Lemma 47.24.1. Let $\varphi : R \to A$ be a finite type homomorphism of Noetherian rings. The functor $\varphi ^!$ is well defined up to isomorphism.

## 47.24 Upper shriek algebraically

For a finite type homomorphism $R \to A$ of Noetherian rings we will construct a functor $\varphi ^! : D(R) \to D(A)$ well defined up to nonunique isomorphism which as we will see in Duality for Schemes, Remark 48.17.4 agrees up to isomorphism with the upper shriek functors one encounters in the duality theory for schemes. To motivate the construction we mention two additional properties:

$\varphi ^!$ sends a dualizing complex for $R$ (if it exists) to a dualizing complex for $A$, and

$\omega _{A/R}^\bullet = \varphi ^!(R)$ is a kind of relative dualizing complex: it lies in $D^ b_{\textit{Coh}}(A)$ and restricts to a dualizing complex on the fibres provided $R \to A$ is flat.

These statemens are Lemmas 47.24.3 and 47.25.2.

Let $\varphi : R \to A$ be a finite type homomorphism of Noetherian rings. We will define a functor $\varphi ^! : D(R) \to D(A)$ in the following way

If $\varphi : R \to A$ is surjective we set $\varphi ^!(K) = R\mathop{\mathrm{Hom}}\nolimits (A, K)$. Here we use the functor $R\mathop{\mathrm{Hom}}\nolimits (A, -) : D(R) \to D(A)$ of Section 47.13, and

in general we choose a surjection $\psi : P \to A$ with $P = R[x_1, \ldots , x_ n]$ and we set $\varphi ^!(K) = \psi ^!(K \otimes _ R^\mathbf {L} P)[n]$. Here we use the functor $- \otimes _ R^\mathbf {L} P : D(R) \to D(P)$ of More on Algebra, Section 15.58.

Note the shift $[n]$ by the number of variables in the polynomial ring. This construction is **not** canonical and the functor $\varphi ^!$ will only be well defined up to a (nonunique) isomorphism of functors^{1}.

**Proof.**
Suppose that $\psi _1 : P_1 = R[x_1, \ldots , x_ n] \to A$ and $\psi _2 : P_2 = R[y_1, \ldots , y_ m] \to A$ are two surjections from polynomial rings onto $A$. Then we get a commutative diagram

where $f_ j$ and $g_ i$ are chosen such that $\psi _1(f_ j) = \psi _2(y_ j)$ and $\psi _2(g_ i) = \psi _1(x_ i)$. By symmetry it suffices to prove the functors defined using $P \to A$ and $P[y_1, \ldots , y_ m] \to A$ are isomorphic. By induction we may assume $m = 1$. This reduces us to the case discussed in the next paragraph.

Here $\psi : P \to A$ is given and $\chi : P[y] \to A$ induces $\psi $ on $P$. Write $Q = P[y]$. Choose $g \in P$ with $\psi (g) = \chi (y)$. Denote $\pi : Q \to P$ the $P$-algebra map with $\pi (y) = g$. Then $\chi = \psi \circ \pi $ and hence $\chi ^! = \psi ^! \circ \pi ^!$ as both are adjoint to the restriction functor $D(A) \to D(Q)$ by the material in Section 47.13. Thus

Hence it suffices to show that $\pi ^!(K \otimes _ R^\mathbf {L} Q[1]) = K \otimes _ R^\mathbf {L} P$ Thus it suffices to show that the functor $\pi ^!(-) : D(Q) \to D(P)$ is isomorphic to $K \mapsto K \otimes _ Q^\mathbf {L} P[-1]$. This follows from Lemma 47.13.10. $\square$

Lemma 47.24.2. Let $\varphi : R \to A$ be a finite type homomorphism of Noetherian rings.

$\varphi ^!$ maps $D^+(R)$ into $D^+(A)$ and $D^+_{\textit{Coh}}(R)$ into $D^+_{\textit{Coh}}(A)$.

if $\varphi $ is perfect, then $\varphi ^!$ maps $D^-(R)$ into $D^-(A)$, $D^-_{\textit{Coh}}(R)$ into $D^-_{\textit{Coh}}(A)$, and $D^ b_{\textit{Coh}}(R)$ into $D^ b_{\textit{Coh}}(A)$.

