The Stacks project

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. $\square$

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 15.82.2. 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 15.74.15 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. $\square$

Proof. Follows from Lemmas 47.15.10 and 47.15.9, $\square$

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 15.60.5 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. $\square$

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 15.82.2. 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 15.60.5 it suffices to construct maps

functorial in $K$. We have

The first equality by More on Algebra, Lemma 15.61.2 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. $\square$

Proof. Special case of Lemma 47.24.5 by More on Algebra, Lemma 15.82.4. $\square$

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 15.60.5. $\square$

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 10.36.3). 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. $\square$

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. $\square$

Proof. (The parenthetical statement follows from More on Algebra, Lemma 15.82.4.) 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 15.82.2. 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 15.60.5. $\square$

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 47.15.3. 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 15.73.1 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 15.60.1). The third equality follows from More on Algebra, Lemma 15.73.1. The fourth because $D_ A \circ D_ A = \text{id}$. The final equality by adjointness again. Thus the result holds by the Yoneda lemma. $\square$