groups-rings-fields

[thorny slate]

no that would be Q(i)

what you wrote is Q

[whole basalt]

Okay the gaussian rationals

That's not the point

I'm talking about a+bj with a,b in Q and j^2=1

[thorny slate]

ah

it's a field isnt it

[whole basalt]

Nope

Not an integral domain

[thorny slate]

oh (j-1)(j+1)

seems to me that <j-1> and <j+1> are both maximal ideals

and the only ones?

[stone fulcrum]

I'm actually interested in those right now myself. What's a local ring? Lol

[whole basalt]

I know that if you do Q [x] / (x^2) you do get a local ring

[thorny slate]

a ring with only one maximal ideal

[whole basalt]

^

[thorny slate]

yeah cuz the ideal there is <x>

but here you have <j-1> and <j+1>

[whole basalt]

I see. That makes sense

Hmm. I'll have to play with those

I haven't looked much into local rings but they seem interesting

The dual numbers in particular

But I'd like to learn more about the split complex numbers at some point.

Anyway thanks

[bleak abyss]

Local rings are supposed to be fairly important

I know a tiny bit from the number theory side of things

Basically if you have a PID that's a local ring, that's called a discrete valuation ring (in this case that unique maximal ideal is also the only prime ideal)

(Third time is the charm lmao)

[thorny slate]

yeah a local dedekind domain is a dvd

dedekind domains are characterized by this property

that they're locally dvds

[bleak abyss]

The principal examples of DVRs are the p-adics, which are ultra important in NT. They turn out to be Dedekind domains but that's trivial because they're PIDs

Though the other way, as jacobian said, is tougher

[stone fulcrum]

Every element is invertible in the split complex, no? Except for 0 I guess

[thorny slate]

(j-1)(j+1) = 0

so those two are not invertible

so they generate ideals

also 2 is in <j-1, j+1>

so it's the whole ring

so they generate maximal ideals

[stone fulcrum]

Fair enough! I have to learn some of this as well

[tulip osprey]

anyone here?

[vestal needle]

can someone explain to me how to do 1a?

[thorny slate]

which part

is it normal?

[vestal needle]

does that mean i have to test every element in Z30

yeah

[thorny slate]

test every element for what

[vestal needle]

using the normal subgroup test

[thorny slate]

well you need to check that for arbitrary g, k, you have gkg-1 in K

there's a fast reason why this holds

in general you can use some tricks, it's enough to check gkg-1 in K for g,k in some set of generators of G and K

but this example is easier

[vestal needle]

so i can just test any random element in G

[thorny slate]

you could but you don't have to in this case

because there's an easy reason why gkg-1 is in K

[vestal needle]

is it because it is abelian

[sonic current]

oh cool, an exercise I can actually understand* jesus christ

[vestal needle]

hi

[sonic current]

I think it's abelian yeah

not sure -- let @thorny slate confirm it

[thorny slate]

yes

so gkg-1 = gg-1 k = k

and so any subgroup of an abelian group is normal

[vestal needle]

are all Z groups abelian then

[sonic current]

i think the multiplicative ones are

[thorny slate]

what's a Z group

[vestal needle]

but none of the Z groups are multiplicative

[sonic current]

nevermind, i'm dumb

[thorny slate]

Z is abelian

and quotients / subgroups of abelian groups are abelian

[vestal needle]

they are all the numbers from 0 to n-1 of Zn

[thorny slate]

those are cyclic groups

[vestal needle]

so like Z30 in this example are all the numbers from 0 to 29

oh lol

so yeah what i meant to ask then is, are all cyclic groups abelian

[sonic current]

yes

it's a known fact!

[thorny slate]

yeah

because are elements are a^k

so

a^m a^n = a^(m+n) = a^n a^m

also they are quotients of Z, which is abelian

[vestal needle]

ok ty

[sonic current]

@thorny slate, since you're here let me ask you a quick one

regarding associate elements

would you say an element a ~ 1 (1 being the multiplicative unity)

implies a | 1?

but in a given ring, this does not imply a = 1 does it

[thorny slate]

sure

yeah it doesn't

even in Z it doesn't

-1 divides 1

[sonic current]

yeah

damn

[thorny slate]

those are the units of the ring

[sonic current]

i have c ~1 and I must prove a ~ b when a = bc

it's quite simple really

is it obvious? can I argue that since c ~ 1 then c is a unit and so a = bc implies a ~b ?

[thorny slate]

sure

[sonic current]

alright

thanks!!

[sonic current]

working on this rn

[thorny slate]

you gotta use the norm

[sonic current]

yes

i know it comes up somewhere

so far i've got that if there are units a and b, then ab = 1

so N(1) = N(a)N(b)

not sure this gets me anywhere relevant

[thorny slate]

well (-1)1 = -1 is not 1

but the norm of a unit has to be a unit

[sonic current]

hmm

let me see if I can work something up

oh I also know a is a unit iff N(a) = 1

[thorny slate]

that makes it straightforward

[sonic current]

so the only possible units are -1 and 1 ?

N(1) = N(a)N(b); since N(a) and N(b) are integers, we reduce the problem to finding which pairs of integers return 1. It's only -1 * -1 and 1 *1

[thorny slate]

i dont get what you're doing

you have to find the norm of a + b sqrt(-n)

[sonic current]

I don't think this is right either or else it's saying that Z(√d) has no units other than -1 and 1

N stands for norm function

the norm is m^2+5n^2

[errant drum]

Is n negative or positive it depends

[sonic current]

it's negative

hmmmmm

so the norm is a^2-5b^2?

[errant drum]

So for example in Z[√3] 2-√3 is a unit

But in Z[√-3] it's just 1,-1,

[sonic current]

no, i was doing it wrong then

i'm working in Z(√-5)

[errant drum]

Yeah so a^2 + 5b^2 = 1

[sonic current]

anyway, the norm is still positive yeah

i still don't get how to relate the norm to the exercise

maybe i'm jumping the gun at N(1) = N(a)N(b)

I should be solving a^2 + 5b^2 = 1?

[errant drum]

Yeah

[sonic current]

hmm!

is there a reliable way to solve this?

my book says something on solving for mod 5 but I don't really get it

[errant drum]

Well that's one way of doing it but much easier is if a^2=1 then b =0

Can you explain why?

[sonic current]

ha, just realised this is an ellipse

[errant drum]

Yeah

[sonic current]

well yeah, 1 + 5b^2 = 1 == 5b^2 = 0 == b = 0?

[errant drum]

Yeah

Now if b is non zero what can you say

[sonic current]

then the solution gets messy?

well

a = √1-5b^2 so no integer solutions?

[errant drum]

I mean if b>0 and b is an integer then b is at least 1 which means we've already overshot

[sonic current]

because 1-5 is negative and no negative square roots on integers?

[errant drum]

Yeah you could reason it that way but if b^2>0 then b^2 is greater than or equal to 1 which means that 5b^2 > 1 so

a^2 + 5b^2 =/= 1

[sonic current]

oh yes! that is way more clean

thank you so much, @errant drum

you guys have been a tremendous help

[errant drum]

Np also I think doing this helps me a lot as well.

