Is the function ring $C^{infty}(M)$ noetherian?











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Let $M$ be a smooth manifold and $C^{infty}(M)$ be its function ring. Is this a noetherian ring?






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    Interesting question.. Share your ideas...
    – user87543
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    Cf. math.stackexchange.com/questions/1239569/…
    – darko
    Sep 9 '16 at 8:27















up vote
9
down vote

favorite
1












Let $M$ be a smooth manifold and $C^{infty}(M)$ be its function ring. Is this a noetherian ring?






differential-geometry commutative-algebra smooth-manifolds noetherian







share|cite|improve this question


















  • 1




    Interesting question.. Share your ideas...
    – user87543
    Sep 9 '16 at 8:16






  • 1




    Cf. math.stackexchange.com/questions/1239569/…
    – darko
    Sep 9 '16 at 8:27













up vote
9
down vote

favorite
1









up vote
9
down vote

favorite
1






1





Let $M$ be a smooth manifold and $C^{infty}(M)$ be its function ring. Is this a noetherian ring?






differential-geometry commutative-algebra smooth-manifolds noetherian







share|cite|improve this question













Let $M$ be a smooth manifold and $C^{infty}(M)$ be its function ring. Is this a noetherian ring?






differential-geometry commutative-algebra smooth-manifolds noetherian




differential-geometry commutative-algebra smooth-manifolds noetherian






share|cite|improve this question













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asked Sep 9 '16 at 8:00









user53216

9651714




9651714








  • 1




    Interesting question.. Share your ideas...
    – user87543
    Sep 9 '16 at 8:16






  • 1




    Cf. math.stackexchange.com/questions/1239569/…
    – darko
    Sep 9 '16 at 8:27














  • 1




    Interesting question.. Share your ideas...
    – user87543
    Sep 9 '16 at 8:16






  • 1




    Cf. math.stackexchange.com/questions/1239569/…
    – darko
    Sep 9 '16 at 8:27








1




1




Interesting question.. Share your ideas...
– user87543
Sep 9 '16 at 8:16




Interesting question.. Share your ideas...
– user87543
Sep 9 '16 at 8:16




1




1




Cf. math.stackexchange.com/questions/1239569/…
– darko
Sep 9 '16 at 8:27




Cf. math.stackexchange.com/questions/1239569/…
– darko
Sep 9 '16 at 8:27










3 Answers
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No, the ring $C^{infty}(M)$ is never noetherian if $dim Mgt 0$.

Indeed consider a strictly decreasing sequence of closed subsets $$M=C_0supsetneq C_1 supsetneq C_2 supsetneq C_3supsetneq cdots$$ (Such a sequence is easy to construct: take the inverse image in a chart of closed concentric balls in $mathbb R^n$.)

A theorem of Whitney ensures that there exist a smooth function $f_jin C^{infty}(M)$ whose zero locus is precisely $C_j$.

Now if you consider the ideal $I_j=Z(C_j)subset C^{infty}(M)$ of functions zero on $C_j$ you get a strictly increasing sequence of ideals of $C^{infty}(M)$ destroying all hopes of noetherianness for that ring: $$I_0 subsetneq I_1 subsetneq I_2 subsetneq I_3 subsetneq cdots subsetneq C^{infty}(M)quad (f_{j+1}in I_{j+1}setminus I_j)$$






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    up vote
    6
    down vote













    I think that this is not true in general.



    Consider the ring $C^{infty}(mathbb{R})$ and the ideal $I$ consisting of smooth functions that vanish on a neighborhood of $0$. I claim that $I$ is not finitely generated.



    Assume by contradiction that $I=(f_{1},ldots,f_{n})$, where each $f_{i}$ vanishes on a neighborhood $V_{i}$ of $0$. Then each element of $(f_{1},ldots,f_{n})$ vanishes on $V:=cap_{i=1}^{n}V_{i}$. Though we can construct smooth functions in $C^{infty}(mathbb{R})$ that vanish on an arbitrarily small neighborhood of $0$, in particular strictly smaller than $V$. So $Ineq (f_{1},ldots,f_{n})$. Hence $I$ is not finitely generated.






    share|cite|improve this answer




























      up vote
      0
      down vote













      Consider the sequene of ideals of $C^{infty}(mathbb {R}) $, for instance, given by:
      $$m mapsto J_m =langle { e, e^x, e^{x^2}, cdots e^{x^m}} rangle $$
      This is an ascending chain of ideals with no upper bound.
      The ring is not Noetherian (neither Artinian).






