groups-rings-fields

[barren sierra]

Consider identity: ℤ -> ℤ

kernel is 0, not a maximal ideal

[coral shale]

💀

[tribal moss]

The kernel of a surjective ring homomorphism onto a field is maximal.

[barren sierra]

Why the 💀

[coral shale]

Z is a fie-

thats a nice fix

[slim kayak]

hehe

[coral steeple]

I have shown that $\Phi: C_n^\times\to\operatorname{Aut}C_n$ given by $a\mapsto \varphi_a$ where $\varphi_a$ is multiplication by $a$ is an injective homomorphism, but I can't show that $\varphi_a(x)=\varphi_a(1)\varphi_a(x)$ (ie $\Phi$ is surjective)

[cloud walrus]

person2709505

[slim kayak]

alwaysh remember the algebraic characterisation of isomorphisms

inverses

[coral steeple]

I tried showing the inverse is an iso but that was harder lol

[slim kayak]

so you want this to be a group homomorphism

[barren sierra]

what you wrote at the end is ax = a * ax?

[slim kayak]

then the inverse of phi_a must be phi_a^-1

[coral steeple]

Uhhh that sounds bad

[slim kayak]

Can you see why those are inverses?

[barren sierra]

am I wrong?

[slim kayak]

no

[barren sierra]

Unless I misunderstand your notation, phi_a(x) = ax and phi_a(1) = a

so you're trying to prove that ax = a^2 x always

which of course isnt working

[rustic crown]

idealy it should be phi_a(x) = phi_a(1+1+1...+1) = x * phi_a(1)

[slim kayak]

should be phi_a(1)x if anything

[coral steeple]

Hmm well the problem is definitely showing Phi is surjective, my alternative way of writing that might just be wrong

[coral steeple]

Yes that's it

[slim kayak]

can you show this?

that phi_a and phi_a^-1 are inverses?

[coral steeple]

I need to check how much of my scratchwork has been nonsense

[tribal moss]

For surjectivity of Phi, you would show that for an arbitrary automorphism psi, you have psi = Phi(psi(1)).

[coral shale]

Man I feel so silly here,

[tribal moss]

It might reduce notational confusion first to show as a lemma that if phi and psi are homomorphisms Cn -> Cn and phi(1) = psi(1) then phi=psi.

[coral shale]

||for every left multiplication by a, there exists a in Cn^x. So like... explicit construction? Am I sillying
nvm ur showing every automorphism is a left multiplication||

oh nvm im sillying

[rustic crown]

shuwi siwwy

[slim kayak]

sillymaxxing

[coral steeple]

I can get the result from this step but I can't see how to prove this...

[delicate orchid]

A homomorphism is uniquely determined by its image on generators, just like a linear map on a basis

Cn is cyclic so the only generator is 1

[barren sierra]

I am so dumb I've spent this whole convo reading C_n as complex numbers lmfao

[dull marsh]

Wait I don't quite understand that, you are saying that a homomorphism is uniquely determined by generators of its image?

[coral steeple]

Image of generators no?

[tribal moss]

Reember that any element of Cn can be written as 1+1+1+...+1 for some appropriate number of ones.

[dull marsh]

Ah okay I see now

[coral steeple]

Argh I've been trying to avoid thinking of multiplication as repeated addition because "that only works in Z". I have been silly

Yep, I see that now, thanks everyone!

[slim kayak]

i mean... thats almost correct too

[tribal moss]

For certain values of "almost".

[slim kayak]

lmao

[coral shale]

what would the n even mean

C^n??

[barren sierra]

That's how I was reading it

I need more coffee

[slim kayak]

roots of unity was what I meant

uh, yeah

[delicate orchid]

The result definitely does NOT hold for the complex numbers lol

[rustic crown]

wild automorphisms of C

also ig just the complex conjugation isn't multiplication by a non-zero

[delicate orchid]

There is ONE automorphism of C. I refuse to hear any different

[coral shale]

{0} is a field, I refuse to hear any different. Now there are 2

oh wait im tripping

but anyways the zero map is clearly an automorphism

[slim kayak]

what?

[coral shale]

in so many ways

[dull marsh]

There is ONE empty set, I refuse to hear any different

[delicate orchid]

… lol?

[slim kayak]

i think he did one of those 1 =2 fake proofs, in which case she is right

[coral shale]

my speed of thought is so fast that no one can understand

[rotund aurora]

the quality of the math here

[delicate orchid]

I’m chatting shite and need to save face

[coral shale]

i aint doin worse than #discussion its ok

[chilly ocean]

0 should be different from 1

[slim kayak]

does every automorphism of C at least fix R?

it fixes Q so...

whoopsie

[coral shale]

are we on about field automorphisms
im so dead

[tribal moss]

No, it doesn't need to fix R.

[slim kayak]

whats an counterexample?

[coral shale]

https://math.stackexchange.com/questions/412010/wild-automorphisms-of-the-complex-numbers

[rustic crown]

if it fixes R, it would be either id or conjugation

[delicate orchid]

They’re all non-constructable

[tribal moss]

That needs the axiom of choice, so can't be exhibited explicitly.

[slim kayak]

i'll reject C for now then

okay so how do i select my maximal ideals now

[coral shale]

all surjections have a right inverse..

[slim kayak]

nuh uh

the index function has no inverse

[coral shale]

hecks that

[delicate orchid]

Indexes stuff

[coral shale]

[slim kayak]

send any element in the disjoint union to it index of the set it came from

[delicate orchid]

Anyway not all epimorphisms are split sweaty. Better luck next time

[coral shale]

are disjoint unions allowed empty sets in them

but ite intresting

[chilly ocean]

What does the "order preserving" mean

[dull marsh]

x <= y implies f(x) <= f(y)

Wait

Well anyway that's the only definition of order-preserving that I know so

[delicate orchid]

Yeah that’s correct

[dull marsh]

Ah I guess the order here is inclusion

[chilly ocean]

So <= indicates inclusion symbol?

[dull marsh]

Yeah

[ \mathfrak x \subseteq \mathfrak y \implies \phi(\mathfrak x) \subseteq \phi(\mathfrak y) ]
In this case

Mathfrak is wild damn

[cloud walrus]

A Lonely Bean

[delicate orchid]

Most sane mathfraker

[slim kayak]

if b contains a, then the image of b contains the image of a

first time seeing math frak ideals other than a, b or p

[dull marsh]

I've seen o used for that too actually

[slim kayak]

rest were all capital letters too i believe

[dull marsh]

I think

[delicate orchid]

I’ve seen capital S

And lower case t, g, h for Lie algebras

[slim kayak]

yeah capital letters are fair game when they look nice

[dull marsh]

Yeah capital S in mathfrak looked like G to me at first

[ \mathfrak S ]

[cloud walrus]

A Lonely Bean

[delicate orchid]

Shame none of them look nice

[slim kayak]

$\mathfrak F$ is I think pretty nice

[cloud walrus]

Kerr

[slim kayak]

nevermind

burn that thing

[dull marsh]

[ \mathfrak M ]
[ \mathfrak V ]

[cloud walrus]

A Lonely Bean

[dull marsh]

Wait what

[slim kayak]

oh right m

maximal ideal time

[rustic crown]

V is cursed

[slim kayak]

how did anyone think thats how you could write v

[cloud walrus]

A Lonely Bean

[dull marsh]

I wanted to show this mb

[slim kayak]

why does V suddenly have pi_1 = Z

[rustic crown]

looks like it

and D becomes simply connected

[dull marsh]

Capital O also looks like D

[rustic crown]

abcpqm are all i use with mathfrak

[delicate orchid]

a is pushing it…

[slim kayak]

at least its simply connected

with mathfrak you could expected it being dissected into the 4 path-components

[rustic crown]

[slim kayak]

$\mathfrak O \mathfrak D$

huh

[delicate orchid]

The 0th homology of mathfrak

[rustic crown]

(how long until a mod sends us to #latex-testing?)

