Finitely generated projective modules are locally free












14












$begingroup$


Let $A$ be a commutative noetherian ring, and let $M$ be a finitely generated projective $A$-module. It is well known and easy to prove that $A$ is locally free in the sense that for every $p inoperatorname{Spec} A$, the module $M_p$ is a free $A_p$-module.



Is it true that projectives are also locally free in the following (more geometric?) sense:




There are elements $f_1,dots,f_n in A$ such that $(f_1,dots,f_n) = 1$, and such that $M_{f_i}$ is a free $A_{f_i}$-module for all $1le i le n$.




Is this true? if so, can you provide a reference or explain how to prove it?



Thanks!






commutative-algebra projective-module







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$endgroup$

















    14












    $begingroup$


    Let $A$ be a commutative noetherian ring, and let $M$ be a finitely generated projective $A$-module. It is well known and easy to prove that $A$ is locally free in the sense that for every $p inoperatorname{Spec} A$, the module $M_p$ is a free $A_p$-module.



    Is it true that projectives are also locally free in the following (more geometric?) sense:




    There are elements $f_1,dots,f_n in A$ such that $(f_1,dots,f_n) = 1$, and such that $M_{f_i}$ is a free $A_{f_i}$-module for all $1le i le n$.




    Is this true? if so, can you provide a reference or explain how to prove it?



    Thanks!






    commutative-algebra projective-module







    share|cite|improve this question











    $endgroup$















      14












      14








      14


      10



      $begingroup$


      Let $A$ be a commutative noetherian ring, and let $M$ be a finitely generated projective $A$-module. It is well known and easy to prove that $A$ is locally free in the sense that for every $p inoperatorname{Spec} A$, the module $M_p$ is a free $A_p$-module.



      Is it true that projectives are also locally free in the following (more geometric?) sense:




      There are elements $f_1,dots,f_n in A$ such that $(f_1,dots,f_n) = 1$, and such that $M_{f_i}$ is a free $A_{f_i}$-module for all $1le i le n$.




      Is this true? if so, can you provide a reference or explain how to prove it?



      Thanks!






      commutative-algebra projective-module







      share|cite|improve this question











      $endgroup$




      Let $A$ be a commutative noetherian ring, and let $M$ be a finitely generated projective $A$-module. It is well known and easy to prove that $A$ is locally free in the sense that for every $p inoperatorname{Spec} A$, the module $M_p$ is a free $A_p$-module.



      Is it true that projectives are also locally free in the following (more geometric?) sense:




      There are elements $f_1,dots,f_n in A$ such that $(f_1,dots,f_n) = 1$, and such that $M_{f_i}$ is a free $A_{f_i}$-module for all $1le i le n$.




      Is this true? if so, can you provide a reference or explain how to prove it?



      Thanks!






      commutative-algebra projective-module




      commutative-algebra projective-module






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 20 '15 at 16:06









      user26857

      39.3k124183




      39.3k124183










      asked Jan 8 '11 at 16:02









      the Lthe L

      2,52812150




      2,52812150






















          3 Answers
          3






          active

          oldest

          votes


















          11












          $begingroup$

          Yes, this is true. See this Math Overflow question for a precise statement and a reference to its proof in Bourbaki's Commutative Algebra.



          This result is also stated in my commutative algebra notes, but the proof is not unfortunately not yet written up there. I certainly hope that this will be remedied soon though, as I will be teaching a course out of these notes starting on Monday. When the proof gets written, I will update this answer with a page number.



