Let $X/k$ be a smooth projective variety over an algebraically closed field $k$. For any divisor $D$ of $X$, is it true that deg$D=0$ $\Rightarrow$ $D$ is principal? The deg function is in the sense of ex 6.2, pg 146 of Hartshorne AG. I know that the converse is true. The reason that I ask this is because I need it to prove the following:
Q1: Is there a prime divisor $D_\infty$, such that for any $f\in k(X)$, there exists $g \in k(X)$ such that $fg$ has poles only at $D_{\infty}$, i.e. $(fg)=\sum n_iD_i-mD_{\infty}$ for $n_i,m>0$?
Also, if the above is true, a related question is:
Q2: Given $(f)=\sum n_i D_i-mD_\infty$, where $n_i,m>0$, does there exist $g \in k(X)$ such that $(gf)=(n_{i_0}-1)D_{i_0}+\sum_{i\neq i_0}n_iD_i-kD_\infty$?
Edited Q3: Is there a prime divisor $D_\infty$, such that for any $f_1, f_2\in k(X)$, there exists $g \in k(X)$ such that $f_1g$ and $f_2 g$ have poles only at $D_{\infty}$?
If $X=C$ is a smooth projective curve (so of dimension one), then the statement that all degree $0$ divisors are principal implies that $C$ is isomorphic to the projective line. For the proof, take two distinct points $x,y \in C$ and consider the divisor $D=[x]-[y]$. Take a function $f$ such that $\operatorname{Div}(f)=D$. Then $f$ defines a morphism $f: C \setminus \{y \} \to \mathbb{A}^1$ (it is a rational function with a pole at $y$) and it can be extended to a morphism $C \to \mathbb{P}^1$ by setting $f(y)=\infty$. This map is nonconstant (hence surjective) and its degree is $1$, so it is an isomorphism.
For $\operatorname{Dim}(X) \ge 2$, I think the definition of the degree of a divisor depends on a choice of projective embedding.