Let $S$ be the polynomial ring $k[x_0,\ldots,x_n]$, $x$ one of the variables $x_i$, $I\subseteq S$ a homogeneous ideal which has a generating set $f_1,\ldots,f_r$ where $\deg_x f_i=0$ for all $i$.

From the short exact sequence $$0\to S/I(-1)\xrightarrow{f} S/I \to S/(x,I)\to 0$$ where the first map $f$ is multiplication with x and the second sends $s+I$ to $s+(x,I)$, I get the long exact sequence $$0\to Hom(S/(x,I),S)\to Hom(S/I,S)\to Hom(S/I(-1),S)\to\ldots$$ $$\to Ext^{m-1}(S/(x,I),S)\to Ext^{m-1}(S/I,S)\xrightarrow{f^*} Ext^{m-1}(S/I(-1),S)\to Ext^m(S/(x,I),S)\to\ldots$$

My aim is to show that $$Ext^m(S/(x,I),S)=\frac{Ext^{m-1}(S/I,S)}{(x)Ext^{m-1}(S/I,S)},$$ so I thought to get there by showing that the induced map $f^\star$ in the above sequence is injective or equivalently: if $J^\bullet$ is an injective resolution of $S$ with boundaries $d^m$ and $\varphi\in Hom(S/I,J^m)$ with $d^m_\star(\varphi)=0$ and $f^\star(\varphi)\in d^{m-1}_\star(Hom(S/I(-1),J^{m-1})),$ then is $\varphi\in d^{m-1}_\star(Hom(S/I,J^{m-1}))$.

Is it true that $f^\star$ on the Ext-modules is an injection? And if yes, how can i show that? Thanks for any help!

  • $\begingroup$ Regarding your first paragraph: if all the $f_i$ have degree zero with respect to all variables, they must be constant. $\endgroup$ Feb 24, 2010 at 13:48
  • $\begingroup$ I mean $x=x_i$ for $\textit{one}$ of the $i$. $\endgroup$
    – Ida B.
    Feb 24, 2010 at 13:50

1 Answer 1


Yes, this is true. Write $R = k[x_0,\ldots,\hat x_i,\ldots,x_n]$ so that $S = R[x_i]$ and $S/I = S \otimes_R R/J$ for an ideal $J$ of $R$. Then there are natural isomorphisms of Ext-groups $$ Ext_S^m(S/I, S) \cong Ext_R^m(R/J,S) \cong S \otimes_R Ext_R^m(R/J,R). $$ The first isomorphism is obtained by applying the natural isomorphism $Hom_S(S \otimes_R -, S) \cong Hom_R(-,S)$ to a resolution of $R/J$. The second is because there is always a map $S \otimes_R Hom_R(-,R) \to Hom_R(-,S)$ that is an isomorphism for finitely generated modules. $R/J$ has a resolution by finitely generated $R$-modules because $R$ is Noetherian, and $S$ is a direct sum of copies of $R$; $Hom$ out of finitely generated modules preserves direct sums.

With respect to this isomorphism, the map $f^*$ is actually given by multiplication by $x_i$ on the $S$ factor, and this is injective.

  • 1
    $\begingroup$ The key isomorphism goes all the way back to Cartan-Eilenberg,Ch. IX, §3, Theorem 1. $\endgroup$ Feb 24, 2010 at 15:12

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