f:G\to H$ is a homomorphism, is it true that $f$ induces a homomorphism $\tilde f: \tilde G\to H$?

[cloud walrus]

Hausdorff

[rustic crown]

yee

a unique one

it's the universal property of abelianization

[median pawn]

how do you prove it tho

[rustic crown]

use universal property of quotients

[median pawn]

alright

[rustic crown]

f([G, G]) = [f(G), f(G)] = 0

[median pawn]

ahh okay! so kerf contains [G,G]

[rustic crown]

yep

[median pawn]

and by the univ property of quotients we are done

[rustic crown]

[hidden haven]

@median pawn unique one satisfying $\tilde f p = f$, not unique in general p is the quotient map

[cloud walrus]

Moldilocks ✓

[winter thorn]

can someone explain quotient groups to me?

[scarlet estuary]

take some equivalence relation

this "partitions" your groups into "classes" which are all related

so like, in the integers for example, congruence modulo 5 partitions the integers into 5 classes

those congruent to 0, those congruent to 1, ... those congruent to 4

[winter thorn]

and these partitions are themselves groups or something like that?

[scarlet estuary]

now construct a group whose elements are these classes

and whose operation is the parent operation of the group

[winter thorn]

seems easy to understand now

thanks

[scarlet estuary]

to finish the example

lets represent the entire class of integers congruent to n by [n]

there are 5 of these, [0], [1], [2], [3], and [4]

we couldve instead wrote [5], [1], [7], [-2], and [13] or whatever

but i think the former way is simpler

now our operation will just be the parent group operation

our parent group was Z under +

so [0] + [1] = [0+1] = [1]

[3] + [4] = [3+4] = [7] = [2]

(since 7 is congruent to 2 mod 5, hence [2] and [7] are the same equivalence class)

[3] + [3] = [6] = [1]

you probably get the idea

[winter thorn]

hmm so we are doing operations on these equivalence classes or is that the case only for this specific example?

[scarlet estuary]

always

if [a], [b] are equivalence classes, we define [a][b] in the quotient group by [a][b] = [ab]

where ab is computed from the operation of the parent group

its a good exercise to check that this is well-defined

[winter thorn]

ah I get it

thanks for explaining this to me

you did it well

[scarlet estuary]

now, a footnote

for notational convenience we often drop the []s

like the quotient group in my example is ℤ/5ℤ

the group of (equivalence classes of) integers modulo 5 under +

but we often write its members simply as 0, 1, 2, 3, 4

instead of [0], [1], etc

and say 3 + 2 = 5 = 0 (mod 5) or whatever

this is purely a notational consideration

so its whatever

but watch out for it

[winter thorn]

I will

[scarlet estuary]

the notation doesnt pose a problem since again

[ab] = [a][b]

so writing ab for the parent group and ab for the quotient group doesnt make a difference

ab in the quotient group is actually [a][b], i.e. [ab]

which by well-definedness is the equivalence class of, well, ab

so you can see we dont run into an issue

[winter thorn]

ye

[scarlet estuary]

so its just a formal note from the construction.

[winter thorn]

i got it