Finite-free local criterion for cohomology and base change

Let $B=A_{\mathfrak{m}}$, a local ring with maximal ideal $\mathfrak{n}=\mathfrak{m}{A}_{\mathfrak{m}}$ and residue field $\kappa =B\mathrm{/}\mathfrak{n}$, and put $J(B)=\mathfrak{n}$ [F4, F5]. Since $K^{\bullet }$ consists of finite free modules, $K_{\mathfrak{m}}^{\bullet }{\otimes }_B\kappa =K^{\bullet }{\otimes }_A\kappa$, so $H^q(K_{\mathfrak{m}}){\otimes }_B\kappa =H^q(K){\otimes }_A\kappa$ and $H^q(K_{\mathfrak{m}}{\otimes }_B\kappa )=H^q(K{\otimes }_A\kappa )$; the map ${\phi }^q$ is unchanged. Every $s\in A\setminus \mathfrak{m}$ maps to a unit of $B$ [F4], so any basis change over $A_s$ is a basis change over $B$; conversely, a computation over $B$ has finitely many matrix entries $a_i\mathrm{/}{s}_i$ with $s_i\in A\setminus \mathfrak{m}$, and with $s=s_1\cdots s_t\in A\setminus \mathfrak{m}$ it is a computation over $A_s$. It therefore suffices to prove statements 1–3 with $(A,\mathfrak{m})$ replaced by the local ring $B$.

The image of ${\phi }^q$ is the image of $\ker d$ in $\ker (d\otimes \kappa )\mathrm{/}\mathrm{im}(e\otimes \kappa )$, so ${\phi }^q$ is surjective if and only if $d^{-1}(\mathfrak{m}{K}^{q+1})=\ker d+\mathrm{im}e+\mathfrak{m}{K}^q\mathrm{.}$ Indeed, by [F1] and [F3] the target is $d^{-1}(\mathfrak{m}{K}^{q+1})\mathrm{/}(\mathrm{im}e+\mathfrak{m}{K}^q)$, while the image is $(\ker d+\mathrm{im}e+\mathfrak{m}{K}^q)\mathrm{/}(\mathrm{im}e+\mathfrak{m}{K}^q)$. Since $\mathrm{im}e\subseteq \ker d$, equality of these two quotients is equivalent to the displayed equality of their numerator submodules.

Under AC [F6] put $r={\dim }_{\kappa }\mathrm{im}(d\otimes \kappa )$. Choose a $B$-basis $g_1,\dots ,g_n$ of $K^{q+1}$ such that $g_1,\dots ,g_r$ reduce modulo $\mathfrak{n}$ to a basis of $\mathrm{im}(d\otimes \kappa )$; this is possible because any $\kappa$-basis of $\mathrm{im}(d\otimes \kappa )$ extends to one of $K^{q+1}\otimes \kappa$ and any basis of the finite free module $K^{q+1}\otimes \kappa$ lifts to a $B$-basis of $K^{q+1}$ [F2, F5]. For $i\le r$ pick $f_i^0\in K^q$ with $d(f_i^0)\equiv g_i(\mathrm{m}\mathrm{o}\mathrm{d}\mathfrak{n}{K}^{q+1})$, which exists by the choice of the $g_i$. Replacing $g_i$ by $d(f_i^0)$ for $i\le r$ leaves a $B$-basis of $K^{q+1}$ [F2, F5], and the elements $f_1^0,\dots ,f_r^0$ are linearly independent modulo $\mathfrak{n}{K}^q$; extend them to a basis $f_1,\dots ,f_m$ of $K^q$ [F2, F5]. In these bases the matrix of $d$ has the block form $M=\left(\begin{array}{cc}{I}_r& C\\ 0& D\end{array}\right),D\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}\mathfrak{n}),$ the congruence for $D$ holding because $r$ was defined as the rank of $d\otimes \kappa$.

