## 15.96 Perfect complexes and the eta operator

In this section we do some algebra to prepare for our version of Macpherson's graph construction, see More on Flatness, Section 38.44. We will use the $\eta _ f$ operator introduced in Section 15.95.

Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a bounded complex of finite free $A$-modules. For each $i$ let $r_ i$ be the rank of $M^ i$ and set

$I_ i(M^\bullet , f) = \text{ideal generated by the } r_ i \times r_ i\text{-minors of } (f, d^ i) : M^ i \to M^ i \oplus M^{i + 1}$

Observe that $f^{r_ i} \in I_ i(M^\bullet , f)$.

Lemma 15.96.1. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ and $N^\bullet$ be two bounded complexes of finite free $A$-modules representing the same object of $D(A)$. Then

$f^ m I_ i(M^\bullet , f) = f^ n I_ i(N^\bullet , f)$

as ideals of $A$ for integers $n, m \geq 0$ such that

$m + \sum \nolimits _{j \geq i} (-1)^{j - i}rk(M^ j) = n + \sum \nolimits _{j \geq i} (-1)^{j - i}rk(N^ j)$

Proof. It suffices to prove the equality after localization at every prime ideal of $A$. Thus by Lemma 15.75.7 and an induction argument we omit we may assume $N^\bullet = M^\bullet \oplus Q^\bullet$ for some trivial complex $Q^\bullet$, i.e.,

$Q^\bullet = \ldots \to 0 \to A \xrightarrow {1} A \to 0 \to \ldots$

where $A$ is placed in degree $j$ and $j + 1$. If $j \not= i - 1, i, i + 1$ then we clearly have equality $I_ i(M^\bullet , f) = I_ i(N^\bullet , f)$ and $m = n$ and we have the desired equality. If $j = i + 1$ then the maps

$(f, d^ i) : M^ i \to M^ i \oplus M^{i + 1} \quad \text{and}\quad (f, d^ i, 0) : M^ i \to M^ i \oplus M^{i + 1} \oplus A$

have the same nonzero minors hence in this case we also have $I_ i(M^\bullet , f) = I_ i(N^\bullet , f)$ and $m = n$. If $j = i$, then $I_ i(M^\bullet , f)$ is the ideal generated by the $r_ i \times r_ i$-minors of

$(f, d^ i) : M^ i \to M^ i \oplus M^{i + 1}$

and $I_ i(N^\bullet , f)$ is the ideal generated by the $(r_ i + 1) \times (r_ i + 1)$-minors of

$(f \oplus f, d^ i \oplus 1) : (M^ i \oplus A) \to (M^ i \oplus A) \oplus (M^{i + 1} \oplus A)$

With suitable choice of coordinates we see that the matrix of the second map is in block form

$T = \left( \begin{matrix} T_1 & 0 \\ 0 & T_2 \end{matrix} \right), \quad T_1 = \text{matrix of first map}, \quad T_2 = \left( \begin{matrix} f \\ 1 \end{matrix} \right)$

With notation as in Lemma 15.8.1 we have $I_0(T_2) = A$, $I_1(T_2) = A$, $I_ p(T_2) = 0$ for $p \geq 2$ and hence $I_{r_ i + 1}(T) = I_{r_ i + 1}(T_1) + I_{r_ i}(T_1) = I_{r_ i}(T_1)$ which means that $I_ i(M^\bullet , f) = I_ i(N^\bullet , f)$. We also have $m = n$ so this finishes the case $j = i$. Finally, say $j = i - 1$. Then we see that $m = n + 1$, thus we have to show that $fI_ i(M^\bullet , f) = I_ i(N^\bullet , f)$. In this case $I_ i(M^\bullet , f)$ is the ideal generated by the $r_ i \times r_ i$-minors of

$(f, d^ i) : M^ i \to M^ i \oplus M^{i + 1}$

and $I_ i(N^\bullet , f)$ is the ideal generated by the $(r_ i + 1) \times (r_ i + 1)$-minors of

$(f \oplus f, d^ i) : (M^ i \oplus A) \to (M^ i \oplus A) \oplus M^{i + 1}$

With suitable choice of coordinates we see that the matrix of the second map is in block form

$T = \left( \begin{matrix} T_1 & 0 \\ 0 & T_2 \end{matrix} \right), \quad T_1 = \text{matrix of first map}, \quad T_2 = \left( \begin{matrix} f \end{matrix} \right)$

Arguing as above we find that indeed $fI_ i(M^\bullet , f) = I_ i(N^\bullet , f)$. $\square$

Lemma 15.96.2. Let $f \in A$ be a nonzerodivisor of a ring $A$. Let $u \in A$ be a unit. Let $M^\bullet$ be a bounded complex of finite free $A$-modules. Then $I_ i(M^\bullet , f) = I_ i(M^\bullet , uf)$.

