groups-rings-fields

[cloud solar]

Let A be a nxn matrix with integer entries, where n>=2 such that a_ii is odd for every i and a_ij = 0 (mod 2^(i-j)) when i>j (below the diagonal). Let B = A^(2^n). Prove that b_ii = 1 (mod 2^i) for every i and b_ij = 0 (mod 2^(min(i,j))) when i≠j.

I proved B = I mod 2 (writing A = I + X mod 2 where X is strictly upper triangular so nilpotent and from binomial theorem it is done)

I proved by direct computation for n=2 case

And i thought maybe it is a good idea to write A as a block matrix s.t. we isolate the last n-1 columns and last n-1 rows and do an induction on n

[candid patrol]

If H, is normal, then S3 is nilpotent, and so, S3 ~ C6

Or you can just says that H = D(S3) which is false

[rocky cloak]

Did you mean B = A^2^i ?

[cloud solar]

Yes

Like

B = A^(2^n)

Where n is the size of the matrix

[rocky cloak]

Then you need some restrictions on i.

If bii = 1 mod 2^i for all i, then that just gives bii = 1

(which can't be true)

[cloud solar]

I dont understand

The element at position i,i in matrix B=A^(2^n) is 1 mod 2^i

Thats a part of what we need to prove

[rocky cloak]

Ah, sorry my reading comprehension is bad
[The two i's are off course the same xD]

[rapid cave]

Let f,g in K[x,y] that are coprime as over K[x,y]_(x,y). Is there a simple way to see that x^n is in <f,g> for some natural number n?

[cloud solar]

No problem :)) i tried like i said writing A as a block matrix and that matrix formed by the intersection of the last n-1 columns and last n-1 rows i called it D. For matrix A = A^(2^0) i called it D_0. For A^(2^k) we call that submatrix D_k. By induction hypothesis (D_0)^(2^(n-1)) has the property we want to prove for A but scaled to n-1. It will be nice to prove D_(n-1) = (D_0)^(2^(n-1)) mod (that rule on the entries)

[rocky cloak]

In a UFD all the height 1 primes are principal, so as f and g are coprime <f, g> is not contained in any height 1 prime. So either <f, g> = <1> or the only prime containing it is (x, y).

In the latter case that means rad<f, g> = (x, y)

[cloud solar]

You may ask why this problem is relevant to this channel. These matrices work like automorphisms of the group Z/2Z + … + Z/2^nZ

So this problem is just equivlent to exp(Aut(Z/2Z + … + Z/2^nZ)) <= 2^n

[cloud solar]

And a thing i found is if we call C_k the column formed by the last n-1 entries of the first column of A^(2^k) we have C_k = 0 mod 2^(k+1) for every k>=0

(Idk if this is relevant but the 2-adic value of each entry of these colums increase in an interesting way. The one for the first element increases by 1 every step k>=0, the second and third by 1 every step k>=1, the fourth,…7th increases by 1 every step k>=2 and so on)

[thorn jay]

Is there a name for groups which have "highly noncommutative normal subgroups" in the sense that [N, M] = 1 => N = 1 or M = 1 for all normal N, M?

[tough raven]

[N, M] ⊆ N ∩ M for normal M, N so your condition (2) would imply that non-trivial normal subgroups have non-trivial intersection (3). If the group is finite (or has finitely many normal subgroups) then (3) implies there is a unique smallest non-trivial normal subgroup N0 (1). Conversely, given (1), (2) is equivalent to [N0, N0] = N0. So at least for finite groups you could look for groups with a unique smallest non-trivial normal subgroup which is perfect.

[ripe harbor]

Thanks!

[azure cairn]

i dont understand isnt this the definition of a partition

like if $S = {1, 2, 3}$ wouldn't a partition be something liike $\Pi = {{1}, {2, 3}}$?

[cloud walrus]

Altanis

[azure cairn]

so what's the difference between $\overline{S}$ and $\Pi$?

[cloud walrus]

Altanis

[glass sinew]

none, structurally

[azure cairn]

also for the uh bracket notation

[glass sinew]

i guess they're trying to illustrate the idea of equivalence classes collapsing to single elements in quotients

[azure cairn]

let's say the equivalence relation returns true if two numbers are of the same parity, so then for $S = {1, 2, 3, 4}$, you have $\Pi = \overline{S} = { {1, 3}, {2, 4} }$. what is the $U$ that generates ${2, 4} \subseteq (\Pi = \overline{S})$?

[cloud walrus]

Altanis

[azure cairn]

is it $[{2, 4}]$ or $[2]$?

[cloud walrus]

Altanis

[kind temple]

the former

according to the author's conventions

the square braces are there to indicate that whatever element we are talking about lives in \bar{S}

to avoid confusion

ironically

[azure cairn]

ah ok

[azure cairn]

i've seen something like [a] to denote the equivalence class C_a in other literature before though

[kind temple]

yes

that is where the notation clashes

[west nacelle]

Do you support IUTT

[vapid vale]

the overarching point is to not consider it as a set inside a set of sets but as a set of elements, whatever the notation is doing is trying to convey that (square instead of curly brackets)

[glass sinew]

we care about the overall structure of the equivalence classes themselves, not the elements they contain

[azure cairn]

alright

but why did he introduce S bar instead of just saying [U] \in \Pi

[vapid vale]

[U] isn't a subset, its a single element

it has the same "type" in S bar as 2 does in S

[azure cairn]

yeah mb i typoed

[glass sinew]

in my experience the bar indicates taking the quotient under the equivalence relation of discourse

[vapid vale]

it is motivating the case where S is a group and S bar is another group with the elements as described

[glass sinew]

consider S = Z and S bar = Z/nZ under the equivalence relation is x ~ y iff x = y mod n, giving the equivalence (residue) classes [1], [2], ... [n-1]

[azure cairn]

ah, i havent formally gotten to quotients yet (although i know wym) so that explains why

[glass sinew]

this should be around the part where they get to it then

[azure cairn]

its in 2 subchapters

[glass sinew]

mud

what text btw? curious

[azure cairn]

artin

[kind temple]

i mean, it is a subset of S...

i would say it signals that we want to think of it as an element of a set

rather than as a subset

[vapid vale]

U is a subset, not [U]

But it doesn’t matter anyway

[azure cairn]

hmm when we say, geometrically, vectors with initial points anywhere in space are equivalent to another vector if it has the same lenght and direction

can it be thought of as all those vectors being in the same equivalence class

where ~ returns true for two vectors if they have the same length and direction

[ripe harbor]

Whatʼs the connection between the centralizer and the normalizer? Is there one? The definitions seem close, just that for one, we consider the whole group and for the other elements

[marble hinge]

One is a subset of another

Subgroup even (because they both are subgroups)

[ripe harbor]

$N_G(H) \coloneqq {g \in G \mid gHg^{-1} = H}$ and $Z_G(H) \coloneqq {h \in G \mid gh = hg \text{ for all $g \in H$}}$

[cloud walrus]

ILikeMathematics

[ripe harbor]

Z_G(H) c N_G(H), right?

Thanks

[marble hinge]

They can be equal too, of course (so not always a proper subset)

[azure cairn]

    We will write the partition as $\bar{S}$, where its elements are the equivalence classes of $S$ (which is, essentially, the partition $\Pi$). When we have some equivalence class $U \subseteq S$, we use $[U]$ to denote this equivalence class as an element of $\bar{S}$ (that is, $[U] \in \bar{S}$).

    In defining $\bar{S}$ by a set of equivalence classes, the fundamental distinction between $S$ and $\bar{S}$ is that $\bar{S}$ permits more elements to be ``equivalent''. If $a \sim b$, then $b \in C_a$ and thus b is ``hidden'' as $[C_a] \in S$. There is no distinction between elements of the same equivalence class under $\bar{S}$.

is my intuition sound?