[slim kayak]

sadly we only live in 3d dimensions, we cant even conceive what part of the mathfrak letters we cant perceive

[topaz solar]

And none are measurable except conjugation iirc

[delicate orchid]

I could measure them

[summer path]

:3

[dull marsh]

What's up with automorphisms of C?

[tribal moss]

C is isomorphic as a field to the algebraic closure of Q(t1,t2,....) where t1, t2, ... are continuum many independent transcendentals.

[dull marsh]

Oh

Can reals be presented in a similar manner?

[tribal moss]

No, there are no nontrivial automorphisms of the reals.

(We can recover the ordering of the reals by defining x >= y iff there exists a such that x = a²+y, and then an automorphism must preserve the Dedekind cut of every real).

[dull marsh]

Looks like I should study field theory more

Interestingly enough I have proven the last thing you mentioned

[tardy hedge]

What yall doin for christmas?

Abstract algebra?

[vernal flume]

Is there an incredibly obvious way to do this?
What I did was:
n=2k+1 for some natural k, logically a^2k=b^2k
a^2=b^2
then a^n-2=b^n-2
a^n a^-2 = b^n b^-2
subbed for n, and a few eliminations later a^-1=b^-1 which proves it.
But this is quite lengthy.

[vernal flume]

christmas got over here 😔

[slim kayak]

uh, yeah

n=2k-1, then a^2k=b^2k but a^2k=a^(2k-1)a=ea=a, same for b

[dull marsh]

Oh wow that seemed to good to be true at first

Because the implication should hold for all groups of odd order

[vernal flume]

Hadn't thought of it that way lul. That considerably shortens the proof. Thank you!

[glossy crag]

I recently was reading about Fraktur, apparently the stereotypical Nazi connection is not entirely correct, because although they supported it initially as TRVE German script, they discontinued it later (in the 40s iirc).

[slim kayak]

with finite orders it can be helpful to keep everything positive yknow?

[vernal flume]

Yup, I'm learning.

[cloud solar]

How to show if R is a commutative ring, a in R nilpotent element then a+x invertible for every invertible x

[stark helm]

Can someone explain what does this lattice mean? I aim to know why the maximal ideal of Z36 is <2> and <3>?

[celest furnace]

Try using a geometric series

[naive whale]

It is a lattice of ideals ordered by inclusion, so the ideal generated by 2, <2> = {0, 2, 4, 6, 8, ..., 34} contains <4> = {0, 4, 8, ..., 32} etc

All the ideals contain <0> = {0} so it's at the bottom

[celest furnace]

1/(a+x) should be equal to x^-1/(1+ax^-1)… is ax^-1 nilpotent?

[naive whale]

An ideal is maximal if there is no proper ideal properly containing it and the diagram shows this clearly, there is no ideal above <2> or <3> except Z36 itself

[rotund aurora]

You can also do it like this ||(x+a)(x-a)=x^2-a^2||, ||(x^2-a^2)(x^2+a^2)=x^4-a^4||, etc. Eventually, you are just left with x^n which is invertible

[celest furnace]

Very clean.

[cloud solar]

So nice

Thanks

[south patrol]

Another way is that a is in every prime ideal. If a + x is not invertible then it is contained in a prime ideal p and so x is also in that prime ideal p, contradiction

[celest furnace]

Why must every ring have a prime ideal that’s not just the whole ring?

[slim kayak]

thats also a pretty useful argument overall

[south patrol]

Well the whole ring isn't prime

[slim kayak]

You cant have both 1+x and x be contained in a prime ideal

[south patrol]

Actually idk what the case is with 0

The 0 ring I mran

Empty spectrum is the convention actually nvm

[slim kayak]

Prime ideals have the small tid bit of not being the whole ring in their definition, easy to miss

[celest furnace]

Ah

[south patrol]

But uh

It follows from Krull's theorem that every non-zero ring has a maximal ideal

Though existence of prime ideals is slightly weaker set theoretically

I prefer to explicitly write an inverse as others have done, bit as Kerr says, the technique is useful

[celest furnace]

Didn’t know this had a name

[south patrol]

tbf there are too many theorems by krull

lol

But yeah it's a standard application of Zorn's 🍋

[slim kayak]

actually not that I think about I dont know why prime ideals should exist

Besides maximal ideals being prime by definition

[south patrol]

there is uh Boolean prime ideal theorem right

[slim kayak]

huh

[naive whale]

We assumed the existence of a nilpotent element, isn't this essentially an additional restriction that the nilradical has a nontrivial element => there is at least one prime ideal?

[south patrol]

yeah it's equivalent to BPI

which is strictly weaker than AC

[slim kayak]

i sure hope the nilradical is non-empty

[south patrol]

equivalent to ultrafilter lemma i.e. every filter contained in an ultrafilter

[south patrol]

To prove that the nilradical is the intersection of primes also needs smth choicy so there is a sort of like

lack of independence between the two

[delicate orchid]

"too big to be maximal" moment

[slim kayak]

maximaler

[topaz solar]

Ultrafilter lemma my beloved

[slim kayak]

if I have pairs $(f/2^n,g/x^m)$ with f and g elements from Z[X], is there some nice way to characterize these pairs when I have the condition that $f x^m = 2^n g$? Seemingly given an polynomial g with the first m coefficients zero this would give me a valid f in that pair. So I have some triple (h,n,m) and the n and m being natural numbers and h some arbitrary polynomial with g=$hx^m$. I must be missing something, right?

[cloud walrus]

Kerr

[south patrol]

real

[dire siren]

@slim kayak you have x^m | 2^n g, so x^m | g, so g=hx^m, for some h in Z[x]
then f=2^n h

[slim kayak]

🤦 thanks...

so its just Z[X] is disguise

[dense flare]

Hi guys, I have a question considering (d). If G is commutative, isn't it necessary for S to be equal to G and hence not satisfying being a proper subset?

[coral spindle]

Indeed G cannot be Abelian if S is to be proper.

[dense flare]

thanks

[glossy crag]

If G=<x> is cyclic, the subgroup of nth powers ought to have index n, but I'm blanking on why it's so.

[rustic crown]

you mean infinite cyclic?

subgroup of C_2 consiting of 4th powers only has index 2

[glossy crag]

No I meant finite.

[slim kayak]

and G has order n?

[glossy crag]

n can be assumed to be strictly less than the size of the group.

[glossy crag]

No, it's whatever else.

E.g. if L is the finite field of size q^n, the subgroup of (q-1)st powers ought to have index q-1.

It comes down to checking the elements 1,x,...,x^{n-1} are inequivalent mod the subgroup, but I'm blanking on that for w/e reason.

[slim kayak]

So you have a cyclic group of order n and then consider the subgroup which consists of k-th powers of the elements of the original cyclic group?

[glossy crag]

Yes.

[slim kayak]

it should have index n/k

[glossy crag]

No it shouldn't, you're confusing that with the subgroup <x^k> of <x>. I misspoke.

[slim kayak]

what is the subgroup of nth powers then

?

[glossy crag]

Yeah ok, the subgroup of nth powers is <x^n> so it has index n (not index n/k), that ought to work.

