Why does a globally generated invertible sheaf admit a global section not vanishing on any irreducible component? Let $L$ be an invertible sheaf on the projective scheme $X$ (of pure dimension $n$). Assume that $L$ is globally generated by the global sections $s_1,\ldots,s_r \in L(X)$.

How do we show that there is some global section $s \in L(X)$ such that its zero locus does not contain any irreducible component of $X$?

 A: Assume $X$ is projective over a field $k$, and the number of irreducible components is smaller than the size of the field (if the field is finite). 
Then you can do this by induction (although I feel like there's probably a much better way to do this).
If $X$ is irreducible, then a global generating set gives you the result immediately.
Now suppose it is true for the union of $m-1$ irreducible components, and $X$ has $m$ irreducible components $Y_1,\dots,Y_m$. Then by the induction hypothesis, we can find a section $s$ whose zero locus doesn't contain $Y_1,\dots,Y_{m-1}$. We then have two cases: either $Z(s) \not \supseteq Y_k$ and we're done, or $Z(s) \supseteq Y_m$.
In this case, pick a global section $t \in L(X)$ for which $Z(t) \not \supseteq Y_k$. We want some $\alpha \in k$ such that $s+\alpha t$ doesn't vanish at every point in $Y_m$. We can try $\alpha = 1$. However, if there exists an $i \in \{0,\dots,m-1\}$ for which $s_y = t_y$ for all $y \in Y_i$, then we need a different $\alpha$. But in the worst case, $s$ and $t$ could be scalar multiples of each other at the stalks of each point in $Y_m$, i.e. we could have $s_y = \alpha_it_y$ if $y \in Y_i$, for $\alpha_i \in k$. Then just pick an $\alpha \in k$ that isn't equal to the $\alpha_i$.
I suppose this could fail over a finite field, but there's probably a better proof anyway...
