By O.U. Schmidt

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8). 1 (A Cyclotomic Reciprocity Law) The prime p is a splitting modulus of Φn (x ) if and only if p ≡ 1 mod n. It turns out that not every polynomial with integer coeﬃcients satisﬁes a reciprocity law, but there is an elegant way to characterize the polynomials with rational coeﬃcients which do satisfy one. If f (x ) is a polynomial of degree n with coeﬃcients in Q then f (x ) has n complex roots, counted according to multiplicity, and these roots, together with Q, generate a subﬁeld of the complex numbers that we will denote by Kf .

2 that x 4 + 4x 2 + 2 and x 4 − 10x 2 + 4 satisfy a reciprocity law, but x 4 − 2 and x 5 − 4x + 2 do not. 2, and if you have an irreducible polynomial with integer coeﬃcients and an abelian Galois group, how do you ﬁnd the congruence conditions which determine its splitting moduli? The answers to these questions are far beyond the scope of what we will do in these lecture notes, because they make use of essentially all of the machinery of class ﬁeld theory over the rationals. We will not even attempt an explanation of what class ﬁeld theory is, except to say that it originated in a program to ﬁnd reciprocity laws which are similar in spirit to the reciprocity laws for polynomials that we have discussed here, but which are valid in much greater generality.

Suppose, on the contrary, that s t bk x k f (x ) = k =0 ck x k k =0 is a factorization of f in Z[x ] with bs = 0 = ct and s and t both less than n. Because a0 ≡ 0 mod q, a0 ≡ 0 mod q 2 and a0 = b0 c0 , one element of the set {b0 , c0 } is ≡ 0 mod q and the other is ≡ 0 mod q. Assume that b0 is the former element and c0 is the latter. As an ≡ 0 mod q and an = bs ct , it follows that bs ≡ 0 ≡ ct mod q. Let m be the smallest value of k such that ck ≡ 0 mod q. Then m > 0, hence m−i am = bj cm−j j =0 for some i ∈ [0, m − 1].