Nonsingular is rational singular I know the following result is well known to the experts, but what are the good references or proofs:
Let $X,Y$ be a smooth variety with projective, birational morphism $f: X \to Y$.  Let $L$ be a line bundle on $Y$, then we have the property that the higher direct image $R^if_*f^* L = 0$ when $i > 0$.
This is equivalent to the statement that nonsingular variety is a rational singular variety.
 A: I think there are two issues here:
1) The "projection formula" tells you that $R^if_*(f^*L)$ is isomorphic to $R^if_*\mathcal{O}_X\otimes L$: see EGA chap. 0 , Prop. 12.2.3 (in EGA III.1).
2) Then the vanishing of $R^if_*\mathcal{O}_X$ for $i>0$. This is one of the many consequences of Hironaka's resolution of singularities: Resolution of singularities of an algebraic variety over a field of characteristic zero, I. Ann. of Math. (2) 79 (1964), 109–203 (the result you need is on p. 144-145).
A: There is another way to view this (and prove the second part) that is probably worth explaining (and mentioning some characteristic $p > 0$ connections).  As abx already pointed out, this problem reduces to computing $R^i f_* O_X$.  You don't need resolution of singularities to prove that vanishing though.
Characteristic zero
In characteristic zero, by Grauert-Riemenschneider vanishing, we have that $R^i f_* \omega_X = 0$ for all $i > 0$ (even if $Y$ is singular).  On the other hand, since the relative canonical divisor of a projective birational map between regular schemes $f : X \to Y$ is always effective (this is true whenever the relative canonical divisor makes sense, including mixed characteristic, see the work of Lipman on pseudo-rational singularities), we have that $f_* \omega_X = \omega_Y$.
In other words, $R f_* \omega_X \simeq \omega_Y$ in the derived category (they are quasi-isomorphic).  Hence by Grothendieck duality for $f$ we see that $O_Y \simeq R f_* O_X$ in the derived category as well.  This implies that $R^i f_* O_X = 0$ for all $i > 0$.
Characteristic $p > 0$
IF $Y$ is smooth, then the same results also holds in characteristic $p > 0$.
See
https://arxiv.org/abs/0911.3599
However, Grauert-Riemenschneider is known to fail in characteristic $p > 0$ for singular $Y$ so you really need $Y$ to at least have mild singularities.  See for instance example 3.11 in
http://arxiv.org/pdf/1212.5105.pdf
It is reasonable to perhaps conjecture that if $Y$ has $F$-regular singularities then Grauert-Riemenschnedier vanishing is true and hence $R^i f_* O_X = 0$ for all $i > 0$ and all resolutions of singularities $f : X \to Y$.
