Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
(Co)chain complexes, abelian Categories, (pre)sheaves, (co)homology in various (possibly highly generalized) settings, spectra, derived functors, resolutions, spectral sequences, homotopy categories. Chain complexes in an abelian category form the heart of homological algebra.
0
votes
Is $C^{\infty}(\mathbb{R}^{m+n})$ a flat module over $C^{\infty}(\mathbb{R}^{m})$?
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 \oti …
2
votes
Is $C^{\infty}(\mathbb{R}^{m+n})$ a flat module over $C^{\infty}(\mathbb{R}^{m})$?
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 generalit …