Definition 15.83.1. Let $R \to A$ be a flat ring map of finite presentation. An object $K$ of $D(A)$ is $R$-perfect or perfect relative to $R$ if $K$ is pseudo-coherent (Definition 15.64.1) and has finite tor dimension over $R$ (Definition 15.66.1).
15.83 Relatively perfect modules
This section is the analogue of Section 15.81 for perfect objects of the derived category. we only define this notion in a limited generality as we are not sure what the correct definition is in general. See Derived Categories of Schemes, Remark 36.35.14 for a discussion.
By Lemma 15.82.8 it would have been the same thing to ask $K$ to be pseudo-coherent relative to $R$. Here are some obligatory lemmas.
Lemma 15.83.2. Let $R \to A$ be a flat ring map of finite presentation. The $R$-perfect objects of $D(A)$ form a saturated1 triangulated strictly full subcategory.
Proof. This follows from Lemmas 15.64.2, 15.64.8, 15.66.5, and 15.66.7. $\square$
Lemma 15.83.3. Let $R \to A$ be a flat ring map of finite presentation. A perfect object of $D(A)$ is $R$-perfect. If $K, M \in D(A)$ then $K \otimes _ A^\mathbf {L} M$ is $R$-perfect if $K$ is perfect and $M$ is $R$-perfect.
Proof. The first statement follows from the second by taking $M = A$. The second statement follows from Lemmas 15.74.2, 15.66.10, and 15.64.16. $\square$
Lemma 15.83.4. Let $R \to A$ be a flat ring map of finite presentation. Let $K \in D(A)$. The following are equivalent
$K$ is $R$-perfect, and
$K$ is isomorphic to a finite complex of $R$-flat, finitely presented $A$-modules.
Proof. To prove (2) implies (1) it suffices by Lemma 15.83.2 to show that an $R$-flat, finitely presented $A$-module $M$ defines an $R$-perfect object of $D(A)$. Since $M$ has finite tor dimension over $R$, it suffices to show that $M$ is pseudo-coherent. By Algebra, Lemma 10.168.1 there exists a finite type $\mathbf{Z}$-algebra $R_0 \subset R$ and a flat finite type ring map $R_0 \to A_0$ and a finite $A_0$-module $M_0$ flat over $R_0$ such that $A = A_0 \otimes _{R_0} R$ and $M = M_0 \otimes _{R_0} R$. By Lemma 15.64.17 we see that $M_0$ is pseudo-coherent $A_0$-module. Choose a resolution $P_0^\bullet \to M_0$ by finite free $A_0$-modules $P_0^ n$. Since $A_0$ is flat over $R_0$, this is a flat resolution. Since $M_0$ is flat over $R_0$ we find that $P^\bullet = P_0^\bullet \otimes _{R_0} R$ still resolves $M = M_0 \otimes _{R_0} R$. (You can use Lemma 15.61.2 to see this.) Hence $P^\bullet $ is a finite free resolution of $M$ over $A$ and we conclude that $M$ is pseudo-coherent.
Assume (1). We can represent $K$ by a bounded above complex $P^\bullet $ of finite free $A$-modules. Assume that $K$ viewed as an object of $D(R)$ has tor amplitude in $[a, b]$. By Lemma 15.66.2 we see that $\tau _{\geq a}P^\bullet $ is a complex of $R$-flat, finitely presented $A$-modules representing $K$. $\square$
Lemma 15.83.5. Let $R \to A$ be a flat ring map of finite presentation. Let $R \to R'$ be a ring map and set $A' = A \otimes _ R R'$. If $K \in D(A)$ is $R$-perfect, then $K \otimes _ A^\mathbf {L} A'$ is $R'$-perfect.
