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.



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