[rustic crown]

you still need n to divide |<x>|

[glossy crag]

I'm fine with that assumption, it does not disturb what I'm doing.

[rustic crown]

Z/nZ, the subgroup of dth powers is dZ/nZ

quotient is Z/dZ

[slim kayak]

how is the order of x not n^2 if <x^n> has order n

[agile burrow]

You can also explicitly count coset representatives

[slim kayak]

huh, oh the number of cosets?

[rustic crown]

hewwo walter

[glossy crag]

That's what I meant, I'm just having trouble explicitly checking 1,x,...,x^{n-1} are inequivalent, lol.

[agile burrow]

Hey det Hope you're doing well

[rustic crown]

if two were equal mod <x^n> then you'll get x^{i-j} in <x^n>, so a smaller power is in the subgroup of n-th powers

[glossy crag]

You're using here that the nth powers are equal to <x^n>, that's what didn't cross my mind at the beginning lol (otherwise it's easy).

[rustic crown]

[glossy crag]

In any case, I'm fine now, thanks.

[rustic crown]

yee

nice tie

oh its a muffler

[agile burrow]

Scarf lol

[rustic crown]

still nice

[agile burrow]

For when it's cold out

Thank you I like the Christmas hat

[rustic crown]

me opened windows and its cold

should close again

[glossy crag]

Just to triple check I'm not still hallucinating, if <x> is finite cyclic and n divides its order, then the subgroup of nth powers is equal to <x^n>, right?

[rustic crown]

yep

[agile burrow]

That's right. You can get that every subgroup is cyclic via like division algorithm or something

[glossy crag]

Yeah I know that, can take it from here.

[rustic crown]

if n didn't divide m = order of group, then it's <x^gcd(m,n)>

[slim kayak]

man this channel sure is active this season, so many new messages

[next obsidian]

I’m a chmonkey

[rustic crown]

I'm a det

[celest furnace]

I am the ULTIMATE CHAD

[tardy hedge]

Kian

[dull marsh]

I am a lonely bean

[chilly ocean]

So what

ㅡㅡㅡㅡ
Why k[x]/I is a field?(I={f(x)|f(0)=0})

[dull marsh]

Doesn't every polynomial get reduced to just the constant term then?

And that should be invertible given k is a field

[slim kayak]

Uh yeah, think about how polynomials look if they have a root at zero

With those you can "reduce" any polynomial in k[x]

[dull marsh]

During a lecture my professor considered the field $\bZ_2[x] / (x^2 + x + 1)$ and got the following tables for addition and multiplication

I am confused because x^2 + x + 1 is just 1 in Z2

[cloud walrus]

A Lonely Bean

[hot lake]

x is not in Z2

[dull marsh]

Yeah sorry I meant the polynomial ring there

I am confused because x^2 + x + 1 is just 1, so how come (x^2 + x + 1) is not the same as Z2[x]?

[hot lake]

How is it 1 ?

[dull marsh]

It's equal to 1 at 0 and 1, should I be considering x to be an element of some larger field?

[next obsidian]

That’s 0 dawg

[hot lake]

Ah it's 1 when evaluated at 0 or 1 ?

[dull marsh]

Yeah

[next obsidian]

And for polynomials, these are formal objects

They aren’t considered equal via equality as a function

[dull marsh]

So my mistake was confusing polynomials with polynomial functions I guess

[hot lake]

Yeah

Also if you use the tables to compute x^2+x+1 where this time x is the class of x in the field you are given, you get 0

[dull marsh]

I see, thanks

[hot lake]

So in that field, the polynomial function of y^2+y+1 isn't 1

[cloud walrus]

Plazzi

[chilly ocean]

Sorry my english math vocabular isn't very good, so I don't know if it's actually called the outer direct product

[cloud walrus]

Plazzi

[cobalt heath]

What is the problem with r(cos phi + i sin phi) being in R+ × S^1 ?

[chilly ocean]

I thought that when we take the direct outer product of two groups, I only combine the original group elements with each other. So for
r_1,r_2 in R and z_1,z_2 in S^1 it should be
(r_1,z_1)•(r_2,z_2)=(r_1×r_2,z_1×z_2)

[cobalt heath]

Yea, it does

I still do not see the issue.

[chilly ocean]

For example how can i "visualize" 10i with the direct product?

[cobalt heath]

Ah, do you mean the identification that (r, e^(i phi)) = r e^(i phi) ?

[chilly ocean]

Yes

[cobalt heath]

I would say this is simply an identification.

One can consider \bC^\times as a direct product of \bR_+ and S^1, after all.

[chilly ocean]

This was the exercise to show that this is true.

[cobalt heath]

Yea, like there is a straightforward way to show that.

[chilly ocean]

And this was the solution, but i'm so confused why we're allowed to say (r,e^(i phi))= r×(e^(i phi))

[cobalt heath]

What was the exact statement of the solution?

[chilly ocean]

Every $z \in \C^{\times}$ can be written as $\ z=r \cdot e^{i\varphi}$ with $r \in \R_+$ and $e^{i \varphi} \in \mathbb{S}^1 \$
This shows
$\C^{\times} = \R_+ \times \mathbb{S}^1 \$

[cloud walrus]

Plazzi

[cobalt heath]

Hmm, this statement is missing "uniquely"

But it checks out, this is different than saying r e^i phi \in R_+ \times S^1.

[chilly ocean]

Oh, i see

I think

[cobalt heath]

Do you recall definition of direct product?

[chilly ocean]

Yes G=U×W with (U,×) and (W,°) (at least two groups)

Afaik G doesn't have to be a group

[cobalt heath]

What about e.g. categorical definition?

[chilly ocean]

categorical definition?

[cobalt heath]

(Doesn't matter by much but it helps if you learned that)

Yea the universal property

[chilly ocean]

I'm not sure if i learned that

[cobalt heath]

Well, then I guess this uses identification instead.

[chilly ocean]

Thank you ♥️

[chilly ocean]

Why this holds(p is ideal)

(p+(x))(p+(y))=p+(xy)?

[daring nova]

it's in their product = p^2 + (x)p + p(y) + (x)(y) = p + (x)(y)
Idk much about ideals, but maybe (x)(y) = (xy) ?

xy y^-1 = x so (xy) contains x
Similarly (xy) contains y

Hence it (xy) contains (x)(y)

The other inclusion is trivial

If your elements are invertible

https://math.stackexchange.com/questions/1267777/product-of-principal-ideals-a-cdot-b-a-b

[chilly ocean]

Why p^2+(x)p+p(y)=p@daring nova

[chilly ocean]

I see now

[dire siren]

not equal, but ⊆

[chilly ocean]

Bezier said they are equal

[rocky cloak]

They're equal modulo p, which is what's relevant

[dire siren]

I think he refered to (x)(y)=(xy)

but the exact same question about (p+(x))(p+(y)) and p+(xy) was asked in my commutative algebra course

the professor incorrectly wrote that they are equal, and she corrected this in another lecture after someone pointed out

[chilly ocean]

Why they are different..

[cobalt heath]

Ideal business is so tricky

[chilly ocean]

p^2+(x)p+p(y)+(x)(y)=3p+(x)(y)=p+(xy)..where's an error

[glossy crag]

If F is a free group, then any exact 1->N->E->F->1 splits, right (just send the free generators of F to their preimages in E)?

[dire siren]

both equalities are wrong

[chilly ocean]

Why (x)p=/=p

@dire siren

[dire siren]

a trivial counterexample is x=0

or, think about Z with p=7Z and x=7

LHS is 49Z, while RHS is 7Z

[chilly ocean]

Whats lhs...

