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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?

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1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – user125056
    Mar 9, 2021 at 20:05
  • $\begingroup$ 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. $\endgroup$
    – Sasha
    Mar 10, 2021 at 6:42

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