[cloud walrus]

Altanis

[marble hinge]

They are both special cases of a stabiliser (but of different group actions)

[noble nexus]

Yes pretty much

equivalence relations are a way of changing the meaning of equality

[azure cairn]

alright ty

so now i see that there's a canonical surjection pi (i saw this in my linalg book and in other literature) that takes S to \bar{S} given an equivalence relation

is this the same thing as S/~?

or wait no thats smth else

[kind temple]

i disagree, but yea, whatever

[vapid vale]

i'm just citing the screenshot 🤷‍♀️

[tardy hedge]

Idk why i dont undeestand the proof of i) <— why do we need both chains to stabilize

If the ascending chain a^-1(Ln) eventually stabilizes, how is that not enough to show (Ln) stabilizes? I guess if something like a^-1(Ln-1) = a^-1(Ln) that doesnt that mean Ln-1 = Ln but idk how the chain in M/N comes into play i guess

[south patrol]

I mean consider case M' = 0 for example

This is just saying the intersections with some submodule (M') stabilise, and that needn't tell you much about the whole things

[tardy hedge]

Yeah

Why with M’’ chains stabilizes then it works?

[wraith cargo]

The idea is basically like

Since the beta(L_n) stabilize, if the L_n keep growing the kernel of the mapping L_n -> beta(L_n) would have to keep growing as well

But the kernel is static

So the L_n's can't keep growing

i.e. the chain has to stabilize

You can argue this formally using the first isomorphism theorem and the lattice theorem

[tardy hedge]

Did this part need M’ noetherian

Thx btw i like your explanation

I am just slow ngl

Lol

[south patrol]

I'd use the five lemma basically

You have sequences 0 -> a^-1(L_n) -> L_n -> a(L_n) -> 0 with maps between them and if the maps between first and 3rd are isos then the middle is

Though here it is much easier as you have submodules everywhere

[wraith cargo]

Uh yes because a is injective so something something this way you can get any possible chain in alpha or something

Like you can take some really stupid example

Take some chain L_n in alpha

Inject it into M

Now you have a chain in M whose inverse image in M' is some generic chain in M'

So every chain in M' appears

So you need M' noetherian

[tardy hedge]

Ill be back

I do need to understand this lol

[chilly radish]

Hi

Ok so do you know the universal property of a free module

[limber thorn]

Hi can someone explain why every R-module is a quotient of a free module?

[limber thorn]

No, I'm not even sure what relation we are quotienting by

[chilly radish]

Alright, what is your definition of a free module

[ocean jolt]

a module which has a basis

[limber thorn]

I have the formal definition but I do not see the intuition which is what I am mainly concerned with

[chilly radish]

The formal definition is generally via the universal property. If F is a free module with basis B, then every map f:B->M with M an R-module extends uniquely to a homomorphism f':F->M

[limber thorn]

From what I understand, they're generalisations of vector spaces, but with objects being from groups, and with scalars being from some ring right? What does having a basis have to add to this?

[ocean jolt]

Interdasting

[chilly radish]

So every vector space is a free module

Over a general ring this is not true

[noble nexus]

A free module is one that is generated by a subset of elements with no relations between them (i.e. a basis) every module is a quotient of a free module because every module can be described by some generating set subject to relations, which is exactly a quotient of a free module

[ocean jolt]

Does this mean in a non-free module you can't use matrices for your linear maps?

nvm

how will you matrix with no basis

[chilly radish]

Yea you generally can't

[limber thorn]

Over a commutative ring right?

[chilly radish]

Any ring

Non commutative also

[limber thorn]

I see, so what does having a basis add to the conversation here

[noble nexus]

you can still use matrices but only when working with R^n (i.e. free modules)

[chilly radish]

So having a basis means that you have a set of generators with no relations between them

[ocean jolt]

yeah but I siad non-free module

[chilly radish]

By quotienting you can introduce relations

[hidden wind]

a related fact which it is probably helpful to think about is that every group is the quotient of a free group

[noble nexus]

ya non free matrices don't work really

[chilly radish]

So by picking a proper basis and a proper submodule of relations we can get any module as a quotient

Let's do a concrete example

[limber thorn]

Wouldn't that imply the ring itself is free?

[chilly radish]

Yes

[limber thorn]

I didn't know that

[limber thorn]

Yes please

[chilly radish]

A ring is always a free module over itself, with basis {1}

[chilly radish]

Consider the Z-module Z/2Z x Z/3Z. This module is generated by the elements (1,0) and (0,1). These have some relations between them, say 2*(1,0)=0, and 3*(0,1)=0. In fact every other relation a linear combination of these two

[tardy hedge]

Wow Shin has True even in his name. That’s badass.

[chilly radish]

We consider the free Z-module Z x Z

[next obsidian]

You can use matrices, kind of

If your module is finitely generated, so it has a finite spanning set, you can map those generators to things using a matrix

But whether it’s well defined is not immediate, you need to respect certain relations

[chilly radish]

This has basis (1,0), (0,1). We can introduce the submodule generated by
{2*(1,0),3*(0,1)}, call it M, then quotienting by M is like introducing the relations, that is,

Z x Z/M is isomorphic to Z/2Z x Z/3Z

The way you do this formally is by defining a map using the basis: Define f:{(1,0),(0,1)}-> Z/2Z x Z/3Z
f(1,0)=(1,0)
f(0,1)=(0,1)

[chilly radish]

You can extend this to a map f':Z x Z -> Z/2Z x Z/3Z

This map will be surjective because it maps generators to generators, thus by the first isomorphism theorem we can get

$\mathbb Z\times \mathbb Z/ker(f') \cong \mathbb Z/2\mathbb Z \times \mathbb Z/3\mathbb Z$

[cloud walrus]

ShiN

[chilly radish]

Yes, i.e. it is spanning and linearly independent

Just like in the case of vector spaces

[limber thorn]

Ok just a minute I want to fully write out the proof you gave previously

[chilly radish]

Alright

I did omit some details such as every linear relation between (1,0) and (0,1) being generated by those two, but those aren't difficult to fill in

Ping me when you're finished

[limber thorn]

@chilly radish

[chilly radish]

Hi

Sorry

Let me rest this

Read

[chilly radish]

Modules can have multiple sets of generators

(1,1)=(1,0)+(0,1)

It's a homomorphism because that's the definition of a free module. Explicitly the homomorphism is just (m,n)-> (m mod 2, n mod 3)

The kernel of the homomorphism you defined is exactly M

[hidden wind]

it’s very straightforward, take a group G, and consider the free group F on the elements of G. there is a canonical surjective morphism F -> G given by simply evaluating the words in G as products in G, this is obviously surjective (every element of G is also a word in G), and by the first isomorphism theorem then G is isomorphic to the quotient of F by the kernel of this map. the elements of this kernel are the words which code the relations in G

[chilly radish]

Here by the Chinese remainder theorem Z/2 x Z/3 is isomorphic to Z/6 so we can equally define this via Z/6Z

But I wanted to give an example with more than one generator

Depending on choice of generating sets you have a lot of presentations of a module in terms of generators and relations

[azure cairn]

why

[kind temple]

it will usually be clear from context

[marble hinge]

Why what? Why they use this notation?

[azure cairn]

ye

[marble hinge]

I think it’s because it’s a generalisation of a congruence notation on numbers. You can kinda do the same but with kernels of homomorphisms, and even write a ≡ b (mod K) where K is the kernel of h.m.

[maiden crater]

so I considered the division of m by some element b ot R

that gave me b=ma+r , but r=0 as N(m) is minimal

so b=ma

so here do I just let b=1, and I'm done?

[ripe harbor]

I saw the notation Z_n recently, for a cyclic group of order n of integers. Apparently, it's defined slightly differently than Z/nZ which is {a + nZ}. [Of course in the end they will be isomorphic] But how is it defined? Do we let Z_n = {0, 1, 2, ..., n} and then define addition as a + b mod n?

[vapid vale]

Yes

[ripe harbor]

Thanks!

[vapid vale]

It’s just a common enough group to deserve its own notation

[noble nexus]

they're both the same really

some sources will just define Z_n to be Z/nZ

[ripe harbor]

Yes they're isomorphic but my book said they're still defined differently

[cursive spindle]

Why do people do this?

