http://dx.doi.org/10.4153/CMB-2010-072-1
Canad. Math. Bull. 53(2010), 667-673
Published:2010-07-26 Printed: Dec 2010
Kazem Khashyarmanesh, Ferdowsi University of Mashhad, Department of Mathematics, Mashhad, Iran
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Abstract
Let $R$ be a commutative Noetherian ring and $\mathfrak{a}$ a proper ideal
of $R$. We show that if $n:=\operatorname{grade}_R\mathfrak{a}$, then
$\operatorname{End}_R(H^n_\mathfrak{a}(R))\cong \operatorname{Ext}_R^n(H^n_\mathfrak{a}(R),R)$. We also
prove that, for a nonnegative integer $n$ such that
$H^i_\mathfrak{a}(R)=0$ for every $i\neq n$, if $\operatorname{Ext}_R^i(R_z,R)=0$ for
all $i >0$ and $z \in \mathfrak{a}$, then
$\operatorname{End}_R(H^n_\mathfrak{a}(R))$ is a homomorphic
image of $R$, where $R_z$ is the ring of fractions of $R$ with
respect to a multiplicatively closed subset $\{z^j \mid j \geqslant
0 \}$ of $R$. Moreover, if $\operatorname{Hom}_R(R_z,R)=0$ for all $z
\in \mathfrak{a}$,
then $\mu_{H^n_\mathfrak{a}(R)}$ is an isomorphism, where $\mu_{H^n_\mathfrak{a}(R)}$
is the canonical ring homomorphism $R \rightarrow \operatorname{End}_R(H^n_\mathfrak{a}(R))$.
© Canadian Mathematical Society, 2013
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