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Brian M. Scott
Brian M. Scott
12:00 AM
@JasonBourne L’Hospital?
 
Old John
Old John
@BrianM.Scott Have you ever looked at the musical score for "La Maja de Goya"? Julian Bream played it amazingly - but the music looks like it isn't any harder to play than stuff in a guitar tutor ...
 
user19161
@BrianM.Scott Yes. I have seen 9000 proofs of that, but guess where the shortest proof is?
 
Brian M. Scott
Brian M. Scott
@OldJohn No, I’ve not $-$ but I have noticed that the easy-looking things are often brutally hard to do really well.
 
math101
math101
Thanks @JasonBourne YOU ARE AWESOME :)
 
user19161
@math101 Oh well, as an exercise you should try to prove my above statement yourself.
 
Brian M. Scott
Brian M. Scott
12:02 AM
@JasonBourne What do you have in mind?
 
Old John
Old John
@BrianM.Scott they certainly can be!
 
user19161
@BrianM.Scott Well, the shortest proof is found in baby Rudin.
 
Brian M. Scott
Brian M. Scott
@JasonBourne That’s certainly no real surprise.
 
Alex J Best
Alex J Best
@JasonBourne Just under a page long in my copy, pretty short indeed!
 
math101
math101
@JasonBourne You know that I am not that great at math, Don't you?
 
user19161
12:09 AM
@math101 Yes, but try to work that out for a while first.
 
Old John
Old John
@math101 That might be true of most of us - If we were really that great at maths, we probably would not be chatting here - we would be preparing our next paper for publication!
(I am excluding the high-rep members from that statement)
 
user19161
@OldJohn Unless you are a banana. A banana is someone great at math but something went wrong somewhere in his life. =)))
 
Old John
Old John
@JasonBourne I never underestimate bananas
 
user19161
@OldJohn Yes, they are very sweet. =)
 
user19161
@math If you still can't figure that out, post another question on the main site. =)
 
Brian M. Scott
Brian M. Scott
12:17 AM
@OldJohn No need to exclude me, at least: I put myself in the very-good-but-not-outstanding category.
 
user19161
@BrianM.Scott Aww, now you are just trying to make us feel bad. =)
 
Brian M. Scott
Brian M. Scott
@JasonBourne I actually did worry a little that it might be taken that way!
 
Old John
Old John
@BrianM.Scott That is a pretty good category - I belong to the not-very-good-and-never-outstanding category ... but I am satisfied with my lot :)
 
robjohn
robjohn
@user58512 I couldn't get my hint to work, so I gave a general expansion.
 
user19161
I have cast 200 downvotes. I must stop now. I will only downvote again if I am downvoted, to keep my rep at a multiple of 5. =)
 
user19161
12:21 AM
But I have cast 2440 upvotes. Yay!
 
Brian M. Scott
Brian M. Scott
I don’t vote as often as I should: I’ve cast only $2643$ votes. But they’re all upvotes.
 
Old John
Old John
That is a pretty good record
 
user19161
@BrianM.Scott Wow, amazingly we have cast the same number of votes to a multiple of 10!
 
user19161
This is a sign that a miracle is about to happen, HAHAHAHAHA
 
user19161
@math Are you still here? I will guide you in the proof.
 
Old John
Old John
12:32 AM
I discovered something interesting last night (at least to users below 10k rep) - if you subscribe to the MSE questions RSS feed, you can get to see deleted questions, as the full question seems to be in the feed
not sure about deleted answers ... yet
@RustynYazdanpour - just curious - is your name of Persian origin?
 
Rustyn Yazdanpour
Rustyn Yazdanpour
Yep
@OldJohn How did you know?
 
Old John
Old John
Just curious - one of my hobbies is learning Persian - I have not got very far yet!
 
Rustyn Yazdanpour
Rustyn Yazdanpour
Ha @OldJohn That is a cool hobby. I took farsi when I was young and I didn't get very far either =(
 
Old John
Old John
@RustynYazdanpour I have a few Iranian friends who meet up from time-to-time to cook Iranian food - and I got accepted as a sort of "honorary Iranian" ... got interested in the language from there
 
Rustyn Yazdanpour
Rustyn Yazdanpour
That sounds like a delicious enterprise. @OldJohn
 
Jonathan O
Jonathan O
12:40 AM
Hey
Anyone here?
 
