Simple extension of $\mathbb Q$ by $e$ What's the smallest Field containing $\mathbb Q$ and $e$, the transcendental number? Is it $\mathbb R$? or a proper subfield of $\mathbb R$?
I understand that the dimension of $\mathbb Q(e)$ over $\mathbb Q$ is infinite because $e$ is transcendental. I just can't imagine how $\pi$ will belong to this simple extension $\mathbb Q(e)$!
 A: Note that $\mathbb R$ is a field containing $\mathbb Q$ and $e$ but not necessarily the smallest possible. $\mathbb Q(e)$ is a much smaller field containing $\mathbb Q$ and $e$ and in fact the smallest subfield of $\mathbb R$ containing both $\mathbb Q$ and $e$. To see this note that $\mathbb Q[e]$ is the smallest ring containing $\mathbb Q$ and $e$ and its quotient field is $\mathbb Q(e)$ which is the smallest field containing $\mathbb Q[e]$.
Determining which irrational numbers $\alpha$ are also in $\mathbb Q(e)$ is related to determining whether or not $\alpha$ and $e$ are algebraically independent over $\mathbb Q$. As mentioned in the comments for $\alpha=\pi$ this is still an open question (we do not even know whether or not $\pi+e$ is irrational)! For other numbers this might be easier to check. Take for example $\alpha=\sqrt2$ or $\beta=e^{1/2}$ from the comments. If there were $x,y\in\mathbb Q(e)$ such that $x^2=\alpha$ and $y^2=e$ then these would be rational functions in $e$. But clearing the denominator in those cases would give polynomials in $e$ which equate to zero, contradicting the transcendence of $e$.
Also: only because $[\mathbb Q(e):\mathbb Q]$ and $[\mathbb R:\mathbb Q]$ are both infinite does not mean that $\mathbb Q(e)=\mathbb R$; we have the same situation for $[\mathbb R:\mathbb Q]$ and $[\mathbb C:\mathbb Q]$ but surely $\mathbb R\ne\mathbb C$. A a matter of fact the extensions are not the same size  in the given case (one is countable while the other is uncountable).