Replacing $f_j$ by $f_j-{\sum }_{i\le r}{c}_{ij}{f}_i$ for $j>r$ (an invertible change of basis) removes the block $C$: if $d(f_j)={\sum }_{i\le r}{c}_{ij}{g}_i+d_j$ with $d_j$ in the span of $g_{r+1},\dots ,g_n$, then $d(f_j-{\sum }_{i\le r}{c}_{ij}{f}_i)=d_j$. Hence, after this change of the basis of $K^q$ alone, the matrix of $d$ is $\left(\begin{array}{cc}{I}_r& 0\\ 0& D\end{array}\right),D\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}\mathfrak{n}),$ and $\ker d$ consists of the vectors $(0,x)$ with $Dx=0$. Writing $K^q=F^{\mathrm{\prime }}\oplus F^{\mathrm{\prime }\mathrm{\prime }}$ for the spans of $f_1,\dots ,f_r$ and $f_{r+1},\dots ,f_m$, we have $d^{-1}(\mathfrak{m}{K}^{q+1})=\mathfrak{m}{F}^{\mathrm{\prime }}\oplus F^{\mathrm{\prime }\mathrm{\prime }}$ and $\ker d=\{0\}\oplus \ker D$, while $\mathrm{im}e\subseteq \ker d\subseteq F^{\mathrm{\prime }\mathrm{\prime }}$ because $d\circ e=0$.

We evaluate step 1.2 in the block form of step 2.1. The condition becomes $\mathfrak{m}{F}^{\mathrm{\prime }}\oplus F^{\mathrm{\prime }\mathrm{\prime }}=\mathfrak{m}{F}^{\mathrm{\prime }}\oplus (\ker D+\mathrm{im}e+\mathfrak{m}{F}^{\mathrm{\prime }\mathrm{\prime }})$, i.e. $F^{\mathrm{\prime }\mathrm{\prime }}=\ker D+\mathfrak{m}{F}^{\mathrm{\prime }\mathrm{\prime }}$; by Nakayama [F2, F6] applied to the finitely generated module $F^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}\ker D$ with the ideal $\mathfrak{n}=J(B)$ [F5] this is equivalent to $F^{\mathrm{\prime }\mathrm{\prime }}=\ker D$, i.e. to $D=0$. Thus ${\phi }^q$ is surjective if and only if, after the constructions of steps 1.3–2.1, the block $D$ vanishes, which is exactly the split form with matrix $\mathrm{diag}(I_r,0)$; this proves statement 1, the basis changes being the ones constructed in steps 1.3–2.1 and the transition between $B$ and $A_s$ being as in step 1.1.

Assume now that over some $A_s$ the differential $d^q$ has matrix $\mathrm{diag}(I_r,0)$ with respect to bases of $K_s^q$ and $K_s^{q+1}$; write $K_s^q=F^{\mathrm{\prime }}\oplus F^{\mathrm{\prime }\mathrm{\prime }}$ accordingly, so that $\ker d^q=F^{\mathrm{\prime }\mathrm{\prime }}$ and $\mathrm{im}e\subseteq F^{\mathrm{\prime }\mathrm{\prime }}$. Then $H^q(K_s)=F^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}\mathrm{im}e$ is finitely generated, and for every $A_s$-algebra $A^{\mathrm{\prime }}$ the differentials of $K^{\bullet }{\otimes }_A{A}^{\mathrm{\prime }}$ are $\mathrm{diag}(I_r,0)$ and $e\otimes 1$, so $H^q(K{\otimes }_A{A}^{\mathrm{\prime }})=\ker (d^q\otimes 1)\mathrm{/}\mathrm{im}(e\otimes 1)=(F^{\mathrm{\prime }\mathrm{\prime }}{\otimes }_{A_s}{A}^{\mathrm{\prime }})\mathrm{/}\mathrm{im}(e\otimes 1)=H^q(K_s){\otimes }_{A_s}{A}^{\mathrm{\prime }},$ the last equality by right exactness of the tensor product [F3]. This proves statement 2.

