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Let $Y$ be a smooth variety of dimension $n$. Then I can get (a representative for) the canonical divisor class $K_Y$ on $Y$ by taking any rational $n$-form $\omega$ on $Y$ and taking its divisor of zeroes and poles, so $K_Y\equiv div(\omega)$.

Now let $f:X \to Y$ be a birational morphism between smooth varieties of dimension $n$, and $\omega$ an $n$-form on $Y$. Then $f^\ast\omega$ is an n-form on $X$, so by the above recipe (take the divisor of any n-form to get a representative for the canonical divisor class) we get $K_X\equiv div(f^*\omega)$. Also $div(f^*\omega)= f^*div(\omega)$ by definition of pullback of a Cartier divisor. So $K_X\equiv f^*K_Y$ which is not true .. I am missing the exceptional divisor. Where is the flaw in my argument?

This question proves $K_X \equiv f^* K_Y + R$ where $R$ is supported on the exceptional divisor, and it might help to spot where I'm going wrong.

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ykm
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It's exactly not true that $div(f^*\omega) = f^*div(\omega)$. (And your claim that it follows from the definition of the pull-back of Cartier divisors is simply not true.)

Let's just consider the case of one point being blown up, and look locally around the point being blown up; let's say we have local coords. $x$ and $y$, and $x = y = 0$ is the point being blown up.

Suppose that $\omega = dx \wedge dy$ locally, so it has trivial divisor (at least in a n.h. of $(0,0)$).

The blow-up map is $(x,y) \mapsto (x,xy),$ and so this differential pulls back to $dx \wedge d(xy) = x dx \wedge dy,$ which now has a non-trivial divisor, namely $x = 0$, which is exactly the exceptional divisor.

This computation shows that indeed $$div(f^* \omega) = f^*div(\omega) +\text{ the exceptional divisor}.$$

Matt E
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    Ok, I see. Thank you! $f^div(\phi) = div(f^\phi)$ for a rational \emph{function} $\phi$, but not for a differential form. – ykm Jun 16 '14 at 04:08