What does it stand fot

But multiplying something with ideal doesn't change it..afaik

[dull marsh]

Left hand side

[chilly ocean]

Do u think I can't study math

[dull marsh]

I think you can

[chilly ocean]

But I remember few things and can't understand anything without explanation

According what I said then IJ=I=J which is not possible

But what was my error

[dull marsh]

Are we still on the question about (x)(y) = (xy)? I think they are equal

[slim kayak]

Yeah

[chilly ocean]

The error was IJ doesn't mean {ij} I guess

[cobalt heath]

Oh, (x)(y) = (xy)? I am somehow confused abt this
(Well, seems obvious now)

[mighty kiln]

Should be true for crings

Probably not rings or crngs tho

[slim kayak]

Works out fine for principal ideals

[elder wave]

These do not exist

☝️

[slim kayak]

It is sufficient to define product ideals as the set of products of elements in (x) and (y), then commute a bit

[dull marsh]

What's a cring?

[mighty kiln]

Commutative ring

[slim kayak]

And non-commutative rings don't exist

[fading field]

it’s true

[mighty kiln]

When freyd-mitchell

[cobalt heath]

Lmaooo

[slim kayak]

Huh?

If my abelian category isn't equivalent to one of modules over crings I don't want it

[dull marsh]

Every ring is commutative and unital

[slim kayak]

a ring without unit is a rng, but that's stochastics so I won't touch that

[mighty kiln]

The adjective for commutative unital ring is "cringy"

[fading field]

is it because it has a younit

[chilly ocean]

Isn't it unity

[mighty kiln]

[slim kayak]

Is a book about commutative algebra a cringe compilation then?

[cobalt heath]

Huh, stochastics?

[elder wave]

Lmao

[elder wave]

@next obsidian owned

[dull marsh]

There are also rigs

[mighty kiln]

Huh

[dull marsh]

Rig = semi-ring (ring minus the requirement of additive inverses)

[slim kayak]

Rings without nunity

What does the n have to do with it

[cobalt heath]

What is ring without 0

[chilly ocean]

There's no such ring

[dull marsh]

A double monoid I suppose

[slim kayak]

An associative quasigroup with monoid structure

A must have really

[cobalt heath]

Ahhh

[naive whale]

a bioid?

[dull marsh]

That makes more sense

[slim kayak]

the additive monoid would have a 0 then

[chilly ocean]

ㅡㅡㅡㅡㅡㅡ
If IJ:={ij}, then IJ=I=J? What did I wrong

[slim kayak]

IJ=(ij)?

with I=(i) and J=(j)?

[chilly ocean]

No {ij}

[slim kayak]

that doesnt make any sense

[jagged gate]

you're doing ideals or what

[slim kayak]

thats not an ideal, an ideal always contains 0

[chilly ocean]

{x|x=ij} I mean

[south patrol]

so {ij}

[slim kayak]

with i in I and j in J? 😭

[chilly ocean]

Yes

[slim kayak]

consider the example of (2)=I and (3)=J

[south patrol]

I would more just ask, what have you done

you have just stated something which is generally false and then asked us what you did wrong

[chilly ocean]

Then J=/=IJ=/=I

[slim kayak]

yuh huh

and IJ consists of all multiples of 6, which are products of elements in (2) and (3)

How did you arrive at I=J=IJ anyways

[chilly ocean]

rI=I is incorrect? rI belongs to I is correct?

@slim kayak

[slim kayak]

rI = I if r is invertible

and $rI \subseteq I$ always

[cloud walrus]

Kerr

[chilly ocean]

ㅡㅡㅡㅡ

[tardy hedge]

Is {1, sqrt(3)} a basis for Q(sqrt(3), sqrt(21)) over Q(sqrt(7))

[south patrol]

Yes I mean Q(sqrt(3),sqrt(21)) = Q(sqrt(3), sqrt(7))

[tardy hedge]

Yea

[south patrol]

Ye

[tardy hedge]

This doesnt seem complicated but for some reason its kinda hurting my brain just a little bit lol

[south patrol]

write k = Q(sqrt(7)) and then a basis for k(sqrt(x)) over k is just {1,sqrt(x)}

:)

[tardy hedge]

Truuuue

So the point here is that {1, sqrt(3)} with scalars from Q(sqrt(7)) generates Q(sqrt(3),sqrt(7))

[south patrol]

ye

and important bit is that the extension is degree 2 as well so any two linearly independent elements form a basis

[tardy hedge]

Answer to 3 is just 12?

[chilly ocean]

I think so

[dire siren]

yes, it's 12

but the proof is not quite straightforward

[chilly ocean]

How to prove

[dire siren]

use [Q(a,b):Q]=[Q(a,b):Q(b)][Q(b):Q]

the first index is not that straightforward, because it requires checking that a polynomial is irreducible over Q(sqrt4), rather than on Q
luckily it's a cubic

oh, wait, I think we can just say that because [Q(a):Q]=3, then 3 divides [Q(a,b):Q]
likewise, since [Q(b):Q]=4, then 4 divides [Q(a,b):Q]
so 12 divides [Q(a,b):Q], so [Q(a,b):Q]>=12

[tardy hedge]

Finite field extensions have all their elements algebraic over the base field. What is the connection between the degree of the field extension and the degree of some element from the field?

[dire siren]

on the other hand [Q(a,b):Q]<=12 (#groups-rings-fields message)
so [Q(a,b):Q]=12

[tardy hedge]

Oh wait

That was a theorem

[tardy hedge]

Its a divisor

Of the degree of field extension

[slim kayak]

Was there some more general method outside of primitive element theorem for this?

[tardy hedge]

It seems like a common trick in this field stuff is constructing some intermediate fields

And forming some sort of bridge between fields

[slim kayak]

Hm but doesnt that come down to using stuff like being relatively prime?

[tardy hedge]

For q4. u in F is algebraic over K because F is a finite extension. The degree of u over K is a divisor of p, it cant be 1 so it must be p. That implies that the set {1, u, … u^(p-1)} is linearly independent in F, so its a basis for F. So F = K(u)

[slim kayak]

Like, what if you replaced the third with an eight root in exercise 3?

[tardy hedge]

Sounds good?

[glossy crag]

Anyone?

[slim kayak]

Yeah

Actually, F is free abelian?

Like this only works too my knowledge when E is abelian, which forces F to be abelian or free abelian in this case

[glossy crag]

No.

[agile burrow]

It seems like it should hold for free groups in general, yeah

[glossy crag]

I expected so, just checking (I'm not so comfortable with free groups).

[agile burrow]

You can also choose to interpret this cohomologically, they certainly split for abelian kernel since H^2 of a free group vanishes

[glossy crag]

That's why I asked this, lol, wanted to show H2 vanishes.

[agile burrow]

There are other ways of showing this as well if you want to think about it topologically, if you care to hear about that

[glossy crag]

I do indeed, I'm not yet versed in how all this relates to topology (e.g. Eilenberg/MacLane's original definitions via aspherical spaces and whatnot).