[sonic current]

what do you do?

math-wise

[errant drum]

I'm a 2nd year student doing maths and economics

In the UK

[sonic current]

i'll see about Z(√-3)

[errant drum]

It's the same king of reasoning basically

[sonic current]

yeah, I just hope I have more possible solutions

it's the same thing?

a = √1-3b^2

very boring exercise

lol

[vestal needle]

hi im trying to do part e

is there a shortcut for this or do i have to do the normal subgroup test for all 8 elements

also U(20) is the group consisting of all numbers smaller than and also relatively prime to 20

beginning with 1

so {1,3,7,9,11,13,17,19}

[thorny slate]

find out which group that is

and what its properties are

it's a group of 8 elements right?

is it abelian?

[vestal needle]

yeah because its multiplication

oh wow

[thorny slate]

there you go

you can also identify which group it is if you want

it's either Z2^3 or Z2 x Z4 or Z8

[vestal needle]

oh ok thx

[thorny slate]

oh or D4 I guess

but its abelian so it cant be D4

[sonic current]

@vestal needle have you learned about isomorphisms yet?

[whole basalt]

@vestal needle -- One other property that can come in really handy is that if a subgroup has half the elements of the original group, it's automatically normal.

See if you can figure out why that is. (What are the left and right cosets?)

(That one fact really helped me identify normal subgroups quickly)

[sonic current]

working in Z(√-5) (norm: a^2+5b^2 for a + b√-5)

p = 2-3√-5

is p irreducible?

I've done a bit of work. I know its norm is 40 which is non-prime, so there's the possibility that it might not be irreducible

so since 40 = 2^3 * 5 i went ahead and equaled the norm to 5; so, a^2+5b^2 = 5 has a solution a = 0 and b = 1. but then, it is impossible to find an element x such that x * √-5 = 2-3√-5. what can I conclude from this?

I think i've exhausted the solutions of a^2+5b^2 = 5

[vestal needle]

@sonic current yeah i have. not my strongest topic tho

@whole basalt thanks for the tip! ill try to figure it out soon

[errant drum]

@sonic current

I'm not sure why you equalled it to 5

I think there's a more elegant solution but N(2-3√-5). = 2^2 + 9 * 5 = 49.

Suppose there were a and b such that ab = 2-3√-5

N(2-3√-5) = N(a)N(b).

So 49 = N(a)N(b)

Now if you had N(a) = 49 and N(b) = 1 that's fine because then it's irreducible. (Same thing for reversing a and b)

Instead you want to make sure you can't have N(a) or N(b)= 7

[vestal needle]

@thorny slate do you know what to do for part f too?

[thorny slate]

yes

here you have to actually check stuff

so you have to check gkg-1 in K for each generator g, k

G is generated by two elements <D, R90>

and K is generated by <R180>

so you just have to check 2 things

[vestal needle]

but D4 isnt generated by anything right

all the rotations can only give you subgroup containing rotations

and all the reflections are their own inverse

[sonic current]

well

can't you get all the rotations out of just one?

and rotate over the rotation?

r rotation, then r^2, r^3, etc.?

[vestal needle]

so the subgroup generated by any reflection is the reflection and the identity R0

yeah you can get all the rotations using R90 and R270

[sonic current]

the reflection has order 2

you just need one

don't you?

[vestal needle]

why do i need only one

[thorny slate]

D4 is generated by 2 elements

rotation and reflection

you only need to check gkg-1 for g in some set that generates the group

[sonic current]

@vestal needle R90 * R90 = R180; R90 * R90 * R90 = R270

[vestal needle]

yeah but isnt it impossible to generate a reflection using R90

im confused lol

[thorny slate]

yes it is

D4 is not generated by any single element

but it is generated by a set of two elements

{rotation, reflection}

[vestal needle]

oh thats what you meant sorry

[sonic current]

yes, I wasn't very clear

you can also have sets generated by two elements!

[slim lily]

whoops haha had my name on that

does anyone want to review/critique my proofs here? 1st semester undergrad algebra

[thorny slate]

asking for someone to read skewed, small, hard to read handwriting and 3 pages of it is a lot

[slim lily]

i swear they were vertical when i took them

[solar prairie]

Yeah, can't read that

[slim lily]

😂

no biggie

[carmine delta]

hello, say i have some ring like idk Z_25, and i want every ideal — is there some hint towards what subrings i should try or do i just brute force everything?

also plz ping in response because there are so many channels here it is easy to get lost clicking around them !

[thorny slate]

every ideal of a ring is a hard question

but for a cyclic one it's easy

the subgroups of Z25 are just Z5

so the ring with that additive group has to be your only ideal

[carmine delta]

thanks, i think that makes sense!

[thorny slate]

probably

dunno

which book are you using

[carmine delta]

i am using judson

it's pretty good

im just not smart

[thorny slate]

if it's hard to understand you can read the same thing in another book

pinter is friendly

[carmine delta]

😀 i will try to find a pdf of this book

also to clarify, judson is not terse at all for the most part, just when it comes to rings

@thorny slate in your example from earlier, can you say that 5Z is an ideal automatically for some reason, or do you still need to do the ideal test?

also this ideal is maximal and prime, right?

ok yes it is

[thorny slate]

you still need to test

yes it's maximal

a lot of subgroups will not be ideals in general

[dense hull]

Does anyone have a good image of types of rings?

One that's like

field \subset euclidean domain \subset PID \subset ... \subset ring?

if that makes sense

[thorny slate]

there's a bunch of types of rings

it's not a single line

the usual you learn in a first course are
field > euclidean domain > pid > ufd > domain > ring

but there's things like dedekind domains and discrete valuation rings

and local rings

and noetherian / artinian rings

[carmine delta]

"a lot of subgroups will not be ideals in general"

oh yeah im aware of that, i was just wondering if there was some theorem like "every subgroup of Z_n /or/ a cyclic group is an ideal"

[thorny slate]

every subgroup of Z_n is an ideal of Z_n yeah

[thorny slate]

yeah

you can

that's correct

that's another description of the group generated by a set S

the set of finite products of its elements and its inverses

[analog patio]

The wiki page on direct sum says the Z_6 = {1,3,5} + {0,2,4} is a direct sum. But i thought the intersection has to be {0} to be a direct sum. what is going on?!

and the sets arent closed under addition...

i dont think thats right

[thorny slate]

yeah

some dumbass wrote that lol

[analog patio]

sweet, im not going crazy

[thorny slate]

the sums aren't even unique, you have 3 + 0 = 1 + 2

should be {0,2,4} + {0,3}

[analog patio]

oooh, didnt think of that. I agree

[vestal needle]

so i have a question about part d. This subset is not a subgroup when you write out all the cyclic subgroups generated by each element but when you do the one-step subgroup test it is a subgroup?

like 3 generates all the elements of Z8

so it cant be a subgroup

[thorny slate]

6+6 = 4

fails

[thorny slate]