          Added: Here is something in the MO answer that I decided was worth a comment here. For finitely generated modules, this stronger version of local freeness is actually equivalent to projectivity, whereas the weaker "pointwise local freeness" is subtly weaker in general.






          share|cite|improve this answer











          $endgroup$









          • 4




            $begingroup$
            Thank you for the reference. For the intrested reader, I should mention that Bourbaki proves the following much stronger result: If $M$ is a finitely presented module and if for some prime $p$, the module $M_p$ is free, then there exists a neighborhood $D(f)$ of $p$ such that $M_f$ is free over $A_f$, so that "being free at a point" implies "being free at a neigborhood of a point".
            $endgroup$
            – the L
            Jan 8 '11 at 16:51






          • 4




            $begingroup$
            @anonymous: Dear anonymous, more generally, if $M,N$ are finitely presented modules and $M_p, N_p$ are isomorphic, then $M_f simeq N_f$ for some $f in A-p$. This is explained in more detail (and generality) in EGA IV-8, where it is proved that (in a certain sense) if ${R_alpha}$ is an inductive system of rings, then the category of f.p. modules over $varinjlim R_alpha$ is the "colimit of the categories of modules over $R_alpha$," which has many useful applications in reducing questions about arbitrary f.p. modules to questions about finitely generated modules over noetherian rings.
            $endgroup$
            – Akhil Mathew
            Jan 15 '11 at 18:40






          • 2




            $begingroup$
            (contd.) For instance, Grothendieck's generic flatness lemma (if $A$ noetherian integral domain, $B$ finite $A$-algebra, $M$ finite $B$-module, then there is $f in A$ such that $M_f$ is flat over $A$) works for finitely presented modules because any such module must "descend" to a finitely presented module over a noetherian (e.g. finitely generated) subring.
            $endgroup$
            – Akhil Mathew
            Jan 15 '11 at 18:41



















          4












          $begingroup$

          For future reference, I have written up a constructive and reasonably self-contained, if somewhat dense, proof (one page). The basic idea is to first verify that idempotent matrices over local rings are equivalent to diagonal matrices with entries $1$ and $0$, thus showing that finitely generated projective modules over local rings are free.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Well, I think there are simpler proofs for finitely generated projective modules over local rings are free. Never mind, +1.
            $endgroup$
            – user26857
            May 8 '15 at 16:40












          • $begingroup$
            @user26857: I would be interested to hear them! Are they constructive in the technical sense that they avoid the use of the law of excluded middle? I only know of my write-up and the one in Mulvey's gem Intuitionistic algebra and representations of rings.
            $endgroup$
            – Ingo Blechschmidt
            May 8 '15 at 16:54










          • $begingroup$
            math.stackexchange.com/questions/147754/…
            $endgroup$
            – user26857
            May 8 '15 at 17:15










          • $begingroup$
            @user26857: Thanks! The problem is that a proof beginning with "Consider a minimal system of generators" is not constructive, since constructively one cannot show that any inhabited subset of the natural numbers contains a minimal element. In some cases though, the problem can be avoided by a clever rewording; I didn't try that in this case hard enough yet.
            $endgroup$
            – Ingo Blechschmidt
            May 8 '15 at 18:03



















          0












          $begingroup$

          This answer shows that if its stalk $M_mathfrak{p}$ at $mathfrak{p}$ is free, then there is an open neighbourhood $mathfrak{p}in D(f)$ on which its value $M_f$ is also free.



          Proof: Reduce to the case that the natural map $Mto M_mathfrak{p}$ is an injection by localising some $f_0in A$. Now use the basis of $M_mathfrak{p}$ to give a surjection $alpha:A^nto M$ whose localisation at $mathfrak{p}$ is the isomorphism $A^n_mathfrak{p}to M_mathfrak{p}$. Thus $ker alpha_mathfrak{p}=0$, so $ker alpha_f=0$ for some $fin A$. Thus $M_fsimeq A^n_f$ is free, completing the proof.



          Thus the quasicompact space $text{Spec}A$ admits a finite cover by $D(f_i)$'s, on each of which $M_{f_i}$ is free. $text{Spec}A=cup D(f_i)$ means precisely that $(f_1,...,f_n)=A$.






          share|cite|improve this answer











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            3 Answers
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            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            11












            $begingroup$

            Yes, this is true. See this Math Overflow question for a precise statement and a reference to its proof in Bourbaki's Commutative Algebra.



