# A proof of quadratic reciprocity law.

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.

• math.stackexchange.com/questions/3056403/… Jan 4 at 23:24
• The third edition of the book (Primes of the Form $x^2+ny^2$) has a solution to this exercise. May 12 at 14:02

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.

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.