groups-rings-fields

[delicate orchid]

Surely you can do it just from the fact that it has to be closed under R-linear combinations of rows/columns (over all matrices in your ideal)

[dull ginkgo]

just used the same proof verbatim that I used for that one other problem and it worked

same for classification of ideals in the triangular matrix rings

[next obsidian]

Ugh

I classified ideals in a matrix ring once

Sucked ass

Took forever to figure it out

[delicate orchid]

Yeah that was overkill

[dull ginkgo]

brother I already did it I am recycling

[next obsidian]

My problem asked me to classify them

[dull ginkgo]

oh

[next obsidian]

And didn’t say what they looked like

[dull ginkgo]

isn't that classifying the underlying ring ideals then

unless it's a subring

[next obsidian]

No I mean

[delicate orchid]

They didn’t say. What they looked like.

[next obsidian]

Writing it like that is the description of what they wanted

I had to discover that result

And prove it

[dull ginkgo]

ah

i am baby

[next obsidian]

Googoogaga bitch

[tough raven]

The one-sided ideal version is nice too.

[dull ginkgo]

Assume R has characteristic nm, both greater than 1, then there is an injection of Z/pmZ into R. Consider the principal ideal (m), which must be (0) or R. It cannot be (0) as R is not 0, thus it must be R. That implies that forall x in R, x = my, thus nx = n(my) = (mn)y = 0, contradicting the minimality of the characteristic

[tough raven]

Is R assumed to be commutative?

[dull ginkgo]

oh, no it isn't

nvm

[tough raven]

Does R have a 1?

[dull ginkgo]

Yes

Unital ring in this case

Same proof

[tough raven]

Usually when it does, one treats "improper ideal" as "contains 1".

[dull ginkgo]

yeah

[tough raven]

Oh, m is in the centre 🤦.

[dull ginkgo]

yeah

Z/char(R)Z embeds into Z(R) lmao

[tough raven]

Yes this works.

[dull ginkgo]

fatal mistatement

every nonzero homomorphism

[tough raven]

Homomorphisms preserve 1 boo

[dull ginkgo]

trivial kernel qed

that's literally all I wrote for that problem

I’m bored

Time to prove that commutative artinian rings are product of locals

[next obsidian]

Zero map ain’t a homomorphism

[cobalt heath]

Ono

[celest furnace]

Thank god

[dull ginkgo]

I think this might need choice lmao

I can work it out if I assert the existence of maximal ideals, or at the very least prime ideals (since primes are maximal in artinian rings)

So yeah just ultrafilter lemma before one of y’all ping me about it again

Artinian rings have finitely many max ideals since the set of finite intersections of them necessarily has a minimal element element, which by definition of minimum under inclusion in this case, the intersection of that minimum with any of the other max ideals must be itself, so it must be contained in the finite collection

Also by def of maximum, they are pairwise coprime too, so you can use Chinese remainder theorem to break up R over that minimum intersection into an product of R/max ideals

I should probably focus on getting to class on time

[rocky cloak]

I would be a little careful with this argument. Since for example it's not true that a noncommutative artinian ring has finitely many maximal ideals. It is true for commutative rings, but this argument doesn't really show it.

What is true is that the intersection of all maximal ideals is equal to a finite intersection of maximal ideals, which is all you need for the remaining.

[rocky cloak]

The fact that a artinian/Noetherian ring has a maximal ideal doesn't necessarily require choice depending on your definition of artinian.

At worst you need dependent choice.

[dull ginkgo]

I meant to imply commutativity, which is why I considered the intersection, not the product for CRT. Sorry fam

[tough raven]

It depends on your formulation of Artinian, but I'm certain pretty sure there is some formulation where it doesn't.

Or put another way, I expect you don't need Choice beyond proving that existence of minimal elements of sets of ideals, stabilisation of descending chains, etc. are equivalent.

[dull ginkgo]

The set of ideals has minimum formulation is fine

the issue is existence of maximal or primes

[tough raven]

But it's not very far to show that there are finitely many maximal ideals from here (commutative case), since M includes M1 intersect … intersect Mn = M1 … Mn => (M being prime) that M includes some Mi => M = some Mi.

[tough raven]

Well, if there are no maximal ideals, then R is a minimal intersection of maximal ideals…

I'm not sure if that's actually acceptable for the rest of the proof though.

[dull ginkgo]

DON'T GIVE IT AWA

[tough raven]

Sorry :(

[dull ginkgo]

But yeah, the intersection of the maximal ideals is nilpotent

observe I^n stabilizes for every ideal I

takes a bit of tinkering to force it to be nilpotent tho

[tough raven]

I should have just added spoiler tags, shouldn't I?
Oh well.

[dull ginkgo]

I just didn't want you to reveal further, I already showed the intersection is nilpotent

[wide brook]

Why is (Z/pZ)^2 where p is prime not a cyclic group?

[coral spindle]

Why do you think it is?

[rocky cloak]

Well, it has order p^2, so if it was cyclic it would need to have an element of order p^2.

And you can just check what the order of the elements are

[wide brook]

Ah I see! But every elements has order p right?

[rocky cloak]

Indeed (except the identity I guess)

[wide brook]

Yeah yeah ok! Thanks for the feedback

[languid trellis]

I am struggling to see how we get to \phi(p^e) = p^e - p^{e-1}. I have tried counting elements that aren't coprime and are less than p^e but I'm lost in the sauce. Would appreciate a nudge

thanks 🙂

[dull ginkgo]

If we have p^k, then the elements coprime to p^k are elements less than p^k that aren't divisible by p. How many are less than p^(k+1)?

How would you "strip" them out?

[languid trellis]

In general, stuff that isn't coprime to p^k is of the form np^j

[dull ginkgo]

You have p^e naturals in [0,p^k]. How many numbers are divisible by p in that interval??

:)

(p * [0,p^k-1])

[languid trellis]

So, we can count p, 2p, 3p, 4p .... (p^{k-1}-1)p

[languid trellis]

Wait what trickery has happened here

[dull ginkgo]

Basically. We have p^k naturals less than or equal to p^k. We need to strip out the numbers divisible by p, all of which are of the form pm, for m in [1,p^(k-1)]. So we strip p^(k-1) numbers from p^k numbers,

[languid trellis]

Oh that's quite clever

[dull ginkgo]

ping deltoid about sieves lmao

[languid trellis]

This is the same idea here but I didn't realise it

[languid trellis]

NOOOO NOT ANT NOOOOOOO

[dull ginkgo]

troll

[languid trellis]

analysis time

eulers totient theorem is pretty cool though

[dull ginkgo]

as much as I want to do my math textbooks I know I need to get shit done for college aaaaaaaaaaaaa

[languid trellis]

I can't help but feel that it's a little "constructed" in this context though. It seems more like a human creation than most, probably because we just define \phi(n) to be all the stuff coprime to n lol

[languid trellis]

yes. log off discord and finish ur shit

[dull ginkgo]

I am in class lol

comp sci class I don't care about, can't work on my project rn

[languid trellis]

log off discord big guy

[dull ginkgo]

trivial kernel qed

[delicate orchid]

holy guacamole!

[dull ginkgo]

sheef

I don't feel like reading the hint. kernel moment

I was about to ask a dumb question and simultaneously answered myself and the problem

[delicate orchid]

u can't just say kernel moment on this one

[sly crescent]

Kernel moment

[dull ginkgo]

mmm class over

[languid trellis]

me when <1> generates R

me when generators determine homomorphisms

[dull ginkgo]

That proof takes one sentence

||a,b ring homomorphisms, H = {x | a(x) = b(x)} is a ring containing S, <S> = H = R||

[dull ginkgo]

LOUD INCORRECT BUZZER

It generates Q

[languid trellis]

chat is this real

[dull ginkgo]

well what characteristic

Hur dur endo Z -> <1>

[languid trellis]

something something yap yap R is a PID

[delicate orchid]

fields are PIDs
WOAH!

