HARBINGERS OF ARTIN’S RECIPROCITY LAW. I. THE CONTINUING STORY OF AUXILIARY PRIMES F. LEMMERMEYER Abstract. In this article we present the history of auxiliary primes used in proofs of reciprocity laws from the quadratic to Artin’s reciprocity law. We also show that the gap in Legendre’s proof can be closed with a simple application of Gauss’s genus theory. Introduction Artin’s reciprocity law, the central result in class field theory, is an isomorphism between some ideal class group and a Galois group, and sends the class of a prime ideal p to the Frobenius automorphism of p. In this series of articles we would like to discuss results that, in the long run, turned out to be related to certain pieces of the quite involved proof of Artin’s reciprocity law. In this article we sketch the history of auxiliary primes. These are prime num- bers whose existence was needed in various proofs of the quadratic as well as the higher reciprocity laws from Kummer to Artin. Subsequent articles will discuss the irreducibility of the cyclotomic equation, Gauss’s Lemma and the transfer map, and finally Bernstein’s reciprocity law, which is the special case of Artin’s reciprocity law for unramified abelian extensions stated by Bernstein long before Artin. Auxiliary primes in connection with proofs of the quadratic reciprocity law have a long history. Legendre’s attempt at proving the reciprocity law failed (see [20, Ch. 1]) because he could not guarantee the existence of certain primes. Gauss’s first proof via induction used auxiliary primes whose existence was secured with the help of an ingenious elementary argument. After Kummer had proved the p-th power reciprocity law in the fields of p-th roots of unity, where p is a regular prime, he gave a proof of the quadratic reciprocity law based on analogous principles which arXiv:1109.1228v1 [math.NT] 6 Sep 2011 also required auxiliary primes; their existence, Kummer claimed, would follow easily from the analytic techniques of Dirichlet. 1. Legendre Legendre attempted to prove the quadratic reciprocity law with the help of his theorem on the solvability of the equation ax2 + by2 = cz2, whose proof relied heavily on the technique of reduction provided by Lagrange: Theorem 1.1. Assume that the integers a,b,c are positive, pairwise coprime, and squarefree. Then the diophantine equation ax2 + by2 + cz2 = 0 (1.1) has non-trivial solutions in Z if the following three conditions are satisfied: bc, ac and ab are squares modulo a, b and c, respectively. − 1 2 HarbingersofArtin’sReciprocityLaw Let us now briefly explain Legendre’s idea for proving the quadrat reciprocity law −1 −1 p p · q q = ( 1) 2 s q − p for distinct odd prime numbers p. Like Gauss, Legendre distinguished eight cases; he denoted primes 1 mod 4 by a and A, and primes 3 mod 4 by b and B. The first of Legendre’s eight≡ cases then states that ≡ b a =1 = =1. a ⇒ b Legendre observes that the equation x2 +ay2 =bz2 is impossible modulo 4; thus the conditions in Thm. 1.1 cannot all be satisfied. But there are only two conditions, b one of which, namely ( a ) = +1, holds by assumption. Thus the second condition ( −a ) = +1 must fail; but then ( −a )= 1 implies ( a ) = +1 as desired. b b − b Next consider the equation Ax2 + ay2 = bz2; it is easily seen to be impossible −aA modulo 4, hence at least one of the three conditions in Thm. 1.1, ( b ) = +1, Ab ab ( a ) = +1, and ( A ) = +1, must fail. Now choose an auxiliary prime b such that b A a ( a )=1and( b )= 1. By what Legendre has already proved, this implies ( b )=1 b − −aA and ( A ) = 1. Then the first condition ( b ) = +1 always holds, and the last − A a two conditions are equivalent to ( a )=+1 and ( A )= 1. Since these cannot hold simultaneously, Legendre concludes that − a A =1 = =1, A ⇒ a as well as A a = 1 = = 1. a − ⇒ A − This result is proved under the assumption that the following lemma holds: Lemma 1.2. Given distinct primes a A 1 mod 4, there exists a prime b satisfying the conditions b 3 mod 4, ( b )=1≡ and≡ ( A )= 1. ≡ a b − As for the existence of these primes b, Legendre remarks On peut s’assurer qu’il y en a une infinit´e; mais voici une d´emon- stration directe qui ´ecarte toute difficult´e.1 He then gives a second proof of this claim, and this time admits Dans cette d´emonstration, nous avons suppos´eseulement qu’il y avois un