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.

$\endgroup$
  • 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 '18 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 '18 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 '18 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 '18 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 '18 at 18:45
0
$\begingroup$

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

$\endgroup$

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.