            This result is also stated in my commutative algebra notes, but the proof is not unfortunately not yet written up there. I certainly hope that this will be remedied soon though, as I will be teaching a course out of these notes starting on Monday. When the proof gets written, I will update this answer with a page number.



            Added: Here is something in the MO answer that I decided was worth a comment here. For finitely generated modules, this stronger version of local freeness is actually equivalent to projectivity, whereas the weaker "pointwise local freeness" is subtly weaker in general.






            share|cite|improve this answer











            $endgroup$









            • 4




              $begingroup$
              Thank you for the reference. For the intrested reader, I should mention that Bourbaki proves the following much stronger result: If $M$ is a finitely presented module and if for some prime $p$, the module $M_p$ is free, then there exists a neighborhood $D(f)$ of $p$ such that $M_f$ is free over $A_f$, so that "being free at a point" implies "being free at a neigborhood of a point".
              $endgroup$
              – the L
              Jan 8 '11 at 16:51






            • 4




              $begingroup$
              @anonymous: Dear anonymous, more generally, if $M,N$ are finitely presented modules and $M_p, N_p$ are isomorphic, then $M_f simeq N_f$ for some $f in A-p$. This is explained in more detail (and generality) in EGA IV-8, where it is proved that (in a certain sense) if ${R_alpha}$ is an inductive system of rings, then the category of f.p. modules over $varinjlim R_alpha$ is the "colimit of the categories of modules over $R_alpha$," which has many useful applications in reducing questions about arbitrary f.p. modules to questions about finitely generated modules over noetherian rings.
              $endgroup$
              – Akhil Mathew
              Jan 15 '11 at 18:40






            • 2




              $begingroup$
              (contd.) For instance, Grothendieck's generic flatness lemma (if $A$ noetherian integral domain, $B$ finite $A$-algebra, $M$ finite $B$-module, then there is $f in A$ such that $M_f$ is flat over $A$) works for finitely presented modules because any such module must "descend" to a finitely presented module over a noetherian (e.g. finitely generated) subring.
              $endgroup$
              – Akhil Mathew
              Jan 15 '11 at 18:41
















            11












            $begingroup$

            Yes, this is true. See this Math Overflow question for a precise statement and a reference to its proof in Bourbaki's Commutative Algebra.



            This result is also stated in my commutative algebra notes, but the proof is not unfortunately not yet written up there. I certainly hope that this will be remedied soon though, as I will be teaching a course out of these notes starting on Monday. When the proof gets written, I will update this answer with a page number.



            Added: Here is something in the MO answer that I decided was worth a comment here. For finitely generated modules, this stronger version of local freeness is actually equivalent to projectivity, whereas the weaker "pointwise local freeness" is subtly weaker in general.






            share|cite|improve this answer











            $endgroup$









            • 4




              $begingroup$
              Thank you for the reference. For the intrested reader, I should mention that Bourbaki proves the following much stronger result: If $M$ is a finitely presented module and if for some prime $p$, the module $M_p$ is free, then there exists a neighborhood $D(f)$ of $p$ such that $M_f$ is free over $A_f$, so that "being free at a point" implies "being free at a neigborhood of a point".
              $endgroup$
              – the L
              Jan 8 '11 at 16:51






            • 4




              $begingroup$
              @anonymous: Dear anonymous, more generally, if $M,N$ are finitely presented modules and $M_p, N_p$ are isomorphic, then $M_f simeq N_f$ for some $f in A-p$. This is explained in more detail (and generality) in EGA IV-8, where it is proved that (in a certain sense) if ${R_alpha}$ is an inductive system of rings, then the category of f.p. modules over $varinjlim R_alpha$ is the "colimit of the categories of modules over $R_alpha$," which has many useful applications in reducing questions about arbitrary f.p. modules to questions about finitely generated modules over noetherian rings.
              $endgroup$
              – Akhil Mathew
              Jan 15 '11 at 18:40






            • 2




              $begingroup$
              (contd.) For instance, Grothendieck's generic flatness lemma (if $A$ noetherian integral domain, $B$ finite $A$-algebra, $M$ finite $B$-module, then there is $f in A$ such that $M_f$ is flat over $A$) works for finitely presented modules because any such module must "descend" to a finitely presented module over a noetherian (e.g. finitely generated) subring.
              $endgroup$
              – Akhil Mathew
              Jan 15 '11 at 18:41














            11












            11








            11





            $begingroup$

            Yes, this is true. See this Math Overflow question for a precise statement and a reference to its proof in Bourbaki's Commutative Algebra.