Z_n

[candid patrol]

There no reason to say that

Z_n = Z/nZ = C_n

[cursive spindle]

When I'm on a competition and my opponent is Z_p

We love playing a guessing game

[noble nexus]

also yeah Z_p is evil

though Z/pZ is kinda long

I usually just write Z/p

[cursive spindle]

Yeah that's fine

[swift root]

this is the way

[tulip otter]

[azure cairn]

How is Z_n defined then

[kind temple]

it’s clarified in that message

[somber goblet]

Z/p is nice

[azure cairn]

oops

[fading acorn]

n-adic integers

which is obviously different integers modulo n

using Z_n for integers mod n is evil

[sonic coral]

localization at n

[swift root]

while Z_p is the p-adics of course

[elfin wraith]

I did think your messages to me earlier were odd

[swift root]

lmao

[elfin wraith]

Seems to be a few accounts getting hacked I may change my password lol

[swift root]

its really weird

it didnt contact all my friends

only a couple

[swift root]

luckily i have all the data i actually care about on an external ssd so i just wiped my computer completely just in case

yeah its good to have backups

[torn thorn]

elementary question. If a group G has order 25, and it has exactly one subgroup of order 5, why would that mean that G is cyclic? The unofficial solution on the internet says that if H is that one subgroup of order 5, then all elements not in H should have order 25. why is that?

[desert verge]

let g be an element not in H. if ord(g) != 25 then ord(g) = 5 and so <g> is another subgroup of G with order 5

[torn thorn]

ahh

wow I feel so stupid. Thanks for answer though

[chilly lantern]

what does groups rings fields mean chat

someone referred me here

i hope it's fields like the cool kind of fields

[rocky cloak]

Groups, rings and fields are the three algebraic structures usually encountered in a first abstract algebra course.

https://en.wikipedia.org/wiki/Group_(mathematics)

https://en.wikipedia.org/wiki/Ring_(mathematics)

https://en.wikipedia.org/wiki/Field_(mathematics)

[chilly lantern]

oh damn

i didn't know algebra was chill like that

or is abstract algebra unrelated to like high school algebra

[rocky cloak]

It's not unrelated, historically they are very related.

But it might not feel very related if you go from hs algebra to uni algebra

hs algebra is basically about studying the specific field the complex numbers (or maybe just the real numbers).

Whereas abstract algebra would like to make more general statements about all fields

[chilly lantern]

im working on learning linear algebra rn

[chilly lantern]

why doesn't it start with the more generalized stuff?

i feel like the specifics make more sense in the context of the bigger picture

[rocky cloak]

That's true, but also generalizations make more sense with examples in mind. So it's a trade off

[chilly lantern]

hmm yeah true

[night jay]

can someone help me, i dont know how to tackle this problem and also, do i need to know the full group and ring theory to compute this?

i have my exam coming up, and i dont have enough time to study group, ring, fields

[rocky cloak]

All you need to know is how to divide in Z/7. Other than that this is just normal polynomial division

In fact you don't even need to know about division, since the leading coefficient of X^2 + 3X is just 1 (which is very easy to divide by)

[night jay]

ughh i hate discreet math, Z/7 is just Z mod 7 right? and what do you mean with leading coefficient of X^2+3X is just 1

[rocky cloak]

1* X^2 + 3*X

First coefficient is 1

[marble hinge]

Haha, and this is following that lengthy appreciation thread for Z/7 notation (as opposed to Z_7 or Z/7Z or C_7)

[elfin wraith]

C_7 is a group, Z_7 is the 7adics or localisation at 7 depending on context, Z/7Z is morally correct but I have too many quotients of Z in my life to bother writing that

Having the notation Z_n for quotients would be lovely but it’s already too overloaded

[rocky cloak]

Nah, Z[1/7] for the localization

C_7 the group, Z/7 the ring / Z-module, F_7 the field.

[elfin wraith]

I can respect that

[tough raven]

ℤ[i]/(1+i) = F_{1+i} 👀

[cursive spindle]

The field with 1+i elements

[rocky cloak]

You know what? I don't hate it

[quiet pelican]

TIL i = 1

[south patrol]

Yeah writing Z_7 for localisation at 7 is smth I have never seen seriously and is rly bad lol

[quiet pelican]

I’ve seen Z_{(7)}
But not Z_7

[tough raven]

Yes, me too!

[cursive spindle]

The guessing game

7-adic or Z/7Z or localization

[marble hinge]

Yes, but no one cares about p-adics or other things until they actually study them

So there is no confusion

[tardy hedge]

Ive never seen abstract algebra homework on an online interface like that lol

[night jay]

The exam is also digital

[tardy hedge]

Lol

Thats interesting

[rocky cloak]

I've seen R_r for a generic R, but not Z_7 yeah

[fleet cairn]

from aluffi grad algebra book, i have one quick question. Isn't it obvious that if f o a' = f o a'' then a' = a'' is true for all functions f and given two random functions a' a''? how can they not be equal, i have been looking at this for almost 30 minutes. i would a different way of thinking abt this or a counter example

[knotty badger]

The key point is that f is fixed here

As a silly example suppose f was a constant function

Then f o alpha would always equal f o alpha’

Even if Alpha and Alpha’ were unequal

The idea is that you should be able to cancel f on the left

[fleet cairn]

always? how?

[knotty badger]

Ok let’s be concrete

Let A = B = N

And f : N -> N be the constant function at zero

If alpha : Z -> N is any function, then f o alpha : Z -> N is the constant function at zero

So f o alpha always equals f o alpha’

The issue is that you can have two different functions that give the same result when you post compose by f

In fact, f being a monomorphism is equivalent to postcomposition by f being injective

[fleet cairn]

ok so what i understood from this is any right composition function with respect to f (f is arbitrary but fixed), is always equal with another right composite with f. so it is in fact not always true that it implies alpha = alpha' (sorry for the atrocious explanation)

[knotty badger]

Mhm

In the category of sets, a function is a monomorphism if and only if it is injective

[fleet cairn]

is my explanation right somewhat?

[knotty badger]

I think it’s along the right lines

The issue is that if f : N -> N is constant, then f o alpha = f o alpha’ always, even if alpha is not equal to alpha’

And indeed f is not injective

Constant functions are almost never injective

[fleet cairn]

with the exception of domain with size 1?

[knotty badger]

Yes

[fleet cairn]

lol thats funny

[fleet cairn]

i dont think f needs to eb constant tho

i think i understood it to be any arbitrary fixed f

[knotty badger]

So “being a monomorphism” is a property that a fixed f can have

A constant function on a domain of size > 1 is not a monomorphism, is my point

But there do exist functions which are monomorphisms

[fleet cairn]

i understand htats it not generally true for all functions f

thanks

[knotty badger]

What is true is the following

If you fix alpha and alpha’ and let f vary

And if f o alpha = f o alpha’ for all f

Then you must have alpha = alpha’

Maybe this is the result you were confusing it with

[fleet cairn]

it def was

😭

[knotty badger]

This is essentially showing the yoneda embedding is faithful

[fleet cairn]

its a ltitle tempting if you look at it at face value

[fleet cairn]

this has somethign to do with the varying f being a left composite

right?

[knotty badger]

Yeah

[fleet cairn]

though i havent fully worked this out in my head

[knotty badger]

The easy way to prove this is to take f = id_A

Then you know that id_A o alpha = id_A o alpha’

Thus alpha = alpha’

[fleet cairn]

but this might show that the range of alpha, going into id_A is the same, but ti doesn't prove that a domain = a' domain

[knotty badger]

Ah yes so you need to assume that

Otherwise it doesn’t make sense to compare f o alpha with f o alpha’

In cat theory usually you’re only allowed to compare morphisms if they have the same domain and codomain

[fleet cairn]

i see its an implicit assumption that domain and codomain are equal

i might be a little pedantic over ts tbh, alright anyways thanks for helping and esp doing more than i had asked (i really do appreciate that)

[knotty badger]

It’s fine, I am cat theory #1 fan after all

[south patrol]

Yeah lol

[maiden crater]

in (a) finding an explicit form won't be direct right

It's onviously not continuous

It's clear I have to use limits at t

Let $A$ be the set of points such that $f(A)=0$. Then a zero divisor of $f$ is \$g= \begin{cases} 0 & x \notin A \ x^2& x \in A\end{cases}$

[next obsidian]

What do you mean explicit form

[maiden crater]

I misunderstood

I was trying to find a zero divisor of 1/f

which is nonsense

[cloud walrus]

wai

[next obsidian]

This function is 0 if f is 0 at 0 and 1 elsewhere

[maiden crater]

hmm, how

[next obsidian]

By the formula

[maiden crater]

okay, consider (x-1/2)

[next obsidian]

What about it

[maiden crater]

which doesn't have an inverse on [0,1]

[next obsidian]

Yeah it’s a zero divisor

[maiden crater]

oh, you mean an indicator

[next obsidian]

I don’t see what that has to do with my point

I’m saying you haven’t shown that any non-invertible function is a zero divisor with that function

[maiden crater]

I don't get how x^2=1 if x is in A

[maiden crater]

Ah

[next obsidian]

I never said it is?

