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For $m>0$ we consider the ring $C^{\infty}(\mathbb{R}^{m})$ of smooth functions on $\mathbb{R}^{m}$. For $n>0$ we consider the projection $\mathbb{R}^{m+n}\to \mathbb{R}^{m}$ hence $C^{\infty}(\mathbb{R}^{m+n})$ is naturally a $C^{\infty}(\mathbb{R}^{m})$ module.

My question is: is it true that $C^{\infty}(\mathbb{R}^{m+n})$ is a flat module over $C^{\infty}(\mathbb{R}^{m})$? I'm not sure if we should take the topology on algebra into consideration.

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  • $\begingroup$ Is it perhaps even a free module? I have no intuition one way or another. $\endgroup$ – Gro-Tsen Dec 15 '19 at 16:50
  • $\begingroup$ My suspicion is that this should work if one uses the topological variants of flatness etc set out in the work of Helemskii and his school -- there might be some relevant work of Ogneva -- but I am away from my reference sources right now $\endgroup$ – Yemon Choi Dec 15 '19 at 21:10
  • $\begingroup$ "Example-Lemma 4.30" of springer.com/gp/book/9780387955438 might be related. $\endgroup$ – Qfwfq Dec 15 '19 at 21:15
  • $\begingroup$ Actually, I see that in an answer to one of your previous questions mathoverflow.net/a/304075 Igor Khavkhine already mentioned Ogneva's work, with a link to the relevant paper(s) and some technical details $\endgroup$ – Yemon Choi Dec 15 '19 at 21:26
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This is just a long comment. Let $X,Y$ be smooth manifolds (here in particular, resp. $\mathbb{R}^n$ and $\mathbb{R}^m$), and let $M$ be a $C^\infty(X)$-module. Then any pair $(m,\phi)\in M\times C^\infty(X\times Y)$ defines a function $Y\to M$, namely $Y\ni c\mapsto \phi(x,c)m\in M$ (which is well defined as for all $c\in Y$, $\phi(x,c)\in C^\infty(X)$). Since this is correspondence is bilinear $M\times C^\infty(X,Y) \to M^Y$, it produces a morphism of $C^\infty(X)$-modules $$M\otimes C^\infty(X,Y) \to M^Y.$$ Is this morphism injective? That is, can one see $M\otimes C^\infty(X,Y) $ as a space of $M$-valued functions on $Y$? If the answer is yes, then any morphism $f:M\to M'$ produces a $f\otimes 1:M\otimes C^\infty(X,Y)\to M'\otimes C^\infty(X,Y)$ which extends to the morphism $M^Y\ni u\mapsto f\circ u\in M'^Y$, which is injective whenever $f$ is injective. Therefore $ C^\infty(X,Y)$ would be a flat $ C^\infty(X)$-module.

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PS: The answer below has gaps, and it is likely incorrect.


Yes, $C^{\infty}(\mathbb{R}^{m+n})$ is a flat $C^{\infty}(\mathbb{R}^{m})$ module. Or, following @Pietro 's comment, with more generality, if $X,Y$ are smooth manifolds, then $C^\infty(X\times Y)$ is a $C^\infty(X)$ flat module. To see this, it is enough to consider $C^\infty(X)$ linear, injective maps $\varphi: K \rightarrow L$, where $K = <r_1,\dots,r_d>_{C^\infty(X)}$ is a finitely generated $C^\infty(X)$ ideal, and $L=C^\infty(X)$ is the ring of scalars.

Notice that in that case, considering that $C^\infty(X)$ is a sub-ring of $C^\infty(X\times Y)$, the tensor operation $.\otimes_{C^\infty(X)} C^\infty(X\times Y) $ applied to an ideal such as $K$ is just the extension of scalars, i.e., in some natural way, $$ K\otimes_{C^\infty(X)} C^\infty(X\times Y) = <r_1,\dots,r_d>_{C^\infty(X\times Y)}, $$ and $$L\otimes_{C^\infty(X)} C^\infty(X\times Y) = C^\infty(X\times Y).$$ Moreover, with these representations, $\bar{\varphi} = \varphi \otimes Id$, the extension of $\varphi$ from $K$ to $K\otimes C^\infty(X\times Y)$, satisfies: $$ \bar{\varphi}( \sum_{i=1}^d r_i\ m_i) = \sum_{i=1}^d \varphi(r_i)\ m_i $$ for any $m_1,\dots, m_d \in C^\infty(X\times Y)$.

We want to verify that, under the assumption the $\varphi$ is injective, $\bar{\varphi}$ is injective. Now, if for some linear combination $\sum_i r_i \ m_i$ we have $\bar{\varphi}(\sum_i r_im_i) = 0$, then $\sum_i \varphi(r_i)\ m_i = 0$ in $C^\infty(X\times Y)$, and so, for all $y \in Y$, $$ \sum_i \varphi(r_i)\ m_i(.,y) = 0 $$ in $C^\infty(X)$, but $\varphi$ being $C^\infty(X)$ linear, we get: $$ \varphi( \sum_i r_i\ m_i(.,y) ) = 0. $$ By injectivity of $\varphi$, $ \sum_i r_i\ m_i(.,y) = 0$

(With @Pietro notation, this is the condition of an element of $M^Y$ having all 0 coordinates applied to the case $M = $ a finitely generated ideal of $C^\infty(X)$)

Since this holds for all $y$, then $\sum_i r_i\ m_i = 0$, which shows injectivity of $\varphi$.

