Nakayama's Lemma and Some Applications

Proof. Let \(x_1\), \(x_2\),\(\dots\), \(x_n\) be a basis of \(M\). Then a endomorphism \(\phi\) is represented by matrix \(\left(\alpha_{ij}\right)\), i.e. \[ \phi(x_i)=\sum_{j=1}^na_{ij}x_j. \] Since \(\phi(M)\subseteq\mf{a}M\), we may assume that \(\alpha_{ij}\in \mf{a}\). Denote by \(\delta_{ij}\) the Kronecker delta. Then the above equality is equivalent to \[\sum_{j=1}^n(\delta_{ij}\phi-\alpha_{ij})\cdot x_j=0.\]

By multiplying both sides the adjugate matrix, we find that \[\det(\delta_{ij}\phi-\alpha_{ij})=0.\]

The desired equality follows by applying the Laplace expansion to the determinant.

Proof. Take \(\phi=\mathrm{id}\) in the Lemma 1, and let \(a=-\sum r_i\), we see that \(m=am\).

Proof. By Lemma 2, there is an element \(r\in \mf{m}\) such that \(m+rm=0\) for all \(m\in M\). Since \(1+r\) is an unit in \(A\), multiplying the inverse \((1+r)^{-1}\), we find that that \(m=0\).

Proof. Let \(N=(f_1,\dots, f_n)\) be the submodule in \(M\). Then we have \[(N+\mf{m}M)/\mf{m}M=M/\mf{m}M.\] It then follows that \(N+\mf{m}M=M\) (by the third isomorphism theorem or the snake lemma). Therefore, \[M/N=(N+mM)/N=m(M/N).\]

Apply Nakayama’s Lemma to \(M/N\), we get \(M/N=0\) and hence \(M=N\).

Proof. Let \(V\) be an affine open neighborhood of \(x\) in \(U\). Then \(\mc{F}\vert_V=\tilde{M}\), \(\mc{F}_x=M\otimes \O_{X,x}\), and \(\mc{F}\vert_x=M\otimes k(x)\), where \(M=\mc{F}(V)\) is a finitely generated \(\O(V)\)-module. Since \(\bar{a}_1\), \(\dots\), \(\bar{a}_n\) generate \(\mc{F}\vert_x\), by Corollary 2, we know that \(a_1\), \(\dots\), \(a_n\) generate \(M=\mc{F}(V)\). Moreover, \(a_1\), \(\dots\), \(a_n\) generates \(\mc{F}_x\).

Those sections define a morphism \(\O\vert_V^n\to \mc{F}\vert_V\) such that the morphism at the stalk level \(\O_x^n\to \mc{F}_x\) is surjective.

By shrinking \(V\), we may assume that \(\O\vert_V^n\to \mc{F}\vert_V\) is surjective (see Lemma 17.9.4 in the Stack Project).

Therefore, \(a_1\), \(\dots\), \(a_n\) generate \(\mc{F}\vert_V\).

Proof. It follows directly from the Lemma 3

Proof. If \(\mc{F}\) is free in a neighborhood of \(x\), then \(e(x)\) is constant over that neighborhood.

Conversely, let \(U\) be a neighborhood of \(x\) such that \(e(y)\) is a constant for any \(y\in U\). Then by shrinking \(U\), we may assume that \[\O\vert_U^e\to \mc{F}\vert_U\] is surjective. Let \(\mc{K}\) be the kernel of this morphism. If \(\mc{K}\neq 0\), then there is a nonzero element \(s\) in \(\mc{K}\). Since \(\mc{K}\subset \O\vert_U^e\) and \(\O\vert_U^e\) is globally generated, we may view \(s\) as a global section of \(\O\vert_U^e\). Since \(X\) is reduced, since X is reduced, the section \(s\) will be non-zero at some generic point \(y\in U\). Localized at \(y\), we get a short exact sequence \[ 0\to \mc{K}_y\to \O_y^e\to \mc{F}_y\to 0, \] where \(\O_y\) is a field because \(y\) is the generic point.

Because \(0\neq s_y\in \mc{K}_y\). There is a contraction by comparing the ranks of the \(\O_y\)-modules in the above short exact sequence.

Therefore, \(\O\vert_U^e\to \mc{F}\vert_U\) is an isomorphism.

