arXiv 1104.45982011-04-24math.NT

与三次扩张相关的积分迹形式

Integral trace forms associated to cubic extensions

Guillermo Mantilla-Soler

备注：: This is a slightly different version from the published one(There are only format differences, and correction of two typos but the content is the same)
发表信息：: Algebra Number Theory 4 (2010), no. 6, 681-699

摘要：

给定一个非零整数$d$，由Hermite定理可知，判别式$d$的立方数域只存在1/2个.然而，可能发生的是，两个非同构的立方域具有相同的判别式。因此很自然地要问，是否存在判别式的自然精化，它完全决定了立方域的同构类。这里我们考虑迹形式$q_K:\text{tr}_{K/\mathbb{Q}}(x^2)|_{O^{0}_{K}}$作为这样的细化。对于基本判别式$d$的三次域，我们证明了Bhargava类群$\Cl(\mathbb{Z}^{2}\otimes\mathbb{Z}^{2}\otimes\mathbb{Z}^{2}; -3d)$中元素$T_K$的存在性，使得$q_K$完全由$T_K$决定.利用Bhargava的一个复合律，证明了当$K$是全实的且具有基本判别式时，$q_K$是完全不变量

英文摘要：

Given a nonzero integer $d$, we know by Hermite's Theorem that there exist only finitely many cubic number fields of discriminant $d$. 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 $q_K:\text{tr}_{K/\mathbb{Q}}(x^2)|_{O^{0}_{K}}$ as such a refinement. For a cubic field of fundamental discriminant $d$ we show the existence of an element $T_K$ in Bhargava's class group $\Cl(\mathbb{Z}^{2}\otimes\mathbb{Z}^{2}\otimes\mathbb{Z}^{2}; -3d)$ such that $q_K$ is completely determined by $T_K$. By using one of Bhargava's composition laws, we show that $q_K$ is a complete invariant whenever $K$ is totally real and of fundamental discriminant

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