Proof. By Lemma 15.64.12 we see that $K \otimes _ A^\mathbf {L} A'$ is pseudo-coherent. By Lemma 15.61.2 we see that $K \otimes _ A^\mathbf {L} A'$ is equal to $K \otimes _ R^\mathbf {L} R'$ in $D(R')$. Then we can apply Lemma 15.66.13 to see that $K \otimes _ R^\mathbf {L} R'$ in $D(R')$ has finite tor dimension. $\square$
Lemma 15.83.6. Let $R \to A$ be a flat ring map. Let $K, L \in D(A)$ with $K$ pseudo-coherent and $L$ finite tor dimension over $R$. We may choose
a bounded above complex $P^\bullet $ of finite free $A$-modules representing $K$, and
a bounded complex of $R$-flat $A$-modules $F^\bullet $ representing $L$.
Given these choices we have
$E^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet , F^\bullet )$ is a bounded below complex of $R$-flat $A$-modules representing $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$,
for any ring map $R \to R'$ with $A' = A \otimes _ R R'$ the complex $E^\bullet \otimes _ R R'$ represents $R\mathop{\mathrm{Hom}}\nolimits _{A'}(K \otimes _ A^\mathbf {L} A', L \otimes _ A^\mathbf {L} A')$.
If in addition $R \to A$ is of finite presentation and $L$ is $R$-perfect, then we may choose $F^ p$ to be finitely presented $A$-modules and consequently $E^ n$ will be finitely presented $A$-modules as well.
Proof. The existence of $P^\bullet $ is the definition of a pseudo-coherent complex. We first represent $L$ by a bounded above complex $F^\bullet $ of free $A$-modules (this is possible because bounded tor dimension in particular implies bounded). Next, say $L$ viewed as an object of $D(R)$ has tor amplitude in $[a, b]$. Then, after replacing $F^\bullet $ by $\tau _{\geq a}F^\bullet $, we get a complex as in (2). This follows from Lemma 15.66.2.
Proof of (a). Since $F^\bullet $ is bounded an since $P^\bullet $ is bounded above, we see that $E^ n = 0$ for $n \ll 0$ and that $E^ n$ is a finite (!) direct sum
\[ E^ n = \bigoplus \nolimits _{p + q = n} \mathop{\mathrm{Hom}}\nolimits _ A(P^{-q}, F^ p) \]
and since $P^{-q}$ is finite free, this is indeed an $R$-flat $A$-module. The fact that $E^\bullet $ represents $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$ follows from Lemma 15.73.2.
Proof of (b). Let $R \to R'$ be a ring map and $A' = A \otimes _ R R'$. By Lemma 15.61.2 the object $L \otimes _ A^\mathbf {L} A'$ is represented by $F^\bullet \otimes _ R R'$ viewed as a complex of $A'$-modules (by flatness of $F^ p$ over $R$). Similarly for $P^\bullet \otimes _ R R'$. As above $R\mathop{\mathrm{Hom}}\nolimits _{A'}(K \otimes _ A^\mathbf {L} A', L \otimes _ A^\mathbf {L} A')$ is represented by
\[ \mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet \otimes _ R R', F^\bullet \otimes _ R R') = E^\bullet \otimes _ R R' \]
The equality holds by looking at the terms of the complex individually and using that $\mathop{\mathrm{Hom}}\nolimits _{A'}(P^{-q} \otimes _ R R', F^ p \otimes _ R R') = \mathop{\mathrm{Hom}}\nolimits _ A(P^{-q}, F^ p) \otimes _ R R'$. $\square$
Lemma 15.83.7. Let $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ be a filtered colimit of rings. Let $0 \in I$ and $R_0 \to A_0$ be a flat ring map of finite presentation. For $i \geq 0$ set $A_ i = R_ i \otimes _{R_0} A_0$ and set $A = R \otimes _{R_0} A_0$.
Given an $R$-perfect $K$ in $D(A)$ there exists an $i \in I$ and an $R_ i$-perfect $K_ i$ in $D(A_ i)$ such that $K \cong K_ i \otimes _{A_ i}^\mathbf {L} A$ in $D(A)$.
