For $n\ge 3, x_{1},...,x_{n} \in \mathbf{Q}^{\ast}$ and $[\mathbf{Q}(\sqrt{x_{1}},...\sqrt{x_{n}}) : \mathbf{Q}] < 2^{n}$ how can we conclude that there are non empty $I \subset \{1,...,n\}$ with $\prod_{i\in I}x_{i}$ in $(\mathbf{Q}^{\ast})^{2}$ ?


HINT $\ $ See my proof here of the following result of Besicovic (Besicovitch).

THEOREM $\ $ Let $\rm\:Q\:$ be a field with $2 \ne 0\:,\:$ and $\rm\ L = Q(S)\ $ be an extension of $\rm\:Q\:$ generated by $\rm\: n\:$ square roots $\rm\ S = \{ \sqrt{a}, \sqrt{b},\ldots \}$ of elts $\rm\ a,\:b,\:\ldots \in Q\:.\:$ If every nonempty subset of $\rm\:S\:$ has product not in $\rm\:Q\:$ then each successive adjunction $\rm\ Q(\sqrt{a}),\ Q(\sqrt{a},\:\sqrt{b}),\:\ldots$ doubles the degree over $\rm\:Q\:,\:$ so, in total, $\rm\: [L:Q] \ =\ 2^n\:.\:$ Hence the $\rm\:2^n\:$ subproducts of the product of $\rm\:S\:$ comprise a basis of $\rm\:L\:$ over $\rm\:Q\:.$

  • $\begingroup$ Wow, this is really nice. The non-emptiness and basis seems clear, however, from your theorem wouldn't it conclude that $(\prod_{i\in I} x_{i})^{2}$ is in $(\mathbf{Q}^{\ast})^{2}$ ?? I appreciate your help. $\endgroup$ – PumaDAce Dec 22 '11 at 22:04
  • $\begingroup$ @PumaDAce No, that's false by hypothesis (take the subset to be S itself). $\endgroup$ – Bill Dubuque Dec 22 '11 at 22:12

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