Proof. Omitted. $\square$

Lemma 15.96.3. Let $A \to B$ be a ring map. Let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a bounded complex of finite free $A$-modules. Assume $f$ maps to a nonzerodivisor $g$ in $B$. Then $I_ i(M^\bullet , f)B = I_ i(M^\bullet \otimes _ A B, g)$.

Proof. The minors of $(f, d^ i) : M^ i \to M^ i \oplus M^{i + 1}$ map to the corresponding minors of $(g, d^ i) : M^ i \otimes _ A B \to M^ i \otimes _ A B \oplus M^{i + 1} \otimes _ A B$. $\square$

Lemma 15.96.4. Let $A$ be a ring, let $\mathfrak p \subset A$ be a prime ideal, and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a bounded complex of finite free $A$-modules. If $H^ i(M^\bullet )_\mathfrak p$ is free for all $i$, then $I_ i(M^\bullet , f)_\mathfrak p$ is a principal ideal and in fact generated by a power of $f$ for all $i$.

Proof. We may assume $A$ is local with maximal ideal $\mathfrak p$ by Lemma 15.96.3. We may also replace $M^\bullet$ with a quasi-isomorphic complex by Lemma 15.96.1. By our assumption on the freeness of cohomology modules we see that $M^\bullet$ is quasi-isomorphic to the complex whose term in degree $i$ is $H^ i(M^\bullet )$ with vanishing differentials, see for example Derived Categories, Lemma 13.27.9. In other words, we may assume the differentials in the complex $M^\bullet$ are all zero. In this case it is clear that $I_ i(M^\bullet , f) = (f^{r_ i})$ is principal. $\square$

Lemma 15.96.5. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a bounded complex of finite free $A$-modules. Assume $I_ i(M^\bullet , f)$ is a principal ideal. Then $(\eta _ fM)^ i$ is locally free of rank $r_ i$ and the map $(1, d^ i) : (\eta _ fM)^ i \to f^ iM^ i \oplus f^{i + 1}M^{i + 1}$ is the inclusion of a direct summand.

Proof. Choose a generator $g$ for $I_ i(M^\bullet , f)$. Since $f^{r_ i} \in I_ i(M^\bullet , f)$ we see that $g$ divides a power of $f$. In particular $g$ is a nonzerodivisor in $A$. The $r_ i \times r_ i$-minors of the map $(f, d^ i) : M^ i \to M^ i \oplus M^{i + 1}$ generate the ideal $I_ i(M^\bullet , f)$ and the $(r_ i + 1) \times (r_ i + 1)$-minors of $(f, d^ i)$ are zero: we may check this after localizing at $f$ where the rank of the map is equal to $r_ i$. Consider the surjection

$M^ i \oplus M^{i + 1} \longrightarrow Q = \mathop{\mathrm{Coker}}(f, d^ i)/g\text{-torsion}$

By Lemma 15.8.9 the module $Q$ is finite locally free of rank $r_{i + 1}$. Hence $Q$ is $f$-torsion free and we conclude the cokernel of $(f, d^ i)$ modulo $f$-power torsion is $Q$ as well.

Consider the complex of finite free $A$-modules

$0 \to f^{i + 1}M^ i \xrightarrow {1, d^ i} f^ iM^ i \oplus f^{i + 1}M^{i + 1} \xrightarrow {d^ i, -1} f^ iM^{i + 1} \to 0$

which becomes split exact after localizing at $f$. The map $(1, d^ i) : f^{i + 1}M^ i \to f^ iM^ i \oplus f^{i + 1}M^{i + 1}$ is isomorphic to the map $(f, d^ i) : M^ i \to M^ i \oplus M^{i + 1}$ we studied above. Hence the image

$Q' = \mathop{\mathrm{Im}}(f^ iM^ i \oplus f^{i + 1}M^{i + 1} \xrightarrow {d^ i, -1} f^ iM^{i + 1})$

is isomorphic to $Q$ in particular projective. On the other hand, by construction of $\eta _ f$ in Section 15.95 the image of the injective map $(1, d^ i) : (\eta _ fM)^ i \to f^ iM^ i \oplus f^{i + 1}M^{i + 1}$ is the kernel of $(d^ i, -1)$. We conclude that we obtain an isomorphism $(\eta _ fM)^ i \oplus Q' = f^ iM^ i \oplus f^{i + 1}M^{i + 1}$ and we see that indeed $\eta _ fM^ i$ is finite locally free of rank $r_ i$ and that $(1, d^ i)$ is the inclusion of a direct summand. $\square$

Lemma 15.96.6. Let $A \to B$ be a ring map. Let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a bounded complex of finite free $A$-modules. Assume $f$ maps to a nonzerodivisor $g$ in $B$ and $I_ i(M^\bullet , f)$ is a principal ideal for all $i \in \mathbf{Z}$. Then there is a canonical isomorphism $\eta _ fM^\bullet \otimes _ A B = \eta _ g(M^\bullet \otimes _ A B)$.

