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Let $X$ be a smooth projective variety of dimension $n$ , and $D$ be a divisor. Suppose the linear system $|D|$ induce a birational map $$f: X -\to Y, $$ and let $H$ be the very ample line bundle such that $f^*H=D$.

I was wondering if the projection formula is still valid in this case? To be precise, I what to have the following two equations: $$D^n = (f^*H)^n=H^n,$$ and $$E\cdot H^{n-1} = 0$$ where $E$ is an exceptional divisor of $f^{-1}$.

Moreover, I was wondering if $Y$ is a normal variety in the above setting?

Any suggestion/comment are very welcome!!!

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I don't think this is true. A simple example is that the self-intersection of a divisor changes when you take its strict transform under a flop. In more detail:

Suppose that $\phi : X \dashrightarrow Y$ is a standard flop between smooth threefolds. Here's what I mean: there is a rational curve $C \subset X$ with normal bundle $\mathcal O(-1) \oplus \mathcal O(-1)$, we blow it up to get $f : Z \to X$ with an exceptional divisor $E$ isomorphic to $\mathbb P^1 \times \mathbb P^1$, which can be blown down along the other ruling via $g : Z \to Y$, contracting $E$ to a curve $C^+$. The induced map $\phi : X \dashrightarrow Y$ is birational, with no exceptional divisors in either direction.

Take $H$ very ample on $Y$, with $D$ its pullback via $\phi$ to $X$ (equivalently, its strict transform). This $D$ has base locus along the flopping curve, and its linear system gives the flop we want.

Now, $f^\ast D = g^\ast H + aE$, where $a = H \cdot C^+$ (which is greater than $0$, since $H$ is ample). Take the top self-intersection of both sides. On the left, we just have $D^3$. On the right, we have $H^3 + 3 a \, (g^\ast H)^2 E + 3 a^2 \, (g^\ast H) E^2 + a^3 E^3 = H^3 -3a^2 \, (H \cdot C^+) + 2a^3 = H^3 - a^3$. So $D^3 \neq H^3$.

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  • $\begingroup$ Thank you! How did you get $f^*D = g^*H + aE$? $\endgroup$ – Li Yutong Feb 21 '14 at 23:20
  • $\begingroup$ @Li Yutong: Since $\phi$ and $\phi^{-1}$ don't have any exceptional divisors, the difference $f^\ast D - g^\ast H$ must be exceptional, and hence just some multiple of $E$. To find the constant, write $f^\ast D = g^\ast H + aE$ and intersect both sides with one of the rulings contracted by $f$. On the left is $0$, and on the right is $H \cdot g_\ast(\gamma) + aE \cdot \gamma$. The normal bundle of $E$ is $(-1,-1)$, so that intersection is $-1$, which gives us the right value of $a$. $\endgroup$ – user47305 Feb 22 '14 at 18:57
  • $\begingroup$ @user47305, if you register an account we can merge both these accounts into it. Email community@stackexchange.com and tell them the account numbers you'd like merged. $\endgroup$ – Scott Morrison Feb 24 '14 at 8:10
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Since you write $D=f^*H$ I think you are assuming that $|D|$ is base point free and $f$ is a morphism (otherwise you must blow up the base locus to achieve this). In this case we have $D^n=deg(f) H^n$ where $deg(f)$ is the generic degree. $E\cdot H^{n-1}=0$ is clear as we can choose generic members $H_i\in|H|$ so that $H_1\cap \ldots \cap H_{n-1}\cap f(E)=\emptyset $. $Y$ is not necessarily normal (just pick a $Y$ singular cubic and $X$ its normalization and $|D|=f^*|O_Y(1)|$. However, if you consider $|tD|$ for $t\gg 0$, then the image is normal.

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    $\begingroup$ I did not assume $|D|$ to be basepoint free, acturally, the case when it has basepoint is what I bothered. Another thing is, $E$ is exceptional divisor of $f^{-1}$. $\endgroup$ – Li Yutong Feb 21 '14 at 16:58

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