[languid trellis]

me when the inverse exists and we have commutativity

[dull ginkgo]

Here’s something for you: Show the complement of the set of units is the union of all ideals

[languid trellis]

doing analysis rn

remind me in 3 hours

[winged void]

Let $n \in \mathbb{Z}^+$ and $a \in \mathbb{Z}$.
\begin{enumerate}
\item Prove: $a \in (\mathbb{Z}/n\mathbb{Z})^* \iff -a \in (\mathbb{Z}/n\mathbb{Z})^*$.
\end{enumerate}

[cloud walrus]

Mootje

[winged void]

Z^+ including zero

its actually trivial but im not so sure

if i can say gcd(a,n) = gcd(-a,n) and finish

[delicate orchid]

prime factorisation in UFDs is only ever unique up to a unit so yes this is true but there's an even easier way

if ba = ab = 1 then what's (-b)(-a) and (-a)(-b)

[winged void]

us also one

[winged void]

but what is here UFDs

what do you mean by that

[dull ginkgo]

if a is coprime to n then (n - a) is coprime to n

[winged void]

is that what UFDs mean

[dull ginkgo]

No,

[winged void]

oh ok

[dull ginkgo]

Why are you bringing up UFD’s for this

[winged void]

not me

but @delicate orchid he was saying something about UFDs but not so sure

what he meant with that

and i do not know what UFDs are

[dull ginkgo]

Doesn’t matter

If you want you can show a coprime to p => (p-a) coprime to p

[winged void]

ohso

[dull ginkgo]

It’s not bad

[delicate orchid]

bit annoyed this message is being ignored

[dull ginkgo]

lol

[winged void]

nope its not ignored

i replied to it

but i was wondering what does UFDs mean

[delicate orchid]

ah so you did, hence -a has an inverse so it's a unit

[winged void]

sure

[languid trellis]

Chat does a have a multiplicative inverse in Z/nZ

[rocky cloak]

UFD = unique factorization domain = ring where you have unique prime factorization.

[winged void]

ah i see

we are now studying only groups actually

but in that case its the set of all gcd(a,n) = 1

[winged void]

thx

[stark helm]

How do you find Gal(E/Q) when E=Q(sqrt(2), 5^(1/3), w)? w=e^i(2pi/3) . I have determined that this group has size 12,but want to find all 12 automorphisms? I think it should map from sqrt(2) to either sqrt(2) or -sqrt(2). map 5^(1/3)->5^(1/3), w-> either w or w^2?( i don't know why i can not collect 12 automorphisms)?

[winged void]

this is also considerd to be a good proof

i guess right

i used your idea but not so sure

but i find it logic

[delicate orchid]

I gave you the full idea

[rocky cloak]

5^(1/3) can map to other things besides just itself

[winged void]

i know but i tried using your idea this proof is my proof correct that way

[delicate orchid]

but this isn't a ring theory question so you should just use whatever definition of Z/nZ* you've been given

[winged void]

well i used the definition of modulo

since it has an inverse etc

can you read it please

[delicate orchid]

ab = 1 mod n => -ab = -1 mod n so something has gone wrong

the point is that (-a)(-b) = ab = 1 mod n

[winged void]

oh yea that is true

so actually what we can say that ab = 1 mod n is the same thing as saying (-a)(-b) = 1 mod nb

and then finish

[delicate orchid]

please put the brackets in, I noticed you neglected to do so in your proof as well

[winged void]

the other way is true as well

i will do so you have good point thanks for the tip

[delicate orchid]

but again, this isn't a ring theory question, so I don't know if you have that (-a)b = -(ab) = a(-b) yet

[winged void]

well we did not get that yet

so actually what we can say that ab = 1 mod n is the same thing as saying (-a)(-b) = 1 mod n but the proof follow from this line right

[dull ginkgo]

well if you know Z/nZ^X is the group of numbers coprime to n mod p

[winged void]

from this line we can conclude that if a is inside then -a is inside as well

[dull ginkgo]

Yeah

[rocky cloak]

I mean, these are integers were talking about

[delicate orchid]

true

I still don't like it

[rocky cloak]

Don't think you need to take ring theory to establish that -1 times -1 is 1

[winged void]

no i do not think so

because then we would not get that question

[dull ginkgo]

(p-1)^2 = p^2 - 2p + 1 = 1 mod p

[winged void]

but @delicate orchid like the completness

i guess

and he has a point

[rocky cloak]

But is your definition of (Z/n)* things with inverse modulo n, or is it things with gcd(a, n) = 1, or can you freely use that they're e equivalent?

[delicate orchid]

yeah just do this and you're fine

[winged void]

but why is it actually that (p-1)^2 = p^2 - 2p + 1 = 1 mod p that this has remainder of 1 when divided by p

[dull ginkgo]

p(p-2) + 1

[winged void]

ohhhhhhh sorry

true

sure

[winged void]

freely use that they are equivalent

but would it make then a problem

[stark helm]

i tried other stuffs a such that a^3=5, but no stuffs can unless cubic roots of uinity, but it seems contradicts homomorphism

[rocky cloak]

I'm not sure I follow what you're saying, but 5^1/3 is a root of x^3 = 5, so an automorphism will map it to other roots of x^3 = 5. Do you know what the roots are?

[stark helm]

I think 5^1/3 *w and 5^1/3 * w^2 can be roots

but I don't think it satisfies homomorphism

[rocky cloak]

That's right. So that gives you 3 choices for where to map 5^1/3

And 2*2*3 = 12, so...

[stark helm]

But w can only map to w or w^2

[rocky cloak]

That's right

So you have two choices for where to map w

Combined with the two choices for sqrt(2) and 3 choices for 5^1/3 that's 12 choices

[stark helm]

I found that I made mistakes on calculatoin, so I misunderstand that it won't satisfy homomorphism

New question: If E=Q( i, 5^1/4), and diagram shows about galois group Gal(E/Q), question:Write the multiplication table for the group? how to write this table?

[coral prawn]

is there an analog to field extensions for rings and groups? it seems kind of intuitive there should be but the operations would be different

[delicate orchid]

ring extensions work in exactly the same way, quotient R[x_1, ..., x_n] by something and you get a new ring

[coral prawn]

oh sweet, yeah i thought so

only difference is the operations are (+, x, -) instead of (+, x, -, /)

does it work for groups too?

[delicate orchid]

there's not really a notion of a "polynomial ring" for groups, by that I mean something with the required universal property blah blah blah

[coral prawn]

oh true

what if you compared groups with adjoined elements though?

[delicate orchid]

extenstions of groups are still a thing though

you just have to use different machinery

for example, you can extend some group G by another group H by just doing G x H

[coral prawn]

oh true

[delicate orchid]

Given a group G and two subgroups H, N < G, you say that G is an extension of H by N if N is a normal subgroup of G, and G/N is isomorphic to H

[coral prawn]

yeah i briefly read about that

[delicate orchid]

it's very different, and the analog for rings is called an "algebra extenstion" iirc

[coral prawn]

does that apply for something like adjoining 1/2 and -1/2 to the integers under addition?

[delicate orchid]

for that you can just do Z[x]/(2x-1)(2x+1) I think

[coral prawn]

oh nice

[delicate orchid]

nah that doesn't work

[coral spindle]

Just 2x+1 I would imagine

[delicate orchid]

Z[x,y]/(2x-1)(2y+1) definitely works

[coral spindle]

Indeed

[delicate orchid]

I don't see how that gets you 1/2

[coral spindle]

-1/2 = -(1/2)...

We are talking about rings yes?

[coral prawn]

oh sorry i meant groups

[delicate orchid]

oh

[coral prawn]

like is there some group extension for adjoining 1/2 and -1/2

[coral spindle]

Indeed the subgroup of Q generated by 1/2 is

uh let's call it G

[delicate orchid]

Z_2 :trollface:

[coral spindle]

There's an exact sequence 1 → Z → G → Z/2Z → 1

[delicate orchid]

0 -> Z -> G -

wow ok

[coral prawn]

oh wow lol

[coral spindle]

So it is an extension of Z/2Z by Z.

One of many lol

[delicate orchid]

first map is an inclusion and the second one sends a+b(1/2) to just b right

[coral spindle]

Yeh

[delicate orchid]

swagger

[coral spindle]

Or you could see it as sending q to 2q

modulo 2, hilariously

[delicate orchid]

true

[coral spindle]

ububububuh but boytjie there's CLEARLY a 2 there so 2q = 0 mod 2

I'm going to shut up

[south patrol]

Real.

[coral spindle]

Rational, really

[south patrol]

I've never understood why G is called an extension of Z/2Z by Z

and not an extension of Z by Z/2

Is there an enlightening pov lol

[delicate orchid]

because, if it was split, the Z/2Z would be acting on the Z via a semidirect product

[south patrol]

cause like you literally have an inclusion Z -> G

[rocky cloak]

I guess just because Z/2 comes first in the notation
Ext(Z/2, Z)

[coral spindle]

I think in fact both ways around have been used at some point, but I find the quotient-by-subgroup bit easier to remember for Ext

what he said

[south patrol]

yeah sure

thanks

[coral spindle]

And Ext is that way around bc of Hom

[south patrol]

ye

[coral spindle]

Once again proving that we should be writing f : B ← A..... f o g my beloved

[rocky cloak]

Or write (x)f

[coral spindle]

Very wise

[delicate orchid]

society...

