Direct limit of directed system of modules commutes with right derived functors of additive, covariant, left...
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Direct limit of directed system of modules commutes with right derived functors of additive, covariant, left exact functor?
Derived functors of torsion functorproving an isomorphism of direct limitsAdditive, covariant functor commutes direct limits, then it commutes with direct sums?Additive exact functors preserve homology of modulesAdditive Exact Functors Commute with HomologyDefining contravariant left/right-exact functor with opposite category?Conservative functor in a category of co-limit preserving functors between $R$ modules and vector spacesCommutative ring as a direct limit of Noetherian ringsOn derived functors of certain $operatorname{Tor}_1$ and $operatorname{Ext}^1$On the natural isomorphism between $I$-torsion functor and direct limit of $mathrm{Hom}$ functor
$begingroup$
Let $R$ be a commutative ring with unity. Let $T: R$-Mod $to R$-Mod be an additive, covariant, left exact functor which commutes with direct limits indexed by directed sets. Let $R^i T$ be the right derived functors of $T$.
Is it true that for any directed system of modules ${M_{alpha}}$ indexed by a directed set, we have $varinjlim (R^i T) (M_alpha)cong (R^iT)(varinjlim M_alpha)$ ?
commutative-algebra modules homological-algebra derived-functors
asked yesterday
user102248user102248
1907
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add a comment |
$begingroup$
Let $R$ be a commutative ring with unity. Let $T: R$-Mod $to R$-Mod be an additive, covariant, left exact functor which commutes with direct limits indexed by directed sets. Let $R^i T$ be the right derived functors of $T$.
Is it true that for any directed system of modules ${M_{alpha}}$ indexed by a directed set, we have $varinjlim (R^i T) (M_alpha)cong (R^iT)(varinjlim M_alpha)$ ?
commutative-algebra modules homological-algebra derived-functors
asked yesterday
user102248user102248
1907
$endgroup$
add a comment |
$begingroup$
Let $R$ be a commutative ring with unity. Let $T: R$-Mod $to R$-Mod be an additive, covariant, left exact functor which commutes with direct limits indexed by directed sets. Let $R^i T$ be the right derived functors of $T$.
Is it true that for any directed system of modules ${M_{alpha}}$ indexed by a directed set, we have $varinjlim (R^i T) (M_alpha)cong (R^iT)(varinjlim M_alpha)$ ?
commutative-algebra modules homological-algebra derived-functors
asked yesterday
user102248user102248
1907
$endgroup$
Let $R$ be a commutative ring with unity. Let $T: R$-Mod $to R$-Mod be an additive, covariant, left exact functor which commutes with direct limits indexed by directed sets. Let $R^i T$ be the right derived functors of $T$.
Is it true that for any directed system of modules ${M_{alpha}}$ indexed by a directed set, we have $varinjlim (R^i T) (M_alpha)cong (R^iT)(varinjlim M_alpha)$ ?
commutative-algebra modules homological-algebra derived-functors
commutative-algebra modules homological-algebra derived-functors
asked yesterday
user102248user102248
1907
asked yesterday
user102248user102248
1907
edited 19 hours ago
user102248
asked yesterday
user102248user102248
1907
asked yesterday
user102248user102248
1907
asked yesterday
user102248user102248
1907
1907
add a comment |
add a comment |
1 Answer
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No. Indeed, this need not even be true for $T$ itself (which is $R^0T$). For instance, if $I$ is an infinite set (and $R$ is nonzero), the functor $T(M)=M^I$ is exact but does not preserve directed colimits, since if $M$ is the direct limit of $(M_alpha)$ then there may be elements of $M^I$ whose coordinates do not all come from any single $M_alpha$.
