groups-rings-fields

[woven delta]

Anyway this discussion is silly

[latent anvil]

I've been thinking about nonassosciative stuff a lot this summer

But it's associative up to a big system of natural isomorphisms so it's still okay

[vestal snow]

finally finds something easy to understand in Atiyah MacDonald (primary decomposition)

The book: "In the modern treatment, [primary decomposition] is no longer a central tool in the theory"

[latent anvil]

lmao I could not understand that section

When I read that I was very relieved

[vestal snow]

Interesting, I thought it was much easier to understand than chapters 2 and 3

[stone fulcrum]

Ch 3 still kicking my ass

[vestal snow]

I feel ya

Though Chapter 2 was imo by far the most difficult

just because it introduced tensor products, exact sequences, and flatness

[elder valley]

maybe this is a silly question. if C is an affine curve, then you can take the projective closure of C by homogenizing to get a projective curve that C embeds into

so in other words, you get a map $\iota : C \rightarrow \overline{C}$ given by something like $\iota(X_1,\ldots, X_n) = (X_1,\ldots,X_n,1)$

[cloud walrus]

Auvera:

[elder valley]

now if you have a morphism of affine varieties $f : C_1 \rightarrow C_2$, my intuition says there should be a "closure" of f which takes the closure of $C_1$ to the closure of $C_2$ in a way that agrees with f

so in other words I would think that there exists an $\overline{f} : \overline{C}_1 \rightarrow \overline{C}_2$ such that you get a commutative diagram $\iota_2 f = \overline{f} \iota_1$

but to get \overline{f} do you just homogenize its components?

because that doesn't seem like it will give you a morphism

[cloud walrus]

Auvera:

Auvera:

[olive mirage]

any non constant morphism of irreducible curves gives an inclusion of their function fields, which then gives you a map on the nicest version of the curves.

(because the function field is associated with the normalization of the curve, so it gets rid of all the singularities)

[elder valley]

hmm. if i understand you're referring to this functor giving an equivalence of categories between curves with morphisms and fields with inclusions. and you're saying if i take my affine map, look at its corresponding map on function fields, and then look at the map of curves corresponding to that, the result is the projective version of the morphism that i'm looking for?

i thought that functor only applied to projective curves

[latent anvil]

@olive mirage why is the projective closure the same as the normalization?

If I close y^2 = x^2(x+1) to zy^2 = x^3 that's still singular at [0 : 0 : 1], right?

Also I don't think projective closure is functorial. I might be wrong though

[elder valley]

i'm not sure how normalization works. i'm guessing it smooths out the curve somehow

[latent anvil]

Actually iirc it's not an even an isomorphism invariant

Z(y - x^2) ≈ Z(y - x^3) ≈ A^1 but their projective closures Z(zy - x^2) and Z(z^2 y - x^3) in P^2 aren't isomorphic. I'm not entirely sure why but that's what the internet says....

[olive mirage]

it's not the same. But at least for my purposes, I don't want ugly signularities, which is what you get when you projectivize

so going curve --> function field --> nice curve is a process that works better for most applications. Most curves don't have nice models in P^2

but you're right that this does not preserve isomorphism, because if you do this process to your favorite nodal cubic, you'll get P^1

(also note that this fails miserably for higher dimensions)

(but the question was about curves specifically)

but to @elder valley normal varieties are smooth in codimension 1, which for curves just means smooth.

[elder valley]

ok, good to know

here's an example from this paper i'm looking at. it's looking at the twisted edwards curve $ax^2 + y^2 = 1 + dx^2y^2$ over some field F

[cloud walrus]

Auvera:

[elder valley]

they give this doubling formula in affine coordinates. how do you projectivize it?

[smoky cypress]

Is $\bR[e^{\frac{2i}{5}}]=\frac{\bR[x]}{\brk{x^5-1}}$?

[cloud walrus]

Whoever:

[solemn rain]

ye

F[a] = F[x]/(p(x)) where a is a root of p(x) no?

[oblique river]

no

x^5 - 1 isn't irreducible

[solemn rain]

fuck lmfao

[oblique river]

also presumably you mean 2*pi*i/5

[solemn rain]

yea yea mb

[oblique river]

haha it wasn't you who typed 2i/5

[solemn rain]

nah i think he meant the root

[stone fulcrum]

Yes that's pretty much correct mo

[solemn rain]

but my bad for not noticing the poly was reducible haha

[stone fulcrum]

That's 99% of cases you'd use it for

[solemn rain]

i cant remember why the pooly must be irreducible

is it because the idieal to be maximal?

hahaha yea kaynex

[oblique river]

you're allowed to quotient by non-irreducible polys

but you just won't get a field

[solemn rain]

yea

F[a] is a field tho so yea

yea i got it

ty

[smoky cypress]

Ah yes

I meant 2πi/5

And yeah I see

Ah a polynomial with real coefficients and of degree >2 is never irreducible

[stone fulcrum]

Yaya all real polys can be factored into quadratics or lower

Book showing a proof for that?

That's a cool one haha

[latent anvil]

the proof is the fundamental theorem of algebra

You can factor any real polynomial over C to get a product of linear terms

complex conjugate pairs have to match up so you're left with a product of real linear terms or real quadratics

[olive mirage]

I would like a proof that doesn't reference C, but I don't know one.

[latent anvil]

I would be surprised if you could give one

It immediately implies the fundamental theorem of algebra, right?

Well it immediately implies it for polynomials with real coefficients and that means C is an algebraic closure of R, from which it follows that C is algebraically closed

[lilac bough]

sorry for the probably stupid question but my prof told us that 2Z/6Z=Z/3Z, but i can't seem to figure out why

quotient groups where just introduced

[steady axle]

can you define a map from 2Z/6Z-> Z/3Z

@lilac bough

[lilac bough]

i don't really understand what 2Z/6Z is to begin with

[steady axle]

how do the elements in 2Z/6Z look like

[lilac bough]

that's what i mean i'm not sure

[steady axle]

If G is a group and H is its normal subgroup then how do elements of G/H look

[lilac bough]

set of cosets

[steady axle]

yes like gH right

[lilac bough]

oh hmm so 2Z is {2n:n\in Z}

[steady axle]

yes

now what happens if we quotient by 6Z

[lilac bough]

hmm i'm not sure

[steady axle]

yes like gH right
so in additive notation they will look like {2n+6Z : n \in Z} wont they?

think about this for some time

[lilac bough]

yes ok i got that

[steady axle]

now you know what equivalence classes are right

[lilac bough]

yes

[steady axle]

so how many elements are there in 2Z/6Z ?

write all of them

[lilac bough]

hmm no sorry i'm not really sure how

this is all pretty new

[steady axle]

i would suggest reread section on quotient groups and maybe revisit equivalence classes

then come back

[lilac bough]

ok thanks i'll do that ty

[vestal snow]

Are the number of terms in a minimal primary decomposition fixed?