            This result is also stated in my commutative algebra notes, but the proof is not unfortunately not yet written up there. I certainly hope that this will be remedied soon though, as I will be teaching a course out of these notes starting on Monday. When the proof gets written, I will update this answer with a page number.



            Added: Here is something in the MO answer that I decided was worth a comment here. For finitely generated modules, this stronger version of local freeness is actually equivalent to projectivity, whereas the weaker "pointwise local freeness" is subtly weaker in general.






            share|cite|improve this answer











            $endgroup$



            Yes, this is true. See this Math Overflow question for a precise statement and a reference to its proof in Bourbaki's Commutative Algebra.



            This result is also stated in my commutative algebra notes, but the proof is not unfortunately not yet written up there. I certainly hope that this will be remedied soon though, as I will be teaching a course out of these notes starting on Monday. When the proof gets written, I will update this answer with a page number.



            Added: Here is something in the MO answer that I decided was worth a comment here. For finitely generated modules, this stronger version of local freeness is actually equivalent to projectivity, whereas the weaker "pointwise local freeness" is subtly weaker in general.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 13 '17 at 12:58









            Community♦

            1




            1










            answered Jan 8 '11 at 16:32









            Pete L. ClarkPete L. Clark

            80.6k9161312




            80.6k9161312








            • 4




              $begingroup$
              Thank you for the reference. For the intrested reader, I should mention that Bourbaki proves the following much stronger result: If $M$ is a finitely presented module and if for some prime $p$, the module $M_p$ is free, then there exists a neighborhood $D(f)$ of $p$ such that $M_f$ is free over $A_f$, so that "being free at a point" implies "being free at a neigborhood of a point".
              $endgroup$
              – the L
              Jan 8 '11 at 16:51






            • 4




              $begingroup$
              @anonymous: Dear anonymous, more generally, if $M,N$ are finitely presented modules and $M_p, N_p$ are isomorphic, then $M_f simeq N_f$ for some $f in A-p$. This is explained in more detail (and generality) in EGA IV-8, where it is proved that (in a certain sense) if ${R_alpha}$ is an inductive system of rings, then the category of f.p. modules over $varinjlim R_alpha$ is the "colimit of the categories of modules over $R_alpha$," which has many useful applications in reducing questions about arbitrary f.p. modules to questions about finitely generated modules over noetherian rings.
              $endgroup$
              – Akhil Mathew
              Jan 15 '11 at 18:40






            • 2




              $begingroup$
              (contd.) For instance, Grothendieck's generic flatness lemma (if $A$ noetherian integral domain, $B$ finite $A$-algebra, $M$ finite $B$-module, then there is $f in A$ such that $M_f$ is flat over $A$) works for finitely presented modules because any such module must "descend" to a finitely presented module over a noetherian (e.g. finitely generated) subring.
              $endgroup$
              – Akhil Mathew
              Jan 15 '11 at 18:41














            • 4




              $begingroup$
              Thank you for the reference. For the intrested reader, I should mention that Bourbaki proves the following much stronger result: If $M$ is a finitely presented module and if for some prime $p$, the module $M_p$ is free, then there exists a neighborhood $D(f)$ of $p$ such that $M_f$ is free over $A_f$, so that "being free at a point" implies "being free at a neigborhood of a point".
              $endgroup$
              – the L
              Jan 8 '11 at 16:51






