RECIPROCITY BY RESULTANT IN k[t] PETE L. CLARK AND PAUL POLLACK Abstract. Let k be a perfect field with procyclic absolute Galois group and containing a primitive nth root of unity. We define a degree n power residue symbol a in the ring k[t], b n show that it is equal to \the character of the resultant Res(b; a)" and deduce a reciprocity law. We are motivated by commonalities between the classical case k = Fq and the novel but very simple case k = R. 1. Introduction 1.1. Quadratic reciprocity in a PID. In this paper we explore quadratic and higher reci- procity laws in the ring k[t] of polynomials over a suitable class of fields k. Here is a simple setup for pursuing abstract algebraic generalizations of quadratic reciprocity: let R be a PID. We say that a; b 2 R are coprime if a and b are nonzero and (a; b) = R. For a coprime a; p 2 R such that (p) is a prime ideal, we define the Legendre symbol p to be 1 if a is a square in the field R=(p) and −1 otherwise. For coprime a; b 2 R, let b = up1 ··· pr × for a unit u 2 R and prime elements p1; : : : ; pr 2 R. We define the Jacobi symbol r a Y a := : b p i=1 i a The value of b does not change if b is replaced with another generator of (b), but this value does in general depend on the chosen generator a of (a). We begin with the following two classical results. Theorem 1 (Quadratic Reciprocity in Z). a) (Gauss) Let p and q be distinct odd prime numbers. Then p p−1 q−1 q = (−1) 2 2 : q p b) (Jacobi) Let a and b be coprime odd positive integers. Then a a−1 b−1 b = (−1) 2 2 : b a Theorem 2 (Quadratic Reciprocity in Fq[t]). Let q be an odd prime power, and let a; b 2 Fq[t] be coprime monic polynomials. Then a q−1 deg a deg b b (1) = (−1) 2 : b a Dedekind stated Theorem 2, when q is prime, in [De57] but did not prove it: he felt that Gauss's fifth proof of Theorem 1 carried over with little change (\the deductions [. ] are so 1 2 PETE L. CLARK AND PAUL POLLACK similar to the ones in the cited treatise of Gauss that no one can escape finding the complete proof.") The first published proof is due to K¨uhne[K¨u02]. 1.2. A low-hanging quadratic reciprocity law. We now give a further simple, but moti- vational, quadratic reciprocity law. Theorem 3 (Quadratic Reciprocity in R[t]). Let a; b 2 R[t] be coprime monic polynomials. Then a b (2) = (−1)deg a deg b : b a In contrast to Theorems 1 and 2, direct calculation suffices to prove Theorem 3: an irreducible × polynomial in R[t] has degree at most 2. For A 2 R we put ( 1 if A > 0; sgn(A) := −1 if A < 0. Evaluation at A gives an isomorphism R[t]=(t − A) −!∼ R. For a 2 R[t] with a(A) 6= 0 we have a = sgn(a(A)) t − A and it follows that for a; b 2 R[t] with a(A)b(A) 6= 0 we have ab a b = sgn(a(A)b(A)) = sgn(a(A)) sgn(b(A)) = : t − A t − A t − A >0 Let Q 2 R[t] be monic irreducible quadratic. Then Q(R) ⊂ R , so for all A 2 R we have Q = sgn(Q(A)) = 1: t − A ∼ In R[t]=(Q) = C every element is a square, so for all a 2 R[t] with a; Q coprime we have a = 1; Q and thus certainly for a; b 2 R[t] with ab; Q coprime we have ab a b = : Q Q Q It follows that both sides of (2) are multiplicative in both a and b, so it suffices to verify the equation when a and b are irreducible. By the above considerations, both sides evaluate to 1 unless a = t − A and b = t − B for A 6= B 2 R, and then a(t)b(t) = sgn(B − A) sgn(A − B) = −1 = (−1)deg a deg b; b(t) a(t) completing the proof. q−1 Theorem 2 looks strikingly similar to Theorem 3. The only difference is that the (−1) 2 × ×2 over Fq is replaced by −1 over R. This can be understood as follows: we have [Fq : Fq ] = × ×2 2 = [R : R ], and thus in either field k we have a unique nontrivial quadratic character { i.e., a unique nontrivial group homomorphism χ: k× ! {±1g. Namely: × q−1 × χ: