Abstract
Given a nonzero integer , we know by Hermite's Theorem that there exist only finitely many cubic number fields of discriminant . However, it can happen that two non-isomorphic cubic fields have the same discriminant. It is thus natural to ask whether there are natural refinements of the discriminant which completely determine the isomorphism class of the cubic field. Here we consider the trace form as such a refinement. For a cubic field of fundamental discriminant we show the existence of an element in Bhargava's class group $\Cl(\mathbb{Z}^{2}\otimes\mathbb{Z}^{2}\otimes\mathbb{Z}^{2}; -3d)$ such that is completely determined by . By using one of Bhargava's composition laws, we show that is a complete invariant whenever is totally real and of fundamental discriminant
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