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Let $A$ be a commutative noetherian ring, $I\subseteq A$ an ideal, $M_\alpha$ be $A$-modules, $\forall\alpha\in J$. It is easily seen that the $I$-torsion commutes with direct sums: $$\Gamma_I(\bigoplus_{\alpha\in J}M_\alpha)=\bigoplus_{\alpha\in J}\Gamma_I(M_\alpha).$$ This is because, those elements in the direct sum annihilated by a power of $I$ also have each of its components annihilated by the same power of $I$, and conversely we can annihilate the direct sum of these components by a large enough power of $I$, too.

Since the local cohomology $H_I^n$ is defined as the right derived functors of $\Gamma_I$, I am wondering whether we can similarly show $$H_I^n(\bigoplus_{\alpha\in J}M_\alpha)\cong \bigoplus_{\alpha\in J}H_I^n(M_\alpha).$$

I have seen some proofs of a more general result about local cohomology commuting with direct limits, but I am looking for a straight-forward proof here.

Thank you very much for your help!

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2 Answers 2

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Yes because homology commutes with direct sums. Alternatively you could use the formulation $$H_{\mathfrak{a}}^{n}(-)\simeq \varinjlim_{t}\text{Ext}_{R}^{n}(R/\mathfrak{a}^{t},-)$$ combined with the fact that $R/\mathfrak{a}^{t}$ is finitely generated to show that local cohomology commutes with all direct limits; in particular it will commute with direct sums.

Edit:

Since $R/\mathfrak{a}^{t}$ is finitely generated, there are isomorphisms $$\text{Ext}_{R}^{n}(R/\mathfrak{a}^{t},\varinjlim_{J}N_{j})\simeq \varinjlim_{J}\text{Ext}_{R}^{n}(R/\mathfrak{a}^{t},N_{j})$$ for any directed system $\{N_{j}\}_{J}$ of modules and $n\geq 0$. Consequently one has isomorphisms $$\begin{align*} H_{\mathfrak{a}}^{n}(\varinjlim_{J}N_{j})&\simeq \varinjlim_{t}\text{Ext}_{R}^{n}(R/\mathfrak{a}^{t},\varinjlim_{J}N_{j}) \\ &\simeq \varinjlim_{t} \varinjlim_{J}\text{Ext}_{R}^{n}(R/\mathfrak{a}^{t},N_{j}) \\ &\simeq \varinjlim_{J} \varinjlim_{t} \text{Ext}_{R}^{n}(R/\mathfrak{a}^{t},N_{j}) \\ &\simeq \varinjlim_{J} H_{\mathfrak{a}}^{n}(N_{j}) \end{align*}$$ for every directed system and $n\geq 0$.

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  • $\begingroup$ Ext only commutes with arbitrary direct sums in the first component, right? Ext commutes with finite direct sums in both components. $\endgroup$ Jul 22, 2020 at 17:53
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    $\begingroup$ @Carlo you have $\text{Ext}^{n}(M,-)$ commuting with arbitrary direct limits in the second argument whenever $M$ has a projective resolution with the first $n+1$ terms finitely generated. $\endgroup$
    – Zeek
    Jul 22, 2020 at 18:17
  • $\begingroup$ I am not aware of this fact. Is it obvious? Or do you have a source? $\endgroup$ Jul 22, 2020 at 18:22
  • $\begingroup$ Thanks for the answer. This really follows from the fact that homology commutes with direct sums. $\endgroup$
    – user771160
    Jul 22, 2020 at 18:39
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    $\begingroup$ @Carlo a proof for the most general rings is Lemma 6.6 in Goebel and Trlifaj's book. $\endgroup$
    – Zeek
    Jul 22, 2020 at 18:42
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Convince yourself first that $H_I^n(M) = \varinjlim_k \operatorname{Ext}_R^n(R / I^k, M);$ then, use the fact that Ext commutes with finite direct sums in the second component, i.e., $$\operatorname{Ext}_R^n(R / I^k, \oplus_{i = 1}^m M_i) \cong \oplus_{i = 1}^m \operatorname{Ext}_R^n(R / I^k, M_i).$$

For the first fact, use the definition of the local cohomology modules as the right-derived functors of $\Gamma_I(M).$ Convince yourself that $\Gamma_I(M) \cong \varinjlim_k \operatorname{Hom}_R(R / I^k, M);$ then, use the facts that (1.) direct limits commute with cohomology and (2.) Ext is the right-derived functor of Hom.

Unfortunately, I am not aware of a more straightforward proof than this.

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  • $\begingroup$ Thank you very much for the answer! $\endgroup$
    – user771160
    Jul 22, 2020 at 18:39
  • $\begingroup$ The straightforward proof is also possible because the ring is noetherian. $\endgroup$
    – user771160
    Jul 22, 2020 at 18:47

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