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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    4












    $begingroup$


    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







    share|cite|improve this question









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      4












      4








      4





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked 10 hours ago









      Ender WigginsEnder Wiggins

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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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
            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











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          1 Answer
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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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
            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
















          5












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
            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














          5












          5








          5





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 9 hours ago

























          answered 10 hours ago









          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
            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 $mathbb{Q}^*$ 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
            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 $mathbb{Q}^*$ 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
          9 hours ago




          $begingroup$
          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
          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


















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