groups-rings-fields

[shy bluff]

Does this have to do with common divisors?

[oblique river]

not really

[shy bluff]

Well (a) is ra, ar for all r in R, (b) is rb, br for all b in R

[oblique river]

yes

our rings are commutative

so you don't have to write ra and ar

[shy bluff]

Then that means that ra = sb for some r and s?

[oblique river]

yes but that's not strong enough

you can say what r is

(a) is contained in (b)

which means a is an element of (b)

which means...

[shy bluff]

a is a multiple or a divisor of b is the only thing that I can think of

[oblique river]

a being a multiple of b and a being a divisor of b are like

opposites

haha

[shy bluff]

I can't keep them straight

[oblique river]

taht's fine, take a second to figure out which one it is

[shy bluff]

Divisors and multiples have lost meaning to me

I think that it means that a is a divisor o f b

Because if a is an element of b then that means that a = rb for... oh no it's a multiple of b

[oblique river]

why don't you try writing it out symbolically

ok great

yes

so if (a) is contained in (b) then a is a multiple of b

what does this tell you about the norm of a and b

[shy bluff]

norm(a) > norm(b)?

[oblique river]

(this is part a of the problem)

[shy bluff]

Like we have that N(a) = N(rb) = N(r)N(b)

So yea norm(a) > norm(b) I think

[oblique river]

yeah but also N(b) divides N(a)

which is part a of the problem

[shy bluff]

oh is that what you were looking for

ok

I see

[oblique river]

I mean I was asking you to state part a :P

but it's fine, all we actually need is N(a) > N(b)

[shy bluff]

Yay

[oblique river]

okay so now if we have a whole chain of (a1) subset (a2) subset (a3) subset ...

then what do we say

[shy bluff]

Uhhh a_i is a divisor of a_j if i > j... maybe directions are mixed up, one sec

Ok so if j > i then a_j divides a_i, and N(a_j) | N(a_i)?

[oblique river]

yes

so what does this say about the sequence N(a1), N(a2), N(a3), etc

[shy bluff]

Each one is divided by the next

or alternatively that we can write N(a_1) as being a product consisting of all the other N(a_i)s?

[oblique river]

no

[shy bluff]

oh

[oblique river]

I'm talking about the sequence

I'm not interested in any particular element

[shy bluff]

Each one is bigger...?

Err sorry

Each successive N(a_i) is smaller?

Like N(a_1) > N(a_2) > ... so on ?

[oblique river]

$\geq$ not $>$

[cloud walrus]

Buncho Bananas:

[shy bluff]

oh sorry

[oblique river]

okay great now there's another thing you know about these norms

they're not just random integers

they're all positive

[shy bluff]

Oh

So the n eventually we'll reach 1?

[oblique river]

so what do we know about this sequence of decreasing postiive integers?

no

[shy bluff]

Well it's decreasing and positive so it's got to stop at some point?

[oblique river]

$2 \geq 2 \geq 2 \geq 2 \geq 2 \geq \cdots$ never reaches 1

[cloud walrus]

Buncho Bananas:

[oblique river]

yes it has to stop at some point

[shy bluff]

Oh so this is what stabilize means

[oblique river]

yes

so now we just need to explain why once the sequence of norms stabilizwes

the sequence of ideals must also stabilize

basically: if a divides b and if N(a) = N(b) then in fact (a) = (b)

because a and b differ by a unit

[shy bluff]

say I never did understand the "Differ by a unit" thing

[oblique river]

for this you need to knwo that something has norm 1 if and only if it's a unit

a = ub

for some unit b

differ in the multiplicative sense

[shy bluff]

hrm

ok

Ok but how can we show that the sequence of norms stabilizes?

[oblique river]

you already did that

Well it's decreasing and positive so it's got to stop at some point?
@shy bluff

[shy bluff]

Oh

Wait so you'd actually write this like "If we write a chain of ideals we see that each one divides the previous one, and so by a) the norms of each ideal divides the norm of the previous ideal, and we have that these norms are strictly positive and decreasing in the chain, so we must stabilize after some finite number of norms/ideals/steps"?

[oblique river]

the norms of the generators

not the norms of the ideals

you know that the nroms of the generators stabilizes

but now you need to prove that the sequence of ideals stabilizes

[shy bluff]

Oh sorry

But lik ethe general idea is there right

Wait wdym

If the norms of the generators stabilize does the sequence of ideals not stabilize

[oblique river]

what?

[shy bluff]

... nevermind

[oblique river]

no you're trying to prove the opposite lol

[shy bluff]

Ok keep going

[oblique river]

if the norms of the generators stabilize

then the sequence of ideals stabilizes

that's the goal of the problem

prove that the sequence of ideals stabilizes

[shy bluff]

Yes

Wait like yea the fact that N(a_{i + 1}) | N(a_i) for each one shows that eventually the norms of the generators stabilize right?

Then I need to show that if the norms stabilize the ideals stabilize

[oblique river]

yes

[shy bluff]

How do you do that?

[oblique river]

why don't you think about it for a bit

I feel like we've made some good progress together

[shy bluff]

Is this what Zorn's lemma is for?

[oblique river]

no

[shy bluff]

Ah

(I am not randomly guessing, I am acutally going thorugh my notes/lecture slides/etc and trying to see how to do this)

[oblique river]

you can just do it directly

and you need this fact I mentioned that something has norm 1 if and only if it's a unit

[shy bluff]

I dont' see what that has to do with anything/can't it continue forever as like 2 like you said and not ever reach 1?

[oblique river]

the a_i aren't going to be the things with norm 1

i'd rather not give you too much more -- I feel like I've helped you out on this problem a lot and I'd rather you spend some more time thinking on yuour own

before I give you any more hints or feedback

[shy bluff]

that's fair

[oblique river]

other people can jump in if they want but i'm gonna take a break from this for now

good luck!

[shy bluff]

thank you

Ok so let's say that the norm stabilizes at some a_i = a_{i + 1}. Then we have to have a_{i + 1} | a_{i}, which means that there exists some b such that a_i = b * a_{i + 1}, but because their norms are the same, we have to have that N(b) = 1? So then they're associates, and because they're associates the ideals that they generate are the same?

[woven delta]

You should try to show that N(b) is 1 if and only if it's a unit

(idk if you have that fact yet)

[shy bluff]

Hrm

Yea I don't think that I have that fact

[woven delta]

But yeah if $a_i = b*a_{i+1}$ where b is a unit then $a_{i+1} = b^{-1}a_i$ so they are the same ideal

Try proving that fact then

[shy bluff]

oke

[cloud walrus]

Liquid:

[woven delta]

Try characterizing the units in Z[alpha]

[shy bluff]

wdym by characterizing

[woven delta]

Like let's say $a+b\alpha$ is an element of $\mathbb Z[\alpha]$. Then when would there exist $c+d\alpha$ with $(a+b\alpha)(c+d\alpha) =1$

[cloud walrus]

Liquid:

[woven delta]

Lol I switched to Q in my head because I've been doing a lot of galois theory lately

Anyway this is just an algebra problem

Ie like high school algebra

You just multiply it out and match the coefficients with 1

Okay so this is actually a hard approach, let's try an easier approach

[shy bluff]

Yea all I got is like ad = -bc and that ac + bd\alpha^2 = 1 lol

[woven delta]

Let's apply the norm to $(a+b\alpha)(c+d\alpha)$

[cloud walrus]

Liquid:

[shy bluff]

sure

That's just the above part

[woven delta]

Since it is 1 we must get $N((a+b\alpha)(c+d\alpha))=1

[shy bluff]

Yea

[woven delta]

So what does that tell us about a+b\alpha

Oh sorry if I'm repeating a part you already did lol

[shy bluff]

Oh no like I mean that you just get N(a + b\alpha) N(c + d\alpha)

Then we get that they have to be multiplicative inverses of each other?

