If the dual of a module is finitely generated and projective, can we claim that the module itself is?When $operatornameHom_R(M,N)$ is finitely generated as $mathbb Z$-module or $R$-module?Prove that if $P$ and $Q$ are projective and finitely generated $R$-modules then $operatornameHom_R(P,Q)$ is projective and finitely generated.On the existence of finitely generated modules with finite injective dimensionA question about finitely generated projective modulesWhen is the localization of a commutative ring a finitely generated projective module?Finitely generated Hom moduleShowing a module is finitely generated and projectiveAn isomorphism concerned about any finitely generated projective moduleprojective module which is a submodule of a finitely generated free moduleDual basis for projective module
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If the dual of a module is finitely generated and projective, can we claim that the module itself is?
When $operatornameHom_R(M,N)$ is finitely generated as $mathbb Z$-module or $R$-module?Prove that if $P$ and $Q$ are projective and finitely generated $R$-modules then $operatornameHom_R(P,Q)$ is projective and finitely generated.On the existence of finitely generated modules with finite injective dimensionA question about finitely generated projective modulesWhen is the localization of a commutative ring a finitely generated projective module?Finitely generated Hom moduleShowing a module is finitely generated and projectiveAn isomorphism concerned about any finitely generated projective moduleprojective module which is a submodule of a finitely generated free moduleDual basis for projective module
$begingroup$
Assume that $R$ is a commutative ring and that $M$ is a (left) $R$-module. Assume also that we know for some reason that $M^*:=mathsfHom_R(M,R)$ is finitely generated and projective as (right) $R$-module. Can we claim that $M$ itself is finitely generated and projective?
It is well-known that the converse is true, but I am able neither to prove nor to disprove the foregoing implication.
Of course, $M^**$ is finitely generated and projective, but in general the canonical morphism $j:Mto M^**$ is not injective, whence I don't know how to use this fact. Can somebody give me a hint, either in proving the statement or in finding a counterexample?
commutative-algebra duality-theorems projective-module
asked 4 hours ago
Ender WigginsEnder Wiggins
865421
$endgroup$
add a comment |
$begingroup$
Assume that $R$ is a commutative ring and that $M$ is a (left) $R$-module. Assume also that we know for some reason that $M^*:=mathsfHom_R(M,R)$ is finitely generated and projective as (right) $R$-module. Can we claim that $M$ itself is finitely generated and projective?
It is well-known that the converse is true, but I am able neither to prove nor to disprove the foregoing implication.
Of course, $M^**$ is finitely generated and projective, but in general the canonical morphism $j:Mto M^**$ is not injective, whence I don't know how to use this fact. Can somebody give me a hint, either in proving the statement or in finding a counterexample?
commutative-algebra duality-theorems projective-module
asked 4 hours ago
Ender WigginsEnder Wiggins
865421
$endgroup$
add a comment |
$begingroup$
Assume that $R$ is a commutative ring and that $M$ is a (left) $R$-module. Assume also that we know for some reason that $M^*:=mathsfHom_R(M,R)$ is finitely generated and projective as (right) $R$-module. Can we claim that $M$ itself is finitely generated and projective?
It is well-known that the converse is true, but I am able neither to prove nor to disprove the foregoing implication.
Of course, $M^**$ is finitely generated and projective, but in general the canonical morphism $j:Mto M^**$ is not injective, whence I don't know how to use this fact. Can somebody give me a hint, either in proving the statement or in finding a counterexample?
commutative-algebra duality-theorems projective-module
asked 4 hours ago
Ender WigginsEnder Wiggins
865421
$endgroup$
Assume that $R$ is a commutative ring and that $M$ is a (left) $R$-module. Assume also that we know for some reason that $M^*:=mathsfHom_R(M,R)$ is finitely generated and projective as (right) $R$-module. Can we claim that $M$ itself is finitely generated and projective?
It is well-known that the converse is true, but I am able neither to prove nor to disprove the foregoing implication.
Of course, $M^**$ is finitely generated and projective, but in general the canonical morphism $j:Mto M^**$ is not injective, whence I don't know how to use this fact. Can somebody give me a hint, either in proving the statement or in finding a counterexample?
commutative-algebra duality-theorems projective-module
commutative-algebra duality-theorems projective-module
asked 4 hours ago
Ender WigginsEnder Wiggins
865421
asked 4 hours ago
Ender WigginsEnder Wiggins
865421
asked 4 hours ago
Ender WigginsEnder Wiggins
865421
asked 4 hours ago
Ender WigginsEnder Wiggins
865421
asked 4 hours ago
Ender WigginsEnder Wiggins
865421
865421
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $operatornameHom_R(M,R)=M^*$ is finitely generated and projective, then $R^ncong M^*oplus N$, so we have
$$
M^**oplus N^*cong R^n
$$
so $M^**$ is finitely generated and projective. Unfortunately, the canonical homomorphism $Mto M^**$ is neither injective nor surjective, in general.
A trivial example is $M=mathbbQ$, with $R=mathbbZ$. You can complicate the situation at will.
