Edit: Assume field K has a positive characteristic.
Let's say I have an algebraic function field $F$ over field $K$ with characteristic > 0 given by a Weirstrass equation $w(x,y)=y^2-x^3-ax-b \in K[x,y]$ in other words $F$ is the fraction field of the quotient ring $K[x,y]/(w)$. $w$ is irreducible. And lets assume that $w$ is smooth.
Now let's denote $t=y+\lambda x + \mu \in K[x,y]$. Assume that the "curve" (line) given by $V_t$ intersects $V_w$ at exactly 2 different places. Therefore I know that the polynomial $w(x,-\lambda x - \mu)$ has only 2 distinct roots, therefore it is not separable because its degree is 3.
I want to show that $F/K(t)$ is not a Galois extension. My guess would be to try to find an element of $F$ such that $w(x,-\lambda x - \mu)$ is a minimal polynomial of that element over $K(t)$. Since I know it is not separable the extension would not be separable.
I also see that $w(x,-\lambda x - \mu)=-y^2+x^2\lambda^2 + 2\lambda \mu x + \mu^2 \in F$ since $y^2 = x^3+ax+b$ in $F$.
Is it the right way? Or should I look for something else then separability. I have spent hours trying to find something but no luck.
Any hints are appreciated.
What is the full, complete, unabridged statement of the exercise? The question as posed is false. Consider $K=\mathbb{C}, a=-1, b=0, \lambda=0, \mu=-\sqrt{2/(3 \sqrt{3})}$. Then $w(x,y) = y^2 - (x^3-x)$ and $t(x,y) = y-\sqrt{2/(3\sqrt{3})}$ intersect in exactly two points: $(2/\sqrt{3}, \sqrt{2/(3 \sqrt{3}))}$ and $(-1/\sqrt{3}, \sqrt{2/(3 \sqrt{3}))}$. However, $F/K(t)$ must be separable, because $K(t)$ has characteristic zero, and every field extension is separable in characteristic zero.