            • 4




              $begingroup$
              @anonymous: Dear anonymous, more generally, if $M,N$ are finitely presented modules and $M_p, N_p$ are isomorphic, then $M_f simeq N_f$ for some $f in A-p$. This is explained in more detail (and generality) in EGA IV-8, where it is proved that (in a certain sense) if ${R_alpha}$ is an inductive system of rings, then the category of f.p. modules over $varinjlim R_alpha$ is the "colimit of the categories of modules over $R_alpha$," which has many useful applications in reducing questions about arbitrary f.p. modules to questions about finitely generated modules over noetherian rings.
              $endgroup$
              – Akhil Mathew
              Jan 15 '11 at 18:40






            • 2




              $begingroup$
              (contd.) For instance, Grothendieck's generic flatness lemma (if $A$ noetherian integral domain, $B$ finite $A$-algebra, $M$ finite $B$-module, then there is $f in A$ such that $M_f$ is flat over $A$) works for finitely presented modules because any such module must "descend" to a finitely presented module over a noetherian (e.g. finitely generated) subring.
              $endgroup$
              – Akhil Mathew
              Jan 15 '11 at 18:41








            4




            4




            $begingroup$
            Thank you for the reference. For the intrested reader, I should mention that Bourbaki proves the following much stronger result: If $M$ is a finitely presented module and if for some prime $p$, the module $M_p$ is free, then there exists a neighborhood $D(f)$ of $p$ such that $M_f$ is free over $A_f$, so that "being free at a point" implies "being free at a neigborhood of a point".
            $endgroup$
            – the L
            Jan 8 '11 at 16:51




            $begingroup$
            Thank you for the reference. For the intrested reader, I should mention that Bourbaki proves the following much stronger result: If $M$ is a finitely presented module and if for some prime $p$, the module $M_p$ is free, then there exists a neighborhood $D(f)$ of $p$ such that $M_f$ is free over $A_f$, so that "being free at a point" implies "being free at a neigborhood of a point".
            $endgroup$
            – the L
            Jan 8 '11 at 16:51




            4




            4




            $begingroup$
            @anonymous: Dear anonymous, more generally, if $M,N$ are finitely presented modules and $M_p, N_p$ are isomorphic, then $M_f simeq N_f$ for some $f in A-p$. This is explained in more detail (and generality) in EGA IV-8, where it is proved that (in a certain sense) if ${R_alpha}$ is an inductive system of rings, then the category of f.p. modules over $varinjlim R_alpha$ is the "colimit of the categories of modules over $R_alpha$," which has many useful applications in reducing questions about arbitrary f.p. modules to questions about finitely generated modules over noetherian rings.
            $endgroup$
            – Akhil Mathew
            Jan 15 '11 at 18:40




            $begingroup$
            @anonymous: Dear anonymous, more generally, if $M,N$ are finitely presented modules and $M_p, N_p$ are isomorphic, then $M_f simeq N_f$ for some $f in A-p$. This is explained in more detail (and generality) in EGA IV-8, where it is proved that (in a certain sense) if ${R_alpha}$ is an inductive system of rings, then the category of f.p. modules over $varinjlim R_alpha$ is the "colimit of the categories of modules over $R_alpha$," which has many useful applications in reducing questions about arbitrary f.p. modules to questions about finitely generated modules over noetherian rings.
            $endgroup$
            – Akhil Mathew
            Jan 15 '11 at 18:40




            2




            2




            $begingroup$
            (contd.) For instance, Grothendieck's generic flatness lemma (if $A$ noetherian integral domain, $B$ finite $A$-algebra, $M$ finite $B$-module, then there is $f in A$ such that $M_f$ is flat over $A$) works for finitely presented modules because any such module must "descend" to a finitely presented module over a noetherian (e.g. finitely generated) subring.
            $endgroup$
            – Akhil Mathew
            Jan 15 '11 at 18:41




            $begingroup$
            (contd.) For instance, Grothendieck's generic flatness lemma (if $A$ noetherian integral domain, $B$ finite $A$-algebra, $M$ finite $B$-module, then there is $f in A$ such that $M_f$ is flat over $A$) works for finitely presented modules because any such module must "descend" to a finitely presented module over a noetherian (e.g. finitely generated) subring.
            $endgroup$
            – Akhil Mathew
            Jan 15 '11 at 18:41