[maiden crater]

this

[next obsidian]

I’m saying if you apply that construction to the function f(x) = 0 when x = 0, f(x) = 1 otherwise

You end up with the 0 function

So that doesn’t show f is a zero divisor because you’re concluding that f•0 = 0

[maiden crater]

so 1 instead of x^2

[next obsidian]

Yes

[maiden crater]

or any constant

okay, cool

tysm

[next obsidian]

Other than 0

But yeah

Or x+1

There’s a gazillion things you could do

[maiden crater]

Yea, I get it

I was stuck on this for 30 minutes because I was trying to find a zero divisor for 1/x 😭

tysm

[next obsidian]

Rip

[supple ice]

https://tenor.com/view/rip-cat-cat-meme-catcoin-gif-9164440438767075157

[craggy oar]

rieap

[ripe harbor]

Why does $e \mathbb Z \leq d \mathbb Z \implies d \mid e$?

[cloud walrus]

ILikeMathematics

[rocky cloak]

Notice that eZ <= dZ means in particular the e is an element of dZ.

What are the elements of dZ?

Nope

[ripe harbor]

Oh wait

{0, \pm d, \pm 2d, ...}

[rocky cloak]

That's right

[ripe harbor]

so e = kd for some k in Z

Ah

Thanks!

[maiden crater]

how do I use this to show

like Z \subseteq Z[√3]

[desert verge]

are you asking for help with the last part?

[maiden crater]

oops

yes

[quiet pelican]

Hint: you only need to find one unit that isn’t 1, -1, then you can deduce there are infinitely many

[maiden crater]

(2+√3)

[quiet pelican]

Now how can you use that, and what you’ve proven already to produce more units

[maiden crater]

(2+√3)(2-√3)=1

so (2a+a√3)(2a-a√3)=1

[quiet pelican]

Hint: ||a power of a unit is a unit||

[quiet pelican]

How?

[maiden crater]

oops

yes

[maiden crater]

ooh, right

🤦

(2+√3)^n≠(2+√3)^m for any n,m

[quiet pelican]

Yup

[maiden crater]

TYSM

really silly but worried I've nto simplified it enough

$\left{a+bi \mid a,b \in \Z;ac-bd=1; ad+bc=0 \right}$

[cloud walrus]

wai

[ashen heron]

what are c and d hmmcat

[maiden crater]

c,d in Z

[ripe harbor]

And why does $e \mathbb Z \leq d \mathbb Z \implies e \mathbb Z/ n\mathbb Z \leq d \mathbb Z/ n \mathbb Z$?

[cloud walrus]

ILikeMathematics

[twilit wraith]

Fourth iso theorem i think

[wraith cargo]

the that fact that the first inclusion implies that d|e

so like

if n|d, then n|e, so both are 0

[ashen heron]

try to see if theres more you can infer about a and b from this

[wraith cargo]

if n|e, then clearly we trivially have 0 < dZ/nZ

so you just have to check what this behaves like for n not dividing e,d

but this is kinda trivial because like

[ripe harbor]

Hm, the chain of reasoning is apparently $$d \mid e \iff e \mathbb Z \leq d \mathbb Z \iff e \mathbb Z/n \mathbb Z \leq n \mathbb Z/n \mathbb Z$$

[cloud walrus]

ILikeMathematics

[maiden crater]

ad/b=-c

[wraith cargo]

take some x in eZ/nZ

then x can be liften to an element x' in eZ

this element is also in dZ

so when you take everything mod n, then x is also in dZ/nZ

so you have that eZ/nZ is a subset of dZ/nZ

but eZ/nZ is a subgroup of Z/nZ so we have your desired inclusion

[ripe harbor]

by mapping x + nZ -> x?

Take $x \in e \mathbb Z/ n \mathbb Z$. Then $x = a + n \mathbb Z$. But because $a \in e\mathbb Z \subseteq d \mathbb Z$, we have $x \in d \mathbb Z/n \mathbb Z$

[cloud walrus]

ILikeMathematics

[ripe harbor]

And then for the other direction, take $a \in e \mathbb Z$. Then $a + n \mathbb Z \in e \mathbb Z/n \mathbb Z \subseteq d \mathbb Z/n \mathbb Z$. But that means $a \in d \mathbb Z$.

[cloud walrus]

ILikeMathematics

[ripe harbor]

Does this look right?

[glass sinew]

correspondence theorem in fact gives that $[d\mbb Z:e\mbb Z]=[d\mbb Z/n\mbb Z:e\mbb Z/n\mbb Z]$

[cloud walrus]

bsharp

[wraith cargo]

I feel like the point was more

[glass sinew]

(if nZ is normal, that is)

[wraith cargo]

how do we prove the correspondence theorem lol

[ripe harbor]

Yeah, I don't have this

[glass sinew]

ah, then my bad

[ripe harbor]

Does what I did look right? These

It seems what Irony did is a little different

[glass sinew]

lgtm

[ripe harbor]

Thanks!

[glass sinew]

(footnote but perhaps to state the obvious, all this does rely on the quotient by nZ being a group, which requires nZ be normal)

[ripe harbor]

Apparently $$(\mathbb Z/n \mathbb Z)^\times = {\text{generators of $\mathbb Z/n \mathbb Z$}} = {a + n \mathbb Z \mid a \in \mathbb Z, \ \gcd(a, n) = 1}.$$ But what is a generator? So apparently for $\mathbb Z/10 \mathbb Z$, $7$ is a generator?

[cloud walrus]

ILikeMathematics

[marble hinge]

It looks like this is confusing additive amd multiplicative groups

[twilit wraith]

A generator k of Z/nZ as an additive group is an element such that every element in Z/nZ can be achieved as a repeated sum of k

It turns out that the elements which can do that must be coprime to n

[marble hinge]

Yes, but he writes multiplicative group first…

[ripe harbor]

We consider Z/nZ as a ring here, sorry for not mentioning that

[twilit wraith]

Well the multiplicative group Z/nZ as a set are the generators of Z/nZ as an additive group

[rocky cloak]

The elements of the multiplicative group are exactly the generators for the additive group (when it comes to Z/n)

[twilit wraith]

Youre considering them as groups

It wouldn't make sense to write the times above if its a ring

Since it already comes with that

[ripe harbor]

the times above means invertible elements wrt multiplication

[twilit wraith]

After all Z/nZ as a ring is just the additive and multiplicative groups mashed together in a sense

[rocky cloak]

I mean R^\times only makes sense for R a ring

[twilit wraith]

True

[marble hinge]

Yeah, just wanted to say that it’s a confusing notation

[twilit wraith]

But the resultant structure is a group

[noble nexus]

It's a bit confusing to write it this way, it happens that the generators of Z/nZ coincide with the invertible elements but shouldn't be used as a definition

I suppose it's true that the generators of a general (commutative) ring (as an ideal) are exactly the invertible elements but idk that's something you prove after

[ripe harbor]

[this is a theorem], so are we looking at the additive generators here?

[marble hinge]

I.e. then in further definition of generators it already uses additive notation. But it’s fine, I think everyone agrees that it’s confusing :)

[twilit wraith]

If you think about it a bit, an element of Z/n is a generator if and only if 1 can be given as a repeated sum of the element

[ripe harbor]

An additive generator?

[twilit wraith]

And because addition extends to multiplication nicely in this structure, the number of times you add the element to get 1 is exactly the multiplicative inverse of the element

[twilit wraith]

Yeah multiplicative generators are a lot worse

[ripe harbor]

Alright

[twilit wraith]

But here I mean additive

[ripe harbor]

=> is clear, why does <= hold?

[noble nexus]

if 1=a+a+\cdots+a, then 1 =na for some n

but then k = (nk)a

for all k

[ripe harbor]

oh

Thanks!!