[woven delta]

So we assumed that to start

And we are showing that they both have norm 1

[shy bluff]

what do you mean?

[woven delta]

Sorry I am proving the direction "is a unit implies norm 1"

[shy bluff]

so we're proving that a unit has norm 1

[woven delta]

Yeah

Do you see why that is?

[shy bluff]

Ok, so if u is a unit, then there exists v such that uv = 1, then we'd have that N(uv) = N(u)N(v) = N(1) = 1

Right?

[woven delta]

Yeah

And if we assume that alpha is sqrt(d) for d negative

[shy bluff]

And as N(u) and N(v) are naturla numbers including 0, that's only true if they're both equal to 1

[woven delta]

Then our norm is nonnegative

(otherwise we can have our norm mapping a unit to -1)

[shy bluff]

Yea

So then N(uv) = N(u)N(v) = N(1) = 1 must have that N(u) = N(v) = 1, as N(x) for all x is defined as being >= 0 right?

[woven delta]

Yeah N(x) is always >= 0 for Z(sqrt(-d))

[shy bluff]

Yea

[woven delta]

Okay so let's do the other direction now

[shy bluff]

"If N(x) = 1, then x must be a unit" yea?

[woven delta]

Yeah

So a nice way for this to work would be to be able to use the norm to find an inverse

Let's think about what the norm is

[shy bluff]

We'd have 1 = a^2 - b^2 \alpha^2

[woven delta]

Oh lol I realized why this is true

[shy bluff]

so 1 = (a + b\alpha)(1 - b\alpha)

?

[woven delta]

Yeah

That's it

Oh sorry

[shy bluff]

?

[woven delta]

That's an a not a 1

[shy bluff]

Pardon?

[woven delta]

so 1 = (a + b\alpha)(1 - b\alpha)
@shy bluff

Replace the (1-b\alpha) with (a-b\alpha)

[shy bluff]

Oh yea sorry

That's supposed to be an a lol

I wrote it dwon as an a on my paper

[woven delta]

That's good

[shy bluff]

I'm not too sure about the next step but I'm going to say that I can go and assume that inverse of N exists

[woven delta]

What do you mean?

[shy bluff]

Uhhh actually I'm not sure

So I have 1 = (a + b \alpha)(a - b \alpha)

[woven delta]

Yes

So N(a+b\alpha) = 1 means that the multiplicative inverse of a+b\alpha exists

Not that the inverse of N exists

[shy bluff]

Wait what

Oh ok

Yea that makes more senes

[woven delta]

Okay so now we have our sequence of principle ideal (a_i)

[shy bluff]

Wait do you mean that the fact that N((a + b\alpha)(a - b\alpha)) = N(1) = 1 means that (a - b\alpha) is a multiplicatiev inverse of (a + b\alpha)?

[woven delta]

Yes

[shy bluff]

Ok good I'm following!

[woven delta]

Okay so now we have our sequence of principle ideal (a_i)
And we showed that the corresponding sequence of norms N(a_i) is eventually constant

[shy bluff]

yes

[woven delta]

What does that tell us about the relationship between $a_i$ and $a_{i+1}$

(you did this before with buncho)

[cloud walrus]

Liquid:

[shy bluff]

They're associates

becaues they differ by a unit

[woven delta]

Okay

Let's spell out the argument though

So we have some b so that $a_i = b a_{i+1}$

[cloud walrus]

Liquid:

[shy bluff]

yes

[woven delta]

(here i is large enough so the norms have already stabilized)

[shy bluff]

Yes

ok good that's what I've written down so far

😅

[woven delta]

So what does that tell us re the norms

[shy bluff]

We have that $N(a_i) = N(a_{i + 1} \cdot b)$

[cloud walrus]

Liria ^(;,;)^:

[shy bluff]

Which means that $N(a_i) = N(a_{i + 1})N(b)$, which would require $N(b) = 1$?

[cloud walrus]

Liria ^(;,;)^:

[woven delta]

Yeah because of the stabilization of the sequence of norms

Write that out too

$N(a_{i+1})=N(a_i)$

[cloud walrus]

Liquid:

[shy bluff]

oh that's true

[woven delta]

Okay so write it out

[shy bluff]

One sec I'm typing this out

[woven delta]

👍

[shy bluff]

If we have a chain of principal ideals, we see that the generator of each successive principal ideal will divide the generator of the principal ideal that comes before it in the chain.

        That is, if we were to have $(a_1) \subseteq (a_2) \cdots$, we would have that $a_2 | a_1, a_{i + 1} | a_i$. By part a, we have that if $x | y$ then we have that $N(x) | N(y)$. 

        Then have that $a_{i + 1} | a_{i}$, we have that $N(a_{i + 1}) | N(a_i)$, which would mean that $N(a_i) \geq N(a_{i + 1})$. Then we see that the norms of the generators of the chain of principal ideals is a decreasing positive sequence of natural numbers. Then it must eventually stabilize at some $a_k$. 

        When it does stabilize we would have that $N(a_k) = N(a_{k + 1}) = N(a_{k + 1}) \cdot 1$,

[cloud walrus]

Liria ^(;,;)^:

[shy bluff]

I'm not quite sure about the last bit

wait taht's not quite right

It should stabilize at N(a_k) not at a_k

[woven delta]

Okay the last line is a bit off

$N(a_{k+1})=N(a_k)$ is good, and you want to use that plus $N(a_k) = N(a_{k+1})N(b)$

[cloud walrus]

Liquid:

[woven delta]

To show that $N(b)$ is 1

[cloud walrus]

Liquid:

[woven delta]

And therefore $b$ is a unit

[cloud walrus]

Liquid:

[shy bluff]

wait why can't we go from $N(a_k) = N(a_{k + 1}) * 1$ to see that $1 = N(b)$ for some unit b?

[cloud walrus]

Liria ^(;,;)^:

[shy bluff]

Or is that what you mean, just written differently

[woven delta]

Well that is true but we can't say $N(a_k) = N(a_{k+1})N(b)$ implies $a_k=a_{k+1}b$

[cloud walrus]

Liquid:

[woven delta]

Because that's not true

We started off with some b so that $a_k= a_{k+1}b$

[cloud walrus]

Liquid:

[woven delta]

And we want to show that $b$ must be a unit

[cloud walrus]

Liquid:

[shy bluff]

oh ok

but doesn't that assume that the chain stabilizes?