I would assume so because the prime components are fixed and each prime component must associate with exactly one primary ideal in a minimal decomposition

[neat ginkgo]

if H and K are subgroups of G, and exactly 1 of them is a normal subgroup, how do i prove H(H U K) is a subgroup of G

[scenic sage]

@neat ginkgo prolly by this

$H(H \cup K) = {hk | h \in H, k \in H \cup K }$

[cloud walrus]

Commander Vimes:

[scenic sage]

which is the same as

${hk | h \in H, k \in H} \cup {hk | h \in H, k \in K}$

[cloud walrus]

Commander Vimes:

[scenic sage]

since H and K are both subgroups and exactly one of them is normal

HK = KH

and HH is obviously subgroup

[neat ginkgo]

ok HH is obv a subgroup regardless of it being normal

but is HK always a subgroup if one of them is normal

[scenic sage]

there is theorem that if H is subgroup of normalizer of K then HK is subgroup

and is K is normal its normalizer is the whole of G

so like then H obviously satisfies

[neat ginkgo]

oh

that explains it then

thanks!

[silk drift]

I am looking for people to self-study with for a 2nd semester Abstract Algebra course

Is this the right place to ask that sort of thing

[chilly ocean]

yea

[solemn rain]

what are you studying

konodioda

[silk drift]

Dummit and Foote

I have finished the first 7 chapters

Well not every problem but a good selection

It's online so I am open to other books

I am also studying Pfaffenberger for Analysis I/II

And am looking for a topology book

[chilly ocean]

see the #books-old and #book-recommendations

[silk drift]

Well I am trying to find a "study buddy"

My last one was only interested in algebra 1 essentially

[silk drift]

Does anyone see any issues with this

one sec i'm finding my proof

Given R commutative, noetherian, the zariski topology is discrete iff R is artinian. FOr the forward direction I have this: I currently have this chain of implications: Spec R discrete->Spec R Hausdorff->dim R=0->R artinian. I just wanted to ask if I am on the right track or did I go wrong somewhere.

Like is this how I "should" be thinking about it

[latent anvil]

I was thinking about this and I came up with a different proof, because I wasn't sure how you got dim 0 => artinian. Spec R is (quasi)compact and discrete so it's finite, so R has finitely many primes. The points are all affine open subsets and they form a disconnection of Spec R, so R is isomorphic to a product of rings R_p for each prime p of R. It suffices to show that if A is a commutative noetherian ring with a unique prime p, then A is artinian.

Clearly p is the nilradical, and p = (f1,...,fn) for some fi in A by noetherianness. Because the fi are in the nilradical, they're nilpotent, and so p^N = 0 for large enough N. We could finish by the structure theorem, but here's an explicit proof:

Consider the SES's of A-modules
0 -> P -> A -> A/p -> 0
0 -> p^2 -> p -> p/p^2 -> 0
...
0 -> p^N -> p^(N-1) -> p^(N-1)/p^N.
Since p^N = 0 is an artinian A-module, it suffices to show each p^i/p^(i+1) is. A-submodules of these are the same as A/p-subspaces, so it suffices to show that they're artinian (i.e. finite dimensional) as A/p vector spaces. But p^i is a finitely generated ideal since A is noetherian, and those same generators give a finite spanning set of p^i/p^(i+1) as an A/p vector space

I like this a little better (ignoring the structure theorem stuff) because there's some geometry to it: the assumptions imply that your space is really just finitely many points, each of which is "easy" to show is artinian

[magic hare]

Can anyone give me an example of an integral domain which is not an example of a atomic/factorization domain?

[latent anvil]

What's a factorization domain?

Every element can be written as a product of irreducibles but no uniqueness guarantee?

This is implied by noetherianness so your ring will look pretty bad

Maybe the subring of k(x, y) generated by k, x, and y/x^n for all n > 1?

It's not clear to me that y/x can be written as a product of irreducibles

[magic hare]

Lol I was writing “how come that x wouldn’t just be irreducible”

Sorry

[latent anvil]

yeah haha mb

I edited to swap the variables at some point

This is the example I know of a non noetherian ring sitting inside a noetherian one

[magic hare]

Lactually same question but for y/x, why wouldn’t that just make it an irreducible

Sorry I have not studied too much algebra

[latent anvil]

You can write it as x * y/x^2

And those should both not be units, I think

[magic hare]

Oh ho ho, stretchin the ol noodle.

Thanks! I’ll look into this

[latent anvil]

There are much simpler examples of integral domains which aren't unique factorization domains, btw

[magic hare]

Oh yes this is much easier for me to conceive of

[latent anvil]

For example k[x, y]/(y^2 - x^3) or Z[sqrt(-5)]

[magic hare]

Without uniqueness though I had a naive thought that if you chose any element in an integral domain and it could not be written as the product of irreducibles, then that would imply it is itself an irreducible and so any integral domain should be trivially a factorization domain without uniqueness, but I think I see where I was going wrong now

I need to bone up on my algebra

[latent anvil]

haha yeah, "not a product of irreducibles" doesn't imply "irreducible"

This almost works though, I think

Well not really

But you should look at the proof of noetherian => factorizations exist

Algebra good

[magic hare]

I like it a lot

I realized that it’s a lot of impenetrable jargon to talk about times tables and it made it a little easier to think about

But I did notice your proof, need to go through it still (it’s late) but I found it a funny coincidence because noetherian rings came up a lot when I was trying to wrap my head around it

[latent anvil]

I'm not really sure what you mean about times tables

[magic hare]

It’s just a bad joke

[chilly ocean]

miwazaki koname chan

[kindred mist]

Can someone explain to me how we get the statements there (that I underlined in red)?

(i.e. how do we know such x_i exist?)

[elder valley]

@kindred mist looks like n applications of the inductive hypothesis

for $1 \leq i \leq n$ define $I_i = \lbrace 1,2,\ldots,n \rbrace - \lbrace i \rbrace$. apply inductive hypothesis on the ideals $\lbrace p_j : j \in I_i \rbrace$

[cloud walrus]

Auvera:

[kindred mist]

wouldn't that be to say that $a \subseteq U_{j \in I_i} p_j$ implies that $a \subseteq p_r$ for some r in $I_j$? but if a is contained in $U_{j \in I_i} p_j$ then how do we get an $x_i$ that's in a but not that union which a is contained in?

[cloud walrus]

𝓒𝓸𝓾𝓷𝓽𝓪𝓫𝓵𝓮:

[elder valley]

they're using the contrapositive statement they write in line 2 of the proof

[kindred mist]

Ok I get it now, tysm!

[grave leaf]

Hello

can i ask a question 🙂

[slow egret]

Ask away

[grave leaf]

How do i solve these type of questions?