nombre premier b de la form 4n 1, qui pouvoit diviser la formule x2 + Ay2.2 − In fact, an equation of the form x2 + Ay2 = abz2 is impossible in nonzero integers since x2 + y2 z2 mod 4 implies that x, y and z must be even. Thus the conditions in Legendre’s≡− Theorem cannot be satisfied. If we choose b prime such A b that ( b ) = 1, then by what we have already proved we know that ( A ) = 1. Since A is a− square modulo a and b, the last condition that ab be a square modulo− − a b a A cannot hold; thus ( A )( A )= 1, and this implies the claim ( A ) = +1. For future reference, let us state− Legendre’s assumption explicitly as 1It is possible to make sure that there are infinitely many such primes; but here is a direct proof that avoids any difficulty. 2In this proof we have only assumed that there be a prime number b of the form 4n − 1 which divides the form x2 + Ay2. September7,2011 FranzLemmermeyer 3 Lemma 1.3. For each prime p 1 mod 4 there exists a prime q 3 mod 4 such that (p/q)= 1. ≡ ≡ − Finally, for proving that b B =1 = = 1, B ⇒ b − Legendre assumes the following Lemma 1.4. For primes q r 3 mod 4 there exists a prime p 1 mod 4 such that (p/q) = (p/r)= 1. ≡ ≡ ≡ − Legendre returns to the problem of the existence of his auxiliary primes at the end of his memoir [18]: on p. 552, he writes Il seroit peut-ˆetre n´ecessaire de d´emontrer rigoureusement une chose que nous avons suppos´ee dans plusieurs endroits de cet article, savoir, qu’il y a une infinit´ede nombres premiers compris dans toute progression arithm´etique, dont le premier terme & la raison sont premier entr’eux . 3. He then sketches an approach to this result which he says is too long to be given in detail. This last remark is problematic in more than one way: Legendre’s Lemma 1.4 certainly follows from the theorem on primes in arithmetic progressions: take a quadratic nonresidues m mod q and n mod r, and use the Chinese Remainder The- orem4 to find an integer c with c 3 mod 4, c m mod q and c n mod r. Then any prime in the progression c +4≡qrk will satisfy≡ Lemma 1.4. ≡ On the other hand it is not clear at all whether Lemma 1.2 and Lemma 1.3 can be deduced from the theorem on primes in arithmetic progression without assuming the quadratic reciprocity law: this was already observed by Gauss in [11, p. 449]. In the second version of his proof in [19], Legendre managed to do with the existence result in Lemma 1.3 alone. 2. Gauss Gauss’s first proof of the quadratic reciprocity law used auxiliary primes that were quite similar to those used by Legendre. Below we will first sketch the main idea behind Gauss’s first proof and then use the techniques from his second proof (genus theory of binary quadratic forms) to prove Legendre’s Lemma 1.3. Gauss’s First Proof. The first proof of the quadratic reciprocity law found by Gauss was based on induction (or descent, as Fermat would have said). Assuming r that the reciprocity law holds for small primes, Gauss assumes that ( q ) = +1, where r and q 1 mod 4 are primea; then x2 r mod q shows that there is an equation of the≡ form e2 = r + fq. Choosing e ≡and f carefully and reducing the equation modulo p, Gauss is able to prove that (q/r) = +1. 3It is perhaps necessary to give a rigorous proof of something that we have assumed in sev- eral places of this article, namely, that there is an infinite number of primes contained in every arithmetic progression whose first term and ratio are coprime ... 4In Legendre’s times, this result was credited to Bachet; the name Chinese Remainder Theorem became common only much later. 4 HarbingersofArtin’sReciprocityLaw This procedure fails if (r/q)= 1 because in this case we do not have an equation like e2 = r + fq to work with. To− get around this problem, Gauss used an auxiliary prime: he found an elementary, but highly ingenious proof of the following Lemma 2.1. For every prime q 1 mod 4 there is a prime p < q with (q/p)= 1. ≡ − By what he already proved, Gauss concluded that (p/q)= 1 as well, hence pr is a quadratic residue modulo q; by carefully studying the equation− e2 = pr + fq and reducing it modulo various primes, Gauss finally proves that (q/r)= 1. The proof of Lemma 2.1 is trivial if one is allowed to use quadratic recipr− ocity: let a < q be an odd quadratic nonresidue modulo q; then there must be a prime number p a with (p/q) = 1.