groups-rings-fields

[past temple]

lubin commented

i dont understand what hes saying tho

<@&286206848099549185>

[past temple]

nvm i figured it out

[next obsidian]

[chilly ocean]

you're welcome

[wild sapphire]

does two elements creating the exact same coset mean the elements must be identical?

[small bison]

Well there’s 1 and 3 in Z/2Z

[wild sapphire]

🤔

I'm trying to see if xH=yH (identical left cosets) implies x^2H=y^2H

I was thinking that maybe the first statement implies they're equal but it might not be that easy 🤔

[small bison]

Should be okay if H is normal

[wild sapphire]

hm, what would happen if it wasn't

[small bison]

Then it may not be true

[cyan marten]

D20 might be a counterexample, if I remember correctly.

[golden pasture]

you're basically asking if y^-1x in H isit true that y^-2x^2 in H

[cyan marten]

If you can find a subgroup H such that there are two elements of order 3 with xy in H, but yx not in H, then that's probably a counterexample.

Up to minor fixes

[vestal snow]

Is there a website where I can divide two polynomials (one of them has variables for the coefficients) and get the remainder and quotient?

[golden pasture]

sage

.quorem

[rigid cave]

Hello! So I'm reading this proof and I am confused with two things which are marked by red in the image. Why is $(h^q)^{p_1^\beta} = 1$ just because $o(x_1) = p_1^\beta$ which $\beta$ as large as possible? (Here I am assuming that $o(x)$ is the order of $x$). Also, how do you know that $o(x_1 x_2 \ldots) = o(x_1)o(x_2)\ldots$? Why does the group have to be abelian for this to happen?

[cloud walrus]

おねえちゃん

[hot lake]

if x is an element with order that is a power of p1, say o(x) = p1^z, then by definition of beta, z <= beta, and so x^(p1^beta) = (x^(p1^z))^(p1^(beta-z)) = 1^(p1^(beta-z)) = 1

they apply that with x = h^q

not exactly but it is true that if the orders are coprime and G is commutative then o(ab) = o(a) o(b)

I mean, in Z/4Z, 1 and 1 have order 4 but 1+1 has order 2 which is different from lcm(4,4)

[rigid cave]

So the theorem holds for every group?

Not only abelian?

[hot lake]

absolutely not

[rigid cave]

Okay! But how can you prove that o(x_1 x_2) = o(x_1) o(x_2)

I can't really see it...

[hot lake]

that's probably a result that is proved earlier in the course

[rigid cave]

I'm not part of that course

[hot lake]

it's not obvious I think

[rigid cave]

I have read the whole paper and nothing is mentioned about this

[hot lake]

so you want to deduce x1^n = 1 from (x1x2)^n = 1 ?

if you have proven the chinese remainder theorem, the result follows from (Z/mnZ) ~= Z/mZ x Z/nZ for coprime m,n

[rigid cave]

I haven't really read about the chinese remainder theorem...

[hot lake]

I think you're forced to use Bézout's identity somewhere

like, if the order of ab is strictly smaller than o(a)o(b)

then o(ab) = xy with x dividing o(a) and y dividing o(b)

then you say y has an inverse mod o(a)

say yz = 1 mod o(a)

[cloud walrus]

slimvesus

[hot lake]

what if o(x1x2) is not a multiple of o(x1) nor o(x2) ?

no, you showed that if (x1x2)^np = 1 then q divides n

no

wait

(x1x2)^ o(x1x2) = 1

so (x1x2)^o(x1x2)p = 1

okay I missed that you apply it with n = o(x1x2)

you used that in the proof that if (x1x2)^np = 1 then q divides n, yes

then I have no idea how your proof is supposed to work

oh right you wrote that

let me reread

it's all good

I'm dumb

[rigid cave]

Ayyyy les goo

But you didn't use the abelian fact

[hot lake]

he did when saying that (x1x2)^np = x2^np

yeah also

[rigid cave]

But why does this need the abelan fact?

[hot lake]

if G is not abelian, you don't have (ab)^n = a^n b^n

and that is used extensively

[rigid cave]

Oh yeah, you use it recursively right?

[cloud walrus]

slimvesus

[rigid cave]

Okay I think I get it now. Thank you so much guys, I really appreciate the help, seriously!

[lethal cipher]

Oh I wasn't thinking about that. It shouldn't make too much of a difference which one you start with. My bad.

[wraith blaze]

In a dihedral group, how do you know that there are n number of reflections?

I mean for example a square has 4 reflections, 1 vertical flip, 1 horizontal flip and 2 diagonal flips.

[lethal cipher]

Okay. I think a good start is to figure out how many reflections there are if the shape has an odd number of vertices

[wraith blaze]

If there are odd number of vertices, then the reflections all are from one corner to the opposite side

[lethal cipher]

And that should be it right?

[wraith blaze]

For even number n/2 corner to corner and n/2 side to side

Ah ok, I just wanted to make sure 😅

[lethal cipher]

Seems like you answered your own question. Glad I could exist 😆

[wraith blaze]

Haha

[glacial bloom]

anyone have any tips on how to do this?

for part a, we just know that r cannot be a multiple of p since p is prime

[viscid pewter]

this doesn't seem right

part b seems wrong to me?

[glacial bloom]

I'm not sure, I'm still new to this abstract algebra stuff

[cloud walrus]

wide liberty time

[carmine fossil]

H is the set of all elements whose order is not divisible by p

[magic owl]

oh lmfao

I thought H was generated by some element of order coprime to p

whoops

Thanks drake lol

Yes this problem is true, @glacial bloom what have you tried

[viscid pewter]

oh, right

[glacial bloom]

my bad I went afk

I honestly don't really know where to start, is cauchy's theorem relevant?

@magic owl

[carmine fossil]

Assume order of G/H is divisible by a prime which is not p,now use Cauchy's theorem to arrive at a contradiction

[shut halo]

Is there any connection between an algebraic number and an algebraic set?