[solemn rain]

what have you tried

what did you think of

@grave leaf

[grave leaf]

am thinking that to prove that F is a field, i need to first prove that its commutative and all elements except 0 is unit

to prove that its commutative
Fx + gx = gx + Fx but how can i prove it

if i dont exactly know what f and g are

[round igloo]

when you are in vertex form and have an certain value as a? how do you graph it

[solemn rain]

@round igloo wrong channel

function addition is commutative

the real question is : is this closed

if i have two continuous functions

is their sum continuous

what about their product

thats what one would think of ig

what do you think

[grave leaf]

you mean, if two cont. functions are closed under closure property of multiplication

[solemn rain]

what

i dont understand

my question is

if i have two continuous functions say f and g

is f+g contiinuous

is fg continuous

this is closure

closed under the operations

then check the other axioms

and i think there is a typo

https://media.discordapp.net/attachments/496784958430380033/730500408715378898/unknown.png

when defining the other operation

does it mean f(x)*g(x)?

[grave leaf]

it says F(g)*g(x)

[solemn rain]

what does f(g) mean if g is a function

[grave leaf]

that independent variable of function f is function g

am i ryt?

[solemn rain]

yea ig

it prob means composite

so the operationw ould be f(g(x))*g(x)

is this closed

if f and g are continuous

is f(g(x))*g(x) continuous

once ur done with closure

check other axioms

does there exist an identity

is it commutative

is every element invertible

zero divisors?

[grave leaf]

one quick question, y=1 is continuous?

[solemn rain]

yes

[grave leaf]

i guess so ryt?

ok

[solemn rain]

now go ahead

answer these questions

[grave leaf]

i'll get back to u in a minute

[solemn rain]

cool

[grave leaf]

@solemn rain F under operator + is Albenian group
but i am struggling to check if its elements are divisor of zero
how can i proceed with that?

[solemn rain]

albenian group

[grave leaf]

abelian

xD

am sorry, i am not strong in this subject

[solemn rain]

try thinking of a counterexample

bro im just joking haha

think of acounterexample

[grave leaf]

its cool

[solemn rain]

to the claim that this is an integral domain

[grave leaf]

this is what i know about divisors of zero

[solemn rain]

ie

find two nonzero continuous functions , f and g

such that f(x) * g(x)=0

[grave leaf]

ahhh that makes sense

[solemn rain]

'*' denotes the mulitplicative operation

multiplicative*

as defined

[grave leaf]

what if
fx = x-2
gx = x
so f(g) is still x-2

does that prove the zero divisor?

if x = 2

since x belongs to [1,2]

@solemn rain

[solemn rain]

yea

its

the operation is f(g)*g(x)

but yea

[grave leaf]

so when fx = x-2
and gx = x
f(g(x)).g(x) = 0
at x=2
but fx is 0 here, i dont understand

[solemn rain]

what

[grave leaf]

f is a zero element here

[solemn rain]

yea

[grave leaf]

so that doesnt not prove divisor of zero case

[solemn rain]

yea yea lmfao

m b

i came across this problem once and i couldnt find the counterexample

[grave leaf]

tis confusion man

i think its not a zero divisor

[solemn rain]

the zero function is the function h(x) such that h(x) = 0 for all x

not what you said

if ur still confused

define your function piece-wisedd

maybe htis makes more sense

@grave leaf get me?

[grave leaf]

the zero function is the function h(x) such that h(x) = 0 for all x
@solemn rain if x != 0 ryt?

[solemn rain]

the zero function is the function which is 0 everywhere

so define af untion

[grave leaf]

H is 0 but x gotta be non 0

[solemn rain]

that is piece-wis'ly' 0 on an interval

and a nonzero on the rest of the interval

f(x) = { x+2 if x is in [1,1.5]

{ 0 if x is in [1.5,2 }

g(x) = 0 if x is in [1,1.5]

-x+2 if x is in [1.5,2]

i htink this should work

got it? @grave leaf

[grave leaf]

i am afraid that i am confused as hell

your f(x) is continuous?

same as with g(x)

i dont feel it

[solemn rain]

check

[grave leaf]

it isnt man, i ploted it roughly

its not continuous

i dont think that F has zero divisors

also i think F failes under unity w.r.t +

[solemn rain]

really

[grave leaf]

i think answer is

https://media.discordapp.net/attachments/496784958430380033/730500408715378898/unknown.png
@solemn rain 2

since its not unity then clearly not field, but also does not have zero divisors

[solemn rain]

the 0 function is continuous

whydo you see unity fails

under addition

[grave leaf]

correct me if i wrong, because am really new

(f+g)(x) . (f+g)(x)^-1 = I

=> (f(x) + g(x)) . (f(x) +g(x))^-1
=> (f(x) + g(x)). (f(x)-1 + g(x)-1) am omitting ^
=>(f(x) + g(x)). (1/f(x) + 1/g(x))

this give (f(x) + g(x))^2 / f(x).g(x) which is not unity

please correct me

@solemn rain

[solemn rain]

what are you doing

what is 'unity'

[grave leaf]

idk

[solemn rain]

im kinda busy now

unity is the existence of an identity

f(x) + I = f(x) for all f(x)

where here is 0

[grave leaf]

unity means a.b = b.a = 1

[solemn rain]

what

this is invertibility

this is not necessarily true

only in a field

invertibility means for every a there exists b such that ab=ba=1

if f(x) is continujous

1/f(x) is continuous

^ prove it first

[solemn rain]

okay fine idc about naming shit

it doesnt matter

if f(x) is continujous
[10:03 PM]
1/f(x) is continuous

[grave leaf]

okay i will solve this from scratch

thanks for the help man

its alot

[solemn rain]

check closure

check invertibility

check integral domain

check commutativity

[grave leaf]

inverse exist

if u mean that by invertibility

such that i get I

[solemn rain]

ok

is it an integral domain

[grave leaf]

for that i need to prove that it doesnt not have any zero divisors

which is the actual bottle neck

[solemn rain]

yea generally

[elder valley]

almost certainly that multiplication is supposed to be defined as (f * g)(x) = f(x) * g(x). it's a typo

otherwise the left hand side is a number and right hand side is a function

[grave leaf]

i am new to this topic and i am literally blown by this question

i feel if i am able to solve this, then i will be understnding rings topic

[elder valley]

i think it will give you a good sense of what a ring is, yes

[grave leaf]

problem is, this question is way too complex to get a google search of

and i dont have any mentor for this topic

[elder valley]

which part are you confused about

[grave leaf]

how to prove that functions f,g does not have any zero divisors

if two functions are continuous how can their product be 0

under x = [1,2]

[elder valley]

zero divisors in this ring would be two continuous functions f and g defined on [1,2] such that f*g=0 as functions

[grave leaf]

i dont think its possible but idk how to prove it

[elder valley]

so meaning f(x)*g(x)=0 for all x in [1,2]

[grave leaf]

how can you go about proving that

if both f and g of x are 1-x?

but that makes f,g =0 at x=1

[dawn kiln]

wait, aren't there examples of that?