Rustyn Yazdanpour
Rustyn Yazdanpour
@JonathanO maybe
 
Jonathan O
Jonathan O
:) Im doing some work on financial math and am stumped
I know what to do but I cant rearrange this formula
 
Old John
Old John
@RustynYazdanpour Yep - the food was often more enticing than the language-learning :)
 
Jonathan O
Jonathan O
I jsut want to know if I need to plot it to solve for interest
it is the annuity to present value formula
 
Old John
Old John
Must get some sleep -goodnight all!
 
Jonathan O
Jonathan O
12:42 AM
A = P[(i(1+i)^n)/((1+i)^n +1)]
or something similar
 
Rustyn Yazdanpour
Rustyn Yazdanpour
@OldJohn night!
 
Jonathan O
Jonathan O
and solve for I
interest rate
 
Brian M. Scott
Brian M. Scott
@RustynYazdanpour I’ve never studied it, but I’ve a wide enough experience with languages in general to recognize that the name almost had to be either Persian or some very closely related language.
@OldJohn G’night!
 
Old John
Old John
@BrianM.Scott G'night
 
Brian M. Scott
Brian M. Scott
@OldJohn That was a fast shuffle!
 
Jonathan O
Jonathan O
12:48 AM
anyone?
 
Ethan
Ethan
How many solutions can $$x^k\equiv b\text{ mod m}$$ have in x, assuming gcd(x,b) divides a?
 
Rustyn Yazdanpour
Rustyn Yazdanpour
@Ethan how come I can't see your latex?
 
Brian M. Scott
Brian M. Scott
@Ethan You can use \bmod{m} to get $x^k\equiv b\bmod m$.
 
anon
anon
@RustynYazdanpour how do you know there's latex there if you can't see it? you mean it isn't parsing. see "LaTex support for chat" pinned on the sidebar to the right.
 
Rustyn Yazdanpour
Rustyn Yazdanpour
@anon ok
 
Ethan
Ethan
12:53 AM
If gcd(x,m) divides b, k is a fixed integer, how many solutions in x are their?
to $x^k\equiv \text{b mod m}$
 
anon
anon
So, "how many solutions x to x^k=b(mod a) are there such that gcd(x,m)|b"? Writing "if gcd(x,m) divides b" would seem to imply x is fixed.
 
Ethan
Ethan
sorry Im not thinking
get rid of that part
 
Rustyn Yazdanpour
Rustyn Yazdanpour
There we go, (latex enabled), thx @anon
 
anon
anon
you're one of the few who were able to get it on the first try :)
 
Rustyn Yazdanpour
Rustyn Yazdanpour
1:10 AM
@anon Lol
 
user19161
@anon I am still waiting for your change of clothes, LOL.
 
anon
anon
le sigh
 
user19161
You can have one for Chinese New Year, it's in two weeks, LOL.
 
Alex J Best
Alex J Best
 
user19161
1:27 AM
@grace Are you around?
 
user19161
1:41 AM
@hpe Did you come here to see what the flags were about?
 
user19161
Hey @people any progress with that proof on my question? =)
 
peoplepower
peoplepower
@JasonBourne Assuming it is the v-shape question, no.
 
user19161
@peoplepower Haha, V shape! Like those swimmer abs eh?
 
peoplepower
peoplepower
@JasonBourne Heh.
What would you call it?
Other than "my question"?
 
user19161
I don't know what those are called.
 
user19161
1:46 AM
Oh, I was talking about the muscles.
 
user19161
For the math also no term.
 
peoplepower
peoplepower
Yes, I did not know I was referencing something until I looked it up.
 
Ethan
Ethan
2:27 AM
hello
 
anorton
anorton
@JasonBourne Sorry--Just got back online. Yes, I do know your real name (if you had your real name as your username a while back)
 
Rustyn Yazdanpour
Rustyn Yazdanpour
 
BenjaLim
BenjaLim
2:42 AM
@Sanchez Hey
are you around?
 
