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I was reading Shafarevich's Basic Algebraic Geometry 1, and I was having a little trouble with the concept of a local equation. I found this link: Explicit example of local equation but I can't quite follow the answer as I don't know what divisors are.

In question 5 (from section 3 - chapter 2), they ask to show that for a cone given by $x^2 + y^2 -z^2$, the generator $L$ defined by $x = 0, y = z$ does not have a local equation in any neighbourhood of $(0,0,0)$.

I guess geometrically, it seems like the idea is showing that no matter which open neighbourhood of $(0,0,0)$ you take, you cannot restrict it in such a manner, that only one equation is needed to describe the line. But I wasn't really able to show this.

Thank you for your help in advance!!

xyzsd3324
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Denote the cone by $X$. The generator $L$ has a local equation in a neighbourhood of $0=(0,0,0)\in X$ iff there is some $f\in\mathcal O_{0,X}$ such that $\mathfrak a_{0,L}=(f)$ (here $\mathcal O_{0,X}$ is a local ring of $X$ at $0$, $\mathfrak a_{0,L}\subset\mathcal O_{0,X}$ is the ideal of $L$). Denote by $\mathfrak m_0\subset\mathcal O_{0,X}$ the maximal ideal of functions vanishing at $0$, by $\bar{f}$ the class of $f$ in $\mathcal O_{0,X}/\mathfrak m_0^2$. If $f$ is a local equation of $L$, then the image of $\mathfrak a_{0,L}$ in the 3-dimensional vector space $\mathfrak m_0/\mathfrak m_0^2$ is generated by $\bar{f}$. But this image has two linearly independent elements $\bar{x}$ and $\overline{y-z}$, hence it cannot be generated by one element.

danneks
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