If $D \equiv 0,1$ mod $4$ a nonzero integer, why is the map given by $\chi: (\mathbb Z / D\mathbb Z)^* \rightarrow \{-1,+1\}, \chi([p]) = (D/p)$ for odd primes $p$ not dividing $D$, a unique homomorhpism? And why is $\chi([-1]) = 1$ if $D > 0$, and $\chi([-1]) = -1$ if $D < 0$?

Why is it unique? And why is it only defined for odd primes $p$ not dividing $D$? (This has been answered)

Can anybody explain in details why it's a homomorphism? I though i had it... but i'm not sure. For the Legendre symbol it seems evident

Should this not follow from $(ab/p) = (a/p)(b/p)$?

How can i show that this homomorphism is well-defined? I'm pretty sure i need this: Why is the Jacobi symbol $(D/m) = (D/n)$ for certain $m,n,D$?

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    $\begingroup$ What do you mean by unique? $\endgroup$ – k.stm Jul 7 '13 at 19:37
  • $\begingroup$ That there is no other homomorphism which for which $\chi([p]) = (D/p)$, with $p$ odd and not dividing $D$, holds. $\endgroup$ – Egon Jul 7 '13 at 19:42
  • $\begingroup$ I wonder if this question is too elementary. If so, i apologize. $\endgroup$ – Egon Jul 7 '13 at 20:12
  • $\begingroup$ @Egon, there are no «too elementary questions» here. $\endgroup$ – Mariano Suárez-Álvarez Jul 7 '13 at 21:35
  • $\begingroup$ @Mariano: That's good to know :-) $\endgroup$ – Egon Jul 7 '13 at 21:49

For any element $[a]$ of $(\mathbb Z/D\mathbb Z)^\times$, there exists a prime $p$ with $p\equiv a\pmod D$ (in fact inifnitely many, according to Dirichlet). Hence necessarily $\chi([a])=\chi([p])=\left(\frac Dp\right)$. This shows uniqueness.

Considering primes dividing $D$ makes no sense, because they don't occur as elements of $(\mathbb Z/D\mathbb Z)^\times$. Or: The Jacobi symbol would be $0\notin\{\pm1\}$.

If $D>0$, why is $\chi([-1])=1$? Let $p$ be an odd prime with $p\equiv -1\pmod D$. If $D$ is odd and $>1$, then $$\left(\frac Dp\right)=(-1)^{\frac{D-1}2\frac{p-1}2}\left(\frac pD\right)=\left(\frac pD\right)=\left(\frac {-1}D\right)=(-1)^{\frac{D-1}2}=1.$$ If $D$ is even, you have to pull out a few factors $\left(\frac 2p\right)$ first; if $4|D$, but $8\not |D$, there are two such factors, hence contribute $(\pm1)^2=1$; otherwise $8|D$ and $p\equiv -1\pmod 8$ implies $\left(\frac2p\right)=1$.

Similar steps can be done to investigate $D<0$, but note that $|D|\equiv 0,3\pmod 4$: In the odd case, with $p\equiv -1\pmod{|D|}$ $$\begin{align}\left(\frac Dp\right)&=\left(\frac{-1}p\right)\left(\frac{|D|}p\right)\\&=(-1)^{\frac{p-1}2}\cdot(-1)^{\frac{|D|-1}2\frac{p-1}2}\left(\frac p{|D|}\right)\\& =\left(\frac{p}{|D|}\right)=\left(\frac{-1}{|D|}\right)=(-1)^{\frac{|D|-1}2}=-1.\end{align}$$

  • $\begingroup$ OK, i understand. Any ideas why $\chi([-1]) = 1$ if $D > 0$, and $\chi([-1]) = -1$ if $D < 0$? I will think about it some more, maybe i'm just too tired to see it :D $\endgroup$ – Egon Jul 7 '13 at 20:19
  • $\begingroup$ Could you please elaborate on the case $D < 0$? $\endgroup$ – Egon Jul 7 '13 at 21:19

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