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I am confused about a couple of steps in the proof of the birational invariance of the geometric genus (Theorem II.8.19 in Hartshorne's Algebraic Geometry).

I shall sketch the proof and highlight my doubts.

Let $X,X'$ be two birationally equivalent nonsingular projective varieties over a field k. Hence there is a birational map $X-->X'$ represented by a morphism $f:V\rightarrow X'$ for some largest open subset $V\subset X$.

Along taxonomic lines, the proof goes like this:

  1. We first prove that $f$ induces an injective map $f^{\ast}:\Gamma(X',\omega_{X'})\rightarrow \Gamma(V,\omega_V)$
  2. Then we prove that the restriction map $\Gamma(X,\omega_X)\rightarrow \Gamma(V,\omega_V)$ is bijective, using the valuative criterion of properness.

From this it follows that $\rho_g(X')\leq \rho_g(X)$, and the reverse inequality follows by simmetry.

In the proof of step 1: the map $f$ induces an isomorphism $U\cong f(U)$ for some open subset $U\subset V\subset X$ and then Hartshorne claims that this implies that f induces an isomorphism $\omega_{V|U}\cong \omega_{X'|f(U)}$. Why is that?

In the proof of step 2: from the valuative criterion of properness it follows that $\textrm{codim }(X\setminus V,X)\geq 2$. In order to prove that $\Gamma(X,\omega_X)\rightarrow \Gamma(V,\omega_V)$ is bijective it suffices to prove it on open sets $U\subset X$ trivializing the canonical sheaf $\omega_{X|U}\cong \mathcal{O}$, namely that $\Gamma(U,\mathcal{O}_U)\rightarrow \Gamma(U\cap V,\mathcal{O}_U\cap V)$ is bijective.

Since $X$ is nonsingular, from the first remark in the previous paragraph we have that $\textrm{codim }(U\setminus U\cap V,U)\geq 2$ and then Hartsorne claims that the result (bijectivity) follows immediately from the fact that for an integrally closed Noetherian domain $A$, we have $A=\bigcap_{\textrm{ht } \mathfrak{p}=1} A_{\mathfrak{p}}$. I do not see this either.

Thanks in advance for any insight.

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    $\begingroup$ Regarding your item 1, the point is that $\omega_V|_U= \omega_U$ and that the canonical sheaf is an isomorphism invariant. Item 2 has to do with what is sometimes viewed as the algebraist's Hartogs theorem: A function defined on the complement of codim $\ge 2$ set of a normal variety extends. This can be reduced to the affine case, where it amounts to the equality $A=\cap A_p$ (over ht 1 primes). I may expand this later, if no one else gives you a more detailed answer. $\endgroup$ – Donu Arapura Mar 15 '12 at 13:07
  • $\begingroup$ Does anyone know why this proof works for the global sections of the canonical divisor, but not for those of the anticanonical divisor? $\endgroup$ – schemer Oct 3 '19 at 5:56
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I am not sure that this proof is correct for the following reason. If $ X^{‘} $ is a non-singular surface and $ X $ is the blow-up of the surface at a non-singular point, then the sub-variety $ X \setminus V $ has codimension one in $ X $. This contradicts the result of step two, which says that $ X \setminus V $ is a closed set of codimension greater than or equal to two. In a thread, I asked about a generalization, and there is a possible revision of this proof that fixes this.

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  • $\begingroup$ To clarify, the result is correct, but the proof may have a mistake in it. $\endgroup$ – schemer Oct 2 '19 at 2:07
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    $\begingroup$ The blowdown is defined on all of $X$, $V=X$. $\endgroup$ – Zach Teitler Oct 2 '19 at 3:26
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    $\begingroup$ Sorry, I didn’t realize that $ V $ was the largest set on which the morphism was defined. I thought it was the set where the morphism is an isomorphism. Thank you Zach. $\endgroup$ – schemer Oct 2 '19 at 7:08

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