# The property of line bundle corresponding to birational map

For a variety, if it has a base point free line bundle, then one can define a morphism from the variety to a $\mathbb{P}^n$. And if a variety is a projective variety(in the sense of Hartshorne), it is equivalent to have a very ample line bundle.

In the same vein, I was wondering if there are properties of line bundle corresponding to birational map. To be precisely:

Let $V$ be a variety, $L$ be a line bundle on $V$. Which conditions are needed to ensure the $map$ defined by $L$ (i.e. using all the basis of $H^0(V,L)$)is a birational map to its image in $\mathbb{P}^n$? Here $n=dim\ H^0(V,L)-1$.

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Note that $L$ doesn't define a unique morphism to a projective space. You need to specify finitely many global sections which generate $L$. –  user18119 Jan 1 '13 at 22:02
@QiL But I guess $L$ will define a morphism up to a PGL(n) action. –  Li Zhan Jan 2 '13 at 1:15
Yes, but in general you don't need to take a basis of $H^0(V, L)$. Any subset of sections generating the sheaf $L$ induces a morphism to some projective space. –  user18119 Jan 2 '13 at 10:46
@Should I require the set defined the morphism is exactly the set of $H^0(V,L)$? –  Li Zhan Jan 2 '13 at 19:16
This is a winy detail, but i think you meant $n = dim H^0(V,L) - 1$ instead of $n = dim H^0(V,L)$.. I though i'd point it out, but its not really important of course.. –  Joachim Jan 3 '13 at 12:41

I don't have a satisfying answer. Anyway here is one.

Let $L$ be a basepoint-free line bundle on a projective variety $V$ over a field $k$. Let $$\pi : V \to \mathbb P(H^0(V, L))\simeq \mathbb P^n$$ be the morphism associated to the complete linear system $|L|$. If we choose a basis $s_0, \dots, s_n$ of $H^0(V, L)$, then set-theoretically $$\pi(v)=[s_0(v):\dots : s_n(v)].$$ If $F(X_0,\dots, X_n)$ is a homogeneous polynomial of degree $m$, then $\pi^{-1}(D_+(F))$ is $V_f$ where $f=F(s_0,\dots, s_n)\in H^0(V, L)^{\otimes m}$ and $$V_f:=\{ v\in V \mid f(v)\ne 0\}.$$ The morphism $V_f\to D_+(F)$ is given by $$k[T_0, \dots, T_n]_{(F)} \to O_V(V_f), \quad P/F^r \mapsto P(s)/F(s)^r.$$

We know that if $L$ is very ample, then for any section $f\in H^0(V, L)$, $V_f$ is an affine open subset and the above homomorphism is surjective. The same is true if we take $f$ in $H^0(V, L)^{\otimes m}$ for any $m\ge 1$.

In general $\pi$ is birational if and only if there exists a dense open subset $U$ of $\mathbb P^n$ such that $\pi: \pi^{-1}(U) \to U$ is a closed immersion (an isomorphism from $\pi^{-1}(U)$ onto $U\cap \pi(X)$). The complement of $U$ is contained in a hypersurface $V_+(F)$ of deree $m$. Shrinking $U$ if necessary, we can suppose $U=D_+(F)$. Then $\pi^{-1}(U)=V_f$ where $f=F(s)$ as above. The condition for $V_f\to U$ to be a closed immersion is: $V_f$ is affine and $O_V(V_f)$ is generated, as $k$-algebra, by $H(s)/f(s)$ where $H$ runs through monomials of degree $m$.

Summarizing the above discussion, we can say that $\pi$ induces a birational morphism from $V$ to its image if and only if there exists $m\ge 1$ and $f\in H^0(V, L)^{\otimes m}$ such that:

1. $V_f$ is affine,
2. $O_V(V_f)$ is generated by $H(s_0,\dots, s_n)/f$ where $H$ are the monomials of degree $m$.

Everything remains true if we replace $H^0(V, L)$ with a vector subspace generating $L$.

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Dear QiL, Thank you so much! I am very happy to have your answer,it clarified a lot thing about birational morphisms. I did not get you in two places in your answer, could you please explain a little bit more? (1)you wrote: $\pi:\pi^−1(U) \to U$ is a closed immersion. Is this because we assume X is proper, and thus $\pi(X)$ is closed? (2) you wrote: The complement of $U$ is contained in a hypersurface $V_+(F)$ of degree $m$. I cannot see why this is true, and moreover, I cannot see why this is used in the following argument? –  Li Zhan Jan 4 '13 at 1:52
Dear @LiZhan: (1) yes because $\pi(X)$ is closed. (2) The complement is defined by a non-zero ideal $I$. Let $F\in I$ be homogeneous and non-zero, then $V_+(F)$ contains this complement. Subsequently the complement is replaced by $V_+(F)$ so $U$ is affine. This implies that $V_f=\pi^{-1}(U)$ is affine. –  user18119 Jan 4 '13 at 13:56