# when a birational morphism is an isomorphism?

Context: Surfaces (smooth projective complex)

The title is way more general than what I'd like to understand. I am trying to understand this question in the following very special situation: let $X$ be a surface and $L$ an ample, base point free and globally generated line bundle such that $\varphi=\varphi_L$ is a birational morphism,

$$\varphi: X\longrightarrow Y:=\varphi(X)\subset\Bbb{P}^N$$

We wonder when is $\varphi$ an embedding. As suggested by Asal B.D. in this SE post, the normality of $Y$ would be enough. However I have no idea how to check normality of $Y$ in this case. Also, I would like to understand the following, possibly trivial, fact:

(Why) would it be enough to show that $H^0(Y,\mathcal{O}(r))=H^0(X,L^{\otimes r})$ for all $r$?

• In your case, $\phi_*\mathcal{O}_X$ would be the normalization of $\mathcal{O}_Y$ and if the above equality on global sections hold, it will easily prove that $Y$ is normal, from the exact sequence,$$0\to \mathcal{O}_Y\to\phi_*\mathcal{O}_X\to \mathrm{coker}\to 0,$$ and tensoring with $\mathcal{O}_Y(r)$. Oct 9 '13 at 14:06

To say that $L$ is ample is equivalent to $X=\mathrm{Proj}\oplus H^0(X,L^{{\scriptscriptstyle\otimes} r})$, while of course $Y=\mathrm{Proj}\oplus H^0(Y,\mathcal{O}(r))$.