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

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?

• Dear Georges, what I meant by "the local ring of $k$ at $0$" is the localization of $k[t]$ at $\langle t \rangle$. I hope this makes sense, or I am missing something obvious.
– Ben
Commented Apr 21, 2020 at 16:31
• Why do you think that $f$ or its restrictions to the lines $L_i$ should vanish at $0$ ? Do you realize that $1\in \mathcal O_{X,(0,0)}$ (as is the case in all rings) ? Commented Apr 21, 2020 at 17:30
• Sorry, I don't understand your question. $f \in \mathcal{O}_{X, (0,0)}$ so it won't vanish at $(0,0)$ by definition. Meanwhile, the $f_i \in \mathcal{O}_{\mathbb{A}^1,(0)}$ so those don't vanish at $(0)$ by definition. $f|_{L_i}$ doesn't vanish at $(0)$ because $f$ doesn't vanish at $(0,0)$.
– Ben
Commented Apr 21, 2020 at 17:50
• You seem not to know what $\mathcal O_{X,(0,0)}$ means. That ring has a $1$, like all rings, and that $1$ does not vanish at $(0,0)$. Read the definition in Shafarevich, page 83, or in any other book for that matter. Commented Apr 21, 2020 at 17:58
• Sorry, my previous comment was very dumb. Every time I said "don't vanish at $0$" should be replaced by "is defined at $0$". Of course there are elements of $\mathcal{O}_{X,(0,0)}$ that vanish at $(0,0)$ (e.g. $0$), and elements that don't (e.g. $1$). Despite my stupid comment, I still don't see how you gleaned from my post that I think every element of $\mathcal{O}_{X,(0,0)}$ vanishes at $(0,0)$.
– Ben
Commented Apr 21, 2020 at 18:04

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 $$k^n$$.
• Don't feel foolish! I've answered this question because I believed your claim was true for several weeks before I read a paper that told me otherwise... I am not sure what Shafarevich means by "determine the local ring". I would think that saying it's the ring $k[x,y]/(xy(x-y))$ localized at $(0,0)$ would make it "determined". Commented Apr 19, 2020 at 20:54