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.


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 $A_m$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.