[woven delta]

Since that tells us that $(a_k) = (a_{k+1})$

[cloud walrus]

Liquid:

[woven delta]

No

Why would that assume that

All I am assuming to get such a b is that the chain is ascending

So $(a_k) \subseteq (a_{k+1})$

[cloud walrus]

Liquid:

[shy bluff]

Well N(a_k) = N(a_{k + 1}) is fine because that's the direction that we're moving in but going from that back to a_k = a_{k + 1} is what we're trying to show isn't it?

[woven delta]

We are trying to show $(a_k) = (a_{k+1})$

[cloud walrus]

Liquid:

[woven delta]

Ie we are looking at principle ideals

Not just the elements

[shy bluff]

I think I need to sleep on this lol

Thank you for the help though

[woven delta]

👍

[oblique river]

liquid it bothers me that you keep sayin "principle" and not "principal"

[woven delta]

Lol

I forget which spelling is which

I guess it's principal

It's weird because when I was a kid a teacher told me that I should remember it because "a principle is ____ your pal"

And I can't remember if the principle is or isn't my pal

I guess he isn't

[oblique river]

you missed it again

lmao

[woven delta]

No this time I am referring to a school principle

[oblique river]

YES

that is a principal

[woven delta]

Oh

Wait

Shit

[oblique river]

lmao

[woven delta]

I guess he is my pal

We didn't have the best relationship tbh

[shy bluff]

... yea that's hard to get around

Why is it called a prinicipal ideal

[charred pewter]

Is it better to prove that a conjugate exists or to prove the equivalent statement that the left and right cosets are equal gH=Hg
Also is <r²>={r²ⁿ, n∈ℤ}?

[cloud walrus]

kxrider:

[charred pewter]

Ty

I'm proving it using the conjugate

[viscid wyvern]

can I ask

what should I know before learning abstract algebra?

I’ve learned enough to figure out that anything else is worthless without this

too many problems

[nova plank]

There aren't any essential prereqs except "mathematical maturity"

[simple valley]

a decent familiarity with proofs and sets

that's about it

you can't really expect "mathematical maturity" from someone who's yet to learn abstract algebra

[viscid wyvern]

oy shush

[simple valley]

no offense

[viscid wyvern]

you’re not entirely wrong

[nova plank]

I suppose it's relative so I should have said "enough mathematical maturity to understand abstract algebra" which is a useless answer

[simple valley]

that's what mathematical maturity means: experience in the various fundamental areas of math

[viscid wyvern]

I’ve been trying to learn other branches of math, even ones that don’t need this first (like real analysis) and I keep coming back to the same issues, ones I think this can help with

[supple marsh]

i thnk learning learning algebra first is helpful

[viscid wyvern]

Are there any good resources to learn abstract algebra?

[supple marsh]

do you know linear algebra?

(on the level of an intro proofs based linear algebra course)

[viscid wyvern]

I took a semester of linear algebra last year

[supple marsh]

i really liked Aluffi's algebra book

[viscid wyvern]

so.... yes?

[somber rivet]

Notchmath, you've been working through Tao's Analysis I right and have gotten through a significant portion of it?

[supple marsh]

its really long but hes so friendly

[somber rivet]

Then you're good

[viscid wyvern]

Yeah I’ve been trying to

the issue is I keep trying to connect his concepts to other concepts I know

[nova plank]

Abstract algebra and real analysis are independent imo. Neither is a prereq for the other.

[viscid wyvern]

to develop a better understanding conceptually

and keep needing algebra to express my connections and confusions

so I think it’s worth taking a break from that and coming to try and get some of this first

[nova plank]

What in real analysis do you feel like you need algebra for?

[viscid wyvern]

well

I find a concept

I apply it to natural numbers

I get myself confused

I try to ask a question in #elementary-number-theory to clear up my understanding

and I have no way to express my question

or even in Linear Algebra I’d do that

[nova plank]

Give a specific example of why you think you need algebra to learn real analysis. I think you probably don't, which is why I'm asking

[viscid wyvern]

I learned about vector spaces, so tried to imagine an infinite-dimensional space describing natural numbers by their prime factorizations, got confused in a way that was confusing my original understanding of vector spaces, and couldn’t formally phrase my question so I could clear up my confusion

and that was in Linear to be clear, not Analysis

but that sort of thing is common

[nova plank]

The natural numbers do not form a vector space

[somber rivet]

you do not need algebra to learn analysis

[viscid wyvern]

well right cause addition didn’t work

well it did

not addition in the numbers, addition in the vectors

[somber rivet]

on the other hand, your experience with tao's analysis should be enough for you to get yourself into an abstract algebra text

[nova plank]

You really don't need algebra to learn real analysis. You are specifically trying to form connections only to the natural numbers, when it might not work, so you're getting confused. A good textbook should contextualize the definitions in analysis already.

[viscid wyvern]

oh not natural numbers sorry positive numbers

I mistyped

so 25/12 would be represented as (-2,-1,2,0,0,0...) but also as (logbase2(25/12),0,0,0,0...) and I got myself confused in a way I don’t exactly remember now

or I was working with basic modularity and wound up in #topology asking about lines on the surface of a torus

that sort of transfer is why I want to do algebra

[nova plank]

You can do what you want, but ending up in topology before studying real analysis won't do you any good. It's good to find interesting questions, but if you don't have the knowledge to understand the answers, you can't go further. Learn algebra too if you want, but don't stop learning analysis because you think you need algebra. You don't.

[viscid wyvern]

that’s my point

I can ask questions but not understand answers

this is how I understand my answers

ugh I wish textbooks weren’t so expensive, if I can’t find an unusually cheap copy I don’t know if I can do anything before I take the actual course

[nova plank]

There are free textbooks online, and alternative means of getting any textbook. But yes, textbooks are ridiculously expensive

[golden pasture]

l i b g e n s c i h u b

[shy bluff]

Ok after sleeping on it I now understand what the ascending chain condition is saying

[shy bluff]

Hrm

To prove that something is not a UFD I just need to find an element that canb e factored in 2 different ways right

[elder valley]

yeah that would do it @shy bluff

[next obsidian]

You could also find something that doesn’t factor into irreducibles at all

[shy bluff]

hrm

I'm still trying to wrap my head around the differenceb etween inrreducible and prime lol

[next obsidian]

Prime says if you divide a product you must divide one

Irreducible says you can’t factor it any further unless one of the things is a unit

In Z both of these properties characterize prime numbers

But irreducible captures the definition of a prime in Z as “a number which is only divided by 1 and itself”

Where as prime captures the thing that “if p divides a•b it divides a or it divides b”

[shy bluff]

Wdym by "You must divide one"

[elder valley]

Where as prime captures the thing that “if p divides a•b it divides a or it divides b”
it means this

[shy bluff]

hrmmm so if p is prime then p|ab means that p|a or p|b but p being irreducibel means that a | p only if a = p or a = 1?

[next obsidian]

Well

Really it means a is p (up to a unit) or a unit

[elder valley]

associate of p or a unit i think

[next obsidian]

Yeah I corrected it

[shy bluff]

oh ok

[next obsidian]

Really it says

If p = ab then one of them is a unit

In Z the only units are +- 1

[shy bluff]

For irreducible, p = ab means that one of those two is a unit

[next obsidian]

And by convention we only look at the positive stuff

Yes

[shy bluff]

And for prime it means that?

