Let $\mathbb{K}\in \{\mathbb{Z},\mathbb{Q},\mathbb{R}\}$ and let $D$ be a base-point free(or ample if it is necessary) $\mathbb{K}$-divisor on a normal projective variety.
I have two questions:
- When $\mathbb{K}=\mathbb{Q}$ or $\mathbb{R}$, can we always take $D^\prime\in |D|_{\mathbb{K}}$ such that $D^\prime$ has arbitary small coefficients?
- When $\mathbb{K}=\mathbb{Z}$ or $\mathbb{Q}$, is there an $E\in |D|_{\mathbb{K}}$ such that $E$ is a prime divisor?
(1) is fully true (as long as we're over an infinite field). Indeed, for any $n$, since $|D|$ is free we can pick $E_1, ... , E_n \in |D|$ distinct and consider $F_n = 1/n\sum E_i$. The coefficients of $F_n$ get arbitrarily small as $n$ gets large and each $F_n$ is $\mathbb{Q}$-linearly equivalent to $D$.
(2) is true in dimensions $\geq 2$ and characteristic $0$ by Bertini's theorem. For a counterexample in the case of curves take $X = \mathbb{P}^1$ and the complete linear series $|\mathcal{O}(2)|$. If $D$ is a $\mathbb{K}$-divisor which is $\mathbb{K}$-linearly equivalent to a divisor in this linear series, we would have that the degree of $D$ is 2, so it cannot be a prime divisor.
Apologies for the substantial edit. I didn't read the word 'free' initially and the old answer was incorrect.