[twilit wraith]

Because 1 is an obvious generator

[ripe harbor]

Right, and if we can generate 1 then we can generate everything else, just like Blake did

Thanks

[twilit wraith]

Yes

[fleet cairn]

by "determine an equivalent relation ... " do they mean there exist a unique equivalence relation such that ... . sorry im not that familiar with common math phrases

[marble hinge]

Ha, I think that’s exactly what @knotty badger covered in her post

https://pseudonium.github.io/2025/09/16/Three_Perspectives_on_Equivalence_Relations.html

[fleet cairn]

😭 ill take a read

[marble hinge]

I think they just mean that you can derive an equivalence relation from a function

I.e. you use a function to build an eq rel out of it

Based on the shared values

[kind temple]

yes, you can think of the phrase “a determines b” as another way of saying “b is a function of a”.

if you ever see “a uniquely determines b”, you parse it as “b is an invertible function of a”

[marble hinge]

If values are the same, then arguments fall into the same eq class

[fleet cairn]

thats definitely clears it up

[marble hinge]

It’s the same I think

[fleet cairn]

wait no

my book fixes the function while she fixes the equivalent relation

😭

[marble hinge]

I.e. both the blog post and your quote is about “giving names to things” :)

And the things that are given the same name end up in the same eq class

f(a) = f(b) - they have the same “name”

So a ~ b

[kind temple]

i think JFK is asking about what they need to prove

[fleet cairn]

"Suppose we start with a set X and have an equivalence relation ~ on it. " (Fixes an equivalence relation)
"To see this, we observe that every function f : A -> B determines an equivalence
relation ~ on A" (Fixes a function)

[marble hinge]

Yeah, but the idea is that they are the same

[fleet cairn]

They are clearly different right?

[marble hinge]

Doesn’t matter what you fix

[kind temple]

the author is not doing this

the function comes first

and it’s not fixed

they say “for every function f, we get a unique eq rel such that a ~ b <—> f a = f b”

[fleet cairn]

both of them say that?

i can see that my book does now lol

im not sure how to interpret hers

[marble hinge]

Yes: “Formally, we could define a relation by: . This has a name within category theory - it’s called the kernel pair associated to our function .”

Ah, FFS

Didn’t copy symbols)

[kind temple]

what phrasing are you confused about?

[marble hinge]

That’s exactly what Aluffi says :)

[fleet cairn]

"Suppose we start with a set X and have an equivalence relation ~ on it. "

[marble hinge]

Read this:

[marble hinge]

Have you read the whole post?)

[kind temple]

sphynx, i’m failing to see your point

[fleet cairn]

no i thought it was just stating the proposition on the top so i didnt read further

[kind temple]

it feels like you are making things unnecessarily complex

[marble hinge]

May be. Up to JFK to decide :)

[fleet cairn]

ok wait all i needed in the first place was a clarification of interpretation of one of allufis paragraphs

i think thats all i need :))

[marble hinge]

Ok, then sorry

[kind temple]

right. which paragraph is it? this doesn’t seem to be all of it

[marble hinge]

I just wanted to give you more context on interplay between eq rels and functions of that type

Via that insightful post

But if you don’t need it or choose to ignore it, that’s fair enough

[fleet cairn]

he tells us to understand intuitively that "this is indeed an equivalence relation" first before he proves it which i haven't done yet

[fleet cairn]

I'm sorry I don't mean any offence, but you mentioned some category stuff and I never learned that so idk what to get from that

categories are next chapter tho ill look forward to it

[marble hinge]

You are reading Aluffi’s book which is full of exactly the same categorical stuff

[fleet cairn]

yes yes but im still on chapter 2 but chapter 3 is where the category stuff begins

all the way until like end of chapter 5 i think

[rocky cloak]

He's telling you to check that it is, no?

[kind temple]

right, so they want you to check that when you are given a function f : A —> B, the eq rel ~_f defined by a ~_f b <=> f a = f b is reflexive, symmetric, and transitive.

using ~_f to emphasize the fact that ~ depends on f/is determined by f

[marble hinge]

Yes, and that post that I linked does exactly that ;)

[fleet cairn]

so is it just involve having the equivalent relation copy all of f(a') = f(a'')? because = already obeys equivalent relation conditions

(sorry for the atrocious explanation)

[kind temple]

yea, idk what to make of this explanation lol but that’s okay.

what this eq rel is doing is identifying the fibers of f

so if y is a point of B in the image of f

[fleet cairn]

then fiber of y is all of domain value such that f(x) = y

[kind temple]

then all points in f^{-1}(y) are equivalent

[fleet cairn]

equivalent in what way?

[kind temple]

in the way that is specified by ~

[fleet cairn]

ohh

[kind temple]

for a and b in f^{-1}(y), f(a) = y = f(b)

[fleet cairn]

my brain is telling me that the fibers are directly the eqiuvalent classes itself

[kind temple]

yes

indeed

[fleet cairn]

but i have no proof

or intuition of it at all

[kind temple]

you should prove it 🙂

it will give you more intuition

maybe try with the example function f : R^2 —> R given by f(x,y) = x^2 + y^2

when do points get identified?

what do the equivalence relations look like?

draw a picture of them.

[fleet cairn]

the set of fibers are disjoint and their unions are the set of the domain, this is probably a relevant fact right?

[kind temple]

yea, it is

actually that is all you need

they partition X and so determine an equivalence relation, the fibers being the equivalence classes

[fleet cairn]

Ok so set of fibers are disjoint and unions is the domain A, therefore its a partition on A, therefore its an equivalence class on A

[kind temple]

basically yea

[fleet cairn]

wait let try to explain each of the mapping

so surjective mapping from A to quotient A

since the quotient are just partition of A, then there is an obvious projective and surjective mapping you can make

[kind temple]

what does it do to an element a in A?

[fleet cairn]

it sends it to an eqiuvalent class in the quotient, that has a

[kind temple]

nice

[fleet cairn]

this means bijection right?

ok it does

im not sure how the quotient can have a bijective mapping to f

[glass sinew]

each element in the image has one fiber above it

[kind temple]

yea, so for this specific eq rel, we know that the fibers of f are the equivalence classes, and so are the points of A/~

given an equivalence class C in A/~, it is non-empty, so there is an element c in C.

how should you figure out the fiber that C corresponds to?

[fleet cairn]

its an equivalent class on A (partition of A) and the surjective fiber are just partition of A as well.

so to figure out what fiber C corresponds to

pick a c in equivalence class c, and prove that it's in a particular fiber?

[kind temple]

hmm. what about f(c)?

[fleet cairn]

hm i dont understand

what about f(c)?

[kind temple]

every other element c’ in C should be equivalent to c, right? so what does c ~ c’ expand to?

[fleet cairn]

1. yes because they are in the same equivalence class
2. c in [c'] or c' in [c]?

[kind temple]

yes, but remember, we are working with a specific equivalence relation

the one determined by f

[fleet cairn]

ok we'll call the first mapping f_1

let me see

[kind temple]

i’m just trying to push you recognize that c ~ c’ is an abbreviation for f(c) = f(c’).

[fleet cairn]

i was about to type some fiber machinery

[fleet cairn]

ok i understood this one

[kind temple]

yea, the question wasn’t meant to be super complex

anyways

can you see how this proves C is a subset of the fiber of f(c)?

[fleet cairn]

reason: this one is true because of the specific projective mapping we use

[kind temple]

yes, but more to the point, just by the way we defined ~

[fleet cairn]

shouldnt big C be a fiber?

[kind temple]

it is

[fleet cairn]

we said C was an eqiuvalence class and therefore it is a fiber

so how can it be a subset

of a fiber

[kind temple]

let me summarize what we just did rq.

1. took a point C of A/~ (so a subset of A)

2. C is non-empty, so we fixed an element c of C.

3. we argued that for all c’ in C, f(c) = f(c’)

4. this means for all c’ in C, c’ is in f^{-1}(f(c)), the fiber of f(c), and so C is a subset of the fiber of f(c).

what is also true is that the fiber of f(c) is a subset of C, and so they are equal: given a point c’ in the fiber of f(c), then by definition, f(c) = f(c’) so c ~ c’ thus c’ is in C.

punchline: for c in C, C = f^{-1}(f(c))

[fleet cairn]

ok understood the subset argument

fiber of f(c) is a subset of C?

ok understood

C and fiber of f(c) are equal

[kind temple]

yes

and so given a point C of A/~, can you see how to map it to a point of im f now?

