2
$\begingroup$

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 $(\mathbf{Q}^{\ast})^{2}$ ?

$\endgroup$
1
$\begingroup$

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 $\rm\:Q\:.$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.