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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 $Y$ ?

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  • $\begingroup$ 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
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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 $f$.

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  • $\begingroup$ 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

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