6 - 6 = 0 isnt

[vestal needle]

i used 6 as a and 3 as b

[thorny slate]

you need to check all a and b

[vestal needle]

ok just making sure

thanks

[carmine delta]

uh this is ugly

give me a sec

very quick question:

this proof of the test $ ab^{-1} \in H$ where H is a subset of G (for a,b in H) implies that H is a subgroup:

First, assume $H \subset G$ is a subgroup, with $a,b \in H$. $b^{-1} \in H$ because H contains all inverses, and $ab^{-1} \in H$ by closure.

all good here

now, going the other way, assume $H \subset G$ which is non-empty, and $ab^{-1} \in H$; we want to show that this implies H is a subgroup, so we want closure, inverses, & identity

(1) identity: my book proves this(?) by claiming that if $g \in H, gg^{-1} \in H$, and if $g \in H, g^{-1}e = g^{-1}$ is also in H. idk how this proves anything, because I'm not sure why we have assurance $g$ is in $H$, or how $g$ being in $H$ implies $g^{-1}e$ is in $H$

(2) inverses --- somehow wrapped up in the thing above

(3) closure: take $h_1, h_2 \in H: h_1(h_2^{-1})^{-1} \in H \implies h_1 h_2 \in H$ I am also lost here because it looks like they are just assuming it is closed in the first step? any help is appreciated!!

ok there is a compile error but the output looks fine to me i think

[cloud walrus]

89_87:
Compile Error! Click the reaction for details. (You may edit your message)

[thorny slate]

if H is nonempty it has some element g

take a = b = g in the test

so that gg-1 = e is in H

now you can pick a = e, b = g and get e g-1 = g-1 in H

now you have inverses for for any h,k in the set pick a = h, b = k-1 and then ab-1 = hk is in H

[carmine delta]

hmm that is very clever, i will try to wrap my head around it! i appreciate the help.

i see what is going on though

i didn't even realize the non-emptyness was that important

[thorny slate]

yeah it's a bit of a slick, hard to see trick

but it's a very compact way of writing subgroup conditions

[whole basalt]

So I'm creating some GRE-subject-test-style quiz questions. Anyone up for telling me if this makes sense / could use work?

---

Let $G$ be a group with $24$ elements, and let $x$ be an element of $G$ such that $x^2=x^{20}$. Which of the following could be the order of $x^{15}$?

(A) $2$ (B) $3$ (C) $4$ (D) $6$ (E) $12$

Answer is supposed to be $2$. The reasoning is that if $x^2=x^{20}$, then $x^{18}=e$, and since the gcd of $24$ and $18$ is $6$, the order of $x$ is could be $1$, $2$, $3$, or $6$. Then, since the gcd of $15$ and $24$ is $3$, the order of $x^{15}$ should be the same as the order of $x^3$, which could only be $1$ or $2$. Does that reasoning make sense?

[cloud walrus]

DMAshura:

[thorny slate]

yes

[whole basalt]

I also think my distractors need work.

[thorny slate]

it has enough distractors

I think it's fine as is

[whole basalt]

Cool.

Thanks. Wanted to make sure it seemed reasonable and I wasn't saying anything stupid. 😃

I assume you took the GRE subject test at some point?

[thorny slate]

ye

last year

[whole basalt]

How did it go? If I may ask

[bleak abyss]

taking the GRE subject test

lmao look at this nerd

[thorny slate]

went well

ezpz

the princeton review book is perfect

[bleak abyss]

Is it? I feel it's good for content but the problems are a lot easier than those on either of the tests I took

[thorny slate]

yeah the problems are on the easy side

you gotta grind the content

[bleak abyss]

To be fair I never took multivariable calculus so at least now I know FTC for integrals over curves or smth

[thorny slate]

lul

[carmine delta]

wow this is a very useful test

makes proving homomorphisms carrying over subgroups very short.

[whole basalt]

@thorny slate I agree on Princeton Review being the best unofficial prep out there

It's definitely got the best overall review, even if it is short on a few minor topics

[bleak abyss]

"short on a few... topics"

tfw still don't know how to integrate on a surface

[whole basalt]

REA is complete garbage

Yeah, that's one of them

Also they have like nothing on ideals

[bleak abyss]

Oh I remember a question on the most recent GRE about them

I guess I'm not technically supposed to say?

[whole basalt]

They make people stay super tight lipped about them, yeah

Because they keep reusing questions

[bleak abyss]

Kind of annoying tbh

[whole basalt]

You can talk super generally though, like what some questions might be about 😛

You know the drill.

[bleak abyss]

It was a question about Artinian rings

(They defined Artinian rings)

[whole basalt]

Jesus

When was this?

[bleak abyss]

Most recent test, in October

[whole basalt]

Wow. That's definitely a new question.

[bleak abyss]

I kinda went in blind because I had done well enough on the first test so this wasn't great on the whole

[vestal needle]

can i get help with part a

[thorny slate]

just check the conditions

remember g is in C(a) iff gag-1 = a

[vestal needle]

and g is another element in G?

[thorny slate]

yeah

[vestal needle]

so is the inverse of g always in the centralizer too?

[thorny slate]

check why

[vestal needle]

i mean it cant be a subgroup without there being an inverse of g right?

[sullen flint]

must fulfill the three axioms

associativity, existence of identity and inverse for each element

am I missing something?

[thorny slate]

right so you have to prove the inverse is there

check why this is true

[carmine delta]

centralizer is subgroup --> subgroup has inverses for all elements ---> g in C(a) ---> g^-1 in C(a)

😉

im pretty sure though if you just think about what a centralizer actually is it follows nicely-ish

[whole basalt]

Is there an algebra book known as "A-M"? Weird question

[thorny slate]

atiyah macdonald

commutative algebra

[whole basalt]

Gotcha

Thanks 😃

[vestal needle]

yeah but im trying to establish that centralizer is a subgroup

this is killing me lol

[carmine delta]

a x = x a --> (a^-1) a x (a^-1 ) = (a^-1) x a (a^-1 )

cancels nicely

x (a^-1 ) = (a^-1) x

so a^-1 in centralizer

i think

i actually think i had this exact question on a test

or something really similar

[vestal needle]

oh i see it now thank you

[vestal needle]

sorry for the bombardment of questions trying to study for final tomorrow, can someone explain to me what part a means in this question

[carmine delta]

uh <a> means subgroup generated by a

idk what V is

[vestal needle]

V is vertical reflection in D4

[carmine delta]

what's confusing you?

[vestal needle]

its two elements separated by a comma

do i have to just find the subgroup generated by both individually or do i find the subgroup generated by R90V

or is it something else

[thorny slate]

both together

the group generated by the set {R90, V}

[vestal needle]

so the latter of what i said

[thorny slate]

no

it's the smalles group containing both R90 and V

[vestal needle]

so that would just be all the rotations plus V then?

[carmine delta]

{all powers of R_90} U {all powers of V} U {all combinations of elements in the previous two sets}?