[south patrol]

uwu

[languid trellis]

society

[stark helm]

If E=Q( i, 5^1/4), and diagram shows about galois group Gal(E/Q), question:Write the multiplication table for the group? how to write this table?

[tardy hedge]

This September, my class will be covering module theory

What prereqs from algebra do i need for that?

It’s mostly ring theory right

[coral spindle]

If you know what a ring and a group is, you're set.

Linear algebra is helpful.

[minor fulcrum]

for the very basics just rings

[last spoke]

@stark helm Well for an automorphism group (e.g. a galois group) the multiplication operation is composition of automorphisms. Since your group has order 8 there are only so many groups that are like that so if you know a decent bit about what those groups are you know what to look for.

How about compare $\phi_3 \circ \phi_5 (5^{1/4})$ to $\phi_5 \circ \phi_3 (5^{1/4})$. Does that tell you anything?

[cloud walrus]

Ryemo🌾

[tardy hedge]

Thank you. How much linear algebra should I review? I took a class a long time ago but I forget a lot of it

I remember some facts about vector spaces

I dont remember much about matrices or stuff like jordan canonical form

[coral spindle]

Honestly man there's no way of telling

[tardy hedge]

Class is covering chapter 10 and 12 of dummit foote

[coral spindle]

The class may or may not need you to know certain things

[tardy hedge]

Ok

[coral spindle]

OK then check that chapter. Doesn't D&F have a prerequisites graph? I used Fraleigh myself.

[stark helm]

the first one maps from 5^1/4 to -5^(1/4) i, the second one maps from 5^1/4 to 5^(1/4)i, these are conjugate. So does it mean we need to find multiplication composition that can map to a pair of conjugate?

[last spoke]

what do those results tell you about the 'multiplication' in this group?

[stark helm]

you mean non-abelian?

[last spoke]

ya thats one thing

do you know about any of the non-abelian groups of order 8?

[stark helm]

how about D4( dihedral group of order 8) ?

And also if I have a and b such that a^4=1 and b^2=1

I guess similar to this property

[last spoke]

Yep, that's one example. Altho it's not enough to specify it that you can find an element of order 2 and a different element of order 4. That's also true of the quaternion group Q8

[languid trellis]

#1217259452621783154

[stark helm]

quaternion grooup is ?

[stark helm]

and also, how can we associate that with finding multiplication group?

[last spoke]

nevermind. My point is that if you can find another group that it looks like then that'll make the multiplication table less tedious to write out. And if you know about an isomorphic group then that'll make the whole composition of automorphisms thing less cumbersome

[stark helm]

OK, then since we have found that this galois group is non-abelian, we can fist confirm that this group is isomorphic to which group?

[coral spindle]

<a, b | a^4 = b^4 = e, ab = b^3a, a^2 = b^2>

Let me latex that.

[stark helm]

composition of automorphism, so we basically need to list like phi3* phi5*phi7...( 5^1/4)?

[coral spindle]

$\langle a, b \mid a^4 = e,\ aba^{-1} = b^{-1},\ a^2 = b^2 \rangle$.

[cloud walrus]

Boytjie

[last spoke]

well no karl, you only need to do that for the pairs. the upshot of the automorphisms forming a group is that they're closed under composition, so you only need to do the pairs phi1(phi1), phi1(phi2), phi2(phi2), etc.

[stark helm]

what does pair means here? i can see that phi1(phi2)=phi2(phi1)=phi2, does it mean pairs?

[last spoke]

multiplication pairs

a*b, b*c, etc.

pairs of elements that you multiply together

you only need to check pairs of elements

[stark helm]

if we don't see closure under multiplication, there are 8^2=64 results. However, because of closure, then there are still 8 pairs, like phi1(phi1)=phi1, phi1(phi2)=phi2, but phi(2)(phi2)=phi1( order of 2). So I don't know if there are 64 pairs or 8 pairs

one stuff I found, this galois group is isomorphic to dihedral group of order 8

[last spoke]

you know it's closed bc it's a group

[stark helm]

so there are 8 pairs right

[last spoke]

ya

well

[stark helm]

cause I found phi2(phi2)=phi1(phi1)

so I am confused

if we only need to write one of them

[last spoke]

there are 8 elements forming 64 pairs, but many of them will have the same answer, so there's 8 pairs up to distinct answer

[stark helm]

really appreciate your help

[last spoke]

the point of the exercise of writing out a multiplication table is to find which ones are the same

you don't know ahead of time before you've figured out the structure of the group

for some of them a(b) and b(a) will not be the same right? because it's non-commutative

so you need to figure out when commutativity fails and how it does

the table will tell you that

[patent girder]

Is there some sort of ternary analogue of a group? The best thing I can come up with is [[abc]de] = [a[bde][cde]] for associativity to carry over the idea of each $x\in G$ inducing a function that is "left multiply by x". Is this like at all interesting? (I chose to avoid [[abc]de] = [a[bcd]e]] = [ab[cde]] since I don't think it'll be as rich)

[cloud walrus]

Zander

[patent girder]

Actually the second thing I give gives rise to a binary operation $a\otimes b:= [ab1]$

[cloud walrus]

Zander

[patent girder]

So yea, definitely I'd want to use the first one

[old hollow]

I’m confused what this abcde stuff is

[celest furnace]

Q8 my beloved

[patent girder]

a,b,c,d,e are elements of a set which I want to equip a ternary analogue of the group product

i've denoted it $[\cdots ]: G^3\to G$

[cloud walrus]

Zander

[stark helm]

First question: does surd root mean root that carries square root? Second question: how can we use induction here?since if we have root that carries square root, like sqrt(r), then we also have root -sqrt(r). Then the third root must be rational .Looks weird to me,so if I have some misunderstanding here?

I think it must have a tower of Q subset Q(sqrt(r1)) subset of Q(sqrt(r2), sqrt(r1))..., so we need to do induction on number of field that forms tower?

[patent girder]

I think you'd use the induction on the height of the tower to show that ||over Q, p(x) has a linear factor (which boils down to a quadratic and linear if i>=1)||

[stark helm]

I am considering if we need to first know that cubic polynomial p(x) with rational coefficient must have its sum of root be rational number here

[stark helm]

at least there are two cases: 1. all three roots are rational numbers 2. there exist a pair of sqrt(r)( namely sqrt(r) and -sqrt(r)), there must left one number that is rational. Do you think this is correct?

i just want to confirm that if surd root refers to square root in this context

[patent girder]

I think it can also be the sum of square roots...hmm

[stark helm]

like a+b*sqrt(r) right

[stark helm]

form that carrieds square root

[patent girder]

Q < Q(\sqrt{2}) < Q(\sqrt{2})(\sqrt{3}) is what I have in mind

[stark helm]

I think basically, sqrt(r) is supposed to be the root, so -sqrt(r) is as well. then cubic polynomial should be the polynomial of product of linear and quadratic.

[patent girder]

Right, but then youd need to show that since youre a cubic a root is either rational or of the form sqrt(r)

In the thing I give, \sqrt{2} + \sqrt{3} is an element, for example

[stark helm]

we first need to determine the automorphism of sqrt(2)+sqrt(3)

[stark helm]

map this element to which

do you think map to -sqrt(2)-sqrt(3) is valid?

[patent girder]

what are you trying to do

[patent girder]

all I've given you is an example as to why this is false

[stark helm]

so surd roots are sum of rational numbers and square roots

[patent girder]

Well just think about what things are valid. Is this one allowed
$Q < Q(\sqrt{2}) < Q(\sqrt{2})(\sqrt[4]{2})$?

[cloud walrus]

Zander

[last spoke]

@patent girder https://math.stackexchange.com/questions/94690/is-anybody-researching-ternary-groups this stack exchange post might interest you

[stark helm]

I think it should be, elements in this tower of field must be surd

[patent girder]

Right! Now my idea for the inductive step basically ignores all of this: ||suppose that p(x) has a root r in B_i, then another root is also in B_i and thus the last root is in also in Q. The base step is obvious; assume now the inductive step and take a root in B_i. If it has a root in some other B_k, then another root is in B_k so that p(x) couldn't have actually has a root in B_i. ||

You need to add in some more details, but I think that's the main idea of the proof

[stark helm]

Base case, Q subset of B1=Q(sqrt(r)), so two roots are in B1, and the third root in Q

[stark helm]

inductive step, assume until Bk when n=k, this rule is true. Then when n=k+1, assume we have one root in B_k+1, and try contradiction, then there might be one root in B_j, which must find another root in B_j, then no way to put rational and inductive fails

[worn spindle]

can someone explain to me what a functorial is

idk what a category is so please dont use that word

but when i google i only get category stuff but im seeing this in my intro algebra class on rings

[coral spindle]

You cannot define functors without the notion of categories.

