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.

**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. $\square$

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## Comments (2)

Comment #3474 by Eamon Quinlan on

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