By algebraic set I mean this https://mathworld.wolfram.com/AlgebraicSet.html

[ivory cosmos]

I'm kind of stuck at having to prove this

I think the idea is that if you have a root p / q

then you know that p | a0, and q | a3, using the assumptions

Then you need to use the fact that a3 + a2 + a1 + a0 is odd, somehow

but I'm not sure what much useful you can get?

Like, you can conclude that exactly one of a2, a1 is even

but I'm not sure where to go from there

[carmine fossil]

Ok, Here's another approach

Let's say f is reducible in Z[x]

Now,assume f=gh where deg g=1,this implies f=(ax-b)(px^2+qx+r)

Since f(0) is odd and leading coefficient is odd this implies a,b are odd numbers

But f(1)=(a-b)k contradicting the assumption

So if f is reducible ,f=c g where c is a constant and g is irreducible polynomial of degree 3,which implies f is irreducible in Q[x]( by properties of UFDs)

[next obsidian]

Vieta’s formula also says that the sum of roots is -a_2/a_3 so that a_2 is a multiple of a_3 by an odd number, so that a_2 is odd

So you can conclude a_1 is even

Dunno if that helps but

¯_(ツ)_/¯

[ivory cosmos]

Well, I understand Drake's approach

[cursive temple]

So by gauss lemma it suffices to look at irreducibility in Z[x]

The given conditions state that the constant term is odd and that the sum of the coefficients is odd

So one of the remaining coefficients is even and the other odd

Now looking mod 2 wins

[ivory cosmos]

right, cause if mod p, you have the same degree, and are irreducible, it's also irreducible

[cursive temple]

?

I mean the polynomial is never 0 mod 2

[ivory cosmos]

yeah, exactly

[cursive temple]

Oh yeah

[ivory cosmos]

from that you can deduce that it's also irreducible in Z

[cursive temple]

This is just drakes proof but with less detail i think

Ok not quite

[ivory cosmos]

no it's substantially different

[cursive temple]

I dont think you can assume that the polynomial is primitive

[ivory cosmos]

you need to use the fact that exactly one of the coefficients is even, and the rest are odd

which implies primitivity

because there's nothing in common at that point

right?

[cursive temple]

But your coefficients can be 3, 6, 9, 27

Or something of the sort

[ivory cosmos]

true

[carmine fossil]

Maybe you can show it's a constant multiple of a primitive polynomial?

[ivory cosmos]

you can always do that 😛

well, if R is a UFD

which Z is

[cursive temple]

But my argument should still hold i think

[ivory cosmos]

I don't think so?

like, how do you conclude that because f has no roots modulo p, that it has no roots at all

without assuming primitivity?

[cursive temple]

F is never 0 mod 2

Isnt that enough

Not quite

[ivory cosmos]

oh, you mean for any integer input

[cursive temple]

Yes

[ivory cosmos]

have you shown that though?

you get x^3 + x + 1 and x^3 + x^2 + 1 modulo 2

which are irreducible in Z/2Z

but uh, they do have roots

oh wait, not integer roots

[cursive temple]

In Z?

[ivory cosmos]

nvmd

never heard of the intermediate value theorem for integers bro?

it's an advanced algebraic geometry technique

[cursive temple]

The thing that you need to note in my argument is that 2 does not divide the gcd of the coefficients

Just put integers in between the other intgers ez

[carmine fossil]

Is that a real AG technique?

[ivory cosmos]

no I'm memeing

you should be able to show primitivity though

given that you end up showing it's irreducible

[cursive temple]

Why though

It isnt iff

[next obsidian]

Okay I have genius plan

Just WLOG the leading coefficient to be 1

By dividing by it

This also won’t work

Nvm

[cursive temple]

I mean the oddness stuff just dies if you make it monic

[next obsidian]

Yeah

Actually hold up

Can you divide by the GCD

That won’t change irreducibility in Q[x]

Will make it primitive

And keeps it in Z[x]

[cursive temple]

Lol true that works

[next obsidian]

Do the other assumptions still hold?

Idk if they do

And don’t want to think about it lol

[cursive temple]

They will im pretty sure

I mean the gcd is odd

So yeah they will

[next obsidian]

Yeah I was thinking about it

You couldn’t make anything even by dividing

Lmao

I was worried about going to 1 or something but 1 is odd lol

[cursive temple]

:D

[next obsidian]

@ivory cosmos

Chmonkey victory

[sturdy marsh]

[next obsidian]

Indeed

[ivory cosmos]

right so like, the line of reasoning is that

Q[x] irred => Z[x] irred

so go ahead and take the liberty of dividing out a gcd

[next obsidian]

Right

[ivory cosmos]

which doesn't change the assumptions about the polynomial being in Z[x], or having the right odd coefficients

then, look at the polynomial mod 2

this doesn't change the degree, and you get either

x^3 + x^2 + 1

or x^3 + x + 1

both of which are irreducible over the Z/2Z

just by trying both elements

hence, the modified polynomial is irreducible in Z, so also in Q

so the original polynomial is irred in Z

[next obsidian]

Yup

[ivory cosmos]

neat

[next obsidian]

I remember my TA told me about this mod p trick

[ivory cosmos]

this one feels nicer than drake's version which does more manipulation of the coefficients

[next obsidian]

Except his was like

To show something was reducible which doesn’t work but he went

“Check it mod small primes, if it’s reducible the same way each time it’s probably reducible”

And I checked and it was, and it gave me an idea as to how it factored lol

[ivory cosmos]

yeah Aluffi mentions this iirc

[next obsidian]

But it’s a neat thing to keep in mind

[ivory cosmos]

where it gives you a likely factorization

[cursive temple]

I remember using this on some comp math problems

[next obsidian]

That makes sense

[ivory cosmos]

this is sort of because Z[x] reducible => Z/pZ[x] reducible

provided modding by p doesn't change the degree

[next obsidian]

Yeah

[ivory cosmos]

so looking at it mod p can give you a shadow of this process

essentially

[next obsidian]