Functoriality is to functors what f(ab) = f(a)f(b) is to homomorphisms of groups.

If you want to know more, you will have to bite the bullet and learn about categories.

[cobalt heath]

Usually these courses discuss what functoriality means in the specitic context.

[old hollow]

Wait what’s the context damn

[teal wind]

Does anyone understand part 1?

[worn spindle]

Well my prof just did lol

I’ve also seen it in my information theory class defined without categories but I just forgot what it is

On page 66 here

[old hollow]

Hm

So it constructs it not explicitly with elements, but just with maps

[worn spindle]

I don’t think that’s the defn of a functorial

[worn spindle]

Nah we’ve done this before and didn’t use the word functorial

[cobalt heath]

They are just using the word in specific context to give future reference

They explain what functoriality means in current context, so that when you learn category theory later, you can understand how it connects.

[old hollow]

What does it mean though

A functor is a mapping between categories

What r the categories

[cobalt heath]

Category of rings, obviously

Basically the Functoriality is that the constructions define endofunctor in Ring.

[mighty kiln]

Group rings can also be considered as a functor Ring × Grp → Ring

And ring of functions Set^op × Ring → Ring

[old hollow]

Ok, so the functor is an endofunctor on Ring

But what’s the actual mapping here

What does R->S giving rise to R[x]->S[x] have to do with endofunctor on Ring

[mighty kiln]

The functor is F1(R) = R[x]

[cobalt heath]

^

[old hollow]

Oh that kinda weird

Hm

But that isn’t surjective right

It’s injective tho right

[mighty kiln]

Hm

[mighty kiln]

I guess it is, depending on the set-thereotic construction

[old hollow]

Ohhhhh wait wait

[mighty kiln]

We rarely talk about injectivity on objects

[old hollow]

So basically R goes to R[x]

[cobalt heath]

Endofunctor does not mean there is a map R -> F(R)

[old hollow]

And any morphism R->S is sent to the morphism R[x] -> S[x]

Right?

[cobalt heath]

Yeah, this is the functoriality

[old hollow]

So functoriality is the ability to give rise to a functor

[mighty kiln]

So this actually gets you a somewhat stronger statement

This translates to the fact that there is a functor F(R, G) = R[G] such that F(R→S, G→H) = (R[G]→S[H])

And the special case of id: G→G shows that if you fix G, then F'(R) = R[G], F'(R→S) = (R[G]→S[G]) is also a functor

[mighty kiln]

Same goes for this: F(X→Y, R→S) = (Fun(Y, R) → Fun(X, S))

Where op means contravariant in X

[crystal vale]

I need to prove that every prime ideal of a Boolean ring is maximal.

So if I is prime ideal in the Boolean ring R.
Then R\I is an integral domain( without unity ) and every element is idempotent.

Then I have a result if S is a Boolean ring with unity and if S is an integral domain then S is isomorphic to Z\ (2Z).

How can I use this result without assuming R has unity?

[swift prawn]

try this for some small examples like D_8 (D_4 depending on convention) and then generalize. You're just writing out the multiplication table

[agile burrow]

Note that Z/2Z is a field

Oh sorry

I didn't see the "without assuming unity"

I'm not sure if it carries over the same

[crystal vale]

Yes I want without assuming unity

Is it true even R has no unity?

[swift prawn]

R is a field

so definitely has unity

unless you meant your boolean ring in which case I don't know

[crystal vale]

How R is field?

[swift prawn]

I thought you were asking about the real numbers

I confused myself mb

[rocky cloak]

Say x is not I. Then
x(y - xy) = xy - xy is in I, so y-xy is in I. Hence x is an identity element in the quotient ring. Thus R/I = Z/2

[crystal vale]

Got it, thank you

I want to show that rad(I) is the intersection of all prime ideals containing I.

I showed that rad(I) is contained in the intersection of all prime ideals containing I but how can I show that intersection of all prime ideals which contains I is contained in rad(I) ?

[dull ginkgo]

I think there’s like 2 ways to go about it

[crystal vale]

How?

[dull ginkgo]

The ideals containing I are in a 1-1 bijection with the ideals of R/I

Look at the intersection in R/I

I'll give you a hint, it uses Zorn's

[stark helm]

how will you show this proof? I can imagine it as x^3-3x-1. and use rrt to confrim no rational root, then how to proceed?

[dull ginkgo]

what are you allowed to use?

[stark helm]

here is the original question, with many parts

[dull ginkgo]

Me on my way to construct the chevyschev polynomials

[stark helm]

what does that mean?

[dull ginkgo]

Inductively you can define a polynomial T_n(x) for every n such that T_n(cos(x)) = cos(nx)

[dire siren]

@stark helm use cos(3x)

[dull ginkgo]

And I was going to use 18 = 2 * 3^2 like a nutcase

But T_3(x) = 4x^3 - 3x

[stark helm]

you mean let theta=3x?

[dire siren]

I mean use the formula for cos(3x) and take x=theta

[dull ginkgo]

Basically what filip said is what I was doing with a taste for the number field

[south patrol]

Nice

[dull ginkgo]

i like chebyschev polys

[south patrol]

Oh cool

Complex numbers real

[south patrol]

Actually there is presumably a nice generating function via that right

[dull ginkgo]

yes

[stark helm]

use the formula for cos(3x),so like trying cos(x+2x)?

[south patrol]

Like ||sum of cos(nx) t^n is real part of sum of [exp(ix) t]^n = 1/(1-(t e^ix) ||

Idk how that's helpful tho lol

[dire siren]

sure, you can deduce it that way

[stark helm]

ok, I will try. How will you solve part c in this diagram?

[dire siren]

it just follows by Vieta

[stark helm]

vieta refers to?

[dull ginkgo]

They are the real part of (x + yi)^n when setting y^2 = 1 - x^2

[dire siren]

https://en.wikipedia.org/wiki/Vieta's_formulas

[stark helm]

one last question, since cos20 can not be rational numbers by RRT, and it can not be surd since no rational number, then it implies cos20 can not be adjoined, so why it still be a root?

I mean, if i have some misunderstanding here

[dire siren]

I don't understand what you are asking

[dull ginkgo]

also 20 degrees is 2pi/18 radians

Therefore $2\cos(\theta) = \zeta_{18} + \zeta_{18}^{-1}$ lo

[cloud walrus]

Request a new nickname

[stark helm]

that means, cos20 not rational, not surd, so it should not be in the tower of a field,right?

because i am confused should a root be either in Q or adjoined

if it is not in Q, it must be adjoined

[dire siren]

that basically means that [Q(cos20):Q]=3

so indeed you cannot construct a tower like in the problem

because [B_t : F]=2^t

[stark helm]

it can't. I actually start confusing a tower Q subset of Q(cos20)

why it can't be a tower

[dire siren]

by their definition of B/F it follows that [B:F]=2
so then [B_t : B_0] = [B_t : B_(t-1)] * [B_(t-1) : B_(t-2)] * ... * [B_1 : B_0] = 2 * 2 * ... * 2 = 2^t

but 3 is not a power of 2

so you can't construct a tower that starts with Q and ends with Q(cos20)

[stark helm]

The thing that I really confused is that why Q subset Q(cos20) not a tower of field, i mean construct a new tower of field

[crystal vale]

Can you explain more ?

[stark helm]

i tied, it works, really appreciate that

[halcyon hull]

How do you impose a Banach structure on a manifold?

[night onyx]

I guess embedd it into a Banach space (which is possible for compact smooth manifolds)

but that doesn't make it a Banach space, just makes it's manifold structure come from that of one

[coral prawn]

just stumbled across this and i love it already https://en.wikipedia.org/wiki/Minimal_polynomial_of_2cos(2pi/n)

trouble is the formula for them is really ugly lol

[coral spindle]

This came up in my research lol

Because this is zeta + zeta^-1 where zeta is an nth root of unity

a primitive* nth root

[coral prawn]

yeah theyre kind of amazing

and the thing is they relate to cyclotomics too

[south patrol]

This also came up in a Galois theory exam and I was big dumb dumb

Well like it was supposed to be easy to show that like this generates the real subfield of Q(zeta_n)

And it's enough to show that Q(zeta_n)/Q(zeta_n + 1/zeta_n) has degree 2

But lol I could not spot the min poly for some reason

[coral spindle]

Min poly hard

I think the trick is to look at the galois group, right

yea

well I mean what else would it be

Wait no shit not the min poly for Q(zeta_n + 1/zeta_n) over Q LMAO

[delicate orchid]

Me when 1/a root of unity is a root of unity

[south patrol]

Yeah lol

[coral prawn]

so this might be a stupid question, but arent more elements of Gal(Q(sqrt(2))/Q) possible by something like $a+b\sqrt2 \to a+(b+1)\sqrt2$?