Proof. Set $N^ i = M^ i \otimes _ A B$. Observe that $f^ iM^ i \otimes _ A B = g^ iN^ i$ as submodules of $(N^ i)_ g$. The maps

$(\eta _ fM)^ i \otimes _ A B \to g^ iN^ i \otimes g^{i + 1}N^{i + 1} \quad \text{and}\quad (\eta _ gN)^ i \to g^ iN^ i \otimes g^{i + 1}N^{i + 1}$

are inclusions of direct summands by Lemma 15.96.5. Since their images agree after localizing at $g$ we conclude. $\square$

Lemma 15.96.7. Let $A$ be a ring. Let $M$, $N_1$, $N_2$ be finite projective $A$-modules. Let $s : M \to N_1 \oplus N_2$ be a split injection. There exists a finitely generated ideal $J \subset A$ with the following property: a ring map $A \to B$ factors through $A/J$ if and only if $s \otimes \text{id}_ B$ identifies $M \otimes _ A B$ with a direct sum of submodules of $N_1 \otimes _ A B \oplus N_2 \otimes _ A B$.

Proof. Choose a splitting $\pi : N_1 \oplus N_2 \to M$ of $s$. Denote $q_ i : N_1 \oplus N_2 \to N_1 \oplus N_2$ the projector onto $N_ i$. Set $p_ i = \pi \circ q_ i \circ s$. Observe that $p_1 + p_2 = \text{id}_ M$. We claim $M$ is a direct sum of submodules of $N_1 \oplus N_2$ if and only if $p_1$ and $p_2$ are orthogonal projectors. Thus $J$ is the smallest ideal of $A$ such that $p_1 \circ p_1 - p_1$, $p_2 \circ p_2 - p_2$, $p_1 \circ p_2$, and $p_2 \circ p_1$ are contained in $J \otimes _ A \text{End}_ A(M)$. Some details omitted. $\square$

Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a bounded complex of finite free $A$-modules. Assume the ideals $I_ i(M^\bullet , f)$ are principal for all $i \in \mathbf{Z}$. Then the maps

$(1, d^ i) : (\eta _ fM)^ i / f(\eta _ fM)^ i \longrightarrow f^ iM^ i/f^{i + 1}M^ i \oplus f^{i + 1}M^{i + 1}/f^{i + 2}M^{i + 1}$

are split injections by Lemma 15.96.5. Denote $J_ i(M^\bullet , f) \subset A/fA$ the finitely generated ideal of Lemma 15.96.7 corresponding to the split injection $(1, d^ i)$ displayed above.

Lemma 15.96.8. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ and $N^\bullet$ be two bounded complexes of finite free $A$-modules representing the same object in $D(A)$. Assume $I_ i(M^\bullet , f)$ is a principal ideal for all $i \in \mathbf{Z}$. Then $J_ i(M^\bullet , f) = J_ i(N^\bullet , f)$ as ideals in $A/fA$.

Proof. Observe that the fact that $I_ i(M^\bullet , f)$ is a principal ideal implies that $I_ i(M^\bullet , f)$ is a principal ideal by Lemma 15.96.1 and hence the statement makes sense. As in the proof of Lemma 15.96.1 we may assume $N^\bullet = M^\bullet \oplus Q^\bullet$ for some trivial complex $Q^\bullet$, i.e.,

$Q^\bullet = \ldots \to 0 \to A \xrightarrow {1} A \to 0 \to \ldots$

where $A$ is placed in degree $j$ and $j + 1$. Since $\eta _ f$ is compatible with direct sums, we see that the map

$(1, d^ i) : (\eta _ fN)^ i / f(\eta _ fN)^ i \longrightarrow f^ iN^ i/f^{i + 1}N^ i \oplus f^{i + 1}N^{i + 1}/f^{i + 2}N^{i + 1}$

is the direct sum of the corresponding map for $M^\bullet$ and for $Q^\bullet$. By the universal property defining the ideals in question, we conclude that $J_ i(N^\bullet , f) = J_ i(M^\bullet , f) + J_ i(Q^\bullet , f)$. Hence it suffices to show that $J_ i(Q^\bullet , f) = 0$ for all $i$. This is a computation that we omit. $\square$

