Loading [MathJax]/extensions/tex2jax.js

The Stacks project

Lemma 26.5.1. Let $R$ be a ring. Let $f \in R$.

  1. If $g\in R$ and $D(g) \subset D(f)$, then

    1. $f$ is invertible in $R_ g$,

    2. $g^ e = af$ for some $e \geq 1$ and $a \in R$,

    3. there is a canonical ring map $R_ f \to R_ g$, and

    4. there is a canonical $R_ f$-module map $M_ f \to M_ g$ for any $R$-module $M$.

  2. Any open covering of $D(f)$ can be refined to a finite open covering of the form $D(f) = \bigcup _{i = 1}^ n D(g_ i)$.

  3. If $g_1, \ldots , g_ n \in R$, then $D(f) \subset \bigcup D(g_ i)$ if and only if $g_1, \ldots , g_ n$ generate the unit ideal in $R_ f$.

Proof. Recall that $D(g) = \mathop{\mathrm{Spec}}(R_ g)$ (see Algebra, Lemma 10.17.6). Thus (a) holds because $f$ maps to an element of $R_ g$ which is not contained in any prime ideal, and hence invertible, see Algebra, Lemma 10.17.2. Write the inverse of $f$ in $R_ g$ as $a/g^ d$. This means $g^ d - af$ is annihilated by a power of $g$, whence (b). For (c), the map $R_ f \to R_ g$ exists by (a) from the universal property of localization, or we can define it by mapping $b/f^ n$ to $a^ nb/g^{ne}$. The equality $M_ f = M \otimes _ R R_ f$ can be used to obtain the map on modules, or we can define $M_ f \to M_ g$ by mapping $x/f^ n$ to $a^ nx/g^{ne}$.

Recall that $D(f)$ is quasi-compact, see Algebra, Lemma 10.29.1. Hence the second statement follows directly from the fact that the standard opens form a basis for the topology.

The third statement follows directly from Algebra, Lemma 10.17.2. $\square$


Comments (0)


Your email address will not be published.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.