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rita
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I think there is no ramification in the case you are interested in.

The surface $\tilde X$ is the minimal ruled surface $\mathbb F_2$. The map $\tilde X \to X$ contracts the $-2$-curve $E$ to a point and maps the fibers of the ruling of $\mathbb F_2$ to lines in $\mathbb P^3$. So at every point of $E$ the map has differential of rank 1 and the kernel of the differential is the tangent space to $E$, because $E$ is contracted.

Since $\tilde C E=1$, the curve $\tilde C$ meetsIf $E$ at only one pointintersects $x$$\tilde C$ transversally, so by the previous remarks, $\tilde C \to C$ has non zero differential at $x$.

I think there is no ramification in the case you are interested in.

The surface $\tilde X$ is the minimal ruled surface $\mathbb F_2$. The map $\tilde X \to X$ contracts the $-2$-curve $E$ to a point and maps the fibers of the ruling of $\mathbb F_2$ to lines in $\mathbb P^3$. So at every point of $E$ the map has differential of rank 1 and the kernel of the differential is the tangent space to $E$, because $E$ is contracted.

Since $\tilde C E=1$, the curve $\tilde C$ meets $E$ at only one point $x$ transversally, so by the previous remarks, $\tilde C \to C$ has non zero differential at $x$.

I think there is no ramification in the case you are interested in.

The surface $\tilde X$ is the minimal ruled surface $\mathbb F_2$. The map $\tilde X \to X$ contracts the $-2$-curve $E$ to a point and maps the fibers of the ruling of $\mathbb F_2$ to lines in $\mathbb P^3$. So at every point of $E$ the map has differential of rank 1 and the kernel of the differential is the tangent space to $E$, because $E$ is contracted.

If $E$ intersects $\tilde C$ transversally, by the previous remarks, $\tilde C \to C$ has non zero differential at $x$.

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rita
  • 6.3k
  • 1
  • 29
  • 39

I think there is no ramification in the case you are interested in.

The surface $\tilde X$ is the minimal ruled surface $\mathbb F_2$. The map $\tilde X \to X$ contracts the $-2$-curve $E$ to a point and maps the fibers of the ruling of $\mathbb F_2$ to lines in $\mathbb P^3$. So at every point of $E$ the map has differential of rank 1 and the kernel of the differential is the tangent space to $E$, because $E$ is contracted.

Since $\tilde C E=1$, the curve $\tilde C$ meets $E$ at only one point $x$ transversally, so by the previous remarks, $\tilde C \to C$ has non zero differential at $x$.