Just to give the flavor, suppose $R$ is a PID and that $M$ is finitely generated. Then $M^*$ is finitely generated and free: you lose all information about the torsion part, when doing the dual.
answered 4 hours ago
egregegreg
185k1486206
$endgroup$
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbbQ^*$ should be finitely generated and projective as $mathbbZ$-module? If I am not mistaken, an element in $mathbbQ^*$ is uniquely determined by its images over the rationals of the form $frac1p$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
4 hours ago
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
3 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
3 hours ago
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
If $operatornameHom_R(M,R)=M^*$ is finitely generated and projective, then $R^ncong M^*oplus N$, so we have
$$
M^**oplus N^*cong R^n
$$
so $M^**$ is finitely generated and projective. Unfortunately, the canonical homomorphism $Mto M^**$ is neither injective nor surjective, in general.
A trivial example is $M=mathbbQ$, with $R=mathbbZ$. You can complicate the situation at will.
Just to give the flavor, suppose $R$ is a PID and that $M$ is finitely generated. Then $M^*$ is finitely generated and free: you lose all information about the torsion part, when doing the dual.
answered 4 hours ago
egregegreg
185k1486206
$endgroup$
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbbQ^*$ should be finitely generated and projective as $mathbbZ$-module? If I am not mistaken, an element in $mathbbQ^*$ is uniquely determined by its images over the rationals of the form $frac1p$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
4 hours ago
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
3 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
3 hours ago
add a comment |
$begingroup$
If $operatornameHom_R(M,R)=M^*$ is finitely generated and projective, then $R^ncong M^*oplus N$, so we have
$$
M^**oplus N^*cong R^n
$$
so $M^**$ is finitely generated and projective. Unfortunately, the canonical homomorphism $Mto M^**$ is neither injective nor surjective, in general.
A trivial example is $M=mathbbQ$, with $R=mathbbZ$. You can complicate the situation at will.
Just to give the flavor, suppose $R$ is a PID and that $M$ is finitely generated. Then $M^*$ is finitely generated and free: you lose all information about the torsion part, when doing the dual.
answered 4 hours ago
egregegreg
185k1486206
$endgroup$
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbbQ^*$ should be finitely generated and projective as $mathbbZ$-module? If I am not mistaken, an element in $mathbbQ^*$ is uniquely determined by its images over the rationals of the form $frac1p$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
4 hours ago
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
3 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
3 hours ago
add a comment |
$begingroup$
If $operatornameHom_R(M,R)=M^*$ is finitely generated and projective, then $R^ncong M^*oplus N$, so we have
$$
M^**oplus N^*cong R^n
$$
so $M^**$ is finitely generated and projective. Unfortunately, the canonical homomorphism $Mto M^**$ is neither injective nor surjective, in general.
A trivial example is $M=mathbbQ$, with $R=mathbbZ$. You can complicate the situation at will.
Just to give the flavor, suppose $R$ is a PID and that $M$ is finitely generated. Then $M^*$ is finitely generated and free: you lose all information about the torsion part, when doing the dual.
answered 4 hours ago
egregegreg
185k1486206
$endgroup$
If $operatornameHom_R(M,R)=M^*$ is finitely generated and projective, then $R^ncong M^*oplus N$, so we have
$$
M^**oplus N^*cong R^n
$$
so $M^**$ is finitely generated and projective. Unfortunately, the canonical homomorphism $Mto M^**$ is neither injective nor surjective, in general.
A trivial example is $M=mathbbQ$, with $R=mathbbZ$. You can complicate the situation at will.
Just to give the flavor, suppose $R$ is a PID and that $M$ is finitely generated. Then $M^*$ is finitely generated and free: you lose all information about the torsion part, when doing the dual.
answered 4 hours ago
egregegreg
185k1486206
edited 3 hours ago
answered 4 hours ago
egregegreg
185k1486206
answered 4 hours ago
egregegreg
185k1486206
answered 4 hours ago
egregegreg
185k1486206
185k1486206
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbbQ^*$ should be finitely generated and projective as $mathbbZ$-module? If I am not mistaken, an element in $mathbbQ^*$ is uniquely determined by its images over the rationals of the form $frac1p$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
4 hours ago
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
3 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
3 hours ago
add a comment |
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbbQ^*$ should be finitely generated and projective as $mathbbZ$-module? If I am not mistaken, an element in $mathbbQ^*$ is uniquely determined by its images over the rationals of the form $frac1p$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
4 hours ago
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
3 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
3 hours ago
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbbQ^*$ should be finitely generated and projective as $mathbbZ$-module? If I am not mistaken, an element in $mathbbQ^*$ is uniquely determined by its images over the rationals of the form $frac1p$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
4 hours ago
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbbQ^*$ should be finitely generated and projective as $mathbbZ$-module? If I am not mistaken, an element in $mathbbQ^*$ is uniquely determined by its images over the rationals of the form $frac1p$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
4 hours ago
2
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
3 hours ago
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
3 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
3 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
3 hours ago
add a comment |
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