Local ring at $(0,0)$ of $n$ lines

Question

This question is motivated by Shafarevich: Basic algebraic geometry in projective space 1, Chapter II, Section 1, Exercise 6.

Let $k$ be an algebraically closed field, and let $X$ be the union of $n$ one-dimensional subspaces $L_1,\dots,L_n\subseteq k^2$. Let $\mathcal{O}_0$ be the local ring of $k$ at $0$. It seems that the local ring $\mathcal{O}_{X, (0,0)}$ of $X$ at $(0,0)$ should be the subring $\mathcal{O}\subseteq \mathcal{O}_0^{\times n}$ consisting of those $(f_1,\dots, f_n)$ for which $f_1(0)=\dots=f_n(0)$. The isomorphism seems to be $f \mapsto (f|_{L_1},\dots,f|_{L_n})$ in one direction and in the other direction you glue together $n$ functions to get a rational function on $X$ regular at $(0,0)$.

... Is this correct? Am I sane?

Answer 1

The map $f\mapsto (f|_{L_1},\ldots ,f|_{L_n})$ is not surjective if $n>2$, which means that in order to glue $n$ functions together you need more conditions than just $f_1(0)=\cdots = f_n(0)$.$\newcommand{\m}{\mathfrak m} \newcommand{\O}{\mathcal O}$ For example, if $x$ is the coordinate of $L$, you can see that $(x,0,0,\ldots )$ is not in the image of your map.

A more geometric reason comes from looking at the tangent spaces. Let $\m$ be the maximal ideal in $\O$, the ideal $\{(f_1,\ldots ,f_n): f_1(0) = \cdots f_n(0) = 0\}$. $\m/\m^2$ is $n$-dimensional: you can check that a basis is $\left\{(0,\ldots ,0,\overset{(i)}x,0,\ldots ,0):1\le i\le n\right\}$, where $x$ is the coordinate of the line.

Meanwhile, if $\m'$ is the maximal ideal of $\O_{X,(0,0)}$, the surjection $ \O_{k^2,(0,0)} \to \O_{X,(0,0)} $ maps $(x,y)/(x,y)^2$ onto $\m'/\m'^2$. Therefore, the dimension of $\m'/\m'^2$ is at most $2$.

I think the ring $\O_0$ you've described is the local ring of the union of the coordinate axes of $k^n$.