I have the following question:

I am reading Serre's book "A Course in Arithmetic" (see http://www.math.purdue.edu/~lipman/MA598/Serre-Course%20in%20Arithmetic.pdf).

On page 75, it is stated that the set $\{p\in \mathbb{P}\ \big|\ (\frac{a}{p})=1\}$ has Dirichlet density $\frac{1}{2}$, if $a$ is an integer which is not a square. Here, $(\frac{a}{p})$ denotes the Legendre symbol.

In the proof, I don't unserstand, why and how it follows from theorem 2 on page 73 that the set of prime numbers verifying the condition $\overline{p}\in H:=$ker$(\chi_a)\leq G(m)$ has for Dirichlet density $\mathbf{the\ inverse\ of\ the\ index\ of}$ $H$ $\mathbf{in}$ $G(m)$.

I would be grateful for any hints.

Thanks for the help!

  • $\begingroup$ Thank you for your comment. Unfortunately, I am still unable to see, why the Dirichlet density is the inverse of the index of $H$ in $G(m)$. Isn't the inequality $\mathscr{P}$ $(\{p\in\mathbb{P}\ |\ p\equiv 1\ \text{mod}\ m\})$ $=\frac{1}{\phi (m)}\neq \frac{1}{2}=\frac{1}{[G(m):H]}$ correct in general? $\endgroup$ Mar 29 '15 at 19:38
  • $\begingroup$ Note that $m=4a$ in the above proof, and the $m$ in Theorem $2$ is not the same $m$. The density is $1/[G(m):H]=1/\phi(4)=1/2$. $\endgroup$ Mar 29 '15 at 20:09
  • $\begingroup$ Thanks again for the help. Sorry, but I'm really at a loss...how does $\phi (4)$ come into play? Why don't we have to use $\phi (m)=\phi (4|a|)$? But we don't know much about $a$... $\endgroup$ Mar 29 '15 at 20:53
  • $\begingroup$ Because, as I said, the $m=4|a|$ is not the $m$ you have to use in Theorem $2$. $\endgroup$ Mar 29 '15 at 21:03
  • $\begingroup$ I'm sorry, but I still don't see, why and how it follows from thm. 2 that the Dirichlet density of $\{p \in \mathbb{P}\ |\ \overline{p}\in H\}$ is given by the inverse of the index of $H$ in $G(m)$. $\endgroup$ Mar 29 '15 at 23:08

$H$ is a subgroup of $G(m)=(\mathbb{Z}/m\mathbb{Z})^*$, which can be thought of as a collection of residue classes modulo $m$. If $H$ has size $k$, then $\bar p$ will be in $H$ iff $p$ is in one of these $k$ residue classes. Each residue class has Dirichlet density $1/\phi(m)$ by Theorem 2. Therefore, the union of these classes has Dirichlet density $k/\phi(m)$. Since $G(m)$ has size $\phi(m)$, this is the reciprocal of the index of $H$ in $G(m)$, which is $\phi(m)/k$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.