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 10 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 10 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 10 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 10 hours ago
Ender WigginsEnder Wiggins
865421
asked 10 hours ago
Ender WigginsEnder Wiggins
865421
asked 10 hours ago
Ender WigginsEnder Wiggins
865421
asked 10 hours ago
Ender WigginsEnder Wiggins
865421
asked 10 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 9 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
9 hours ago
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
9 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
9 hours ago
add a comment |
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1 Answer
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1 Answer
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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 9 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
9 hours ago
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
9 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
9 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 9 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
9 hours ago
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
9 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
9 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 9 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 9 hours ago
egregegreg
185k1486206
edited 9 hours ago
answered 9 hours ago
egregegreg
185k1486206
answered 9 hours ago
egregegreg
185k1486206
answered 9 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
9 hours ago
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
9 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
9 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
9 hours ago
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
9 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
9 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
9 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
9 hours ago
2
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
9 hours ago
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
9 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
9 hours ago
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
9 hours ago
add a comment |
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