Equivalent definitions for normal variety Can anyone show (or provide a reference) that an algebraic variety $X$ is normal iff every finite, birational morphism $f:Y\rightarrow X$ is an isomorphism. 
More importantly, can you describe your intuition that connects these two viewpoints.

Here is a proof of the equivalence:
Suppose that $Y$ is normal, and let $f: X \to Y$ be finite.   Let Spec $A$ be an
open affine subset of $Y$.  Since $f$ is finite, we have that $f^{-1}($Spec $A)$
is an open affine subset Spec $B$ of $X$, with $A \to B$ finite.
Since $f$ is birational, $A$ and $B$ have the same fraction field, say $K$.
Thus $B$ is a finite, and in particular integral, $A$-subalgebra of $K$.
Since $Y$ is normal, $A$ is integrally closed (by definition of normal),
and thus actually $B = A$.  Thus $f$ is an isomorphism.
If $Y$ is not normal, then as you-sir-33433 notes, the normalization of $Y$
is a finite birational morphism that is not an isomorphism.

My intuition for this statement is essentially just to remember this proof,
in the following capsule form:  birational means same field of fractions; finite means the affine rings are being extended integrally; normal means the affine rings are integrally closed.