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

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        active

        oldest

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        up vote
        15
        down vote



        accepted










        No, the ring $C^{infty}(M)$ is never noetherian if $dim Mgt 0$.

        Indeed consider a strictly decreasing sequence of closed subsets $$M=C_0supsetneq C_1 supsetneq C_2 supsetneq C_3supsetneq cdots$$ (Such a sequence is easy to construct: take the inverse image in a chart of closed concentric balls in $mathbb R^n$.)

        A theorem of Whitney ensures that there exist a smooth function $f_jin C^{infty}(M)$ whose zero locus is precisely $C_j$.

        Now if you consider the ideal $I_j=Z(C_j)subset C^{infty}(M)$ of functions zero on $C_j$ you get a strictly increasing sequence of ideals of $C^{infty}(M)$ destroying all hopes of noetherianness for that ring: $$I_0 subsetneq I_1 subsetneq I_2 subsetneq I_3 subsetneq cdots subsetneq C^{infty}(M)quad (f_{j+1}in I_{j+1}setminus I_j)$$






        share|cite|improve this answer

























          up vote
          15
          down vote



          accepted










          No, the ring $C^{infty}(M)$ is never noetherian if $dim Mgt 0$.

          Indeed consider a strictly decreasing sequence of closed subsets $$M=C_0supsetneq C_1 supsetneq C_2 supsetneq C_3supsetneq cdots$$ (Such a sequence is easy to construct: take the inverse image in a chart of closed concentric balls in $mathbb R^n$.)

          A theorem of Whitney ensures that there exist a smooth function $f_jin C^{infty}(M)$ whose zero locus is precisely $C_j$.

          Now if you consider the ideal $I_j=Z(C_j)subset C^{infty}(M)$ of functions zero on $C_j$ you get a strictly increasing sequence of ideals of $C^{infty}(M)$ destroying all hopes of noetherianness for that ring: $$I_0 subsetneq I_1 subsetneq I_2 subsetneq I_3 subsetneq cdots subsetneq C^{infty}(M)quad (f_{j+1}in I_{j+1}setminus I_j)$$






          share|cite|improve this answer























            up vote
            15
            down vote



            accepted







            up vote
            15
            down vote



            accepted






            No, the ring $C^{infty}(M)$ is never noetherian if $dim Mgt 0$.

            Indeed consider a strictly decreasing sequence of closed subsets $$M=C_0supsetneq C_1 supsetneq C_2 supsetneq C_3supsetneq cdots$$ (Such a sequence is easy to construct: take the inverse image in a chart of closed concentric balls in $mathbb R^n$.)

            A theorem of Whitney ensures that there exist a smooth function $f_jin C^{infty}(M)$ whose zero locus is precisely $C_j$.

            Now if you consider the ideal $I_j=Z(C_j)subset C^{infty}(M)$ of functions zero on $C_j$ you get a strictly increasing sequence of ideals of $C^{infty}(M)$ destroying all hopes of noetherianness for that ring: $$I_0 subsetneq I_1 subsetneq I_2 subsetneq I_3 subsetneq cdots subsetneq C^{infty}(M)quad (f_{j+1}in I_{j+1}setminus I_j)$$






            share|cite|improve this answer












            No, the ring $C^{infty}(M)$ is never noetherian if $dim Mgt 0$.

            Indeed consider a strictly decreasing sequence of closed subsets $$M=C_0supsetneq C_1 supsetneq C_2 supsetneq C_3supsetneq cdots$$ (Such a sequence is easy to construct: take the inverse image in a chart of closed concentric balls in $mathbb R^n$.)

            A theorem of Whitney ensures that there exist a smooth function $f_jin C^{infty}(M)$ whose zero locus is precisely $C_j$.

            Now if you consider the ideal $I_j=Z(C_j)subset C^{infty}(M)$ of functions zero on $C_j$ you get a strictly increasing sequence of ideals of $C^{infty}(M)$ destroying all hopes of noetherianness for that ring: $$I_0 subsetneq I_1 subsetneq I_2 subsetneq I_3 subsetneq cdots subsetneq C^{infty}(M)quad (f_{j+1}in I_{j+1}setminus I_j)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 9 '16 at 8:40









            Georges Elencwajg

            118k7180328




            118k7180328






















                up vote
                6
                down vote













                I think that this is not true in general.



                Consider the ring $C^{infty}(mathbb{R})$ and the ideal $I$ consisting of smooth functions that vanish on a neighborhood of $0$. I claim that $I$ is not finitely generated.