**Proof.**
Choose a factorization $R \to P \to A$ as in the definition of $\varphi ^!$. The functor $- \otimes _ R^\mathbf {L} : D(R) \to D(P)$ preserves the subcategories $D^+, D^+_{\textit{Coh}}, D^-, D^-_{\textit{Coh}}, D^ b_{\textit{Coh}}$. The functor $R\mathop{\mathrm{Hom}}\nolimits (A, -) : D(P) \to D(A)$ preserves $D^+$ and $D^+_{\textit{Coh}}$ by Lemma 47.13.4. If $R \to A$ is perfect, then $A$ is perfect as a $P$-module, see More on Algebra, Lemma 15.77.2. Recall that the restriction of $R\mathop{\mathrm{Hom}}\nolimits (A, K)$ to $D(P)$ is $R\mathop{\mathrm{Hom}}\nolimits _ P(A, K)$. By More on Algebra, Lemma 15.70.14 we have $R\mathop{\mathrm{Hom}}\nolimits _ P(A, K) = E \otimes _ P^\mathbf {L} K$ for some perfect $E \in D(P)$. Since we can represent $E$ by a finite complex of finite projective $P$-modules it is clear that $R\mathop{\mathrm{Hom}}\nolimits _ P(A, K)$ is in $D^-(P), D^-_{\textit{Coh}}(P), D^ b_{\textit{Coh}}(P)$ as soon as $K$ is. Since the restriction functor $D(A) \to D(P)$ reflects these subcategories, the proof is complete.
$\square$

Lemma 47.24.3. Let $\varphi $ be a finite type homomorphism of Noetherian rings. If $\omega _ R^\bullet $ is a dualizing complex for $R$, then $\varphi ^!(\omega _ R^\bullet )$ is a dualizing complex for $A$.

Lemma 47.24.4. Let $R \to R'$ be a flat homomorphism of Noetherian rings. Let $\varphi : R \to A$ be a finite type ring map. Let $\varphi ' : R' \to A' = A \otimes _ R R'$ be the map induced by $\varphi $. Then we have a functorial maps

for $K$ in $D(R)$ which are isomorphisms for $K \in D^+(R)$.

**Proof.**
Choose a factorization $R \to P \to A$ where $P$ is a polynomial ring over $R$. This gives a corresponding factorization $R' \to P' \to A'$ by base change. Since we have $(K \otimes _ R^\mathbf {L} P) \otimes _ P^\mathbf {L} P' = (K \otimes _ R^\mathbf {L} R') \otimes _{R'}^\mathbf {L} P'$ by More on Algebra, Lemma 15.58.4 it suffices to construct maps

functorial in $K$. For this we use the map (47.14.0.1) constructed in Section 47.14 for $P, A, P', A'$. The map is an isomorphism for $K \in D^+(R)$ by Lemma 47.14.2. $\square$

Lemma 47.24.5. Let $R \to R'$ be a homomorphism of Noetherian rings. Let $\varphi : R \to A$ be a perfect ring map (More on Algebra, Definition 15.77.1) such that $R'$ and $A$ are tor independent over $R$. Let $\varphi ' : R' \to A' = A \otimes _ R R'$ be the map induced by $\varphi $. Then we have a functorial isomorphism

for $K$ in $D(R)$.

**Proof.**
We may choose a factorization $R \to P \to A$ where $P$ is a polynomial ring over $R$ such that $A$ is a perfect $P$-module, see More on Algebra, Lemma 15.77.2. This gives a corresponding factorization $R' \to P' \to A'$ by base change. Since we have $(K \otimes _ R^\mathbf {L} P) \otimes _ P^\mathbf {L} P' = (K \otimes _ R^\mathbf {L} R') \otimes _{R'}^\mathbf {L} P'$ by More on Algebra, Lemma 15.58.4 it suffices to construct maps

functorial in $K$. We have

The first equality by More on Algebra, Lemma 15.59.2 applied to $R, R', P, P'$. The second equality because $A$ and $R'$ are tor independent over $R$. Hence $A$ and $P'$ are tor independent over $P$ and we can use the map (47.14.0.1) constructed in Section 47.14 for $P, A, P', A'$ get the desired arrow. By Lemma 47.14.3 to finish the proof it suffices to prove that $A$ is a perfect $P$-module which we saw above. $\square$

Lemma 47.24.6. Let $R \to R'$ be a homomorphism of Noetherian rings. Let $\varphi : R \to A$ be flat of finite type. Let $\varphi ' : R' \to A' = A \otimes _ R R'$ be the map induced by $\varphi $. Then we have a functorial isomorphism

for $K$ in $D(R)$.