Sanchez
Sanchez
2:52 AM
@BenjaLim, hi
 
BenjaLim
BenjaLim
@Sanchez oh thank god you're here
Consider Qiaochu's answer here @Sanchez
How did he determine the factorisation of $2$ in $mathcal{O}_L$?
$L = \Bbb{Q}(\sqrt{2},\sqrt{-13})$
and looking at his second answer below, $\alpha = (\sqrt{2} + \sqrt{-26})/2$
 
Sanchez
Sanchez
What do you not understand about his derivation? @BenjaLim
 
BenjaLim
BenjaLim
@Sanchez how did he determine the splitting of $2$ in $\mathcal{O}_L$?
@MarianoSuárez-Alvarez Hi.
 
Sanchez
Sanchez
@BenjaLim, $(2) = (\sqrt{2})^2$
 
BenjaLim
BenjaLim
@Sanchez yes
 
Sanchez
Sanchez
2:59 AM
So it suffices to factorize $(\sqrt{2})$
 
BenjaLim
BenjaLim
but the thing I don't understand is
 
Sanchez
Sanchez
To do that, look at $O_L/(\sqrt{2})$
 
BenjaLim
BenjaLim
I get that $\mathcal{O}_L / (\sqrt{2}) \cong \Bbb{F}_2[\alpha]$
yes
 
Sanchez
Sanchez
That's unlikely, what makes you think so?
Don't forget that $\alpha$ and $\sqrt{2}$ may not be linearly independent
 
BenjaLim
BenjaLim
@Sanchez Hmm that is correct
@Sanchez Ok we know that $\mathcal{O}_L = \Bbb{Z}[\sqrt{2},\alpha]$
So we are trying to identify
$\Bbb{Z}[\sqrt{2},\alpha]/(\sqrt{2})
@Sanchez This seems to me like we are doing
 
Sanchez
Sanchez
3:02 AM
yes
 
BenjaLim
BenjaLim
$\Bbb{Z}[\alpha][x]/(x^2 - 2) / (x,x^2 - 2)/(x^2 - 2)$ @Sanchez
Which by the third isomorphism theorem should give me $\Bbb{Z}[\alpha][x]/(x,x^2 - 2)$
 
Sanchez
Sanchez
ok, let me see
 
BenjaLim
BenjaLim
@Sanchez and further $(x,x^2 - 2) = (x,2)$
 
Sanchez
Sanchez
Yes
 
BenjaLim
BenjaLim
So $\mathcal{O}_L \cong \Bbb{Z}[\alpha][x]/(x,2)$
 
Sanchez
Sanchez
3:06 AM
Yes
 
BenjaLim
BenjaLim
which is $\Bbb{Z}[\alpha]/(2)$
 
Sanchez
Sanchez
no
 
BenjaLim
BenjaLim
why not?
 
Sanchez
Sanchez
$\alpha$ and $x$ are not independent here
 
BenjaLim
BenjaLim
$x$ is just an indeterminate?
 
Sanchez
Sanchez
3:07 AM
hm, ok, go on first
 
BenjaLim
BenjaLim
@Sanchez Even Qiaochu says it here
 
Sanchez
Sanchez
go on.
 
BenjaLim
BenjaLim
But if it is true that $\mathcal{O}_L/(\sqrt{2}) \cong \Bbb{F}_2[\alpha]$
how does he then say that $\mathcal{O}_L/(\sqrt{2}) \cong \Bbb{F}_2[\alpha]/(\alpha-1)^2$?
 
Sanchez
Sanchez
Your $\alpha$ is not a free variable
 
BenjaLim
BenjaLim
yes.
 
Sanchez
Sanchez
3:11 AM
$\alpha^2 = -6 + \sqrt{-13}$
 
BenjaLim
BenjaLim
yes that is correct
 
Sanchez
Sanchez
wait
 
Ethan
Ethan
@Sanchez How much complex analysis do I need to learn about Dirichlet characters?
 
Sanchez
Sanchez
@Ethan, zero
it is an algebraic concept.
 
BenjaLim
BenjaLim
@Ethan could you please not be so rude as to interrupt and on going conversation?
 
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
3:12 AM
that approach to learning, akin to an accountant doing book-keeping, is very weird
just study complex analsys
 
Ethan
Ethan
@Sanchez Are you sure it looks like I need to know about roots of unity
 
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
you will no tget very far without working knowledge of complex analsysis
roots of unity, on the other hand, are quite unrelated
 
Ethan
Ethan
@mariano Whats 'working knowladge'
Where can I pick up that kind of knowledge?
 
