Splitting field's Galois group of $p(X)=X^4+(t^3+1)$ over $\mathbb{Q}(t)$ where $t$ transcendent over $\mathbb{Q}$ I would like to describe $$G=\text{Gal}(X^4+(t^3+1)/\mathbb{Q}(t)).$$
However, I have serious problems always when I am working with polynomials in let's say $K(t)[x]$. In my mind, things do not work as they do in $K[x]$ and I do not know why.
I have tried to find all roots of $p(X)=X^4+(t^3+1)$ in order to describe its splitting field.
I know $$X^4+(t^3+1) = (X^2 - i\sqrt{t^3+1})(X^2+i\sqrt{t^3+1})$$
and since this two polynomials are not in $\mathbb{Q}(t)$, I can conclude that $p(X)$ is irreducible over $\mathbb{Q}(t)$. If I try to solve as always, I would get
$$e^{\frac{2\pi ik}{4}}i\sqrt[4]{t^3+1} = e^{\frac{2\pi ik}{4}}e^{\frac{\pi i}{2}}\sqrt[4]{t^3+1} = e^{\frac{2\pi i(k+1)}{4}}\sqrt[4]{t^3+1}$$
are its roots for $k=1,...,4$. From what I have conclude its splitting field would be $$E=\mathbb{Q}(\sqrt[4]{t^3+1}, i).$$
I do not know if I have done things correctly til here. I would appreciate any feedback you have.
Now, since $\alpha = \sqrt[4]{t^3+1}$ is root of $p(X)$ and $p(X)$ is irreducible over $\mathbb{Q}(t)$, $[\mathbb{Q}(\alpha):\mathbb{Q}(t)]=4$. On the other hand, $[\mathbb{Q}(\alpha,i):\mathbb{Q}(\alpha)]=2$ since $q(X)=X^2+1$ is the irreducible of $i$ over $\mathbb{Q}(\alpha)$.
Since $G$ is the splitting field of $p(X)$ and we are in $ch(\mathbb{Q}(t))=0$, the extension is separable. It is also normal since it is finite, and thus algebraic. Hence, it is Galois extension.
Then,
$$|\text{Gal}(X^4+(t^3+1)/\mathbb{Q}(t))| = [\mathbb{Q}(t):E]=8.$$
Moreover, I know $$\text{Gal}(X^4+(t^3+1)/\mathbb{Q}(t)) \cong H \leq \mathbb{S}_4.$$
From what I guess that $$\text{Gal}(X^4+(t^3+1)/\mathbb{Q}(t)) \cong D_8.$$
QUESTION


*

*Is my reasoning correct?

*Why do I need $t$ transcendent over $\mathbb{Q}$?

*Do you normally proceed as if you were working in any $K$?

*What do you take into account differently when you are working in $K(t)$?
 A: 1 and 2. Your reasoning to prove irreducibility of $p$ is a bit short, to my opinion.
Hint. Instead, try to use Eisenstein criterion for a suitable PID. This is where you need $t$ to be transcendental over $\mathbb{Q}$. 
Your guess for the Galois group is correct, but you will have to justify it.
Hint. What are the groups of order $8$, up to isomorphism ? What are  the groups of order $8$ having a subgroup which is not normal ? 
If this hint does not ring a bell, you could just compute the automorphisms, and try to find a element of order 4 and an element of order $2$ satisfying the appropriate relation. 


*Yes.

*Nothing.
A: You reasoning is reasonably accurate. Below I collected a few isolated remarks, comments and attempts at answers to your questions.


*

*A weak link in your argument is that of irreducibility of $p(X)$ over $\Bbb{Q}(t)$. You looked at one factorization over an extension field, but in theory the other ways of pairing up the roots may work better. Compare with the following. Just because we can write $$g(x):=x^4+5x^2+4=(x^2+3i x-2)(x^2-3i x-2),$$ where the factors are not in $\Bbb{Q}[x]$ we cannot conclude that the polynomial is irreducible over $\Bbb{Q}$. Indeed, $g(x)=(x^2+1)(x^2+4)$.

*Instead, I would use Eisenstein's criterion. It applies to the field of fractions of any UFD. Instead of the more common UFD $D=\Bbb{Z}$ we should use the fact that a polynomial ring over a field $D=\Bbb{Q}[t]$ is also a UFD, and that $\Bbb{Q}(t)$ is its field of fractions. Undoubtedly you can justify why the linear polynomial $t+1$ is an irreducible element of $\Bbb{Q}[t]$. Let's look at the coefficients of
$$
f(X)=X^4+(t^3+1).
$$ 
The leading coefficient is not divisible by $t+1$, but all the other coefficients are:
$t^3+1=(t+1)(t^2-t+1)$. Furthermore, $t^2-t+1$ does not vanish at $t=-1$, so we can conclude that $t^3+1$ is not divisible by $(t+1)^2$. Therefore Eisenstein's criterion with $p=t+1$ works, and gives us the irreducibility of $f(X)$ – first over $\Bbb{Q}[t]$ and then, by the analogue of Gauss's lemma, also over the field $\Bbb{Q}(t)$.

*You need $t$ to be transcendental to conclude that $\Bbb{Q}[t]$ is isomorphic to the polynomial ring and hence a UFD. Otherwise we would not know that $t+1$ is irreducible (it might be a unit for all we know). And Eisenstein's criterion would not be available in a non-UFD anyway (it may have a variant that would work, but then we would need to work harder).

*More generally, I always treat a rational function field $K(t)$, $K$ some field, as the field of fractions of the polynomial ring $K[t]$. The machinery is a bit different, but also similar. The above example use of Eisenstein is somewhat typical. In many cases we needs tools from elementary algebraic geometry to deal with finite extension fields of $K(t)$, as such fields can be used to describe curves over $K$.

*The Galois group is, indeed, isomorphic to the dihedral group of the eight symmetries of the square. I would denote it $D_4$ ($D_n$= symmetries of the regular $n$-gon), but I have heard that Dummit & Foote and some other heretics denote it $D_8$. Anyway, the Galois group has eight elements. It is isomorphic to the Galois group of $X^4-2$ over $\Bbb{Q}$.