[agile burrow]

Sure, so you might be familiar with K(G, 1)s? These are topological spaces whose fundamental group is G and whose higher homotopy groups vanish. It turns out the integral cohomology of G agrees with the integral cohomology of a K(G, 1). One way to see this is that the universal cover of a K(G, 1) is a contractible space (by Whitehead's theorem), so it's cellular chain complex is a resolution of Z. Furthermore, the action of G (viewed as the fundamental group of the base space) on the universal cover makes the cellular chain groups of the universal cover into free ZG-modules. Thus, the cellular chain complex of the universal cover is a free ZG resolution of Z, so you can use it to compute group cohomology

In practice, a common way to construct a K(G, 1) is to start with a contractible space X that G acts on freely and properly discontinuously (and I'm probably forgetting some adjectives or something). Then by covering space theory, the quotient X/G is a K(G, 1). In the case of free groups, an easier way is just to invoke Van Kampen on the wedge of n circles. This space has fundamental group F_n and it's universal cover is contractible, so it's a K(G, 1). Then one can see that it's higher cohomology groups vanish since it's a 1-dimensional CW complex.

[glossy crag]

Neat.

[agile burrow]

Admittedly I'm not too sure how the story works when you want to do homology with coefficients if you have a non-trivial action. I think you have to look at local coefficients or something which I'm not familiar with.

[glossy crag]

Wait, when you say H^n(F,A) vanish do you assume that F acts trivially? H^2(F,A), at least, should vanish regardless of the kind of action.

[agile burrow]

It should vanish regardless in general

[slate tide]

Yeah local coefficients is where it’s at, the bundle of groups picture in hatchers chapter 3 is pretty nice when talking about this

[agile burrow]

If I'm remembering correctly it's because you can interpret H^n(F, A) as Ext_{ZF}(Z, A) and since the integral cohomology vanishes in degrees higher than 2, Z has projective dimension 1 as a ZF-module so all the higher Ext groups vanish regardless of A.

[agile burrow]

Oh ok, I'll check this out

[low wyvern]

Can I please have a hint to get me started as I am unsure where to proceed from gcd(h,k) = 1 ...

[rustic crown]

do you know lagrange's theorem?

[low wyvern]

I do, but I never thought to use it because I believe this question comes before that in my lec notes.

Ill have another go thanks

[low wyvern]

Ok so, so far I have that since h and k divide |G|, I can write; for some x,y in Z, we have hx=|G| and ky=|G|. Thus, by the fact that gcd(h,k) = 1 and by Bezout, we have hx + ky = 1........ [I think now I need to use properties of cosets perhaps, but as I still need to revise them I am unsure atm. Can you please see if I am on the correct lines and I will come back to it when I have reviewed cosets.]

[south patrol]

this is much more complicated than it needs to be

[low wyvern]

oh

[south patrol]

you don't need to do anything with elements, for example

[low wyvern]

I haven't done anything with the elements

[south patrol]

Oh sorry i misread

But I mean no Bezout

[low wyvern]

ok

[rustic crown]

btw G need not be finite

only H and K are finite subgroups

[low wyvern]

Ye, Idk how that would hinder me though, but ill keep it in mind

[rustic crown]

you can't talk about h and k dividing |G|

[low wyvern]

oh ye lol. Well then, my whole argument goes to shit lmao

[rustic crown]

what can you say about |H n K|?

[low wyvern]

idk, am I missing a theorem or somet

[south patrol]

Hint: H cap K is a group

[low wyvern]

oh

[rustic crown]

it's a subgroup of both H and K 🙈

[low wyvern]

loll

[slim kayak]

Two ways really. If you know that intersections of groups are groups and lagrange that way, or you just ask yourself:
can x^p = x^q = e for p and q coprime?

[south patrol]

oh i guess that would be a way to use bezout lol

[rustic crown]

well that still uses Lagrange

[south patrol]

but ye

[slim kayak]

the second step?

[rustic crown]

to show order of element divides order of group

[slim kayak]

idk you can do order of elements before establishing lagrange. I think we did it that way? But entirely sure

[rustic crown]

ah sure ><

[slim kayak]

Like, the order can be defined as a minimum of n s.t x^n=e, then clearly n has to divide p and q

[rustic crown]

yea but then one has to do work and show such an n exists

which is like a pigeon hole thingy

[slim kayak]

for finite groups no

its a subset of the natural numbers, so well-ordered

[rustic crown]

yea but why non-empty

[slim kayak]

if no such n exists then <x> generates an infinite subgroup

x^n+1 is distinct from all the power of x before it, if it wasnt then you would obtain an k s.t. x^k=e

[south patrol]

well you only care about x in H, say

[rustic crown]

and this is pigeon hole right 🙈

[slim kayak]

idk maybe?

its either a finite set has an infinite subset or there exist elements s.t x^k=e

[south patrol]

i confusion now lol

[low wyvern]

Lagranges theorem apparently works for groups of infinite order?

[south patrol]

ye

[rustic crown]

yea but infinite cardinalities aren't so rigid

[low wyvern]

I see

[slim kayak]

what does lagrange for infinite groups even say really?

[south patrol]

i mean lagrange is just counting that |G| = |G/H| |H|

which doesn't use finiteness

[low wyvern]

Well, I have to go and have my dinner now as family are around. I appreciate the help and ill see what I can figure out when I am back 🙂

[coral steeple]

Real functions with compact (set-theoretic) support are not a ring, but those with compact closed support are right?

[rustic crown]

so like, you have to use group property somewhere to argue that the repeating thingy is a cycle and not a shape like ρ
you could start like x, x^2, x^3, x^4, and then repeat like x^2, x^3, x^4, x^2, x^3, x^4, ....
so {x^n : n>0} apriori can be finite and never hit e.

(i'm just emphasizing trivial stuff at this point)

anyway, i would say by pigeon hole you have a<b such that x^a = x^b, then since inverses are a thing, x^(b-a) = e gives the required thingy

[slim kayak]

Take x from a finite group G. There exists a k s.t. x^k=e:
Suppose there exists no such k. For all n x^n+1 is not in {x^1,..., x^n}, otherwise there exists some m<n s.t. x^m=x^n+1 -> x^n+1-m=e. A contradiction.
So G (a finite set) contains an infinite subset. A contradiction again

[south patrol]

well sure

[slim kayak]

I mean yeah, the practicallity resides completely in whether you had to show things like these before lagrange

[south patrol]

ye

[south patrol]

compact closed support definitely

[rustic crown]

yep, its usually a common thing for an elementary number theory course :p

[slim kayak]

real functions with the base space not hausdorff seems like they would be rare so its probably overlooked

compactness on its own isnt stable under intersection

but I am too hausdorff pilled to give you a counterexample

[delicate orchid]

I bet you it is

[coral steeple]

Wait is this the case in general but true in R? I thought it was unstable under union

(Sorry about the analysis in #groups-rings-fields lol)

[delicate orchid]

It shouldn’t be hard to prove it’s true in R^n using the characterisation of compact subspaces as closed and bounded

Under union it’s definitely not stable though yeah

[slim kayak]

yeah but that uses hausdorffness

[delicate orchid]

All topological spaces are hausdorf

[slim kayak]

R is nice enough where compact is closed

hausdorffpilled

[coral steeple]

Ahh ok I was confused. Thanks

[slim kayak]

all rings are absolutely flat

[rustic crown]

a simple example is real line with two origins say 0 and 0'.
look at the subspace X = {1/n : n > 0} then X u {0} and X u {0'} are both compact, but their intersection isn't.

[slim kayak]

ooooh

bringing out the Certified Topology Classics™️ I see

[rustic crown]

🙈

[slim kayak]

yeah please avoid this crankery in the future

it isnt hausdorff, it doesnt exist

[rustic crown]

Spec A :c

[slim kayak]

Yes?

Hausdorff like I said

[rustic crown]

we're in algebra channel :3

at least affine schemes are still qcqs

[slim kayak]

qcqs?