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
what you get when you study it
 
BenjaLim
BenjaLim
when you study more foundational material.
 
anon
anon
3:14 AM
roots of unity are more or less an elementary number theory topic, you don't need to know anything about complex derivatives or contour integration or Riemann surfaces etc. for that (i.e. "complex analysis")
 
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
but dirichlet characters show up in analytic number theory
it is a canonical example of complex analsysis put to use
 
BenjaLim
BenjaLim
@Sanchez yes?
 
Mariano Suárez-Alvarez
Mariano Suárez-Alvarez
question like «how much do I need to know about X to know Y?» are silly
just study X and Y
 
anon
anon
thinking one needs complex analysis to work with roots of unity is like thinking one needs vector calculus to add vectors
 
Sanchez
Sanchez
@BenjaLim, I find it easier to do the calculation by myself once, so here's what I would do.
Consider $\mathbb{Z}[\alpha] \to \mathbb{Z}[\sqrt{2},\alpha]/(\sqrt{2})$
 
BenjaLim
BenjaLim
3:16 AM
yes
 
Sanchez
Sanchez
and look at the kernel
2 lies in the kernel
 
BenjaLim
BenjaLim
that is right.
the kernel I think is $(\sqrt{2}) \cap \Bbb{Z}[\alpha]$
 
Sanchez
Sanchez
and by $2\alpha - \sqrt{2} = \sqrt{-26}$ and square it
we also get $\alpha^2 + 7$ is in the kernel
i.e. $\alpha^2 - 1$ is in the kernel
 
BenjaLim
BenjaLim
Ah ok yes.
Right.
 
Sanchez
Sanchez
I think this is exactly the kernel, but let me see how to do it rigorously
 
BenjaLim
BenjaLim
3:23 AM
@Sanchez second isomorphism theorem
 
Sanchez
Sanchez
huh?
 
BenjaLim
BenjaLim
we can use it no?
If the kernel is $I$ say
 
Sanchez
Sanchez
Oh ok, so if $\alpha - 1 \in (\sqrt{2})$ then one can show that $\alpha \in $\mathbb{Q}(\sqrt{-2})$, absurd.
 
BenjaLim
BenjaLim
then $\Bbb{Z}[\alpha]/\left( (sqrt{2}) \cap \Bbb{Z}[\alpha] \right)\cong (\Bbb{Z}[\alpha] + I)/I$
 
Sanchez
Sanchez
so the kernel is exactly $(\alpha-1)^2$
 
BenjaLim
BenjaLim
3:25 AM
wait waa
 
Sanchez
Sanchez
Oh okay. I only said that $\alpha^2-1$ lies in kernel. I didn't show that its factor does not lie in it.
To be rigorous I need to show that $\alpha - 1$ does not lie in it.
 
BenjaLim
BenjaLim
right.
 
Sanchez
Sanchez
Yeah, so anyway, now you can see that the kernel is exactly $(2, (\alpha-1)^2)$
 
BenjaLim
BenjaLim
@Sanchez why is $(\alpha - 1)^2 = \alpha^ 2 - 1$?
 
Sanchez
Sanchez
sorry, it was a typo
That line should have been "so the kernel is exactly $(2,(\alpha-1)^2)$
It's clear that $(2,(\alpha-1)^2) = (2,\alpha^2-1)$
 
BenjaLim
BenjaLim
3:28 AM
ok.
@Sanchez ok before that one thing is
we determined that the kernel contains $2$ and $\alpha^2 - 1$
and hence contains $2$ and $(\alpha- 1)^2$ @Sanchez
 
Sanchez
Sanchez
Yes.
 
BenjaLim
BenjaLim
ok but how come if we show that $(\alpha -1)$ is not in the kernel
 
Sanchez
Sanchez
and $\mathbb{Z}[\alpha]/ker$ is isomorphic to $\mathbb{Z}[\sqrt{2},\alpha]/(\sqrt{2})$
If you look at the structure of $\mathbb{Z}[\alpha]/(2,(\alpha-1)^2) \cong \mathbb{F}_2[x]/(x-1)^2)$ has only three ideals, $(0)$, $(x-1)$, and the whole ring. Pulling back, it corresponds to $(2,(\alpha-1)^2)$, $(2,\alpha-1)$, and the whole ring. These are the only candidates for the kernel.
 
