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I've encountered this assertion and I'm wondering how it is proved. (Here, a multiple point is defined as a point whose local ring is not a DVR, [EDIT] and a curve is a variety whose function field has transcendence degree 1 over the base field).

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(You should perhaps make explicit what is a curve for you...)

There is an open non-empty set of smooth points in a curve. The multiple points are therefore in the complement, and hence are finitely many.

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An irreducible algebraic (or even algebraic with no multiple components) curve $\cal C$ has certainly a finite number of multiple points, since these are the points that $\cal C$ has in common with other curves.

On the other hand, it's easy to come up with real analytic curves with infinitely many multiple points.

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    Indeed, if $c$ is the number of components (no multiple) and $n$ the degree of the curve, one has $\sum_{P} m_{P}(m_{P}-1) \leq \frac{(n-1)(n-2)}{2} + c-1 \leq \frac{n(n-1)}{2}$. This is obtained as a consequence of Bézout's theorem in the treatment of Fulton ($\textit{Algebraic curves}$).2016-08-31