Let $X$ be an $n$ dimensional smooth projective variety and $\mathcal{F}$ a globally generated rank $n$ locally free sheaf on $X$. 



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> Question: Is the locus of sections $s\in H^0(\mathcal{F})$ having a positive dimensional zero locus Zariski closed?
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My idea: denote by $s\_1,\dots,s\_{m}$, $m=h^0(\mathcal{F})$, a basis for $H^0(\mathcal{F})$. Then we get a map
$$G:X\to Gr(m-n,m),$$
where $Gr(m-n,m)$ denotes the Grassmannian of $m-n$ planes in $\mathbb{C}^m$. This map sends a point $x\in X$ to subspaces formed by those $\underline{\lambda}=(\lambda\_1,\dots,\lambda\_m)\in\mathbb{C}^n$ for which $s\_{\underline{\lambda}}(x):=\sum\_{i=1}^m\lambda\_i s\_i(x)=0$. Since $\mathcal{F}$ is globally generated this is indeed an $m-n$ plane. Then we have the following
$$X\stackrel{G}{\to} Gr(m-n,m)\stackrel{\pi\_2}{\leftarrow} \mathcal{U}\stackrel{\pi\_1}{\to} \mathbb{C}^m$$
where $\mathcal{U}$ is the universal subspace bundle. Then the zero locus of $s\_{\underline{\lambda}}$ is
$$G^{-1}\pi\_2\pi\_1^{-1}(\underline{\lambda}).$$
Now I would want to use some sort of upper semicontinuity result to conclude, but I cannot quite figure out how. 


An idea for how to finish the above argument, or an alternative argument, or a counterexample to the statement, would all be appreciated.



Locus of global sections with positive dimensional zero locus is Zariski closed?
I put my comment as an answer. Yes, the result is true. Here is a standard argument: in $\mathbb{P}(H^0(\mathcal{F}))\times X$, consider the subvariety $Z$
 of pairs $([s],x)$ such that $s(x)=0$.
This is clearly closed, hence projective, and you are looking at the sublocus of $\mathbb{P}(H^0(\mathcal{F}))$ where the fibers of the projection $Z\rightarrow \mathbb{P}(H^0(\mathcal{F}))$
 have positive dimension. This is closed by semi-continuity of the dimension of the fibers (Chevalley's theorem, see e.g. EGA IV 13.1.5). 