BenjaLim
BenjaLim
the map $\Bbb{Z}[\alpha] \to \Bbb{Z}[\sqrt{2},\alpha]$ may not be surjective.
 
Sanchez
Sanchez
No, but $\mathbb{Z}[\alpha] \to \mathbb{Z}[\sqrt{2},\alpha]/(\sqrt{2})$ is.
 
BenjaLim
BenjaLim
3:31 AM
uhhhh....
how come?
ok
I think I see it
all the polynomial expressions in $\Bbb{Z}[\alpha,\sqrt{2}]$
 
Sanchez
Sanchez
er, because any element in $\mathbb{Z}[\sqrt{2},\alpha]$ is of the form $a + b\alpha + c\sqrt{2} + d\sqrt{2}\alpha$ for $a,b,c,d \in \mathbb{Z}$
 
BenjaLim
BenjaLim
when we quotient out by $(\sqrt{2})$
 
Sanchez
Sanchez
Yes
 
BenjaLim
BenjaLim
the root 2 guys all get killed
@Sanchez still I don't get why the kernel is exactly what you wrote.
 
Sanchez
Sanchez
ok
 
user19161
3:36 AM
@anon What a lovely new picture you have!
 
Sanchez
Sanchez
So $\mathbb{Z}[\alpha] \to \mathbb{Z}[\sqrt{2},\alpha]/(\sqrt{2})$ is surjective, with kernel containing $(2, (\alpha-1)^2)$
 
BenjaLim
BenjaLim
yes.
 
Sanchez
Sanchez
So we study the ring $\mathbb{Z}[\alpha]/(2,(\alpha-1)^2)$ - the kernel would map to an ideal of this quotient by correspondence theorem.
 
BenjaLim
BenjaLim
yes that is true.
ah ok I see waht you mean now.
 
Sanchez
Sanchez
But this ring is isomorphic to $\mathbb{F}_2[x]/(x-1)^2$ which has only three ideals, $(0), (x-1)$ and the whole ring.
Pull back, they are $(2,(\alpha-1)^2)$, $(2,\alpha-1)$, and the whole ring.
 
BenjaLim
BenjaLim
3:38 AM
right. got it.
 
Sanchez
Sanchez
Cool.
 
BenjaLim
BenjaLim
sometimes I get confused when there are so many quotients flying around.
right so indeed we have identified the kernel with $(2,(\alpha-1)^2) = (2,\alpha - 1)^2$
 
Sanchez
Sanchez
Yes
 
BenjaLim
BenjaLim
ok.
so applying the correspondence theorem again
@Sanchez hmm I think there is a problem
we can't conclude that the factorisation of $\sqrt{2}$ from here is $((\alpha-1)^2,2)$
 
Sanchez
Sanchez
OK
 
BenjaLim
BenjaLim
3:42 AM
the problem is that $\sqrt{2}$ is not an element of the ring $\Bbb{Z}[\alpha]$
 
Sanchez
Sanchez
so, you were looking at $O_L/(sqrt{2})$
 
BenjaLim
BenjaLim
right
 
Sanchez
Sanchez
Primes that divide $\sqrt{2}$ contain $\sqrt{2}$
so a prime lying over $\sqrt{2}$ corresopnds to a prime ideal of that quotient
 
BenjaLim
BenjaLim
yes.
what is a prime ideal?
 
Sanchez
Sanchez
but now you know $O_L/(\sqrt{2})$ is just $\mathbb{Z}[\alpha]/(2,(\alpha-1)^2)$
and the only prime in this quotient is $(2,(\alpha-1)$
 
BenjaLim
BenjaLim
3:43 AM
ok and the primes in here are just $(2, (x-1)^2)$ and $(2,x-1)$
 
Sanchez
Sanchez
No, $(2,(x-1)^2)$ is not prime.
 
BenjaLim
BenjaLim
wait it is not an integral domain?
 
Sanchez
Sanchez
it's not.
 
BenjaLim
BenjaLim
ah ok wtf
$(\alpha - 1) (\alpha - 1)= 0$
 
Sanchez
Sanchez
If the quotient is a domain, then $\sqrt{2}$ is a prime to start with.
 
BenjaLim
BenjaLim
3:44 AM
facepalm.......
 