[next obsidian]

Prime is the dividing a product

You divide one of the two

[shy bluff]

hrmmm

[next obsidian]

p | ab then p | a or p | b

[shy bluff]

Ok I see that these are two different things

I just ened to wrap my head around it some more lol

[next obsidian]

In a UFD they’re the same tho

Prime is always irreducible tho

[shy bluff]

Ok

oh wait I think I see

I think I wasl ooking at it the wrong way

The example of 6 = 2 * 3 = (1 + sqrt(-5))(1 - sqrt(-5)) in Z[sqrt(-5)] helped I think

Because then I can see that what you're saying is that 2 is irreducible because it can only be expressed as being 2 = ua where one of the two is a unit, but that 2 does not divide (1 + sqrt(-5)) or (1 - sqrt(-5)) individually

but it divides their product?

[next obsidian]

Umm yeah I guess that’s a way to see it yeah

So this shows 2 isn’t prime in there

Even tho it’s irreducible

[shy bluff]

ok I see

I think that I just nedede the example lol

[next obsidian]

And this happens only because

It isn’t a UFD

Or well, that isn’t strictly true

But it’s necessary

Because in a UFD the two are the exact same thing

Irreducibility is like saying you can’t break it down anymore

Prime is about it dividing other stuff

[shy bluff]

Yea

I think I see it

Wait question, why can we say that 2 is irreducible here

[next obsidian]

You have to show it

The easiest way is that you have a norm

And it’s multiplicative (just grab the norm from C)

You can show that since 2 has norm 2 that it couldn’t be broken down further

This is the same thing you need to do to show 1 +- sqrt(-5) is irreducible but there the norm isn’t quite enough

You need to use the fact that it has nonzero imaginary part as well

The idea is that if 2 = ab

Then a and b are of the form c + dsqrt(-5)

If d ≠ 0 then that element has norm at least sqrt(5) which already makes it too big to possibly multiply into 2

So now you know a and b are just integers

And 2 is irreducible there (in Z)

The same sorta thing works for 3 but you need to do a bit more since sqrt(5) < 3

[shy bluff]

Huh

That's a cool arguement

[next obsidian]

https://www.overleaf.com/read/vbccvpybydrz

In here is a proof of Z[i] being a Euclidean domain and Z[sqrt(-5)] not being a UFD

It’s in week 6

[shy bluff]

Oh hey I get the first 2 questions on that week0 stuff

Ohh ey it's the chinese remainder theorm thing from last week

[next obsidian]

So you only need to show that 2 is irreducible

That one of (1 +- sqrt(-5)) is irreducible

Then show those two aren’t associate

Because that’s enough to imply it isn’t a UFD since then 2 and the (1 +- sqrt(-5)) are different irreducible factors, even up to units

[woven delta]

Is there a nice universal property for composite field extensions?

[next obsidian]

It’s the smallest field containing both

My only guess would be that like two morphisms from the two like, building blocks to the same target which “agree” in some way factors through it??

[woven delta]

Like a nice one that comes to mind that that for field extensions G, G' over F, the composite GG' is the initial object in the category of field which are field extensions of G and G' where the extending maps extends extension from F to G and G'

[next obsidian]

This is basically just saying it’s a push out lol

[woven delta]

Sure

[next obsidian]

Oh sure I see what you’re saying too

But I don’t think what I said is true

Or it might be? I don’t think that makes it a push out though

[woven delta]

Idk tbh, am walking rn

[next obsidian]

Haha

[woven delta]

So will think more later

[next obsidian]

I have weak field theory tbh

Both times I learnt it I felt like I was really shaky

[wind parrot]

To me the compositum of two fields K and L can really only be reasonably defined if both are taken to be subfields of some larger field F (otherwise, how would you define the compositum of R and Q[X]/(X^3-2) for instance). And since any subfield of F that contains both K and L contains KL, it seems to me like KL is the categorical coproduct of K and L in the category of subfields of F, with inclusions as morphisms.

@woven delta

[woven delta]

I don't want to refer to a larger field

Which you can avoid

[knotty mason]

How is it avoided?

[shy bluff]

UFDs are euclidean domains right

[nova plank]

Euclidean domains are UFDs.

[shy bluff]

Oh other way around?

[nova plank]

Not all UFDs are Euclidean domains

[shy bluff]

Euclideans contain UFD

ok

thank you

[nova plank]

Np

Set of Euclidean domains is included in the set of Principal Ideal Domains which is included in the set of UFDs. And both of those inclusions are strict

[next obsidian]

Fields are included in Euclidean domains as well

Strictly

[woven delta]

Lol "set"

[shy bluff]

So every element of a UFD can be written as a product of irreducibles, and that product is unique?

[golden pasture]

up to units

[shy bluff]

Err yea

[nova plank]

That is usually the definition of a UFD, so yes

[vestal snow]

For the last part, the book says that if $\sum i_k j_k = 1$, then the $i_k$ generate I

[cloud walrus]

The formerly edible banana:

[vestal snow]

How do I prove this?

[next obsidian]

You just show it manually

I forget quite how I did it but I think I just showed that every element can be written as a sum of products of the i_k and other stuff

[vestal snow]

Lol you just threw the definition at me

But it helped

I was looking for some slick trick, but it turned out pretty simple

[shy bluff]

Are gcd domains also ufds?

Or is it the other wayaround

[elder valley]

see the chain of ring inclusions on this page @shy bluff https://en.wikipedia.org/wiki/Unique_factorization_domain

[shy bluff]

Hrm

ok

So now I'm trying to prove this,

I know how this works for prime numbers in the integers

But how does this work for a finite field?

Err

Err I see that if K is infinite it's the same thing as Euclid's proof

But what about when it's finite?

Good question

I'm not sure

From my understanding, Euclid's proof (for integers) is that if you have n prime numbers, you can set q = p_1...p_n + 1, and that q is either prime or not prime

If q is prime then we have a prime not in the list
If q is not prime then q must be divisible by a prime, but this prime must also divide 1

I'm not sure how you'd extend this to work on an arbitrary field or polynomial ring ovre a field

I'm making the assumption that I can treat an infinite field the same way as I can nte integers

oh wait

ok I think that was my problem

I forgot that K[x] is a ufd

And to double check, irreducibles in a UFD are also prime yes

[next obsidian]

yes

[shy bluff]

So S is the set of all possible porducts of elements of S_o with units?

[elder valley]

yes

[shy bluff]

Also, another question, as u is a unit, and S is multiplicatively closed, the inverse of u is also in S?

[elder valley]

S contains all units because you can choose k=0 and u as any unit

[shy bluff]

Yea

Ok that's what I thought

Oh so u is any unit?

