I'm now painfully studying abstract algebra. I want to ask two minor questions to clarify my concepts, though they would be somehow silly. We know that $\pi$ is transcendental over $\Bbb Q$, hence $\Bbb Q[\pi]$ is an integral domain, and $\Bbb Q[\pi]\subsetneq\Bbb Q(\pi)$. Then I wonder if there is a field $F^\star$ such that $\Bbb Q(\pi)\subsetneq F^\star\subsetneq \Bbb R$? And is there any relationship between $\Bbb Q(\pi)$ and say, $\Bbb Q(e)$? ($e$ is the Euler constant)

  • $\begingroup$ Since $\pi$ is transcendental you can add an irrational algebraic $\mathbb{Q}(\pi)(\sqrt{2})$ and be sure the field grew more. $\mathbb{Q}(\pi)$ and $\mathbb{Q}(e)$ are isomorphic by sending $\pi$ to $e$ and rationals to themselves. $\endgroup$ – orole Jan 15 '18 at 15:22
  • 4
    $\begingroup$ Not silly at all. These are the sorts of questions you should be asking yourself all the time. $\endgroup$ – fredgoodman Jan 15 '18 at 15:25
  • $\begingroup$ Are you familiar with cardinality? $\endgroup$ – Noah Schweber Jan 15 '18 at 15:35
  • $\begingroup$ @NoahSchweber Yes, I have learnt it before. $\endgroup$ – Eric Jan 15 '18 at 15:37
  • $\begingroup$ @Eric Well, what do you know about the cardinality of $\mathbb{Q}(\pi)$ versus $\mathbb{R}$? $\endgroup$ – Noah Schweber Jan 15 '18 at 15:40
  1. $\mathbb{Q}(\pi)\varsubsetneq\mathbb{Q}\bigl(\sqrt\pi\bigr)\varsubsetneq\mathbb R$.
  2. $\mathbb{Q}(\pi)\simeq\mathbb{Q}(e)$.
  • 1
    $\begingroup$ +1 First part of this answer gives more than what is asked. By iterating, that is repeatedly taking square roots, one can get an infinite tower of fields sandwiched between $\mathbf{Q}(\pi)$ and $\mathbf{R}$ $\endgroup$ – P Vanchinathan Jan 16 '18 at 2:45

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.