[cloud walrus]

sam

[coral prawn]

for $a,b \in Q$

[cloud walrus]

sam

[south patrol]

Elements of the Galois group must send roots of polynomials over Q to other roots

In particular this doesn't preserve roots of x^2 - 2

Or note this sends 0 to sqrt(2) and hence isn't a ring homomorphism

[south patrol]

But yeah this is key w Galois grp

[coral prawn]

oh i didnt know that was a condition

good to know thank you

[south patrol]

Well

It's not an extra condition

It's a property of any element of the Galois group

You can check by direct computation

[coral prawn]

oh ok

so it is still just 2 elements then or am i completely an idiot

[south patrol]

The former is correct

The latter is therefore false

[coral prawn]

awesome

[winged void]

I have small question

@south patrol

im not really sure if there is an intution behind

it

[south patrol]

Ye feel free to ask

[winged void]

but why when we say for example $[a] or \overline{a}$ then we say that this is equivalent to $x = a mod z $

[cloud walrus]

Mootje

[winged void]

for example until now i do not understand why if we want to prove that [a] + [b] = [a+b]

is well defined

then i need to say that a_1 = a_2 mod n and then say the same b_1 = b_2 mod n and use them to prove well defined

me as mathmaticien or somoene who is going to be a mathmaticien why should he do that

and what is here well defined really mean

[winged void]

thanks

this is my question

[south patrol]

Sure, it's a good question.

Well the point is like concretely we are thinking about Z/nZ so we have Z and identify x ~ y if x = y mod n.

We want to define an addition on Z/nZ, I.e. add two equivalence classes. Ideally what you want is to say you can add x and y by writing x= [a] and y = [b] and letting x + y = [a+b]. But this definition involves choosing representatives a and b of the equivalence classes x and y. So to show the definition makes sense, you need to show that your choices of a and of b don't matter

[winged void]

sure i agree with you that the choice do not matter of a and b

because then it will be a problem

but the thing that im asking [a] means actually a mod n

why cannot i say that for example a mod n + b mod n = a+b mod n

or i mean is equivalen to a mod n + b mod n congruent to a + b mod n

in the proof they do this

and the thing that i do not get it how [a] is similar to saying a_1 congruent a_2 mod n

or maybe im missing something

in the definition of [a] or \overline{a}

[winged void]

correct me if im wrong

please

[south patrol]

a_1 and a_2 are two representatives of the same class

And you're checking you get the same equivalence when u add each

[winged void]

sure i agree on that

oh i see what you mean

its here a must

because when i say for example

just what i suggested

a_1 mod n + a_2 mod n = (a_1 + a_2) mod n

then actually i did not check that the choices

[south patrol]

Yeah

[winged void]

but when i use two representative of the same class and them together and get that result

that mean it does not matter which represen i choose

i will get the same result

that is the intuition behind it

i see it now

thanks alot!

[south patrol]

Nppp

Like in some sense the definition is a bit "naughty" in that you are choosing things arbitrarily

But then you show the choice didn't do anything

[winged void]

exactly

[south patrol]

But yeah in future, when people say stuff is "well-defined" it usually means "doesn't depend on choices of equivalence classes"

I guess sometimes it means for example that the domain actually maps to the codomain or things like that

[coral prawn]

wait sorry back to this

would a function that adds one to b when b isnt 0 work?

[south patrol]

No, well firstly it won't send roots to roots

It also sounds very ad hoc so unlikely to be a ring hom (and indeed it will still not be additive)

[coral prawn]

so when you say it has to send roots to roots, do you mean roots of the same polynomial?

as in there might not exist a polynomial with roots bsqrt2 and (b+1)sqrt2

[south patrol]

if $K/F$ is a field extension and $\sigma$ a field auto of $K$ fixing $F$ and if $p(x) \in F[x]$ is a poly with a root $\alpha \in K$ then $p(\sigma(\alpha)) = \sigma(p(\alpha)) = 0$

[cloud walrus]

Süßkartoffel

[south patrol]

So in particular for this case, if g is in galois group then g(sqrt(2)) = \pm sqrt(2)

[coral prawn]

i might be confused but sqrt2 and 2sqrt2 are both roots in Q(sqrt2), doesnt the b->b+1 map send roots to roots?

[south patrol]

They aren't both roots of x^2 - 2

[coral prawn]

ohhh so it does depend on the polynomial

that makes more sense

wait but couldnt they both be roots of something else?

[south patrol]

Sure but i'm sayign this is necessary

[chilly ocean]

[coral prawn]

im a little lost lol

[south patrol]

I'm saying like if sigma is anything as above, then for every polynomial p in F, sigma must send any root of p to another of p

so in particular for p(x) = x^2 -2, sigma must send sqrt(2) to +/- sqrt(2)

[coral prawn]

OH it just clicked

[south patrol]

this is jut by calculation

[coral prawn]

so that applies for every polynomial then?

[south patrol]

like uh lemme write s lol

if a^2 = 2 then s(a)^2 = s(a^2) = s(2) = 2

[coral prawn]

that helps a ton thanks lol

[south patrol]

np

[lavish sigil]

Suppose that we have a finitely generated and free $\mathbb{Z}$-module $M$ with a basis $b_1,\dots,b_r\in M$. Suppose that we also have a bilinear (in the sense of $\mathbb{Z}$-modules) form $B:M\times M\to\mathbb{Z}$ which in this basis is represented by a tridiagonal matrix with ones on the "side diagonals", and a sequence of integers $d_1,\dots,d_r$ on the main diagonal. If $M$ has another basis $b_1',\dots,b_r'\in M$ with this property, but with another sequence of integers $d_1',\dots,d_r'$, does it follow that this is the same integers but possibly in another order?

[cloud walrus]

Gustav

[lavish sigil]

In other words, are these integers uniquely determined by the bilinear form $B$ itself, up to ordering?

[cloud walrus]

Gustav

[sonic coral]

if i wanted to go play with the inverse galois problem is there any strategy to it? i was going to try and find an irreducible polynomial that has galois group of a transitive subgroup of Sn for n<= 7 but im not really sure what intuition there is other than just playing with it and finding patterns

i know i can look at the discriminate and what not for n <= 4 but there’s like 19 other transitive subgroups to go through

[tranquil musk]

Im hoping somebody can help me show that: the kernel of the evaluation map E[a] is a maximum ideal of R[x], where the evaluation map maps R[x] (polynomials with real coefficients) to the real numbers.

So far I have that the kernel of the evaluation map is (x-a)p(x) and Im not sure how to finish the question from here

[chilly radish]

Generally solutions to the inverse galois problem use techniques from other areas of math, in particular algebraic geometry and more advanced algebra. I doubt you'll be able to do much with just the tools of basic galois theory

[sonic coral]

hm okay thank you

[chilly radish]

That's not to say you shouldn't play around with it! Just don't expect too much. I would maybe look up some early literature on these types of problems, those might use more elementary methods

Or maybe a historical overview

There are surely some of those online

[last spoke]

@tranquil musk there’s something special about the quotient of a ring by a maximal ideal, do you know what it is?

[tranquil musk]

so is it the ideal is maximal iff R/I is a field

Im just struggling to write it all out

am I correct in my initial calculation that the kernel of the evaluation map is (x-a)p(x)?

[last spoke]

Ya

So what does R[x]/(x-a) look like? Writing (x-a) for the ideal gen’d by x-a

[tranquil musk]

well im just trying to understnand the concept of quotient rings rn

but why does the 'p(x)' in '(x-a)p(x)' get dropped?

and isnt there some easier way to prove it is a maximal ideal by showing there are no ideals properly contained between I and R?

[last spoke]

The elements of the quotient ring R/I are additive cosets I + r for r in R. I just don’t write the multiplied p(x) because it’s redundant for describing the ideal (by definition they’re closed under multiplication by r in R).

Maybe there’s another easy way to do it, but if you have a good understanding of what a quotient ring is then I think doing it by asking whether R[x]/(x-a) is a field is pretty concise

[tranquil musk]

So would you say that I gotta show that R[x]/(x-a) is commutative under the multiplication rule?

you got anyway you would recommend to show that its a field?

[sage wagon]

Pls help: how do you simplify quotients of free abelian group presentations? E.g. <a,b,c>/<a+b-c, a-b+c>, why isn't that equal to <a+b-c,a-b+c,c>/<a+b-c, a-b+c>=<c>? Is it because c is not independent of a+b-c and a-b+c so you can't just quotients those out? Are there any articles/resources that explain how to do this?

