## Nefness of $h-e$ in the blowup of $\mathbb{P}^n$

Let $S$ be the blow up of $\mathbb{P}^n$ in a point $P$. Let $h$ be the class of the pullback of an hyperplane of $\mathbb{P}^n$ and $e$ the class of the exceptional divisor. Why is the divisor $l=h-e$ nef? Thank you very much!

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We can choose a representative of $h$ which is the preimage of a hyperplane $H$ passing through $P$. This preimage is then equal to $\tilde{H} + E$, where $\tilde{H}$ is the proper transform of $H$ and $E$ is the exceptional divisor. Thus $\tilde{H}$ is a representative for $h - e,$ and is an effective divisor.

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 Not every effective divisor is nef. – jvp Jan 29 2010 at 22:37 Thanks. To complete the argument, see jvp's answer. – Emerton Jan 30 2010 at 2:08

Not every effective divisor is nef. By definition a divisor is nef if it intersects every curve non-negativelly. For instance the exceptional divisor $E$ is effective but not nef as it intersects any line contained in it negatively.

To see that $\tilde H$ is nef one can use Emerton's argument to show that the linear system $|\tilde H|$ is free from base points since it contains all the strict transforms of hyperplanes through $P$. So given a curve $C \subset S$ we can choose among these strict transforms one which does not contain $C$ to show that $\tilde H \cdot C \ge 0$.

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As a complement to JVP's answer, here is a direct proof to that $\tilde{H}\cdot C\geq 0$.

Note that nefness is numerically invariant. To check the nefness of $h-e$, we only need to show that for any irreducible curve $C$ in the blowing-up, the intersection number $(h-e)\cdot C$ is nonnegative. If $C$ is not contained in $e$. then the image of $C$, denoted by $D$, is still a curve. In this case, by projection formular, $(h-e)\cdot C=H\cdot D\geq 0$, where $H$ is any hyperplane in $\mathbb{P}^n$. In the case $C$ is contained in $e$, $h\cdot C =0$ however $-e\cdot C =-\deg N_{e/X}|C=1$, where $N_{e/X}$ is the normal bundle of the exceptional divisor in the blowing up $X$. Therefore $h-e$ is nef.

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