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How to show that for a non-principal real Dirichlet character mod $k$, there are infinitely many primes $q$ such that $\chi(q)=1$ and infinitely many primes $q'$ such that $\chi(q')=-1$.

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Are you familiar with Dirichlet's Theorem on primes in arithmetic progression? – Gerry Myerson Dec 13 '11 at 4:50
Yes, what are you suggesting? – Rob Dec 13 '11 at 4:52
If a number $J$ is congruent to 1 modulo $k$, what do you reckon $\chi(J)$ is going to be? – Gerry Myerson Dec 13 '11 at 6:08
Since Bruno has given a complete answer, it seems worth expanding on Gerry Myerson's comments: if $\chi$ is non-trivial, there exists some $a$ mod $k$ such that $\chi(a) = -1$, and of course $\chi(1) = 1$. Dirichlet's theorem gives infinitely many $q$ congruent to $1$ mod $k$, and infinitely many $q'$ congruent to $a$ mod $k$. QED. – Matt E Dec 13 '11 at 6:13
Yes, it's much easier to prove this directly with Dirichlet's Theorem! It's always fun to go back to first principles though :) – Bruno Joyal Dec 13 '11 at 8:10
up vote 8 down vote accepted

As others have pointed out, this is easy to prove with Dirichlet's theorem. However, since you used the (analytic-number-theory) tag, here's a proof from first principles:

According to a the "main lemma" of Dirichlet, for any nontrivial character $\chi$, the $L$-function $L(s, \chi)$ is nonzero and holomorphic at $s=1$.

Hence this means $$0 <\prod_{p}(1-\chi(p)p^{-1})^{-1} < \infty$$ hence

$$-\infty < \sum_{p}\log(1-\chi(p)p^{-1}) = \sum_{p}\sum_{n\geq 1}\frac{\chi(p^n)}{np^n}< \infty.$$

Now the only non-negligible thing in this sum is the sum of the terms when $n=1$, i.e.




Now since $$\sum_{p}\frac{1}{p}$$ diverges, there must be infinitely many $p$ for which $\chi(p)=1$ and infinitely many $p$ for which $\chi(p)=-1$. (Note that there are only finitely many $p$ for which $\chi(p)=0$.) $\: \: \square$

This observation leads to the formulation of one of the great theorems of modern number theory, the Chebotarev density theorem.

Addendum: here is a plot of the partial sums $\displaystyle \sum_{p \leq p_n}\frac{\chi(p)}{p}$ for $\chi = \left(\frac{\cdot}{5}\right)$:

enter image description here

The behaviour is quite complex and gives an idea of the rather uniform distribution of $\chi(p)$. For comparison, here is an example of partial sums $\displaystyle \sum_{p \leq p_n}\frac{f(p)}{p}$ where $f$ takes random uniformly distributed values in $\{-1,1\}$:

enter image description here

And finally, here is a picture of a baby duck.

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LOL for the baby duck. +1 – Patrick Da Silva Dec 13 '11 at 23:26

By the way, one can prove in an utterly elementary way that there must be infinitely many primes $q$ with $\chi(q)=-1$. Given any set of primes $q_1,\dots,q_r$ with $\chi(q_i)=-1$, let $Q=kq_1\cdots q_{r-1}+q_r$. Then $\chi(Q)=\chi(q_r)=-1$, so there must be some prime $q$ dividing $Q$ with $\chi(q)=-1$. But $Q$ is not divisible by any of the primes $q_1,\dots,q_r$, so $q$ must be a new prime. This shows no finite list is complete, so there must be infinitely many.

Well, I've swept one little thing under the rug; I have to know that there is at least one prime $q_1$ with $\chi(q_1)=-1$. But that's easy; we're told $\chi$ is not the principal character, so there is some number $m$ with $\chi(m)=-1$, so some prime factor $q_1$ of $m$ has $\chi(q_1)=-1$.

No doubt there is a similar argument for $\chi(q)=1$, but I'm not seeing it.

EDIT: Here's the start of an argument for $\chi(q)=1$.

Write $\chi_n$ for the non-principal real Dirichlet character modulo $n$.

Given any finite set of primes $q_i$ with $\chi_k(q_i)=1$, multiply them all together, square the result, and add $k$; call the result $Q$. Then $Q$ is divisible by some new prime, call it $p$. We have $\chi_p(-k)=1$.

Now $\chi_p(-k)=\chi_p(-1)\chi_p(k)$. If $p\equiv1\pmod4$, then $\chi_p(-1)=1$, and, by quadratic reciprocity, $\chi_p(k)=\chi_k(p)$, so $\chi_k(p)=1$, and we are done - we have proved the finite list wasn't exhaustive.

If $p\equiv k\equiv3\pmod4$, then $\chi_p(-1)=-1$, and $\chi_p(k)=-\chi_k(p)$, so again $\chi_k(p)=1$, and we are done.

But I don't see the trick that will handle $p\equiv3\pmod4$, $k\equiv1\pmod4$.

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