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Let $X$ be a normal variety and $D$ be a Cartier divisor, suppose $\sigma, \delta$ are two basepoint free linear systems in $|D|$, then we have two morphisms defined by these two linear system: $$\phi_{\sigma}: X \to \mathbb{P}^n,\qquad \phi_{\delta}: X \to \mathbb{P}^m. $$ My question is: are the images $\phi_{\sigma}(X), \phi_{\delta}(X) $ the same (maybe after a normalization)?

Li Yutong
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  • As Jake Levinson's answer shows, the images are not the same. But the Stein factorisations of the two morphisms are the same. (I think this is explained in Lazarsfeld's book, but I don't have time to check just now.) –  Dec 15 '14 at 17:13

1 Answers1

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No. For example, let $X$ be a smooth quartic surface in $\mathbb{P}^3$ and let $L$ be the restriction of $\mathcal{O}(1)$.

Now $L$ is very ample, and its complete linear system is the (isomorphic) embedding of $X$. On the other hand, if we project from a point $p$ not on $X$, the corresponding linear system is still basepoint free, and gives a surjective map $X \to \mathbb{P}^2$. So the images are not isomorphic (from the adjunction formula, the canonical divisor on $X$ is trivial, so $X \not\cong \mathbb{P}^2$.)

Jake Levinson
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  • If I understand correctly, you construct $X \hookrightarrow \mathbb{P}^3 \to \mathbb{P}^2$, but the pullback of $\mathcal{O}{\mathbb{P}^2}(1)$ is not $\mathcal{O}{\mathbb{P}^3}(1)$, so they are not corresponds to the same linear system $L$. – Li Yutong Nov 08 '14 at 22:37
  • @LiYutong: Hi Li, $\mathcal{O}{\mathbb{P}^3}(1)$ and $\mathcal{O}{\mathbb{P}^2}(1)$ will both pull back to the same line bundle $L$ on $X$. You can see this directly -- pull back a hyperplane (i.e. a line) from $\mathbb{P}^2$ first to $\mathbb{P}^3$ (so it becomes a $2$-plane), then to $X$. Either way, the divisor you get on $X$ is the same. – Jake Levinson Nov 08 '14 at 23:54
  • The pull back of a line won't be a 2-plane. In fact, the $f^*\mathcal{O}_{\mathbb{P}^2}(1)$ will be a nef divisor but not ample - this is because the morphism $f$ contracts curves. – Li Yutong Nov 09 '14 at 01:15
  • @LiYutong What is $f$? If $f$ is the projection from a point (which incidentally is not a morphism $\mathbb{P}^3 \to \mathbb{P}^2$, but only a rational map; but is a morphism $X \to \mathbb{P}^2$ since $p \notin X$) then $f$ does not contract any curves on $X$. Indeed $f: X \to \mathbb{P}^2$ is a finite map, a Noether normalization of $X$. – Jake Levinson Nov 09 '14 at 02:13
  • @LiYutong: Also, the (closure of the) preimage of a line of $\mathbb{P}^2$ in $\mathbb{P}^3$ from projection from a point is indeed a 2-plane. You may wish to look up 'projection from a point' as it is important for understanding incomplete linear systems. (Incidentally on $\mathbb{P}^n$, a nontrivial divisor is nef if and only if it is ample...) – Jake Levinson Nov 09 '14 at 02:22