Given $K_0, L_0 \in D(A_0)$ with $K_0$ pseudo-coherent and $L_0$ finite tor dimension over $R_0$, then we have
\[ \mathop{\mathrm{Hom}}\nolimits _{D(A)}(K_0 \otimes _{A_0}^\mathbf {L} A, L_0 \otimes _{A_0}^\mathbf {L} A) = \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathop{\mathrm{Hom}}\nolimits _{D(A_ i)}(K_0 \otimes _{A_0}^\mathbf {L} A_ i, L_0 \otimes _{A_0}^\mathbf {L} A_ i) \]
In particular, the triangulated category of $R$-perfect complexes over $A$ is the colimit of the triangulated categories of $R_ i$-perfect complexes over $A_ i$.
Proof. By Algebra, Lemma 10.127.6 the category of finitely presented $A$-modules is the colimit of the categories of finitely presented $A_ i$-modules. Given this, Algebra, Lemma 10.168.1 tells us that category of $R$-flat, finitely presented $A$-modules is the colimit of the categories of $R_ i$-flat, finitely presented $A_ i$-modules. Thus the characterization in Lemma 15.83.4 proves that (1) is true.
To prove (2) we choose $P_0^\bullet $ representing $K_0$ and $F_0^\bullet $ representing $L_0$ as in Lemma 15.83.6. Then $E_0^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (P_0^\bullet , F_0^\bullet )$ satisfies
\[ H^0(E_0^\bullet \otimes _{R_0} R_ i) = \mathop{\mathrm{Hom}}\nolimits _{D(A_ i)}(K_0 \otimes _{A_0}^\mathbf {L} A_ i, L_0 \otimes _{A_0}^\mathbf {L} A_ i) \]
and
\[ H^0(E_0^\bullet \otimes _{R_0} R) = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(K_0 \otimes _{A_0}^\mathbf {L} A, L_0 \otimes _{A_0}^\mathbf {L} A) \]
by the lemma. Thus the result because tensor product commutes with colimits and filtered colimits are exact (Algebra, Lemma 10.8.8). $\square$
Lemma 15.83.8. Let $R' \to A'$ be a flat ring map of finite presentation. Let $R' \to R$ be a surjective ring map whose kernel is a nilpotent ideal. Set $A = A' \otimes _{R'} R$. Let $K' \in D(A')$ and set $K = K' \otimes _{A'}^\mathbf {L} A$ in $D(A)$. If $K$ is $R$-perfect, then $K'$ is $R'$-perfect.
Proof. We can represent $K$ by a bounded above complex of finite free $A$-modules $E^\bullet $, see Lemma 15.64.5. By Lemma 15.75.3 we conclude that $K'$ is pseudo-coherent because it can be represented by a bounded above complex $P^\bullet $ of finite free $A'$-modules with $P^\bullet \otimes _{A'} A = E^\bullet $. Observe that this also means $P^\bullet \otimes _{R'} R = E^\bullet $ (since $A = A' \otimes _{R'} R$).
