$operatorname{Ext}^bullet_R(R/rR,M)$ and $0 to A[r] to B[r] to C[r] to A/rA to B/rB to C/rC to 0$












2












$begingroup$


$newcommand{Ext}{operatorname{Ext}}newcommand{Hom}{operatorname{Hom}_R}$Let $R$ be a commutative ring with unity and $r in R$.



Applying snake lemma to the following diagram:



$$begin{array}{c} 0 & longrightarrow & A~~ & longrightarrow & B~~ & longrightarrow & C~~ & longrightarrow & 0 \ && downarrow r && downarrow r && downarrow r \ 0 & longrightarrow & A~~ & longrightarrow & B~~ & longrightarrow & C~~ & longrightarrow & 0 end{array}$$



We obtain the exact sequence $0 to A[r] to B[r] to C[r] to A/rA to B/rB to C/rC to 0$.




Now, assuming that $r$ is not a zero-divisor, $0 to R xrightarrow r R (to R/rR to 0)$ is a projective resolution of $R/rR$, and applying $Hom(-,M)$ for $M in textbf{$R$-Mod}$ gives $0 to Hom(R,M) xrightarrow{f mapsto (x mapsto f(rx))} Hom(R,M) to 0$, i.e. $0 to M xrightarrow r M to 0$, so $Ext^1_R(R/rR, M) = M/rM$ and $Ext^n_R(R/rR, M) = 0$ for $n>1$, matching the exact sequence above.




If $r$ is a zero-divisor though, then according to this answer we do not get $M/rM$, but we need to apply $operatorname{Hom}_R(-,M)$ to the exact sequence $0 to rR to R to R/rR to 0$ to get



$$0 to M[r] to M to Hom(rR,M) to Ext^1_R(R/rR,M) to 0 to cdots$$



Now $Hom(rR,M) cong { m in M cdot forall c in R[r], c cdot m = 0}$ which I will notate as $M[R[r]]$.



So the exact sequence tells us $Ext^1_R(R/rR,M) cong M[R[r]]/rM$ instead of $M/rM$.



The exact sequence also tells us $Ext^{n+1}_R(R/rR,M) cong Ext^n_R(rR,M)$ which may or may not be useful in computing $Ext^bullet_R(R/rR,M)$.




My question is: what is the role of the exact sequence $0 to A[r] to B[r] to C[r] to A/rA to B/rB to C/rC to 0$ in the grand scheme of things involving $Ext^bullet_R(R/rR,-)$? If possible, I would also like to generalize to non-commutative ring with unity $R$ with $r in Z(R)$.






homological-algebra







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Please clarify the notation $A[r]$. I suppose it means ${xin Amid rx=0}$, sometimes denoted by $(0:_Ar)$, but this is only a supposition.
    $endgroup$
    – user26857
    Jan 3 at 9:26


















2












$begingroup$


$newcommand{Ext}{operatorname{Ext}}newcommand{Hom}{operatorname{Hom}_R}$Let $R$ be a commutative ring with unity and $r in R$.



Applying snake lemma to the following diagram:



$$begin{array}{c} 0 & longrightarrow & A~~ & longrightarrow & B~~ & longrightarrow & C~~ & longrightarrow & 0 \ && downarrow r && downarrow r && downarrow r \ 0 & longrightarrow & A~~ & longrightarrow & B~~ & longrightarrow & C~~ & longrightarrow & 0 end{array}$$



We obtain the exact sequence $0 to A[r] to B[r] to C[r] to A/rA to B/rB to C/rC to 0$.




Now, assuming that $r$ is not a zero-divisor, $0 to R xrightarrow r R (to R/rR to 0)$ is a projective resolution of $R/rR$, and applying $Hom(-,M)$ for $M in textbf{$R$-Mod}$ gives $0 to Hom(R,M) xrightarrow{f mapsto (x mapsto f(rx))} Hom(R,M) to 0$, i.e. $0 to M xrightarrow r M to 0$, so $Ext^1_R(R/rR, M) = M/rM$ and $Ext^n_R(R/rR, M) = 0$ for $n>1$, matching the exact sequence above.




If $r$ is a zero-divisor though, then according to this answer we do not get $M/rM$, but we need to apply $operatorname{Hom}_R(-,M)$ to the exact sequence $0 to rR to R to R/rR to 0$ to get



$$0 to M[r] to M to Hom(rR,M) to Ext^1_R(R/rR,M) to 0 to cdots$$



Now $Hom(rR,M) cong { m in M cdot forall c in R[r], c cdot m = 0}$ which I will notate as $M[R[r]]$.



So the exact sequence tells us $Ext^1_R(R/rR,M) cong M[R[r]]/rM$ instead of $M/rM$.



The exact sequence also tells us $Ext^{n+1}_R(R/rR,M) cong Ext^n_R(rR,M)$ which may or may not be useful in computing $Ext^bullet_R(R/rR,M)$.




My question is: what is the role of the exact sequence $0 to A[r] to B[r] to C[r] to A/rA to B/rB to C/rC to 0$ in the grand scheme of things involving $Ext^bullet_R(R/rR,-)$? If possible, I would also like to generalize to non-commutative ring with unity $R$ with $r in Z(R)$.






homological-algebra







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Please clarify the notation $A[r]$. I suppose it means ${xin Amid rx=0}$, sometimes denoted by $(0:_Ar)$, but this is only a supposition.
    $endgroup$
    – user26857
    Jan 3 at 9:26
















2












2








2





$begingroup$


$newcommand{Ext}{operatorname{Ext}}newcommand{Hom}{operatorname{Hom}_R}$Let $R$ be a commutative ring with unity and $r in R$.



