# In what degrees does Ext(S/(f),S) vanish?

Let $S=k[x_0,...,x_n]$ be the polynomial ring over a field $k$ and $f\in S$ non-zero and homogeneous. Is it true that $Ext^m(S/(f),S)$ is zero?

This would help me to show that $Ext^m(S/fI,S)\cong Ext^m(S/I,S)(\deg f)$ for $m\geq 2$ and a homogeneous ideal $I$ of codim $\geq 2$. I tried the following approach: Applying the long exact sequence of $Ext$ to the exact sequence of graded $S$-modules $$0\to S/I\xrightarrow{\cdot f} S/fI(\deg f)\xrightarrow{\tau}S/(f)(\deg f)\to 0,$$ where $\tau$ is the canonical morphism $s+fI\mapsto s+(f)$, brings $$\ldots\to Ext^m(S/(f)(\deg f),S)\to Ext^m(S/fI(\deg f),S)\to Ext^m(S/I,S)\to$$
$$Ext^{m+1}(S/(f)(\deg f),S)\to Ext^{m+1}(S/fI(\deg f),S)\to Ext^{m+1}(S/I,S)\to\ldots$$
and two zeros on the left would suffice.

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It is not true, $Ext^1(S/(f), S)\neq 0$ as Ben pointed out. However, to prove what you want $Ext^m(S/I,S)\cong Ext^m(S/fI,S)(deg(f))$ for $m\geq 2$, just note that $$Ext^m(S/I,S) = Ext^{m-1}(I,S)$$ for any $I$, any $m\geq 2$ (using $0 \to I \to S\to S/I \to 0$) and $I(-deg(f)) \cong fI$ as $S$-modules.
Thanks for the answer! Just a few remarks, the $\deg f$ hat to be replaced by a $-\deg f$ and i think that the first sequence is in fact exact: $$S/I\to S/fI, s+I\mapsto fs+fI,$$ has kernel 0. – Ida B. Feb 12 '10 at 12:26
Consider the exact sequence $0 \to S(-\mathrm{deg}\; f) \to S \to S/(f) \to 0$ (where the first map is multiplication by $f$) and take its long exact sequence of $\mathrm{Ext}$ groups. Since both $S$ and $S(-\mathrm{deg}\; f)$ are free $S$-modules, their higher $\mathrm{Ext}$ groups vanish, and you get $\mathrm{Ext}^m(S/(f), S) = 0$ for all $m \geq 2$. In addition, it is clear that $\mathrm{Ext}^0(S/(f), S) = \mathrm{Hom}(S/(f), S) = 0$.