An exercise in Galois theory dealing with function fields Our professor gave an example in class: compute the Galois group of $ x^4-2x^2-1$ over $\mathbb{ Q}$. So the main steps are: $\alpha:=\sqrt{1+\sqrt{2}}$, and observe that the splitting field is $\mathbb{ Q}(\alpha,i)$. Now the degree of the extension is 8, since $ i$ isn't real. Then after computing the relations of the Galois elements we see that the group is exactly dihedral.
O.K. So then he left us an exercise with some more delicate technicalities.
Let $ K=\mathbb{C}(t)$, the function field of one variable. Let $u$ be some nonzero complex number, and consider the polynomial $x^4+2utx^2+ t\in K[x]$ ($t$ is the variable of the function field).
Compute its Galois group.
I try to immitate the example in class. Then we let $\alpha$ be a root of the polynomial ( it's irreducible thanks to Eisenstein criterion). We can similarly obtain that the splitting field is $ K(\sqrt{t}, \alpha)$. But this time I found it hard to explain why $\sqrt{t}\notin K(\alpha)$. I'm even not sure whether this depends on the value of $ u$.
Could anyone give me some hint? Thanks a lot in advance!
 A: Some elements, not a solution
If $\sqrt t \in K(\alpha)$, it exists $p_0, p_1, p_2, p_3 \in K = \mathbb C(t)$ such that
$$\sqrt t= p_0 + p_1 \alpha + p_2 \alpha^2 +p_3 \alpha^3$$ with $$\alpha^4 + 2 ut \alpha^2 + t=0.$$
Hence $\alpha$ is a root of $q(x) = (p_0 + p_1 x + p_2 x^2 +p_3 x^3)^2 -t \in K[x]$ and also of $p(x) =  x^4 + 2 ut x^2 + t \in K[x]$. Performing the long polynomial division of $q$ by $p$ in $K[x]$ you get
$$q(x) = p(x) b(x) + r(x)$$ where the degree of $r$ is equal to 3 at most and $r(\alpha) = 0$. You can conclude of that that the 4 coefficients of $r$, which belong to $K(t)$ are all equal to zero.
In the end, the question is: "are those 4 equations possible or not?".
Note: sympy polynomial may help to perform the long polynomial division if you know it and are lazy at performing it by hand.
Doing so, you'll get that $\sqrt t \in K(\alpha)$ if and only if the following equations in $p_0, p_1, p_2, p_3 \in K$ have a solution
$$\begin{cases}
p_{0}^{2} - 2p_{1} p_{3} t - p_{2}^{2} t + 2 p_{3}^{2} t^{2} u - t &= 0\\
p_{0} p_{1} - p_{2} p_{3} t &= 0\\
2p_{0} p_{2} + p_{1}^{2} - 4 p_{1} p_{3} t u - 2 p_{2}^{2} t u + 4 p_{3}^{2} t^{2} u^{2} - p_{3}^{2} t &= 0\\
p_{0} p_{3} + p_{1} p_{2} - 2 p_{2} p_{3} t u &= 0
\end{cases}$$
... and I don't see an immediate argument to state that those equations are not compatible.
