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Number fields with the same zeta function are said to be arithmetically equivalent. Arithmetically equivalent fields share much of the same properties; for example, they have the same degrees, discriminants, number of both real and complex valuations, and prime decomposition laws (over Q). They also have isomorphic unit groups and determine the same normal closure over Q[7]. Strangely enough, it has been shown (for example [4], or more recently [6] and [7]) that this does not imply that arithmetically equivalent fields are isomorphic. Just recently [3] B. De Smit and R. Perlis showed that arithmetically equivalent fields do not even necessarily have the same class number. In this short note we take this recent result of De Smit and Perlis, and a well-known fact from algebraic number theory and use them to show that the integral normset (that is, the set of integral norms from a ring of algebraic integers to Z) uniquely determines a larger class of extensions of Q than the splitting set does. (In this paper, we use the standard convention that “equality of splitting sets” means that the symmetric difference is finite. That is to say, if X and Y are sets of prime ideals, then we say that they are equal if their symmetric difference is finite). Good background information on the splitting set can be obtained in [1] or [5], and information on the normset can be found in [2].

A remark on arithmetic equivalence and the normset