Taking stalks is itself a colimit, and colimits commute with colimits.

This is exactly the same proof as the next lemma, minus the reduction to being an affine scheme.

WLOG, we can assume that \(X\) is affine, so \(X = \operatorname {Spec}A\), and furthermore we can assume \(A\) is noetherian. Indeed, take an open affine noetherian cover as guarenteed by local noetherianity. Then \(x\) is contained in some neighborhood \(U = \operatorname {Spec}A\) in the cover. If we show the theorem for \(\operatorname {Spec}A\), then we have shown it for \(X\).

Now, as \(\mathcal{F}\) is a coherent sheaf on \(A\), \(\mathcal{F} \cong \tilde{M}\) for some finitely generated module \(M\) on \(A\). \(x\), being a point of \(\operatorname {Spec}A\), is/corresponds to a prime ideal \(\mathfrak {p}\) in \(A\). Now, our assumption about freeness of the stalk says that \(M_{\mathfrak {p}} \cong A_{\mathfrak {p}}^{\oplus n}\) for some \(n\). Indeed, \(M_{\mathfrak {p}}\) is the stalk of \(\tilde{M}\) and \(A_{\mathfrak {p}}\) is the local ring of \(\operatorname {Spec}A\) at \(\mathfrak {p}\). Let \(m_{1} , \ldots , m_{n}\) the a basis/free generating set for \(M_{\mathfrak {p}}\). In other words, \(m_{i}\) is the image (along some isomorphism \(A_{\mathfrak {p}}^{\oplus n}\)) of \((0, \ldots , 1, \ldots , 0) \in A_{\mathfrak {p}}^{\oplus n}\) where the \(1\) is in the \(i\)-th place. Since the \(m_{i}\) are elements of the stalk \(\mathcal{F}_{x} \cong M_{\mathfrak {p}}\), we can choose a neighborhood \(U^{\prime }\) of \(x\) with representatives \(m_{1}^{\prime }, \ldots , m_{n}^{\prime }\), whose images in the stalk are the \(m_{i}\). Indeed, we can do this individually for each \(m_{i}\), and since there are finitely many, we can choose \(U^{\prime }\) a neighborhood which dominates each of them in the diagram (i.e. a neighborhood which is contained in all of them, for example the intersection). Moreover, we can choose \(U^{\prime }\) to be affine by taking a smaller neighborhood (\(X\) is a scheme). If we prove the theorem for \(U^{\prime }\), then we have proven it for \(U\), so we can assume that \(U = U^{\prime }\). Aside: we can see that the \(m_{i}^{\prime }\) here aren’t zero, because there is a ring map from the sections over \(U^{\prime }\) to the stalk, and the images are nontrivial in the stalk.

Let \(x_{1}, \ldots , x_{k}\) be a finite generating set for \(M\). Now, we have the following equations in \(M_{\mathfrak {p}}\):

Aside: if \(\frac{x_{i}}{1} = 0\), then all the \(a_{ij}\) are zero and the \(b_{ij}\) are \(1\) and the rest of the proof is not affected.

Using the characterization of when elements of the localization are zero (i.e. that they are \(s\)-torsion for some \(s \in A \smallsetminus \mathfrak {p} \)), we have the equations

for some \(t_{i} \in A \smallsetminus \mathfrak {p}\), where the sum takes place in in \(M\) (which is a module over \(A\)). Note that we must have the factor of \(\prod _{}^{} b_{ij} \) as one multiply “top and bottom” by this term to put the element \(x_{i} - \sum _{}^{} \frac{a_{ij}}{b_{ij}}m_{i} \) into the form \(\frac{p}{q}\) with \(p,g \in M\).

Let \(b \colonequals \prod _{i}^{} t_{i} \prod _{i,j}^{} b_{ij}\). We know that the \(\frac{x_{i}}{1}\) generate \(M_{b}\) as an \(A_{b}\)-module by the characterization of elements in the localization as fractions (given an element of \(M {b}\), one has an equation in \(M\) for its numerator, and then one over its denominator is in \(A_{b}\)). Thus, the equations above and the fact that \(A \smallsetminus \mathfrak {p}\)-torsion is the kernel of the localization map show that \(m_{i}^{\prime }\) generate \(M_{b}\).

Thus, we have that the following map

is surjective (where the \(1\) is in the \(i\)-th place). Let the Kernel of the above map be denoted \(K\). We want to consider the exact sequence

As \(A\) is noetherian, \(A_{b}\) is noetherian, and thus so is \(A_{b}^{\oplus n}\). Thus, \(K\), which is a submodule of \(a_{b}^{\oplus n}\), is finitely generated, say by \(k_{1}, \ldots , k_{\ell }\).

Now, we apply the localization functor to the above exact sequence. However, we note that the second (nontrivial) map is indeed the same map as the isomorphism we gave in the first place. Thus, we conclude that \(K_{\mathfrak {p}} = 0\). This means, by the characterization of the kernel of a localization map, that there are \(s_{i} \in A \smallsetminus \mathfrak {p}\) such that \(s_{i}k_{i} = 0\) in \(A\) for all \(i\). (the same is true if we replace \(A\) with \(A_{b}\), one can use whichever ring is more convenient). Now, if we let \(b^{\prime } = b \prod _{i}^{} s_{i} \), then we see that by the equations above and the characterization of the kernel of the localization as torsion, that \(K_{b^{\prime }} = 0\). This means, by the exact sequence above and/or the fact that kernels commute with localization, that \(A_{b^{\prime }}^{\oplus n} \cong M_{b^{\prime }}\), and we have what we want.

\(X\) has an affine cover of finite type \(k\)-algebras. Thus, by hilbert basis, it is (locally) noetherian. Apply the previous theorem.