Consider $C,C'$ two cubics in the plane with $C$ smooth. There are $9$ basepoint on the linear system generated by $C$ and $C'$ so if we blow them we get a map $X \to\mathbb P^1$, where $X$ is $\mathbb P^2$ blown-up at 9 points. Now is my question : how to compute $K_X$ ? I saw that $K_X = - C$. But I don't understand how to get it. Is the following argument correct ? $K_X = -c_1(X) = -c_1(f^*\mathbb P^2) = - f^*c_1(\mathbb P^2) = - f^*3H$ where $H$ is an hyperplane section. Now $3H \sim C$ and $f^*C$ it the strict transform of $C$. Thanks in advance for any remarks !
Compute a canonical divisor
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algebraic-geometry
1 Answers
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Here is a formula you should learn (say from Hartshorne). If $Y$ is a smooth surface, $p\in Y$ is a point and $\pi:X\to Y$ the blow up of $p$ and $E$ the exceptional divisor, then $K_X=\pi^*K_Y+E$.
In your case, $K_X=f^*K_{\mathbb{P}^2}+\sum E_i$. Since $K_{\mathbb{P}^2}=-C$, we get $K_X=-f^*C+\sum E_i$ and thus $K_X$ is the negative of the proper transform of $C$. ($K_X=-C$ is not meaningful, since $C$ is not a curve on $X$).
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0Just for be sure I did understand well : $f^* K_X = f^*C = \tilde{C} + \sum E_i$ and thus by your formula $K_X = -f^*C + \sum E_i = - \tilde{C}$, with $\tilde{C} =$ proper transform of $C$ ? – 2017-01-02
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0What is $f^*K_X$? $f^*$ applies to things on $\mathbb{P}^2$, not on $X$. – 2017-01-02
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0Yes it was a typo. Sorry I though you were writing $C$ for the proper transform of $C$ but now with your edit this is all clear, really thanks a lot ! – 2017-01-02