Let $X$ be a smooth projective variety, and $\phi: X\dashrightarrow \mathbb P^n$ a rational map given by some complete linear system $|D|$ where $D$ is a divisor. Is the image variety $\phi(X)\subset \mathbb P^n$ normal?
Image of a rational map given by a complete linear system
algebraic-geometry
Related Solutions
The i.e. part is a bit misleading. It is true that there exists $m$ so that $|mD|$ defines a map birational to its image, but this is not necessarily the case for every $m$ such that the map associated to $|mD|$ is generically finite to its image.
For example, if $X$ is a smooth curve of genus $2$, the complete linear system $|K|$ associated to the canonical divisor gives a hyperelliptic cover of $\mathbf{P}^1$. However, $K$ is ample so sufficiently high powers define embeddings.
Let's prove that if $D$ is big, there is some $m > 0$ so that $|mD|$ defines a map that is birational to its image. It suffices to show that the map is generically one-to-one to its image.
(This is following 1.4.2 from dFEM14.)
Let $D$ be big. Then, for some $m$, we may write $mD = A + E$ where $A$ is ample and $E$ is effective. Replacing $m$ by a sufficiently high multiple, we may then assume $mD - E$ is very ample.
Now, let $|W_m|$ be the image of $|mD - E|$ in $|mD|$. Then, $|W_m|$ and $|mD|$ define rational maps to projective space differing by a linear projection $\mathbf{P}(H^0(X, \mathcal{O}(mD))) \dashrightarrow \mathbf{P}(W_m)$. But now, away from $E$, the map defined by $|W_m|$ agrees with the one defined by $|mD - E|$, which is an embedding. Hence, the map given by $|W_m|$ is generically one-to-one to its image.
But now, this implies that $X \dashrightarrow \mathbf{P}(H^0(X, \mathcal{O}(mD)))$ is also generically one-to-one to its image, and the result follows.
EDIT: Let $X \dashrightarrow Y$ be a dominant map of varieties. We claim that $X$ and $Y$ are of the same dimension if and only if this map is generically finite.
Suppose they are of the same dimension. Harthsorne's exercise II.3.22 implies that a general fiber is finite. Hence, the map is generically finite.
Conversely, if $X \dashrightarrow Y$ is dominant and generically finite, then $K(Y) \subset K(X)$ is finite so their transcendence degrees over $k$ must be the same, and hence the dimensions of $X$ and $Y$ are the same.
EDIT2: Let's show that if $\varphi: X \dashrightarrow Y$ is a generically one-to-one dominant rational map of projective varieties, it is birational. To do this, we reduce to the case that $\varphi$ is regular(ie. everywhere defined).
Let $\Gamma$ be the graph of $\varphi$, so there is a birational map $\Gamma \to X$, and a regular morphism $\overline{\varphi}: \Gamma \to Y$ dominating $\varphi$. But then, $\varphi$ being birational is equivalent to $\overline{\varphi}$ being birational. We may then assume $\varphi$ is everywhere defined.
In this case, we can find an open subset $U \subset Y$ so that $\varphi^{-1}(U) \to U$ is bijective and projective (as the base change of $\varphi$, a projective morphism). Hartshorne's exercise III.11.2 implies this map is finite of degree 1, and hence birational.
Best Answer
Not necessarily.
For example, take a general curve $C$ over $\mathbb{C}$ of genus not a triangular number, and $P\in C$ a point. Let $D=nP$ with $h^0(D)=3$. Then $\phi\colon C\dashrightarrow\mathbb{P}^2$ extends to $C\to\mathbb{P}^2$ and obviously $\phi(C)$ cannot be smooth (since it would violate the degree-genus formula).