[elder valley]

well those are real numbers, so the only thing you can really say based on that statement is that f(x) = 0 or g(x)=0 for each x

[grave leaf]

if f,g are 0 then they are not zero divisors isnt it

integral domain proved

[elder valley]

the function 0 is not a zero divisor because zero divisors need to be nonzero

[dawn kiln]

what about this

those are both continuous

[elder valley]

if you're having trouble proving something it may be a sign that it's not true

[grave leaf]

what about this
@dawn kiln if u think about it,
say red is f, blue is g

i cant take values of f(x) and g(x) when it is at x axis

[dawn kiln]

what do you mean "can't take values"

[grave leaf]

and when they are not at x-axis, their product is not 0

[dawn kiln]

at every point along the interval [1, 2], their product is 0

[grave leaf]

because they are 0 at x axis,

[dawn kiln]

those are functions that take in an x value

[grave leaf]

to prove that f,g are zero divisors they must be non zero

[dawn kiln]

f(1.2) = 0, g(1.2) = -0.6

[grave leaf]

to prove that f,g are zero divisors they must be non zero
@grave leaf this

[dawn kiln]

look carefully at what the definition of 0 is here

it's not saying that the function is 0 for some portion of its interval

the function has to be the "0 function", meaning f(x) = 0 for all points in [1, 2]

there's only one of those

[grave leaf]

zero divisors
there exist some ring R for which
ab = 0, where a,b belongs to R and a,b is not 0

[dawn kiln]

here, 0 is the (unique) function h(x) = 0

my examples for red/blue (or f and g) are non-zero

because there is at least one point in [1, 2] for which f(x) and g(x) =/= 0

[grave leaf]

what about this
@dawn kiln my point is at any given point of x belonging to [1,2]
either f or g is 0
which doesnt holds the theorem

[dawn kiln]

yes it does

we don't care about the value of f and g at various points, we care about what function they actually are

it doesn't matter that f happens to be 0 for a bit of it, that's irrelevant

f has to be zero for the whole interval for it to be the zero element

[grave leaf]

i am having hard time understanding u, because of the zero divisor theorem

[elder valley]

f=g if and only if f(x)=g(x) for all x

so because f is nonzero for some x, the function f is different from the 0 function

[dawn kiln]

ok, so the definition of a zero divisor is: an element a in my ring, such that for some element b in the ring, ab = 0, but neither a nor b are 0

[dawn kiln]

here, 0 is a function, not a number

[grave leaf]

sorry i forgot to crop

[dawn kiln]

your argument is like saying that A and B aren't zero divisors, because in all four positions in each matrix, there is a 0 in one or the other (which is not the definition)

[grave leaf]

A,B are not 0 but A.B = 0

so A,B are zero divisors

[dawn kiln]

yep

[grave leaf]

but in your graph
f,g are 0 at any point of interval

[dawn kiln]

and in the same way, the blue/red functions i posted have a couple of zeros, but neither of them are the 0 function

[grave leaf]

theorem says
both a and b shouldnt be 0

[dawn kiln]

neither f nor g are zero though...

[elder valley]

the red function is only 0 on [1,1.5]

[grave leaf]

neither f nor g are zero though...
@dawn kiln pick any x
either f is zero or g is zero
or both is zero at x=1.5

[dawn kiln]

ok, you're getting confused here

between definitions of 0

you're thinking of values of the function

[grave leaf]

yes

[dawn kiln]

values of functions are not elements in our ring

so they don't concern us

when i say 0, i mean the function, which i'll write as 0(x) = 0

the 0 on the left is a function, the 0 on the right is a value

now, im saying because neither f nor g are 0(x), the definition of a zero-divisor applies

forget about the values

they satisfy f(x)g(x) = 0(x) (forall x in [1, 2])

[grave leaf]

if that is true sir then why does value of elements in that matrix example matters?

[dawn kiln]

they don't

just like the individual elements in the matrix aren't important, the individual values of the function aren't either

[grave leaf]

am having hard time digesting this

forget about the values
@dawn kiln

i dont understand why 😦

[dawn kiln]

ok, let me put it this way one final time - if our ring contains two elements (ie. functions) f(x) and g(x), such that f(x)g(x) = 0(x), but f(x) isn't 0(x) and g(x) isn't 0(x)

then it contains zero divisors

do you agree that f and g in the example i gave are not equal to 0(x)

[grave leaf]

why are you writing it as 0(x) instead of 0? i read ur above mesg as well thats why am asking

[dawn kiln]

im writing 0(x) to emphasise that it's a function

a function that takes in an x value and returns 0

(i appreciate for anyone reading this that im playing a bit fast and loose with f vs f(x) btw lol)

[elder valley]

there are two kinds of 0. there is the real number 0, and there is the function zero defined to be zero(x)=0 for all x. we write zero as 0 also

[dawn kiln]

^

[grave leaf]

if a zero function returns 0 all the time, i am afraid to ask this, why bother mentioning it as a function

[elder valley]

because the ring elements are functions, not numbers

[dawn kiln]

because it still counts as a function, even if it's boring

[grave leaf]

i am trying to make sense in my brain, i dont mean to bother you guys

[dawn kiln]

yeh it's alright

[elder valley]

the number 0 is not an element of your ring F. but the function zero:[1,2] -> R is

[grave leaf]

i am getting it now

so if F is set of functions and f belongs to F
f is also an element of F ryt
so does 0

isnt it

[dawn kiln]

yes, if you mean 0 as a function, not a number

[grave leaf]

what i understand from u guys is, that 0 is an element but not a function so it doesnt matter for our F set

[dawn kiln]

no no no

0 is a function

[elder valley]

it's exactly the same as 0 as a number versus the zero matrix

[dawn kiln]

^

[grave leaf]

this makes it more confusing, dont you think?

[elder valley]

it's best to distinguish them with different symbols, otherwise you'll get confused

say "0" is the number and "zero" is the function

[dawn kiln]

thing is, is abstract algebra tends to relax with notation

[elder valley]

yes, but when you first learn it it's best to be careful with these things

[dawn kiln]

yep

you'll get used to the idea of 0 and 1 representing different ideas depending on the group/ring/field you're working with

[grave leaf]

so our set F hase zero divisors

[dawn kiln]

yes

[grave leaf]

therefore its neither field

not an integral domain

[dawn kiln]

unfortunately not

[grave leaf]

F is not commutative

because of multiplier operator

this makes it a non commutative ring

D is the answer isnt it

why did i waste my time if F is not even commutative
A,B,C filters out right away

[dawn kiln]

hold up

not commutative?

[grave leaf]

you see (f o g)(x) != (g o f)(x)

[dawn kiln]

why not

[grave leaf]

f(g(x)).g(x) != g(f(x)).f(x)

[dawn kiln]

what

where'd you get that

[grave leaf]

its given that (f o g)(x) = f(g).g(x)

[dawn kiln]

yes

[grave leaf]

thats why, f(g) != g(f)

[dawn kiln]

bro

wait

lmao i didn't even see that it was f(g) lmao, i apologise

but yeh

[elder valley]

it's certainly a typo

[grave leaf]

i dont think it is

[elder valley]

is it from a book?

