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. $\square$

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