[fleet cairn]

which mapping was f again

is it the A to B one?

is there not smt wrong with having im of f

just before the codomain it came from?

[kind temple]

no there is not because we are still defining ~f

im f is a subset of B so there is a natural inclusion into B as well

[fleet cairn]

so does the mapping of im of im f

involve projection

in which it makes it injective

[kind temple]

it’s not called a projection in this case, it is called an inclusion

because it includes im f into B

[fleet cairn]

is there smt important about the inclusion?

[kind temple]

how do you think it should be defined?

[fleet cairn]

identity function

so then it becomes injective

[kind temple]

right, it doesn’t move any points. it is the identity function of B restricted to the image of f

that’s all for now

[kind temple]

but back to my question, so we have been given a point C of A/~.

we know that for c in C, C = f^{-1}(f(c)).

how can we take C to a point of im f?

[knotty badger]

Oh yeah nice

[fleet cairn]

i dont seem to see any obvious way to map this or even yet guarantee bijection

[kind temple]

no

[fleet cairn]

im acutally losing it

[kind temple]

haha that’s okay, i’ve been there.

i have essentially put the answer here

[fleet cairn]

didnt our previous f refer to the first mapping which is f_1

the f(c)

[kind temple]

f : A —> B hasn’t changed at all, it’s still the same one that we started with

[fleet cairn]

ok now nothing makes sense because i was thinking of the c, c', C and f(c) discussion as the function f that refers to the first mapping

😰

[kind temple]

😭 that’s okay. maybe take a short break and let somebody else try later

[fleet cairn]

ok thanks for helping and sorry for the time waste

[kind temple]

no not a waste

im sure ive brainwashed you just a little bit, and thats all u really need to start understanding lol

but somebody else may explain in a way that makes it click for you

[fleet cairn]

that reminds me of the times when a book may accidentally or maybe purposely leave out a clarification or important detail

you work it out yosuelf fustrated

then found the problem it had so you have the best feeling of releif and satisfaction ever

[knotty badger]

Hi I am here now

Wassup

[fleet cairn]

wsp

[karmic moat]

deeZ/nutZ

[knotty badger]

What’s the problem you’re trying

[kind temple]

oh btw, i think it’s worth thinking about this @fleet cairn

[fleet cairn]

trying to understand this

[knotty badger]

Oh pog

[fleet cairn]

ok ill look into it a bit

[knotty badger]

I think the things you want for this are the universal properties of subobjects and quotient objects

[marble hinge]

I think it would help to just look at a simple example of a function on {1,2,3} that is neither subjective nor injective. And see how it decomposes

[fleet cairn]

the onlty machinery i have is equivalence class, partition, the 3 function properties and thats it

[marble hinge]

Like f(1) =1, f(2)=1, f(3)=3

[knotty badger]

Ok I can try explaining the universal property of subobjects if you want

[fleet cairn]

so decomposes is like a thing you do on a set?

[fleet cairn]

sure

[knotty badger]

Consider f(x) = x^2

This is a function from the reals to the reals

[fleet cairn]

yep def

[knotty badger]

But it also works as a function from the reals to [0, infinity)

Because squares are always nonnegative

[fleet cairn]

yeah it make sense

[knotty badger]

Strictly speaking, those aren’t the same function

They have different codomains

[fleet cairn]

yes i understand

even tho range are the same

[knotty badger]

One is R -> R, the other is R -> [0, infinity)

Mhm

Actually have you watched jjk

[fleet cairn]

S1 yeah

is jjk gonna be of use here 😭

[knotty badger]

Ok then you’ll be familiar with the concept of a domain expansion right

[fleet cairn]

yes

[knotty badger]

Here we’re doing the opposite

A codomain contraction

We want to shrink the codomain of our function from R to [0, infinity)

[fleet cairn]

Ok i definitely understand that a function is in a way different because the codomain is changed

despite the functions (a set) are the same

[knotty badger]

Yeah

[fleet cairn]

ok understood

[knotty badger]

It turns out you quite often want to do this codomain contraction

You have a function into a set X

[marble hinge]

No, decompose function f into surjective and injective functions composition — the thing Aluffi talks about in your diagram

[knotty badger]

But really, the range lies in a subset S

And you want to shrink the codomain to S

A good example js defining groups

[fleet cairn]

but is this codomain contraction a trivial one or like one where the range is maintained

[fleet cairn]

i see

[knotty badger]

The range is the same, you just change the codomain

[fleet cairn]

Okok

[knotty badger]

Like you have matrix multiplication M_2(R) x M_2(R) -> M_2(R)

[fleet cairn]

im not familiar with LA sorry

[knotty badger]

Then you restrict to just invertible 2x2 matrices

[fleet cairn]

los angelas

[knotty badger]

Ah ok so then

You have a binary operation on Y

So Y x Y -> Y

You want to take a subset G of Y that forms a group under this operation

Restricting to G gives you an operation G x G -> Y

But really you want to shrink the codomain to G as well

[marble hinge]

Come on, let him just do this :)

[knotty badger]

That’s exactly what you need “closure” for

[marble hinge]

A simple example of a function on 3-elements set

[fleet cairn]

im still in chapter 2 of allufi so im that familiar with groups

[knotty badger]

You want that if g_1 and g_2 are in G, so is g_1 * g_2

[fleet cairn]

ohhh i understand the intention now

[marble hinge]

Why we need groups and matrices for such a simple idea

[knotty badger]

This says that the image of G x G -> Y is actually contained in G

[marble hinge]

The idea is that any function can be decomposed in surjective, bijective and injective functions

[knotty badger]

And that allows you to shrink your codomain to G

[marble hinge]

No need for any category theory whatsoever

Or group tgeory

Or matrix theory

[knotty badger]

That’s true

[fleet cairn]

As in for any fixed function f: A -> C, there is an intermediate process f: A -> B -> C such it makes f injective, surjective and bijective

[marble hinge]

You just need to know what a function is, how to compose them and what is inj, bij, surj

Maybe eq class too, since he defines it like that

That’s it, and you can do example on {1,2,3}

and maybe it will clear up things a bit

[karmic moat]

it doesn't make f inj/surj/bij, but you can write f as the composition of inj, surj, bij functions

[fleet cairn]

should i be given a function first tho? not just a set

[marble hinge]

Of course

I gave you the func

[marble hinge]

Here

[fleet cairn]

f(1) = 1
f(2) = 1
f(3) = 3

so the question is

[marble hinge]

Thus func is neither inj

Nor surj

Nor bij

A funky function

[vestal trench]

a funktion

[marble hinge]

You need to make it a composition of good functions

Good means inj or surj or bij

[merry summit]

does anyone have a simple to understand proof for the criterion of gaussian integers mod p being a field. (4 divides p-3)

[fleet cairn]

ok and now we decompose it where the intermediate process is inj/surj or bij?

f: A -> C is given

f o g: A -> B - > C

[marble hinge]

What is “intermediate process”?

[fleet cairn]

like the middle of the decomposition

[vestal trench]

you aren't composing anything with your given f

youre finding three functions that compose to give you f

[fleet cairn]

oh wrong notation

[marble hinge]

It’s a function, not a “process”. Do you know how to compose functions?

[fleet cairn]

f: A-> B
g: B -> C
f o g: A -> C

[marble hinge]

Like say square composed with logarithm?

Ok

[fleet cairn]

log(x^2) im not sure the order u intend

[marble hinge]

So now you need to express f as a composition of three good functions

f = g . h . k

[fleet cairn]

Ok

[marble hinge]

Can you do it manually?

For that f func?

[fleet cairn]

what is required of those functions?

k h g

[vestal trench]

look at this screenshot you sent

it describes the properties of those three functions

[marble hinge]

They have to be injective, surjective or bijective

And yes, it’s in the screenshot

[fleet cairn]

i thought we were still talking abotu decomposition in general

oh alright

[marble hinge]

Yes

I think you forgot what you are trying to prove

Or understand

[fleet cairn]

I thought you are like making me go through a pedagogically step

[marble hinge]

I just wanted to make it more concrete for you

With a concrete function

[fleet cairn]

Yeah im sorry for misunderstanding

[marble hinge]

And concrete domain/ranges

But it looks like I am running of steam like others :)

[fleet cairn]

sorry guys ill figure it out myself

[ripe harbor]

What's your favorite way to motivate group actions? How would you describe why we care about them?