[grave leaf]

no its not

suppose its a type, then this is indeed commutative

and answer would be C in that case

[elder valley]

yeah

[grave leaf]

where should i study abstract algebra

do you guys have any source

i need to study ideals and subrings

thats left

[elder valley]

depends what methods of learning are effective for you i suppose

could try youtube or just reading. i'd definitely suggest doing lots of practice problems and discussing them as we did above

[dawn kiln]

personally don't rate youtube

http://abstract.pugetsound.edu/download/aata-20170805.pdf this is a decent textbook that i used

but i would suggest you spend a decent chunk of time on the basics

since rings get complex very quickly

[grave leaf]

its pretty depressing because group theory is a vast topic but i need to prepare other topics as well

because in my exam atmost 2 questions will come from group theory

[dawn kiln]

i personally hate courses that shoehorn in rings and fields before giving groups a really solid treatment

[grave leaf]

this makes studying depressing because i wont enjoy the subject

i gotta read a whole textbook for 4% of questions that will appear in my exam

[dawn kiln]

the textbook i linked gives a solid 180 pages dedicated to nothing but groups

[grave leaf]

i hate education system

[dawn kiln]

yeah that's annoying lol

do you have a syllabus on you?

[grave leaf]

you wont believe how abstract it is

lemme show u

i study basic Group theory then comes question from premutation groups

[dawn kiln]

permutations and symmetry is not even a big area of group theory, it's pretty much the biggest area

[grave leaf]

am gonna read that book of yours, not sure if i will be making notes because i dont have much time

[dawn kiln]

https://homepages.warwick.ac.uk/~maseap/teaching/aa/aanotes.pdf

this is a set of lecture notes from warwick

covers much less stuff, but might be more suitable for how much your course covers

and is written in a much more appetising way

[grave leaf]

thanks man, i'll read it ^

[dawn kiln]

idk how your year is structured, but definitely worth looking at some more groups stuff

[grave leaf]

all of this is nothing but generalized form of maths which is kinda hard to read

[dawn kiln]

that's the point of abstract algebra ;)

[grave leaf]

idk how your year is structured, but definitely worth looking at some more groups stuff
@dawn kiln this is an enterance test not semester exam

[dawn kiln]

the second link is a more gentle introduction to the area

[grave leaf]

this is why syllabus is huge

[dawn kiln]

wait for what

like a university?

[grave leaf]

yes

[dawn kiln]

which one, if you don't mind me asking?

[grave leaf]

i'll tell u in personal message

[thorn delta]

let H and K be subgroups of G and suppose that HK = KH. I know we have that HK is a subgroup, but we also always have that H and K are normal in HK, right?
K(HK) = K(KH) = KH = HK = (HK)K
and similarly for H. I'm not missing anything, am I?

[latent anvil]

@thorn delta looks good to me. I'd never thought of this before

[thorn delta]

yea, lang seems to be using it here. He never really mentioned why we can form HK/K, so just making sure lol

[elder valley]

@thorn delta it's 2 pages before that

[thorn delta]

dang, i forgot about that. wait but he doesn't say that K is normal in KH there. Since KH = HK, doesn't it just follow from symmetry tho?

[elder valley]

he just swapped the roles of H and K

[thorn delta]

ik that.... I am claiming further that both subgroups should be normal in KH. Lang doesn't mention it, so i feel like I am missing something.

i.e. H is normal in KH since H(KH) = H(HK) = HK = KH = (KH)H.
I don't see any reason why you can't repeat the exact same proof with K instead of H.

[latent anvil]

K is a subgroup of N_H iff H is a subgroup of N_K iff HK = KH right?

[thorn delta]

wait a second... showing K(KH) = (KH)K is not the same as showing K is normal in HK 🤦‍♂️ . The correct reasoning is:

Suppose K is contained in the normalizer of H. Then kH = Hk (K is normal in H) so it follows that KH = HK. Now, H is normal in KH because (kh)H = k(hH) = k(Hh) = (kH)h = (Hk)h = H(kh).

Now, you can't repeat this proof to show that K is normal in HK because H is not necessarily (afaik) normal in K.

[latent anvil]

you're right, I'm sorry for not pointing it out earlier. You get k h k^(-1) = h k' k^(-1) but maybe k = k' so you don't end up in H

And the thing I said about normalizers is wrong, take any semidirect product I think

[bleak abyss]

Alright fuckers

Time for SL_2(Z)

Specifically I never showed the map SL_2(Z)->SL_2(Z/NZ) is onto so let's do that

[solemn rain]

what about SL_2(Z)

[bleak abyss]

Specifically I never showed the map SL_2(Z)->SL_2(Z/NZ) is onto so let's do that
@bleak abyss this

[cloud walrus]

Daminark:

[latent anvil]

dami why isn't that obvious

oh I see, because the determinant is only one mod n

Carry on

[solemn rain]

how is this map defined

[bleak abyss]

Reduce the entries of the matrix

[solemn rain]

ok

[stone fulcrum]

Does that always map into SL2(Z/NZ)?

[latent anvil]

yes

[bleak abyss]

Yeah determinant is a polynomial

[latent anvil]

you have a ring homomorphism

Determinant is a polynomial

[stone fulcrum]

Gotchya yes okay I see it

[solemn rain]

idk why isnt this obvious

whats with the determinant :d

[bleak abyss]

Here's the problem

If you just try to do the naive thing

Pick a representative of a,b,c,d

ad-bc is 1................................. mod N

Doesn't mean it's 1

To be in SL_2(Z) you gotta be 1

[solemn rain]

lmfaaao

yea

[bleak abyss]

Okay so Keith Conrad here is saying something and I'm not too sure

So I'll talk it out

[latent anvil]

what happens if you just expand (a + pN)(d + qN) - (b + rN)(c + sN)?

[solemn rain]

yea maybe something cool happens

[cloud walrus]

Daminark:

[latent anvil]

hmm

sounds possibly maybe?

[bleak abyss]

🤷

[latent anvil]

You want xa + y(b + kN) = 1 right? Maybe you choose use the equation saying your matrix is determinant 1 to find x, y, k?

or maybe it's simpler than that idk

[latent anvil]

Okay yeah I see it

let ad - bc = 1 + kN

nvm I'm being dumb

oof

yeah I don't see the CRT thing, sorry

you want something which is b mod N and 1 mod a I guess?

[bleak abyss]

Not necessarily 1 mod a, just a unit really

[latent anvil]

yeah that's true

[bleak abyss]

I guess okay when a = 0 we're fucked already and have to do something else

https://people.reed.edu/~jerry/332/mfms.pdf

Okay here are some problems let's do these

Exercise 1 is lol but

So obv gcd(c,d,N)=1 because 1 = ad - bc - kN

[latent anvil]

right

[bleak abyss]

I'll post here so no tab switching

[latent anvil]

ty I couldn't open that pdf for some reason

[cloud walrus]

Daminark:

[bleak abyss]

If N was 0 mod p then p\mid gcd(c,d,N). And obviously t only involves primes that d excludes

So gg

So yeah d' is coprime with c

And this kind of thing works with Conrad's argument too

And his computation just settles it

So a=0 now

[bleak abyss]

Okay back and uh a = 0 actually is bothersome

[cloud walrus]

Daminark:

[bleak abyss]

Oh wait right we can choose a to be a multiple of N

I was like wait do b and c have to be \pm 1?!