Even if $T$ preserves directed colimits, its derived functors may not. For instance, let $A=k[S]$ be a polynomial ring over a field $k$ with an infinite set $S$ of variables, let $I=(S)$ be the ideal generated by all the variables, and let $R=A/I^2$. Let $N=R/(s)$ for some variable $sin S$ and consider the functor $T=operatorname{Hom}(N,-)$. Since $N$ is finitely presented, $T$ preserves directed colimits. To compute the derived functors $R^iT=operatorname{Ext}^i(N,-)$ we take a minimal free resolution of $N$ which has the form $$to R^{oplus S}to Rstackrel{s}to Rto Nto 0$$ (where "minimal" means every map in the resolution is $0$ mod $I$). If $M$ is any $R$-module which is annihilated by $I$, we then see that $$operatorname{Ext}^2(N,M)cong operatorname{Hom}(R^{oplus S},M)cong M^S.$$ Since $S$ is infinite, $Mmapsto M^S$ does not preserve directed colimits, so $R^2T=operatorname{Ext}^2(N,-)$ does not preserve directed colimits.
If you assume that $T$ preserves directed colimits and $R$ is Noetherian, then it is true. As a sketch of a proof, if $M$ is a colimit of a directed system $(M_alpha)$, then we can construct an injective resolution of $M$ as a directed colimit of injective resolutions of the $M_alpha$, using the fact that directed colimits of injective modules are injective since $R$ is Noetherian. (This step is nontrivial, since we can't actually cobble injective resolutions of all the $M_alpha$ into a diagram that commutes on the nose in any obvious way. One way to handle this is to reduce to the case that the system $(M_alpha)$ is indexed by an ordinal and is cocontinuous, so you can build a commutative diagram of injective resolutions by transfinite induciton.) We then see that computing $R^iT(M)$ using this injective resolution is the same as computing $R^iT(M_alpha)$ using their injective resolutions and then taking the colimit.
answered yesterday
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
I'm very sorry for not mentioning how $T$ behaves with direct limits ... can you please refer to my slightly edited question now ..? Thanks ..
$endgroup$
– user102248
19 hours ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
No. Indeed, this need not even be true for $T$ itself (which is $R^0T$). For instance, if $I$ is an infinite set (and $R$ is nonzero), the functor $T(M)=M^I$ is exact but does not preserve directed colimits, since if $M$ is the direct limit of $(M_alpha)$ then there may be elements of $M^I$ whose coordinates do not all come from any single $M_alpha$.
Even if $T$ preserves directed colimits, its derived functors may not. For instance, let $A=k[S]$ be a polynomial ring over a field $k$ with an infinite set $S$ of variables, let $I=(S)$ be the ideal generated by all the variables, and let $R=A/I^2$. Let $N=R/(s)$ for some variable $sin S$ and consider the functor $T=operatorname{Hom}(N,-)$. Since $N$ is finitely presented, $T$ preserves directed colimits. To compute the derived functors $R^iT=operatorname{Ext}^i(N,-)$ we take a minimal free resolution of $N$ which has the form $$to R^{oplus S}to Rstackrel{s}to Rto Nto 0$$ (where "minimal" means every map in the resolution is $0$ mod $I$). If $M$ is any $R$-module which is annihilated by $I$, we then see that $$operatorname{Ext}^2(N,M)cong operatorname{Hom}(R^{oplus S},M)cong M^S.$$ Since $S$ is infinite, $Mmapsto M^S$ does not preserve directed colimits, so $R^2T=operatorname{Ext}^2(N,-)$ does not preserve directed colimits.
If you assume that $T$ preserves directed colimits and $R$ is Noetherian, then it is true. As a sketch of a proof, if $M$ is a colimit of a directed system $(M_alpha)$, then we can construct an injective resolution of $M$ as a directed colimit of injective resolutions of the $M_alpha$, using the fact that directed colimits of injective modules are injective since $R$ is Noetherian. (This step is nontrivial, since we can't actually cobble injective resolutions of all the $M_alpha$ into a diagram that commutes on the nose in any obvious way. One way to handle this is to reduce to the case that the system $(M_alpha)$ is indexed by an ordinal and is cocontinuous, so you can build a commutative diagram of injective resolutions by transfinite induciton.) We then see that computing $R^iT(M)$ using this injective resolution is the same as computing $R^iT(M_alpha)$ using their injective resolutions and then taking the colimit.
answered yesterday
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
I'm very sorry for not mentioning how $T$ behaves with direct limits ... can you please refer to my slightly edited question now ..? Thanks ..