            4












            $begingroup$

            For future reference, I have written up a constructive and reasonably self-contained, if somewhat dense, proof (one page). The basic idea is to first verify that idempotent matrices over local rings are equivalent to diagonal matrices with entries $1$ and $0$, thus showing that finitely generated projective modules over local rings are free.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Well, I think there are simpler proofs for finitely generated projective modules over local rings are free. Never mind, +1.
              $endgroup$
              – user26857
              May 8 '15 at 16:40












            • $begingroup$
              @user26857: I would be interested to hear them! Are they constructive in the technical sense that they avoid the use of the law of excluded middle? I only know of my write-up and the one in Mulvey's gem Intuitionistic algebra and representations of rings.
              $endgroup$
              – Ingo Blechschmidt
              May 8 '15 at 16:54










            • $begingroup$
              math.stackexchange.com/questions/147754/…
              $endgroup$
              – user26857
              May 8 '15 at 17:15










            • $begingroup$
              @user26857: Thanks! The problem is that a proof beginning with "Consider a minimal system of generators" is not constructive, since constructively one cannot show that any inhabited subset of the natural numbers contains a minimal element. In some cases though, the problem can be avoided by a clever rewording; I didn't try that in this case hard enough yet.
              $endgroup$
              – Ingo Blechschmidt
              May 8 '15 at 18:03
















            4












            $begingroup$

            For future reference, I have written up a constructive and reasonably self-contained, if somewhat dense, proof (one page). The basic idea is to first verify that idempotent matrices over local rings are equivalent to diagonal matrices with entries $1$ and $0$, thus showing that finitely generated projective modules over local rings are free.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Well, I think there are simpler proofs for finitely generated projective modules over local rings are free. Never mind, +1.
              $endgroup$
              – user26857
              May 8 '15 at 16:40












            • $begingroup$
              @user26857: I would be interested to hear them! Are they constructive in the technical sense that they avoid the use of the law of excluded middle? I only know of my write-up and the one in Mulvey's gem Intuitionistic algebra and representations of rings.
              $endgroup$
              – Ingo Blechschmidt
              May 8 '15 at 16:54










            • $begingroup$
              math.stackexchange.com/questions/147754/…
              $endgroup$
              – user26857
              May 8 '15 at 17:15










            • $begingroup$
              @user26857: Thanks! The problem is that a proof beginning with "Consider a minimal system of generators" is not constructive, since constructively one cannot show that any inhabited subset of the natural numbers contains a minimal element. In some cases though, the problem can be avoided by a clever rewording; I didn't try that in this case hard enough yet.
              $endgroup$
              – Ingo Blechschmidt
              May 8 '15 at 18:03














            4












            4








            4





            $begingroup$

            For future reference, I have written up a constructive and reasonably self-contained, if somewhat dense, proof (one page). The basic idea is to first verify that idempotent matrices over local rings are equivalent to diagonal matrices with entries $1$ and $0$, thus showing that finitely generated projective modules over local rings are free.






            share|cite|improve this answer











            $endgroup$



            For future reference, I have written up a constructive and reasonably self-contained, if somewhat dense, proof (one page). The basic idea is to first verify that idempotent matrices over local rings are equivalent to diagonal matrices with entries $1$ and $0$, thus showing that finitely generated projective modules over local rings are free.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited May 8 '15 at 15:35

























            answered May 8 '15 at 15:01









            Ingo BlechschmidtIngo Blechschmidt

            1,385815




            1,385815












            • $begingroup$
              Well, I think there are simpler proofs for finitely generated projective modules over local rings are free. Never mind, +1.
              $endgroup$
              – user26857
              May 8 '15 at 16:40