But a needn't be 0 in Z

Well I guess either a isn't 0 or c isn't 0

So if a is 0, choose d' coprime to c which isn't 0. Then cx + d'y = 1. Then a' = a + y(1-(ad' - bc)) and b' = b + x(1-(ad'-bc))

[cloud walrus]

Daminark:

[bleak abyss]

Okay woo

And okay I should compute the size of SL(2,Z/NZ) but I'll do that later for now I'd rather make some progress in the book

[latent anvil]

Let S be a graded, commutative ring

[cloud walrus]

shamrock:

[latent anvil]

Is $S_0 \oplus \bigoplus_{d > k} S_d$ a subring for any k?

[cloud walrus]

shamrock:

[latent anvil]

I think it is but I've never seen anyone talk about it so I'm a little suspicious...

In fact you should be able to get a subring by summing over any subset of N closed under addition and containing 0 (i.e. any submonoid), right?

[vestal snow]

Can I skip the exercises of chapter 4 (primary decomposition) of Atiyah-MacDonald?

[steady axle]

you can ofc

[vestal snow]

The text says that primary decomposition isn't a central tool in the modern theory

[steady axle]

what you want to know is should you

[vestal snow]

Yeah my bad

[steady axle]

was just messing

[vestal snow]

@chilly ocean do you still have all hw problems that you were assigned from that book?

I would really appreciate if you could share it with me

I'm thinking of doing those when I first read through the text, and doing the rest after

Thanks much

I'll probably be done by it by then

I'm trying to get it done by the end of summer

For sure

[latent anvil]

@vestal snow I had a similar goal in winter and asked my prof about this

Whether I could just skip that chapter

He said it would be fine, primary decomposition is worth learning about at some point but not critical on a first pass

Stuff about integral extensions is way more important

[bleak abyss]

I've heard that Atiyah-Macdonald's treatment of the stuff is apparently outdated

And that "associated primes" are more a thing nowadays

[latent anvil]

I've plugged it before and I'll plug it again: undergraduate commutative algebra by reid is great and treats associated primes (iirc)

[olive mirage]

I have never found commutative algebra especially useful without the accompanying geometry.

[So as a result, I find Eisenbud the most enlightening to read. I only use Atiyah Macdonald when I actively don't want to learn commutative algebra, and just want to see that the theorem I want is true.]

[But boy is it great for taht]

[bleak abyss]

Alright lemme just compute the order of SL(2,Z/NZ)

Does CRT apply here? To split into prime powers?

[olive mirage]

Yes

that problem is in my algebra packet, haha

[bleak abyss]

Oh wait a sec

Nvm for a sec I thought that was a dox lol

https://people.reed.edu/~jerry/332/mfms.pdf

This came up earlier in this chat

[olive mirage]

I'm not super public with my identity, but it is no well kept secret 😛

haha yeah, I was gonna say, or like, the world's easiest googling, or...

[bleak abyss]

That is also true. Yeah on my end... anyone who knows me immediately recognizes the username

Because D(Amin)ark

[bleak abyss]

Alright back so now let's compute the order of SL(2,Z/p^nZ)

This might be induction. Like if n=1 we can just say it's |GL(2,Z/pZ)|/(p-1)

And the order of GL(2,Z/pZ) is (p^2 - 1)(p^2 - p)

So the order of SL(2,Z/pZ) is p(p+1)(p-1) = p^3 - p = p^3(1-1/p^2)

And we know SL(2,Z/p^nZ) surjects onto SL(2,Z/p^{n-1}Z)

Because we already proved that even SL(2,Z) does so lol

Just gotta compute the kernel of that map

[cloud walrus]

Daminark:

Daminark:

[solemn rain]

is there anything that i can say about an element in a matrix group from the determinant

[cloud walrus]

Daminark:

Daminark:

Daminark:

[bleak abyss]

Woo

[opaque granite]

any book recommendations for a first group theory book?

[bleak abyss]

A general algebra book is probably what you want. Artin, Jacobson, or D&F

[scarlet estuary]

controversial topic

https://discordapp.com/channels/268882317391429632/716264872018706443/718931426346795099

[bleak abyss]

I am filled with joy by the popularity of my algebra books guide

[opaque granite]

which one do you like

[bleak abyss]

How's your linear algebra?

[opaque granite]

elementary

bad

[bleak abyss]

Artin

Artin basically does LA from the ground up

Along with the algebra

[opaque granite]

sick

[bleak abyss]

(And he does it more theoretically which is useful)

[opaque granite]

is there a free pdf online?

[bleak abyss]

I think the best copy is probably on library genesis

[opaque granite]

https://www.amazon.ca/Algebra-2nd-Michael-Artin/dp/0132413779 i this the book?

[bleak abyss]

Yup

[opaque granite]

thx mate

[chilly ocean]

i like that $1000 new copy

[elder valley]

what's the reasoning for Lang being "king"? i've never looked at it, just wondering what makes it so good

[thorn delta]

I think it has to do with the sheer amount of information in the book. Kind of like the Rudin trilogy, I think it is sort of canonized as the "gold standard" for literature on algebra.

[scarlet estuary]

yeah its coverage is immense and very clean

not the best pedagogically perhaps

but a fantastic reference text and definitely "the old standard"

it's also one of the first textbooks that frames undergraduate algebra in a truly "modern" way

and, well, it's written by fucking lang himself

so it makes sense that it became the standard

[bleak abyss]

Is Lang a research hotshot? Or do you just mean like, because he's a bigshot author?

[latent anvil]

Later in his life, Lang was an HIV/AIDS denialist. He claimed that HIV had not been proven to cause AIDS and protested Yale's research into HIV/AIDS.
the more you know!

I was just looking at his Wikipedia page

[bleak abyss]

Since I feel like only his Algebra book is as big as it is

[latent anvil]

His number theory book is big I think

[bleak abyss]

Maybe Algebraic Number Theory

Sniped

Also I guess mayyyyyyybe complex analysis is like, in the running

Though it seems strictly less common than like, S&S and Ahlfors

[scarlet estuary]

i never understood hiv/aids denialism

i know this isnt really #groups-rings-fields but like

whats the motivation

is it just some form of homophobia that i'm too straight to understand the context of

i can understand homophobes downplaying AIDS, but just saying its not caused by HIV at all seems... like a really weird angle

i know that some of them use it to promote alternative medicine (i.e. grift people) but i can't imagine lang having that motivation

[opaque granite]

Guys this is a typo right?

I'm not just horrendously stupid or something right?

[scarlet estuary]

which part

[opaque granite]

The matrix multiplication

[woven delta]

[2x1][1x3][3x2]?