It’s just a necessary condition that it would be reducible in each quotient

[ivory cosmos]

right

[next obsidian]

So if it is, chances are it might be in Z[x]

[ivory cosmos]

if you do cryptography tricks like this become important

[cursive temple]

Sounds a bit like some local-global

[ivory cosmos]

because you make heuristic algorithms that cheat and then correct things

[next obsidian]

Kinda. Idk if there’s like a rigorous thing for it, like over all p

Obviously at some point p gets larger than any coefficient

But I don’t think that implies it alone

[ivory cosmos]

I'm sure you can find pathological examples

[cursive temple]

Isnt hasses principle the posterchild for these types of results

[next obsidian]

I don’t really know number theory

Haha

If you turn your head to p-adics you get Hensel’slemma

If you assume some stuff

[cursive temple]

Oh yeah hensel

Cool stuff

[ivory cosmos]

I mean yeah aren't p-adic numbers about this?

[next obsidian]

I think there’s a local to global for like

Special cases

Something like quadratics or sometbinf?

I know it fails for elliptic curves

Ah but that’s about solutions

Also it’s not about quadratics I realize

I think it’s about how many variables you adjoin

Once again my noob number theory knowledge strikes again

[ivory cosmos]

I must be missing something, cause I don't really see how you can use the criterion here?

like, you need a prime ideal containing 1 and 4

so that leaves you with (1)

but then that contains the top coefficient

[cursive temple]

substitute x+1 into x

[ivory cosmos]

umm, and then multiply out?

[cursive temple]

yes

,wa (x+1)^6+4(x+1)^3+1

shit

whats the call

[ivory cosmos]

yeah I see

you don't have to multiply everything out

just realize that all the middle terms end up with 2 as a factor

and the last term is 2

the top term is 1

so you can use (2)

[cursive temple]

oh yeah

i just put it into wa and noticed that (3) works as well

[ivory cosmos]

so uh, what's a nice and neat proof that f(x) irred <=> f(x + 1) irred

[cursive temple]

i mean

well lets see

contraposition sounds good

[ivory cosmos]

actually this is simple

[cursive temple]

ye

[viscid pewter]

sub

[ivory cosmos]

f(x) = g(x)h(x)

f(x + 1) = g(x + 1)h(x + 1)

in the other direction

f(y) = g(y)h(y)

[cursive temple]

does f(x) irred <=> f(ax+b) irred

probably yes

[ivory cosmos]

f(y - 1) = g(y - 1)h(y - 1)

[ivory cosmos]

I don't think so unless you can divide out a

[cursive temple]

just btw the same trick can be used to show that the pth cyclotomic polynomials are irreducible

[ivory cosmos]

you need it to be an invertible operation I think

[cursive temple]

well yeah

but in Q and a and b rationals

a nonzero

[ivory cosmos]

are there more general invertible functions?

I wonder how far you can stretch this reasoning

I think the general formula is that

if phi is an isomorphism of rings, then f(x) irred <=> f(phi(x)) is irreducible

using an abuse of notation 😛

[shut halo]

Can someone help me interpret this notation?

The original definition I was given was this

Does $V(f,g) = {(x,y) \in \mathbb{C}^2 | f(x,y) = 0, g(x,y) =0 }$?

[cloud walrus]

snypehype

[next obsidian]

yes

[viscid pewter]

quick sanity check: does the first part of this question not contradict the third

[next obsidian]

what's the third part

according to you

and the first part I guess

If first part is |xy| divides the lcm

and the third is that |xy| is not the lcm, then no

|xy| could be smaller

[viscid pewter]

oh not equal

i'm a fool

[next obsidian]

consider x and x^-1

oh haha

[viscid pewter]

yeah i got it instantly for some reason i thought it was just saying the opposite of the first part

well that's a fun time

[next obsidian]

oh haha

[bleak crystal]

i was told to ask the question here

why is Q not isomorphic to R

and why is Z⁴ × Z⁴ not isomorphic to Z¹⁶ (4 and 16 as a subscript)

[scarlet estuary]

as groups?

or as what

[viscid pewter]

what have you tried

[bleak crystal]

I'm on the introductory part of isomorphisms at the moment

i guess I'm not exactly sure where to start

i haven't really... tried? anything

the last one i did was really easy, since M(R) isn't communtative so it couldn't be an isomorphism anyway

[viscid pewter]

so what is isomorphism, isomorphism means everything about the groups behave the same way

and all their elements

[bleak crystal]

right

[viscid pewter]

so what are some ways in which elements can be different

so like if one is abelian and one isn't

[bleak crystal]

i don't know what that means

[scarlet estuary]

are these structures groups

i feel like that hasnt been answered

the question is very different if these are vector spaces

or fields or w/e

[bleak crystal]

rings

[viscid pewter]

uhhh a group is abelian if for all x, y in G xy = yx

[scarlet estuary]

okay, then abelian is a different criterion

[viscid pewter]

ok well that's not a thing here then

[scarlet estuary]

no, abelian rings are a thing (and also an isomorphism invariant), it just means something else

[viscid pewter]

yeah, the thing i was saying doesn't quite apply...

orders of elements

[bleak crystal]

should i be looking at the irrational elements of R?

[viscid pewter]

if there are elements of a certain order in one but not in another they can't be the same

similarly if orders of the groups are different it's a no-brainer, but that's not incredibly helpful here

[scarlet estuary]

anyway, if ℚ and ℝ were isomorphic, that would mean a bijection exists between them

[bleak crystal]

right

how do i prove there's not

[scarlet estuary]

are you familiar with infinite cardinalities?

in particular countable vs uncountable

this is a fairly standard diagonalization argument if so

[bleak crystal]

not particularly, at least we haven't talked about it in class yet

[viscid pewter]

i mean that stuff's more basic set theory

[scarlet estuary]

hm

if you dont want a cardinality argument

i guess you could reason something like this

[bleak crystal]

they're both uncountable, right?

