# Desingularisation of curves (Lorenzini-Invitiation to Arithmetic Geometry, chap 6,ex 7)

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.