$endgroup$
– user102248
19 hours ago
add a comment |
$begingroup$
No. Indeed, this need not even be true for $T$ itself (which is $R^0T$). For instance, if $I$ is an infinite set (and $R$ is nonzero), the functor $T(M)=M^I$ is exact but does not preserve directed colimits, since if $M$ is the direct limit of $(M_alpha)$ then there may be elements of $M^I$ whose coordinates do not all come from any single $M_alpha$.
Even if $T$ preserves directed colimits, its derived functors may not. For instance, let $A=k[S]$ be a polynomial ring over a field $k$ with an infinite set $S$ of variables, let $I=(S)$ be the ideal generated by all the variables, and let $R=A/I^2$. Let $N=R/(s)$ for some variable $sin S$ and consider the functor $T=operatorname{Hom}(N,-)$. Since $N$ is finitely presented, $T$ preserves directed colimits. To compute the derived functors $R^iT=operatorname{Ext}^i(N,-)$ we take a minimal free resolution of $N$ which has the form $$to R^{oplus S}to Rstackrel{s}to Rto Nto 0$$ (where "minimal" means every map in the resolution is $0$ mod $I$). If $M$ is any $R$-module which is annihilated by $I$, we then see that $$operatorname{Ext}^2(N,M)cong operatorname{Hom}(R^{oplus S},M)cong M^S.$$ Since $S$ is infinite, $Mmapsto M^S$ does not preserve directed colimits, so $R^2T=operatorname{Ext}^2(N,-)$ does not preserve directed colimits.
If you assume that $T$ preserves directed colimits and $R$ is Noetherian, then it is true. As a sketch of a proof, if $M$ is a colimit of a directed system $(M_alpha)$, then we can construct an injective resolution of $M$ as a directed colimit of injective resolutions of the $M_alpha$, using the fact that directed colimits of injective modules are injective since $R$ is Noetherian. (This step is nontrivial, since we can't actually cobble injective resolutions of all the $M_alpha$ into a diagram that commutes on the nose in any obvious way. One way to handle this is to reduce to the case that the system $(M_alpha)$ is indexed by an ordinal and is cocontinuous, so you can build a commutative diagram of injective resolutions by transfinite induciton.) We then see that computing $R^iT(M)$ using this injective resolution is the same as computing $R^iT(M_alpha)$ using their injective resolutions and then taking the colimit.
answered yesterday
Eric WofseyEric Wofsey
188k14216346
$endgroup$
$begingroup$
I'm very sorry for not mentioning how $T$ behaves with direct limits ... can you please refer to my slightly edited question now ..? Thanks ..
$endgroup$
– user102248
19 hours ago
add a comment |
$begingroup$
No. Indeed, this need not even be true for $T$ itself (which is $R^0T$). For instance, if $I$ is an infinite set (and $R$ is nonzero), the functor $T(M)=M^I$ is exact but does not preserve directed colimits, since if $M$ is the direct limit of $(M_alpha)$ then there may be elements of $M^I$ whose coordinates do not all come from any single $M_alpha$.
Even if $T$ preserves directed colimits, its derived functors may not. For instance, let $A=k[S]$ be a polynomial ring over a field $k$ with an infinite set $S$ of variables, let $I=(S)$ be the ideal generated by all the variables, and let $R=A/I^2$. Let $N=R/(s)$ for some variable $sin S$ and consider the functor $T=operatorname{Hom}(N,-)$. Since $N$ is finitely presented, $T$ preserves directed colimits. To compute the derived functors $R^iT=operatorname{Ext}^i(N,-)$ we take a minimal free resolution of $N$ which has the form $$to R^{oplus S}to Rstackrel{s}to Rto Nto 0$$ (where "minimal" means every map in the resolution is $0$ mod $I$). If $M$ is any $R$-module which is annihilated by $I$, we then see that $$operatorname{Ext}^2(N,M)cong operatorname{Hom}(R^{oplus S},M)cong M^S.$$ Since $S$ is infinite, $Mmapsto M^S$ does not preserve directed colimits, so $R^2T=operatorname{Ext}^2(N,-)$ does not preserve directed colimits.