[scarlet estuary]

if we had such a bijection f from R to Q, then we must have f(2) = f(sqrt(2) * sqrt(2)) = f(sqrt(2))*f(sqrt(2)), but also f(2) = f(1+1) = f(1) + f(1)

[viscid pewter]

they are not

[bleak crystal]

oh?

[scarlet estuary]

by definition of a ring homomorphism, f(1) must be 1

[viscid pewter]

Q is countable

[scarlet estuary]

so f(1) + f(1) = 2

[bleak crystal]

that's interesting

[scarlet estuary]

hence 2 = f(sqrt(2)) * f(sqrt(2))

[bleak crystal]

it needs to be from Q to R

[scarlet estuary]

meaning there must be some rational number f(sqrt(2)) that squares to 2

[viscid pewter]

i mean intuitively it's just like N^2 right

[scarlet estuary]

but this is impossible, hence a contradiction

[bleak crystal]

"show that the first ring is not isomorphic to the second"

[viscid pewter]

minus a few things

.>

it's an equivalence relation

[scarlet estuary]

the existence of an isomorpism A -> B implies the existence of an isomorphism B -> A

and vice versa

[viscid pewter]

if one way can't work the other way can't work

[scarlet estuary]

we call this map an "inverse"

[bleak crystal]

that's

a good point

i don't really understand this class and neither does anyone else, i just feel behind at all times

[scarlet estuary]

so thats the non-cardinality-based argument id use

[bleak crystal]

i like your argument

it makes a lot of sense

could we talk about the cardinality as well?

[scarlet estuary]

2 = 1+1 = f(1) + f(1) = f(1+1) = f(2) = f(sqrt(2) * sqrt(2)) = f(sqrt(2)) * f(sqrt(2))

but we know the square roots of 2 are irrational

so this is impossible

[viscid pewter]

so Q has same cardinality as N

[bleak crystal]

why?

[viscid pewter]

and N has different cardinality to R

ahhh i can't really be bothered

there's things all over the web

that'll explain it better

[scarlet estuary]

https://math.stackexchange.com/questions/7643/produce-an-explicit-bijection-between-rationals-and-naturals

you can think of it as sort of "Zigzagging"

[viscid pewter]

except fiddly

[scarlet estuary]

set up something like this

clearly it enumerates all positive rationals

[bleak crystal]

right

i think this is over my head

[scarlet estuary]

now apply this sort of process

[bleak crystal]

why?

[scarlet estuary]

so f(1) = 1/1, f(2) = 1/2, f(3) = 2/1, f(4) = 1/3, f(5) = 2/2, f(6) = 3/1, f(7) = 1/4...

[bleak crystal]

oh i see

[scarlet estuary]

except theres a problem: some rationals will appear multiple times

for example, 1/1 = 2/2 = 3/3

1/2 = 2/4

etc

[bleak crystal]

true

[scarlet estuary]

(in fact, all rationals will appear infinitely many times)

but tehres a "fix" to this

if a rational already appeared in the list

just "skip" it

[bleak crystal]

how does that change the cardinality?

there's still infinitely many, right?

[scarlet estuary]

like so

if we apply this process infinitely, we've given an explicit bijection between the natural numbers and all positive rational numbers

now to extend this to the negative rational numbers as well

just "intermix"

so rather than having:

f(1) = 1
f(2) = 1/2
f(3) = 2
f(4) = 1/3

we'd have:

f(1) = 1
f(2) = -1
f(3) = 1/2
f(4) = -1/2
f(5) = 2
f(6) = -2
f(7) = 1/3
f(8) = -1/3

and this gives us all positive and negative rationals

[bleak crystal]

alright, well, this is definitely over my head now, if it wasn't before

[native schooner]

Ahahah

[scarlet estuary]

now all we're missing is the only rational number that is neither positive or negative

i.e. 0

[viscid pewter]

it's just fiddly

basically it's the cool diagram

[scarlet estuary]

but we can just make f(1) equal to 0 and "shift over" everything else

[bleak crystal]

none of what you're saying connects to the original idea, to me

[scarlet estuary]

so f(1) = 0, f(2) = 1, f(3) = -1, f(4) = 1/2, and so on

[viscid pewter]

this is a bijection

[bleak crystal]

i don't see the relationship

[viscid pewter]

we're showing a bijection between the things

everything is covered

[scarlet estuary]

i'm demonstrating a bijection between the rationals and naturals

[viscid pewter]

and one-to-one

[native schooner]

He is showing that there is a bijection

[bleak crystal]

why?

[viscid pewter]

be more specific

[scarlet estuary]

this means they have the same cardinality

by definition of cardinality

[viscid pewter]

why what

[native schooner]

Because if there is than it isn't

Isomorphic

[bleak crystal]

so?

[scarlet estuary]

like okay

let me be a bit more explicit

here's the chain of reasoning:

[bleak crystal]

how is Q countable?

[viscid pewter]

so that grid is Q

[scarlet estuary]

i just demonstrated a bijection with N

[viscid pewter]

and you can go over them one at a time with N

[native schooner]

Q is countable

[viscid pewter]

with the zigzag

so they're the same

[scarlet estuary]

pause for a second

let me explain whats going on

[native schooner]

You cannot write down all the R numbers

[viscid pewter]

tfw no one pauses except you

[bleak crystal]

i would like to see you write down all the rational numbers

[scarlet estuary]

We want to show it's impossible for ℚ and ℝ to be isomorphic as rings.

In order for two rings to be isomorphic, they must have the same cardinality; that is to say, there must be a bijection between them.

One way to show ℚ and ℝ are NOT isomorphic, then, is to show they CANNOT have a bijection between them. (This won't always work for all pairs of rings, but it does work in this case).

To do this, we can do the following:

1. Show ℚ and ℕ have a bijection between them.
2. Show ℕ and ℝ do NOT have a bijection between them.
3. Conclude that ℚ and ℝ CANNOT have a bijection between them.

