I've been trying to prove the following proposition:
Let $(X,\mathcal{O}_X)$ be a quasi-compact ringed space which is locally isomorphic to an affine variety over an algebraically closed field $k$.
Let $x\in X$ and $U\subseteq X$ be an affine open set of $X$ with $x\in U$. Then for $A=\Gamma(U,\mathcal{O}_X)$ with maximal ideal $m=\{f\in A\mid f(x)=0\}$, we have $A_m\cong\mathcal{O}_{X,x}$, the stalk of $\mathcal{O}_X$ at $x$.
I've can prove the case where $U$ is irreducible:
We have $A_m=\{\frac{f}{g}\mid f,g\in\Gamma(U,\mathcal{O}_X),g(x)\neq0\}$. Let $\varphi:A_m\to\mathcal{O}_{X,x}$ send $\frac{f}{g}$ to the germ $(D(g),\frac{f}{g})$.
To show that $\varphi$ is surjective, take any $(H,h)\in\mathcal{O}_{X,x}$, where $H\subseteq X$ is open, $x\in H$ and $h\in\Gamma(H,\mathcal{O}_X)$. We have $H\cap U=\cup_{i=1}^nD(g_i)$ for some $g_i\in\Gamma(U,\mathcal{O}_X)$, with $h\mid_{D(g_i)}=\frac{f_i}{g_i}\mid_{D(g_i)}$ for some $f_i\in\Gamma(U,\mathcal{O}_X)$. Then $(H,h)=(D(g_1),\frac{f_1}{g_1})=\varphi(\frac{f_1}{g_1})$ since $h$ and $\frac{f_1}{g_1}$ agree on $\cap_{i=1}^nD(g_i)$, which is non-empty since $U$ is irreducible.
For injectivity, if we have $\varphi(\frac{f_1}{g_1})=\varphi(\frac{f_2}{g_2})$, then we know that $\frac{f_1}{g_1}$ and $\frac{f_2}{g_2}$ agree on some non-empty open $W\subseteq D(g_1)\cap D(g_2)$. Then $W\subseteq V(f_1g_2-f_2g_1)$, so by the irreducibility of $U$ we have that $f_1g_2=f_2g_1$ on all of $U$, so $\frac{f_1}{g_1}=\frac{f_2}{g_2}$ in $A_m$.
However this argument makes heavy use of the irreducibility of $U$, and I can't seem to generalise it to the case where $U$ might not be irreducible.
Best Answer
For surjectivity, we know there exist some $f,g\in\Gamma(U,\mathcal{O}_X)$ such that $x\in D(g)\subseteq H\cap U$ and $h\vert_{D(g)}=\frac{f}{g}\vert_{D(g)}$, and so $\varphi(\frac{f}{g})=(H,h)$, we don't need to mess around with the full cover.
For injectivity, since $\frac{f_1}{g_1}$ and $\frac{f_2}{g_2}$ agree on some open $W\subseteq D(g_1)\cap D(g_2)$, we know they agree on some $D(h)\subseteq W$ with $h\in\Gamma(U,\mathcal{O}_X)$ and $x\in D(h)$. Then $f_1g_2-f_2g_1$ on $D(h)$, so $h(f_1g_2-f_2g_1)=0$ everywhere. Since $h(x)\neq0$ we have $h\notin m$, so this shows that $\frac{f_1}{g_1}=\frac{f_2}{g_2}$ in $A_m$.