$\mathbb{Q}(\pi, i\pi)$ over $\mathbb{Q}$ Is $\mathbb Q(\pi,i\pi):\mathbb Q$ a simple extension?
 A: Suppose there exists a $x\in\mathbb Q(\pi,i)$ such that $\mathbb Q(\pi,i)=\mathbb Q(x)$. Since $\mathbb Q(\pi,i)$ is infinite over $\mathbb Q$, $x$ must be trascendental. Lüroth's theorem, then, tells us that every subfield of $\mathbb Q(x)$ properly containing $\mathbb Q$ is itself a simple trascendental extension of $\mathbb Q$. But this then applies to $\mathbb Q(i)\subseteq\mathbb Q(x)$. Of course, this is absurd.
A: Let's see. Assume otherwise, thus $\mathbb Q(\pi,i\pi) = \mathbb Q(\pi,i) = \mathbb Q(x)$ for some x. This means that they are equivalent as modules over $\mathbb Q$, and thus $i = a_nx^n + \cdots + a_0$ for some polynomial with $n \geq 1$, and so we have that $(a_nx^n + \cdots + a_0)^2 + 1 = 0$ is a rational polynomial of nonzero degree with $x$ as a root, thus $x$ is algebraic. But $\pi$ is transcendental, and since the algebraic numbers are a field we cannot write $\pi$ as a polynomial in $x$, contradicting the assumption that $\mathbb Q(\pi,i) = \mathbb Q(x)$. So the answer is no.
A: No.  Let $x \in \mathbb Q(\pi,i\pi) = \mathbb Q(i,\pi).$  If $x$ is algebraic, then $\mathbb Q(x)$ is finite over $\mathbb Q$, hence an algebraic extension, and so does not contain the transcendental element $\pi.$  On the other hand, if $x$ is transcendental over $\mathbb Q$, then $\mathbb Q(x)$ is isomorphic to the field
of rational functions in one variable over $\mathbb Q$.  The algebraic closure of
$\mathbb Q$ in this extension is equal to $\mathbb Q$ itself, and so this field does not contain $i$.
Thus $\mathbb Q(\pi,i\pi)$ is not of the form $\mathbb Q(x)$ for any of its elements $x$, and so is not a simple extension of $\mathbb Q$.
