Suppose that $X$ and $Y$ are smooth projective varieties which are birationally equivalent. I would like to have that $$\textrm{deg} \ \textrm{td}(X) = \textrm{deg} \ \textrm{td}(Y).$$ Invoking the Hirzebruch-Riemann-Roch theorem, this boils down to showing that $$ \chi(X,\mathcal{O}\_X) = \chi(Y,\mathcal{O}\_Y).$$ 


This is probably a basic fact. A stronger statement is apparently shown in *Birationale Transformation von linearen Scharen auf algebraischen Mannigfaltigkeiten* by van der Waerden. The only problem is that I can't seem to find it in that article (probably because I don't read that well German).


For $\dim X =2$ one can prove this as Hartshorne does as follows.


Any birational transformation of nonsingular projective curves can be factored into a sequence of monoidal transformations and their inverses. For such a monoidal transformation, the result follows from Proposition 3.4 in Chapter V of Hartshorne. 


Does this work in the general case?



Is the Euler characteristic a birational invariant
If you are willing to stick to characteristic zero, then you can assume that there is actually a morphism $f\colon X\longrightarrow Y$ realizing the birational equivalence (reason: look at the graph $\Gamma\subset X\times Y$ realizing the birational equivalence and take its closure, use resolution of singularities to resolve $\Gamma$, and then replace $X$ by $\Gamma$). In this case, $f\_{\*}\mathcal{O}\_{X}=\mathcal{O}\_{Y}$, and all higher direct images are zero, the Leray spectral sequence then implies that the Euler characteristics are equal. 


More generally, if $Y$ has *rational singularities* and $f\colon X\longrightarrow Y$ is a proper birational map, with $X$ smooth, then $f\_{\*}\mathcal{O}\_{X}=\mathcal{O}\_Y$ and all higher direct images are zero (this is the definition of rational singularities) and so the same conclusion follows. Smooth varieties have rational singularities! (The computation for smooth varieties is necessary to show that the definition makes sense, i.e., that checking that this property holds for one resolution $X$ implies that it holds for all resolutions).