[native schooner]

Yes @chilly ocean good definition

[bleak crystal]

that makes some sense

not entirely lost but i still think i need to tape a baseball glove on my head

then i could catch it if it goes over

[scarlet estuary]

my argument was a way to show that ℚ and ℕ have a bijection between them

we construct a way of "listing" the rationals

[bleak crystal]

i see

[scarlet estuary]

and by "indexing" this "list" with naturals

we get a bijection

[bleak crystal]

again, still not entirely following, but following more than 0

[scarlet estuary]

now, the fact that ℕ and ℝ do NOT have a bijection is very well-known

and due to Cantor

this means that ℚ and ℝ cannot have a bijection either

since otherwise we could simply go:

ℕ → ℚ → ℝ

each step being a bijection

giving us a bijection ℕ to ℝ, which we know cant exist

[bleak crystal]

how about part 2 of my question? Zmod4 × Zmod4 to Zmod16

[viscid pewter]

so there's a list of things

f(1) = 1
f(2) = -1
f(3) = 1/2
f(4) = -1/2
etc.
so this gives us a bijection map bc we can do
1 -> 1
2 -> -1
3 -> 1/2
4 -> -1/2
etc.

everything in Q is covered

and nothing is covered more than once

[scarlet estuary]

for that, ℤ₄ × ℤ₄ and ℤ₁₆ actually have the same size

so we cant use a cardinality argument there

but we can use some other facts which an isomorphic should respect

[viscid pewter]

so this is an opportunity to just think very simply about the order of elements thing i was talking about

[bleak crystal]

oh?

[viscid pewter]

so what's the largest order of an element in z16 gonna look like

[bleak crystal]

do you mean 15?

or...

[scarlet estuary]

for example: there's an element in ℤ₁₆ called "5" such that none of these:

5
5+5
5+5+5
5+5+5+5
are equal to 0

[viscid pewter]

depression

where did you get 15 from

[scarlet estuary]

can you find such an element in ℤ₄ × ℤ₄?

[bleak crystal]

it's the largest element

[viscid pewter]

no

[bleak crystal]

the class containing 15

[viscid pewter]

nope

[native schooner]

No what

[viscid pewter]

stop guessing

[scarlet estuary]

kaisheng i think its clear theyre unfamiliar with what "order" means

[viscid pewter]

so the order of an element x is the smallest n where x^n = 1

[scarlet estuary]

uh

[viscid pewter]

wait, rings

[scarlet estuary]

remember

[viscid pewter]

urgh

[scarlet estuary]

we're working with rings here

[bleak crystal]

no

[scarlet estuary]

right, so ℤ₄ × ℤ₄ and ℤ₁₆ cannot be isomorphic

[native schooner]

But how do you prove that there is none

[bleak crystal]

I'm missing the intermediate step there

[viscid pewter]

if there is an element with order n in one group and no elements with order n in another, then they can't be isomorphic

[scarlet estuary]

okay first off

if f is a ring homomorphism then

f(0) = 0

[viscid pewter]

arghhh

[scarlet estuary]

where 0 is the additive identity in both rings

are you familiar with that result?

[bleak crystal]

yes

[scarlet estuary]

okay, so this means that

for any element a in ℤ₄ × ℤ₄, we have that

a + a + a + a = (0, 0)

so f(a+a+a+a) = f((0, 0)) = 0

does that make sense?

[bleak crystal]

yes

[scarlet estuary]

but then f(a+a+a+a) = f(a) + f(a) + f(a) + f(a) = 0

[bleak crystal]

if it were a bijection, right?

[scarlet estuary]

yes, if f were a bijective isomorphism

but thats actually a contradiction

since we KNOW there's an element in ℤ₁₆

[bleak crystal]

so since that's not the case in Z16, it can't be an isomorphism

since it can't be a bijection

am i jumping the gun

[scarlet estuary]

(specifically 5)

[native schooner]

Ahhh i get it

[scarlet estuary]

such that 5 + 5 + 5 + 5 is NOT equal to 0

which means that 5 CANNOT be in the image of f

since for all a, f(a) + f(a) + f(a) + f(a) = 0

this means f is not a surjection

hence f cannot be a bijection

a contradiction.

[the only catch here is that i "skipped" actually demonstrating that a + a + a + a = (0, 0) for all a in ℤ₄ × ℤ₄]

[bleak crystal]

i think it's obvious since Z4 is small

[scarlet estuary]

[this can be done by hand; in fact, it follows from the analogous property for ℤ₄ alone]

yeah, its easy to check

basically, the general principle at play here is that

isomorphism means the rings are "the same", just with elements relabelled

so we want to look at rings and find something thats "different"

[bleak crystal]

so i understand and follow the logic, though I don't exactly know how to phrase that in an argument. here's what I've got so far:

[scarlet estuary]

in this case, i found that every element of ℤ₄ × ℤ₄ satisfies a+a+a+a = (0, 0) the additive identity

but this is not true for every element of ℤ₁₆

so i found a difference between the two rings, and now it just fell on demonstrating why this means there cant exist an isomorphism

[bleak crystal]

f(a+a+a+a)= 0 for all a in Z4. In Z16, there is an element 5 for which this is not the case. thus f is not onto and thus not a bijection, therefore it cannot be an isomorphism

[scarlet estuary]

youre skipping the intermediary there

[bleak crystal]

yeah i know

it feels like I'm missing something

like i said, I'm a fucking idiot

[viscid pewter]

mood

[native schooner]

f(1)
must be the identity (1,1) in Z4 x Z4 since f is a homomorphism of rings with identity

[scarlet estuary]

assume f is an isomorphism from Z_4 x Z_4 to Z_16.

a + a + a + a = 0 for all a in Z_4, and hence a+a+a+a = (0, 0) for all elements a of Z_4 x Z_4.

since homomorphisms map additive identities to additive identities, we must therefore have f(a+a+a+a) = f(0, 0) = 0.

but then 0 = f(a+a+a+a) = f(a)+f(a)+f(a)+f(a) for all a in Z_4 x Z_4.

since f is an isomorphism, it should be surjective; that is, ANY element y of Z_16 can be written as f(a) for some a in Z_4 x Z_4. in particular, this means ANY element y of Z_16 satisfies y + y + y + y = f(a) + f(a) + f(a) + f(a) = 0.

but this isnt true! for example, take y = 5. then 5 + 5 + 5 + 5 = 20 = 4, which is not the additive identity of Z_16.

thus we have our contradiction; f cannot be an isomorphism.