[rustic crown]

quasi-compact and quasi-separated

[slim kayak]

no they are compact

All rings are absolutely flat, so all Spec A are hausdorff

[rustic crown]

[slim kayak]

actually they are equivalent, kinda funny

[balmy belfry]

I was doing an exercise in my book and was wondering if this logic is ok

exercise: suppose G is a finite abelian guppy. Show that the product of all the elements has order 1 or 2

G = {a_1, a_2, …, a_n}

Ok so I went with induction

So my first part were I feel weird is

In the base case

I used base 3

[mighty kiln]

How do you induction

[dull marsh]

How will you do the inductive step

[balmy belfry]

Only group with bae 3 is {e,a,a^—}

So obviously that has order 1

So I said “it has order 1, so it has order 1 or 2”

Can I do this?

It feels cheating

[mighty kiln]

Yes but how do you induction

[balmy belfry]

Wait but

Let’s say I do an induction proof for real numbers

[dull marsh]

You will need to find a subgroup of order n in a group of order n+1, that's not possible for like infinitely many n

[balmy belfry]

And I change the base case to end with “x is in R, so it is in C”

[mighty kiln]

???

[balmy belfry]

Why

[tribal moss]

In fact that' only possiible if n=2.

[dull marsh]

Lagrange's theorem

[balmy belfry]

Oh

😭

[dull marsh]

When n=1 maybe?

Or perhaps you meant the implication n-1 -> n

[balmy belfry]

Daaang it 😭😭😭

[mighty kiln]

The key is very simple

It may just involve a slight bit of clevering

[balmy belfry]

So wait what’s the key

Aw

[tribal moss]

One take-home point here: When you think, "oh, I'll try induction", your first consideration should be, "do I have a way to make an induction step work". Base cases can always be figured out once you know that the step works.

[balmy belfry]

I thought I did

I didkd consider to check

If my subset was a group

lol

😭😭😭

[balmy belfry]

What is it then

[tribal moss]

Clevering = multiplying the elements in an order that makes most of them telescope away.

[balmy belfry]

Telescope?

[slim kayak]

i am so proud of you for leaving discussy 🥰

[balmy belfry]

lol

I didn’t take analysis yet tho

Not even a little

[mighty kiln]

AMUKH SPELLES ANALYSIS

[balmy belfry]

LOL

ITS NOT DISCUSSY 😭😭😭

[slim kayak]

oh that exercise

[balmy belfry]

Yeh

Is it bad

[slim kayak]

no its fine

[balmy belfry]

Ok

[tribal moss]

For example, if you sum the elements of Z/7Z, instead of
0 + 1 + 2 + 3 + 4 + 5 + 6 (mod 7)
try doing
1 + 6 + 2 + 5 + 3 + 4 + 0 (mod 7)
from left to right...

[slim kayak]

damn

tropo was fast

[balmy belfry]

wtf

Am I even redy for this how would I have though if this

[slim kayak]

Please try to work on specific examples

[balmy belfry]

So this is for odds

[slim kayak]

try it with Z/Z8 and Z/9Z

[balmy belfry]

This is the same as 3*7

Wait

Yeah that’s 0

Wait what

Oh

0 has order 1

So maybe odd groups have order 1

And even have order 2

[slim kayak]

please try to work on more examples yourself

[mighty kiln]

Counterexample: (Z/2Z)² has order 1

[balmy belfry]

Even groups always have an involution right

[tribal moss]

As I wrote it here, it looks like it's only for cyclic groups -- I'll leave it for you to figure out how to state a corresponding principle for arbitrary finite abelian groups.

[balmy belfry]

Ok

Wait what’s an example of an asyuc group

Ok wait nvm S3

[tribal moss]

"asyuc"?

[balmy belfry]

Is S2 also?

[slim kayak]

the product of two non-relatively prime cyclic group?

[tardy hedge]

To show K(u^2) = K(u) (u is algebraic over K..) whats the best way to approach it? u^2 must be algebraic over K since u^2 is in K(u), so the only way these two fields are equal is if degree of u^2 over K is the same as degree of u over K right

[balmy belfry]

Acyclic

Typo

[slim kayak]

K(u) contains u^2 tho?

[tardy hedge]

Oh

[tribal moss]

S3 is not cyclic, but it's not abelian either.
An example of an abelian group that is not cyclic would be Z/5Z × Z/5Z.

[tardy hedge]

K(u^2) may not contain u ?

[slim kayak]

yeah

but like

Q(-1) is deffo not the same as Q(i)

[tardy hedge]

Yeah

[slim kayak]

so whats your hypothesis for u?

[tardy hedge]

Before i say that, why doesnt this work

Degree of u^2 over K must be same as degree of u over K? Why not

[slim kayak]

works any pair of cyclic groups as long as they have a gcd > 1 no?

by bezout

[balmy belfry]

Guys I think i figured something out

I think I finished it, but idk if I’m crank

[tardy hedge]

If they are the same as vector spaces they must have same dimension right

[slim kayak]

lets hear it

[rocky cloak]

If two vector spaces are equal, they also have the same dimension yes

[slim kayak]

wdym by equal anyways

[mighty kiln]

Isomorphic?

[slim kayak]

dimensionality is well-defined lol

[rocky cloak]

I mean equal, as in literally the same

[tribal moss]

Yeah.

[slim kayak]

I am getting the dot treatment 😭

[tardy hedge]

Whats the dot treatment lol

[languid trellis]

...

[slim kayak]

Yes.
has a different vibe than
Yes
that type of thing

[mighty kiln]

Ok.

[slim kayak]

i was being /jk anyways

[mighty kiln]

Sure.

[languid trellis]

im not joking.

[tribal moss]

The second one must be written by someone who habitually doesn't use correct punctuation in their chat posts.

[tardy hedge]

Hahahahaha

[slim kayak]

Perchance.

[balmy belfry]

So if G is even |G| = 2n for some n.
Let the elements be a_1,…,a_2n ok
The product is a_1 a_2 … a_2n
If I saure it, a_1^2 a_2^2 … a_2n^2 right
So iirc the order of a product is the LCM of the orders
And the order of a power is the power / gcd(power, order of base)

So the orders of a_j ^2 for some j is 2/[gcd(2,|a_j|)]

So the order of the product is the LCM of all of those.

[tardy hedge]

Dot treatment makes me 😢

[mighty kiln]

camel Case Go Brrrr.

[balmy belfry]

There is more I was just lagging

[slim kayak]

I feel like thats wrong

[tribal moss]

the order of a product is the LCM of the orders
This is not true.

[slim kayak]

Why wouldnt this work for any collection of elements a_i?

[balmy belfry]

Because they need to commute

[balmy belfry]

Wait actually?

but like

[south patrol]

If they commute

[rocky cloak]

It's at most the lcm

Could be less

[south patrol]

Wait ye

[balmy belfry]

if a^4 = e and b^k = 5 then won’t (ab)^20 = e?

[south patrol]

Lol

[tribal moss]

Take any element x of order n != 1. Then x^-1 also has order n, but x^-1 x has order 1, which is not the lcm of n and n.

[balmy belfry]

Huh

[dull marsh]

x^n = e doesn't imply O(x) = n

[slim kayak]

i believe the exercise had being abelian in its hypothesis

[balmy belfry]

Oh yeah

Right

Shit

[tribal moss]

||hint!||

[slim kayak]

Amukh, this isnt exactly the type of exercise where you use a lot of machinery. Its one about making a critical observation

[balmy belfry]

Wait

Ok but the gcd part is ok right

And I think all the gcds would be 1s and 2s

And let’s say I replaced = LCM with <= LCM

And the LCM of a bunch of ones and twos is 2

Wait

No it’s not

Is it

Yes it is 😭

[slim kayak]

G is abelian, so if you have abcd you can make it abdc, or bcad or any mix.
Now suppose you have a product that includes all elements of your group. What would be a clever way of using this re-ordering property?