            • $begingroup$
              @user26857: I would be interested to hear them! Are they constructive in the technical sense that they avoid the use of the law of excluded middle? I only know of my write-up and the one in Mulvey's gem Intuitionistic algebra and representations of rings.
              $endgroup$
              – Ingo Blechschmidt
              May 8 '15 at 16:54










            • $begingroup$
              math.stackexchange.com/questions/147754/…
              $endgroup$
              – user26857
              May 8 '15 at 17:15










            • $begingroup$
              @user26857: Thanks! The problem is that a proof beginning with "Consider a minimal system of generators" is not constructive, since constructively one cannot show that any inhabited subset of the natural numbers contains a minimal element. In some cases though, the problem can be avoided by a clever rewording; I didn't try that in this case hard enough yet.
              $endgroup$
              – Ingo Blechschmidt
              May 8 '15 at 18:03


















            • $begingroup$
              Well, I think there are simpler proofs for finitely generated projective modules over local rings are free. Never mind, +1.
              $endgroup$
              – user26857
              May 8 '15 at 16:40












            • $begingroup$
              @user26857: I would be interested to hear them! Are they constructive in the technical sense that they avoid the use of the law of excluded middle? I only know of my write-up and the one in Mulvey's gem Intuitionistic algebra and representations of rings.
              $endgroup$
              – Ingo Blechschmidt
              May 8 '15 at 16:54










            • $begingroup$
              math.stackexchange.com/questions/147754/…
              $endgroup$
              – user26857
              May 8 '15 at 17:15










            • $begingroup$
              @user26857: Thanks! The problem is that a proof beginning with "Consider a minimal system of generators" is not constructive, since constructively one cannot show that any inhabited subset of the natural numbers contains a minimal element. In some cases though, the problem can be avoided by a clever rewording; I didn't try that in this case hard enough yet.
              $endgroup$
              – Ingo Blechschmidt
              May 8 '15 at 18:03
















            $begingroup$
            Well, I think there are simpler proofs for finitely generated projective modules over local rings are free. Never mind, +1.
            $endgroup$
            – user26857
            May 8 '15 at 16:40






            $begingroup$
            Well, I think there are simpler proofs for finitely generated projective modules over local rings are free. Never mind, +1.
            $endgroup$
            – user26857
            May 8 '15 at 16:40














            $begingroup$
            @user26857: I would be interested to hear them! Are they constructive in the technical sense that they avoid the use of the law of excluded middle? I only know of my write-up and the one in Mulvey's gem Intuitionistic algebra and representations of rings.
            $endgroup$
            – Ingo Blechschmidt
            May 8 '15 at 16:54




            $begingroup$
            @user26857: I would be interested to hear them! Are they constructive in the technical sense that they avoid the use of the law of excluded middle? I only know of my write-up and the one in Mulvey's gem Intuitionistic algebra and representations of rings.
            $endgroup$
            – Ingo Blechschmidt
            May 8 '15 at 16:54












            $begingroup$
            math.stackexchange.com/questions/147754/…
            $endgroup$
            – user26857
            May 8 '15 at 17:15




            $begingroup$
            math.stackexchange.com/questions/147754/…
            $endgroup$
            – user26857
            May 8 '15 at 17:15












            $begingroup$
            @user26857: Thanks! The problem is that a proof beginning with "Consider a minimal system of generators" is not constructive, since constructively one cannot show that any inhabited subset of the natural numbers contains a minimal element. In some cases though, the problem can be avoided by a clever rewording; I didn't try that in this case hard enough yet.
            $endgroup$
            – Ingo Blechschmidt
            May 8 '15 at 18:03




            $begingroup$
            @user26857: Thanks! The problem is that a proof beginning with "Consider a minimal system of generators" is not constructive, since constructively one cannot show that any inhabited subset of the natural numbers contains a minimal element. In some cases though, the problem can be avoided by a clever rewording; I didn't try that in this case hard enough yet.
            $endgroup$
            – Ingo Blechschmidt
            May 8 '15 at 18:03











            0












            $begingroup$

            This answer shows that if its stalk $M_mathfrak{p}$ at $mathfrak{p}$ is free, then there is an open neighbourhood $mathfrak{p}in D(f)$ on which its value $M_f$ is also free.