[thorny slate]

^

[vestal needle]

oh i see

i didnt think about that last one

ok so its all the rotations plus V plus D since R90V=D and VR90= D'

basing that off this table btw

[whole basalt]

Inspired by last night's discussion, I'm trying to make sense of noetherian and artinian rings

So lemme see if I can phrase this in my own words and have it make sense

A ring $R$ is noetherian if any sequence of ideals $I_1\subseteq I_2\subseteq\cdots$ reaches some point where that sequence has to become constant, as in there exists some $N$ after which $I_n = I_N$ for all $n>N$. And then for artinian it's the other way around, as in any sequence $I_1\supseteq I_2\supseteq\cdots$ reaches some point where it becomes constant.

Does that seem correct?

[cloud walrus]

DMAshura:

[tulip osprey]

Can someone recommend a good abstract algebra book or online resource? I just finished a Linear Algebra class and want to teach myself some abstract algebra. So maybe something fairly introductory

[carmine delta]

Judson is online and free (but really everything is free if you know where to look 😉 )

pinter is highly recommended but i don't like it

[uncut girder]

Saracino was my first book

[chilly ocean]

J.J.Rotman is OK, alas not free...

[tulip osprey]

Yeah. It doesn't HAVE to be free. Just something well written that's good for someone learning by themselves and not in a classroom.

[chilly ocean]

http://www-users.math.umn.edu/~garrett/m/algebra/ is ok

[tulip osprey]

I don't mind paying. I just want something that explains concepts well and has lots of examples and worked problems .

[chilly ocean]

Dummit&Foote is still popular

[tulip osprey]

Thanks guys.

Pretty exciting. What are some key aspects from linear algebra or other maths that are really essential to be able to start diving into abstract algebra? I felt pretty comfortable with proofs, at least understanding them if not writing them. Once we got to inner products I was a bit uncomfortable, such as implications of orthogonal sets to linear independence or really why we care about them at all. But I don't know how important that stuff is, even though I'll probably come back to it before moving on to AA.

[carmine delta]

knowing how to do proofs is ridiculously important in AA, but knowing those specific things is less so

[uncut girder]

AA is a beginners course

Not much prerequisites

[tulip osprey]

That's good.

[uncut girder]

Just jump into it with a friendly book

[vestal needle]

can someone explain why S3 is not cyclic

[stone fulcrum]

Because there's no element in S3, that when multiplied by itself over and over, will span all of S3

The only cyclic groups are Zn, and Z

[vestal needle]

right thanks!

[stone fulcrum]

Np!

They're only talking left cosets of H

One such coset would be ax, where x is an element of H
Another such coset would be by, where y is an element of H

[chilly ocean]

what does this mean

[stone fulcrum]

Wait, no. Let me resay all that

A coset of H is gH, where H is all elements from H. And g is a choice from the group.

Let's say you create a coset using an element "a". This coset is aH.

Let's say you create a coset from an element b. This coset is bH.

Let's say aH and bH have an element in common. Then there's at least one case where ax = by, where x, y ∈ H

Because x and y need not be equal

Just both from H

[shut anvil]

Hello, could somebody help with my problem

[stone fulcrum]

Take an element "a". Multiply it by every element of H. This will be a bunch of elements, call it a coset.

Take an element "b". Multiply it by every element of H. This will be a bunch of elements, call it another coset.

If these two cosets have an element in common, then there must have been some ax, where x is in H, and some by, where y is in H, such that ax = by

We don't necessarily know that x = y

[shut anvil]

Consider the demand function q = −3p + 150. Find the point elasticity when (a) p = 10.
(b) p = 40.
(c) Why is demand elastic when p is high and inelastic when p is low, in this example?

[hot bay]

Wrong channel fam @shut anvil

[whole basalt]

Did anyone see my noetherian/artinian thing earlier

[hot bay]

Wait Yes I will try to think of an example to help u here

Assuming I understand your problem

So i can’t write this down in full because I’m on mobile

But say you want cosets of {0,3,6} in Z_9

You probably gonna write them all out

Find some have the same elements

[errant drum]

@chilly ocean

Not necessarily. For example 4 * 4 and 3 * 3 = 2 in Z7

Integers modulo 7

I've never see it written that way but sure

[raw moth]

Z_7 and 7Z aren't the same thing

[hot bay]

^^^

[stone fulcrum]

The subgroup is 7Z. The group the cosets generate is Z7

[errant drum]

Oh so I was right

@chilly ocean

Okay in Z7

4 * 3 = 1 * 5

[stone fulcrum]

Mind you, we are talking about cosets, not the group it forms. These two things are very similar

[errant drum]

Btw when you say Z7 does it refer to the additive group or the multiplicative?

I think you'd need to account for the multiplicative one not having zero right?

So is it Z7/{0}

But then it looks like a quotient group

[hot bay]

Consider Z_12; what are the cosets of H = {0,4,8} ?

1 + H = {1, 5,9}
2 + H = {.....
.
.
.
9 + H = {9, 1, 5} = 1+H

So aH = bH does not imply a = b

[errant drum]

^

Took the words right out of my mouth

[hot bay]

❤️

[errant drum]

💙

So take Z7 and take H = Z7

Then 4 * 4 = 3 * 3

You can make the subgroup the whole group

If you want a more explicit example in the symmetric group S3 you can divide it in to even and odd permutations
The the coset (12)odd = 132 (even)

And (12)(13) = (132)(132)

[vestal needle]

hey quick question, the inverse of (56)(13) would be itself because its just 2-cycles right?

[errant drum]

You have to exchange the order so it's (13)(56)

[vestal needle]

ok thanks yeah i was wondering if i had to chance the order or not

[stone fulcrum]

@chilly ocean
Anyway, cosets partition the group. They either share no elements in common, or are the same coset. That's the important thing to take away

[earnest valley]

It doesn't matter, they are disjoint.

[errant drum]

Partitions as in P(n)

Ohhh

Nvm

[whole basalt]

So if anyone still feels like scrolling back up to my question, all I'm looking for is yes or no ;P

[errant drum]

I don't think I've done enough group theory @whole basalt

[whole basalt]

On whether my definition of noetherian/artinian rings makes sense

[errant drum]

The equivalence classes partition the set

[whole basalt]

@yes: "partition" means a set has been split into nonoverlapping parts. "Equivalence classes" are subsets where everything in them is equivalents according to some defined equivalence relation..

[errant drum]

^ that's probably better than anything I could write up

[whole basalt]

When you define an equivalence relation on a set, by doing so you split up ("partition") that set into equivalence classes.

If your equivalence relation on the set of people is "has the same birthday as", then each equivalence class contains all the people born on some particular day.

My equivalence class contains all the people born on March 26.

Since no person can be in two equivalence classes (since that would mean you have two birthdays!), this equivalence relation has partitioned the set of people.

[raw moth]

you can use the transitivity of an equivalence relation to show that an element of a set can't be in two different equivalence classes

[vestal needle]

Can someone explain this to me before my final

[whole basalt]

@vestal needle I might start off with playing with what other elements get sent to.

What is φ(14), for example?

[vestal needle]

Is it 7*14 mod 15

[stone fulcrum]

Remember, φ is a homomorphism, so it splits over addition.

φ(14) = φ(7 + 7) = φ(7) + φ(7) = 12

[vestal needle]

Oh ok

[whole basalt]

By playing with those some more you can get a feel for what φ is doing

[stone fulcrum]

Always true are also:
φ(1) = 1
φ(x¯¹) = φ(x)¯¹

[earnest valley]

Additive or multiplicative group ?