**Proof.**
Special case of Lemma 47.24.5 by More on Algebra, Lemma 15.77.4.
$\square$

Lemma 47.24.7. Let $A \xrightarrow {a} B \xrightarrow {b} C$ be finite type homomorphisms of Noetherian rings. Then there is a transformation of functors $b^! \circ a^! \to (b \circ a)^!$ which is an isomorphism on $D^+(A)$.

**Proof.**
Choose a polynomial ring $P = A[x_1, \ldots , x_ n]$ over $A$ and a surjection $P \to B$. Choose elements $c_1, \ldots , c_ m \in C$ generating $C$ over $B$. Set $Q = P[y_1, \ldots , y_ m]$ and denote $Q' = Q \otimes _ P B = B[y_1, \ldots , y_ m]$. Let $\chi : Q' \to C$ be the surjection sending $y_ j$ to $c_ j$. Picture

By Lemma 47.14.2 for $M \in D(P)$ we have an arrow $\psi ^!(M) \otimes _ B^\mathbf {L} Q' \to (\psi ')^!(M \otimes _ P^\mathbf {L} Q)$ which is an isomorphism whenever $M$ is bounded below. Also we have $\chi ^! \circ (\psi ')^! = (\chi \circ \psi ')^!$ as both functors are adjoint to the restriction functor $D(C) \to D(Q)$ by Section 47.13. Then we see

where we have used in addition to the above More on Algebra, Lemma 15.58.4. $\square$

Lemma 47.24.8. Let $\varphi : R \to A$ be a finite map of Noetherian rings. Then $\varphi ^!$ is isomorphic to the functor $R\mathop{\mathrm{Hom}}\nolimits (A, -) : D(R) \to D(A)$ from Section 47.13.

**Proof.**
Suppose that $A$ is generated by $n > 1$ elements over $R$. Then can factor $R \to A$ as a composition of two finite ring maps where in both steps the number of generators is $< n$. Since we have Lemma 47.24.7 and Lemma 47.13.2 we conclude that it suffices to prove the lemma when $A$ is generated by one element over $R$. Since $A$ is finite over $R$, it follows that $A$ is a quotient of $B = R[x]/(f)$ where $f$ is a monic polynomial in $x$ (Algebra, Lemma 10.35.3). Again using the lemmas on composition and the fact that we have agreement for surjections by definition, we conclude that it suffices to prove the lemma for $R \to B = R[x]/(f)$. In this case, the functor $\varphi ^!$ is isomorphic to $K \mapsto K \otimes _ R^\mathbf {L} B$; you prove this by using Lemma 47.13.10 for the map $R[x] \to B$ (note that the shift in the definition of $\varphi ^!$ and in the lemma add up to zero). For the functor $R\mathop{\mathrm{Hom}}\nolimits (B, -) : D(R) \to D(B)$ we can use Lemma 47.13.9 to see that it suffices to show $\mathop{\mathrm{Hom}}\nolimits _ R(B, R) \cong B$ as $B$-modules. Suppose that $f$ has degree $d$. Then an $R$-basis for $B$ is given by $1, x, \ldots , x^{d - 1}$. Let $\delta _ i : B \to R$, $i = 0, \ldots , d - 1$ be the $R$-linear map which picks off the coefficient of $x^ i$ with respect to the given basis. Then $\delta _0, \ldots , \delta _{d - 1}$ is a basis for $\mathop{\mathrm{Hom}}\nolimits _ R(B, R)$. Finally, for $0 \leq i \leq d - 1$ a computation shows that

for some $c_1, \ldots , c_ d \in R$^{2}. Hence $\mathop{\mathrm{Hom}}\nolimits _ R(B, R)$ is a principal $B$-module with generator $\delta _{d - 1}$. By looking at ranks we conclude that it is a rank $1$ free $B$-module.
$\square$

Lemma 47.24.9. Let $R$ be a Noetherian ring and let $f \in R$. If $\varphi $ denotes the map $R \to R_ f$, then $\varphi ^!$ is isomorphic to $- \otimes _ R^\mathbf {L} R_ f$. More generally, if $\varphi : R \to R'$ is a map such that $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is an open immersion, then $\varphi ^!$ is isomorphic to $- \otimes _ R^\mathbf {L} R'$.

**Proof.**
Choose the presentation $R \to R[x] \to R[x]/(fx - 1) = R_ f$ and observe that $fx - 1$ is a nonzerodivisor in $R[x]$. Thus we can apply using Lemma 47.13.10 to compute the functor $\varphi ^!$. Details omitted; note that the shift in the definition of $\varphi ^!$ and in the lemma add up to zero.