Proof. By Lemma 3, it suffices to show that there is a closed point \(y\) in the closure of every point \(x\) of \(X\). Indeed, let \(U\) be an open neighborhood of \(y\) such that \(\mc{F}\vert_U\) is generated by it globally sections. Since \(y\in U\), we see \(x\in U\). Otherwise, if \(x\in U^c\), then \(y\in\overline{\{x\}}\subset U^c\) which implies a contradiction.

The fact that every point in \(X\) has a closed point in its closure follows from the quasi-compactness of \(X\).

Since \(X\) is non-empty quasi-compact, it admits an irredundant finite open cover \(X=U_1\cup \cdots \cup U_r\). Restrict to \(\overline{\{x\}}\subset X\), we get an irredundant finite open cover \(\overline{\{x\}}=V_1\cup \cdots \cup V_s\). Because a prime ideal is always contained in a maximal ideal. In each affine open set, there is a closed point. Let \(x_1\) be a close point in \(V_1\). If it is a closed point of \(\overline{\{x\}}\), then we are done. Otherwise, let \(x_2\in \overline{\{x_1\}}\) such that \(x_2\neq x_1\) and \(x_2\not\in V_1\), where the closure is taken in the closed subset \(\overline{\{x\}}\) but not in \(V_1\). With loss of generality, we may assume that \(x_2\in V_2\). Repeating this procedures, as the open cover is finite, we know that there must be a closed point \(y\) in \(\overline{\{x\}}\) which is also closed in \(X\).

Proof. If \(\mc{F}\) is locally free of rank 1, then \(\mc{G}=\sHom_{\O_X}(\mc{F},\O_X)\) satisfies that \[ \mc{F}\otimes \sHom_{\O_X}(\mc{F},\O_X)=\sHom_{\O_X}(\mc{F},\mc{F})\O_X \] because \(\mc{F}\) is locally free.

Conversely, for each point \(x\in X\), we get \[ \mc{F}_x\otimes_{\O_x}\mc{G}_x=\O_x. \] Tensor with \(k(x)=\O_x/\mf{m}_x\), we get \[ \mc{F}\vert_x\otimes_{k(x)}\mc{G}\vert_x=\mc{F}_x\otimes_{\O_x}k(x)\otimes_{k(x)}k(x)\otimes_{\O_x}\mc{G}_x=k(x). \] Note that \(\mc{F}\vert_x=\mc{F}_x/\mf{m}_x\mc{F}_x\) is a vector field over the residue field \(k(x)\). Therefore, \(\mc{F}\vert_x\) has rank 1 at every point \(x\in X\). By Corollary 4, \(\mc{F}\) is locally free of rank 1.

Proof. From the statement, we are expect that \(N=p_*(L\otimes M^{-1})\). Indeed, this is true. First, because \(X\) is complete and \(L_y\otimes M_y^{-1}\) is trivial, then \(p_*(L\otimes M^{-1})_y=k(y)\) for any \(y\in Y\). By Grauert’s Theorem, we know that \(p_*(L\otimes M^{-1})\) is locally free and moreover of rank 1.

We now show that the natural morphism \[ \alpha: p^*p_*(L\otimes M^{-1})\to L\otimes M^{-1} \] is an isomorphism. Let \(q\) be the restriction of \(p\) on the fiber \(X_y\). Then by diagram chasing and the assumption, we have an isomorphism along the fiber \(X_y\). \[ (p^*p_*(L\otimes M^{-1}))\vert_y=q^*q_*((L\otimes M^{-1})\vert_y)=\O_{X_y} \to (L\otimes M^{-1})\vert_y=\O_{X_y}. \] Consequently, for each closed point \((x, y)\in X\times Y\), the morphism \[ p^*p_*(L\otimes M^{-1})\vert_{(x, y)}\to \O_{X\times Y}\vert_{(x,y)} \] is an isomorphism.

Therefore, by Nakayama lemma, \(p^*p_*(L\otimes M^{-1})\to O_{X\times Y}\) is surjective. Because \(p^*p_*(L\otimes M^{-1})\) is of the rank 1, then the surjective morphism \(p^*p_*(L\otimes M^{-1})\to O_{X\times Y}\) must be an isomorphism, i.e. \(p^*p_*(L\otimes M^{-1})\cong O_{X\times Y}\).

Proof. Let \(N\) be the line bundle on \(Y\) such that \(L=M\otimes p^*N\). Over \(\{x\} \times Y\), we have \(L_x\cong M_x \otimes (p^∗N)_x\). Therefore, \((p^∗N)_x\cong N\) is trivial, which implies that \(p^*N\) is trivial and \(L\cong M\).