            Proof: Reduce to the case that the natural map $Mto M_mathfrak{p}$ is an injection by localising some $f_0in A$. Now use the basis of $M_mathfrak{p}$ to give a surjection $alpha:A^nto M$ whose localisation at $mathfrak{p}$ is the isomorphism $A^n_mathfrak{p}to M_mathfrak{p}$. Thus $ker alpha_mathfrak{p}=0$, so $ker alpha_f=0$ for some $fin A$. Thus $M_fsimeq A^n_f$ is free, completing the proof.



            Thus the quasicompact space $text{Spec}A$ admits a finite cover by $D(f_i)$'s, on each of which $M_{f_i}$ is free. $text{Spec}A=cup D(f_i)$ means precisely that $(f_1,...,f_n)=A$.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              This answer shows that if its stalk $M_mathfrak{p}$ at $mathfrak{p}$ is free, then there is an open neighbourhood $mathfrak{p}in D(f)$ on which its value $M_f$ is also free.



              Proof: Reduce to the case that the natural map $Mto M_mathfrak{p}$ is an injection by localising some $f_0in A$. Now use the basis of $M_mathfrak{p}$ to give a surjection $alpha:A^nto M$ whose localisation at $mathfrak{p}$ is the isomorphism $A^n_mathfrak{p}to M_mathfrak{p}$. Thus $ker alpha_mathfrak{p}=0$, so $ker alpha_f=0$ for some $fin A$. Thus $M_fsimeq A^n_f$ is free, completing the proof.



              Thus the quasicompact space $text{Spec}A$ admits a finite cover by $D(f_i)$'s, on each of which $M_{f_i}$ is free. $text{Spec}A=cup D(f_i)$ means precisely that $(f_1,...,f_n)=A$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                This answer shows that if its stalk $M_mathfrak{p}$ at $mathfrak{p}$ is free, then there is an open neighbourhood $mathfrak{p}in D(f)$ on which its value $M_f$ is also free.



                Proof: Reduce to the case that the natural map $Mto M_mathfrak{p}$ is an injection by localising some $f_0in A$. Now use the basis of $M_mathfrak{p}$ to give a surjection $alpha:A^nto M$ whose localisation at $mathfrak{p}$ is the isomorphism $A^n_mathfrak{p}to M_mathfrak{p}$. Thus $ker alpha_mathfrak{p}=0$, so $ker alpha_f=0$ for some $fin A$. Thus $M_fsimeq A^n_f$ is free, completing the proof.



                Thus the quasicompact space $text{Spec}A$ admits a finite cover by $D(f_i)$'s, on each of which $M_{f_i}$ is free. $text{Spec}A=cup D(f_i)$ means precisely that $(f_1,...,f_n)=A$.






                share|cite|improve this answer











                $endgroup$



                This answer shows that if its stalk $M_mathfrak{p}$ at $mathfrak{p}$ is free, then there is an open neighbourhood $mathfrak{p}in D(f)$ on which its value $M_f$ is also free.



                Proof: Reduce to the case that the natural map $Mto M_mathfrak{p}$ is an injection by localising some $f_0in A$. Now use the basis of $M_mathfrak{p}$ to give a surjection $alpha:A^nto M$ whose localisation at $mathfrak{p}$ is the isomorphism $A^n_mathfrak{p}to M_mathfrak{p}$. Thus $ker alpha_mathfrak{p}=0$, so $ker alpha_f=0$ for some $fin A$. Thus $M_fsimeq A^n_f$ is free, completing the proof.



                Thus the quasicompact space $text{Spec}A$ admits a finite cover by $D(f_i)$'s, on each of which $M_{f_i}$ is free. $text{Spec}A=cup D(f_i)$ means precisely that $(f_1,...,f_n)=A$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 24 '18 at 17:01

























                answered Dec 24 '18 at 16:55









                MeowMeow

                600314




                600314






























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