A proof of quadratic reciprocity law.

Question

I am reading Cox's Primes of the Form $x^2+ny^2$ and solving Exercise 1.13, which depends on Lemma 1.14.

Lemma 1.14. If $D\equiv 0, 1\pmod{4}$ is a nonzero integer, then there is a unique homomorphism $\chi:(\mathbb{Z}/D\mathbb{Z})^*\to \{\pm 1\}$ such that $\chi([p])=(D/p)$ for odd primes $p$ not dividing $D$. Furthermore, $$ \chi([-1])=\left\{ \begin{array}{ll} 1 & \text{when }D>0, \\ -1 & \text{when }D>0 \\ \end{array} \right. $$

Exercise 1.13. We will assume that Lemma 1.14 holds for all nonzero integers $D\equiv 0,1 \pmod{4}$, and we will prove quadratic reciprocity and the supplementary laws.

(a) Let $p$ and $q$ be distinct odd primes, and let $q^*=(-1)^{(q-1)/2}q$. By applying the lemma with $D=q^*$, show that $(q^*/\cdot)$ induces a homomorphism from $(\mathbb{Z}/q\mathbb{Z})^*$ to $\{\pm 1\}$. Since $(\cdot /q)$ can be regarded as a homomorphism between the same two groups and $(\mathbb{Z}/q\mathbb{Z})^*$ is cyclic, conclude that the two are equal.

My goal is proving that $(\cdot /q)$ and $(q^*/\cdot)$ are not trivial. That is, there exists $a, b\in (\mathbb{Z}/q\mathbb{Z})^*$ such that $(a/q)=(q^*/b)=-1$.

I have no problem with $(\cdot /q)$.

Suppose that $(\mathbb{Z}/q\mathbb{Z})^*=\langle g\rangle$. I am trying to show that $(q^*/g)=-1$.

I have proven the case $q\equiv 1\pmod{4}$, but my argument uses the Generalized Quadratic Reciprocity Law, which I am not supposed to use it because this exercise is a proof of that law. Here is my argument: since $(q/g)(g/q)=(-1)^{(q-1)/2 \cdot (g-1)/2}=1$, $(q/g)$ and $(g/q)$ have the same sign. Since $(g/q)=g^{(q-1)/2}=-1 \pmod{q}$, we have $(g/q)=-1$. I am stuck here.

Answer 1

When $q\equiv 3 \pmod 4, $ the lemma says $(q^*/\cdot)$ is not trivial; hence, $(q^*/\cdot)=(\cdot/q).$

Now we prove $(q/\cdot)$ is nontrivial when $q\equiv 1 \pmod 4.$ Note that it is enough to check on odd primes ( it is a homomorphism and we can write $-1\equiv 2q-1, 2\equiv q+2$ which are product of odd primes).

Suppose $(q/\cdot)$ is trivial. We know $(\cdot/q)$ is not trivial. Therefore we can find an odd prime $p$ such that $(q/p)=1,$ but $(p/q)=-1.$

If $p\equiv 3 \pmod 4,$ we have $(p^*/\cdot)=(\cdot/p)$ by the discussion above. Therefore, $$ (q/p)=(p^*/q)=(-1/q)(p/q)=(-1)^{(p-1)/2}(p/q)=-1. $$

If $p\equiv 1 \pmod 4$, then $(p/\cdot)$ is nontrivial by $(p/q)=-1.$ Therefore $(p/\cdot)=(\cdot /p).$ So $$ (q/p)=(p/q)=-1. $$

In both cases, we prove $(q/p)=-1$ which contradicts to the assumption. So we conclude $(q/\cdot)$ is nontrivial.

Answer 2

The first thing that comes to my mind is the following:

If $q\equiv3 (mod \ 4)$, then the lemma trivially implies that $\chi(-1)=-1$. The harder part is showing that $\chi([a])=-1$ for some $[a]\in(\mathbb{Z}/q\mathbb{Z})^*$.

Suppose for contradiction that it doesn't and $\chi$ is trivial. Then $\left(\frac{q}{p}\right)=1$ for all primes $p$. By local-global principle this implies that $x^2-q$ is solvable over $\mathbb{Q}$, which is certainly false.

Apologies for such an overkill, a simpler solution does not appear to me.