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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)

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  • $\begingroup$ 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
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    $\begingroup$ 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
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  1. $\mathbb{Q}(\pi)\varsubsetneq\mathbb{Q}\bigl(\sqrt\pi\bigr)\varsubsetneq\mathbb R$.
  2. $\mathbb{Q}(\pi)\simeq\mathbb{Q}(e)$.
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    $\begingroup$ +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

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