What is the splitting field of $x^3 - \pi$? What is the splitting field of $x^3 - \pi$? Is it $\mathbb R(\sqrt[3] \pi, \xi_3)$
or $\mathbb Q(\sqrt[3] \pi, \xi_3)$? (where $\xi_3$ denotes the third root of unity)
It is a polynomial over $\mathbb R[x]$, so I guess it must be $\mathbb R(\sqrt[3] \pi, \xi_3)$, but I never saw such an extension.
 A: Since $\sqrt[3]\pi$ is already an element of $\mathbb R$ and $\xi_3=-\frac12\pm i\frac{\sqrt{3}}2$, the splitting field is simply $\mathbb C$. In fact, $\mathbb R$ and $\mathbb C$ are the only candidates for algebraic extensions of $\mathbb R$.
A: As commented, the question is missing an essential piece of information, the ground field.
To get a somewhat non-trivial question, the ground field should probably be $\mathbb Q(\pi)$. Now the following reasoning works:
Any $a\in\mathbb Q(\pi)$ has the form $a = \frac{f(\pi)}{g(\pi)}$ with $f,g\in \mathbb Q[x]$, $g\neq 0$. If $a^3 = \pi$, then $f(\pi)^3 - \pi g(\pi)^3 = 0$.
So $\pi$ is a root of the polynomial $f(x)^3 - xg(x)^3\in\mathbb Q[x]$. Since $g\neq 0$, this is not the zero polynomial. This is a contradiction, because $\pi$ is known to be transcendental over $\mathbb Q$.
So $x^3 - \pi$ is irreducible in $\mathbb Q(\pi)[x]$. Let $K$ be its splitting field. We have that $[K : \mathbb Q(\pi)]$ divides $3! = 6$, and by the irreducibility, $[K : \mathbb Q(\pi)]$ is a multiple of the degree $3$. From $\mathbb Q(\pi) \subset \mathbb R$ and the fact that $x^3 - \pi$ has zeros in $\mathbb C\setminus\mathbb R$, the complex conjugation incudes an Galois automorphism of order $2$, so $[K : \mathbb Q(\pi)]$ is also a multiple of $2$. This shows $[K : \mathbb Q(\pi)] = 6$.
By making use of the roots $\zeta^i\sqrt[3]{\pi}$ ($\zeta$ a primitive third root of unity; $i=0,1,2$) of $x^3 - \pi$, $K$ can be represented explicitly as
$$K = \mathbb Q(\sqrt[3]{\pi},\zeta\sqrt[3]{\pi}) = \mathbb Q(\sqrt[3]{\pi}, \zeta).$$
By the degree of the extension, none of these generators can be dropped (nor replaced by $\pi$).
