Effective Cartier divisors corresponding to a line bundle

Question

Let $k$ be a field; $X$ a projective scheme over $k$; $\mathcal A$ an ample $\mathcal O_X$-module. Sometimes we want to apply induction on dimension by finding an effective Cartier divisor $D$ such that $\mathcal O_X(D) \cong \mathcal A$. However, the problem is, whether $D$ always exists, i.e., a regular section of $\mathcal A$ exists, under some assumptions on $\mathcal A$, for example, if we assume that $\mathcal A$ is generated by global sections, is it true that then $D$ exists? If it is not true, then maybe we have to assume that $X$ is integral, then any section of $\mathcal A$ is regular. But now we run into another problem, whether $D$ can be chosen to be integral again? If $D$ is not integral, then we cannot apply the induction hypothesis, then maybe we also have to assume that $X$ is smooth in order to apply the Bertini theorem to make sure $D$ is again smooth and integral. But sometimes smoothness is too strong for $X$ to hold, if there is a more suitable way to deal with this kind of problem.

Answer 1

If $\mathcal A$ is generated by global sections and $k$ is infinite then a regular section must exist. For each irreducible component of $X$, global sections vanishing on that component form a positive-codimension subspace of global sections. For non-reduced $X$, global sections vanishing on the induced reduced subscheme of any associated prime also find a positive-codimension subspace.

Over an infinite field, we may avoid arbitrarily many positive codimension subspaces with a single vector.

If $k$ is finite then this is not true. We could take $X$ to be the union of all $\mathbb F_q$-rational lines in $\mathbb P^2$ and $\mathcal A = \mathcal O(1)$. Then using the ideal sheaf of $X$ we can check that all global sections of $\mathcal A$ are the obvious global sections of $\mathcal O(1)$ on $\mathbb P^2$, each of which vanishes on a line.

But probably, in the situation you're thinking of, it's fine to assume $k$ is infinite.