I'm reading Serre's A course in Arithmetic and I have the following question about Proposition 14 in Chapter VI. It uses this version of Dirichlet's theorem: Let $m\ge 1$, $(a,m) = 1$. Let $P_a$ be the set of prime numbers such that $p\equiv a \mod m$. The set $P_a$ has density $1/\phi(m)$.
Proposition 14 Let $a$ be an integer which is not a square. The set of prime numbers $p$ such that $\bigl(\frac{a}{p}\bigr) = 1$ has density $\frac{1}{2}$.
His proof goes like this: WLOG $a$ is square free. Let $m=4|a|$ and $\chi_a$ be the unique character $\mathbb{Z}/m\mathbb{Z}^\times$ such that $\chi_a(p) = \bigl(\frac{a}{p}\bigr)$ for all prime numbers $p$ not dividing $m$. Note that $\bigl(\frac{a}{p}\bigr) = 1$ iff $p \in \ker\chi_a$. Using Dirichlet's theorem, $[\mathbb{Z}/m\mathbb{Z}^\times: \ker \chi_a]$ is equal to the density of the primes satisfying this condition. $\square$
I have three questions about this:
- Why can we assume a is square free?
- How do we apply Dirichlet? In particular, we selected $m = 4|a|$ and $m$ is not coprime with $a$, so I'm not sure how the theorem is applied.
- How is the index equal to the density of the primes satisifying the condition? (I don't quite understand the last sentence of the proof)
Why can you take $a$ squarefree: just look at an example, say $a = 45 = 3^2 \cdot 5$. The value of $(\frac{45}{p})$ equals the value of $(\frac{5}{p})$ as long as $p \not= 3$.
Since $\chi_a$ is a nontrivial quadratic character, it takes two values and those values are given by congruence conditions: the kernel of $\chi_a$ is a subgroup of $(\mathbf Z/m\mathbf Z)^\times$ with index 2, so the kernel has size $\varphi(m)/2$. The primes in each unit class has density $1/\varphi(m)$, so the primes that reduce mod $m$ to a congruence class in $\ker \chi_a$ have density $(\varphi(m)/2))/\varphi(m) = 1/2$.