Explicit example of local equation I'm reading Shafarevich's Basic Algebraic Geometry 1 and struggling to understand local equations of a sub variety in a neighbourhood. In particular, let $X=\mathbb{V}(y^2-xz)\subset \mathbb{P}^2$ and $Y=\{[4:2:1]\}$, and say that we want to find a local equation of $Y$ in a neighbourhood of $p=[1:1:1]$.
So (according to the definition) we need to find $\pi\in k[X']$ such that $\mathfrak{a}_{Y'}=(\pi)$ where $Y'=Y\cap X'$ and $X'$ is some affine neighbourhood of $p=[1:1:1]$.
I know this should be simple, but I just keep going round in circles.
 A: Take, for example $Y = \mathbb P^1$ and suppose $f$ collapses $X$ to $0 \in Y$. Consider the divisor $D = 1 \cdot [\infty]$. Since $Y$ is a smooth variety, $D$ is a locally principal divisor. This means for every $y \in Y$, I can find a neighborhood $U$ of $y$ and a regular function $\phi$ on $U$ such that the divisor of $\phi$ equals $D$ restricted to $U$. For instance in our example, I should be able to find a neighborhood $U$ of $0$ and a regular function $\phi$ on $U$ whose divisor gives $D \mid_U$. We can take $U = \mathbb P^1 - 0$ and $1$ on $U$. $D\mid_U$ is empty, and the constant function 1 has no zeros or poles. So $1$ cuts out the divisor $D$ locally.
A: Your projective variety $X$ is living in $\mathbb P^2$, say with coordinates $[x, y, z]$. It is an important fact that $\mathbb P^2$ can be covered by three open copies of $\mathbb A^2$, by setting $x$, $y$, or $z$ to $1$. We call these $U_i$, for $i=x, y, z$ (usually we index by $x_i$). So simply consider the intersections $Y \cap U_i$, for each $i$. These are (isomorphically affine varieties in $A^2$, cut out by the homogeneous polynomial of $X$, evaluated at $x$, $y$, or $z=1$. In particular, your point $[1, 1, 1]$ lives in all three neighborhoods, so locally the point $[1:1:1]$ corresponds to $(1, 1)$ in $U_x$ sitting in the variety $V(y^2-z) \subset \mathbb A^2$.
