# Degenerations of hyperelliptic coverings

Take six distinct points $$p_1,\dots,p_6\in\mathbb{P}^1$$ and consider the double covering $$f:C\rightarrow \mathbb{P}^1$$ ramified over $$p_1,\dots,p_6\in\mathbb{P}^1$$. Then $$C$$ is a smooth curve of genus two.

Can we degenerate $$C$$ to a singular rational curve or to a union of smooth rational curves by collapsing some of the $$p_i$$ together?

If $$p_1 = p_2 \ne p_3 = p_4 \ne p_5 \ne p_6$$ then the normalization of the double cover branched at the divisor $$D = \sum_{i=1}^6 p_i$$ is a smooth irreducible rational curve. If also $$p_5$$ and $$p_6$$ collide, the normalization of the double cover is the union of two smooth rational curves.
• Thank you very much. By embedding the curve in $\mathbb{P}(1,1,3)$ I can see that in the second case the curve degenerates to a union of two rational curves and that in the first case we get an irreducible curve with two singular points. But why is this last curve rational?
• The double covering branched along the zero locus of a polinomial $f(x)$ is given by the equation $y^2 = f(x)$. If $f(x)$ has a root of multiplicity 2 (say at $x = 0$) then (etale) locally the double covering is given by $y^2 = x^2$. So, locally, it has two branches and each maps isomorphically to the base. Therefore, after normalization the morphism is not ramified over this point. Thus, when $f$ has degree 6 with two roots of multiplicity 2 and two roots of multiplicity 1, the normalization is a double covering branched only at two points, hence rational. Mar 10, 2021 at 6:42