Is an irreducible algebraic curve over $\mathbb{C}$ uniquely determined by a local parametrization? Let $F(X, Y)$ be an irreducible polynomial of $\mathbb{C}[X, Y]$ with $F(0, 0) = 0$ and non-singular there, and let $(X(t), Y(t))$ be a local analytical parametrization of the corresponding curve, that is, $F(X(t), Y(t)) = 0$ for $t$ in a neighborhood of $0$, with $X(0) = Y(0) = 0$ and $X'(0) \neq 0$.
If, for some polynomial $G(X, Y)$, we have $G(X(t), Y(t)) = 0$ for $t$ in the neighborhood in question, does then $G(X, Y)$ necessarily belong to the ideal generated by $F(X, Y)$ ? In other words, can the whole curve $F(X, Y) = 0$ be reconstructed from its local snippet?
And for an arbitrary algebraically closed field, is there a well-known characterization for a subset of an irreducible algebraic curve to define the curve already ? Is it for instance enough for such a subset to be infinite ?
 A: The result that decides your problem is the following: let $f,g \in \mathbb{C}[x,y]$ be two non-associated irreducible polynomials. Then the set of common zeros of $f$ and $g$ is finite.
Proof: we can see $f,g$ as polynomials in $\mathbb{C}(x)[y]$.
Suppose one of them is constant: then (for instance) $f=f_0(x)=x-x_0$ ($f$ irreducible) and $g=g(x,y)$. If $f$ and $g$ have infinitely many common roots, this means that $g(x_0,\cdot)=0$. Then $x-x_0$ divide all the coefficients of $g(x,y) \in \mathbb{C}[x][y]$, so that $f|g$ and thus $f,g$ are proportional.
Otherwise, $f$ and $g$ are two non-proportional irreducible elements of $\mathbb{C}[x,y]$ so they are irreducible non-proportional in $\mathbb{C}(x)[y]$ so coprime. Thus there are $a,b \in \mathbb{C}(x)[y]$ such that $af+bg=1$. Multiplying by all denominators, it means that there is some nonzero $p \in \mathbb{C}[x] \cap (f,g)$. Similarly, there is some nonzero $q \in \mathbb{C}[y] \cap (f,g)$. Then, if $(x,y)$ is a root of $f$ and $g$, $p(x)=0$ and $q(y)=0$, which leaves finitely many possibilities. QED.
A: The result you need is the following: in a irreducible complex  variety every open subset in the standard topology is Zariski dense. Therefore, if two irreducible complex varieties $X$, $X'$  (say inside some bigger space) are such that $X\cap X'$ has non-empty interior in both $X$ and $X'$ in the standard topologies, then $X = X'$.
I think you can find a proof of the statement above in the book of Mumford, Algebraic Geometry I.
