Closure of a number field with respect to roots of a cubic Consider the following property of subfields ${\mathbb K}$   of ${\mathbb C}$ :
$$
\text{Any polynomial of degree 3 with coefficients in} \ {\mathbb K} \ \text{has a root in } {\mathbb K} \ \ \ \  (*)
$$
Thus, $\mathbb R$ itself satisfies (*), but it is easy to see that there are many other subfields with this property.
If ${\mathbb K} \subseteq {\mathbb R}$ satisfies $(*)$, does ${\mathbb K}(i)$ satisfy $(*)$ also ?
This question was inspired to me by that other one.
 A: I have a counter-example.
Consider the integer polynomial
$$ f(x) = (x^3 + x + 1)^2 + 1 $$
$f$ is irreduible over $\mathbb{Q}$ and factors into two cubic polynomials over $\mathbb{Q}(i)$. Let $g(x)$ be one of those polynomials$.
Using sage, I compute that the splitting field $E$ of $f$ is degree 72: exactly what you would expect due to the factorization $f = g \bar{g}$. Of particular note is that the Galois group has no subgroups of order 24, and thus the splitting field contains no cubic subfields.
Now suppose $K$ is any field constructed by starting with $\mathbb{Q}$ and iteratively making cubic extensions (transfinite iteration, unless you carefully order the extensions).
Since $K$ has no degree 3 extensions, every cubic polynomial over $K$ has a root in $K$. Also, every finitely generated subfield $L \subseteq K$ satisfies $[L:\mathbb{Q}] = 3^n$, and contains a subfield of degree 3 over $\mathbb{Q}$
In particular, let $L = E \cap K$. Because $E$ has no subfields of degree $3$, we must have $E \cap K = \mathbb{Q}$. However, this would imply $E \cap K(i) = \mathbb{Q}(i)$, and thus $g$ is irreducible over $K(i)$.
A: Let $K$ be a perfect field such that every cubic polynomial over $K$ has a root in $K$. Let $i \in \overline{K}$ be a solution to $x^2+1 \in K$, if $i \in K$ there is nothing to prove so we assume not. We claim every cubic polynomial over $K(i)$ also takes a root. Suppose not then we can find cubic polynomial $p(x) \in K(i)[x]$ such that $p(x)$ is irreducible over $K(i)[x]$. Let $L$ be the splitting field of p(x), then $L$ is also an extension field of $K$. In particular $[L:K]=6,12$. In both cases the Sylow 2-subgroup of $\mathrm{Gal}(L/K)$ has index $3$, so its fixed field is a degree $3$ extension of $K$. This is impossible, so every cubic polynomial over $K(i)$ has a root. Notice that this proof works for any degree $2$ extension of $K$. 
It seems likely to me that such a field does not have any extensions of degree divisible by $3$. But I don't think it's true that every group whose order is divisible by $3$ has a subgroup of index $3$, so this technique will likely not generalize. 
Edit: It was pointed out in the comments that this approach is incorrect because $L/K$ need not be Galois. So you have to look at normal closure $E$ of $L$ over $K$. In order to salvage this approach in this case one would need to show that every transitive subgroup of $S_6$ whose order is divisible $3$ and $2$ has a subgroup of index $3$. I'll leave this up in case its useful to anyone else attempting to solve the problem. 