For that example would it be equal to <a,b,c>/<2a,2(b-c)>? It's supposed to be (Z/2Z)xZ but not sure how to get that..

[sage wagon]

Oh I think it's because they need to form bases of 'integer points', not just R3

[crystal vale]

If G = S_n and X = {1,2,.....,n } and G acts on X so mapping G -> Sym(X) is a bijective mapping?

[languid trellis]

g |-> g_L is a bijection yes

[crystal vale]

Okay, thank you

[crystal vale]

If stab(a) = stab (B) , a is an element of G and B is subset of G, then B contains a ?

[languid trellis]

Let S_n act on itself by conjugation, n >3. So, the centre is trivial. Take two disjoint non-identity elements a,b in S_n. Then stab(a)=stab(b) but a neq b.

A_n is simple for n geq 5. Let A_6 act on itself by conjugation. The centre is trivial, and it contains no normal subgroups. Let G be the subgroup of A_6 generated by (12). Let a be an element not in this subgroup. Then stab(a) = stab(G) but a is not an element of G.

In general, this seems unlikely because stab(a) and stab(B) are two different objects: stab(a) = {g \in G s.t. ga = g}, in other words, the group action stabilises a specific element, wheras stab(B) = normaliser of B = {g \in G s.t. gB = B}. We only require that it remains in the same subset/subgroup, it need not stabilise any specific element, in fact.

wait actually I'm not sure stab(a) = stab(G) in the second example. My first example still works however.

[crystal vale]

Okay, thank you , I understand the first example, I am not sure about the second one

If G acts transitive on the finite set A. And Stab(a) is a maximal subgroup of G then how can I show that they act as primitively?

Got it, thank you

[coral prawn]

Back to this one last time (I understand now, I’m just playing devil’s advocate) what if we found a function that sometimes acts as one existing function and sometimes acts as another? Example: a+bsqrt2 -> a-bsqrt for rational a but INTEGER b; this would be the identity for non integer b values, but conjugate the radical for integer values (a kind of combination of the 2 existing functions)

Would that even count as its own unique function? It’s kind of dependent on the other 2

[south patrol]

I mean something so ad hoc is not going to be a ring homomorphism

Note any element of Gal(Q(sqrt(2))/Q) is Q-linear and is fixed by where it sends sqrt(2)

[rocky cloak]

s(a + bsqrt(2)) = s(a) + s(bsqrt(2)) = a + s(b)s(sqrt(2)) = a + b s(sqrt(2))

So s(sqrt(2)) determines the whole shabang

[coral prawn]

Oh I keep forgetting about the ring homomorphisms lol

That makes sense

[hidden wind]

if i'm asked to compute the permutation (12m)(12k)^2... what's a nice way to get around splitting into cases on m and k?

[south patrol]

Lol im sconfused by your notation

Do you mean like (1,2,m)(1,2,k)^2

[sly rain]

What groups in introduced in a general intro to abstract algebra book (either in main body or introduced in the exercises) are not toy examples, but play an important role in some other, not too obscure, math-field or youre likely to see again regularly when you study stuff beyond contents of such a book?

Context: wew lads hating on dihedral groups and me not recognizing some groups that are introduced but not motivated (e.g. some matrix groups)

One way to answer could be:
I study topic X and we care about group(s) Y, because of reason Z

[proven halo]

the matrix groups like GL(n, R), SL(n, R), O(n), SO(n), U(n), etc are example of Lie groups and they occur a lot when you care about continuous symmetries. They show up in physics, as well as differential geometry, etc. For example SO(3) is the group of proper rotations in R^3 so it comes up whenever you care about object orientations in 3d

S_n (the group of permutations on {1, ..., n}) occurs everywhere (and by Cayley's theorem, any finite group can be viewed as a subgroup of S_n for some n)

if you study Galois theory you'll run into a lot of other finite groups as the galois groups of polynomials

dihedral groups are kind of like discrete versions of O(2)

they are rotations and reflections but restricted to a finite "angle step"

[errant shadow]

What's the picture you have in mind when you think of modules?

[night onyx]

the symmetric group is actually really important considering how early you learn about it

[proven halo]

so dihedral groups would show up a lot when studying lattices and other systems with "discrete" properties

[night onyx]

it comes up in analysis/integrating over compact groups of matrices (maybe a more niche/difficult application)

[proven halo]

symmetric (S_n) and alternating groups (A_n) show up in the definition of determinants (the even and odd permutation sign expansion)

[rustic crown]

was i summoned

hewwo walter

[agile burrow]

hi det, hope you're well

[rustic crown]

yee

i hope you're uwu too :3

[agile burrow]

i'm alright

[summer path]

Hi det

[agile burrow]

I took an extended break from AG so I have to relearn some things

[summer path]

How do you summon a det

[rustic crown]

hiii tubu

[rustic crown]

det relearing some homological alg, det always forget stuff ><

but each time det relearns something, det's understanding gets uwu'er

[rustic crown]

you talk about determinants or eevees when det randomly checks discord

[summer path]

[south patrol]

Hom algies

[rustic crown]

homie alg

[summer path]

commie alg and homie alg

[rustic crown]

[delicate orchid]

I don’t hate dihedral groups, they’re a wonderful class. I hate how they’re introduced and you were forced to do exercises related to them before you had any machinery (if you are who you think you are anyway )

[rustic crown]

how would wew introduce them?

[delicate orchid]

After group actions

[sly rain]

Thats fair, what Type of machinery are you talking about for example?
Also not Sure what you mean by the content in the (...)

[delicate orchid]

Quotients and homomorphisms

At the minimum

[rustic crown]

right, you don't really know what presentations are without free groups and quotients

[delicate orchid]

S_n is much better as an introductory non-abelian example because it’s far more combinatorial

[delicate orchid]

Ok not that far lol, free groups would go pretty close to the end of a course

[rustic crown]

oh ><

[delicate orchid]

(Not to say a non-rigorous treatment of presentations wouldn’t do wonders in a first course)

[languid trellis]

D_n is when something acts on an n-gon (?)

how the hell ar eyou defining d_n in terms of group actions

am i missing something

[delicate orchid]

We’re not. But why would you care about D_n before group actions

[rustic crown]

D_n is subgroup of GL_R(C) generated by conjugation and zeta_n

[languid trellis]

oh I see what you're saying

[delicate orchid]

This would be a good alternative too

One of my undergrad profs loved writing groups as matrix groups.

I wonder if there’s a field that studies how to do that

[rustic crown]

representation theory >.<

[delicate orchid]

(I presume you mean GL_2(R) as well)

[languid trellis]

sorry what do you mean by GL_R(C)?

[rustic crown]

Aut_R(C) maybe better? R-linear automorphisms of C

[delicate orchid]

Oh you mean Au- yeah ok

[rustic crown]

after choosing basis same as GL_2(R)

[languid trellis]

mhm

[rustic crown]

yee, i like this now that i think about it

[delicate orchid]

It bypasses my main problem - early students don’t really have a rigorous way of proving the relators in Dn

Now u can just do it with matrix multiplication

[languid trellis]

I can relate. I've supposedly seen a proof of this but it's still magic

[delicate orchid]

If you use the action of Dn on an n-gon to write the group elements as permutations u will notice that reflections are products of 2-cycles and rotations are n-cycles

You can prove the relators like this as well

[sly rain]

Aluffi does free groups pretty early in his first group chapter then ... then presentations during the "homomorphism theorems" then group actions.
The way they where introduced didnt make them feel harder than the rest (or do you think they get introduced at the end for different reasons?)

[summer path]

i thought det liked rep theory

[dull ginkgo]

It sucks yes

[chilly ocean]

is there a group such that the order is a continuous function?

[coral spindle]

The order is a cardinality, so how could it be a continuous function?

If you mean the order of each element, well yes but not in an interesting way, since we could choose the topology on the group G and the naturals however we wish.

You're not going to get super interesting results here because the function sending each element to its order has a discrete domain.

[chilly ocean]

I am talking about this order
https://en.wikipedia.org/wiki/Order_(group_theory)

[coral spindle]

In mathematics, the order of a finite group is the number of its elements.

Darling

[languid trellis]

Order the elements of your group, g_1, ... , g_n. Then the function g_i -> |g_i| N-> N is cts as every point is isolated

lul

[nova holly]

log has always been my favorite number

[chilly ocean]

but you mean a continuous from discrete elements right?

[coral spindle]

What is a 'discrete element'

[languid trellis]

its continuous in the standard way

[chilly ocean]

integers for example

[dull ginkgo]

You’d need to define what “continuity” is in this case

[coral spindle]

The order of an element is defined for any group, not merely finite ones.