[elder valley]

yeah it says u is in R^x

[shy bluff]

Ah I thought that was a specifci u

[elder valley]

it would be chosen outside of the set notation if that were that case

prior to defining S

[shy bluff]

Yea ok

1 counts as a unit too right

[elder valley]

yes

[shy bluff]

oke

Thank you

[elder valley]

lol in this wiki page one of the references is just a proof https://en.wikipedia.org/wiki/Irreducible_element

[shy bluff]

If I have a localization? Rinvg of fractions? Whatever they callit, S^-1R, with f/g being a unit in it, then it requires that g/f in S^-1R right

Wait nevermind

that equal ssign is in .. D right? like they're equal in the sense that they're equal according to the ring that they come from?

[elder valley]

it's in whatever set the products "re" and "sd" land in

[shy bluff]

hrm

Can things divide only if they're in the same set?

Yep, looks like it...

ok, so if I have f/g of S^{-1}R, with f/g being a unit, then I can say that there exist a/b in S^{-1}R such that (f/g) * (a/b) = 1/1, and that we have to have that f, a in R, g, b in S right?

[timid hull]

Does the affine group over integers Aff(Z) have any nice properties ? Like it is also called "infinite dihedral like group" . I came across this while preparing for finals . Like it has properties like quotient of Aff(Z) is Aff(Z) itself or Dn for some n and stuff . I am wondering does it have any other nice properties like this ...

[next obsidian]

I’m curious what the affine group even is?

[timid hull]

Like they are 2x2 matrices generated by

$\begin{pmatrix}
1 & 1 \
0 & 1
\end{pmatrix}$ , $\begin{pmatrix}
-1 & 0 \
0 & 1
\end{pmatrix}$

[cloud walrus]

WhyamIsohot?:

[woven delta]

Isn't that SL(2, Z)?

[timid hull]

uhhh

one of em has determinant -1

[woven delta]

Oh oof

Nice

[devout crow]

hello i have a (probably basic) group theory question

let $G$ be a group and $M = {g_1,...,g_n}$ be a minimal generating set for $G$. is it true that $M$ can be constructed by choosing $g_{i+1}$ if $g_{i+1} \notin \langle g_1,...,g_i \rangle$ for $1 < i < n$ ?

[cloud walrus]

xdres:

[devout crow]

if so, why? if not, why not?

[next obsidian]

I’m really confused

For one I don’t know what “constructed means”

But furthermore if {g1,..,gn} is a generating set for G

Wait umm

Ohhh

I think I see

[wooden ember]

How can g_i+1 not be an element of <g1,g2,...>

[next obsidian]

Lol

Because it only goes up to i

Not to n

[wooden ember]

Oh yeah oops

[next obsidian]

Yeah I had the same thing

I feel like this might actually not be true

But I can’t justify it

For abelian case I can

Because then it’s a Z-module and torsion means you won’t have a basis

[wooden ember]

Im not experienced enough ill leave this to you

[next obsidian]

And I think what he’s describing is basically incrementally building a basis

Like I feel like a minimal generating set might still have relations

Because it isn’t free

Yeah xdres my gut says no

But honestly this seems really hard and I’m not feeling up to the task ;w;

[devout crow]

no worries :)

[woven delta]

Yeah I don't think this is true

I don't think you can even get a finite generating set this way, much less a minimal generating set

Probably you can show this using the free group on 2 generators

Like this really follows from the fact you can embed the free group on 3 generators into the free group on 2 generators

You can keep partitioning up the free group on 2 generators G so that at stage n you can always find an element $g_n$ so that $g_n \notin <g_1, ..., g_{n-1}>$ and another free group on 2 generators inside G so that that subgroup intersects $<g_1, ..., g_n>$ trivially

[cloud walrus]

Liquid:

[woven delta]

So this kind of idea doesn't really work

[elder valley]

a counter example is Z/4Z, where you have generating set {1}, but {2,1} is also a generating set

[woven delta]

Yeah obviously the minimality is bad

But also I think this sort of idea will produce a generating set in the finitely generated abelian case

[elder valley]

or in any group choose g_1 as the identity

[woven delta]

In the non abelian case it won't even do that

[next obsidian]

No I think you misinterpret it

I thought you have a given minimal generating set

And you try to build up the group adding one element of that minimal generating set at a time

So in the case of Z/4Z the minimal generating set is {1}

So g_1 has to be 1

I thought this was basically saying that if {g1,...,gn} is a minimal generating set, is g_i not in {g_1,...,g_{i - 1}}

This isn’t necessarily true for modules, since if you aren’t free your minimal generating set has relations among the generators

[woven delta]

Oh okay the statement said "can M be constructed" so I assumed that we were trying to construct a minimal generating set

[elder valley]

yeah that's what i thought too. like similar to the algorithm of constructing a basis for vector spaces

otherwise "constructed" seems like a strange word to use

[next obsidian]

Yeah, I mean

I sort of thought the same thing, but strictly speaking g_i is already in scope

Tbh I'm not sure what they originally intended

[supple marsh]

Somewhat related is that there are theorems about small random subsets of a finite group giving generating sets. E.g. see theorem 1 here : https://www.math.ucla.edu/~pak/papers/randk14.pdf

and discussion afterwards

[shy bluff]

ok so for part c here, I'm trying to show it in the direction of "if x is associate with y... x is irreducible". What I"ve got so far is that if x is associates with y then there exists a unit u such that $x = uy = u\frac{f}{1} = \frac{u}{1}f = af$, where $a$ is also a unit

[cloud walrus]

Liria ^(;,;)^:

[shy bluff]

Bu tI don't know how to go from that to irreducible

Or how to use the fact that f is irreducible?

My thought is that like we assume that x is not irreducible, then x = cd for some c, d non unit

The nyou have that cd = af, where f a is a unit and f is irreducible

But not really sure where to go from there

[shy bluff]

Anybody?

[elder valley]

i think i see a messy proof

[shy bluff]

my proof being messy? seems likely

[elder valley]

first off x and y are associates in S^-1 R, so your u is a fraction

[shy bluff]

oh

[elder valley]

no i mean the proof i just thought of is messy

[shy bluff]

ah

do we just let y = f/1, and then let u be a unit and get uf/1?

[elder valley]

elements in that ring have the form a/b, where a is in R and b is in S. and the units are the ones where a is also in S

[shy bluff]

Yea

[elder valley]

maybe call your unit u/v, where u and v are in S

so you have x=uf/v

[shy bluff]

sure

[elder valley]

you want to show that x is irreducible, so suppose it factors into the product of 2 fractions

[shy bluff]

Err wait you'd have (u/v)(f/1) right?

[elder valley]

yeah and then multiply them together

[shy bluff]

Yea

suppose it factors into like (c/d)(a/b) or something right

[elder valley]

yeah, then clear denominators to get an equation in R

or use the equivalence relation, whichever way you think about it

[shy bluff]

sure, then you get that acv = ufbd

[elder valley]

right. so b,d,u,v are all in S, and a,c are just in R

[shy bluff]

yea

Do u and v also have to be units?

I think they do

[elder valley]

no, just in S

[shy bluff]

oh they're units in S but not units in general?

