Let $D$ be a base-point-free divisor on a normal projective variety $X$, and let $Y$ be the image of the morphism $f_{D}:X\rightarrow Y$ induced by $D$. Assume that $f_D$ is birational.

Now, let $X(D)=Proj\left(\bigoplus_{k\geq 0}H^{0}(X,kD)\right)$. Is $X(D)$ the normalization of $Y$ ?

  • $\begingroup$ No, that is not true. Let $X$ be a hyperelliptic curve. Let $D$ be the unique basepointfree divisor class of degree $2$. Then $Y$ is $\mathbb{P}^1,$ and the morphism $f_D$ is the quotient by the hyperelliptic involution. Yet $X(D)$ equals $X$, not $Y$. $\endgroup$ Jan 20 '18 at 14:51
  • $\begingroup$ maybe something is wrong in your question. Say $X$ is a curve maps to $Y=\mathbb{P}^1$ and take $D$ to be pull back of a point on $Y$. Then $X(D)=X$ since $D$ is ample. $\endgroup$
    – Chen Jiang
    Jan 20 '18 at 14:54
  • $\begingroup$ I forgot to specify that $f_D:X\rightarrow Y$ is birational. Sorry. I edited the question. $\endgroup$
    – J. Ross
    Jan 20 '18 at 14:59

The modified assertion is true. For a field $k$, for every proper $k$-scheme $X$, for every $k$-morphism $$f:X\to \mathbb{P}^n,$$ the irreducible curves in $X$ that are contracted by $f$ are precisely the irreducible curves having degree $0$ with respect to the invertible sheaf $\mathcal{L}:=f^*\mathcal{O}(1)$. These are also the curves that are contracted by the natural morphism $$g:X \to X(\mathcal{L}),$$ where $X(\mathcal{L})$ is $\text{Proj} \bigoplus_{d\geq 0} H^0(X,\mathcal{L}^{\otimes d})$. Thus, the natural $k$-morphism $$h:X(\mathcal{L})\to f(X)$$ is a finite morphism. If $X$ is normal, then also $X(\mathcal{L})$ is normal. If also $h$ is birational, e.g., this holds if $f$ is birational, then $h$ is the normalization of the image of $f$.

  • $\begingroup$ Thank you very much for the answer. There is just one thing I do not understand yet. How do we know that $h:X(\mathcal{L})\rightarrow f(X)$ is exactly the normalization? Do you think that there may exist a birational morphism (which is not an isomorphism) $g:X(\mathcal{L})\rightarrow f(X)^{\nu}$ such that $\nu\circ g = h$ where $\nu:f(X)^{\nu}\rightarrow f(X)$ is the normalization? $\endgroup$
    – J. Ross
    Jan 20 '18 at 20:45
  • $\begingroup$ Since $X(\mathcal{L})$ is normal, the dominant morphism $h$ factors through the normalization of $f(X)$. Since $h$ is finite, the morphism from $X(\mathcal{L})$ to the normalization of $f(X)$ is also finite. By Zariski's Main Theorem, a birational, finite morphism between normal varieties is an isomorphism. $\endgroup$ Jan 20 '18 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.