Let \(X\) be a noetherian scheme, and let \(\mcf\) be a coherent sheaf.
If the stalk \(\mcf_x\) is a free \(\OO_x\)-module for some point \(x \in X\), then there is a neighborhood \(U\) of \(x\) such that \(\mcf|_U\) is free.
\(\mcf\) is locally free if and only if its stalks \(\mcf_x\) are free \(\OO_x\)-modules for all \(x \in X\).
\(\mcf\) is invertible, i.e. locally free of rank \(1\), if and only if there is a coherent sheaf \(\mcg\) such that \(\mcf \tensor \mcg \cong \OO_X\). This justifies the terminology: it means that \(\mcf\) is an invertible element of the monoid of coherent sheaves under \(\tensor\).
Solution
All sheaf tensor products are over \(\OO_X\).
1. For part (a), a free stalk of rank \(r\) extends to a free sheaf of rank \(r\) on a neighborhood of that point.
Choose an affine neighborhood \(V=\Spec A\) of \(x\), and let \(\mathfrak p\) be the prime corresponding to \(x\). Since \(X\) is noetherian and \(\mcf\) is coherent, there is a finite \(A\)-module \(M\) with \(\mcf|_V\cong\widetilde M\) [@Har10a, Proposition II.5.4]. Then \(\mcf_x\cong M_{\mathfrak p}\). Its free rank \(r\) is finite because it is finitely generated over \(A_{\mathfrak p}\).
Represent each member of a basis of \(M_{\mathfrak p}\) as \(m_i/s_i\), with \(m_i\in M\) and \(s_i\notin\mathfrak p\). Multiplying the individual basis vectors by the units \(s_i\) shows that \(m_1/1,\ldots,m_r/1\) are also a basis. Define \[ u:A^r\longrightarrow M,\qquad (a_i)\longmapsto\sum_i a_im_i. \] The localized map \(u_{\mathfrak p}\) is an isomorphism. Let \(K=\ker u\) and \(C=\operatorname{coker}u\). Both are finite: \(C\) is a quotient of \(M\), and \(K\) is a submodule of the finite module \(A^r\) over the noetherian ring \(A\). Exactness of localization gives \(K_{\mathfrak p}=C_{\mathfrak p}=0\).
For every member of finite generating lists of \(K\) and \(C\), choose an element of \(A\setminus\mathfrak p\) annihilating it. The product \(s\) of these finitely many elements lies outside \(\mathfrak p\) and annihilates both modules. Thus \(K_s=C_s=0\), and \(u_s:A_s^r\to M_s\) is an isomorphism. On \(U=D_V(s)\), it induces \(\mcf|_U\cong\OO_U^r\). If \(r=0\), the same argument applies to \(u:0\to M\) and gives \(\mcf|_U=0\).
2. Part (b) holds.
An isomorphism \(\mcf|_U\cong\OO_U^r\) induces \(\mcf_x\cong\OO_{X,x}^r\) for every \(x\in U\) by passing to stalks. Consequently local freeness implies freeness of every stalk. Conversely, if every stalk is free, step 1 supplies a free neighborhood at each point. These neighborhoods cover \(X\), proving local freeness.
3. For the forward implication of part (c), a tensor inverse of \(\mcf\) is\[ \boxed{\mcg=\dualof{\mcf}=\sheafhom_{\OO_X}(\mcf,\OO_X)}. \]
Assume \(\mcf\) is locally free of rank one. The evaluation morphism \[ \mcf\otimes\dualof{\mcf}\longrightarrow\OO_X, \qquad s\otimes\lambda\longmapsto\lambda(s) \] is defined compatibly on local sections, as in P-AGH251DUALSHEAF. On a trivializing open set with frame \(e\), the coordinate functional \(\dualof{e}\) is a frame of \(\dualof{\mcf}\), and evaluation sends \(e\otimes \dualof{e}\) to \(1\). It is therefore an isomorphism on each such open set, hence on \(X\). The sheaf \(\dualof{\mcf}\) is locally free of rank one as well, so its local modules over noetherian affine charts are finite and it is coherent. The evaluation morphism is independent of the chosen frames; the frames only verify that it is an isomorphism.
4. Conversely, a coherent tensor inverse forces \(\mcf\) to be locally free of rank one.
Suppose \(\mcf\otimes\mcg\cong\OO_X\) with \(\mcg\) coherent. Fix \(x\in X\), put \(R=\OO_{X,x}\), and write \(\mathfrak m\) for its maximal ideal and \(\kappa=R/\mathfrak m\) for its residue field. The modules \(M=\mcf_x\) and \(N=\mcg_x\) are finite over \(R\). Taking stalks and then tensoring with \(\kappa\) gives \[ M\otimes_R N\cong R, \qquad (M/\mathfrak mM)\otimes_\kappa(N/\mathfrak mN)\cong\kappa. \] The product of the dimensions of these finite-dimensional vector spaces is one, so each dimension is one. By Nakayama's lemma, \(M\) is cyclic. Every element of \(\Ann_R M\) annihilates \(M\otimes_R N\cong R\), so \(\Ann_R M=0\). For a generator \(m\in M\), the surjection \(R\to M\), \(a\mapsto am\), has kernel \(\Ann_R m=\Ann_R M=0\). Hence \(M\cong R\). This holds for every \(x\), and the rank-preserving construction of step 1 supplies rank-one trivializations around all points. Thus \(\mcf\) is invertible.
5. Q.E.D.
Steps 1 and 2 prove parts (a) and (b). Steps 3 and 4 prove both implications of part (c).