[elder valley]

they're units in S^-1R

or u/v is a unit i mean

[shy bluff]

Err yea ok

[elder valley]

so now use that R is a UFD. b,d,u,v all factor into products of elements from S_0

[shy bluff]

oh I see

[elder valley]

and also use that f is not in S_0, or associate to anything in there

what you're aiming for is that a/b or c/d is a unit in S^-1R, which means either a or c is in S

[shy bluff]

Ah so the no nthe left side you get that v is in S, and that on the righ tside, ubd is in S as well

And then you can divide out all the common factors between them?

ERr cancel out?

[elder valley]

you could...i'm not sure it helps

[shy bluff]

hrm

[elder valley]

like they could have no common factors to begin with

[shy bluff]

that's true

[elder valley]

maybe use an analogy with Z. if f=2 and S is all odd primes, and b,d,u,v are all odd numbers...what do you conclude about a,c

[shy bluff]

Wait question, by part b there don't we have that f/g is a unit in S^-1R if and only if f in S?

[elder valley]

yeah

we used that with u.v

u/v

[shy bluff]

oh right ok

[elder valley]

actually i didn't even read (a) or (b) xD

[shy bluff]

lol

hrmmm well if f = 2 then we have that bduf is also even

Then the product of acv has to be even?

[elder valley]

yep

v is odd also

[shy bluff]

Err yea

oh that's a cool arguement

[elder valley]

what's the conclusion then

[shy bluff]

either a or c has to be even

[elder valley]

yup

although we wanted a or c in S

[shy bluff]

hrm

In this analogy, S is the odds

that's not necessarily true though is it

[elder valley]

odd primes

[shy bluff]

Err yea, S is odd primes

Wait

[elder valley]

sorry S_0 is odd primes

[shy bluff]

S is the product of odd primes

Yea lol

[elder valley]

S is odds, you're right

[shy bluff]

So bduf is an even number, which means that acv must be even

As v is in S

Then either a or c is even

[elder valley]

yes

[shy bluff]

but that says nothing about the other?

[elder valley]

now look at the RHS again. ufbd with u,b,d in S

or rather 2ubd

[shy bluff]

ubd is still in S

Pardon?

[elder valley]

you have prime factorization in Z, so ubd is a product of a bunch of odd primes

[shy bluff]

Yes

[elder valley]

so 2ubd=2 * 3^(n_1) 5^(n_2)...

[shy bluff]

Yes

[elder valley]

and LHS has to have that same factorization

[shy bluff]

Ah

Oh and because f is not an associate to any elemnt of S_0 we can't just have v = f with a, c units?

[elder valley]

yeah v is in S and f is not

[shy bluff]

f is not associate to any element of S_0

[elder valley]

that's equivalent to f not in S i think

[shy bluff]

yea

ok so then we get something like

Let v = p_i... p_k

Then we get that bduf = (p_1... p_j)f = ac(p_i... p_k)

And then we can divide out all the common irreducible (if any), leaving us with some number of irreducible that are not common betwee ntehm

And as f is not an element of S

We have to have that ac = whatever primes left over right?

[elder valley]

i don't think they all have to cancel

[shy bluff]

not all

[elder valley]

from the RHS

here's a better reasoning for the Z case

we want that a or c is in S={odd numbers}. so suppose they're both even

you have 4 dividing acv=2ubd, but u,b,d are all odds

[shy bluff]

Oh I see

[smoky cypress]

How do I prove this: Let $G_1,G_2$ be simple groups, then the only normal subgroups of $G_1\times G_2$ are $G_1\times G_2$, $1$, or a group isomorphic to $G_1$ or $G_2$

Oh I think I got it

[oblique river]

it's not true as stated

@smoky cypress

[smoky cypress]

Oh wait yeah

[cloud walrus]

Whoever:

[oblique river]

that's more believable

[woven delta]

What do you mean "a group isomorphic to G_1 or G_2"?

[smoky cypress]

The normal subgroup of G_1 x G_2

Then I have no idea how to prove this

But lemme give it a try

[shy bluff]

@elder valley cant' I just write that if we have $acv = f(ubd)$, where $ubd, v in S$, if we assume that neither a nor c in S, we would have that $v = ubd$, and thus $f = ac$, but $f$ is irreducible, so this is a contradiction, and so either a or c in S?

[cloud walrus]

Liria ^(;,;)^:

[elder valley]

how do you get v=ubd

[shy bluff]

Assume neither a nor c in S

Then neither a, c, f in S

Then we have that v and ubd in S

So we need v = ubd?

Like the two sides have to have the same prime factorization

Or does that not make sense

[elder valley]

i don't see it

you can use that there's only 1 f in the factorization of the RHS, so there can be only 1 in the LHS

all the other irreducible factors are from S_0

[shy bluff]

Yea the two sides have to have the same factorization because they're in a ufd

[elder valley]

yes but only 1 f. f can't go into both a and b

[shy bluff]

Wdym

[elder valley]

when you prime factorize a and b, f can only be present in 1 of those factorizations

because there's only 1 on the right side

[shy bluff]

Wait it's acv = fbdu

What do you mean by "f can only be present in one of those factorizations"

[elder valley]

like you write both sides of the equation into a product of irreducibles, and the product has to be the same by uniqueness. but there's only 1 f on the right

just like with the Z case there was only one 2

[upbeat juniper]

Then I have no idea how to prove this
@smoky cypress maybe for every subgroup of G1 x G2 you can consider the projections onto G1/G2?

[smoky cypress]

Thanks

[shy bluff]

Like because ti's uniquely factorized, we have that $fbdu = f(gf_1 \cdots f_k) = acv$ for some unit $g$, $f_1, \cdots f_k \in S_0$

[cloud walrus]

Liria ^(;,;)^:

[shy bluff]

And then if we assume that both a, c not in S, then a, c can't be written as being the product of elements of S_0

So then we'd have f = ac, v = gf_1... f_k

[elder valley]

you don't get that f=ac. like a could be f*f_1, which is not in S

you can try using that f irreducible -> f is prime

maybe that's easier

[shy bluff]

wait why?

[elder valley]

because you're in a UFD

[shy bluff]

Like if I have that both f, a, c not in S

[elder valley]

v could be 1

[shy bluff]

Yea but then if v is 1 then i require that ac = gf_1... f_k

err

acv = f(gf_1... f_k), but if v = 1 then we just have ac = f(gf_1... f_k)

[elder valley]

yes. but that doesn't mean ac=f

[shy bluff]

It doesnt', it just means that one of ac is in S

[elder valley]

i'm getting lost in all this

[shy bluff]

So am I 😅

what do you mean by "try using that f is irreducible -> f is prime"

[elder valley]

you have that f divides acv

and f is prime

really you just want that f doesn't divide both a and c

[shy bluff]

hrmmm

f divides acv and f is prime

As f not in S_0 we have that f does not divide v?

[elder valley]

right

suppose it divides a then

[shy bluff]

hrm

That says nothing about c or a though? all that it means is that there's something that we can multiply f by to get a?

