Fano Variety and Blow up

Question

Let $Y$ be a Fano threefolds and $\pi : X \rightarrow Y$ be the blowup along a nonsingular subvariety $Z$ of $Y$ with $\mbox{codim(Z)} > 1$. Is $X$ a Fano threefold?

Answer 1

The question already has a satisfactory answer from Sasha, but I just want to point out that you don't need to know any big facts like finiteness of topological types of Fanos to see the answer is negative in general. All you need is the canonical bundle formula

$$K_Y = \pi^* K_X + (c-1) E$$ where $E$ is the exceoptional divisor and $c$ is the codimension of the blowup centre.

Using this, you can show, for example, that when you blow up two points in $\mathbf P^3$, the proper transform $C$ of the line joining the two points satisfies $K_Y \cdot C =0$, showing that $K_Y$ is not ample.

If you want an example with connected centre, let $C$ be a smooth curve in $\mathbf P^3$ which possesses a 4-secant line $l$, meaning a line which intersects it transversely in 4 points. Let $\pi: Y \rightarrow \mathbf P^3$ be the blowup along $C$. According to the canonical bundle formula above we have $$ -K_Y = 4(\pi^* H) - E$$ where $H$ is the hyperplane class on $\mathbf P^3$. Now let $\tilde{l}$ be the proper transform on $Y$ of the 4-secant line $l$. Then $\pi^* H \cdot \tilde{l}=H \cdot l = 1$ (using the projection formula) and $E \cdot \tilde{l}=4$ since they meet transversely in 4 points. Therefore we get $-K_Y \cdot \tilde{l}=0$, showing that $Y$ is not Fano.

Answer 2

Not in general. For instance, there is only finitely many topological types of Fano 3-folds, but with blowups you can create as many topological types as you want.