[balmy belfry]

Ok fuck

How did I now see that

Not

I can put them next to their inverse

OHH

AND IF IRS EVEN

[slim kayak]

but why does that work?

[balmy belfry]

THERE IS AN INVOLUTION AND THAYS WHERE THE 2 COMES IN

I’m actually stoopid

[balmy belfry]

Everyone has an inverse

[slim kayak]

you can do it once, maybe twice? Can you do this with all element inverse pairs?

[balmy belfry]

Wait is that it

[slim kayak]

Could it happen that you actually use up someones inverse when doing this?

[balmy belfry]

Wdym

Use up?

It’s unique

Nobody is cheating on their fiend

Friend

[mighty kiln]

Can you perform the procedure for, say, (Z/2Z)²

[balmy belfry]

What’s the squared

Just regular Cartesian?

[mighty kiln]

Nvm

[slim kayak]

I am asking for you to make sure that sets of the form {a,a^-1} form a partition

[low wyvern]

This was the correct way to think about it, but for some reason I would never have seen it, otherwise it was actually really simple lmao...

[tribal moss]

I think Kerr's (very nice) point is that what you have is not just abcd, but abcdabcd, because your goal is to show that abcd has order 1 or 2.

[balmy belfry]

They don’t if the order is rven right

[slim kayak]

think about it!

[balmy belfry]

They don’t

Because of the a^2 = e element

[slim kayak]

think more!

[balmy belfry]

But it’s fine I’ll pair him with e

[slim kayak]

with {a,a^-1} I mean the literal set

so if a^2=e then {a,a^-1}={a}

[balmy belfry]

Yes

So throw e in with it

So it is a partition

[tribal moss]

But you still have two copies of a in the original product.

[balmy belfry]

No I don’t

Wdym

[tribal moss]

Yes, you do.

[slim kayak]

i dont see why you would either

[tribal moss]

^

[balmy belfry]

abcdxe

x is the element st x^2 = e

I’ll part it like this

[tribal moss]

There might be many of those.

[balmy belfry]

Why

[slim kayak]

||the idea is, to me at least, that all elements with arent their own inverse cancel in the product. Leaving a product of self-inverse elements, but that product has order 2 since G is abelian||

[balmy belfry]

😭😭😭😭

[balmy belfry]

But there is never an even number of them

Errr wait

Yeah

Never an even number of them

[rocky cloak]

Now I'm curious what are you actually trying to prove?

[tribal moss]

That's what I thought of first too, but it's easier to say: ||We want to show that abcd...xyz has order 1 or 2, which is the same as showing abcd...xyzxabcd...xyz=1. So just pair each element in the first list with its inverse from the second list,|| and then we don't have to worry about left-overs.

[ashen heron]

#groups-rings-fields message

[tribal moss]

(I thought you were already hinting at that).

[slim kayak]

||but the left overs are the thing that explains why the product has either order 1 or 2?||

||no leftovers -> product is just e. Else it has order 2||

[balmy belfry]

I’m impressed

[tribal moss]

No, what explains that the product has order 1 or 2 is that (the product)·(the product) is the identity!

[balmy belfry]

With myself for not clicking them

[slim kayak]

its a tiny bit more constructive is what I mean

like logically either method is equivalent

[balmy belfry]

So is my partitioning work

[slim kayak]

what I mean is that in your case its less clear why there should be situations where it has order 2 or oder 1. The conclusion is that it has order at most 2

[mighty kiln]

???

[balmy belfry]

lol I

[tribal moss]

Are you suggesting that "at most 2" could be something else than 2 or 1??

[balmy belfry]

Am missing an entire convo

[slim kayak]

at most 2 could mean that it always has order 1 or 2

[mighty kiln]

No, keep looking at the math

[tribal moss]

No, because it it 1 for the trivial group and 2 for C2.

[balmy belfry]

Ok

[mighty kiln]

Has order 1 or 2 includes that too

[balmy belfry]

Well if the group order is odd

The inverses match up

And get me eee…e and then I multiply by e

That’s order 1

[mighty kiln]

Why

[slim kayak]

yes

the difference is that in one you would have to bring up an example to see why both cases can happen. I am just explaining a slight preference to one proof 😭

[mighty kiln]

Tho the solution should not be dependent on parity of group order

[balmy belfry]

Oh

[slim kayak]

if your group has even order the product must have order 2. For odd orders it can be either

but thats a different problem entirely

[mighty kiln]

(Z/2Z)²

[rocky cloak]

If your group has odd order the product is the identity

For even it depends

[balmy belfry]

So I said the right thing but the wrong why?

[mighty kiln]

Well do you know why this is the case

[rocky cloak]

You were right, just didn't justify it I guess

[mighty kiln]

(But this is actually tangential to the solution)

(there is a construction that works for any group)

[balmy belfry]

Ok let me try fresh

Take all the elements of the group that don’t have order 1 or 2

That product will be e, because they have a unique inverse that isn’t itself

You can put them next to eachother and they cancel

That’s true so far cuz comm

The product is e, so I’ll throw it away

[tribal moss]

For odd orders, the product is always the identity -- Lagrange says there are no elements of order 2.

[balmy belfry]

Then the product of all elements is the product of elements with orders 1 and 2

The product of elements of orders 1,2 is <= lcm(1,2)

<= 2

That’s it

[slim kayak]

I see the situation in (Z/2Z)^2 now... there exist elements with order 2 (well, any) and the sum of say (0,1) and (1,0) happens to be the third (1,1) so the product of self-inverses can still reduce to 0...

[slim kayak]

yeah

[balmy belfry]

So wait this doesn’t work?

Why

[tribal moss]

Why do you think it doesn't work?

[balmy belfry]

Because nobody said it does

And my thing doesn’t sound the same as ur guyses

[slim kayak]

your thing is that you reduced it to a product of elements of order 2, and yeah, that has order 1 or 2

[delicate orchid]

#discussion message
I win

[balmy belfry]

So I did it?

[delicate orchid]

also damn this is like the third time this question has been asked within the last 2 weeks

[balmy belfry]

Or is there anything else I need to do

[tribal moss]

I think that is fine.

Next up, Wilson's theorem.

[delicate orchid]

https://tenor.com/view/wilson-castaway-sea-lost-gif-6160664

[balmy belfry]

What’s wilsons theorem

[delicate orchid]

a number is prime if and only if (n-1)! = n-1 mod n

[balmy belfry]

;-;

Did Wilson make it

[crystal turtle]

#chill

[summer path]

Mod gets told to post gifs in chill

[next obsidian]

Ouch

[rustic crown]

chmonkey

[summer path]

det

[rustic crown]

tubu

[slim kayak]

If n is the product of pairwise coprime numbers $n_1$ to $n_k$, and defining $m_i$ as $n/n_i$. Then does there for all a and r exist some k s.t. this holds?

$m^{r}_i k + n\mathbb{Z}= a + n\mathbb{Z}$

[cloud walrus]

Kerr

[rotund aurora]

no

a could be coprime with n, while m_i no

[stark helm]

I am wondering what does D(x) mean? does it mean a field of quoteint of D that represent f/g as f and g is in D and g !=0?