                Assume by contradiction that $I=(f_{1},ldots,f_{n})$, where each $f_{i}$ vanishes on a neighborhood $V_{i}$ of $0$. Then each element of $(f_{1},ldots,f_{n})$ vanishes on $V:=cap_{i=1}^{n}V_{i}$. Though we can construct smooth functions in $C^{infty}(mathbb{R})$ that vanish on an arbitrarily small neighborhood of $0$, in particular strictly smaller than $V$. So $Ineq (f_{1},ldots,f_{n})$. Hence $I$ is not finitely generated.






                share|cite|improve this answer

























                  up vote
                  6
                  down vote













                  I think that this is not true in general.



                  Consider the ring $C^{infty}(mathbb{R})$ and the ideal $I$ consisting of smooth functions that vanish on a neighborhood of $0$. I claim that $I$ is not finitely generated.



                  Assume by contradiction that $I=(f_{1},ldots,f_{n})$, where each $f_{i}$ vanishes on a neighborhood $V_{i}$ of $0$. Then each element of $(f_{1},ldots,f_{n})$ vanishes on $V:=cap_{i=1}^{n}V_{i}$. Though we can construct smooth functions in $C^{infty}(mathbb{R})$ that vanish on an arbitrarily small neighborhood of $0$, in particular strictly smaller than $V$. So $Ineq (f_{1},ldots,f_{n})$. Hence $I$ is not finitely generated.






                  share|cite|improve this answer























                    up vote
                    6
                    down vote










                    up vote
                    6
                    down vote









                    I think that this is not true in general.



                    Consider the ring $C^{infty}(mathbb{R})$ and the ideal $I$ consisting of smooth functions that vanish on a neighborhood of $0$. I claim that $I$ is not finitely generated.



                    Assume by contradiction that $I=(f_{1},ldots,f_{n})$, where each $f_{i}$ vanishes on a neighborhood $V_{i}$ of $0$. Then each element of $(f_{1},ldots,f_{n})$ vanishes on $V:=cap_{i=1}^{n}V_{i}$. Though we can construct smooth functions in $C^{infty}(mathbb{R})$ that vanish on an arbitrarily small neighborhood of $0$, in particular strictly smaller than $V$. So $Ineq (f_{1},ldots,f_{n})$. Hence $I$ is not finitely generated.






                    share|cite|improve this answer












                    I think that this is not true in general.



                    Consider the ring $C^{infty}(mathbb{R})$ and the ideal $I$ consisting of smooth functions that vanish on a neighborhood of $0$. I claim that $I$ is not finitely generated.



                    Assume by contradiction that $I=(f_{1},ldots,f_{n})$, where each $f_{i}$ vanishes on a neighborhood $V_{i}$ of $0$. Then each element of $(f_{1},ldots,f_{n})$ vanishes on $V:=cap_{i=1}^{n}V_{i}$. Though we can construct smooth functions in $C^{infty}(mathbb{R})$ that vanish on an arbitrarily small neighborhood of $0$, in particular strictly smaller than $V$. So $Ineq (f_{1},ldots,f_{n})$. Hence $I$ is not finitely generated.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 9 '16 at 8:39









                    studiosus

                    1,577713




                    1,577713






















                        up vote
                        0
                        down vote













                        Consider the sequene of ideals of $C^{infty}(mathbb {R}) $, for instance, given by:
                        $$m mapsto J_m =langle { e, e^x, e^{x^2}, cdots e^{x^m}} rangle $$
                        This is an ascending chain of ideals with no upper bound.
                        The ring is not Noetherian (neither Artinian).






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          Consider the sequene of ideals of $C^{infty}(mathbb {R}) $, for instance, given by:
                          $$m mapsto J_m =langle { e, e^x, e^{x^2}, cdots e^{x^m}} rangle $$
                          This is an ascending chain of ideals with no upper bound.
                          The ring is not Noetherian (neither Artinian).






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Consider the sequene of ideals of $C^{infty}(mathbb {R}) $, for instance, given by:
                            $$m mapsto J_m =langle { e, e^x, e^{x^2}, cdots e^{x^m}} rangle $$
                            This is an ascending chain of ideals with no upper bound.
                            The ring is not Noetherian (neither Artinian).






                            share|cite|improve this answer












                            Consider the sequene of ideals of $C^{infty}(mathbb {R}) $, for instance, given by:
                            $$m mapsto J_m =langle { e, e^x, e^{x^2}, cdots e^{x^m}} rangle $$
                            This is an ascending chain of ideals with no upper bound.
                            The ring is not Noetherian (neither Artinian).







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 28 at 16:15









                            Leonhardt Euler

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                            11






























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