[fleet cairn]

its late anyways too

[marble hinge]

No worries, it’s fine to struggle with some things!

[karmic moat]

https://kconrad.math.uconn.edu/blurbs/grouptheory/gpaction.pdf

[fleet cairn]

ive been chatting in this channel for like almost 2 horus straight on ts problem i think

😭

i think aluffi kinda suddenly dropped ts without building up that well

we went from some natural projection A x B to A and to B

to randomly this monster of a diagram

[marble hinge]

Maybe you should switch to his other book, at least consider

What’s your reason of reading Alg Ch0?

[fleet cairn]

it says its self contained

[marble hinge]

Cat theory allure?

[fleet cairn]

thats basically it

i think some algebra books required

nubmer theory

or maybe analysis

[karmic moat]

the map $\tilde f$, defined below in the screenshot, sends an equivalence class $[a]_\sim$ to $f(a)$. check three things

1. this map is well-defined
2. this map is injective
3. this map is surjective
   just check them one by one

[cloud walrus]

anamono

[karmic moat]

i'll give you 1 to give you an idea

[marble hinge]

There are many abstract algebra books without any specific reqs other than mathematical maturity

Ch 0 is a graduate level book

[karmic moat]

suppose $[a]\sim = [b]\sim$ in $A/\sim$. this means that $a$ and $b$ are in the same equivalence class. but remember that the equivalence is literally defined by $a \sim b$ if and only if $f(a) = f(b)$. so $\tilde f([a]\sim) = f(a) = f(b) = \tilde f([b]\sim)$

[cloud walrus]

anamono

[marble hinge]

May be on the gentle side but still

[karmic moat]

so \tilde{f} is well-defined

[fleet cairn]

I actually understood this thanks

[karmic moat]

now show that it's injective: if $\tilde{f}([a]\sim) = \tilde{f}([b]\sim)$, how do you conclude $[a]\sim = [b]\sim$?

[cloud walrus]

anamono

[karmic moat]

recall that a function g is injective if g(a) = g(b) implies a = b

[fleet cairn]

i could try the contrapositive it looks more less intimidating

[karmic moat]

dont overthink it

[fleet cairn]

yes

[karmic moat]

what is $\tilde{f}([a]\sim)$? what is $\tilde{f}([b]\sim)$?

[cloud walrus]

anamono

[fleet cairn]

One of the ranges of \tilde{f} that's what we know

and to see if it injective

[karmic moat]

the map \tilde{f} is given to you

i'm basically just asking you to tell me explicitly what \tilde{f}([a]_~) is

hint: it's f(a)

[fleet cairn]

~f([a]~) is f(a) for any ~[a], so its clearly bijective?

can i make a statement like this

[karmic moat]

no we're in the process of showing it's bijective lol

bijective means injective + surjective

so we're showing it's injective right now

so if $\tilde{f}([a]\sim) = \tilde{f}([b]\sim)$ then that means $f(a)=f(b)$, right?

[cloud walrus]

anamono

[fleet cairn]

yep

oh and the equivalence classes are the same

[karmic moat]

if f(a) = f(b) can you conclude [a]_~ = [b]_~?

yes

so [a]_~ = [b]_~, thus the function is injective

now surjectivity

[fleet cairn]

ok and for surjective, prove that for all values of the codomain, there exist a x such that f(x)

[karmic moat]

what does an element of im(f) look like?

as in, what is the definition of im(f)

[fleet cairn]

it is a particular subset of B

[karmic moat]

yeah but what is its definition lol

[fleet cairn]

All values of y such that there exist x such that y = f(x)

[karmic moat]

yes

so im(f) = {f(x) | x in X}

[fleet cairn]

yep

[karmic moat]

so if i'm given $f(a)$, an element of $\operatorname{im}f$, can you find an element $x \in A/\sim$ such that $\tilde{f}(x) = f(a)$?

[cloud walrus]

anamono

[noble nexus]

Exploiting symmetry is a central theme in math, so if you have some kind of mathematical object with symmetry it is useful to study its symmetries in the abstract. The connection between an abstract group and the objects that have symmetries described by that group is exactly group actions

really group actions motivate group theory

not the other way around

[fleet cairn]

I'm not sure about this?

[vapid vale]

^

i gave a talk about this xd

[karmic moat]

recall the definition of \tilde{f}

$\tilde{f}([a]_\sim) = f(a)$, so $x$ should be...?

[cloud walrus]

anamono

[fleet cairn]

yea but does that imply that f(a) also is defiend as ~f([a]~)

x is the equivalence class of a?

[karmic moat]

other way around, \tilde{f}([a]_~) is defined as f(a)

[karmic moat]

yes, x = [a]_~

[fleet cairn]

yeye that what i meant

[karmic moat]

so we've shown that \tilde{f} is well-defined (and thus a function) and we've also shown that it's both injective and surjective, which means...?

[fleet cairn]

it is bijective

[karmic moat]

yes

[fleet cairn]

so now we have mapping of im im f to B

[karmic moat]

so from what youve said before, you understand that A -> A/~ is surjective
you now know that A/~ -> im f is bijective

can you see why im f -> B, the inclusion, is injective?

[fleet cairn]

yes

its not im f though, its mapping of image of ~f

which is subset of im f

[karmic moat]

wdym?

[fleet cairn]

OH ITS BIJECTIVE

therefore its im f

[karmic moat]

im talking about this map here

[fleet cairn]

i was thinking about talking about range of im f

[karmic moat]

yea we just showed the image of tilde{f} is im f

do you understand why the inclusion mapping im f -> B is injective?

do you know what the inclusion mapping is?

[fleet cairn]

its an identity function but the domain is smaller

[karmic moat]

yeah it's just the identity function

do you see why it's injective?

[fleet cairn]

so therefore its injective

yes

[karmic moat]

yeah

[fleet cairn]

because inclusion function is always injective

[karmic moat]

yes

so that gives you the three functions

all that's left to show is that f is actually equal to the composition of these three functions, which i leave to you

[fleet cairn]

is the 3 properties of the function the importnat part

or the equivalence class

whats the main important idea to get out of this

[karmic moat]

the important part is that f breaks into the composition of a surjective, bijective, and injective function

the equivalence class is the tool you use to show this

[fleet cairn]

oh thats actually cool af

i think itll make more sense if it talked direclty abt partition instead of equivalence class

iff that was even possible tho

[karmic moat]

im guessing the reason you look at A/~ is because later on you'll see statements of the form "A/~ is isomorphic to im f" for some appropriate "equivalence relation" ~ (spoiler alert: for groups/rings/modules, replace ~ with ker f)

[fleet cairn]

this result is kinda shocking but not so neat

[karmic moat]

this result is one of the most based results ever

[fleet cairn]

ill see about that 😈

[karmic moat]

it's so based it gets its own name: the first isomorphism theorem

[fleet cairn]

the fact that it sends im f to the codomain of f, idk it just trips me off

[vestal trench]

even if its self-contained, this is from the prerequisite material so its stuff you are hopefully somewhat familiar with or at least wont have that much trouble figuring out.

his other algebra book (notes from the underground) is also very good and has a gentler pace/assumes less prior knowledge, so you might consider trying it out and seeing if it suits u better

[glass sinew]

notes from underground is rings-first right? seems interesting

[vapid vale]

to expand on blake's answer which is absolutely correct, every axiom of a group becomes completely obvious when you consider that groups are listing functions on some thing X which preserves some structure/information that X has (i.e. functions are always associative, doing nothing surely preserves structure, and preserving structure/information means you can undo what you did)

so then knowign that groups are an inevitability, the question becomes motivating "structure/information preserving actions" (aka symmetries) as a thing to care about. for historical motivations, you can think about roots of polynomials, in which you roughly motivate galois theory (but goes back to the study of symmetric polynomials dating back to vieta etc) or projective geometry with the erlangen program. and then the reason why we care about these symmetries is that they've resulted in incredible mathematical advancements so they have some merit

for non-historical pedagogical motivations, you can just note that basically every group that you've seen is in some way defined via its action (on a polygon, on a "clock", on a finite set, on a vector space, etc)

[vestal trench]

yeah its rings-first

i think its a really good pedagogical approach actually

like you just start by talking about the integers for a few lectures which everyone is familiar with

instead of just giving the definition of a group and convincing people they should care

and then you actually appreciate what a group is and how much you can still get out of a small amount of structure

[ripe harbor]

Thanks Blake and hk!

