Let $S\subset \mathbf{P}^2$ be a quadric, and let $X \subset \mathbf A^3$ be the cone over $S$ with vertex $P$.
As a practice, I want to calculate $$(K_X+L)|_L,$$ where $L$ is a rule of $X$. My progress is as follows:
Let $\pi : \widetilde X \rightarrow X$ be a resolution of $P\in X$
First note that
- these are $\mathbf Q$-Cartier divisors, so the expression makes sense ($K_X$ is Cartier, and so is $2L$);
- $\pi_* \pi^* D = D, D \in \text{Pic}X$;
- $K_{\widetilde X} = \pi^*K_X + E, E$ the exceptional divisor;
- $\pi^*D.E=0, D\in \text{Pic}X.$
We then go ahead and calculate
\begin{align} (K_X+L)|_L &= \frac{1}{4}\pi_*\pi^*((2K_X+2L).(2L))\\ &= \frac{1}{4}\pi_*(\pi^*(2K_X+2L).\pi^*(2L)) \tag{1}\\ &= \pi_*((\pi^*K_X + \pi^*L).\pi^*L) \\ &= \pi_*((K_\widetilde X - E + \widetilde L + \frac{1}{2}E).\widetilde L) &(L^2=\frac{1}{2}P)\\ &= \pi_*(K_\widetilde L - \frac{1}{2}Q) &(Q \in \pi^{-1}P) \\ &= K_L - \frac{1}{2}P. \end{align}
A different approach, which does not use (1), would be $L^2=\frac{1}{2}P$ and $\pi_*K_Y=K_X$, so
\begin{align} K_X.L &= \frac{1}{2}\pi_*K_Y.2L \\ &= \frac{1}{2}\pi_*(K_Y.2\pi^*L) &\text{(using projection formula instead)}\\ &= \frac{1}{2}\pi_*(K_Y.2\widetilde L + K_Y.E) &(\widetilde L^2 =0)\\ &= \frac{1}{2}\pi_*(2K_\widetilde L - 2Q) & (\text{since }0=\pi^*K_X, E^2 = -2Q,Q\in E) \\ &= K_L - P, \end{align} which leads to the same result.
Now, I am told the result is actually $K_L + \frac{1}{2}P$. Did I go wrong somewhere, or is this information incorrect?
Thanks.