[opaque granite]

What

[scarlet estuary]

,w {{1},{2}}{{1, 0, 1}}{{2,0},{1,1},{0,1}}

[opaque granite]

I'm sry i don't understand

101

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Namington#1983

[opaque granite]

Yeahhh

[woven delta]

I stated the dimensions of the matrix

[opaque granite]

Yeah it's uhh not correct right

[woven delta]

I assumed your problem was with the dimensions

[opaque granite]

O sry

I see

[woven delta]

Guess not

[scarlet estuary]

anyway uh

the numbers they get are wrong

but the principle is correct

[opaque granite]

Ok

[scarlet estuary]

(AB)C = A(BC) for matrices

[opaque granite]

Thx

[shy bluff]

wait why is this true?

I see that we get that (x+1)^2 = x^2 + 1

but why does that mean that it's not prime?

Because you can factor it?

[scarlet estuary]

yes

[shy bluff]

Ah

So any time that you have a polynomial and it's factorable in a ring, that tells you that that polynomial is not prime in that ring?

[stone fulcrum]

Is that not irreducible?

[scarlet estuary]

whenever you have a factorable element

we call p "prime" if p|ab implies p|a or p|b; but if you write ab = (x+1)^2 then p doesnt necessarily divide a or b

we could have a = b = x+1

which p does not divide

in UFDs, primes and irreducibles coincide

[shy bluff]

Uhhh I don't think we're there yet, we just introduced what a prime ideal is

[stone fulcrum]

Ah right, and Z2[x] is a PID and UFD

[scarlet estuary]

theres an equivalent definition from prime ideals

but basically the idea of prime elements of a ring is to behave "similarly" to prime numbers in the integers

(there's an argumetn that irreducible elements do this better but thats a side-tangent)

if you can factor something, and factorization is unique, that means it's not really behaving like a prime number, is it?

since it can be factored

[shy bluff]

ayea

ok I think I get it

[bleak abyss]

Alright lemme find the index of the subgroup of SL(2,Z/NZ) consisting of upper triangular matrices

Well we can find the size

Because it's just

Top left corner is a unit in Z/NZ

[latent anvil]

[bleak abyss]

And then top right is anything in Z/NZ

Bottom row is then determined lmao

Y u thonk me shamrock

[latent anvil]

I thought "well we can find the size" was funny for some reason

big think

[bleak abyss]

Oh I thought you meant I was wrong or something

And I'm like

[latent anvil]

Oh no it looks right to me

nφ(n)

[bleak abyss]

Yup

[latent anvil]

actually, is it some kind of weird semidirect product?

Not sure

[bleak abyss]

Could be but I just need the size atm

[latent anvil]

yee sorry for interrupting

[bleak abyss]

Uh oh

[old hollow]

Why is g^|G| = 1 (mod n)

[latent anvil]

well uh

what do you mean by mod n?

[old hollow]

I mean

[latent anvil]

g^|G| is some element of a group G

Right?

[old hollow]

The multiplicative group of integers mod n, g is a generator/primitive root of n

[latent anvil]

ah

I see

So this isn't really something about the multiplicative group of the integers mod n

I mean it's known as fermat's little theorem in elementary number theory

But it's really a statement in group theory

[old hollow]

I looked up a proof of fermat's little theorem and lagrange's theorem was mentioned, but idk how lagranges theorem relates to exponentiation

[latent anvil]

Oh you know Lagrange's theorem?

Given an element g, consider the cyclic subgroup generated by g

[old hollow]

Isn't lagrange's theorem like the order of h divides the order of g

And h is generated by an element in g

[latent anvil]

If g is an element of some group G, define <g> = {1,g,g^(-1), g^2, g^(-2),...}

This is a subgroup of G

Does that make sense?

[old hollow]

Yeah

[shy bluff]

Is Z a maximal ideal of itself?

[latent anvil]

No liria, maximal ideals have to proper (i.e. not the whole thing)

[scenic sage]

Isn't lagrange's theorem like the order of h divides the order of g
@old hollow yep

[latent anvil]

okay so @old hollow, the order of <g> divides the order of G

[scenic sage]

that order of any subgroup of G divides |G|

[latent anvil]

what's the order of <g>?

[shy bluff]

Ah ok

Thank you

[old hollow]

The number of elements in the group generated by g

[scenic sage]

yes, but also in cyclic group

this number has precise meaning

[latent anvil]

yeah, how many is that

[old hollow]

If <g> = G then G is cyclic

[latent anvil]

how many elements are in the subgroup generated by g I mean

Sure, but we might not have <g> = G

[old hollow]

Oh

I mean, doesn't it depend on the group

[latent anvil]

that's a good point, it depends on G and g

But we can say something about it in terms of g

[old hollow]

Uhhh

Hm

[latent anvil]

If G is finite, the elements of that subgroup are like {1, g, g^2,...,g^n} up to some finite n, right?

[old hollow]

Yes

[latent anvil]

If it kept going forever then G would be infinite

What's the largest n such that g,g^2,...,g^n are all distinct?

[scenic sage]

$Z^+$ is example of infinite cyclic group

[cloud walrus]

Commander Vimes:

[scenic sage]

and $Z_n$ is finite cyclic group

[cloud walrus]

Commander Vimes:

[old hollow]

Yeah the set of positive integers are generated by using the group operation repeatedly on 1

Ok I think I understand lagrange's theorem, I don't understand though how it relates to the fact that g^phi(n) = 1 (mod n)

[latent anvil]

well I'm trying to get at that lol

[old hollow]

Oh sorry lol

[latent anvil]

It all comes down to the question I posed above

About <g>

[old hollow]

What's the largest n such that g,g^2,...,g^n are all distinct?

[latent anvil]

yeah

this will be the size of <g>

Do you see why?

[old hollow]

Oh yeah lol

My misunderstanding though is why specifically g^n always equals 1

[latent anvil]

g^n won't always equal 1

I mean I'm not sure what n is when you say that

in the multiplicative group of Z/nZ? Because it's not true that g^n = 1 there

[bleak abyss]

@scenic sage doesn't really help when you have people giving linearly independent explanations, probably best to just let Category Disliker do his thing and if he's failing maybe then jump in

[old hollow]

If g is a generator of G, then |<g>| = |G| = n

Oh whoops sorry

[scenic sage]

sure, sloth

[old hollow]

I meant why is g^|<g>| = 1 (mod n)

Since phi(n) gives the order of <g>, then rephrased my question is

Why does g^phi(n) = 1 (mod n) always?

[latent anvil]

so what I'm saying is that g might not be a generator of G

But you can say something about |<g>| anyways

Forget about (Z/nZ)^* for a sec

[old hollow]

Ok

[latent anvil]

I'm just talking about groups in general

[old hollow]

Another example that's not multiplicative:
Consider Z/6Z with +
[3] generates only [0] and [3] so <3> = {[0], [3]}

Where [n] is the congruence class

[latent anvil]

I'm not really sure what you're saying here

[3] = [0]

And <[3]> = {[0]}

[old hollow]

Oh whoops I'm stupid

Lemme edit my previous, I meant mod 6

[latent anvil]

[1] won't be in there

You want [3]

But yeah it's good to work out examples

[old hollow]

Yeah I'm stupid, i edited again lol

[latent anvil]

When you're looking at algebra stuff

So what's the order of <g> in this case?

