In what degrees does Ext(S/(f),S) vanish? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T06:19:18Z http://mathoverflow.net/feeds/question/14996 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14996/in-what-degrees-does-exts-f-s-vanish In what degrees does Ext(S/(f),S) vanish? Ida B. 2010-02-11T14:22:10Z 2010-02-12T13:14:35Z <p>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?</p> <p>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 $$\ldots\to Ext^m(S/(f)(\deg f),S)\to Ext^m(S/fI(\deg f),S)\to Ext^m(S/I,S)\to$$<br /> $$Ext^{m+1}(S/(f)(\deg f),S)\to Ext^{m+1}(S/fI(\deg f),S)\to Ext^{m+1}(S/I,S)\to\ldots$$<br /> and two zeros on the left would suffice.</p> http://mathoverflow.net/questions/14996/in-what-degrees-does-exts-f-s-vanish/14997#14997 Answer by Alberto García-Raboso for In what degrees does Ext(S/(f),S) vanish? Alberto García-Raboso 2010-02-11T15:24:43Z 2010-02-11T15:58:12Z <p>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 $\mathrm{Ext}^0(S/(f), S) = \mathrm{Hom}(S/(f), S) = 0$.</p> http://mathoverflow.net/questions/14996/in-what-degrees-does-exts-f-s-vanish/15000#15000 Answer by Hailong Dao for In what degrees does Ext(S/(f),S) vanish? Hailong Dao 2010-02-11T15:45:18Z 2010-02-12T13:14:35Z <p>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. </p>