Lemma 26.6.2. Let $X$ be a locally ringed space. Let $f \in \Gamma (X, \mathcal{O}_ X)$. The set

is open. Moreover $f|_{D(f)}$ has an inverse.

Lemma 26.6.2. Let $X$ be a locally ringed space. Let $f \in \Gamma (X, \mathcal{O}_ X)$. The set

\[ D(f) = \{ x \in X \mid \text{image }f \not\in \mathfrak m_ x\} \]

is open. Moreover $f|_{D(f)}$ has an inverse.

**Proof.**
This is a special case of Modules, Lemma 17.24.10, but we also give a direct proof. Suppose that $U \subset X$ and $V \subset X$ are two open subsets such that $f|_ U$ has an inverse $g$ and $f|_ V$ has an inverse $h$. Then clearly $g|_{U\cap V} = h|_{U\cap V}$. Thus it suffices to show that $f$ is invertible in an open neighbourhood of any $x \in D(f)$. This is clear because $f \not\in \mathfrak m_ x$ implies that $f \in \mathcal{O}_{X, x}$ has an inverse $g \in \mathcal{O}_{X, x}$ which means there is some open neighbourhood $x \in U \subset X$ so that $g \in \mathcal{O}_ X(U)$ and $g\cdot f|_ U = 1$.
$\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)