MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Example 2.2.19 of

Lazarsfeld, Positivity in Algebraic Geometry I,

I found the following statement:

Let $D$ be a divisor on an irreducible projective variety $X$. Then $D$ is nef and big if and only if there exists an effective divisor $N$ such $D-\frac{1}{k}N$ is ample for $k\gg 0$.

It is clear to me that $D$ nef and big implies that there exists an effective divisor $N$ such $D-\frac{1}{k}N$ is ample for $k\gg 0$. It is quite clear that $D-\frac{1}{k}N$ ample implies $D$ big.

I can not understand why $D-\frac{1}{k}N$ ample implies $D$ nef. Perhaps I am just misunderstanding the staments. We may write $\epsilon = \frac{1}{k}$ and $D-\epsilon N$. Does the statement mean $D-\epsilon N$ is ample for some $\epsilon$ small enough or for any $0< \epsilon\ll 1$ ?

share|cite|improve this question
I would vote for the latter, as otherwise there are obvious counterexamples (like ample plus exceptional $E$ becomes back ample after subtracting $\frac1k(kE)$). – Alex Degtyarev May 15 '14 at 20:37
up vote 1 down vote accepted

$D-\frac{1}{k}N$ ample for any $k\gg 0$, that is $D-\epsilon N$ ample for any $0<\epsilon\ll 1$ ample, implies that $D$ is nef. This is because $Nef(X)$ is the closure of $Amp(X)$. If $C$ is an irreducible effective curve then $(D-\epsilon N)\cdot C = D\cdot C-\epsilon N\cdot C >0$. Therefore $D\cdot C>\epsilon N\cdot C$ for any $0<\epsilon\ll 1$, that is $D\cdot C\geq 0$, and $D$ is nef.

On the other hand $D-\epsilon N$ for some $\epsilon >0$ implies that $D$ is big by Corollary 2.2.7 of Lazarsfeld, Positivity in Algebraic Geometry I. Nowever $D$ is not nef in general. For instance, let us consider $X = \mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1}(-3))$. The anti-canonical divisor is $-K_{X}=2C_0+(e+2)F$, where $C_0$ is the section and $F$ is the fiber. Now, if $\frac{1}{3}<\epsilon < 2$ then $$D:=-K_{X}-\epsilon C_0 = (2-\epsilon)C_0+(e+2)F$$ is ample. Indeed the cone of curves is generated by $C_0$ and $F$. We have $D\cdot C_0 = 3\epsilon -1 > 0$ because $\epsilon >\frac{1}{3}$, and $D\cdot F = 2-\epsilon >0$ because $\epsilon < 2$.

However, $-K_{X}\cdot C_0 = -1$, and $-K_X$ is not nef. By the way, this construction works with $X = \mathbb{P}(\mathcal{O}_{\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1}(-e))$ for any $e\geq 3$.

share|cite|improve this answer

Here's another take on the second half of the above answer, with a bit less detail to worry about. Take your favorite effective but not nef divisor $E$, on a variety $X$. Let $A$ be ample, and take $D = A + mE$ with $m$ pretty big. Then $D$ is big, and $D - mE$ is ample, but $D$ isn't nef because it's negative on whatever curve $E$ is negative on (asuming you picked $m$ big enough). If your favorite effective but not nef divisor happens to be the negative section on $F_3$ (why not), you get Cobian's example.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.