Unibranch points of a classical variety

Assume the Axiom of Choice. Let $X$ be an irreducible classical variety over an algebraically closed field with normalization $\nu \text{\hspace{0.17em}⁣}:X^{\nu }\to X$ (Normalization of a classical variety by gluing affine normalizations). A point $x\in X$ is unibranch when the fibre ${\nu }^{-1}(x)$ is a single point (Images and fibres of a regular map).

The fibre is finite because $\nu$ is a finite morphism (Normalization is finite, surjective and birational), so unibranchness is the statement that this finite fibre has exactly one element. In the curve case the points of ${\nu }^{-1}(x)$ correspond to the local branches of $X$ at $x$, so a unibranch point is one with a single branch in this normalization sense; over an arbitrary algebraically closed field this wording makes no reference to complex analytic topology.

Recorded properties. Every normal point is unibranch: over the normal locus the normalization restricts to an isomorphism (The normalization is an isomorphism over the normal locus), so the fibre over a normal point is a single point, and a point of $X$ is normal exactly when its local ring is an integrally closed domain (Normal points and normal varieties). The converse fails: the cusp on the companion page is unibranch but not normal, and the node is not unibranch. Unibranchness is a property of the point alone, read in the normalization; for a reducible reduced variety one uses the full disjoint-union normalization of Normalization of a classical variety by gluing affine normalizations, counting preimages on every component through $x$. A point on two distinct components therefore cannot be unibranch.