1
$\begingroup$

I am trying to show that any non-square $a$ is a quadratic non-residue modulo an infinite number of primes, and this is my argument so far:

There is a well-defined group homomorphism $\chi : (\mathbb{Z}/4a\mathbb{Z})^{\times} \to \{\pm 1\}$ satisfying the relation $\chi(p) = \left(\frac{a}{p}\right)$ for all primes $p$ not dividing $4a$. This is because if $q = p + 4ak$ is also prime:

If $a$ is odd,

$\begin{align*} \chi(q) &= \left(\frac{a}{q}\right) = \left(\frac{q}{a}\right) (-1)^{\frac{q-1}{2}\frac{a-1}{2}} = \left(\frac{p + 4ak}{a}\right) (-1)^{\frac{p+4ak-1}{2}\frac{a-1}{2}}\\ &=\left(\frac{p}{a}\right) (-1)^{\frac{p-1}{2}\frac{a-1}{2}}\\ &= \left(\frac{a}{q}\right) = \chi(q), \end{align*}$

and if $a = 2b$ is even (so $q=p+8bk$),

$\begin{align*} \chi(q) &= \left(\frac{2}{q}\right)\left(\frac{b}{q}\right) = \left(\frac{2}{p+8bk}\right)\left(\frac{b}{p}\right) \text{ (as above)}\\ &= \left(\frac{2}{p}\right)\left(\frac{b}{p}\right) = \chi(p). \end{align*}$

Let us suppose that this homomorphism is not trivial. Then there exists a number $d \mod 4a$ such that $\chi(d) = -1$. If $d = p_1...p_k$ is a product of primes then $-1 = \chi(p_1)...\chi(p_k)$ and so there exists a class $p_i$ with $\chi(p_i) = \left(\frac{a}{p_i}\right) = -1.$

From here it is easy: the arithmetic progression $p_i + 4ak$ contains infinitely many primes by Dirichlet (clearly gcd$(p_i,4a) = 1$ by construction) and so there are infinitely many primes modulo which $a$ is a quadratic non-residue.

So, my question: Is it easy to show that $\chi$ is in fact non-trivial? If so, I would greatly appreciate a proof. Further, if I have made any mistakes in the rest of my proof I would like to know.

$\endgroup$
12
  • 1
    $\begingroup$ interesting comment in Cox that, if we demand $a$ odd and squarefree, then the kernel is a subgroup of index 2. in the first edition, page 18 books.google.com/… $\endgroup$
    – Will Jagy
    Jul 26, 2018 at 18:20
  • 1
    $\begingroup$ he says a proof is in Weil Number Theory: An Approach through History, which I have $\endgroup$
    – Will Jagy
    Jul 26, 2018 at 18:22
  • 1
    $\begingroup$ the same thing, really. the kernel is where the character $\chi$ takes value $1,$ the (only) other coset is where it is $-1.$ Once there is any element with the character equal to $-1,$ we get an entire coset, same cardinality as the kernel $\endgroup$
    – Will Jagy
    Jul 26, 2018 at 18:29
  • 1
    $\begingroup$ try $a$ odd prime and see what happens. I guess throw in $a = -p.$ Then try $a = pq$ I suspect the squarefree part is a help, although Weil's Appendix I is five pages 287-291, and the part you want is on the final page. I think the point may be that things are fairly concrete when $a$ is an odd prime. Put briefly, how do we know an odd prime has any quadratic nonresidues? Probably something about Fermat's little theorem $\endgroup$
    – Will Jagy
    Jul 26, 2018 at 18:41
  • 1
    $\begingroup$ found it, odd prime $p$ and target $b,$ there are two solutions to $x^2 \equiv b$ if any. Therefore the set of residues, the image of the squaring map, is no bigger than half $p$ $\endgroup$
    – Will Jagy
    Jul 26, 2018 at 18:45

1 Answer 1

-1
$\begingroup$

First, we can reduce to the case where $a$ is squarefree: let $a = s^2n$ for squarefree $n>1$ (since $a$ is not a square). Then for any prime $p > a$, $\left(\frac{a}{p}\right) = \left(\frac{s^2}{p}\right)\left(\frac{n}{p}\right) = \left(\frac{n}{p}\right)$.

So the list of primes satisfying $\left(\frac{n}{p}\right) = -1$ and the corresponding list for $\left(\frac{a}{p}\right) = -1$ are eventually identical.

So suppose $a = q_1q_2...q_k$ is squarefree. Choose $e_1,e_2,...,e_k \in \{\pm 1\}$ so that $\prod e_i = -1$, and note that for each $i$ the equation $\left(\frac{q_i}{p}\right) = e_i$ is implied by $p \equiv f_i \mod q_i$ for some $f_i$ (or if $q_i = 2$, $p \equiv f_i \mod 8$) by quadratic reciprocity.

By the Chinese Remainder Theorem these congruences have a solution $p \equiv f \mod 4a$, and hence $\left(\frac{a}{f}\right) = \prod \left(\frac{q_i}{f}\right) = -1$ as required.

Since $f$ doesn't divide any of the $q_i$ by construction, gcd$(f,a) = 1$, and so the rest of the proof in the question carries over.

If I've made any mistakes, don't hesitate to point them out. :D

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .