If the dual of a module is finitely generated and projective, can we claim that the module itself is?
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Assume that $R$ is a commutative ring and that $M$ is a (left) $R$-module. Assume also that we know for some reason that $M^*:=mathsf{Hom}_R(M,R)$ 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
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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 $M^*:=mathsf{Hom}_R(M,R)$ 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
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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 $M^*:=mathsf{Hom}_R(M,R)$ 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
$endgroup$
Assume that $R$ is a commutative ring and that $M$ is a (left) $R$-module. Assume also that we know for some reason that $M^*:=mathsf{Hom}_R(M,R)$ 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 yesterday
Ender WigginsEnder Wiggins
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1 Answer
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If $operatorname{Hom}_R(M,R)=M^*$ 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 $M=mathbb{Q}$, with $R=mathbb{Z}$. 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.
$endgroup$
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbb{Q}^*$ should be finitely generated and projective as $mathbb{Z}$-module? If I am not mistaken, an element in $mathbb{Q}^*$ is uniquely determined by its images over the rationals of the form $frac{1}{p}$ for $p$ prime, isn't it?
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– Ender Wiggins
yesterday
2
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@EnderWiggins The dual is $0$. Don't confuse notations.
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– egreg
yesterday
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
yesterday
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
If $operatorname{Hom}_R(M,R)=M^*$ 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 $M=mathbb{Q}$, with $R=mathbb{Z}$. 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.
$endgroup$
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbb{Q}^*$ should be finitely generated and projective as $mathbb{Z}$-module? If I am not mistaken, an element in $mathbb{Q}^*$ is uniquely determined by its images over the rationals of the form $frac{1}{p}$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
yesterday
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
yesterday
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
yesterday
add a comment |
$begingroup$
If $operatorname{Hom}_R(M,R)=M^*$ 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 $M=mathbb{Q}$, with $R=mathbb{Z}$. 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.
$endgroup$
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbb{Q}^*$ should be finitely generated and projective as $mathbb{Z}$-module? If I am not mistaken, an element in $mathbb{Q}^*$ is uniquely determined by its images over the rationals of the form $frac{1}{p}$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
yesterday
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
yesterday
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
yesterday
add a comment |
$begingroup$
If $operatorname{Hom}_R(M,R)=M^*$ 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 $M=mathbb{Q}$, with $R=mathbb{Z}$. 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.
$endgroup$
If $operatorname{Hom}_R(M,R)=M^*$ 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 $M=mathbb{Q}$, with $R=mathbb{Z}$. 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.
edited yesterday
answered yesterday
egregegreg
185k1486206
185k1486206
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbb{Q}^*$ should be finitely generated and projective as $mathbb{Z}$-module? If I am not mistaken, an element in $mathbb{Q}^*$ is uniquely determined by its images over the rationals of the form $frac{1}{p}$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
yesterday
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
yesterday
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
yesterday
add a comment |
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbb{Q}^*$ should be finitely generated and projective as $mathbb{Z}$-module? If I am not mistaken, an element in $mathbb{Q}^*$ is uniquely determined by its images over the rationals of the form $frac{1}{p}$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
yesterday
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
yesterday
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
yesterday
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbb{Q}^*$ should be finitely generated and projective as $mathbb{Z}$-module? If I am not mistaken, an element in $mathbb{Q}^*$ is uniquely determined by its images over the rationals of the form $frac{1}{p}$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
yesterday
$begingroup$
Thanks egreg but I am afraid I didn't get your point. Why $mathbb{Q}^*$ should be finitely generated and projective as $mathbb{Z}$-module? If I am not mistaken, an element in $mathbb{Q}^*$ is uniquely determined by its images over the rationals of the form $frac{1}{p}$ for $p$ prime, isn't it?
$endgroup$
– Ender Wiggins
yesterday
2
2
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
yesterday
$begingroup$
@EnderWiggins The dual is $0$. Don't confuse notations.
$endgroup$
– egreg
yesterday
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
yesterday
$begingroup$
Oh, you're right, my bad. I realized it just know. Thanks.
$endgroup$
– Ender Wiggins
yesterday
add a comment |
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