Let $N^1(X)=Div(X)/Num(X)$ be the Neron-Severi group of a smooth projective variety $X$. It is known that $N^1(X)$ is a free Abelian group of finite rank, say $r$. Are there ample (or nef) divisors $D_1,\dots, D_r$ that provide a $\mathbb{Z}$-basis of $N^1(X)$?
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1$\begingroup$ Yep - pick a primitive ample class $H$ and $r-1$ other divisors $E_1,\cdots E_{r-1}$. Then take $D_i=E_i+nH$ and $E_r=H$ for sufficiently large $n$. $\endgroup$– dhyCommented Dec 24, 2019 at 3:42
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$\begingroup$ Small quibble, I think you want that last equation to read $D_r = H$. $\endgroup$– Tabes BridgesCommented Dec 25, 2019 at 1:01
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$\begingroup$ I filled in the details in dhy's answer. Here is a statement that I can prove. Let us choose $E_1,\dots ,E_r$ to be a $\mathbb{Z}$-basis of $NS^1(X)$, then after changing the order of $E_1,\dots,E_r$ if necessary, we can find an ample basis of the following form: $D_i=E_i-k_i E_r+nH$ ($1\leq i\leq r-1$) and $D_r=H$, where $k_i$ are some integers and $n$ is sufficiently large. $\endgroup$– TodorCommented Dec 26, 2019 at 3:11
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