Let $K/\mathbb Q$ be the function fiels defined by the affine curve $y^2=D(x)$ with $D(x)=x(x-1)(x-2)$. I want to prove that for any irreducible polynomial $Q\in\mathbb Q[X]$ inert in $K$, there is no $H\in\mathbb Q[X]$ such that $y+H$ has a zero at $\mathfrak Q$, where $\mathfrak Q$ is a place of $K$ above $Q$. I did that: Let $\alpha\in\overline{\mathbb Q}$ such that $Q(\alpha)=0$ and $H\in\mathbb Q[X]$. Then $y+H$ has a zero at $\mathfrak Q$ if and only if (I am not sure here) $y(\alpha)+H(\alpha)=0$, that is $y(\alpha)=-H(\alpha)$. Then $H^2(\alpha)=D(\alpha)$. And $\alpha$ is a root of the polynomial $H^2-D$, that is $Q$ divides $H^2-D$. But $Q$ is inert, so the polynomial $T^2-D$ is irreducible in $O_{\mathfrak Q}/\mathfrak Q$, where $O_{\mathfrak Q}$ is the ring of functions of $K$ regular outside $\mathfrak Q$. So, one has a contradiction. So such polynomial $H$ can exist.
Is it correct? If yes, can one do shorter?
Thanks in advance