[bleak crystal]

man i so rarely see anyone use qed

i just draw a square

[scarlet estuary]

does that argument make sense?

[viscid pewter]

tombstone gang

halmos gang

[bleak crystal]

let me go through it slowly

[scarlet estuary]

i skipped a little bit of reasoning in this line:

a + a + a + a = 0 for all a in Z_4, and hence a+a+a+a = (0, 0) for all elements a of Z_4 x Z_4.

the in-between step is:

we can write all elements of Z_4 x Z_4 as pairs (a, b) for a, b in Z_4. but (a, b) + (a, b) + (a, b) + (a, b) = (a+a+a+a, b+b+b+b) = (0, 0) for any a, b in Z_4.

and again, i still havent actually justified that "a + a + a + a = 0 for all a in Z_4"

but this can be done by hand: 0+0+0+0=0, 1+1+1+1=4=0, and so on up to 3

[native schooner]

Z16s additive order is 16

Z4 x Z4s is not

[scarlet estuary]

yes, thats a more concise way to phrase my argument

[native schooner]

Ah you are amazing man

[scarlet estuary]

"additive order is an isomorphism invariant. Z_16's maximum additive order is 16, attained by, say, 5. Z_4 x Z_4's maximum isomorphism order is 4."

QED

but that's "skipping" some steps of the justification

[bleak crystal]

if i understood what thay meant

yes your argument makes sense

[scarlet estuary]

it's the kind of argument i'd accept late into a ring theory course (since all the verification is mundane), but not in the first month

[bleak crystal]

i wish i understood algebra

[viscid pewter]

no you don't

[bleak crystal]

uhhhhh

it's march

[viscid pewter]

abandon sanity, all ye who enter here

[bleak crystal]

first month?

[scarlet estuary]

well idk how long youve been working with ring isomorphisms specifically

[bleak crystal]

yeah every time i get on this discord i feel fucking stupid

uhhh...a week

[viscid pewter]

mood

[scarlet estuary]

that's my point

perhaps i phrased it poorly

[bleak crystal]

i left this server a while back because i felt so stupid every time i got on here

i ended up dropping that real analysis class like an hour before the first test

i asked my teacher if we could talk after class and she literally just walked away without saying anything except sorry I'm busy

anyway, irrelevant

i wish i understood math

man i started algebra in 7th grade

[scarlet estuary]

also yeah, in hindsight, looking at f(1, 1) specifically is a bit quicker/simpler of an argument

not by much

it's the exact same idea

[bleak crystal]

like, fuck

we all start somewhere, but we never stop anywhere

[scarlet estuary]

but theres less reasoning required

[chilly ocean]

hey, I've been trying to ask a question here but saw that the discussion kept continuing, can I ask it now?

[viscid pewter]

ppl stop

we just don't hear from them

and that's valid

[native schooner]

Man

[viscid pewter]

ask

[native schooner]

I feel stupod at

[scarlet estuary]

0 = f(0) = f(4) = f(1+1+1+1) = f(1)+f(1)+f(1)+f(1) = (1, 1) + (1, 1) + (1, 1) + (1, 1) = (0, 0)

thats a far faster version

of my above argument

[native schooner]

Differential geometry

[scarlet estuary]

though i still like my above argument since i think it demonstrates more general principles

[bleak crystal]

ABELIAN

sorry

[chilly ocean]

anyone got any tips for this one?

[native schooner]

Yes yours is better

But mine is super short haha

[scarlet estuary]

yeah

[viscid pewter]

consider the order of G/N, and all the groups of that order?

hmm

yeah

it doesn't have to be cyclic

i think it has to be either z9 or z3xz3

oh right and i see how to get H now i think

[bleak crystal]

sorry this is an aside, but did i mention I'm taking this class as part of a high school math education sequence? fucking insane

i don't see why I need ring theory to teach calculus

but uh yeah

[viscid pewter]

hahahahahha

you really don't

[native schooner]

Sylow theorems ?

[bleak crystal]

<-slow brain

[chilly ocean]

I got no idea how to do this one I'm so lost

[native schooner]

Bruh

[viscid pewter]

i gave you a hint

[native schooner]

I feel mentally handicapped

Lmao

[viscid pewter]

o_0

watch this

[chilly ocean]

alright lemme think for a second

[viscid pewter]

q8/1

[native schooner]

What if you loon at the normalizer of the intersection of
two subgroups of order 9??

Does that help lmao

[bleak crystal]

jesus no wonder only a third of americans have a degree

[native schooner]

Nah bro

[bleak crystal]

I'm gonna follow up with more questions after Mx. cs31 is done

[native schooner]

Most degrees are bs

Drama and ahit

[bleak crystal]

i despise you

[native schooner]

Yea ahah

[bleak crystal]

bad fucking pun

[native schooner]

Well stem degrees are great

[viscid pewter]

maths degrees are bs

[bleak crystal]

stem is overrated as fuck

[native schooner]

Nah

[viscid pewter]

what is this abstract algebra tomfoolery anyway

[bleak crystal]

actually I'm getting a BA

[viscid pewter]

spending years messing around with Q8 and R/Q and crap

[bleak crystal]

BA is just a BS with foreign language, BAs are strictly better

[native schooner]

Trade schools are better than shitty college degrees

[viscid pewter]

how is that useful

[native schooner]

Idk

[bleak crystal]

not if you're a teacher

[viscid pewter]

answer: it really isn't

[solemn rain]

its cool

for some

[native schooner]

