Free generators for the localization of a module












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Let $A$ a commutative ring with 1 and $M$ a finitely generated $A$-module. Assume the localization $M_p$ is a free $A_p$-module for some prime ideal $p$. Is it true that there is some set of generators of $M$ whose image in $M_p$ freely generate $M_p$? Feel free to assume $A$ is Noetherian.



Fancy way to ask the same question: is there a surjective map $A^{oplus n} to M$ which induces an isomorphism $A_p^{oplus n} to M_p$?



The reason I'm asking this is: I was able to reduce some Exercise in Hartshorne (II.5.7) to this statement, and all solutions I've found online treat this step as it's obvious but I just don't see it.






algebraic-geometry commutative-algebra







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$endgroup$

















    3












    $begingroup$


    Let $A$ a commutative ring with 1 and $M$ a finitely generated $A$-module. Assume the localization $M_p$ is a free $A_p$-module for some prime ideal $p$. Is it true that there is some set of generators of $M$ whose image in $M_p$ freely generate $M_p$? Feel free to assume $A$ is Noetherian.



    Fancy way to ask the same question: is there a surjective map $A^{oplus n} to M$ which induces an isomorphism $A_p^{oplus n} to M_p$?



    The reason I'm asking this is: I was able to reduce some Exercise in Hartshorne (II.5.7) to this statement, and all solutions I've found online treat this step as it's obvious but I just don't see it.






    algebraic-geometry commutative-algebra







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Let $A$ a commutative ring with 1 and $M$ a finitely generated $A$-module. Assume the localization $M_p$ is a free $A_p$-module for some prime ideal $p$. Is it true that there is some set of generators of $M$ whose image in $M_p$ freely generate $M_p$? Feel free to assume $A$ is Noetherian.



      Fancy way to ask the same question: is there a surjective map $A^{oplus n} to M$ which induces an isomorphism $A_p^{oplus n} to M_p$?



      The reason I'm asking this is: I was able to reduce some Exercise in Hartshorne (II.5.7) to this statement, and all solutions I've found online treat this step as it's obvious but I just don't see it.






      algebraic-geometry commutative-algebra







      share|cite|improve this question









      $endgroup$




      Let $A$ a commutative ring with 1 and $M$ a finitely generated $A$-module. Assume the localization $M_p$ is a free $A_p$-module for some prime ideal $p$. Is it true that there is some set of generators of $M$ whose image in $M_p$ freely generate $M_p$? Feel free to assume $A$ is Noetherian.



      Fancy way to ask the same question: is there a surjective map $A^{oplus n} to M$ which induces an isomorphism $A_p^{oplus n} to M_p$?



      The reason I'm asking this is: I was able to reduce some Exercise in Hartshorne (II.5.7) to this statement, and all solutions I've found online treat this step as it's obvious but I just don't see it.






      algebraic-geometry commutative-algebra




      algebraic-geometry commutative-algebra






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      asked Jan 21 at 21:16









      Marco FloresMarco Flores

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          No. For instance, if $p$ and $q$ are two different maximal ideals of $A$ and $M=A/q$, then $M_p=0$ so we would have to have $n=0$ but no map $A^0to M$ can be surjective.



          This result is not needed to prove Exercise II.5.7, though, since you only need such a map $A^nto M$ locally (i.e., after localizing at some element $fin Asetminus p$.). As a sketch of a proof, you can take a basis of $M_p$ and lift each element of this basis to $M_f$ for some $fin Asetminus p$. This gives a homomorphism $A_f^nto M_f$ which is an isomorphism after localizing at $p$. To make this homomorphism surjective, note that its cokernel is a finitely generated module whose localization at $p$ is $0$, so you can find some $gin Asetminus p$ which annihilates the cokernel, and then the map $A_{fg}^nto M_{fg}$ obtained by also localizing at $g$ will be surjective. (Assuming $A$ is Noetherian, you can also do the same thing for the kernel and find a localization at an element of $Asetminus p$ which makes the map an isomorphism.)






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            $begingroup$

            No. For instance, if $p$ and $q$ are two different maximal ideals of $A$ and $M=A/q$, then $M_p=0$ so we would have to have $n=0$ but no map $A^0to M$ can be surjective.



