We use cookies to distinguish you from other users and to provide you with a better experience on our websites. Close this message to accept cookies or find out how to manage your cookie settings.
We use cookies to distinguish you from other users and to provide you with a better experience on our websites. Close this message to accept cookies or find out how to manage your cookie settings.
Let 
$R$ be a commutative Noetherian ring and 
$\mathfrak{a}$ a proper ideal of $R$. We show that if 
$n\,:=\,\text{grad}{{\text{e}}_{R}}\,\mathfrak{a}$, then 
$\text{En}{{\text{d}}_{R}}(H_{\mathfrak{a}}^{n}(R))\,\cong \,\text{Ext}_{R}^{n}(H_{\mathfrak{a}}^{n}(R),\,R)$. We also prove that, for a nonnegative integer 
$n$ such that 
$H_{\mathfrak{a}}^{i}(R)\,=\,0$ for every 
$i\,\ne \,n$, if 
$\text{Ext}_{R}^{i}({{R}_{z}},\,R)\,=\,0$ for all 
$i\,>\,0$ and 
$z\,\in \,\mathfrak{a}$, then 
$\text{En}{{\text{d}}_{R}}(H_{\mathfrak{a}}^{n}(R))$ is a homomorphic image of $R$, where 
${{R}_{z}}$ is the ring of fractions of $R$ with respect to a multiplicatively closed subset 
$\{{{z}^{j}}|j\,\ge \text{0}\}$ of $R$. Moreover, if 
$\text{Ho}{{\text{m}}_{R}}({{R}_{z}},R)=0$ for all $z\,\in \,\mathfrak{a}$, then 
${{\mu }_{H_{\mathfrak{a}}^{n}(R)}}$ is an isomorphism, where ${{\mu }_{H_{\mathfrak{a}}^{n}(R)}}$ is the canonical ring homomorphism 
$R\,\to \,\text{En}{{\text{d}}_{R}}(H_{\mathfrak{a}}^{n}(R))$.
