Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Set $Z = V(I) \subset \mathop{\mathrm{Spec}}(A)$. Recall that (47.8.3.1) is the functor
In fact, there is a natural transformation of functors
Namely, given a complex of $A$-modules $K^\bullet $ the canonical map $R\Gamma _ I(K^\bullet ) \to K^\bullet $ in $D(A)$ factors (uniquely) through $R\Gamma _ Z(K^\bullet )$ as $R\Gamma _ I(K^\bullet )$ has $I$-power torsion cohomology modules (see Lemma 47.8.1). In general this map is not an isomorphism (we've seen this in Lemma 47.8.4).
Lemma 47.10.1. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal.
the adjunction $R\Gamma _ I(K) \to K$ is an isomorphism for $K \in D_{I^\infty \text{-torsion}}(A)$,
the functor (47.8.3.1) $D(I^\infty \text{-torsion}) \to D_{I^\infty \text{-torsion}}(A)$ is an equivalence,
the transformation of functors (47.10.0.1) is an isomorphism, in other words $R\Gamma _ I(K) = R\Gamma _ Z(K)$ for $K \in D(A)$.
Proof. A formal argument, which we omit, shows that it suffices to prove (1).
Let $M$ be an $I$-power torsion $A$-module. Choose an embedding $M \to J$ into an injective $A$-module. Then $J[I^\infty ]$ is an injective $A$-module, see Lemma 47.3.9, and we obtain an embedding $M \to J[I^\infty ]$. Thus every $I$-power torsion module has an injective resolution $M \to J^\bullet $ with $J^ n$ also $I$-power torsion. It follows that $R\Gamma _ I(M) = M$ (this is not a triviality and this is not true in general if $A$ is not Noetherian). Next, suppose that $K \in D_{I^\infty \text{-torsion}}^+(A)$. Then the spectral sequence
(Derived Categories, Lemma 13.21.3) converges and above we have seen that only the terms with $q = 0$ are nonzero. Thus we see that $R\Gamma _ I(K) \to K$ is an isomorphism.
Suppose $K$ is an arbitrary object of $D_{I^\infty \text{-torsion}}(A)$. We have
by Lemma 47.8.2. Choose $f_1, \ldots , f_ r \in A$ generating $I$. Let $K_ n^\bullet = K(A, f_1^ n, \ldots , f_ r^ n)$ be the Koszul complex with terms in degrees $-r, \ldots , 0$. Since the pro-objects $\{ A/I^ n\} $ and $\{ K_ n^\bullet \} $ in $D(A)$ are the same by More on Algebra, Lemma 15.94.1, we see that
Pick any complex $K^\bullet $ of $A$-modules representing $K$. Since $K_ n^\bullet $ is a finite complex of finite free modules we see that
where $(K_ n^\bullet )^\vee $ is the dual of the complex $K_ n^\bullet $. See More on Algebra, Lemma 15.73.2. As $(K_ n^\bullet )^\vee $ is a complex of finite free $A$-modules sitting in degrees $0, \ldots , r$ we see that the terms of the complex $\text{Tot}((K_ n^\bullet )^\vee \otimes _ A K^\bullet )$ are the same as the terms of the complex $\text{Tot}((K_ n^\bullet )^\vee \otimes _ A \tau _{\geq q - r - 2} K^\bullet )$ in degrees $q - 1$ and higher. Hence we see that
for all $n$. It follows that
Thus we see that the map $R\Gamma _ I(K) \to K$ is an isomorphism. $\square$
Lemma 47.10.2. Let $A$ be a Noetherian ring and let $I = (f_1, \ldots , f_ r)$ be an ideal of $A$. Set $Z = V(I) \subset \mathop{\mathrm{Spec}}(A)$. There are canonical isomorphisms in $D(A)$. If $M$ is an $A$-module, then we have similarly in $D(A)$.
\[ R\Gamma _ I(A) \to (A \to \prod \nolimits _{i_0} A_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_ r}) \to R\Gamma _ Z(A) \]
\[ R\Gamma _ I(M) \cong (M \to \prod \nolimits _{i_0} M_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} M_{f_{i_0}f_{i_1}} \to \ldots \to M_{f_1\ldots f_ r}) \cong R\Gamma _ Z(M) \]
Proof. This follows from Lemma 47.10.1 and the computation of the functor $R\Gamma _ Z$ in Lemma 47.9.1. $\square$
Lemma 47.10.3. If $A \to B$ is a homomorphism of Noetherian rings and $I \subset A$ is an ideal, then in $D(B)$ we have where $Y = V(IB) \subset \mathop{\mathrm{Spec}}(B)$.
\[ R\Gamma _ I(A) \otimes _ A^\mathbf {L} B = R\Gamma _ Z(A) \otimes _ A^\mathbf {L} B = R\Gamma _ Y(B) = R\Gamma _{IB}(B) \]
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)