Ahh teaching is nice

[viscid pewter]

not unless you're literally going into maths

[bleak crystal]

you need a teaching license to teach, which requires a degree

[native schooner]

They pay mathematicians very well

[bleak crystal]

usually an education degree

[native schooner]

Not teachers

Applied mathematics

[vital fossil]

huh

[vital fossil]

oh

[native schooner]

Actuarial maths

[viscid pewter]

hmmmmmmmmmmmmm

[native schooner]

You can make

100k easily

[viscid pewter]

oh well actuaries

pfft

[vital fossil]

well is relative

[native schooner]

Well in the US

[bleak crystal]

32g/yr is the minimum in my state

[vital fossil]

but giving a number is good

[native schooner]

Ahahaha

[vital fossil]

LOL

[native schooner]

Come

Coke

[bleak crystal]

meth in indiana

we had an hiv outbreak in southern indiana when mike fucking hitler banned needle exchange program overnight

[native schooner]

Jeez

Why r they banning

Needle exchanges

Anyone can quickly prove the navier stokes for me?

[bleak crystal]

mike pence? electric fence pence?

[native schooner]

[bleak crystal]

because he's a piece of shit that's why

[native schooner]

Lmaoo

[viscid pewter]

more like marks and spencers amirite?

[bleak crystal]

anyway

ring theory just to teach high schoolers algebra I and II

makes sense

[native schooner]

@chilly ocean namington casuall solves the riemann hypothesis

[bleak crystal]

and i get to look forward to less than 40 grand a year! yippee!

[native schooner]

Casually

[bleak crystal]

oh well at least I'm making a difference

[native schooner]

Terence Tao learnt from namington

[viscid pewter]

have you tried giving up and becoming a personal tutor

[native schooner]

We need namington memes

[bleak crystal]

TAO

[cloud walrus]

mirzathecutiepie

[viscid pewter]

tau > pi

[native schooner]

$\psi$ moment

[cloud walrus]

[daddy]Adika

[bleak crystal]

$fuck$

[cloud walrus]

b1gchungu5

[native schooner]

Same

[bleak crystal]

you're welcome

[native schooner]

Looks badass

[bleak crystal]

yeah it does

f(uck)

[viscid pewter]

i'm a fan of phi

classic, aesthetic

[bleak crystal]

um?

[cloud walrus]

mirzathecutiepie

[bleak crystal]

$\zeta$

fuck

[viscid pewter]

you are not blessed

[cloud walrus]

b1gchungu5

[bleak crystal]

oh yeah

that letter is fucking SICK

[cloud walrus]

mirzathecutiepie

[daddy]Adika
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[native schooner]

Oh

Is it spelled

Kszi

[viscid pewter]

zeta looks cool but is also a pain to do

chi?

xi?

[bleak crystal]

xi

[viscid pewter]

epsilon

[bleak crystal]

$\xi$

[cloud walrus]

b1gchungu5

[native schooner]

Yess xi

[hot lake]

$\Xi$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[native schooner]

Sorry i am hungarian

We say kszi

[bleak crystal]

习

[next obsidian]

$\frac{\underline{\overline{\Xi}}}{\overline{\Xi}}$

[cloud walrus]

Chmonkey didnt get into Columbia

[hot lake]

$\frac{\overline \Xi} \Xi$

[cloud walrus]

Zef Klop 🍃 🌿 🌻

[hot lake]

aw Im too slow

[next obsidian]

[bleak crystal]

$\frac{\overline \zeta} \xi$

[cloud walrus]

b1gchungu5

[next obsidian]

big chungus in lowmath!?!?!?

[bleak crystal]

this is my sorority

i am the chungus

me too

i have shit due tomorrow

[next obsidian]

me too

except not actually

[bleak crystal]

i have to finish this fucking algebra

[next obsidian]

what do u have to do

[bleak crystal]

and then ed psych to read

[next obsidian]

algebra is ez

you just do the things

and then they work out

[bleak crystal]

yeah true

just do the stuff

[next obsidian]

unlike analysis where you do the things

but then it doesn't work

this meme brought to you by: I am not as good at analysis gang

[bleak crystal]

R is an integral domain. assume the division algorithm works fine in R[x]. prove R is a field

[next obsidian]

Uh so

How much thoery do you know haha

[bleak crystal]

i have No Idea why the DA matters here

something with inverses or units i bet

[next obsidian]

It implies R[x] is a euclidean domain

So like

if you know a lot of theory you get that R[x] is dim 1, so that R is dim 0, and any dim 0 ID is a field

[bleak crystal]

wayment

[next obsidian]

but I'm gonna guess this doesn't mean anything to you

lmfao

[bleak crystal]

dim

[next obsidian]

krull dimension

[native schooner]

Algebra is 100x easier than geometry for me

[bleak crystal]

linear algebra

sounds familiar

[next obsidian]

longest length of prime ideals

[bleak crystal]

oh nvm

[next obsidian]

sorry of chains of primes

yeah

haha

[native schooner]

Anyoen else hates differential geometry

[bleak crystal]

yes i do

[native schooner]

Same

[bleak crystal]

never taken it but

i HATE it

[next obsidian]

Okay so actually

I have an argument for you

[bleak crystal]

yes please

[next obsidian]

If you have the division algorithm

[bleak crystal]

am i right about units

[next obsidian]

then it's a euclidean domain, so in particular a PID

[bleak crystal]

or something

what's that

[next obsidian]

fuck

lmfao

[bleak crystal]

euclidean domain

[next obsidian]

it's literally

[viscid pewter]

it's where euclid lives

[next obsidian]

an integral domain where the division algorithm works

hahaha

[bleak crystal]

oh

[next obsidian]

Okay so

[bleak crystal]

yeah I'm gonna guess I'm not allowed to use that

[next obsidian]

You want to show that every ideal is principal

so every ideal is generated by 1 thing

[bleak crystal]

"1+1=2 because 2=1+1"