Lemma 15.96.9. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a bounded complex of finite free $A$-modules. Assume $I_ i(M^\bullet , f)$ is a principal ideal for all $i \in \mathbf{Z}$. Consider the ideal $J(M^\bullet , f) = \sum _ i J_ i(M^\bullet , f)$ of $A/fA$. Consider the set of prime ideals

\begin{align*} E & = \{ f \in \mathfrak p \subset A \mid \mathop{\mathrm{Ker}}(d^ i \bmod f^2)_\mathfrak p \text{ surjects onto } \mathop{\mathrm{Ker}}(d^ i \bmod f)_\mathfrak p \text{ for all }i \in \mathbf{Z}\} \\ & = \{ f \in \mathfrak p \subset A \mid \text{the localizations }\beta _\mathfrak p \text{ of the Bockstein operators are zero}\} \end{align*}

Then we have

1. $J(M^\bullet , f)$ is finitely generated,

2. $A/fA \to C = (A/fA)/J(M^\bullet , f)$ is surjective of finite presentation,

3. $J(M^\bullet , f)_\mathfrak p = 0$ for $\mathfrak p \in E$,

4. if $f \in \mathfrak p$ and $H^ i(M^\bullet )_\mathfrak p$ is free for all $i \in \mathbf{Z}$, then $\mathfrak p \in E$, and

5. the cohomology modules of $\eta _ f M^\bullet \otimes _ A C$ are finite locally free $C$-modules.

Proof. The equality in the definition of $E$ follows from Lemma 15.95.7 and in addition the final statement of that lemma implies part (4).

Part (1) is true because the ideals $J_ i(M^\bullet , f)$ are finitely generated and because $M^\bullet$ is bounded and hence $J_ i(M^\bullet , f)$ is zero for almost all $i$. Part (2) is just a reformulation of part (1).

Proof of (3). By Lemma 15.96.5 we find that $(\eta _ fM)^ i$ is finite locally free of rank $r_ i$ for all $i$. Consider the map

$(1, d^ i) : (\eta _ fM)^ i / f(\eta _ fM)^ i \longrightarrow f^ iM^ i/f^{i + 1}M^ i \oplus f^{i + 1}M^{i + 1}/f^{i + 2}M^{i + 1}$

Pick $\mathfrak p \in E$. By Lemma 15.95.8 and the local freeness of the modules $(\eta _ fM)^ i$ we may write

$\left((\eta _ fM)^ i / f(\eta _ fM)^ i\right)_\mathfrak p = (A/fA)_\mathfrak p^{\oplus m_ i} \oplus (A/fA)_\mathfrak p^{\oplus n_ i}$

compatible with the arrow $(1, d^ i)$ above. By the universal property of the ideal $J_ i(M^\bullet , f)$ we conclude that $J_ i(M^\bullet , f)_\mathfrak p = 0$. Hence $I_\mathfrak p = fA_\mathfrak p$ for $\mathfrak p \in E$.

Proof of (5). Observe that the differential on $\eta _ fM^\bullet$ fits into a commutative diagram

$\xymatrix{ (\eta _ fM)^ i \ar[d] \ar[r] & f^ iM^ i \oplus f^{i + 1}M^{i + 1} \ar[d]^{\left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)} \\ (\eta _ fM)^{i + 1} \ar[r] & f^{i + 1}M^ i \oplus f^{i + 2}M^{i + 2} }$

By construction, after tensoring with $C$, the modules on the left are direct sums of direct summands of the summands on the right. Picture

$\xymatrix{ (\eta _ fM)^ i \otimes _ A C \ar[d] \ar@{=}[r] & K^ i \oplus L^ i \ar[r] \ar[d] & f^ iM^ i \otimes _ A C \oplus f^{i + 1}M^{i + 1} \otimes _ A C \ar[d]^{\left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)} \\ (\eta _ fM)^{i + 1} \otimes _ A C \ar@{=}[r] & K^{i + 1} \oplus L^{i + 1} \ar[r] & f^{i + 1}M^ i \otimes _ A C \oplus f^{i + 2}M^{i + 2} \otimes _ A C }$

where the horizontal arrows are compatible with direct sum decompositions as well as inclusions of direct summands. It follows that the differential identifies $L^ i$ with a direct summand of $K^{i + 1}$ and we conclude that the cohomology of $\eta _ fM^\bullet \otimes _ A C$ in degree $i$ is the module $K^{i + 1}/L^ i$ which is finite projective as desired. $\square$

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