[chilly ocean]

then I did not understand that

what do you mean by that

[languid trellis]

I'm just meming because every functions are continuous at isolated points, though its not super interesting

[chilly ocean]

that's not the same as saying that the integers are continuous?

for example

[languid trellis]

It doesn't make sense to ask if a set is continuous

[dull ginkgo]

Blud what are you waffling about

[coral spindle]

Guys just read the wikipedia page it explains everything

[chilly ocean]

I am trying to find a example on a group where the order of a group is a continuous function

[coral spindle]

Think of any finite group

Tadah, the order is a continuous function

[languid trellis]

Wow I just read the Wikipedia page and I feel no more enlightened

[dull ginkgo]

A continuous function in respect to what

[coral spindle]

(With the obvious topologies on the group)

[coral spindle]

(It was a joke about the fact that I tried to clarify which kind of order was meant, but all I got in answer was a wikipedia page)

[languid trellis]

ah lol

[chilly ocean]

today I was talking with
with my abstract algebra teacher and asked if it is possible for the order to be a Limit .

so , with a infinite numbers of operations it converges to the identity element

[dull ginkgo]

[coral spindle]

Yes, for example 3 = lim_{x → infty} 3

What you're talking about is a dynamics problem at heart

This requires information about the topological properties of the group

[chilly ocean]

i.e: a binary operation such that for some a that belongs to the set of some group it happens that
a * a * a * a * a * a ..... until infinity = e

[dull ginkgo]

I am so fucking confused anyway

Chinese remainder theorem

[chilly ocean]

so he told me that a neccesary condition was that the ord must be a continuous function in order to be able to introduce a limit there

by "ord" I mean the function ord

[languid trellis]

i might start the exercises for that section now

[chilly ocean]

order

[dull ginkgo]

I am stuck with my family and cannot maybe

*math

[languid trellis]

spend time with them

[chilly ocean]

so he said that the problem was finding a non trivial group
but that just was a question I made

not really related to the class material

he show me an example where in the real numbers without 0 , that was not possible.

with multiplication as binary operation

[winter shore]

Sanity check, for a finitely generated module M over a pid, we have that M = R^Rank M + R/a_i. Does this imply that the least number of generators must be rank M + number of a_i's?

It seems in principle you could get R/a_i + R/a_j from a single kernel R -> R, lowering the number of generators by one

[dull ginkgo]

You’re gonna be far ahead of me, I have physics and human studies work to do tonight

[languid trellis]

I'm not doing any more tonight i'll let you catch up

[winter shore]

he's lying to get ahead of you

ignore your family, they will die eventually, math is eternal

[dull ginkgo]

I am attempting an actual proof of CRT because I never actually proved it for general rings lmao

First as a base, let A and B be proper ideals of the ring R, such that A + B = R
We want to find a y such that y = x_1 + ca = x_2 + cb for some a in A, b in B

Then 0 = (x_1 - x_2) + c(a - b) => c(a - b) = (x_2 - x_1), and we can find some a b such that a - b = 1, so
c = (x_2 - x_1) thus we have found our c, and y = x_1 + (x_2 - x_1)a = x_2 + (x_2 - x_1)b

Thus: y = x_1 (mod A) and y = x_2 (mod B)

unsure if we can repeat that off rip with y and some x_3 and another ideal coprime with the first two off rip, let me check

[rocky cloak]

You're right that if ai and aj are relatively prime, then R/ai (+) R/aj = R/aiaj

What you might do is put the module into Smith normal form. Meaning that ai divides ai+1. In which case the minimal number of generators is indeed the number of ais

[languid trellis]

Is the idea in this exercise to notice Z/pZ is the prime ring of F. Then we can consider the multiplicative subgroup of each ring, so (Z/pZ)* is isomorphic to a subgroup of F \ {0}, so p-1 | |F| -1.

[dull ginkgo]

yes

[south patrol]

Would be twice as readable if it were "divides" instead of "|"

[south patrol]

This is great yes

Though if you know about the prime ring it is almost immediate that the size is a power of p so i find that amusing

I mean that in the sense it is an interesting pedagogical choice

[languid trellis]

All I know about prime rings is that they're all Z or Z/kZ

[south patrol]

Yeah and for a field it's Z or Z/pZ

[languid trellis]

yeah

[dull ginkgo]

We haven't gotten to a lot of "deeper" field stuff and how they're solutions to X^q - X

[south patrol]

But idk if the "we shall see later" is another exercise so I probabl shouldn't say anything aha

[dull ginkgo]

I am sorta stuck on the inductive generalization of CRT

let me try typing it in chat hopefully so while I'm typing it I realize I'm an idiot

I'll spoiler it for Swiftee

[south patrol]

Have you got n=2 then?

[dull ginkgo]

yes that's my base case

[south patrol]

Cool

[dull ginkgo]

this

[south patrol]

Well base can be n=1 lel

But hmm

Tbh the proof I remember learning doesn't use induction

[dull ginkgo]

Jacobson's phrasing of it is a bit different

[dull ginkgo]

oh seriously?

huh

not strong either?

[south patrol]

You just write down a map and show it is injective and surjective

Well

If the rings are commutative

[dull ginkgo]

loud incorrect buzzer

[south patrol]

The non commutative case is painful and idk if anyone uses it

Lol

[dull ginkgo]

the mf isn't

[south patrol]

Well actually depends

[rocky cloak]

If you want to do it inductively, prove that if C is relatively prime to both A and B, then it's also relatively prime to A\cap B

[dull ginkgo]

no assumption of commutativity

[dull ginkgo]

OH SHIT

[south patrol]

Ah yes I did this too

It's just funny because the assumptions are like

[dull ginkgo]

i did that already lol

i forgort

[south patrol]

If you do the normal proof for CRT for comm rings, the assumptions here are exactly what you need for the proof to work iirc

[rocky cloak]

The noncommutative case is the same as the commutative, no?

[south patrol]

I think the non comm is way weaker

[rocky cloak]

How so?

[south patrol]

Iirc

Uhhh

Wait nah im wrong

Wait no

Like here they have that intersection

But normally you can just do smth weaker right like pairwise coprime or smth

Maybe im wrong tho

[rocky cloak]

Isn't pairwise coprime the usual condition?

Or what is the "stronger" thing?

[dull ginkgo]

I think it can be weakened to C being relatively prime to AB

[south patrol]

Yes but in Miz's statement there is a stronger assumption

[rocky cloak]

Oh, I see. Yeah that shouldn't be necessary either way I don't think

[south patrol]

From what i can find online it seems the general form with pairwise coprime ideals only works for commutative rings, tho I don't know a counterexample

Hm

[rocky cloak]

Wikipedia has it for arbitrary rings

https://en.m.wikipedia.org/wiki/Chinese_remainder_theorem

[dull ginkgo]

I want to show that if A,B,C are coprime ideals then C is coprime to AB

AB is a subring of A \cap B so if I can show that I'm golden

let me try an idea out

actually i wonder if both AB and BA are coprime to C

wait bruh

[languid trellis]

For this part of the exercise: Let a1, a2 \in R. Then a1 = i1 + i2 for some i1 \in I_1, i2 \in I_2. WTS that there exists an a such that a = a1 (mod I_1) i.e. a - a1 \in I_1. a - i1 - i2 \in I_1. Can we not just take a = -i_2?

[rocky cloak]

You also want a = a2 mod I2

[languid trellis]

Oh right I see.

[languid trellis]

are we not just saying that there exist cosets such that a1 + I1 \cap a2 + I2 is nonempty?

wait

[rocky cloak]

You can think of it that way if you like

[languid trellis]

for any two elements we pick, those cosets intersect

I wonder if going down the route of some partition argument may be useful

[rocky cloak]

Maybe, though I think the easiest route is just to directly construct a

[languid trellis]

if a1 = i1 + i2, a2 = i1' + i2', then for a - i1 - i2 \in I1, a - i1' - i2' \in I2, take a = i2 + i1'

nice

why is this called the chinese remainder theorem?

how is this related to remainders

[rocky cloak]

You're doing stuff mod I

So you're looking at the remainder

The original formulation has the ring being the integers

[languid trellis]

I suppose when I think of "remainders" i think of specifically mod n, n an integer. I guess we can take Z as R and mZ and nZ as our ideals if gcd(m,n) = 1

then, for any two integers j,k we have a = j (mod m), a = k (mod n). I'm struggling to see hwo this is illuminating tbh

like, sometimes stuff has the same remainder?