[whole basalt]

I'm not so sure I agree with φ(1) = 1 in this context, @stone fulcrum .

φ(identity) = identity, but in this case that means φ(0) = 0.

[stone fulcrum]

Oh derp. True

[whole basalt]

(This is why I don't like calling the identity "1" but "e".)

[vestal needle]

So

φ(x) = φ(x+ 7) = φ(x)+ φ(7)

[stone fulcrum]

I can't see where you got φ(x) = φ(x + 7)

[vestal needle]

I just replaced the 7 in your equation with x

[whole basalt]

I'd start looking at what φ does to more multiples of 7, and look for a pattern

[errant drum]

^

[earnest valley]

Try finding phi of 1 with 0=50=49+1.

[whole basalt]

Like now that you know φ(7) and φ(14), you could figure out φ(21), etc

And in particular yeah, φ(49) will tell you a lot, because 49 = -1 in Z_50.

[errant drum]

Once you know phi(1) you can figure everything else out

[whole basalt]

And once you know φ(-1), you can very quickly figure out φ(1)

[vestal needle]

Ok so phi(21) would be 12+6 then

[whole basalt]

Yup, but mod 15

[vestal needle]

And that’s 18

Mod 15 is 3

[whole basalt]

Since the codomain is Z_15

Right!

Find what φ(49) is

[vestal needle]

And phi(49) is 7*(phi)7?

[whole basalt]

Should be, yes. Mod 15. What do you get?

[vestal needle]

Well 7*6 is 42 mod 15 is 12

[whole basalt]

Great. So now you have φ(49) = 12.

Do you understand why in Z_50, 49 is basically the same as -1?

[vestal needle]

Well 50 is basically 0 because of modular arithmetic

[whole basalt]

Yup

[vestal needle]

And so I guess 49 is one less and -1 is one less than 0 so makes sense

[whole basalt]

So, now you can use the fact @stone fulcrum mentioned: since φ is a homomorphism, it should preserve inverses.

So since φ(49) = 12, you should have φ(inverse of 49 in Z_50) = inverse of 12 in Z_15.

What does that give you?

[vestal needle]

3?

[whole basalt]

φ(?) = 3

[vestal needle]

Phi(1) = 3?

[whole basalt]

Yup!

That basically unlocks your entire φ

Because you can now easily find out φ of any element of Z_50.

Can you describe now what φ(x) is, in general?

@yes: Correct on equivalence class, not so much on a partition

[vestal needle]

Lol now I’m stuck

[whole basalt]

An example of a partition of the set {1,2,3,4,5,6,7,8,9,10} is
{ {1,5,6}, {4,8}, {2,3,9}, {10} }

A partition of a set is a way of splitting the set up into disjoint subsets

[hot bay]

I think he said that?

[whole basalt]

Correct

There are many ways to partition a set

But one way to do so is to define an equivalence relation

And that will automatically partition your set

@vestal needle Well, you've got φ(1) = 3, φ(2) = 6, φ(3) = 9, φ(4) = 12, φ(5) = 0. See a pattern?

[vestal needle]

Phi(x) is phi(x)^-1 mod 15?

[whole basalt]

No

If I just gave you f(1) = 3, f(2) = 6, f(3) = 9, f(4) = 12, what would you guess my function f does to a number?

[vestal needle]

Add 3

[whole basalt]

2+3 is not 6

[vestal needle]

Oh multiply by 3 mb

[whole basalt]

Right

So you'd have f(x) = 3x

Your φ does almost the same thing, except φ(5) = 0

How could you account for that?

[vestal needle]

It’s because it’s mod 15 right

[whole basalt]

Yup

So φ(x) = ...

[vestal needle]

3x mod 15

[whole basalt]

Bingo.

That's your answer to part (a).

[vestal needle]

Awesome

[whole basalt]

In order to get there, you had to play with φ(7) = 6, to deduce what φ does to the rest of Z_50.

(And now φ(7) = 6 makes sense --- φ(7) = 3(7) mod 15 = 21 mod 15 = 6.)

Do you understand how to do the rest?

[vestal needle]

No not really lol

[whole basalt]

Okay, well here are some definitions that'll help you.

The image of φ is everything in Z_15 that φ is able to map to.
The kernel of φ is everything in Z_50 that maps to the identity (0) in Z_15.

[vestal needle]

Oh thanks that helps a lot actually

[errant drum]

I'd say start with the kernel it's easier

[whole basalt]

@errant drum It's either-or. We've already done enough with φ that one can just see the image right off the bat.

I actually like the idea starting with the image for this problem, because it reveals the pattern for what the kernel will be.

[vestal needle]

So the image would include 0,3,6,9, and 12 in this case

[whole basalt]

Yup

😃

[vestal needle]

And the kernel would include 0,5,10,15,20,25,30,35,40,45

[whole basalt]

Excellent

Then for (d), can you figure out everything that maps to 3?

(Hint: Your kernel is a big clue.)

[vestal needle]

It’s like all the elements in the kernel plus 1 right

[whole basalt]

Yes

(That's called a coset of the kernel)

[vestal needle]

Wow thanks a bunch dude you’re a lifesaver

[whole basalt]

No problem 😄

[vestal needle]

Is it left or right coset btw

[whole basalt]

Both, because Z_50 is abelian

[vestal needle]

Ok ok thanks

[chilly ocean]

Cosets are either equal or disjoint

This is true

3Z + (3Z+1) + (3Z+2) + (3Z+3) = Z

btw

[lone hedge]

3Z+3 is just 3Z

that's why

[chilly ocean]

Tbh

Try to prove cosets are either disjoint or equal actually

[lone hedge]

sorry I gave the answer

beri big sorry

[chilly ocean]

Yeah

Exactly

Anyway

Disjoint union just indexes where the elements in the union came from.

So

{1,2,3} U+ {1,2} = {(1,0),(2,0),(3,0),(1,1),(2,1)}

It's a set index

Are you not doing disjoint union?

Hm

t!wiki disjoint union

I don't know what you're being taught

[fossil mango]

📖 | ** https://en.wikipedia.org/wiki/Disjoint_union **

[chilly ocean]

Much clearer eh?

[cloud walrus]

creamy shits:

[chilly ocean]

,$ \bigsqcup_{i\in\mathscr{I}} = \bigcup_{i\in\mathscr{I}}{(x,i):x\in A_i}

,$ \bigsqcup_{i\in\mathscr{I}} A_i= \bigcup_{i\in\mathscr{I}}{(x,i):x\in A_i}

:l

[cloud walrus]

Jichael Mackson:

[chilly ocean]

@chilly ocean It's not associative

So it depends which one you do first.

Kk

So evaluate first one on the left

[cloud walrus]

creamy shits:

[chilly ocean]

{(1,0),(2,0),(2,1),(3,1)} U+ {3,4}}

= {((1,0),0),((2,0),0),((2,1),0),((3,1),0),(3,1),(4,1)}

Gotta nest those elements boi

[lone hedge]

lol

[chilly ocean]

Hang on, did I goof here.