[elder valley]

yeah, so write a=f*t for some t

then substitute and cancel f

[shy bluff]

Oh then we get tcv = gf_1... f_k

[elder valley]

the prime factorization of t,c,v must then be some sub-product of the RHS

so you have c in S

[shy bluff]

wait why must c be in S?

and also, you're assuming that t also not in S right

[elder valley]

all of them will be

[shy bluff]

oh right

ok I see

Because it equals to that mess

and there's no longer any factors that aren't in S

[elder valley]

yeah exactly

[shy bluff]

Ok It hink that I was trying to say something similar but just worded really badly

[elder valley]

that's what i meant by f is only in one of a or c, because there's only 1 f to cancel on the right

[shy bluff]

Thank you

Ok I tryt og oand prove it the other way myself

[charred pewter]

Okay uh, I need help getting all the elements for D₆

I have D₆={e,r,r²,r³,r⁴,r⁵,s,rs,r²s,r³s,r⁴s,r⁵s}

Is that right?

(r is for rotations and s is for reflections)

And then for the subgroup, I have <r³>={e,r³}

I'm asking mostly because the first time I looked for the elements of D₆, I had trouble finding the 12th element but I think that's fixed now

[bleak abyss]

Yeah, in general for D_{2n} you have n rotations, given by r^k for k = 0,...,n-1

(r^0 = e here)

And n reflections, given by r^k s

[shy bluff]

just to clarify for this one, for part c, one direction is "If x is irreducible, show that y = f/1 is irreducible" right

[charred pewter]

Wait what

Are we looking at D_2n here or D_n

[bleak abyss]

Oh right

I forgot what your notation was

Yeah D_n

I tend to prefer saying D_n has 2n elements but for some reason I thought you were using D&F's notation

[charred pewter]

Also I thought rⁿ=e

[bleak abyss]

Where D_{2n} has n elements

r^n = r^0 = e

Either way, you can index 1 to n or 0 to n-1 here

[charred pewter]

Alright

So assuming I have the elements for D₆ and <r³> right, I'm kind of confused on how to find the cosets

For the Cayley table I mean

Say, for the left cosets, do I multiply <r³> from the left by an element of D₆ not in <r³> to find cosets?

[bleak abyss]

You want to find representatives for each coset

So for example

r is equivalent to r^4 and that's it, right?

So you have a coset {r,r^4}

You also have {r^2,r^5}

These two cosets multiply to what? To the identity coset {e,r^3}

[charred pewter]

Since |D₆|=12 and |<r³>|=2, we have 6 distinct cosets right?

And I can find them using that method?

[nova plank]

Yeah, multiplying from the left by an element of D6 not in <r^3> also works.

So you want to multiply by every element of D6 that is not already in one of the cosets you have

[charred pewter]

Yeah, that's how I remember doing cosets, although the way Sloth did it was the equivalent of multiplying r*<r³> and r² * <r³>

[nova plank]

So, say multiply by r then you get r<r> = {r, r^4}. Now since you've gotten r^4 here too, you don't have to multiply by it too. So multiply by r^2 then you get {r^2, r^5} and you don't have to multiply by r^5 either since you'll get the same coset

Yes, just wanted to make sure you know what you said is fine

[charred pewter]

Thanks, Luna

Also sorry if I hijacked the chat, I thought you had deoccupied it, Liria.

[shy bluff]

Question: Can we divide units? Like say that u is a unit, do we have that v | u?

Oh dwbi

Idm, I use this chat way too much lol

[charred pewter]

lol

[nova plank]

@shy bluff Yes, no reason you can't

[shy bluff]

Units divide everything in the ring

[nova plank]

Units divide every other element, so a unit is at least divisble by every other unit

[shy bluff]

hrm

Ok but we can't necessarily say that it's divisible by a non-unit right

[nova plank]

You can't assume it is, but if you can find an element q such that u = vq then it doesn't matter whether they are units or not

[shy bluff]

Yea ok

thank

[nova plank]

Np

[charred pewter]

For my problem, once I have the 6 cosets, how would I go about building the Cayley table?

Referring to this problem:

[thin flume]

Try listing the 6 cosets on the top row of the table and the left column of the row

What are the 6 cosets you got btw?

@charred pewter

[thin flume]

Alright, have you seen a Cayley Table before?

[charred pewter]

Yes

[thin flume]

So, the top row of the table and the left column of the row consists of the element of the set we're talking about, in this case it's the 6 cosets you take the pic of

Now, try writting them by order to the table,

||By order, I mean,
<r^3> | r<r^3> | r^2<r^3> | s<r^3> | rs<r^3> | r^2s<r^3>
<r^3>
r<r^3>
r^2<r^3>
s<r^3>
rs<r^3>
r^2s<r^3>||

[charred pewter]

In the order I found them in the pic?

[thin flume]

Yeah

Actually, it's not really needs to be by order, but the way you write it must be the same

Say you write it like a,b,c,d,e,f on the horizontal side

then you need to write it a,b,c,d,e,f on the vertical side too

[charred pewter]

Okay,

[thin flume]

After writing the appropriate elements on the Cayley Table, you fill the table by multiplying the correspondences elements

By multiplying, I mean with the operation of the group has too

After done doing that, you'll have the Cayley Table for D_6/<r^3>

Does my explanation clear enough? Sorry it's not clear enough :/

[charred pewter]

By multiplying two cosets with their corresponding elements, can you show for example how to multiply <r^3> and r<r^3>

I'm kinda confused by what you mean

[thin flume]

I mean, multiplying them like this

[charred pewter]

Oh, yeah

Can I assume the identity coset <r^3> is going to act like e?

[thin flume]

Yeah

I mean, <r^3> is the identity coset after all

[charred pewter]

So for example when I multiply r<r^3> by r^2<r^3> I can just write it in the cayley table as a multiplication of the two cosets?

[thin flume]

Yeah

[charred pewter]

Oh, okay

I thought I had to find something it's equal to lol

[thin flume]

wdym lol

[charred pewter]

Like r<r^3>*r^2<r^3> is how I'd write it in the cayley table

[thin flume]

Yeah that's correct

But

You can simplify it more no?

[charred pewter]

That's the part that's confusing to me

[thin flume]

They are cosets right? Which is on the group of D_6/<r^3> too, therefore it's a quotient group.. So when we have a<r^3>b<r^3>, then it's operation will be coset multiplication (which is the operation of quotient group), which is equal to ab<r^3>

Also we know it's a quotient group because <r^3> is a subgroup of D_6

and D_6 is cyclic, therefore it's abelian too

and because <r^3> is subgroup of D_6, and D_6 is cyclic, then <r^3> is a normal subgroup of D_6, which leads us into getting D_6/<r^3> is a quotient group

Then, to answer your question of how to simplify "$r\langle r^3 \rangler^2 \langle r^3 \rangle$", since the operation is coset multiplications, we can get that $r \langle r^3 \rangler^2 \langle r^3\rangle =r^3 \langle r^3 \rangle=\langle r^3 \rangle$

[cloud walrus]

Wilston Lynx:

[thin flume]

@charred pewter

Wait, I might be wrong a bit, is D_6 a cyclic group or abelian?

Hmm, no, D_6 is not abelian

Wait

Ugh, I'm sorry man but I'm not quite sure myself what's the operation of D_6/<r^3> is :/

[charred pewter]

Well aren't all cyclic groups abelian?

[thin flume]

Yeah, but D_6 isn't cyclic right?