In the general case note that $R' \otimes _ R R' = R'$. Hence the result follows from the base change results above. Either Lemma 47.24.4 or Lemma 47.24.5 will do. $\square$

Lemma 47.24.10. Let $\varphi : R \to A$ be a perfect homomorphism of Noetherian rings (for example $\varphi $ is flat of finite type). Then $\varphi ^!(K) = K \otimes _ R^\mathbf {L} \varphi ^!(R)$ for $K \in D(R)$.

**Proof.**
(The parenthetical statement follows from More on Algebra, Lemma 15.77.4.) We can choose a factorization $R \to P \to A$ where $P$ is a polynomial ring in $n$ variables over $R$ and then $A$ is a perfect $P$-module, see More on Algebra, Lemma 15.77.2. Recall that $\varphi ^!(K) = R\mathop{\mathrm{Hom}}\nolimits (A, K \otimes _ R^\mathbf {L} P[n])$. Thus the result follows from Lemma 47.13.9 and More on Algebra, Lemma 15.58.4.
$\square$

Lemma 47.24.11. Let $\varphi : A \to B$ be a finite type homomorphism of Noetherian rings. Let $\omega _ A^\bullet $ be a dualizing complex for $A$. Set $\omega _ B^\bullet = \varphi ^!(\omega _ A^\bullet )$. Denote $D_ A(K) = R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet )$ for $K \in D_{\textit{Coh}}(A)$ and $D_ B(L) = R\mathop{\mathrm{Hom}}\nolimits _ B(L, \omega _ B^\bullet )$ for $L \in D_{\textit{Coh}}(B)$. Then there is a functorial isomorphism

for $K \in D_{\textit{Coh}}(A)$.

**Proof.**
Observe that $\omega _ B^\bullet $ is a dualizing complex for $B$ by Lemma 47.24.3. Let $A \to B \to C$ be finite type homomorphisms of Noetherian rings. If the lemma holds for $A \to B$ and $B \to C$, then the lemma holds for $A \to C$. This follows from Lemma 47.24.7 and the fact that $D_ B \circ D_ B \cong \text{id}$ by Lemma 47.15.2. Thus it suffices to prove the lemma in case $A \to B$ is a surjection and in the case where $B$ is a polynomial ring over $A$.

Assume $B = A[x_1, \ldots , x_ n]$. Since $D_ A \circ D_ A \cong \text{id}$, it suffices to prove $D_ B(K \otimes _ A B) \cong D_ A(K) \otimes _ A B[n]$ for $K$ in $D_{\textit{Coh}}(A)$. Choose a bounded complex $I^\bullet $ of injectives representing $\omega _ A^\bullet $. Choose a quasi-isomorphism $I^\bullet \otimes _ A B \to J^\bullet $ where $J^\bullet $ is a bounded complex of $B$-modules. Given a complex $K^\bullet $ of $A$-modules, consider the obvious map of complexes

The left hand side represents $D_ A(K) \otimes _ A B[n]$ and the right hand side represents $D_ B(K \otimes _ A B)$. Thus it suffices to prove this map is a quasi-isomorphism if the cohomology modules of $K^\bullet $ are finite $A$-modules. Observe that the cohomology of the complex in degree $r$ (on either side) only depends on finitely many of the $K^ i$. Thus we may replace $K^\bullet $ by a truncation, i.e., we may assume $K^\bullet $ represents an object of $D^-_{\textit{Coh}}(A)$. Then $K^\bullet $ is quasi-isomorphic to a bounded above complex of finite free $A$-modules. Therefore we may assume $K^\bullet $ is a bounded above complex of finite free $A$-modules. In this case it is easy to that the displayed map is an isomorphism of complexes which finishes the proof in this case.

Assume that $A \to B$ is surjective. Denote $i_* : D(B) \to D(A)$ the restriction functor and recall that $\varphi ^!(-) = R\mathop{\mathrm{Hom}}\nolimits (A, -)$ is a right adjoint to $i_*$ (Lemma 47.13.1). For $F \in D(B)$ we have

The first equality follows from More on Algebra, Lemma 15.69.1 and the definition of $D_ B$. The second equality by the adjointness mentioned above and the equality $i_*((D_ A(K) \otimes _ A^\mathbf {L} B) \otimes _ B^\mathbf {L} F) = D_ A(K) \otimes _ A^\mathbf {L} i_*F$ (More on Algebra, Lemma 15.58.1). The third equality follows from More on Algebra, Lemma 15.69.1. The fourth because $D_ A \circ D_ A = \text{id}$. The final equality by adjointness again. Thus the result holds by the Yoneda lemma. $\square$

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