By Henry B. Mann.

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**Example text**

Let R' denote the integral closure of the Dedekind ring R in a finite-dimensional extension L of the quotient field K of R . Let p be a nonzero prime ideal of R . Suppose there is an element 6 such that the integral closure of R, in L is R,[B]. LetJ'(X) be the minimal polynomial of 0 over K . Letf(X) denote the polynomial obtained by reducing the coefficients off'(X) modulo p. Suppose f ( X ) = g,( X p. * . g , ( X Y is the factorization off(X) as a product of the distinct irreducible polynomials 33 7.

X,) = det IT(xixj)I is called the discriminant of the basis xI..... x,. If we select the xiin R', then x i x j is in R ' s o T ( x i x j )isin R . ,x, range over all possible bases of L / K which lie in R', the discriminants generate an ideal of R which we shall call the discriminant ideal of R' over R . We denote this ideal by A or A ( R ' / R ) . We begin the study of the discriminant by showing it can be determined by localization. 1 Lemma. Let S be a multiplicative set in R . Then A(Rs'/Rs) = A ( R ' / R ) s .

Let p i = q in R . I) = pyif'. Consider now the case with K = Q = rational field and R = Z = rational integers. L and R' have the same meanings as above. For any ideal 91 of R', N (91) is an ideal in Z which is necessarily a principal ideal, say N (91) = Zm = (m)for some integer m. If we require that m 2 0 then m is uniquely determined. Let us denote the integer m by N(91)so that the norm of an ideal 91 # 0 is now a positive integer. I. 6 Proposition. I) is equal to the number of elements in the ring R'/21.