Let $I = \mathop{\mathrm{Ker}}(R' \to R)$. Then $I^ n = 0$ for some $n$. Choose $[a, b]$ such that $K$ has tor amplitude in $[a, b]$ as a complex of $R$-modules. We will show $K'$ has tor amplitude in $[a, b]$. To do this, let $M'$ be an $R'$-module. If $IM' = 0$, then
\[ K' \otimes _{R'}^\mathbf {L} M' = P^\bullet \otimes _{R'} M' = E^\bullet \otimes _ R M' = K \otimes _ R^\mathbf {L} M' \]
(because $A'$ is flat over $R'$ and $A$ is flat over $R$) which has nonzero cohomology only for degrees in $[a, b]$ by choice of $a, b$. If $I^{t + 1}M' = 0$, then we consider the short exact sequence
\[ 0 \to IM' \to M' \to M'/IM' \to 0 \]
with $M = M'/IM'$. By induction on $t$ we have that both $K' \otimes _{R'}^\mathbf {L} IM'$ and $K' \otimes _{R'}^\mathbf {L} M'/IM'$ have nonzero cohomology only for degrees in $[a, b]$. Then the distinguished triangle
\[ K' \otimes _{R'}^\mathbf {L} IM' \to K' \otimes _{R'}^\mathbf {L} M' \to K' \otimes _{R'}^\mathbf {L} M'/IM' \to (K' \otimes _{R'}^\mathbf {L} IM')[1] \]
proves the same is true for $K' \otimes _{R'}^\mathbf {L} M'$. This proves the desired bound for all $M'$ and hence the desired bound on the tor amplitude of $K'$. $\square$
Lemma 15.83.9. Let $R$ be a ring. Let $A = R[x_1, \ldots , x_ d]/I$ be flat and of finite presentation over $R$. Let $\mathfrak q \subset A$ be a prime ideal lying over $\mathfrak p \subset R$. Let $K \in D(A)$ be pseudo-coherent. Let $a, b \in \mathbf{Z}$. If $H^ i(K_\mathfrak q \otimes _{R_\mathfrak p}^\mathbf {L} \kappa (\mathfrak p))$ is nonzero only for $i \in [a, b]$, then $K_\mathfrak q$ has tor amplitude in $[a - d, b]$ over $R$.
Proof. By Lemma 15.82.8 $K$ is pseudo-coherent as a complex of $R[x_1, \ldots , x_ d]$-modules. Therefore we may assume $A = R[x_1, \ldots , x_ d]$. Applying Lemma 15.77.6 to $R_\mathfrak p \to A_\mathfrak q$ and the complex $K_\mathfrak q$ using our assumption, we find that $K_\mathfrak q$ is perfect in $D(A_\mathfrak q)$ with tor amplitude in $[a - d, b]$. Since $R_\mathfrak p \to A_\mathfrak q$ is flat, we conclude by Lemma 15.66.11. $\square$
Lemma 15.83.10. Let $R \to A$ be a ring map which is flat and of finite presentation. Let $K \in D(A)$ be pseudo-coherent. The following are equivalent
$K$ is $R$-perfect, and
$K$ is bounded below and for every prime ideal $\mathfrak p \subset R$ the object $K \otimes _ R^\mathbf {L} \kappa (\mathfrak p)$ is bounded below.
Proof. Observe that (1) implies (2) as an $R$-perfect complex has bounded tor dimension as a complex of $R$-modules by definition. Let us prove the other implication.
Write $A = R[x_1, \ldots , x_ d]/I$. Denote $L$ in $D(R[x_1, \ldots , x_ d])$ the restriction of $K$. By Lemma 15.82.8 we see that $L$ is pseudo-coherent. Since $L$ and $K$ have the same image in $D(R)$ we see that $L$ is $R$-perfect if and only if $K$ is $R$-perfect. Also $L \otimes _ R^\mathbf {L} \kappa (\mathfrak p)$ and $K \otimes _ R^\mathbf {L} \kappa (\mathfrak p)$ are the same objects of $D(\kappa (\mathfrak p))$. This reduces us to the case $A = R[x_1, \ldots , x_ d]$.
Say $A = R[x_1, \ldots , x_ d]$ and $K$ satisfies (2). Let $\mathfrak q \subset A$ be a prime lying over a prime $\mathfrak p \subset R$. By Lemma 15.77.6 applied to $R_\mathfrak p \to A_\mathfrak q$ and the complex $K_\mathfrak q$ using our assumption, we find that $K_\mathfrak q$ is perfect in $D(A_\mathfrak q)$. Since $K$ is bounded below, we see that $K$ is perfect in $D(A)$ by Lemma 15.77.3. This implies that $K$ is $R$-perfect by Lemma 15.83.3 and the proof is complete. $\square$
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