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  • $\begingroup$ I am outside my comfort zone when it comes to Frechet algebras and modules rather than Banach ones, but in extension-of-scalars arguments don't we need to ise a suitable completed tensor product? Even at the vector space level, my gut feeling is that $C^\infty({\bf R})\otimes C^\infty({\bf R})$ is all of $C^\infty({\bf R}^2)$, as this seems similar to claiming that all continuous linear maps $C^\infty({\bf R})\to C^\infty ({\bf R})'$ have finite rank... $\endgroup$ – Yemon Choi Dec 15 '19 at 21:09
  • $\begingroup$ @Yemon Choi: According to Nestruev (see my previous comment in the OP), $\mathcal{C}^\infty (M\times N)$ is isomorphic to the "smooth envelope" (3.36) of the tensor product over $\mathbb{R}$ of the two algebras. There, "$\mathcal{C}^\infty$-closed" is defined by means of a property (3.32) that reminds me of the $\mathcal{C}^\infty$-schemes by Joyce. $\endgroup$ – Qfwfq Dec 15 '19 at 21:49
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    $\begingroup$ Maybe I'm missing some obvious fact, but the last line is not quite clear to me: If, for all $y\in Y$, one has $\sum_i r_i m_i(.,y)=0$ as an element of $K$, that is $\sum_i r_i m_i$ induces the zero function, that is it is $0$ as an element of $K^Y$, why can we conclude that $\sum_i r_i m_i=0$ as an element of $K\otimes C^\infty(X\times Y)$? $\endgroup$ – Pietro Majer Dec 15 '19 at 23:50
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    $\begingroup$ Translating into algebraic language: Set $R := C(X), S := C(X\times Y), \varphi: K \hookrightarrow R$ for $K \triangleleft R$. Then $R \le S$ and what you are saying is that there is a commutative diagramm $$K \otimes_R S \xrightarrow{\varphi \otimes id} R\otimes_R S$$ $$\cong\, \downarrow \qquad\qquad \downarrow\ \,\cong$$ $$KS \quad \xrightarrow[\bar{\varphi}] \quad S$$ Why is the left vertical arrow an isomorphism? $\endgroup$ – tj_ Dec 16 '19 at 5:27
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    $\begingroup$ I don't see how your argument uses in any way properties of the ring of smooth functions. Is the corresponding question known for rings of continuous functions on (locally) compact Hausdorff spaces $X$ and $Y$? This is perhaps in the comfort zone of Yemon Choi. $\endgroup$ – Jochen Wengenroth Dec 16 '19 at 9:46
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Let $X,Y$ be smooth manifolds; let $\Im = <r_1,\dots,r_d>_{C^\infty(X)}$ be a finitely generated ideal of $C^\infty(X)$. Then $\Im \otimes_{C^\infty(X)} C^\infty(X \times Y) = \{ \sum_{i=1}^d r_i \otimes g_i \ | \ g_1,\dots,g_d \in C^\infty(X \times Y)\}$. I claimed elsewhere that the condition $\sum_{i=1}^d r_i \otimes g_i = 0$ in $\Im \otimes C^\infty(X \times Y)$ is equivalent to $\sum_{i=1}^d r_i g_i = 0$ in $C^\infty(X \times Y)$. That is not the case. The correct statement is:

The following conditions are equivalent:

i) $\sum_{i=1}^d r_i g_i = 0$ in $C^\infty(X \times Y)$
ii) $k(\sum_{i=1}^d r_i \otimes g_i) = 0$ for all $k \in \Im$

Proof $i) \Rightarrow ii)$ If $\sum_{i=1}^d r_i g_i = 0$ and $k \in \Im$ then $0 = k \otimes (\sum_{i=1}^d r_i g_i ) = \sum_{i=1}^d k \otimes (r_i g_i) =$
$ \sum_{i=1}^d r_i \ (k \otimes g_i) = \sum_{i=1}^d (r_i k) \otimes g_i = k (\sum_{i=1}^d r_i \otimes g_i) $

$ii) \Rightarrow i)$ Take $k = r_j$, and assume $0 = r_j( \sum_{i=1}^d r_i \otimes \ g_i) =\sum_{i=1}^d r_i \otimes ( r_j \ g_i) $. Fix $a \in Y$, and consider the bilinear map $e : \Im \times C^\infty(X \times Y) \to C^\infty(X)$ defined as $ e(k,g) = k(x)g(x,a)$. If $\bar{e}$ is the corresponding map on $\Im \otimes C^\infty(X \times Y)$, then $0 = \bar{e}( \sum_{i=1}^d r_i \otimes ( r_j \ g_i) ) = \sum_{i=1}^d \bar{e}(r_i \otimes ( r_j \ g_i)) =$
$ \sum_{i=1}^d r_i \ r_j \ g_i(.,a) = r_j ( \sum_{i=1}^d r_i \ g_i(.,a) )$

Since this holds for all $j$, at any point $x \in X$ where some $r_j $ does not vanish, $\sum_{i=1}^d r_i(x) \ g_i(x,a) =0$, and the same holds trivially at any point $x$ where all $r_j$'s vanish.

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  • $\begingroup$ As far as I see this does not answer question but rather shows that and why your previous answer is not conclusive. I would suggest that you edit your first answer and delete this one. $\endgroup$ – Jochen Wengenroth Dec 18 '19 at 8:27

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