Is the function ring $C^{infty}(M)$ noetherian?
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
asked Sep 9 '16 at 8:00
user53216
9651714
add a comment |
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
asked Sep 9 '16 at 8:00
user53216
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
add a comment |
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
asked Sep 9 '16 at 8:00
user53216
9651714
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
asked Sep 9 '16 at 8:00
user53216
9651714
asked Sep 9 '16 at 8:00
user53216
9651714
asked Sep 9 '16 at 8:00
user53216
9651714
asked Sep 9 '16 at 8:00
user53216
9651714
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
add a comment |
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
add a comment |
3 Answers
3
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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: $$I_0 subsetneq I_1 subsetneq I_2 subsetneq I_3 subsetneq cdots subsetneq C^{infty}(M)quad (f_{j+1}in I_{j+1}setminus I_j)$$
answered Sep 9 '16 at 8:40
Georges Elencwajg
118k7180328
add a comment |
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.
answered Sep 9 '16 at 8:39
studiosus
1,602713
add a comment |
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).
answered Nov 28 at 16:15
Leonhardt Euler
11
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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: $$I_0 subsetneq I_1 subsetneq I_2 subsetneq I_3 subsetneq cdots subsetneq C^{infty}(M)quad (f_{j+1}in I_{j+1}setminus I_j)$$
answered Sep 9 '16 at 8:40
Georges Elencwajg
118k7180328
add a comment |
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: $$I_0 subsetneq I_1 subsetneq I_2 subsetneq I_3 subsetneq cdots subsetneq C^{infty}(M)quad (f_{j+1}in I_{j+1}setminus I_j)$$
answered Sep 9 '16 at 8:40
Georges Elencwajg
118k7180328
add a comment |
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: $$I_0 subsetneq I_1 subsetneq I_2 subsetneq I_3 subsetneq cdots subsetneq C^{infty}(M)quad (f_{j+1}in I_{j+1}setminus I_j)$$
answered Sep 9 '16 at 8:40
Georges Elencwajg
118k7180328
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: $$I_0 subsetneq I_1 subsetneq I_2 subsetneq I_3 subsetneq cdots subsetneq C^{infty}(M)quad (f_{j+1}in I_{j+1}setminus I_j)$$
answered Sep 9 '16 at 8:40
Georges Elencwajg
118k7180328
answered Sep 9 '16 at 8:40
Georges Elencwajg
118k7180328
answered Sep 9 '16 at 8:40
Georges Elencwajg
118k7180328
answered Sep 9 '16 at 8:40
Georges Elencwajg
118k7180328
118k7180328
add a comment |
add a comment |
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.
answered Sep 9 '16 at 8:39
studiosus
1,602713
add a comment |
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.
answered Sep 9 '16 at 8:39
studiosus
1,602713
add a comment |
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.
answered Sep 9 '16 at 8:39
studiosus
1,602713
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.
answered Sep 9 '16 at 8:39
studiosus
1,602713
answered Sep 9 '16 at 8:39
studiosus
1,602713
answered Sep 9 '16 at 8:39
studiosus
1,602713
answered Sep 9 '16 at 8:39
studiosus
1,602713
1,602713
add a comment |
add a comment |
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).
answered Nov 28 at 16:15
Leonhardt Euler
11
add a comment |
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).
answered Nov 28 at 16:15
Leonhardt Euler
11
add a comment |
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).
answered Nov 28 at 16:15
Leonhardt Euler
11
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).
answered Nov 28 at 16:15
Leonhardt Euler
11
answered Nov 28 at 16:15
Leonhardt Euler
11
answered Nov 28 at 16:15
Leonhardt Euler
11
answered Nov 28 at 16:15
Leonhardt Euler
11
11
add a comment |
add a comment |
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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