tern and euclid - 2017-01-18

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euclid

16:39

@arctictern hi Tern:)

how are you?

i was waiting for you here these days

euclid

16:50

@arctictern please tern

euclid

17:17

and this is true?: $o(\chi)=d iff $\chi^d=\chi_0$

arctic tern

@euclid no

we've already talked about these things

for example, $\chi_0^d=\chi_0$ but $\chi_0$ does not have order $d$

@euclid originally I said n=g^k where g was a primitive root mod p, so k is an integer mod p-1, which k mod d only makes sense if d|(p-1). however, if we instead interpret k as an arbitrary integer from the get-go we can prove the multiplicativeness of ramanujan sums as I did in the answer.

euclid

@arctictern please tell me again for sure

arctic tern

@euclid I did just tell you again.

euclid

k is ind_p(n).true? why k is mod p-1?

arctic tern

also, I don't know where you're getting this from - maybe you have to do some kind of student research, maybe this is for class, whatever - but I don't have time to walk you through your entire journey. I help numerous people with isolated problems, not long term one-on-one things.

@euclid because g has order p-1 so g^k only depends on k mod p-1. I'm saying it's easier to just pretend k is a normal integer though.

euclid

17:25

@arctictern i dont want to disturb you sir.

but i am confused about order of a character now

i dont know other one for helping me

arctic tern

it's the same as talking about the order of a root of unity. consider for example 6th roots of unity as depicted in the complex plane. there are 6 in total. among them are 2 roots of order 6, 2 roots of order 3, one root of order 2, and one root of order 1

@euclid no worries

euclid

so what does mean $o(\chi)=d$ exactly? in a group we now order of a member is the smallest integer than $\chi^d=\chi_0$ as identity member of group

arctic tern

@euclid yes, o(X)=d means d is the smallest positive integer for which X^d=X_0

for example, consider -1. the fact that (-1)^6=1 does not mean -1 has order 6. instead, we go through the powers (-1)^1, (-1)^2, ... and the first time we see 1 we record that value of d as the order.

in other words, the statement "(-1)^6=1 iff -1 has order 6" is wrong, because (-1)^6=1 is true but "-1 has order 6" is false

euclid

@arctictern you are right. also we know that $\chi_0(n)=1$ for (n, p)=1. so why we cant say $\chi^d=1$?

arctic tern

we can say $\chi_0^d=\chi_0$, this is true

we cannot say $o(\chi)=d\iff \chi^d=1$, this is false

euclid

17:43

it means members of order d are subset of members that $\chi^d=1$?

arctic tern

a proper subset, yes

let's consider integers mod 6 under addition. they all satisfy 6x=0 (the additive version of X^d=1). but that doesn't mean they all have order 6. the elements 0,1,2,3,4,5 have order 1,6,3,2,3,6 respectively.

euclid

and the set of $\chi^d=1$ makes a cyclic group again?

arctic tern

yes, the set of characters $\chi$ satisfying $\chi^d=\chi_0$ (when the modulus $p$ is prime) is cyclic

euclid

you told that the set of $\chi^d=1$ has 6 members.true?

@arctictern in your answer about equality

arctic tern

@euclid I there are 6 sixth roots of unity above

@euclid what?

euclid

17:55

Q: a formula involving order of Dirichlet characters, $\mu(n)$ and $\varphi(n)$

Let $p$ a prime number, ${q_{_1}}$,..., ${q_{_r}}$ are the distinct primes dividing $p-1$, ${\mu}$ is the Möbius function, ${\varphi}$ is Euler's phi function, ${\chi}$ is Dirichlet character $\bmod{p}$ and ${o(\chi)}$ is the order of ${\chi}$. How can I show that: $$\sum\limits_{d|p - 1} {\f...

arctic tern

if you mean my comment on the other answer, then yes, that is where I said there are $d$ solutions to $\chi^d=\chi_0$

I also said this many times in chat with you I believe

euclid

@arctictern i meant your answer

arctic tern

not that it matters, but I don't see me saying how many solutions $\chi^d=\chi_0$ has anywhere in my answer

euclid

so what does mean $\psi^0,\psi^1....$

arctic tern

Please quote me correctly.

euclid

18:03

Let's show f is multiplicative. First off, let g be a generator for (Z/pZ)× and write n=gk, then let ψ be a generator for the group {χ:χd=1}, in which case we may say o(χ)=d⟺χ=ψe for a unit e mod d.

arctic tern

yes.

$\chi$ has order $d$ if and only if $\chi=\psi^e$ for some unit $e$ mod $d$

every $\chi$ is expressible as $\psi^r$ for some $r$ (since $\psi$ is a generator for the group of all solutions to $\chi^d=\chi_0$), and if $r$ shared any divisors with $d$ then $\psi^r$ would not have order $d$ (and conversely if $r$ shares no divisors with $d$ then $\psi^r$ does have order $d$)

anyway have to leave

euclid

ok.thank you very much @arctictern

so i have to change it in that answer?

arctic tern

huh?

my answer is correct

euclid

but you used X^d=X_0 instead of X^d=1

arctic tern

we were using 1 to refer to $\chi_0$. IIRC you started that convention and I rolled with it

it's sensible since $\chi_0$ is the identity element of the character group and $\chi_0(u)=1$ for all integer $u$ mod $p$