It remains to discuss finite projectivity. Keep the split form of step 4.1, so that $H^q(K_s)=\mathrm{coker}(e^{\mathrm{\prime }}:K_s^{q-1}\to F^{\mathrm{\prime }\mathrm{\prime }})$ with $e^{\mathrm{\prime }}$ the composite of $e$ with the projection onto $F^{\mathrm{\prime }\mathrm{\prime }}$. Right exactness [F3] shows that $\mathrm{coker}(e^{\mathrm{\prime }})$ commutes with every base change if $e^{\mathrm{\prime }}$ can be written as $\mathrm{diag}(I_{r^{\mathrm{\prime }}},0)$ in suitable bases of $K_s^{q-1}$ and $F^{\mathrm{\prime }\mathrm{\prime }}$: then the cokernel is the free module on the remaining basis vectors.

Suppose $H^q(K_s)=\mathrm{coker}(e^{\mathrm{\prime }})$ is finite projective over $A_s$. The surjection $F^{\mathrm{\prime }\mathrm{\prime }}\to H^q(K_s)$ splits, so $\mathrm{im}(e^{\mathrm{\prime }})$ is a finite projective direct summand of $F^{\mathrm{\prime }\mathrm{\prime }}$; the surjection $K_s^{q-1}\to \mathrm{im}(e^{\mathrm{\prime }})$ then splits, and its kernel is also finite projective. Localize at $\mathfrak{m}$: all these finite projective summands, including the complementary copy of $H^q(K_s)$ in $F^{\mathrm{\prime }\mathrm{\prime }}$, become finite free over the local ring $B=A_{\mathfrak{m}}$ by the basis-lifting and Nakayama argument of [F2]. Bases of the summands and their inclusions and projections involve finitely many matrix entries and inverse determinants in $B$; clear their denominators and the finitely many matrix equalities over a further $A_{s^{\mathrm{\prime }}}$ with $s^{\mathrm{\prime }}\mathrm{\notin }\mathfrak{m}$ (as in step 1.1). The resulting bases of $K_{s^{\mathrm{\prime }}}^{q-1}$ and $F_{s^{\mathrm{\prime }}}^{\mathrm{\prime }\mathrm{\prime }}$ exhibit $e^{\mathrm{\prime }}$ as $\mathrm{diag}(I_{r^{\mathrm{\prime }}},0)$ on that neighbourhood, as required for the converse to step 5.1.

Finally, the criterion of statement 1 applied with $q$ replaced by $q-1$ to the map $e=d^{q-1}:K^{q-1}\to K^q$ says that ${\phi }^{q-1}$ is surjective if and only if, after shrinking, $e$ has split form $\mathrm{diag}(I_{r^{\mathrm{\prime }}},0)$ with respect to bases of $K^{q-1}$ and $K^q$. If $K^{q-1}=0$ then $H^{q-1}(K)=0=H^{q-1}(K{\otimes }_A\kappa )$ and ${\phi }^{q-1}$ is the zero map of the zero module, hence surjective; this covers the degree $-1$ convention. Combining with steps 5.1 and 6.1, finite projectivity of $H^q(K_s)$ is equivalent to surjectivity of ${\phi }^{q-1}$ after shrinking: if $e^{\mathrm{\prime }}=\mathrm{diag}(I_{r^{\mathrm{\prime }}},0)$ in bases of $K_s^{q-1}$ and $F^{\mathrm{\prime }\mathrm{\prime }}$, then adjoining the basis of $F^{\mathrm{\prime }}$ puts the matrix of $e$ into $\mathrm{diag}(I_{r^{\mathrm{\prime }}},0)$ after reordering the basis of $K_s^q$, which is the criterion for ${\phi }^{q-1}$; conversely ${\phi }^{q-1}$ surjectivity produces such bases and step 6.1 gives projectivity. This proves statement 3 and completes the proof.