[stone fulcrum]

Because 3Z and 3Z + 3 are the same set

Wait, that may not matter

[chilly ocean]

The index tbh

I think you got it right first time though

A U+ B U+ C =/= (A U+ B) U+ C tho

Hm this is kind of a weird operation

So yeah, stick with your original answer

Principal difference is the ordered pair with a set index though

So {1} U+ {1} U+ {1} U+ {1} = {(1,0),(1,1),(1,2),(1,3)} =/= {1} = {1} U {1} U {1} U {1}

The union is literally disjoint

[stone fulcrum]

Yeah I have no idea what this question is asking

[errant drum]

Show how Lagrange's theorem follows and then reverse the arrows

[stone fulcrum]

@chilly ocean
You can use cosets to partition any group into disjoint sets, each with equal size.

Therefore, the group order must be some integer multiple of that coset order, which is the same as the subgroup order.

|H| × number of cosets = |G|

Number of cosets = |G|/|H|

|H| cleanly divides |G|

[bleak abyss]

Okay so, let's say $k$ is a field, $S\subset k[x_1,\ldots,x_n]$. Then define $V(S) = {(a_1,\ldots,a_n) \in k^n \mid f(a_1,\ldots,a_n) = 0\ \forall f\in S}$

[cloud walrus]

Daminark:

[whole basalt]

So ... S is a set of polynomials, and V(S) is the set of all points where all polynomials in S vanish?

[bleak abyss]

Yup

So, if you know that some set of polynomials vanish on a set of points, you know the ideal they generate will also vanish on that same set

[thorny slate]

it's the set of solutions to the equations given by those polynomials

[whole basalt]

And the ideal they generate is all polynomials that are divisible by at least one of those polynomials

I can see why a series I"m about to watch on all this is called "nonlinear algebra"

[bleak abyss]

So, Hilbert basis theorem guarantees for us that $k[x_1,\ldots,x_n]$ is Noetherian, so its ideals are finitely generated. So for any $S \subset k[x_1,\ldots,x_n]$, there exists finitely many polynomials $f_1,\ldots,f_N$ such that $V(S) = V(f_1,\ldots,f_N)$.

[cloud walrus]

Daminark:

[whole basalt]

Translation: You can find a finite basis set of polynomials for S?

(As long as you're only working in finitely many variables)

[bleak abyss]

Yup

[whole basalt]

Great.

(I apologize if that will get annoying, but I understand things a lot better when I do my best to translate things into ideas not using algebraic notation XD)

[bleak abyss]

Okay I am now back!

But yeah so, you have what are called radical ideals

Given an ideal I, an element is in rad(I) if some power of it is in I. The idea is that if you take the vanishing set of a square of a polynomial, that's just the vanishing set of that polynomial. Hilbert's Nullstellensatz says that if you take an ideal of polynomials, take the vanishing set, and then ask what are all the polynomials which vanish on that set, it'll be the radical of the original ideal, so it turns out you can always write V(S) = V(I) for a radical ideal I

Now, the actual theorem of substance I know about Noetherian rings is that radical ideals are finite unions of prime ideals, and it turns out prime ideals have vanishing sets which cannot be written as unions of finitely many proper subsets that are themselves vanishing sets (terminology: irreducible varieties)

So it turns out varieties are finite unions of irreducible varieties

[whole basalt]

What is a "radical ideal" in particular? I understand what a radical of an ideal is

[bleak abyss]

Oh whoops

Sorry, a radical ideal is an ideal which is equal to its radical

[whole basalt]

Ahh okay

So for example 2Z is a radical ideal of Z, whereas 4Z isn't because its radical is 2Z

[bleak abyss]

Yup

[whole basalt]

So the Nullstellensatz is basically saying that you can reduce a set of polynomials to a "more basic" set with the same properties of where they'll vanish

[bleak abyss]

In the case of polynomial rings, the radical of (x^2) is (x), say

[whole basalt]

Got it

So, radical ideals being finite unions of prime ideals ... in Z aren't the only prime ideals radical ideals themselves?

[bleak abyss]

In general, when you're just dealing with k[x], then any ideal is principal, so factorize the polynomial $p = (x-a_1)^{n_1} \ldots (x-a_k)^{n_k}$, then $(p) = ((x-a_1)\ldots (x-a_k))$

[cloud walrus]

Daminark:

[whole basalt]

So we'd need something with a bit more structure to see them as a union

[bleak abyss]

Oh sorry not union, I meant intersection

Okay that was an epic goof lmao

[whole basalt]

I was gonna say, I'm not used to unions of ideals being a thing I care about XD

So for Z, again, you're saying (21) would be a radical ideal?

Since it's the intersection of (3) and (7)

[bleak abyss]

Yup

[whole basalt]

Makes sense

So going back to polynomials

It's basically saying that a given variety can be written as a finite intersection of those irreducible varieties, which are its linear factors

[bleak abyss]

Not exactly, it's worth noting that the vanishing locus of an intersection is actually the union of vanishing loci

Basically, if we demand that a set of points vanishes on a larger set of polynomials, it's a requirement that's harder to meet, so fewer points satisfy it

So passing to vanishing sets is order reversing

[whole basalt]

So $V\big({(x-1)^2(x+2)^3,(x-4)(x-1)(x+2)^5}\big)=V\big({x-1,x+2}\big)$ for example

[cloud walrus]

DMAshura:

[bleak abyss]

Looks good!

[whole basalt]

And that would equal $V({x-1})\cup V({x+2})$

[cloud walrus]

DMAshura:

[bleak abyss]

Yeah. Assuming we're thinking about these as elements of $k[x]$, as opposed to $k[x,y]$ or something, that's ${1,2}$

[cloud walrus]

Daminark:

[whole basalt]

Yeah, I'm working in one variable for now because it's easier for me to understand 😛

So this whole thing is basically a reeeeeally rigorous way to describe the theory of equation-solving and zero-loci of polynomials

[bleak abyss]

Yup

Artinian rings come up in a context I understand less, more non-commutative algebra

You have a theorem called Artin-Wedderburn which states that artinian semisimple rings are actually direct sums of matrix rings over division rings. I know in the case of the group algebra k[G] (where k is a field whose characteristic doesn't divide |G|), Mashke's theorem tells you it's semisimple, and the isomorphism of Artin-Wedderburn is explictly given by a generalization of the Fourier transform to finite groups

But that's about as much as I can say with any form of confidence

[whole basalt]

Okay

I understand it a lot better than I did before so there's certaily that 😃

Thank you!

[bleak abyss]

No problem!

[chilly ocean]

Trying to show that for any pair of subgroups $H,K$ of $G$ that $Z(H)\times Z(K) = Z(H\times K)$, while I can show this follows from the definition of a center the question demands that I provide a double inclusion proof.

[cloud walrus]

Jichael Mackson:

[chilly ocean]

So a bit lost on how you do that without doing "hey look at the definition of a center" and slowing it down to a double inclusion.

[delicate chasm]

Double inclusion is the standard way I think?

I suppose you're doing an if and only if proof?

