Equivalent definitions for normal variety

Question

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.

Answer 1

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.

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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.

Answer 2

There are a few things going on here. First of all, we should say something like "isomorphism with an open subset". Then one implication is just a form of what is usually called Zariski's main theorem.

The other implication follows from normalization itself. If $X$ is not normal, then $\tilde{X}\to X$ is a finite, surjective, birational map which is not an isomorphism.

As for intuition, I would say that I always associate normality with regularity in codimension $1$ (this is part of Serre's criterion*). Since a birational morphism is an isomorphism away from a set of codimension $1$, it seems reasonable that we would first look to normality when deciding if such a morphism can be extended.

* Specifically, a variety is normal if and only if its singular locus has codimension at least $2$, plus an additional technical condition (it is enough to be a hypersurface in something smooth).