Let $X$ be a projective algebraic variety over an algebraically closed field $k$ and $\mathcal{L}$ be a line bundle over $X$. Suppose we have a decomposition $X=U\sqcup Z$, where $U$ is an open subscheme of $X$ and $Z$ is its complement with reduced structure. Suppose that the restriction of $\mathcal{L}$ to each of them is nef, then can we deduce that $\mathcal{L}$ is a nef line bundle over $X$? (Assume $Z$ has positive dimension)
Note that if $C\subset X$ is a "testing" curve, then 3 cases may happen:
$C\subset U$, then the degree of $\mathcal{L}$ has non-negative degree over $C$ according to our condition;
$C\subset Z$, same analysis as above;
$C$ intersects both $U,Z$ non-trivially. Geometrically thinking, if I am working over $\mathbb{C}$, then I may slightly move the curve such that $C\cap Z$ has dimension $1$ rather than $0$. Then I may see that $\mathcal{L}$ has nonnegative degree on $C\cap U$ and $C\cap Z$, hence their union. But it seems that this is only some intuition, not argument.
I would like to ask whether $\mathcal{L}$ is still nef? And how to give a valid argument?
No. Consider, for instance, $$ X = \mathbb{P}^1 \times Z = (\mathbb{A}^1 \times Z) \sqcup Z $$ and let $L$ be the pullback of a line bundle on $\mathbb{P}^1$. Then the restriction of $L$ to both $U$ and $Z$ is trivial, hence nef. But on the entire $X$ it is nef only if the original line bundle on $\mathbb{P}^1$ is nef.