Skip to main content
MathOverflow

Return to Answer

added 1 characters in body
Source Link
abx
abx
  • 38k
  • 3
  • 86
  • 146

For a module of finite length $M$ over a principal ideal domain $A$, we have $\mathrm{Hom}(M,A)=0$, and $\mathrm{Ext}^1(M,A)$ (the dual of $A$, see Bourbaki, Algebra VII, §4, No. 9) is non-canonically isomorphic to $M$. Therefore $\chi (\mathrm{Hom}(M,A))=-\mathrm{length}\,(\mathrm{Ext}^1(M,A))=-\chi (M)$$\chi (\mathrm{RHom}(M,A))=-\mathrm{length}\,(\mathrm{Ext}^1(M,A))=-\chi (M)$.

For a module of finite length $M$ over a principal ideal domain $A$, we have $\mathrm{Hom}(M,A)=0$, and $\mathrm{Ext}^1(M,A)$ (the dual of $A$, see Bourbaki, Algebra VII, §4, No. 9) is non-canonically isomorphic to $M$. Therefore $\chi (\mathrm{Hom}(M,A))=-\mathrm{length}\,(\mathrm{Ext}^1(M,A))=-\chi (M)$.

For a module of finite length $M$ over a principal ideal domain $A$, we have $\mathrm{Hom}(M,A)=0$, and $\mathrm{Ext}^1(M,A)$ (the dual of $A$, see Bourbaki, Algebra VII, §4, No. 9) is non-canonically isomorphic to $M$. Therefore $\chi (\mathrm{RHom}(M,A))=-\mathrm{length}\,(\mathrm{Ext}^1(M,A))=-\chi (M)$.

Source Link
abx
abx
  • 38k
  • 3
  • 86
  • 146

For a module of finite length $M$ over a principal ideal domain $A$, we have $\mathrm{Hom}(M,A)=0$, and $\mathrm{Ext}^1(M,A)$ (the dual of $A$, see Bourbaki, Algebra VII, §4, No. 9) is non-canonically isomorphic to $M$. Therefore $\chi (\mathrm{Hom}(M,A))=-\mathrm{length}\,(\mathrm{Ext}^1(M,A))=-\chi (M)$.