Let $k=\mathbb{C}(t)$ be the field of rational functions in one variable. Find the Galois group over $k$ of the polynomial $f(x)=x^3+x+t$.
My approach: I've made some progress on this problem. Obviously, $f(x)$ is irreducible over $k$. Otherwise, it should have a root in $\mathbb{C}[t]$ and this root must divide $t$ and hence has form $c_0$ or $c_ot$ but none of these will be a root of $f(x)$.
Let $K$ its splitting field over $k$ then I showed that $K/k$ is normal and separable and hence it Galois extension.
Let's calculate the discriminant of this polynomial. It is equal to $$\Delta_f=-4-27t^2=i^2(4+27t^2).$$
But how to show that $4+27t^2$ is not square in $\mathbb{C}(t)$?
Suppose it is square then $$4+27t^2=\left(\dfrac{a_{n+1}t^{n+1}+\dots+a_1t+a_0}{b_nt^n+\dots+b_1t+b_0}\right)^2$$ where $a_i,b_i \in \mathbb{C}$.
And as you see I do not know what should I do further?
Suppose that $27t^2+4$ is a square in $\mathbb{C}(t)$ then $\exists p(t),q(t)\in \mathbb{C}[t]$ such that $27t^2+4=\left[\dfrac{p(t)}{q(t)}\right]^2$. Also we can assume that $\text{gcd}(p(t),q(t))=1$. Easy to note that $p(t)$ and $q(t)$ must have degree $n$ and $n-1$, repectively. Since these polynomials have coefficient in $\mathbb{C}$ then $p(t)=a_n(t-\alpha_{1})\dots (t-\alpha_{n})$ and $q(t)=b_{n-1}(t-\beta_{1})\dots (t-\beta_{n-1})$. Since they are coprime then $\beta_j\neq \alpha_i$. We get the following: $$27t^2+4=\left[\dfrac{a_n(t-\alpha_{1})\dots (t-\alpha_{n})}{b_{n-1}(t-\beta_{1})\dots (t-\beta_{n-1})}\right]^2,$$
$$(27t^2+4)[b_{n-1}(t-\beta_{1})\dots (t-\beta_{n-1})]^2=[a_n(t-\alpha_{1})\dots (t-\alpha_{n})]^2.$$
Plugging $t=\beta_1$ we get that LHS is $0$ but RHS $\neq 0$. Is this correct reasoning?