Fq 7! {±1g; a 7! a 2 ; χ: R 7! {±1g; a 7! sgn(a): RECIPROCITY BY RESULTANT IN k[t] 3 Thus if k is either Fq or R, then for coprime monic polynomials a; b 2 k[t] we have a b = χ(−1)deg a deg b : b a 1.3. Reciprocity by resultant. Here is another way to look at the proof of Theorem 3: for coprime monic a; b 2 R[t], let A (resp. B) be the \split part" of a (resp. b) { i.e., the largest monic divisor of a that has only real roots. Then the above considerations show a A = : b B If we write out A = (t − α1) ··· (t − αr);B = (t − β1) ··· (t − βs); then using the bimultiplicativity of Jacobi symbols established above, we get 0 1 A Y t − αi Y = = sgn @ (βj − αi)A = sgn Res(B; A); B t − βj 1≤1≤r; 1≤j≤s 1≤i≤r; 1≤j≤s where Res(B; A) 2 R[t] is the resultant of the polynomials B and A. This motivates us to examine the connection between Jacobi symbols and resultants for all coprime monic a; b 2 R[t]. Let α; β 2 C n R be such that α2 = fβ; βg and let A 2 R. Then we have t − A sgn Res((t − α)(t − α); t − A)) = sgn (α − A)(α − A) = 1 = ; (t − α)(t − α) (t − α)(t − α) sgn Res(t − A; (t − α)(t − α)) = sgn (A − α)(A − α) = 1 = ; t − A and sgn(Res((t − α)(t − α); (t − β)(t − β))) = sgn((α − β)(α − β)(α − β)(α − β)) (t − β)(t − β) = 1 = : (t − α)(t − α) Because Res(a; b) is also bimultiplicative, this establishes the following: Theorem 4. Let a; b 2 R[t] be coprime monic polynomials. Then a (3) = sgn Res(b; a): b For any field k and monic a; b 2 k[t], we have the (obvious!) primordial reciprocity law (4) Res(b; a) = (−1)deg a deg b Res(a; b): We observe that (3) and (4) immediately imply (2). It is natural to ask: does the analogous identity hold in Fq[t]? Indeed it does: Theorem 5. Let a; b 2 Fq[t] be coprime monic polynomials. Then a q−1 (5) = Res(b; a) 2 : b 4 PETE L. CLARK AND PAUL POLLACK We observe that (5) and (4) immediately imply (1). Ore gave a proof of Theorem 2 centered around (5) in 1934 [Or34]. Several years earlier, Schmidt had proved Theorem 2 by an equivalent approach [Sc27], but without drawing attention to the fact that the expressions appearing in his proof could be described as resultants. (Both authors treat not only quadratic reciprocity, but the higher reciprocity law described below in Theorem 6.) We believe that Ore's decision to make Theorem 5 explicit was a wise one; indeed, one of the main points of this note to is demonstrate that (5) is a harbinger of a more general phenomenon. Contemporary expositions (e.g. [R, Ch. 3]) seem to follow Schmidt rather than Ore, so that Theorem 5 is no longer well known. The present authors learned of Theorem 5 from a more recent paper of Hsu [Hs03], who seems to have independently rediscovered it. And now the plot thickens: already in 1902, K¨uhne gave a higher reciprocity law in Fq[t]. + 1 For this, let n 2 Z be such that n j q − 1: equivalently, Fq contains the nth roots of unity. × × ×n Let µn ⊂ Fq be the subgroup of nth roots of unity. Then [Fq : Fq ] = n, and the map × q−1 χn : Fq ! µn; a 7! a n × ×n induces an isomorphism Fq =Fq ! µn. Now for coprime a; p 2 Fq[t] with p irreducible, we define the nth power residue symbol a qdeg p−1 : n = χn;Fq[t]=(p)(a mod p) = a ; p n a this extends by bimultiplicativity to a symbol b n defined for all a; b 2 Fq[t] n f0g. Then we have the following result: Theorem 6. Let q be a prime power, and let n j q − 1 be a positive integer. Let a; b 2 Fq[t] be coprime monic polynomials. Then: a a) (Ore) b n = χn(Res(b; a)). q−1 a deg a deg bb n deg a deg bb b) (K¨uhne) b n = χn(−1) a n = (−1) a n. Again we observe that via the primordial law (4), Theorem 6a) implies Theorem 6b). 1.4. Statement of the Main Theorem. This brings us to a more precise goal: to generalize this \reciprocity by resultant" to k[t] for other fields k.