[dull ginkgo]

swiftee I'm gonna join vc

[languid trellis]

im going to bed soon sorry

[dull ginkgo]

Two questions:
What are ideals in Z?
What is Bezout Lemma?

that'll hopefully make things connect

[languid trellis]

ideals in Z is anything of the form kZ. Bezout's lemma says that the equation ax + by = 1 has a solution x,y if a and b are coprime.

So coprime ideals generate Z?

[dull ginkgo]

y e s

that's coprimality

[rocky cloak]

Well you have a bijection between numbers modulo mn and pairs of numbers modulo m and n respectively.

That means that to solve an equation modulo mn you can split it into to equations modulo m and modulo n, which might be much easier to solve

Comes up in basic number theory

[dull ginkgo]

A + B = R implies A and B are coprime

bezout lemma is CRT for Z

[languid trellis]

could you elaborate on the bijection between numbers modulo mn and pairs mod m, n respectively. The only thing I know about modulo mn stuff is that mnZ = mZ \cap nZ if m,n are coprime

[winged void]

I have small question I’m confused regarding something in group theory

What is now the difference between $\overline{a}$ and a

[cloud walrus]

Mootje

[winged void]

Because they use them in a book in a way as if they are the same

But they are not

[dull ginkgo]

Also I feel like this is implying R is commutative?

[rocky cloak]

Like say I want to solve the equation

x^2 = 1 (mod 15)

Then Chinese remainder theorem says that x is a solution iff

x^2 = 1 (mod 3)
and
x^2 = 1 (mod 5)

And further that if I can find any pair of solutions to these two equations, then they combine into a unique solution of the mod 15 equation.

x^2 = 1 (mod 3) has two solutions (1 and 2) and x^2 = 1 (mod 5) has two (1 and 4).

So x^2 = 1 (mod 15) has 4 solutions.

(1, 4, 11, 14)

[rocky cloak]

Why do you say that?

[winged void]

From where did you find 11 and 14

Because I’m also learning about chines remainder theorem

[dull ginkgo]

oh my fucking god dude it's so fucking easy I am a MORON

[rocky cloak]

There is some nontrivial work in going the other direction, but CRT guarantees it's unique at least

[winged void]

True

I have a question rehearing something else @rocky cloak hopefully you can help because I really hate gcd and mod

[dull ginkgo]

let A, B, C, be pairwise coprime.
a_1 + b_1 = a_2 + c_1 + b_2 + c_2 = 1

So 1 = c_1 + a_2 = c_1 + (c_2 + b_2)a_2 = (c_1 + c_2a_2) + b_2 a_2 = 1
c_1 + c_2a_2 is in C
so C + BA = R

[languid trellis]

So, in this case are we taking our base ring to be Z/15Z, and our ideals to be Z/3Z, Z/5Z resp. I take a1 = a2 = 1. Then there is an element a in Z/15Z such that a = 1 (mod 3), a = 1 (mod 5). I'm slightly lost how we justify the line "Then Chinese remainder theorem says that x is a solution iff ....". Is the wikipedia page on CRT any good because I feel like I'm using up too much time by asking a question of the form "why do we care about CRT" lol

[dull ginkgo]

R = C + A = C + (C + B)A = C + CA + BA = C(R + A) + BA = CR + BA = C + BA LOL

[rocky cloak]

So if the ring is Z, and the ideals are 3Z and 5Z, then CRT says there is an isomorphism between

Z/15Z and Z/3Z x Z/5Z

In other words every pair in Z/3Z x Z/5Z corresponds to an element in Z/15Z and vica versa

[winged void]

Let $n \in \mathbb{Z}_{>0}$ and $a \in \mathbb{Z}$.
\begin{enumerate}[(a)]
\item Prove: $\overline{a} \in (\mathbb{Z}/n\mathbb{Z})^* \Leftrightarrow \overline{-a} \in (\mathbb{Z}/n\mathbb{Z})^*$.
\item Prove: If $n$ is odd, then $\overline{a} = \overline{-a} \Leftrightarrow \overline{a} = \overline{0}$.
\end{enumerate}

[cloud walrus]

Mootje
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[winged void]

well the first one i prove it

the second one if n is odd then that

i did not know how to prove

[dull ginkgo]

HI SHIN

[chilly radish]

Hi

[languid trellis]

Oh, and intuitively this is an isomorphism because, by the construction in the proof, our choice of a depends on a1, a2.

[rocky cloak]

Hint: ||you can rearrange a = -a into 2a = 0||

[dull ginkgo]

Interestingly this is a weaker condition

[winged void]

well thats what im not getting

what is now the diffrence between a with overline

[languid trellis]

I might read the wikipedia page, this seems interesting

[winged void]

and a without overline

[languid trellis]

a with the overline is an element of the quotient structure

[rocky cloak]

The overline refers to the equivalence class of a in Z/nZ

[dull ginkgo]

If, for ideals A B C, that A is coprime to C, and B is coprime to C, then C is coprime to BA and coprime to AB

A and B need not be coprime?

huh

[winged void]

so then a is an element of the equivalence class of [a ]

can i see it like that

or in other word a is an element of \overline {a}

am i right @rocky cloak

oh sorry for penging you

[rocky cloak]

Sure

[winged void]

alright i will try the hint

[languid trellis]

I'm gonna go to sleep, thanks for your time as always jagr. I appreciate it

[chilly radish]

A=B

[dull ginkgo]

Let $A, B, C$ be ideals in ring $R$ \
Assume $a + c_1 = b + c_2 = 1$ where $a \in A, b \in B, c_1 \in C, c_2 \in C$ \
Then $1 = c_1 + a = c_1 + (c_2 + b)a = c_1 + c_2a + ba$, but $c_1 + c_2a \in C$ and $ba \in BA$ \
Likewise $1 = c_1 + a = c_1 + a(c_2 + b) = c_1 + ac_2 + ab$, but $c_1 + ac_2 \in C$ and $ab \in AB$\
This implies that if $A$ and $B$ are both coprime to $C$, then $C$ is coprime to both product ideals $AB$ and $BA$, both within $A \cap B$

[winged void]

i do not know why it feels that my proof is wrong

[dull ginkgo]

@chilly radish is this valid?

[winged void]

this is what i have

it feels as if im playing or something

im not convinced

[chilly radish]

All ideals are two-sided?

Yea this is fine

[cloud walrus]

Request a new nickname

[dull ginkgo]

yes for CRT

[chilly radish]

I'm not sure why you need this for CRT

[dull ginkgo]

induction step

[rocky cloak]

2a = 0 thus a = 0 is sort of a big unjustified step

[chilly radish]

Oh yea

Sure

This works yes

[winged void]

Well then what do you suggest ?

[dull ginkgo]

yipee

[rocky cloak]

Just unpack the definition of what it means that two things are equal in Z/nZ

[dull ginkgo]

I revoke my statement this isn't that bad

[winged void]

Well that mean that it divides their difference

Or congruent to each other

I guess that’s what you kean

Mean

[rocky cloak]

Indeed, and then you had this hypothesis of n being odd

[chilly radish]

Oh wait sorru

That's only in comm rings

[winged void]

I will try that

One second

[dull ginkgo]

yeah I had a short circuit moment

pretty sure CRT in the noncom case just still considers intersection, which will contain the products afaik

[chilly radish]

Yes

Intersection is the correct operation to consider since it acts like GCD

[rocky cloak]

I guess it's true in the noncommutative case as well if product means AB + BA

[dull ginkgo]

By induction that shows if I_1...I_n are pairwise coprime, that I_n is coprime to the intersection of the other I_m before it

oh i didn't need to show that

lol whatever still jotting it down

[winged void]

@rocky cloak i get stuk then

i guess there is a part that i do not fully understand or something

but not so sure

[dull ginkgo]

what

[rocky cloak]

So 2a = 0 (mod n) should mean that n divides 2a

[winged void]

but why can i say that

i mean $\overline{2a}$ is not the same as $2a$

[rocky cloak]

Because that's the definition

[cloud walrus]

Mootje

[winged void]

am i right

because \overline{2a} like has many elemnts right

[rocky cloak]

Overline {2a} just means 2a modulo n

[winged void]

for example $\overline{0} mod 3$ has many elemnt s

[cloud walrus]

Mootje

[winged void]

if you get what i mean

like 0 3 6 9 etc

[dull ginkgo]

For $\mathbb{Z}_n$, $\overline{x} = {y \in \mathbb{Z} : n \mid (x - y)}$

[cloud walrus]

Request a new nickname

[winged void]

well sure agree on that but here y is arbitrair

[dull ginkgo]

so what if the set $\overline{x} = \overline{-x}$?

[winged void]

but here we use like the classes as elements

[cloud walrus]

Request a new nickname

[hidden wind]

yes