If a divisor $D$ satisfies that $D^{2}=1$, is it true that the morphism induced by $|D|$ is birational? Let $X\subset \mathbb{P}^{5}$ be a non-degenerate algebraic surface. Let us suppose that $D\subset X$ is a curve such that $D^{2}=1$. I would like to know if the rational map induced by the complete linear system $|D|$
$$
X\dashrightarrow \mathbb{P}(H^{0}(X,\mathcal{L}(D))),
$$
where $\mathcal{L}(D)$ is the sheaf of modules associated to $D$, is birational.
If I am not wrong, under these hypothesis
$$
\mathcal{L}(D)\otimes \mathcal{L}(D)\simeq \mathcal{O}_{X}(1).
$$
Does it help? Do we need any more hypothesis? Do we need less?
 A: Here are two things I can show that are close to what you want:
Claim. Let $X$ be a normal projective surface, and let $D \subset X$ be an irreducible curve such that $D^2 > 0$. Then,
$$X \dashrightarrow \mathbf{P}(H^0(X,\mathcal{L}(mD))$$
is birational onto its image for some $m > 0$.
Proof. $D$ is nef since it is an irreducible curve on a surface with positive self-intersection. But then, $D^2 > 0$ also implies $D$ is big by [Lazarsfeld, Thm. 2.2.16], and so for some power of $D$, we have that $X \dashrightarrow \mathbf{P}(H^0(X,\mathcal{L}(mD))$ is birational onto its image by the theorem on Iitaka fibrations [Lazarsfeld, Thm. 2.1.33]. $\blacksquare$
Claim. Let $X$ be a smooth projective surface, and let $D$ be a divisor such that $D^2 > 0$. Then, either $D$ or $-D$ is big, and the conclusion of the previous claim holds for either $D$ or $-D$.
Proof. Fix an ample $H$. Then, $D.H \ne 0$, for if $D.H = 0$, then by the Hodge index theorem [Hartshorne, V, Thm. 1.9], we would have $D^2 < 0$. After possibly replacing $D$ by $-D$, we can assume without loss of generality that $D.H > 0$. We then claim that $h^2(X,\mathcal{L}(mD)) = 0$ for $m \gg 0$. By Serre duality, we have
$$h^2(X,\mathcal{L}(mD)) = h^0(X,\mathcal{L}(K-mD)),$$
and if $h^0(X,\mathcal{L}(K-mD)) \ne 0$, there is an effective curve linearly equivalent to $K - mD$, and so $K-mD.H > 0$. But this is a contradiction for $m$ large enough, since $D.H > 0$.
Now, applying the Riemann–Roch theorem [Hartshorne, V, Thm. 1.6] to $mD$, we have
$$h^0(X,\mathcal{L}(mD)) \ge \frac{1}{2} m^2D^2 - \frac{1}{2}mD.K + 1 + p_a(X).$$
Thus, $h^0(X,\mathcal{L}(mD))$ grows quadratically as a function of $m$, and so $D$ is big by [Lazarsfeld, Lem. 2.2.3], and we conclude as in the other Claim. $\blacksquare$