Sanchez
Sanchez
Pulling back to $O_L$, the only prime that contains $\sqrt{2}$ is $(\sqrt{2},\alpha-1)$
 
BenjaLim
BenjaLim
@Sanchez yes.
 
Sanchez
Sanchez
and that $\sqrt{2} = (\sqrt{2},\alpha-1)^2$
 
BenjaLim
BenjaLim
@Sanchez and then we apply ramification theory from there to get the inertia degrees and blah.
 
Sanchez
Sanchez
No need.
You already know what the quotient is
 
BenjaLim
BenjaLim
3:46 AM
ah $2$ is totally ramified in the biquadratic extension.
 
Sanchez
Sanchez
inertia degree should be 1.
 
BenjaLim
BenjaLim
yes.
 
Sanchez
Sanchez
Yes
 
BenjaLim
BenjaLim
ok let me summarise again @Sanchez
we have determined that $\mathcal{O}_L /(\sqrt{2}) \cong \Bbb{Z}[\alpha]/(2, (\alpha - 1)^2 )$
any prime containing $\sqrt{2}$
would be a prime of $\Bbb{Z}[\alpha]/(2,(\alpha-1)^2)$
However this last ring is $\Bbb{Z}[x]/(2, (x-1)^2, f(x))$ where $f$ is the minimal polynomial of $\alpha$.
$f = x^4 +12x^2 + 49$
 
Sanchez
Sanchez
Why?
 
BenjaLim
BenjaLim
3:50 AM
I just used the third isomorphism theorem.
 
Sanchez
Sanchez
Hmm okay
$f$ is probably redundant, but yeah.
 
BenjaLim
BenjaLim
@Sanchez what I want to say from here is that $(2, (x-1)^2,f(x)) = (2,(x-1)^2)$
 
Sanchez
Sanchez
Yes.
 
BenjaLim
BenjaLim
wait I'm doing long division
@Sanchez Macaulay2 says they are equal :D
o1 = R

o1 : PolynomialRing

i2 : I = ideal (2,(x-1)^2,x^4 + 12*x^2 + 49)

2 4 2
o2 = ideal (2, x - 2x + 1, x + 12x + 49)

o2 : Ideal of R

i3 : J = ideal (2,(x-1)^2)

2
o3 = ideal (2, x - 2x + 1)

o3 : Ideal of R

i4 : I == J

o4 = true

i5 :
 
Sanchez
Sanchez
Well they are definitely equal
 
BenjaLim
BenjaLim
3:56 AM
ok
 
Sanchez
Sanchez
$x^4+12x^2+49 \equiv x^4-1$ mod 2.
Actually $\equiv (x-1)^4$ mod 2.
 
BenjaLim
BenjaLim
oh wtf
of course
yes
ok @Sanchez so $\Bbb{Z}[\alpha]/(2,(\alpha - 1)^2) \cong \Bbb{Z}_2[x]/(x-1)^2$
 
Sanchez
Sanchez
yes
 
BenjaLim
BenjaLim
and the primes in here are just $(x-1)$
 
Sanchez
Sanchez
Yes
 
BenjaLim
BenjaLim
3:59 AM
pulling back to $\Bbb{Z}[\alpha]/(2, (\alpha - 1)^2)$
that is just $(2,\alpha - 1)$
and so the only prime containing $\sqrt{2}$ in $\Bbb{Z}[\alpha,\sqrt{2}]$ is
 
Ethan
Ethan
@anon
 
anon
anon
yes
 
Ethan
Ethan
of the n roots of unity $\phi(n)$ of them are primitive right?
 
BenjaLim
BenjaLim
@Ethan that is the definition.
 
