# The local ring at a point as an inverse limit of regular functions

Let $$\mathbb{F}$$ be an algebraically closed field and $$X$$ an affine algebraic set. Fix $$x \in X$$. For an open set $$U \subset X$$ let $$\mathcal{O}_X(U)$$ denote the collection of regular functions on $$U$$. Given two open neighborhoods $$U \subset V$$ of $$x$$ then restriction defines a map of algebras $$\mathcal{O}_X(V) \to \mathcal{O}_X(U)$$. The family of algebras $$\mathcal{O}_X(U)$$ as $$U$$ runs over all open neighborhoods of $$x$$ is an inverse system with respect to these restriction maps. I want to show that $$\mathcal{O}_x:= \mathbb{F}[X]_{M_x}$$ is the inverse limit of this inverse system in the category of $$\mathbb{F}$$ algebras. Here $$M_x \subset F[X]$$ is the maximal ideal of functions which vanish at $$x$$.

In order to show $$\mathcal{O}_x$$ is this inverse limit I want to find $$\pi$$, a collection of maps $$\{\pi_U : U \subset X \text{ an open neighborhood of }x\}$$ such that for each open set $$U$$ we have $$\pi_U:\mathcal{O}_x \to \mathcal{O}_X(U)$$. Further I need $$(\mathcal{O}_x,\pi)$$ to be a cone over my inverse system. However I'm struggling even to define a map $$\pi_U:\mathcal{O}_x \to \mathcal{O}_X(U)$$. It seems that it really suffices to define $$\pi_U$$ for $$U = X$$ where $$\mathcal{O}_X(X) = \mathbb{F}[X]$$ as, for $$(\mathcal{O}_x,\pi)$$ to be a cone we must have $$\pi_U\left(\frac{f}{g}\right) = \pi_X\left(\frac{f}{g}\right)\bigg{|}_{U}$$ for any open set $$U$$.

After thinking about this for a bit, I think I might be misinterpreting something but I'm really not sure where. Any help would be appreciated.

• Maybe I'm mistaken, but I think you've got things backwards. You should want to show that $\mathcal{O}_{x}$ is the colimit of your system, and so you want maps $\mathcal{O}_{X}(U) \to \mathcal{O}_{x}$ for each open $U$ containing $x$. (Accordingly, given a coherent system of maps out of the $\mathcal{O}_{X}(U)$s, you want a universal map out of $\mathcal{O}_{x}$.) Sep 27, 2019 at 3:00
• You're totally right. That makes infinitely more sense. Sep 27, 2019 at 3:08

We first define a collection of maps $$\pi_U:\mathcal{O}_X(U) \to \mathcal{O}_x$$ so that $$(\mathcal{O}_x, \pi)$$ is a cocone from $$\{\mathcal{O}_X(U)\}$$. Given $$f \in \mathcal{O}_X(U)$$, by definition of $$\mathcal{O}_X(U)$$ we can find $$g,h \in \mathbb{F}[X]$$ so that $$f = \frac{g}{h}$$ on a neighborhood of $$x$$ with $$h(x) \neq 0$$. If this is the case then let $$\pi_U(f) = \frac{g}{h}$$. I claim $$\pi_U$$ is well-defined. First note that since $$h(x) \neq 0$$, $$\frac{g}{h} \in \mathcal{O}_x$$. Next suppose that $$f = \frac{g_1}{h_1}$$ on $$x \in U_1 \subset U$$ and $$f = \frac{g_2}{h_2}$$ on $$x \in U_2 \subset U$$. Then we hae $$\frac{g_1}{h_1} = \frac{g_2}{h_2}$$ on $$U_1 \cap U_2$$. For a function $$s \in \mathbb{F}[X]$$ let $$X_s := X \backslash \mathcal{V}(s)$$. We know these principle open sets for a basis for the topology on $$X$$ so that we can assume that $$\frac{g_1}{h_1} = \frac{g_2}{h_2}$$ on $$X_s$$ with $$x \in X_s$$. We then have $$s(g_1h_2 - g_2h_1) = 0$$ on $$X$$ and thus this is equality in $$\mathbb{F}[X]$$. Since $$x \in X_s$$, $$s(x) \neq 0$$ and $$\frac{g_1}{h_1} = \frac{g_2}{h_2}$$ in $$\mathcal{O}_x$$. This shows $$\pi_U$$ is well-defined.
Next we check that $$(\mathcal{O}_x, \pi)$$ is a cocone from $$\{\mathcal{O}_X(U)\}$$. Note that if $$V \subset U$$ then $$\pi_{V}(f|_{V}) = \pi_U(f)$$ as, if $$f = \frac{g}{h}$$ in a neighborhood $$W \subset U$$ of $$x$$ then $$f|_{V} = \frac{g}{h}$$ in $$W \cap V$$, a neighborhood of $$x$$.
Finally we need to check $$(\mathcal{O}_x,\pi)$$ is universal with respect to cocones from $$\{\mathcal{O}_X(U)\}$$. Let $$(Y,\sigma)$$ be such a cocone. Given $$\frac{f}{g} \in \mathcal{O}_x$$ we have $$\frac{f}{g} \in \mathcal{O}_X(X_g)$$. Define $$\alpha:\mathcal{O}_x \to Y$$ by $$\alpha\left(\frac{f}{g}\right) = \sigma_{X_g}\left(\frac{f}{g}\right)$$. We need to check that $$\alpha$$ is well-defined. Suppose $$\frac{f_1}{g_1} = \frac{f_2}{g_2}$$ in $$\mathcal{O}_x$$. Then these two functions are equal on $$X_{g_1g_2}$$ so that, as $$(Y, \sigma)$$ is a cocone,
$$\sigma_{X_{g_1}}\left(\frac{f_1}{g_1}\right) = \sigma_{X_{g_1\cdot g_2}}\left(\frac{f_1}{g_1}\bigg{|}_{X_{g_1\cdot g_2}}\right) = \sigma_{X_{g_1\cdot g_2}}\left(\frac{f_2}{g_2}\bigg{|}_{X_{g_1\cdot g_2}}\right) = \sigma_{X_{g_2}}\left(\frac{f_2}{g_2}\right).$$
Further $$\alpha \circ \pi_U = \sigma_U$$. To see $$\alpha$$ is unique note that if $$\alpha'\circ \pi_U = \sigma_U$$ then $$\frac{f}{g} \in \mathcal{O}_x$$ then
$$\alpha'\left(\frac{f}{g}\right) = \alpha' \left( \pi_{X_g}\left(\frac{f}{g}\right)\right) = \sigma_{X_g}\left(\frac{f}{g}\right) = \alpha\left(\frac{f}{g}\right).$$