Galois number fields that have the imaginary unit.

Question

$\newcommand\Q{\mathbb Q}$Is it possible to find an irreducible polynomial $f\in \Q[x]$ of degree $4$ such that the following holds:

1. All roots of $f$ are non-real
2. The splitting field of $f$, $K_f$, contains "the" imaginary unit $\mathrm i$
3. The maximum real subfield of $K_f$, i.e. $K_f\cap \mathbb R$, is not Galois over $\Q$
4. $\mathrm{Gal}(K_f/\Q)\cong A_4$

I have the feeling that this cannot be done but I'm not sure how to prove this. Item 3. suggests that $K_f$ should not be some sort of cyclotomic extension but that is all I have figured. Can it be some radical extension? there is not much room to work with because I have only degree $4$.

In general, I would also like to have an intuition what the implication of having the imaginary unit is in a Galois number field. Will the Galois group have some specific property or will there be some characterizations for such fields?

Edit: I managed to find an example that satisfies 1,3 and 4. For instance $f=x^4 - x^3 - 3x + 4$.

Answer 1

Items 2 and 4 are incompatible. Assume that $$ \Bbb{Q}\subset \Bbb{Q}(i)\subset K_f, $$ where $K_f$ is Galois over $\Bbb{Q}$, $G=Gal(K_f/\Bbb{Q})\simeq A_4$. By Galois correspondence $\Bbb{Q}(i)$ is the fixed field of a subgroup $H\le G$. Because $[\Bbb{Q}(i):\Bbb{Q}]=2$ it follows that $|H|=|A_4|/2=6$.

But $A_4$ has no subgroups of order six.