Applying snake lemma to the following diagram:



$$begin{array}{c} 0 & longrightarrow & A~~ & longrightarrow & B~~ & longrightarrow & C~~ & longrightarrow & 0 \ && downarrow r && downarrow r && downarrow r \ 0 & longrightarrow & A~~ & longrightarrow & B~~ & longrightarrow & C~~ & longrightarrow & 0 end{array}$$



We obtain the exact sequence $0 to A[r] to B[r] to C[r] to A/rA to B/rB to C/rC to 0$.




Now, assuming that $r$ is not a zero-divisor, $0 to R xrightarrow r R (to R/rR to 0)$ is a projective resolution of $R/rR$, and applying $Hom(-,M)$ for $M in textbf{$R$-Mod}$ gives $0 to Hom(R,M) xrightarrow{f mapsto (x mapsto f(rx))} Hom(R,M) to 0$, i.e. $0 to M xrightarrow r M to 0$, so $Ext^1_R(R/rR, M) = M/rM$ and $Ext^n_R(R/rR, M) = 0$ for $n>1$, matching the exact sequence above.




If $r$ is a zero-divisor though, then according to this answer we do not get $M/rM$, but we need to apply $operatorname{Hom}_R(-,M)$ to the exact sequence $0 to rR to R to R/rR to 0$ to get



$$0 to M[r] to M to Hom(rR,M) to Ext^1_R(R/rR,M) to 0 to cdots$$



Now $Hom(rR,M) cong { m in M cdot forall c in R[r], c cdot m = 0}$ which I will notate as $M[R[r]]$.



So the exact sequence tells us $Ext^1_R(R/rR,M) cong M[R[r]]/rM$ instead of $M/rM$.



The exact sequence also tells us $Ext^{n+1}_R(R/rR,M) cong Ext^n_R(rR,M)$ which may or may not be useful in computing $Ext^bullet_R(R/rR,M)$.




My question is: what is the role of the exact sequence $0 to A[r] to B[r] to C[r] to A/rA to B/rB to C/rC to 0$ in the grand scheme of things involving $Ext^bullet_R(R/rR,-)$? If possible, I would also like to generalize to non-commutative ring with unity $R$ with $r in Z(R)$.






homological-algebra







share|cite|improve this question









$endgroup$




$newcommand{Ext}{operatorname{Ext}}newcommand{Hom}{operatorname{Hom}_R}$Let $R$ be a commutative ring with unity and $r in R$.



Applying snake lemma to the following diagram:



$$begin{array}{c} 0 & longrightarrow & A~~ & longrightarrow & B~~ & longrightarrow & C~~ & longrightarrow & 0 \ && downarrow r && downarrow r && downarrow r \ 0 & longrightarrow & A~~ & longrightarrow & B~~ & longrightarrow & C~~ & longrightarrow & 0 end{array}$$



We obtain the exact sequence $0 to A[r] to B[r] to C[r] to A/rA to B/rB to C/rC to 0$.




Now, assuming that $r$ is not a zero-divisor, $0 to R xrightarrow r R (to R/rR to 0)$ is a projective resolution of $R/rR$, and applying $Hom(-,M)$ for $M in textbf{$R$-Mod}$ gives $0 to Hom(R,M) xrightarrow{f mapsto (x mapsto f(rx))} Hom(R,M) to 0$, i.e. $0 to M xrightarrow r M to 0$, so $Ext^1_R(R/rR, M) = M/rM$ and $Ext^n_R(R/rR, M) = 0$ for $n>1$, matching the exact sequence above.




If $r$ is a zero-divisor though, then according to this answer we do not get $M/rM$, but we need to apply $operatorname{Hom}_R(-,M)$ to the exact sequence $0 to rR to R to R/rR to 0$ to get



$$0 to M[r] to M to Hom(rR,M) to Ext^1_R(R/rR,M) to 0 to cdots$$



Now $Hom(rR,M) cong { m in M cdot forall c in R[r], c cdot m = 0}$ which I will notate as $M[R[r]]$.



So the exact sequence tells us $Ext^1_R(R/rR,M) cong M[R[r]]/rM$ instead of $M/rM$.



The exact sequence also tells us $Ext^{n+1}_R(R/rR,M) cong Ext^n_R(rR,M)$ which may or may not be useful in computing $Ext^bullet_R(R/rR,M)$.




My question is: what is the role of the exact sequence $0 to A[r] to B[r] to C[r] to A/rA to B/rB to C/rC to 0$ in the grand scheme of things involving $Ext^bullet_R(R/rR,-)$? If possible, I would also like to generalize to non-commutative ring with unity $R$ with $r in Z(R)$.






homological-algebra




homological-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 19:42









Kenny LauKenny Lau

19.9k2159




19.9k2159








  • 1




    $begingroup$
    Please clarify the notation $A[r]$. I suppose it means ${xin Amid rx=0}$, sometimes denoted by $(0:_Ar)$, but this is only a supposition.
    $endgroup$
    – user26857
    Jan 3 at 9:26
















  • 1




    $begingroup$
    Please clarify the notation $A[r]$. I suppose it means ${xin Amid rx=0}$, sometimes denoted by $(0:_Ar)$, but this is only a supposition.
    $endgroup$
    – user26857
    Jan 3 at 9:26










1




1




$begingroup$
Please clarify the notation $A[r]$. I suppose it means ${xin Amid rx=0}$, sometimes denoted by $(0:_Ar)$, but this is only a supposition.
$endgroup$
– user26857
Jan 3 at 9:26






$begingroup$
Please clarify the notation $A[r]$. I suppose it means ${xin Amid rx=0}$, sometimes denoted by $(0:_Ar)$, but this is only a supposition.
$endgroup$
– user26857
Jan 3 at 9:26












0






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