[next obsidian]

that's not what the divison algorithm is hahaha

[bleak crystal]

i am aware

[next obsidian]

sol ike

[bleak crystal]

uh

what's an ideal

[next obsidian]

if you can show that every ideal is principal

holy crap

they really want you to do this like

without any theory

RIP

[bleak crystal]

yes

[native schooner]

The proof fot 1+1=2

Is 129pages long

[bleak crystal]

that's the point of the class

[next obsidian]

there goes anotehr one

uhh

[native schooner]

Lmao

[solemn rain]

an ideal is a subring closed under multiplication from the rings elements

an ideal is principioal if its 'generated' by one element

[next obsidian]

it's not a ring if you require rings to have 1

[solemn rain]

in the sense that (a) = {na|n in R}

just multiples of a

[next obsidian]

okay so

[solemn rain]

that what it means to be generated

[bleak crystal]

wetterbern's theorem is neat

[next obsidian]

you need to show every element has an inverse okay so

let's try it this way

okay so

do you think you could show that

[native schooner]

And there is a 360 page one

[next obsidian]

the degree(f(x)g(x)) is the sum of the degrees of f and g

[native schooner]

Some idea of the scope and comprehensiveness of the “Principia” can be gleaned from the fact that it takes over 360 pages to prove definitively that 1 + 1 = 2. Today, it is widely considered to be one of the most important and seminal works in logic since Aristotle's “Organon”.

[next obsidian]

because of the integral domain-ness

[viscid pewter]

unfortunately the syntax is absolute trash

[bleak crystal]

ha, he spelled octagon wrong

[native schooner]

Lmao

[next obsidian]

so do the following

show that deg(fg) = deg(f) + deg(g)

[bleak crystal]

how does that make it a field

[next obsidian]

then this tells you the inverse of any degree 0 element must be another degree 0 element if it exists

[bleak crystal]

oh

yeah

true

[next obsidian]

So now

Umm

now I wanted to apply the division algorithm to somehow like

get that every constant has an inverse inside of R[x]

then the thing I said above

[bleak crystal]

yeah i have no fucking idea why the DA matters

[next obsidian]

implies the inverse is also a constant

so that it has an inverse in R

[bleak crystal]

that's why i have nothibg written

[next obsidian]

cuz the constants is just R

Like

[bleak crystal]

I'm not really following

[next obsidian]

Okay so

let c be a constant

aka an element of R

let's just say we knew c^-1 exists in R[x]

well what if c^-1 wasn't a constant?

what if it includes liek an x term or something

the thing I said above says that's not possible

so that c^-1 is a constant

[bleak crystal]

then c would have to be deg1

[next obsidian]

deg 0

oh right

if it had a linear term

so the point is my little thing I said means

[bleak crystal]

which we said it wasn't

[next obsidian]

if you can find an inverse for c inside of R[x]

you get an inverse inside of R

[bleak crystal]

right

[next obsidian]

so you can try to find inverses to elements of R inside R[x]

[bleak crystal]

how do we know we can find one

using the DA

[next obsidian]

so now you probably have to do some clever shit with the division algorithm

so I think you want to

divide 1

by c

so like

you can write

1 = cq(x) + r

that's what the division algorithm says

Show that r = 0

[bleak crystal]

r(x)

[next obsidian]

right

But if you can show r(x) = 0

then q(x) is an inverse inside of R[x]

which we just argued has to be a constant

[bleak crystal]

how can i guarantee r is zero

[next obsidian]

Idk, probably something like

1 is degree 0

like

r is supposed to be degree less than the thing you divided by no?

or like

c is degree 0

so shouldn't r be degree < 0

aka has to be 0

or something like that

[bleak crystal]

wait so q(x)=c^-1 right

[next obsidian]

yea

if you can show r(x) = 0

[bleak crystal]

so r(x) has to be deg0

[next obsidian]

well no

q(x) is only c^-1

after you show that r(x) = 0

because if it weren't then

well...

q(x)c = 1 - r(x)

[bleak crystal]

we're supposing that q(x) is the inverse right

[next obsidian]

so it isn't an inverse

no

You show that r(x) = 0

[bleak crystal]

what are we supposing

[next obsidian]

you aren't supposing anything

[bleak crystal]

we have to let something be something

[next obsidian]

you apply the division algorithm

to get the statement

1 = cq(x) + r(x) for r(x) like having degree < degree(c) I think

then show that r(x) = 0

then you get the equality

1 = cq(x) so that q(x) = c^-1

so is also in R

then c is invertible in R

[bleak crystal]

and this make it a field...

because c has an inverse

[next obsidian]

yeah

[bleak crystal]

interesting

[next obsidian]

I mean one thing I didn't say is

c is nonzero

[bleak crystal]

I'm like 75% following

[next obsidian]

but that's so you can even apply the division algorithm

Like what is your statement of the division algorithm

you should be able to make this work

Anyway I have to do stuff so I can't explain anymore

This probably works

so do what you gotta do to make it work

[bleak crystal]

so suppose c^-1 = q(x) or

I'm missing the starting point

[next obsidian]

bro

c is in R

by division algorithm

you get

1 = cq(x) + r(x)

right?

that's what it says?

you haven't supposed anything other than c is in R and I guess nonzero

then you prove that r(x) = 0

this proves q(x) = c^-1

so that q(x) is in R

you shouldn't be assuming anything about q(x) really

[bleak crystal]

interesting

so just by da alone, 1= cq + r

[next obsidian]

yeah

I mean

probably, I don't know what your definition of DA is

but if it's anything like the one I know

you should be able to get that

[bleak crystal]

i see

[next obsidian]

anyway gtg

[bleak crystal]

have fun

man

let c in R ≠ 0.
1=cq(x)+r(x)
cq(x)=1-r(x).

since deg(c)=0, deg(r)<0
r(x)=0.
thus q(x)=c^-1 in R[x]
so R is a field

what's missing?

[pine patio]

is Z^2 = <(1 0), (0 1)> ?

yes i mean Z x Z. i want to show that it's not cyclic

so do i just note that (1, 0) \neq (0, 1) ?