## 15.94 Perfect complexes and the eta operator

In this section we discuss when the operator $\eta _ f$ introduced in Section 15.93 transforms a perfect complex into a perfect complex and what you can say about the result. This material is somewhat nonstandard.

Lemma 15.94.1. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a complex of $A$-modules such that $f$ is a nonzerodivisor on all $M^ i$. Assume

1. $M^ i$, $M^{i + 1}$ free of ranks $r_ i, r_{i + 1}$,

2. the $r_ i \times r_ i$ minors of $(f, d^ i) : M^ i \to M^ i \oplus M^{i + 1}$ generate a principal ideal.

Then $(\eta _ fM)^ i$ is locally free of rank $r_ i$ and the canonical map $(\eta _ fM)^ i \to f^ iM^ i \oplus f^{i + 1}M^{i + 1}$ is the inclusion of a direct summand.

Proof. Observe that $I$ contains the element $f^{r_ i}$. Hence $I = (g)$ where $g$ divides a power of $f$. Consider the complex of finite free $A$-modules

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

which becomes split exact after localizing at $f$. By construction $(\eta _ fM)^ i$ is isomorphic to the kernel of the second map. Hence $(\eta _ fM)^ i$ is the kernel of the map

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

By Lemma 15.8.8 the finite module $Q$ can locally be generated by $r_{i + 1}$ elements. On the other hand we have $Q_ f \cong M^{i + 1}_ f$ is free of rank $r_{i + 1}$. Hence $Q$ is locally free of rank $r_{i + 1}$. Thus $Q$ is a projective module and the surjection above is split. $\square$

Lemma 15.94.2. Let $A \to B$ be a ring map. Let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a complex of $A$-modules. Assume

1. $f$ maps to a nonzerodivisor $g$ in $B$,

2. $f$ is a nonzerodivisor on $M^ i$,

3. $M^ i$ is finite free of rank $r_ i$,

4. the $r_ i \times r_ i$ minors of $(f, d^ i) : M^ i \to M^ i \oplus M^{i + 1}$ generate 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 15.94.1. Since their images agree after localizing at $g$ we conclude. $\square$

Lemma 15.94.3. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a complex of $A$-modules such that $f$ is a nonzerodivisor on all $M^ i$. For $i \in \mathbf{Z}$ the following are equivalent

1. $\mathop{\mathrm{Ker}}(d^ i \bmod f^2)$ surjects onto $\mathop{\mathrm{Ker}}(d^ i \bmod f)$,

2. $\beta : H^ i(M^\bullet \otimes _ A f^ iA/f^{i + 1}A) \to H^{i + 1}(M^\bullet \otimes _ A f^{i + 1}A/f^{i + 2}A)$ is zero.

These equivalent conditions are implied by the condition $H^{i + 1}(M^\bullet )[f] = 0$.

Proof. The equivalence of (1) and (2) follows immediately from the definitions. If $\beta \not= 0$, then $H^{i + 1}(M^\bullet )[f] \not= 0$ because $\beta$ factors through a map $H^ i(M^\bullet \otimes f^ iA/f^{i + 1}A) \to H^{i + 1}(M^\bullet \otimes f^{i + 1}A)$ (see discussion following the construction of the Bockstein operators). $\square$

Lemma 15.94.4. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a complex of $A$-modules such that $f$ is a nonzerodivisor on all $M^ i$. If $\mathop{\mathrm{Ker}}(d^ i \bmod f^2)$ surjects onto $\mathop{\mathrm{Ker}}(d^ i \bmod f)$, then the canonical map

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

identifies the left hand side with a direct sum of submodules of the right hand side.

Proof. With notation as in Remark 15.93.4 we define a map $t^{-1} : Z^ i \to (\eta _ fM)^ i / f(\eta _ fM)^ i$. Namely, for $x \in M^ i$ with $d^ i(x) = f^2y$ we send the class of $x$ in $Z^ i$ to the class of $f^ ix$ in $(\eta _ fM)^ i / f(\eta _ fM)^ i$. We omit the verification that this is well defined; the assumption of the lemma exactly signifies that the domain of this operation is all of $Z^ i$. Then $t \circ t^{-1} = \text{id}_{Z^ i}$. Hence $t^{-1}$ defines a splitting of the short exact sequence in Remark 15.93.4 and the resulting direct sum decomposition

$(\eta _ fM)^ i / f(\eta _ fM)^ i = Z^ i \oplus B^{i + 1}$

is compatible with the map displayed in the lemma. $\square$

Lemma 15.94.5. 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 $I \subset A$ with the following property: a ring map $A \to B$ factors through $A/I$ 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 $I$ 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 $I \otimes _ A \text{End}_ A(M)$. Some details omitted. $\square$

Lemma 15.94.6. Let $A$ be a ring and let $f \in \mathfrak m_ A$ be a nonzerodivisor. Let $M^\bullet$ be a complex of $A$-modules such that $f$ is a nonzerodivisor on all $M^ i$. Assume

1. $M^ i$ is finite free of rank $r_ i$ and $M^ i = 0$ for $|i| \gg 0$,

2. the $r_ i \times r_ i$ minors of $(f, d^ i) : M^ i \to M^ i \oplus M^{i + 1}$ generate a principal ideal

for all $i \in \mathbf{Z}$. Consider the set of prime ideals

$E = \{ f \in \mathfrak p \subset A \mid \mathop{\mathrm{Ker}}(d^ i \bmod f^2)_\mathfrak p \text{ surjects onto } \mathop{\mathrm{Ker}}(d^ i \bmod f)_\mathfrak p \text{ for all }i \in \mathbf{Z}\}$

There exists a finitely generated ideal $f \in I$ such that $I_\mathfrak p = fA_\mathfrak p$ for all $\mathfrak p \in E$ and such that the cohomology modules of $\eta _ f M^\bullet \otimes _ A A/I$ are finite free.

Proof. By Lemma 15.94.1 we find that $(\eta _ fM)^ i$ is free of rank $r_ i$ for all $i$. Consider the map

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

of Lemma 15.94.4. By Lemma 15.94.1 this map is split injective. Let $I_ i \subset A$ be the ideal containing $f$ such that the ideal $I_ i/fA \subset A/fA$ is the one found in Lemma 15.94.5 for the displayed arrow. We set $I = \sum _{i \in \mathbf{Z}} I_ i$. Since almost all $I_ i = A$ this is a finitely generated ideal.

Pick $\mathfrak p \in E$. By Lemma 15.94.4 and the 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 above. By the universal property of the ideal $I_ i$ we conclude that $(I_ i)_\mathfrak p = fA_\mathfrak p$. Hence $I_\mathfrak p = fA_\mathfrak p$ for $\mathfrak p \in E$.

To finish the proof we need to show the assertion about cohomology. To see this, 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 $A/I$, the modules on the left are direct sums of direct summands of the summands on the right. Picture

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

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 $\eta _ fM^\bullet \otimes _ A A/I$ in degree $i$ is the module $K^{i + 1}/L^ i$ which is finite projective as desired. $\square$

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