Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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 $\rm\:Q\:.$

share|cite|improve this answer
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. – PumaDAce Dec 22 '11 at 22:04
@PumaDAce No, that's false by hypothesis (take the subset to be S itself). – Bill Dubuque Dec 22 '11 at 22:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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