We know that there are exactly 27 lines on a smooth cubic surface. And it comes from $ \mathbb{P}^{2}$ blow up at six general points , say $p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}$. Those exceptional lines are 6 out of 27.
Let $L_{ij}$ be the line in $\mathbb{P}^{2}$ through $p_{1}$ and $p_{2}$, then its strict transform is a line. There are 15 lines out of 27.
Let $C_{i}$ be the plane conic through 5 points, then we get another 6 lines.
These are lines on the cubic surface.
However, let’s denote $X$ is the cubic surface, $\pi: X\to \mathbb{P}^{2}$ is the blowing up. We know by the property of blowing up, $\pi$ is isomorphism always 6 points. That is $\pi: Y\to Z:=\mathbb{P}^{2}-\{p_{1},p_{2}, p_{3}, p_{4}, p_{5}, p_{6}\}$ is an isomorphism, where $Y$ is the preimage of $Z$.
Now we pick a general line $L$ in $\mathbb{P}^{2}$ which does not pass any $p_{i}$, then $\pi^{-1}(L) \to L$ should be isomorphic. Thus $\pi^{-1}(L)$ should also be a line. Thus it seems $X$ should have many lines which is a contradiction.
Could you tell me where I made mistakes? Thanks a lot.
Best Answer
The preimage of a line isn't necessarily a line. Consider, for example, the paraboloid surface $\{x^2 + y^2 = z\}$ in $\mathbb{A}^3$. This doesn't contain any lines; however, the projection onto the $xy$-plane is an isomorphism, and of course $\mathbb{A}^2$ contains many lines. Likewise, in your example, just because a rational curve on a cubic surface projects to a straight line in $\mathbb{P}^2$ doesn't mean it's a straight line on the cubic surface embedded in $\mathbb{P}^3$.