            This result is not needed to prove Exercise II.5.7, though, since you only need such a map $A^nto M$ locally (i.e., after localizing at some element $fin Asetminus p$.). As a sketch of a proof, you can take a basis of $M_p$ and lift each element of this basis to $M_f$ for some $fin Asetminus p$. This gives a homomorphism $A_f^nto M_f$ which is an isomorphism after localizing at $p$. To make this homomorphism surjective, note that its cokernel is a finitely generated module whose localization at $p$ is $0$, so you can find some $gin Asetminus p$ which annihilates the cokernel, and then the map $A_{fg}^nto M_{fg}$ obtained by also localizing at $g$ will be surjective. (Assuming $A$ is Noetherian, you can also do the same thing for the kernel and find a localization at an element of $Asetminus p$ which makes the map an isomorphism.)






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              $begingroup$

              No. For instance, if $p$ and $q$ are two different maximal ideals of $A$ and $M=A/q$, then $M_p=0$ so we would have to have $n=0$ but no map $A^0to M$ can be surjective.



              This result is not needed to prove Exercise II.5.7, though, since you only need such a map $A^nto M$ locally (i.e., after localizing at some element $fin Asetminus p$.). As a sketch of a proof, you can take a basis of $M_p$ and lift each element of this basis to $M_f$ for some $fin Asetminus p$. This gives a homomorphism $A_f^nto M_f$ which is an isomorphism after localizing at $p$. To make this homomorphism surjective, note that its cokernel is a finitely generated module whose localization at $p$ is $0$, so you can find some $gin Asetminus p$ which annihilates the cokernel, and then the map $A_{fg}^nto M_{fg}$ obtained by also localizing at $g$ will be surjective. (Assuming $A$ is Noetherian, you can also do the same thing for the kernel and find a localization at an element of $Asetminus p$ which makes the map an isomorphism.)






              share|cite|improve this answer









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                1












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                $begingroup$

                No. For instance, if $p$ and $q$ are two different maximal ideals of $A$ and $M=A/q$, then $M_p=0$ so we would have to have $n=0$ but no map $A^0to M$ can be surjective.



                This result is not needed to prove Exercise II.5.7, though, since you only need such a map $A^nto M$ locally (i.e., after localizing at some element $fin Asetminus p$.). As a sketch of a proof, you can take a basis of $M_p$ and lift each element of this basis to $M_f$ for some $fin Asetminus p$. This gives a homomorphism $A_f^nto M_f$ which is an isomorphism after localizing at $p$. To make this homomorphism surjective, note that its cokernel is a finitely generated module whose localization at $p$ is $0$, so you can find some $gin Asetminus p$ which annihilates the cokernel, and then the map $A_{fg}^nto M_{fg}$ obtained by also localizing at $g$ will be surjective. (Assuming $A$ is Noetherian, you can also do the same thing for the kernel and find a localization at an element of $Asetminus p$ which makes the map an isomorphism.)






                share|cite|improve this answer









                $endgroup$



                No. For instance, if $p$ and $q$ are two different maximal ideals of $A$ and $M=A/q$, then $M_p=0$ so we would have to have $n=0$ but no map $A^0to M$ can be surjective.



                This result is not needed to prove Exercise II.5.7, though, since you only need such a map $A^nto M$ locally (i.e., after localizing at some element $fin Asetminus p$.). As a sketch of a proof, you can take a basis of $M_p$ and lift each element of this basis to $M_f$ for some $fin Asetminus p$. This gives a homomorphism $A_f^nto M_f$ which is an isomorphism after localizing at $p$. To make this homomorphism surjective, note that its cokernel is a finitely generated module whose localization at $p$ is $0$, so you can find some $gin Asetminus p$ which annihilates the cokernel, and then the map $A_{fg}^nto M_{fg}$ obtained by also localizing at $g$ will be surjective. (Assuming $A$ is Noetherian, you can also do the same thing for the kernel and find a localization at an element of $Asetminus p$ which makes the map an isomorphism.)







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Jan 21 at 22:05









                Eric WofseyEric Wofsey

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                188k14216346






























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