Sanchez
Sanchez
@BenjaLim, yes.
 
anon
anon
4:02 AM
more or less
 
Ethan
Ethan
how is that the definition?
if some of the integers 0,1,2,...n-1 were not coprime to n
then they wouldn't be nth roots of unity
that seems to follow from the definition
of an nth root of unity
 
user19161
Hey @math I am going to sleep now.
 
anon
anon
the nth roots of unity form a cyclic group of order n under multiplication, a root of unity is primitive iff it generates said group, and an element of Z/nZ is a generator iff its representative is coprime to n, of which there are phi(n) such numbers.
 
math101
math101
@JasonBourne My head was killing so I decided to take a break and watch a movie
 
BenjaLim
BenjaLim
@Sanchez hmmm why does $(2,\alpha - 1)$ become $(\sqrt{2},\alpha -1 )$
in $\Bbb{Z}[\sqrt{2},\alpha - 1]$?
 
math101
math101
4:04 AM
Stupid decision
 
Ethan
Ethan
What does it mean to be say the 'distinct primitive nth roots of unity' arn't all primitive nth roots of unity distinct?
 
user19161
@math101 Ah, did you watch my movie, the Bourne trilogy?
 
Ethan
Ethan
wikipedia says distinct?
 
Sanchez
Sanchez
@BenjaLim, you are pulling back to $O_L/(\sqrt{2})$ first
Therefore when you pullback to $O_L$, you add a $\sqrt{2}$
 
anon
anon
@Ethan (a) where, (b) why would one use the singular "a" to refer to the plural "roots"?
 
math101
math101
4:05 AM
@JasonBourne no lol
 
Ethan
Ethan
my bad
 
BenjaLim
BenjaLim
@Sanchez ?
 
user19161
@math101 How about my other movie, Good Will Hunting? Have you watched it before?
 
BenjaLim
BenjaLim
hold on I am replying to keith conrad on main.
 
math101
math101
@JasonBourne I guess I gotta start watching the right movies :)
 
user19161
4:06 AM
@anon I looked up aeria gloris and it means heavenly glory, seems to be anime related.
 
anon
anon
saw that coming an hour ago
 
user19161
@math101 Yes, then you will see me. =)
 
Sanchez
Sanchez
@BenjaLim, I'm saying that when I say $(2,(\alpha-1)^2)$, it's really an ideal of $O_L/(\sqrt{2})$. When you pull this back to $O_L$, it's the ideal $(\sqrt{2},2,(\alpha-1)^2) = (\sqrt{2},(\alpha-1)^2)$
 
math101
math101
@JasonBourne I am gonna watch Good Will Hunting next :)
 
BenjaLim
BenjaLim
@Sanchez ok. I think I get it now.
@Sanchez thanks :D
 
user19161
4:08 AM
@math101 Anyway, good night! Yawn...
 
math101
math101
Gnight :)
 
Sanchez
Sanchez
@BenjaLim, sure :)
 
BenjaLim
BenjaLim
@Sanchez whoops I misread the $4$ in the subscript as a superscript.
 
Ethan
Ethan
@anon Would it make sense to say a kth root of unity is a primitive nth root of unity iff k and n are coprime?
 
anon
anon
no
 
Ethan
Ethan
4:11 AM
why?
doesn't k and n being coprime imply its a primitive nth root of unity
 
anon
anon
it's not true. like, ever.
 
Ethan
Ethan
because if they had a common factor
it would cancle in the fraction ontop of the exponential
and it would be some smaller primitive ath root of unity
where a is the denominator in reduced form of that fraction
 
BenjaLim
BenjaLim
@Sanchez hmm actually determining the prime decomposition of $2$ was a lot simpler than I thought in my question
@Sanchez But I learnt a lot from the discussion with you.
 
Sanchez
Sanchez
@BenjaLim, I see.
That's nice :)
 
anon
anon
@Ethan: Suppose x is both a kth and nth root of unity where (k,n)=1. Then by Bezouts there exist integers r,s such that rk+ns=1. Hence x=x^(rk+ns)=(x^k)^r*(x^n)^s=1^r*1^s=1, i.e. x=1, which is not a primitive root of unity.
 
peoplepower
peoplepower
4:14 AM
@Ethan Perhaps you mean, "the kth power of a primitive nth root of unity is a primitive nth root of unity iff k and n are coprime".
 
anon
anon
Yes, that is almost surely what he means.
 
Ethan
Ethan
wait I thought they were $e^(2pik/n)$ for k=0,1,2,3,..n-1
I exclude 0?
 
anon
anon
yes, those are the nth roots of unity
you exclude k=0 (and k= anything else not coprime to n) when you want primitive roots, yes
 
Ethan
Ethan
thats what i thought i said
if the k and n in the fraction arn't coprime
 
anon
anon
no, you talked about something being an nth root and a $k$-th root
 
Ethan
Ethan
4:16 AM
doesn't that imply its not a primitive nth root of unity
 
anon
anon
yes, it does imply that
 
BenjaLim
BenjaLim
@Sanchez I should go now to work alone. Thanks again. I have learned very much from you.
 
