Normalization

Remarks

The scheme $\tilde{X}$ is normal. For every normal integral scheme $Z$ and every dominant morphism $f \colon Z \to X$, there is a unique morphism $g \colon Z \to \tilde{X}$ with $f = \nu \circ g$. Example: let $A = k[x, y]/(y^2 - x^3) \cong k[t^2, t^3]$ over a field $k$, with $x = t^2$ and $y = t^3$. Its fraction field is $k(t)$, and $t$ is integral over $A$ since $t^2 = x$; also $k[t]$ is integrally closed, so $\tilde{A} = k[t]$. The normalization of the cusp is $\mathbb{A}^1_k \to \operatorname{Spec} A$, $t \mapsto (t^2, t^3)$. Non-example: the identity morphism of the cusp is not its normalization, since the cusp is not normal and $A \ne \tilde{A}$. If $X$ is normal, then $\nu$ is an isomorphism.