[old hollow]

The order is 2, right

[latent anvil]

Yup

Vaguer question, do you have a sense of "why" this is?

[old hollow]

And the order of Z/6Z with + is gonna be 6 right

[latent anvil]

What is it about [3] that makes <[3]> have order 2?

[old hollow]

I can explain why 3 generates 0 and 3, but I don't know why it magically always divides n

[latent anvil]

sure

Maybe we can try some more examples

What about <[2]>?

See if you can spot a pattern

[old hollow]

0, 2, 4. So the order is 3

Sorry I'm on mobile, doing the brackets and stuff is a bit difficult lol

[latent anvil]

no worries

[old hollow]

Oh my god

Oh my god

Is it cause 6/2 is 3 and 6/3 is 2

Bruh moment

[latent anvil]

This is true, but not exactly what I was thinking

[old hollow]

Frick

Yeah division doesn't translate to the multiplicative group

[latent anvil]

But yeah you can fit exactly 3 multiples of 2 in {0, 1,2,3,4,5} so 6/2 = 3

okay I feel like I should give you a hint

[old hollow]

Ok

[latent anvil]

What's the order of [3]? What about [2]?

[old hollow]

Order of 3 is infinite since it's a congruence class

Do u mean <[3]>

[latent anvil]

So there's some weirdness in how we use the term order

If I have some arbitrary set X, it's not meaningful to say "the order of X is blah blah blah"

In particular, if I'm thinking of [3] as {0, 3, -3, 9, -9,...} I won't use the word order

I might say "size" or "cardinality"

But [3] is an element of the group Z/6Z, and I can ask about the order of an element of a group

Does that make sense?

[old hollow]

What is the order of an element though

[latent anvil]

oh

[old hollow]

I thought order was for sets

[latent anvil]

The order of g is the smallest nonnegative number n such that g^n = 1, or infinity if there's no such n

[old hollow]

Hold up huh

I thought order just related to any arbitrary group or set. Its definition relies on exponentiation?

[latent anvil]

order means two things, both involving a group

If G is a group, the order of G means the cardinality (i.e. size) of G, just as a set

If g is an element of some group G, then the order of g means what I posted above

[kindred rivet]

and its not really exponentiation

[latent anvil]

exponentiation here means multiplication in the group G

[old hollow]

What if the group doesn't have multiplication as its binary operation, what if it's like addition

[kindred rivet]

doing addition

multiple times

[old hollow]

Would it be multiplication if the group was additive

[kindred rivet]

by itself

[latent anvil]

when we say multiplication in group theory, we just mean the group operation

[kindred rivet]

for example in Z/3Z the order of [2] would be 3

[latent anvil]

I might write the statement that every group has inverses as "for any g in G, there's some h in G such that gh = 1", even though the operation on a particular group G is addition and not multiplication

So the order of a congruence class [k] in (Z/nZ, +) is actually the smallest number a such that a[n] = 0

exponentiation (i.e. repeated multiplication) is interpreted as multiplication by an integer (i.e. repeated addition)

So in Z/6Z, the order of [2] would be 3

Because [2] isn't the identity, and it's still not the identity if you add it to itself, but if you add it to itself three times you get the identity

[scenic sage]

Hm, is it correct that if cyclic group G acts on some set A then it is enough to define action for any generator of G on A, say x, and then just repeat induced permutation?

[bleak abyss]

Yeah

More generally, a G action on A is the same as a homomorphism G->Sym(A)

And once you know what a homomorphism does to generators you're good

[old hollow]

Ohhh thank you @latent anvil, I think I understand it a bit, using the group operation repeatedly on an element to reach the identity is what defines the generator's order

Oh god what is a homomorphism

For example, exp(a + b) = exp(a) * exp(b) has something to do with a homomorphism right?

[delicate bloom]

do you know what a group is @old hollow

[old hollow]

Uhhh I think so, isn't it a set with a binary operation that has closure, associativity, identity, and inverse

[scarlet estuary]

yeah so

the real numbers form a group under addition (this is easy to see)

and the positive reals form a group under multiplication

now, a homomorphism is a structure-preserving map

the idea is that it "preserves" the operation from one group to the next

so if * is the binary operation in one group, and % is the operation in another

a group homomorphism from the first group to the second is a function that satisfies exp(a * b) = exp(a) % exp(b)

note that the group operations in this context make sense; exp is a function from the first group to the second, a*b is a member of teh first group, and exp(a) and exp(b) are members of the second group (so exp(a) % exp(b) makes sense)

anyway, the idea is that the structure is, well, preserved

the operation * is "preserved" by the operation

now let's take the examples i gave earlier

of the reals under + and the positive reals under *

exp is a homomorphism between these groups

from the first to the second

since it satisfies exp(a+b) = exp(a) * exp(b)

• is the operation of the first group (teh reals under addition), * is the operation of the second group (the positive reals under multiplication)

and this map preserves structure between the groups

[old hollow]

Ohhhhhhh my god

[scarlet estuary]

now, you probably know that exp with this domain/codomain is invertible - i.e. a bijection

so we can actually make a stronger statement

its not just a homomorphism

it's an isomorphism

[old hollow]

Wait so a "homomorphism" is really a function that maps one group to another sort of

[scarlet estuary]

it's a function from one group to another that preserves structure

if you have a two-way (invertible) homomorphism (an "isomorphism")

[old hollow]

Is a map another word for function?

[scarlet estuary]

this is just a way of saying your groups are "basically the same"

labels might be changed

[old hollow]

Is an isomorphism one-to-one or bijective

[scarlet estuary]

but the structure itself is the same

the relation between things is the same

an isomorphism is bijective

[old hollow]

Ok I see

[scarlet estuary]

and yes, map is another term for function

(though it often refers specifically to functions between different structures)

[old hollow]

Ok, thank you so much

One question

[scarlet estuary]

anyway that was a major crash-course explanation

this is an important topic so its worth reading about and doing exercises in

[old hollow]

Ok

I don't quite understand what is meant by the =~ symbol for isomorphism

[scarlet estuary]

it just represents that an isomorphism exists between two structures

i.e. they are isomorphic

[old hollow]

Like what is the exact meaning of it, if you were reading it aloud

Oh lol

[scarlet estuary]

or "essentially the same but with labels changed"

like an obvious example is

the integers form a group under +, right?

[old hollow]

Yeah

[scarlet estuary]

...-3, -2, -1, 0, 1, 2, 3, ...

well, so do the multiples of 10

...-30, -20, -10, 0, 10, 20, 30...

and if we do some sample computations:

4 + (-2) = 2
40 + (-20) = 20

you can see how these structures feel, well, the same

i mean, we relabelled the numbers, yes

but all the operations are the same

the structure of the groups are basically identical

[old hollow]

Oh my god that makes so much sense

[scarlet estuary]

we say these are "isomorphic"