The double inclusion method says:
If (h,k) is in the centre of H\times K then (stuff) which implies h is in the centre of H and k is in the centre of K.
If h is in the centre of H and k is in the centre of K then (stuff) which implies (h,k) is in the centre of H \times K

[chilly ocean]

@delicate chasm Okay making the point that it should just be done elementwise helps. Thanks.

[delicate chasm]

👍🏽

[whole basalt]

Just a thought I had ... $\bR/\bZ$ very much resembles a finite cyclic group of rotations, but it doesn't have a single generator. Is there any word for the "almost cyclic" behavior it shows?

[cloud walrus]

DMAshura:

[thorny slate]

what do you mean?

it is very far from being cyclic

maybe you're talking about a topological property? irrational numbers have dense orbits

[whole basalt]

I suppose it is more of a topological property than an algebraic one

Though if you did Q/Z you would get that "finite wraparound" property

[thorny slate]

yeah every element has finite order

[whole basalt]

I'm just trying to find the right word ;P

[thorny slate]

the right word is "every element has finite order"

[whole basalt]

To compare it with how the finite cyclic groups are also all rotations

It's a "rotation-y sort of group"

Oh wait. There's a name for the group of rotations in 2D over some field

Is it SO (2,F) I think?

[thorny slate]

yeah

[whole basalt]

So what I'm thinking is that finite subgroups of SO (2,F) are cyclic

[thorny slate]

no way

there's other finite groups in there

[whole basalt]

Like what?

[thorny slate]

like uhhh

yeah ok

[whole basalt]

How could you have a finite group of rotations be anything but cyclic

[thorny slate]

you're right

[whole basalt]

Okay

[thorny slate]

you need 3d to have other groups

[whole basalt]

So what I was thinking about was really mostly about living inside a particular group

I can dig that

[sick acorn]

if G is isomorphic to either Z or Z/nZ, is it fair to say that G is cyclic?

[raw moth]

yes

[sick acorn]

hmm, okay--thanks!

[covert vector]

what about 3-categories

do people do special cases of that

also at what point do people stop really looking at the k-category for some fixed k

like maybe at 5-category there's obscure nifty examples

then after that they resort to inductively defining n-cat

[delicate chasm]

I have no idea

But there are infinity cats

and they are important

[calm token]

can someone help with linear algebra ?

i'm a bit confused

how do you do change of basis :v ?

[earnest valley]

More context.

[calm token]

i'll send u an example

its says here that B is a basis of [ E11,E12,E21,E22] and C is a basis of [A,B,C,D]

A B C D is all a 2x2 matrix

find P c<-b

[earnest valley]

You have that $X=PX^\prime$ where $X^\prime$ is the new coordinates.

[sonic current]

@calm token you need to work out a matrix P, which will be the change-of-basis matrix

[thorny slate]

what kind of group structure do you want

show 26

I assume the hint captures the idea though

uhh

the idea doesn't really work as stated

oh no it does

ok

so fix a point O in the parabola

and define A + B to be the point obtained by doing the same thing

you take the line AB

and you draw a parallel line through O

and the point that crosses the parabola is the point A+B

the point that isn't O

26 is kinda different

is this the usual group law on elliptic curves @bleak abyss

anyway you got what I said?

if C was the fixed point, yeah

but you probably want the fixed point to be O

yeah

I mean you don't need some C

you want a fixed point

and you already have O

in the other case there's a C because it's degree 3

so any line will intersect A, B and some C

yeah but like

the line should be parallel you know

[visual sable]

Draw a parallel line to AB that passes through O, it's not tangent (except if A and B have same y, but then yiu can just say in this case A+B = 0)

@chilly ocean

[thorny slate]

if it's tangent at O then the point is just O

but this does not look tangent

you need to like

zoom in

and you see the intersection

lol

try drawing a line with more slope

i mean make the line AB have more slope

put A lower

[visual sable]

Rip

Btw jacobian is there a reason to take a parallel line ?

[stone fulcrum]

This is the most obvious way to get an identity and inverse. Hopefully associativity happens too

[cloud walrus]

creamy shits:

[stone fulcrum]

What's (a, a²)?

Oh you're using vectors

[thorny slate]

@visual sable dunno

you can look at the equation maybe and see what it means

I don't know elliptic curves

so I can't give a good intuition of what's going on

[stone fulcrum]

You know the slope of the line is
(b² - a²) / (b - a) = a + b

So you want
(a + b)x = x²
x = a + b

Wow, it's literally just "a + b"

Which is obvi associative

a ⨁ b = a + b
This is going to be an easy isomorphism to set up

[cloud walrus]

creamy shits:

[stone fulcrum]

Oh yeah I guess we're working in ℝ² aren't we

[cloud walrus]

creamy shits:

creamy shits:

[stone fulcrum]

(A + B) + C = ((a + b) + c, ((a + b)² + c)²)

Which is weird, yeah

Or wait, no you're right

Hmm. There's something weird and un-groupy about that. I'll think about it

[cloud walrus]

creamy shits:

[stone fulcrum]

φ(a, a²) = a

Oh yeah that's better

[cloud walrus]

creamy shits:

creamy shits:

[marsh rose]

what exactly is an eigen vector

And how do I find it

Ive been watching videos and I dont get

[vale folio]

An eigenvector of a linear transformation is any vector that is only scaled under the transformation (assuming real eigenvalue)

normally a linear transformation that maps from a vector space to the same vector space can be thought to take an arbitrary vector and scale and rotate it, and an eigenvector is a vector that isn't rotated

[marsh rose]

so what happens when I have an imaginary eigenvalue

[errant drum]

If you've worked imaginary matrices/vectors use it if not then ignore it.

[marsh rose]

alright

Thanks

So when I find my eigen values how do I use them to solve for eigenvectors

[errant drum]

It depends on how much linear algebra you've done. The quickest way is to sub in lambda and find the null spacr

[marsh rose]

alright

[errant drum]

If not Av = Lv

[marsh rose]

Thats what it is

Null space

In my notes all wrote was solve for space

lol thanks

[errant drum]

Np

[marsh rose]

One last question

What is the quickest way to determine if a matrix is invertible

Right now what I do is try to use ERO to try to get it to the identity matrix

But is there an easier way for me to determine it

[chilly ocean]

the determinant of the matrix is not 0

also it must be a square matrix

n x n

[marsh rose]

alright

thanks

[sick acorn]

why don't you try first starting by writing f as a linear combination of the standard basis vectors and x_i

[sage sparrow]

@sick acorn

[sick acorn]

tbh the question confuses me a bit

[sage sparrow]

hmm

[unborn light]

Can anyone explain me this picture?

@slow egret

[raw moth]

not without more context

I think I can figure out the original question but I really cba to have to do that

[errant drum]

There have to be some restrictions on alpha and beta without know what they are it's sort of pointless explaining an inequality where we know nothing about the symbols in the inequality

[raw moth]

I'm guessing alpha and beta are the solutions to some quadratic relating to some lucas sequence from the question but I'm not certain

x^2 - 3x - 2 = 0 I guess

[errant drum]

Oh I see

[raw moth]

but ya some more details about the question would be nice

[agile linden]

holy fuck this looks complicated

[whole basalt]

Are there any rings of finite characteristic other than Z_p^n and its associated polynomial rings?

