Let \(X\) be a noetherian scheme and \(\mcf\) a coherent sheaf on \(X\). Consider the function \[ \varphi(x) = \dim_{k(x)} \mcf_x \tensor_{\OO_x} k(x), \] where \(k(x) = \OO_x / \mfm_x\) is the residue field at \(x\). Use Nakayama’s lemma to prove the following.
The function \(\varphi\) is upper semi-continuous, i.e. for any \(n \in \ZZ\) the set \(\ts{x \in X \st \varphi(x) \geq n}\) is closed.
If \(\mcf\) is locally free and \(X\) is connected, then \(\varphi\) is a constant function.
Conversely, if \(X\) is reduced and \(\varphi\) is constant, then \(\mcf\) is locally free.
(Reducedness in part (c))
Let \(k\) be a field, put \(R=k[\varepsilon]/(\varepsilon^2)\), and let \(\mcf\) on \(X=\Spec R\) be the sheaf associated to \(R/(\varepsilon)\). The scheme \(X\) has one point, so \(\varphi\) is constant with value one. The sheaf is coherent since \(R\) is noetherian and its module is finite. It is not locally free: its stalk is nonzero and is annihilated by the nonzero element \(\varepsilon\), whereas every nonzero free \(R\)-module has zero annihilator.
Solution
The fibers in the definition of \(\varphi\) are finite-dimensional because \(\mcf\) is coherent.
1. If \(\varphi(x)=r\), then \(x\) has an affine neighborhood \(U\) and a surjection \(\OO_U^r\to\mcf|_U\).
Take an affine neighborhood \(V=\Spec A\) of \(x\) with \(\mcf|_V\cong\widetilde M\), where \(M\) is a finite \(A\)-module [@Har10a, Proposition II.5.4]. Let \(\mathfrak p\) correspond to \(x\). Choose elements \(m_1,\ldots,m_r\in M\) whose images form a basis of \(M_{\mathfrak p}/\mathfrak pM_{\mathfrak p}\): each basis vector can be represented by a fraction from \(M\), and multiplication by its nonzero denominator only rescales that basis vector. By Nakayama's lemma, their images generate \(M_{\mathfrak p}\).
Let \(u:A^r\to M\) send the standard basis to these elements. The finite module \(C=\operatorname{coker}u\) satisfies \(C_{\mathfrak p}=0\). For each element of a finite generating list of \(C\), choose an annihilator outside \(\mathfrak p\). Their product \(s\notin\mathfrak p\) annihilates \(C\), so \(u_s:A_s^r\to M_s\) is surjective. The associated-sheaf map is the asserted surjection on \(U=D_V(s)\). When \(r=0\), this construction gives \(\mcf|_U=0\).
2. The function \(\varphi\) is upper semi-continuous, proving part (a).
Fix \(n\in\ZZ\) and \(x\) with \(\varphi(x)=r<n\). On the neighborhood \(U\) from step 1, taking the stalk at \(y\in U\) and tensoring with \(k(y)\) gives a surjection \[ k(y)^r\longrightarrow\mcf_y\otimes_{\OO_{X,y}}k(y). \] Thus \(\varphi(y)\le r<n\) throughout \(U\). It follows that \(\{x:\varphi(x)<n\}\) is open, and its complement \(\{x:\varphi(x)\ge n\}\) is closed.
3. If \(\mcf\) is locally free, then \(\varphi\) is locally constant; connectedness makes it constant, proving part (b).
On an open set with \(\mcf|_U\cong\OO_U^r\), every fiber is \(k(y)^r\), so \(\varphi(y)=r\) for all \(y\in U\). Hence each set \(\varphi^{-1}(r)\) is open. Its complement is the union of the other such sets and is also open, so \(\varphi^{-1}(r)\) is closed as well. If \(X\) is nonempty and connected, the fiber containing any chosen point must be all of \(X\). For empty \(X\) the constancy assertion is vacuous.
4. If \(X\) is reduced and \(\varphi\) is constant, then \(\mcf\) is locally free, proving part (c).
Write the constant value as \(r\) and fix \(x\in X\). Step 1 gives an affine neighborhood \(U=\Spec B\) and a surjective module homomorphism \[ u:B^r\longrightarrow N \] whose associated sheaf is a surjection \(\OO_U^r\to\mcf|_U\). The ring \(B\) is reduced because \(X\) is reduced. For each \(\mathfrak q\in\Spec B\), the induced map \[ u\otimes_B k(\mathfrak q):k(\mathfrak q)^r\longrightarrow N\otimes_B k(\mathfrak q) \] is surjective between vector spaces of the same finite dimension \(r\), hence is an isomorphism.
Let \((b_1,\ldots,b_r)\in\ker u\). Its image in \(k(\mathfrak q)^r\) maps to zero under this isomorphism, so every \(b_i\) has zero image in \(k(\mathfrak q)\). The kernel of \(B\to k(\mathfrak q)\) is \(\mathfrak q\), and therefore \(b_i\in\mathfrak q\) for every prime \(\mathfrak q\). The intersection of all prime ideals is the nilradical, which is zero in the reduced ring \(B\) [@Har10a, Chapter II, §2]. Thus every \(b_i\) is zero and \(u\) is injective. It follows that \(u\) is an isomorphism and \(\mcf|_U\cong\OO_U^r\). The construction applies at each point, giving local freeness.
5. Q.E.D.
Steps 2, 3, and 4 prove parts (a), (b), and (c), respectively.