Deprecated: $wgMWOAuthSharedUserIDs=false is deprecated, set $wgMWOAuthSharedUserIDs=true, $wgMWOAuthSharedUserSource='local' instead [Called from MediaWiki\HookContainer\HookContainer::run in /var/www/html/w/includes/HookContainer/HookContainer.php at line 135] in /var/www/html/w/includes/Debug/MWDebug.php on line 372
On the first and second problems of Hartshorne on cofiniteness - MaRDI portal

On the first and second problems of Hartshorne on cofiniteness (Q6600713)

From MaRDI portal





scientific article; zbMATH DE number 7909493
Language Label Description Also known as
English
On the first and second problems of Hartshorne on cofiniteness
scientific article; zbMATH DE number 7909493

    Statements

    On the first and second problems of Hartshorne on cofiniteness (English)
    0 references
    0 references
    10 September 2024
    0 references
    Let \(R\) be a commutative noetherian ring, \(\mathfrak{a}\) an ideal of \(R\), and \(M\) an \(R\)-module. For any \(i \geq 0\), the \(i\)th local cohomology module of \(M\) with respect to \(\mathfrak{a}\) is given by \N\[\NH^{i}_{\mathfrak{a}}(M) \cong \underset{n\geq 1}\varinjlim \operatorname{Ext}^{i}_{R}\left(R/ \mathfrak{a}^{n},M\right).\N\]\NHartshorne defines an \(R\)-module \(M\) to be \(\mathfrak{a}\)-cofinite if \(\operatorname{Supp}_{R}(M)\subseteq \operatorname{Var}(\mathfrak{a})\) and \(\operatorname{Ext}^{i}_{R}\left(R/ \mathfrak{a},M\right)\) is a finitely generated \(R\)-module for every \(i\geq 0\). Then he asks the following questions:\N\N\begin{itemize}\N\item For which ideal \(\mathfrak{a}\) of \(R\), is the local cohomology module \(H^{i}_{\mathfrak{a}}(M)\) \(\mathfrak{a}\)-cofinite for every finitely generated \(R\)-module \(M\) and every \(i\geq 0\)?\N\item For which ideal \(\mathfrak{a}\) of \(R\), is the category of \(\mathfrak{a}\)-cofinite \(R\)-modules abelian?\N\end{itemize}\N\NThe author of this paper shows that if \(\mathfrak{a}\) is an ideal of \(R\) which satisfies the condition of the first question, then it also satisfies the condition of the second question. He further provides an example to show that the converse does not hold in general.
    0 references
    0 references
    0 references
    0 references
    0 references

    Identifiers