I've been solving some problems from my Galois Theory course, and need help with this one:
For $a$ a root of $X^3-2$, find $\mathrm{Aut}_{\mathbb Q(i)}(\mathbb Q(i,a))$ and $\mathrm{Aut}_{\mathbb Q(a)}(\mathbb Q(i,a))$.
The work I've done so far:
First I noticed that since $a$ is root of $X^3-2$, then $a\in\{\sqrt[3]2,\sqrt[3]2\omega,\sqrt[3]2\omega^2 \}$ with $\omega=\frac{1}{2}(-1+i\sqrt3)$. Given this, it's clear that $\mathbb{Q}(\sqrt[3]2,\sqrt[3]2\omega)=\mathbb Q(\sqrt[3]2,i\sqrt3)$ is the splitting field of $X^3-2$ over $\mathbb Q$ (I'm not sure if this is useful for this problem).
First, I find $\mathrm{Aut}_{\mathbb Q(a)}(\mathbb Q(i,a))$: since the extension $\mathbb Q(i,a)/\mathbb{Q}(a)$ has degree 2, then it's normal, hence $\mathbb Q(i,a)$ is the splitting field of $X^2+1$ over $\mathbb Q(a)$, implying that there are two different $\mathbb Q(a)$-automorphisms: $$\sigma_1(i)=i, \quad \sigma_1(\lambda)=\lambda, \forall\lambda\in\mathbb{Q}(a)$$ $$\sigma_2(i)=-i, \quad \sigma_2(\lambda)=\lambda, \forall\lambda\in\mathbb{Q}(a)$$
Now, I need to find $\mathrm{Aut}_{\mathbb Q(i)}(\mathbb Q(i,a))$, and this is where I'm stuck. I know it must be verified that $\sigma(\lambda)=\lambda$, $\forall\lambda\in\mathbb Q(i)$, and that $\sigma(a)=\{\text{ root of $X^3-2$ }\}$, but I don't know how to find which of that 3 possible asignations is valid and how to justify it correctly, since I think (I may be wrong) that $\mathbb Q(i,a)$ is not a splitting field of $X^3-2$ over $\mathbb Q(i)$ in general.
How can I end this problem? Is what I've already done correct? Any help will be appreciated, thanks in advance.