Calculating the Galois group of a quartic

I've been asked to work out the Galois group of $\displaystyle t^4-2t^2-3 \in \mathbb{Q}[x]$ which I think I can do but then it says "You need not show that any homomorphisms you use are homomorphisms, but should explain why they are members of the Galois group. Write down the subgroups of the Galois group and the intermediate fields of the relevant field extension, and explain why your lists are complete." which has thrown me a bit, not really sure what I'm meant to put?

To work out the Galois group of $\displaystyle t^4-2t^2-3 \in \mathbb{Q}[x]$ I have said that as $\displaystyle t^4-2t^2-3 = (t^2+1)(t^2-1)$ which are two irreducible quadratics, then the splitting field $\displaystyle f(x)$ is the compositum of $\displaystyle \mathbb{Q}(i)$ and $\displaystyle \mathbb{Q}(\sqrt{3})$. This is a biquadratic extension and thus the Galois group is $\displaystyle V_{4}$. Is this right and any help deciphering what else I'm meant to do would be appreciated.

Is it reffering to the fact that the Klein group $\displaystyle V_4$ is the product of $\displaystyle \mathbb{Z}_2$x$\displaystyle \mathbb{Z}_2$ which are the galois groups of the two quadratics I have. How do I show they are complete?

Thanks for any advice you can offer!

Re: Calculating the Galois group of a quartic

The problem said "Write down the subgroups of the Galois group". What are the subgroups of V_{4}? What are the corresponding "intemediate fields"?

Re: Calculating the Galois group of a quartic

Say there are 4 permutations $\displaystyle I, P_1, P_2, P_3 $ with $\displaystyle P_3=P_1P_2$ then there are five subgroups:

$\displaystyle \{I\}$, corresponding to $\displaystyle \mathbb{Q}(i, \sqrt{3})$

$\displaystyle \{I,P_1\}$, corresponding to $\displaystyle \mathbb{Q}(\sqrt{3})$ As this fixes $\displaystyle \sqrt{3}$ say

$\displaystyle \{I,P_2\}$, corresponding to $\displaystyle \mathbb{Q}(i)$ As this fixes $\displaystyle i$ say

$\displaystyle \{I,P_3\}$, corresponding to $\displaystyle \mathbb{Q}(i\sqrt{3})$ As this fixes $\displaystyle i\sqrt{3}$

and $\displaystyle \{I,P_1,P_2,P_3\} $corresponding to $\displaystyle \mathbb{Q}$

I think thats right??