Sanchez
Sanchez
Sure. I also learn from discussion with you too :)
 
BenjaLim
BenjaLim
@Sanchez WTF
I am so stupid
I forgot we can write $\Bbb{Q}(\sqrt{2},\sqrt{-3}) = \Bbb{Q}(\sqrt{-3},\zeta_3)$
WTF
 
peoplepower
peoplepower
@BenjaLim I must be a moron then.
 
BenjaLim
BenjaLim
4:27 AM
@peoplepower because if we write it like this then the problem is trivial
 
anon
anon
 
Ethan
Ethan
Why is the sum of the primitive nth roots of unity equal to $\mu(n)$ anon?
 
anon
anon
magic
actually I don't think I know why offhand
 
BenjaLim
BenjaLim
@Sanchez I posted an answer to my latest question.
 
anon
anon
ah, mobius inversion
Let $s(n)$ denote the sum of primitive $n$th roots of unity. Define $\ell(n)=\sum_{d\mid n}s(d)=\sum_{k=0}^{n-1}\zeta_n^k=\delta_n$. By mobius inversion, $s(n)=\sum_{d\mid n}\mu(d)\ell(n/d)=\mu(n)$.
you might be interested in ramanujan sums, as well
 
Ethan
Ethan
4:40 AM
how did you get the first equality
the divisors are equal to that sum?
$\sum_{d \mid n}s(d)=\sum_{k=0}^{n-1}...$
 
anon
anon
every nth root is a primitive dth root for exactly one d|n (namely d is the order of n)
so the nth roots and the disjoint union of primitive dth roots for divisors d|n are in bijective correspondence
 
Ethan
Ethan
you lost me at disjoint
 
anon
anon
disjoint just means the sets being union'd don't share any elements, i.e. a primitive nth root is not equal to a primitive mth root when n,m are distinct
 
Ethan
Ethan
o
oh I see if d divides n, then its an (n/d)th root of unity
 
anon
anon
err, namely d is the order of n should read namely d is the order of the root
 
Ethan
Ethan
4:45 AM
what do you mean by order?
 
anon
anon
in a group, the order of an element is the least nonnegative integer such that applying the element to itself that many times gives you the identity (which here means the least nonnegative d such that x^d=1)
 
Ethan
Ethan
thanks
 
deezy
deezy
5:37 AM
A function f(t, y) is said to satisfy a Lipschitz condition in the variable y on a set $D \subset \mathbb R^2$ if ... (etc).

Does the $R^2$ mean a two dimensional space of real numbers?
 
anon
anon
yes
2
 
BenjaLim
BenjaLim
@WillJagy hey
@WillJagy Keith Conrad is very pro.
 
peoplepower
peoplepower
Is this free star day?
 
Poseidonium
Poseidonium
6:04 AM
what is free star day?
2
 
peoplepower
peoplepower
6:32 AM
Wow, I was being a moron when I tried learning category theory before learning any math. My notes from then comprises a conglomerate of blunders.
 
 
1 hour later…
Gustavo Bandeira
Gustavo Bandeira
7:51 AM
@peoplepower Really?
I thought that it could be enlightening to try to learn it before learning math.
 
peoplepower
peoplepower
@GustavoBandeira I was not looking for enlightenment, I just thought it was possible.
 
Gustavo Bandeira
Gustavo Bandeira
8:06 AM
@peoplepower I'm aware that category theory is used in some kind of descriptive order for some mathematical objects, I was told that CT deals with complicated mathematical objects and the study of these objects was needed for CT.
 
peoplepower
peoplepower
@GustavoBandeira One idea that CT focuses on is rigorously converting structure-preserving maps via functors. So there is a functor between the category of finite-dimensional vector-spaces linked by linear maps and the category of integers with $m\times n$ matrices being the maps.
In general, it is the properties of an object determined by structure-preserving maps into and out of it that is of interest rather than its internal structure.
But I am just a beginner, and this is what I have picked up so far.
 
 
3 hours later…
user19161
10:57 AM
Free star day or star free day?
 
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