Endlich erzeugter Modul/Lokal freie Garbe/Charakterisierungen von lokal/en/Fakt/Beweis

Beweis

${}(1)\Rightarrow (2)$. This is a specialization.
${}(2)\Rightarrow (3)$. We fix a maximal ideal ${}{\mathfrak {m}}$. By assumtion there exists an ${}R_{\mathfrak {m}}$-isomorphism

$\varphi \colon {\left(R_{\mathfrak {m}}\right)}^{r}\longrightarrow M_{\mathfrak {m}}.$

We may write the image of the ${}i$th standard vector as

${}\varphi (e_{i})={\frac {m_{i}}{g_{i}}}\,$

with ${}m_{i}\in M$ and ${}g_{i}\in R\setminus {\mathfrak {m}}$. Let ${}g=g_{1}{\cdots }g_{r}$ and consider the situation on ${}D(g)$. The above given isomorphism ${}\varphi$ is defined over ${}D(g)$, that is we have a homomorphism

$\psi \colon (R_{g})^{r}\longrightarrow M_{g}$

which induces ${}\varphi$ in the localization at ${}{\mathfrak {m}}$, but ${}\psi$ may not be an isomorphism over ${}D(g)$. Let ${}v_{1},\ldots ,v_{m}$ be a generating system for the module ${}M$. Since ${}\psi$ induces a surjection, there exist ${}u_{j}={\frac {a_{j}}{h_{j}}}\in {\left(R_{\mathfrak {m}}\right)}^{r}$ mapping to ${}v_{j}$. Since the denominators ${}h_{j}$ do not belong to ${}{\mathfrak {m}}$, we may replace ${}g$ by ${}gh_{1}\cdots h_{m}$ and then we may assume that ${}\psi$ is surjective. Finally, let ${}N$ be the kernel of (this new) ${}\psi$. Since ${}\varphi$ is injective, we know that ${}N_{\mathfrak {m}}=0$. Since ${}R$ is noetherian, ${}N$ is finitely generated and so again there will be a ${}f$, ${}f\not \in {\mathfrak {m}}$, such that ${}N_{f}=0$. So further shrinking shows that there is an isomorphism ${}\psi \colon (R_{f})^{r}\rightarrow M_{f}$ for some ${}f$, ${}f\not \in {\mathfrak {m}}$.

So far we have shown that for every maximal ideal ${}{\mathfrak {m}}$ there exists an open neighborhood ${}{\mathfrak {m}}\in D(f_{\mathfrak {m}})$ such that ${}M_{f_{\mathfrak {m}}}$ is free of rank ${}r$. Hence

$\bigcup _{{\mathfrak {m}}{\text{ maximal ideal }}}D(f_{\mathfrak {m}})$

contains all maximal ideals and therefore all prime ideals, so it is an open cover of ${}\operatorname {Spek} {\left(R\right)}$. This means that ${}{\left(f_{\mathfrak {m}}:\,{\mathfrak {m}}{\text{ maximal}}\right)}$ is the unit ideal, and so finitely many of these generate already the unit ideal.
${}(3)\Rightarrow (4)$. Since the elements generate the unit ideal, the corresponding open subsets ${}D(f_{j})$, ${}j=1,\ldots ,k$, cover ${}\operatorname {Spec} {\left(R\right)}$. That ${}M_{f_{j}}$ is a free ${}R_{f_{j}}$-module of rank ${}r$ means that ${}{\tilde {M}}{|}_{D(f_{j})}\cong {\tilde {(R_{f_{j}})^{r}}}\cong {\mathcal {O}}_{D(f_{j})}^{r}$. So ${}{\tilde {M}}$ is locally free.
${}(4)\Rightarrow (1)$. Let ${}{\mathfrak {p}}\in \operatorname {Spek} {\left(R\right)}$ be a prime ideal. The local freeness means that we have an open covering ${}X=\bigcup _{i\in I}U_{i}$ such that ${}{\tilde {M}}{|}_{U_{i}}$ is free of rank ${}r$. So there exists an index ${}i$ such that ${}{\mathfrak {p}}\in U_{i}$. We may shrink ${}U_{i}$ to a smaller open neighborhood of ${}{\mathfrak {p}}$ and then assume that ${}U_{i}=D{\left(f\right)}$ and that ${}{\tilde {M_{f}}}\cong {\tilde {M}}{|}_{D(f)}$ is free of rank ${}k$. But then also the localization ${}M_{\mathfrak {p}}$ is free of rank ${}r$.

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