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Given a nonsingular complete curve over algebraically closed $\bar{k}$, which is interpreted as a field $\bar{k}(X)$ of transcendence degree 1 and its set of valuations trivial on $\bar{k}$, we may choose elements $x,y$ such that $x$ is trancendental, and $\bar{k}(X)=\bar{k}(x)(y)$. Letting $F$ be the homogenization of the minimal polynomial of $y$ over $\bar{k}(x)$, using these coordinates we get a function from the set of valuations to the points of the projective curve $X_F(\bar{k})$, given by mapping a valuation $v$ to the unique point $P$ of $X_F(\bar{k})$ such that $O_P \subset O_v$ and inclusion of maximal ideals.

The first question is to show that if $x,y$ are in $O_v$, then the valuation $v$ is mapped to is $(x(v):y(v):1)$, where $x(v),y(v)$ are the cosets of $x,y$ in $O_v / M_v\cong \bar{k}$.

For this, my thought was that if $\psi =G/H$ with $H(x(v):y(v):1)\neq 0$, then we get the coset of $H(x,y,1)$ in $O_v / M_v$ is nonzero, so we can conclude $G/H$ is defined at $v$, in the sense of having nonnegative valuation.

I feel okay with this, and the homogenisation/dehomogenisation is just from our isomorphism of fields $\bar{k}(X)\cong \bar{k}(X_F)$, but the next part I am feeling stuck, which is to show that if $y\notin O_v, x/y\in O_v$, then the point is $((x/y)(v):1:0)$.

Any assistance with how to think about this would also be very much appreciated.

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