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Let $I$ be a ideal of $\Omega(M)$ ($M$ a manifold) that is locally generated by $p$ independent $1$-forms $\omega_1,...,\omega_p$. Set $\omega = \omega_1 \wedge ...\wedge \omega_p$ Then $I$ is a differential ideal iff $$d\omega = \alpha\wedge\omega $$for some $1$-form $\alpha$.

The $\implies$ direction is straightforward, since $d\omega_i = \sum \eta_j \wedge \omega_i$ for some $1$-forms $\eta$. The other direction has got me stuck. I thought about trying to prove that $d\omega_i \in I$, but I don't see how to use this hypothesis since it seems like I"m losing a lot of data.

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  • $\begingroup$ I've never seen such a claim before, and it seems wrong to me. Consider $\omega_1 = dz-y\,dx$ [the usual suspect] and $\omega_2 = dz$ on $\Bbb R^3-\{y=0\}$. Then $\omega=\omega_1\wedge\omega_2$ satisfies your condition, but $\omega_1$ most definitely does not. $\endgroup$ – Ted Shifrin Jun 29 '18 at 22:26
  • $\begingroup$ When you say $\omega_1$ does not satisfy the condition, do you mean $d\omega_1 \notin I$? But $d\omega_1 = dx\wedge dy = (\frac{1}{y}dy)\wedge (\omega_1 - \omega_2)$. Unless you mean something else? $\endgroup$ – Yotis Jul 1 '18 at 4:16
  • $\begingroup$ I was being idiotic, thinking $d\omega_i\in (\omega_i)$, rather than $d\omega_i\in I$. See my answer now. $\endgroup$ – Ted Shifrin Jul 1 '18 at 6:08
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Here's the idea for the proof. Locally, extend $\omega_1,\dots,\omega_p$ to a basis $\omega_1,\dots,\omega_p,\omega_{p+1},\dots,\omega_n$ for the $1$-forms on $M$. For simplicity, I'll take $p=2$ and $n=4$, but you can work out the general case with a bit more notation. Let $I=(\omega_1,\omega_2)$. Then, working mod $I$, we have \begin{align*} d\omega_1 &= f\omega_3\wedge\omega_4 \pmod I \\ d\omega_2 &= g\omega_3\wedge\omega_4 \pmod I. \end{align*} Now $d(\omega_1\wedge\omega_2) = f\omega_3\wedge\omega_4\wedge\omega_2 - g\omega_1\wedge\omega_3\wedge\omega_4$. This expression is of the form $\alpha\wedge\omega_1\wedge\omega_2$ for some $1$-form $\alpha$ if and only if $f=g=0$, as desired.

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