# Birational Invariants of Regular Surfaces

Let $$X,Y$$ surfaces (so $$2$$-dimensional proper $$k$$-schemes) which are regular and birational and denote by $$f: X – \ – \to Y$$ the corresponding rational birational morphism between $$X$$ and $$Y$$.

My aim is to show that for regular surfaces the dimensions of cohomology groups are birational invariants; therefore $$\dim_k H^i(X,O_X)= \dim_k H^i(Y,O_Y)$$ for $$i=0,1$$.

In order to do it I tried following consideration:

It is well know that if $$b: B:=Bl_z(Z) \to Z$$ is the blowing up of regular surface $$Z$$ at $$z \in Z$$, then the dimensions of cohomology groups are conserved; i.e. invariant.

If $$f$$ would be a “classical” morphism (so not only rational) then according to a factorization theorem (which one I don’t know more; does anybody know it’s name?) then $$f$$ factorize into a finite sequence of successively blowing ups

$$f: X= Y_n \to Y_{n-1} \to … \to Y_0=Y$$

where $$Y_{k+1}= Bl_{Z_i}(Y_k)$$ is the blow up of the previous one.

The problem here is that our $$f$$ is just a rational map.

My idea was to try to construct following diagram (D)

where $$\Gamma := \overline{\Gamma_f} \subset X \times Y$$ where $$\Gamma_f$$ also rational defined graph morphism via the pull back on $$id_Y:Y \to Y$$ along $$f$$. This induces rational projection $$\tilde{pr_X}: \Gamma_f – \ – \to X$$.

My goal would be that after taking closure $$\Gamma := \overline{\Gamma_f} \subset X \times Y$$ the rational morphism $$\tilde{pr_X}$$ would induce “classical” map $$pr_X:\Gamma \to X$$ making the diagram (D) commutative. Then – if $$pr_X,pr_Y$$ are birational and $$\Gamma$$ regular – I can apply the factorization theorem to $$pr_X,pr_Y$$ and ontain the desired result.

And exactly this is the point: Which properties does this closure $$\Gamma$$ inherit? Stays it regular, proper and birational to $$X,Y$$? Why?

The problem is that I’m not sure what control over $$\Gamma_f$$ I have after taking the closure in $$X \times Y$$.