Is $\Bbb Q(\sqrt 2, e)$ a simple extension of $\Bbb Q$? 
My general question is to find, if this is possible, two real numbers $a,b$ such that $K=\Bbb Q(a,b)$ is not a simple extension of $\Bbb Q$.
  $\newcommand{\Q}{\Bbb Q}$

Of course $a$ and $b$  can't be both algebraic, otherwise $K$ would be a separable ($\Q$ has characteristic $0$) and finite extension, which has to be simple. So I tried with $\Q(\sqrt 2, e)$ but any other example would be accepted.
The field $\Q(\sqrt 2, e)$ has transcendence degree $1$ over $\Q$, but I'm not sure if this imply that it is isomorphic to $\Q(a)$ for some transcendental number $a$ (the fact that two fields have the same transcendence degree over another field shouldn't imply that the fields are isomorphic).
I'm not sure about the relation between the algebraic independence of $a$ and $b$, and the fact that $\Q(a,b)/\Q$ is a simple extension.
Notice that $\Q(\pi, e)$ is probably unknown to be a simple extension of $\Q$.
Thank you for your help!
 A: The extension $\mathbb{Q}(\sqrt{2},e)\supset\mathbb{Q}$ is not simple. If $\mathbb{Q}(u)=\mathbb{Q}(\sqrt{2},e)$, then $\mathbb{Q}(u)$ is infinite-dimensional over $\mathbb{Q}$, so $u$ is transcendental. But then $\mathbb{Q}(u)$ is purely transcendental over $\mathbb{Q}$ while $\mathbb{Q}(\sqrt{2},e)$ is not.
A: You have to show that
$$
\mathbb{Q}(X) \subsetneq \mathbb{Q}(e,\sqrt{2})
$$
for any $X \in \mathbb{Q}(e,\sqrt{2})$.


*

*If $X$ is algebraic, then $[\mathbb{Q}(X) : \mathbb{Q}]$ is finite while $[\mathbb{Q}(e,\sqrt{2}): \mathbb{Q}]$ is infinite.

*If $X$ is not algebraic, then it is transcendental. It suffices to show that $\mathbb{Q}(X)$ does not contain a square root of $2$. Since $\mathbb{Q}(X)$ is isomorphic to the fraction field of polynomials, you need to show that there do not exist polynomials $p(X)$, $q(X)$ with rational coefficients such that
$$
\left( \frac{p(X)}{q(X)} \right)^2 = 2.
$$
Can you take it from here?
A: Call $K=\Bbb{Q}(\sqrt{2})$. Then $\Bbb{Q}(\sqrt{2} , e)$ is isomorphic to $K(x)$, the fraction field of $K[x]$. If this were a simple extension of $\Bbb{Q}$, it would be isomorphic to $\Bbb{Q}(x)$, the fraction field of $\Bbb{Q}[x]$. So
$$\Bbb{Q}(x) \cong K(x)$$
But this contradicts the fact that $X^2-2 \in \Bbb{Q}(x)[X]$ has no root in $\Bbb{Q}(x)$.
