Let $L$ be a line bundle (or more generally any sheaf) on $X$. Obviously the condition 'globally generated' implies 'base point free'. Is the converse always true?
Edit
Base point free means for any $x\in X$ we can find a global section $s$ of $L$ such that $s_x=0$ (or $s_x\notin m_x$?).
Globally generated means there exists a surjection of sheaves $\mathcal O_X^{\oplus I}\to L$.
And I didn't clarify the assumptions because I want to know about in how general this is true. For example, is this true for $X$ is a scheme and $L$ any sheaf? If not, what condition should be added?
This is true for line bundles. If $L$ is basepoint-free, then for each $x\in X$ choose a global section $s(x)$ such that the image of $s(x)$ in the fiber at $x$ is nonzero, and consider the map of sheaves $\mathcal{O}_X^{\oplus X}\to L$ that is given by $s(x)$ on the $x$ coordinate. This map is then surjective on each stalk (since $L$ is a line bundle so each stalk is generated by any element whose image in the fiber is nonzero), and thus surjective.
It is not true for general sheaves. For instance, let $M$ be your favorite sheaf that is not globally generated and let $L=\mathcal{O}_X\oplus M$. Then $L$ is basepoint-free because $\mathcal{O}_X$ is, but is not globally generated because $M$ is not. (The converse is also not true for general sheaves: for instance, the zero sheaf is globally generated, but is not basepoint-free!)