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 10.36.3). 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

$x^ i \delta _{d - 1} = \delta _{d - 1 - i} + b_1 \delta _{d - i} + \ldots + b_ i \delta _{d - 1}$

for some $c_1, \ldots , c_ d \in R$1. 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. $\square$

[1] If $f = x^ d + a_1 x^{d - 1} + \ldots + a_ d$, then $c_1 = -a_1$, $c_2 = a_1^2 - a_2$, $c_3 = -a_1^3 + 2a_1a_2 -a_3$, etc.

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