The operation we used for defining Aut(Z/nZ) as group was regular function composition, but we consider Z/nZ as group with the operation as multiplication here, so the RHS is multiplication and the left is composition.

[ripe harbor]

So how does that fit?

[raven belfry]

yes that is composition, (f circ g) (1+nZ) what does g(1+nZ) look like? maybe m + nZ for some nonzero m? can we break m+nZ into a nice product of things in Z/nZ and use the fact that f has nice algebraic properties?

[ripe harbor]

Ah so that is right, thanks. Yeah, (fg)(1 + nZ) = f(g(1 + nZ)) = f(m + nZ) = f(m) + f(nZ) = m * f(1) + f(nZ)

And f(nZ) = 0, right (mod n)?

[raven belfry]

well waht is f(m) tho right?

[ripe harbor]

f(m) = f(1) * m

Ah wait, we need to stay in Z/nZ with the arguments of f

[raven belfry]

yeah, also think about what allows you to say that your results arent gonna be 0

[ripe harbor]

OK, consider everything mod n here to simplify notation (1 = 1 + nZ), then: f(1)g(1) = f(1 * g(1)) = f(g(1)) = (fg)(1)

That should be it

Thanks!

[marble hinge]

On the other hand, it's simpler to deal with just one operation. And one can always use integers mod N as an example of a group, if one likes dealing with integers so much. Also, it feels like those rings-first courses depend much on elementary number theory, i.e. Aluffi introduces primes, fundamental theorems of arithmetic, GCD, and all of that to motivate rings. And technically this is also something that people may or may not know. So I am not entirely convinced that rings first is pedagogically better.

[raven belfry]

how is f(1)g(1) = f(1*g(1))

[ripe harbor]

Because f is a group homomorphism

It is f(1) + f(1) + ... + f(1) = f(1 + 1 + ... + 1) = f(g(1))

[marble hinge]

I personally feel that it's better to start with one operation, explore what you can do, and then extend with another operation and distribution between the two, and see what that adds/changes, and what can you achieve with a richer structure

but I am not an Abstract Algebra teacher, so that's not a very well informed opinion :)

[ripe harbor]

It's interesting how Aluffi apparently does things, I haven't ever used the book but apparently category theory is right away introduced? Did you read it?

[marble hinge]

Chapter 0? Or Notes from Underground?

[ripe harbor]

I think the CT one is Chapter 0..?

[marble hinge]

I read parts of Notes from Underground

Notes from Underground is rings-first, Chapter 0 is groups-first, AFAIR

[raven belfry]

i dont get it

[marble hinge]

but yeah, both use category theory to some degree

I think Aluffi's books are nice, he likes to pose questions, answers them, he is chatty in a good way. And also somewhat unusual point of view, so I plan to read Chapter 0 at some point.

[ripe harbor]

Denote $\ov 1 \coloneqq 1 + n \mathbb Z$ for readability. \ $f(\ov 1)$ and $g(\ov 1)$ are in $(\mathbb Z/n \mathbb Z)^\times$. So $$f(\ov 1) g(\ov 1) = \underbrace{f(\ov 1) + \cdots + f(\ov 1)}{g(\ov 1) \text{ times}} = f(\underbrace{\ov 1 + \cdots + \ov 1}{{g(\ov 1) \text{ times}}}) = f(g(\ov 1))$$

[raven belfry]

is f not a homomorphism of a multiplicative group

[cloud walrus]

ILikeMathematics

[raven belfry]

im not sure the approach i was hinting towards really works btw

i think we need to use more information

[rocky cloak]

No, its an automorphism of Z/n

[ripe harbor]

I think in the case of elements of Aut(Z/nZ), we consider Z/nZ as additive group, so the group homomorphisms in there are wrt addition. In the case of (Z/nZ)^\times we consider it as multiplicative because well 0 isnt in there anymore and we need to pick 1 as neutral element

Jagr correct me if I'm wrong

[copper kestrel]

hey all :] i'm trying to show Z[x] isnt a PID so i'm considering <2, x> and trying to show its an ideal, but i'm not sure how to represent an element of <2,x> to show a<2,x> subset <2, x>

[ripe harbor]

Ah, so Ch0 is like a second course

[wraith cargo]

if <2,x> was principal, then 2 = af(X) and x = bf(x) for some a,b in Z[x]

show that this is only possible with f(X) = 1

but <2,x> clearly isn't the whole ring

so we have a contradiction i.e. <2,x> can't be principal

[copper kestrel]

i see! ty :]

i forgot about that lol

[raven belfry]

ah i think i was being silly, the presented proof should be fine then

[copper kestrel]

does this make sense?

ignore the fact im doing this on my notes app LMAOOOOO

[chilly radish]

You don't need to do a proof by contradiction here twice (first supposing Z[x] is a PID then supposing again that (2,x) is principal. You can just suppose BWOC that (2,x) is principal, reach a contradiction then conclude directly that Z[x] is not a PID). Other than that it's good

Oh one thing

[copper kestrel]

🫡

i ended up changing it a little bit

(i just changed the thing you asked in a different image but i dont want to spam lol

[chilly radish]

You mixed up a and b at the end there

[copper kestrel]

oh rats yeah ur right

fixed

[chilly radish]

Also, it IS possible that p(x)=-1

[copper kestrel]

rats

[chilly radish]

It's still the same thing since -1 is invertible, just in that case a=-1 as well

[copper kestrel]

can i just add a sentence that says we can note that p(x) = -1, but then b = -x and a = -2 and we still get that <p(x)> = Z[x]?

[chilly radish]

Err sorry b=-x, a=-2

[copper kestrel]

p(x) has the possibility to be -1 maybe?

[chilly radish]

It's just that as it stands what you wrote is "x has coefficient 1, thus p(x)=1" and this is untrue

[copper kestrel]

ah

[chilly radish]

You could write ap(x)=2 implies either
a=-2,p=-1 or a=2,p=1 and continue from there

[chilly radish]

Wait no thos is wrong

A priori you can have a=-1,p=-2 or a=1,p=2 as well

But p has to be invertible by bp=x thus it is 1 or -1

Ok to conclude:
You could simply use ap=2 to conclude p is of degree 0, then by comparing coefficients in x=pb you get that p is invertible in Z, so it has to be 1 or -1

[copper kestrel]

sick

ty shiN!

[chilly radish]

No problem!

[next obsidian]

Read this later:

You can also do this by showing any nonzero prime ideal of a PID is a maximal ideal.

Proof: Let (p) be a prime ideal, then by definitions we p is a prime element. Suppose (p) is contained in a proper ideal (m). Now m | p so p = km, but a prime is irreducible, so one of k or m is a unit. Can’t be m, so it must be k, so p is associate to m and (p) = (m). This says (p) is maximal.

To apply to this proof, note that (2) is prime in Z[x] but it’s contained in (2,x) so it isn’t maximal

[ripe harbor]

via its action
But its not really a group acting on a set by itself, is it? We still need to define/find a whole new function G x X -> X

[next obsidian]

This fact can be summed up as saying PIDs are 1-dimensional (maximal height of primes is 1, height is the longest chain of primes ending in a given prime [counting from 0, so it goes (0) < (p) is length 1])

Turns out for a ring that dim R[x] > dim R, so this says a polynomial ring over a PID R is a PID iff R is a field (the other direction is the fact Euclidean algorithm works).

Fun fact: dim R < dim R[x] <= 2dim R I believe, or maybe the upper bound was 2dimR + 1. If R is Noetherian then dim R[x] = dim R + 1 always

[copper kestrel]

turns out i solved the wrong problem

[next obsidian]

lol

[copper kestrel]

i only had to prove if a given norm was a euclidean norm

despair

sharp says i can still use it and just add a sentence saying so it this norm cant make it a euclidean domain

[next obsidian]

Which I guess is just ED => PID

Yeah as long as you know the proper theorems

[copper kestrel]

i think i'll just show that x+ 1 = (2x + 1)q(x) + r implies q(x) not in Z[x]

since q(x) has to be of degree 0, and x = q2x implies q = 1/2, so q not in Z