Oh, I guess matrices.

How mean is this?

[stone fulcrum]

II isn't true?

Colour me fooled

[whole basalt]

I'm ... actually still going back and forth on that one

I suppose matrices aren't an integral domain even over a finite field

If we're talking polynomials over a field, then yeah the units of the polynomial ring are the units of the field

But I'm trying to think of if there are any other possible examples of a finite-characteristic integral domain for which the units aren't just the units of some field

So II might be true after all

[stone fulcrum]

Let me look it up, I might be misremembering

[whole basalt]

I know that the multiplicative group of a finite field is certainly cyclic

But finite-characteristic?

[solar wyvern]

@whole basalt are you sure?

[stone fulcrum]

Oh right that's for fields

Yeah that's probs not true for integral domains

But I'm not sure

[solar wyvern]

what about F_{2^3}

does that work

[whole basalt]

The units of F_8 are isomorphic to Z_7

[solar wyvern]

um

F_9

[whole basalt]

Z_8 😛

In general $F_{p^k}^\times\cong \mathbb{Z}_{p^k-1}$

[cloud walrus]

DMAshura:

[bleak abyss]

Oh what are we discussing?

[solar wyvern]

not sure if I'm thonking

[whole basalt]

I'm trying to figure out if an integral domain with finite characteristic forces the units to form a cyclic group

Not necessarily a finite integral domain, because that's a finite field and it's definitely true

[bleak abyss]

This is probably false for the algebraic closure of F_p(t)

[whole basalt]

Oh wow. That's a hell of a thought.

[bleak abyss]

Or like, take any uncountable field of finite characteristic

Because the group of units is uncountable

But cyclic groups are finite or countable

[whole basalt]

I don't think I've ever run into those

But just a thought ... how about rational functions over F_p?

Those do form a field, but an infinite one

And there's no way that the units could be cyclic

That seems like a simple counterexample if I'm right

[bleak abyss]

Actually I wonder if there's any infinite field work cyclic group of units. I'm gonna say probably not

If such a field exists it has to be char 2 since otherwise -1 fucks you up

[solar wyvern]

@bleak abyss isn't Aut(C_{2^{k+2}}) noncyclic?

[whole basalt]

Explain, @bleak abyss ?

What about -1 fucks you up?

[bleak abyss]

It's order 2 but the group of units has to be Z, which has no element of order 2

[whole basalt]

Ohhhhh I gotcha

That makes sense

. . . so, $\bF_3(x)$ is a counterexample to the claim

[cloud walrus]

DMAshura:

[whole basalt]

Does that seem right?

[bleak abyss]

Yup

[whole basalt]

Beautiful.

SO I WAS RIGHT YAY

This was the context. I was asking how mean this is. 😛

I think the answer is now shown to be "kinda mean, maybe".

[solar wyvern]

@bleak abyss why are the units cyclic again for finite field? feeling really dumb

[bleak abyss]

Hmm, I is obviously false and thus III need not be checked

[whole basalt]

I isn't "obviously" false if a person misremembers a key fact about finite integral domains

[bleak abyss]

Fair but I dunno how hard that's emphasized, while polynomial rings just kill it

[whole basalt]

Difference between "finite" and "finite characteristic"

So yeah, if you know your examples of rings then you can come up with counterexamples

[bleak abyss]

@solar wyvern uh so there's the first proof I saw which is really really overpowered, you want that?

[solar wyvern]

sure

[bleak abyss]

Okay so, let's say A is a finite abelian group such that there are fewer than n elements whose order is n

Then A is abelian

Proof: structure theory of modules

[whole basalt]

Also yeah, good point, if you realize that I is false you don't actually need to check III. But that's fine with me, heh. If a person realizes that, then they're tackling the test right 😛

[bleak abyss]

Good point

@solar wyvern I meant that A is cyclic

[solar wyvern]

ah

[bleak abyss]

And maybe it's the number of elements whose order divides n

[solar wyvern]

think there's an easier proof

[bleak abyss]

There's one which uses less machinery and doesn't start off assuming you're abelian

But I find this easier to remember because it's easy once you know a fact that's important enough everyone should know

[solar wyvern]

tru

[bleak abyss]

But yeah so, once you have this, you say okay, a finite subgroup of the multiplicative group of a field is gonna be an abelian group where the n-torsion is solutions to x^n - 1

There are fewer than n of those

So you satisfy the above and are done

[solar wyvern]

oh wait Z_9 isn't a field

[bleak abyss]

Yeah it isn't

[solar wyvern]

me big dumdum

[bleak abyss]

It happens to everyone

Anyway I'm heading to sleep so see you

[solar wyvern]

might, thanks dami

[whole basalt]

@solar wyvern There is a field with 9 elements. But it isn't Z_9.

Instead it looks like Z_3[α], where α is the root of an irreducible quadratic over Z_3.

Like x² + x + 2.

[solar wyvern]

quadratic I think

[whole basalt]

. . . whoops. XD

Yeah.

If it were cubic you'd have F_27.

I make that mistake too often.

3^degree, not 3*degree.

[solar wyvern]

okay well g night

[whole basalt]

G'night indeed

[cloud walrus]

creamy shits:

creamy shits:

creamy shits:

[bleak abyss]

Yeah you're right

[stone fulcrum]

@chilly ocean
I has idea

Pick any set of elements from [0, 1). Is there a single element that generates them all?

Hrm. On second thought maybe not

[lyric falcon]

are you still trying to prove all subgroups of that are cyclic?

[stone fulcrum]

Yeah, that sounds about right.
Let H be a subgroup of G
(insert property of ℝ/ℤ here)
Therefore H has a generator and is cyclic

But we also need a group property since there's clearly not always a generator for some random set

The inverse of x is 1 - x

Yes, since x + (1 - x) = 0

[oblique river]

sorry im just hopping in; is the question whether or not all subgroups of R/Z are cyclic?

[stone fulcrum]

Yus that

[oblique river]

what about Q/Z

[stone fulcrum]

The reals that repeat on every integer

So 0.3 ≡ 1.3
because 1.3 - 0.3 ∈ ℤ

You can imagine it as the points on a circle

@oblique river
Note this is the group exactly:

[oblique river]

yeah I know what R/Z is

maybe I/m still a bit confused about the question ^^;

[stone fulcrum]

Okay, so it is R/Z and I'm not drinking

[oblique river]

by Q/Z I mean the subgroup consisting of those rational x

[stone fulcrum]

The question wanted a proof that there was no non-cyclic subgroups, but Q/Z is clearly one

I'm wondering if I did something wrong, lol. Am I seeing this right @chilly ocean

[oblique river]

what is true is that there are no noncyclic finite subgroups

or: "all finite subgroups are cyclic"

[stone fulcrum]

Derp, yes, we seemed to have dropped "finite" along the way.

So yes, we want to prove all finite subgroups can be generated with a single element

[oblique river]

aha! in that case, thinking about Q/Z is still useful