Is $\Bbb Q(\sqrt 2, e)$ a simple extension of $\Bbb Q$?

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.

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?

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 $\Bbb{Q}(x)$.