If you assume that $T$ preserves directed colimits and $R$ is Noetherian, then it is true. As a sketch of a proof, if $M$ is a colimit of a directed system $(M_alpha)$, then we can construct an injective resolution of $M$ as a directed colimit of injective resolutions of the $M_alpha$, using the fact that directed colimits of injective modules are injective since $R$ is Noetherian. (This step is nontrivial, since we can't actually cobble injective resolutions of all the $M_alpha$ into a diagram that commutes on the nose in any obvious way. One way to handle this is to reduce to the case that the system $(M_alpha)$ is indexed by an ordinal and is cocontinuous, so you can build a commutative diagram of injective resolutions by transfinite induciton.) We then see that computing $R^iT(M)$ using this injective resolution is the same as computing $R^iT(M_alpha)$ using their injective resolutions and then taking the colimit.
answered yesterday
Eric WofseyEric Wofsey
188k14216346
$endgroup$
No. Indeed, this need not even be true for $T$ itself (which is $R^0T$). For instance, if $I$ is an infinite set (and $R$ is nonzero), the functor $T(M)=M^I$ is exact but does not preserve directed colimits, since if $M$ is the direct limit of $(M_alpha)$ then there may be elements of $M^I$ whose coordinates do not all come from any single $M_alpha$.
Even if $T$ preserves directed colimits, its derived functors may not. For instance, let $A=k[S]$ be a polynomial ring over a field $k$ with an infinite set $S$ of variables, let $I=(S)$ be the ideal generated by all the variables, and let $R=A/I^2$. Let $N=R/(s)$ for some variable $sin S$ and consider the functor $T=operatorname{Hom}(N,-)$. Since $N$ is finitely presented, $T$ preserves directed colimits. To compute the derived functors $R^iT=operatorname{Ext}^i(N,-)$ we take a minimal free resolution of $N$ which has the form $$to R^{oplus S}to Rstackrel{s}to Rto Nto 0$$ (where "minimal" means every map in the resolution is $0$ mod $I$). If $M$ is any $R$-module which is annihilated by $I$, we then see that $$operatorname{Ext}^2(N,M)cong operatorname{Hom}(R^{oplus S},M)cong M^S.$$ Since $S$ is infinite, $Mmapsto M^S$ does not preserve directed colimits, so $R^2T=operatorname{Ext}^2(N,-)$ does not preserve directed colimits.
If you assume that $T$ preserves directed colimits and $R$ is Noetherian, then it is true. As a sketch of a proof, if $M$ is a colimit of a directed system $(M_alpha)$, then we can construct an injective resolution of $M$ as a directed colimit of injective resolutions of the $M_alpha$, using the fact that directed colimits of injective modules are injective since $R$ is Noetherian. (This step is nontrivial, since we can't actually cobble injective resolutions of all the $M_alpha$ into a diagram that commutes on the nose in any obvious way. One way to handle this is to reduce to the case that the system $(M_alpha)$ is indexed by an ordinal and is cocontinuous, so you can build a commutative diagram of injective resolutions by transfinite induciton.) We then see that computing $R^iT(M)$ using this injective resolution is the same as computing $R^iT(M_alpha)$ using their injective resolutions and then taking the colimit.
answered yesterday
Eric WofseyEric Wofsey
188k14216346
edited 18 hours ago
answered yesterday
Eric WofseyEric Wofsey
188k14216346
answered yesterday
Eric WofseyEric Wofsey
188k14216346
answered yesterday
Eric WofseyEric Wofsey
188k14216346
188k14216346
$begingroup$
I'm very sorry for not mentioning how $T$ behaves with direct limits ... can you please refer to my slightly edited question now ..? Thanks ..
$endgroup$
– user102248
19 hours ago
add a comment |
$begingroup$
I'm very sorry for not mentioning how $T$ behaves with direct limits ... can you please refer to my slightly edited question now ..? Thanks ..
$endgroup$
– user102248
19 hours ago
$begingroup$
I'm very sorry for not mentioning how $T$ behaves with direct limits ... can you please refer to my slightly edited question now ..? Thanks ..
$endgroup$
– user102248
19 hours ago
$begingroup$
I'm very sorry for not mentioning how $T$ behaves with direct limits ... can you please refer to my slightly edited question now ..? Thanks ..
$endgroup$
– user102248
19 hours ago
add a comment |
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