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Consider a smooth projective variety $X$ over $\mathbb{C}$ such that $X$ has models over $\mathbb{Z}[1/N]$ and $X_p=X_{\mathbb{Z}[1/N]}\times \text{Spec}(\mathbb{F}_p)$ is also a smooth projective variety.

Now let $L$ be a line bundle over $X_{\mathbb{Z}[1/N]}$ and $L_\mathbb{C}$ and $L_p$ be the corresponding line bundle over $X=X_{\mathbb{Z}[1/N]}\times\text{Spec}(\mathbb{C})$ and $X_p$ respectively.

How do the properties of $L_\mathbb{C}$ (ample, nef, big, effective) relate to the properties of $L_p$?

For example, if $L_\mathbb{C}$ is nef does this imply that $L_p$ is nef?

Any references or starting points would be greatly appreciated.

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up vote 6 down vote accepted

No, that is not true: the property of being nef, ample, etc. is not stable under specialization (although it is stable under generization). For instance, begin with $Y=\mathbb{P}^2_{\mathbb{Z}}$, i.e., $\text{Proj} \ \mathbb{Z}[s,t,u]$ together with its natural projection $$\pi:Y\to \text{Spec}\ \mathbb{Z}.$$ Consider the homogeneous ideal $$I = \langle st(t-s),s(u-pt),t(u-ps),u(t-s),u(u-ps) \rangle.$$ The corresponding zero scheme is the union of three disjoint sections of $\pi$, namely $[1,0,0]$, $[0,1,0]$ and $[1,1,p]$. Let $\nu:X\to Y$ be the blowing up of $Y$ along $I$.

There is a canonically defined pullback map of invertible sheaves, $$f:\nu^*\omega_{Y/\mathbb{Z}} \to \omega_{X/\mathbb{Z}},$$ which identifies $\nu^*\omega_{Y/\mathbb{Z}}$ with $\omega_{X/\mathbb{Z}}(-\underline{E})$ for a unique Cartier divisor $\underline{E}$ on $X$, the exceptional divisor of $\nu$. Now consider the invertible sheaf $L = (\nu^*\mathcal{O}_Y(2))(-\underline{E})$.

Over $\mathbb{Z}[1/p]$, the invertible sheaf $L$ is nef, and even globally generated. Essentially this is because the associated ideal $I[1/p]$ is generated by the homogeneous generators of degree $2$; indeed, the one cubic generator $st(t-s)$ is already in the ideal generated by the quadratic generators, $$st(t-s) = \frac{1}{p}\left(s[t(u-ps)]-t[s(u-pt)]\right).$$ However, in characteristic $p$ this is not nef: the strict transform of the line $Z(u)$ has intersection number $-1$ with $L$.

Edit. Replace "openness" claim by "stable under generization", see comments below.

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I'm not sure quite what you mean by your parenthetical remark in the first line, but nefness is not an open property--see e.g. – Daniel Litt Aug 15 '13 at 16:37
I should remark, the example I linked to is for an $\mathbb{R}$-divisor (theorem 1.2); I don't know of an honest example for a Cartier divisor, but I don't see why nefness should be open in that case either. – Daniel Litt Aug 15 '13 at 16:52
@Daniel: "Open" should be "stable under generization". Let $(X_R,\mathcal{O}(1))$ be a projective, flat $R$-scheme, where $R$ is a DVR. Let $\mathcal{L}_R$ be an invertible sheaf on $X_R$. Assume that $\mathcal{L}_0$ is nef on the closed fiber $X_0$. Then for every positive integer $N$, $\mathcal{L}_0^{\otimes N}(1)$ is ample on $X_0$ by Kleiman's criterion. Thus, by openness of ampleness, $\mathcal{L}_R^{\otimes N}(1)$ is ample for every positive integer $N$. Thus, $\mathcal{L}_R$ is nef. – Jason Starr Aug 15 '13 at 19:17

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