Because I saw that D is integral domain, which means D is commutative ring. and it implies that D(x) is clearly commutative, but I am not sure if D(x) means f/g where f and g is in D

[coral spindle]

When D is a field, then D(x) is indeed the field of rational functions. It's less standard to write the same thing when D is merely a domain, although you could simply define it as Frac(D[x]) which is isomorphic to Frac(D)(x).

[hidden haven]

In case the question is just about the statement in the screenshot, they say D[x] instead of D(x), and D[x] is the ring of polynomials with coefficients in D.

D(x) is the field of fractions of D[x] as Byotty said

[coral spindle]

Ah I hadn't appreciated the confusion may have been D[x] vs D(x)...

[hidden haven]

silly bitty

[coral spindle]

You sure did show me, moudlylocks

[stark helm]

OK, I think I have got it, really appreciate that, I didn't realize the difference between D(x) vs D[x]

[chilly ocean]

ㅡㅡㅡㅡㅡㅡㅡㅡ

[slim kayak]

wtf is a hcf

[chilly ocean]

If a=(4), b=(6), then ab=(24)?

[chilly ocean]

hcf=gcd

Highest common factor

[slim kayak]

sure

[chilly ocean]

Why ab=/=(2)?

[slim kayak]

ab is the set of elements of the form pq with p from a and q from b

[chilly ocean]

a and b are ideals there

[slim kayak]

so they look like 4x and 6y respectively, which gets you 24xy

[chilly ocean]

ㅡㅡㅡㅡㅡㅡㅡㅡ

I can't get notifications in mobile for any chat

[south patrol]

what is hcf/gcd in german lol is it hgt or smth

or ggt

[chilly ocean]

Y r u curious of that

[south patrol]

ggt nice

oh kerr lol

[chilly ocean]

Can u receive notifications for a chat without @mention@slim kayak

[hidden haven]

Those are disabled for large servers and there's no option to enable those

Literally 1984

[slim kayak]

yeah?

[south patrol]

lol that was a response to autist

like i'm saying you not knowing about hcf made me wonder what german term is

but ye it's ggt

wasn't sure if it could be hgt (for höchster) lol but yes

[slim kayak]

yeah no, ours translates the most literally to gcd

[south patrol]

lol ye

[slim kayak]

hcf looks cheaper somehow idk

[south patrol]

lol ye

gcd feels more nicely rounded/full

[slim kayak]

hcf sounds like some weird chemical you spray on stuff

[south patrol]

hcl lol

[chilly ocean]

I can't eng

So what does that mean

[barren sierra]

a and b are ideals

[chilly ocean]

Left implies right or vice versa

[barren sierra]

oh

if right then left

if provided the fact that a + b = (1), then a cap b = ab

[chilly ocean]

I misunderstood as left then right

[barren sierra]

nah nah (in fact there's an easy counterexample to that)

[south patrol]

a = b = 0

[barren sierra]

boooo spoiling the exercise

also ||just a = 0 suffices as long as b != (1) I think||

[white oxide]

oh i remember this exercise

[delicate orchid]

WHAT

[barren sierra]

?

why the caps

[cobalt heath]

Careful, there is a member nicked hsf in this server /j

[chilly ocean]

ㅡㅡㅡㅡㅡㅡ
Does RI=I hold? Cause 1i=i

[slim kayak]

yes

[chilly ocean]

Then it doesn't need to be a+b=(1)

Always (a+b)(a cap b)=a cap b

[next obsidian]

No

[chilly ocean]

(a+b) may not contain 1?

[next obsidian]

That’s equivalent

To being (1)

I feel like we’ve had this conversation before

[chilly ocean]

Ah I mistook cause there's various definitions of ring

Ideal is ring but may not contain 1 as long as I understood

[south patrol]

this book assumes rings have unit

(like most / almost all books on algebra afaik?)

[chilly ocean]

Unit or unity

[south patrol]

hm i mean "a multiplicative identity"

[hidden haven]

You can only have units once you have a unity

[stark helm]

if D is integral domain and f(x) and g(x) is in D[x], is it correct to state that g(x) must be the factor of f(x)

[tardy hedge]

no?

[stark helm]

Could you help me translate what does this sentence mean?

I am considering if f and g are in D and nothing more, then g is not factor of f? and g will be the factor of f once there exist h in D such that f=g*h?

[slim kayak]

Given f and g, then g is a factor of f if condition

said condition is that there exists some h s.t. f=gh

[stark helm]

So if there are only f and g in D[x], then g is not factor of f unless we have addition h in D[x] right?

[slim kayak]

the "if f and g in D[x}" is just introducing a pair of polynomials

[stark helm]

so i am considering if we can say any element in D[x] must be the divisor of other element in D[x]

[slim kayak]

"If A and B are numbers, we say that A divides B if there exists some C so that AC=B"

[stark helm]

What does the divisor x-a of f(x) have to do with a is zero, I mean why it can imply (x-a) as factor of f(x)?

[slim kayak]

If f(a)=0 then for g(x)=f(x+a) we have g(0)=0. What can you say about the coefficients of g?

Then consider g(x-a)

[stark helm]

then g(x-a)=f(2a)*g(0)=0?

[slim kayak]

No?

[stark helm]

I am confused about why we have g(x)=f(a-x)*g(0)=0?

[slim kayak]

We don't?

There, fixed for clarity

g(x-a)=f((x-a)+a)=f(x+a-a)=f(x)

Write g as a_0 x^0 + ...

[stark helm]

shouldn't f(a-(x-a))=f(2a-x)?

[slim kayak]

Ight

g(x)=f(x+a)

Fixed up now

[stark helm]

g(x-a)=f(x) and we have g(x-a)=a0*(x-a)^0+..., so f(x) will have factor x-a?

f(a)=0 is just the assumption right to represent that a is a zero of polynomial?

[slim kayak]

Yeah

You only need to figure out why a0 must be 0

[hidden kite]

If someone is free and willing to do so, could you help me check if my proof is valid here? I checked it myself and I think it should work.

Also, if you have some insight that could make the link between (ii) and associativity intuitively clearer, please do share it --- I made it work, but I feel that I still don't have the intuition for this.

[chilly ocean]

Why a_n+b=(1)?

[rocky cloak]

Because there is an element in b which is 1 modulo an

[chilly ocean]

Thats true but why then an+b=(1)

[rocky cloak]

If an ideal contains 1, then it's equal to (1)

[chilly ocean]

Thx
ㅡㅡㅡㅡㅡㅡㅡ

[chilly ocean]

How can phi(x)=(1,0,...,0)?

I think it should be (1+a1,a2,...,an)

[dull marsh]

It looks like they meant that, yeah

[static glen]

Notation

Abuse of*

[cobalt heath]

Technically not even an abuse, since each A/a_i is a ring there

0, 1 makes sense as an element of general ring

[static glen]

well abuse is not precisely defined but writing the same symbols for elements from different sets is kinda abuse in my eyes

[cobalt heath]

Hmm, I see. I would not call using 0 and 1 for different rings "abuse" tho

There are way more severe abuses

[balmy belfry]

I am getting ghost pings here

@wispy urchin is it you

[wispy urchin]

what

[balmy belfry]

Ok sorry

[tardy hedge]

[K(u) : K] is odd number, show that K(u^2) = K(u)

By using minimal polynomial stuff i have that some a0u^(-1)+a1u is in K(u^2)

Can i do anything from that?

[south patrol]

just think in terms of degrees of fields

over one another

[tardy hedge]

Yeah thats how it was proved in book

But im just wondering how to do it if u just go more basic