I made a mistake by stating that it's cyclic earlier

[charred pewter]

Confused here too

Yeah, it's not abelian

cyclic*

[knotty mason]

if you think of D_6 as the rotations and flipping of a hexagon

you can think of D_6/<r^3> as the rotation and flipping of a hexagon where the vertices are labelled 1,2,3,1,2,3 instead of 1,2,3,4,5,6

if you join up the numbers 1,2,3 you get 2 triangles rather than 1 hexagon

in fact D_6/<r^3> = D_3

[charred pewter]

D_3 is still non abelian

So the quotient group operation Wilston said I could do doesn't apply, right?

How else could I find what r<r^3>*r^2<r^3> is equal to?

[thin flume]

you can think of D_6/<r^3> as the rotation and flipping of a hexagon where the vertices are labelled 1,2,3,1,2,3 instead of 1,2,3,4,5,6
@knotty mason I personally doesn't understand how this come up? Can you elaborate more?

D_3 is still non abelian
@charred pewter The thing Skymoo mentioned earlier was meant to point us that D_6/<r^3>=D_3, so I think by making the cayley table for D_6/<r^3> is basically the same as making the cayley table for D_3

[charred pewter]

Wait so I wasted time with those cosets?

[thin flume]

So the elements we need to operate are the ones from the D_3

Wait so I wasted time with those cosets?
I think--

I personally still not sure how one could come up with D_6/<r^3>=D_3 tho?

[charred pewter]

Same here

[thin flume]

and by equal here does that mean they're isomorphic or basically the same group?

Might need anyone to elaborate about this more lol

Sorry if I turned out not help you much Aur :/

[charred pewter]

No it's okay, you did help

[wind parrot]

you can still just use the cosets you found

you didn't waste your time

[charred pewter]

Do you know how to find the product of two cosets if the group isn't abelian

That's where we were stuck

[wind parrot]

to mulitply cosets, use a<r^3>* b<r^3> = ab<r^3>

[charred pewter]

Even if it doesn't form a quotient group?

[thin flume]

to mulitply cosets, use a<r^3>* b<r^3> = ab<r^3>
@wind parrot this only applies when the quotient part is a normal subgroup right?

Because iirc that operation is only well-defined when the quotient part is a normal subgroup

[wind parrot]

it is normal

[charred pewter]

Yeah, I just checked, it's normal

[thin flume]

Wait, how did you guys know it's normal?

By Definition?

[charred pewter]

Yes, I checked that it's closed under conjugation

I don't know how he knew but he was right

[thin flume]

Ahh my bad then

[charred pewter]

The good news is that we can straight up just do a<r^3>*b<r^3>=ab<r^3> now

in the cayley table

Anyway, thank you both for the help 👍

[knotty mason]

@thin flume sure I can

@charred pewter you didn't waste time doing cosets, that's an important thing to learn and practice

[charred pewter]

I'm glad I did them now that you mention it

I definitely needed the practice

[knotty mason]

D_6 is rotation and flip symmetries of the hexagon, the next one is identifying opposite sides which is the same as quotienting by <r^3>. the next is the same thing but making it clearer (hopefully) why it is isomorphic to D_3

I think we can argue in general that if tau is the flip, D_n / tau = C_n. and D_n / r^d = D_{n/d}

[plucky flicker]

is every leading ideal radical? i mean, we have an I ideal in the polynomial ring K[x_1, ..., x_n], and the leading ideal is defined to be the L(I) generated by the leading monomials of polynomials in I. My question is, that is this L(I) a radical ideal?

[timid hull]

@knotty mason D_n/tau=C_n is not true ....

Any non-trivial quotient of a dihedral group is always dihedral or Z/2Z or Z/2ZxZ/2Z iirc ...

[thin flume]

Even if it's being true, we can't just assume they're the same right? I mean, afaik Isomorphic means that they have the same structure, and by that, the elements that has same order are correspondences to each other, not being necessarily the same right?

[hot lake]

quotienting by group elements ?

[knotty mason]

@timid hull ah thank you

[plucky flicker]

is every leading ideal radical? i mean, we have an I ideal in the polynomial ring K[x_1, ..., x_n], and the leading ideal is defined to be the L(I) generated by the leading monomials of polynomials in I. My question is, that is this L(I) a radical ideal?

[shy bluff]

Hrm... For a quotient ring over polynomials, does irreducible => we cant' find roots in the original ring?

Ah nevermind

[shy bluff]

Reading through Dummit and Foote, I wish that my prof had followed the ordering more..

[smoky cypress]

How do I show that if $G$ is a finite group, $A,B\subseteq G$, and $|A|+|B|>|G|$, then $G=AB$

[cloud walrus]

Whoever:

[smoky cypress]

Can someone give me a hint

A,B are just subsets, not necessarily subgroups

[oblique river]

Form A^(-1) = {a^(-1) | a in A}. What can you say about A^(-1) \cap B? can you adapt this argument to work "for any group element"?

(that last sentence won't make sense until you figure out what the first sentence is telling you)

[smoky cypress]

Ah I see

So let $g\in G$ then $\brc{a\inv g:a\in A}$ and $A$ have same cardinality, so it must overlap with $B$, or there exists elements $a,b$ such that $a\inv g=b$

[cloud walrus]

Whoever:

[smoky cypress]

then g=ab and G is a subset of AB

Got it

Thanks

[oblique river]

👍

any time you have some kind of assumption like |A| + |B| > |C| then some kind of overlapping argument might be helpful

[smoky cypress]

I see

[old roost]

,iam adv

[cloud walrus]

Your roles have been updated!

[next obsidian]

Wait how the bell

Did this person give themselves adv

In an advanced channel

😳

[smoky cypress]

No

What happened was he gave himself adv in another channel

Removed it in this channel

Then got it again in another channel

[stone fulcrum]

That represents a rotation by π/4 radians on the unit circle

How many such rotations until you get back to identity?

[charred pewter]

7 more?

Is the identity 0 or 2pi or pi/4

[stone fulcrum]

You tell me! What number, when multiplied with a complex number, doesn't change that complex number?

[charred pewter]

1

[stone fulcrum]

But you're correct, 8 rotations in total is a full rotation

And yes, the number that you'd rotate back to is 1

The subgroup is of order 8!

[charred pewter]

Oh that's neat

I have literally never encountered this kind of problem before

I found it weird that my prof put it on homework

It said it was wrong 😕

Now I'll never know because it's one attempt only lmao

[stone fulcrum]

Dafuq

[charred pewter]

I put 40320 but that's the same thing as 8!

[stone fulcrum]

Oh

[charred pewter]

it should have accepted it

[stone fulcrum]

Oh no

[charred pewter]

?

[stone fulcrum]

oh god

[smoky cypress]

LMAO

[charred pewter]

LMFAO

I THINK I KNOW

You meant 8

[stone fulcrum]

Rofl OOPS

[charred pewter]

LOL

[stone fulcrum]

Common math chat mistake I suppose

Suddenly factorial

[charred pewter]

I'm laughing so hard rn lmao

I thought you meant factorial

[bleak abyss]

https://www.reddit.com/r/unexpectedfactorial/

[stone fulcrum]

Welp next time you'll know what the answer should be haha