If and only if criterion for something to be a differential ideal Let $I \subset \Omega^*(M)$ be a ($2$-sided) ideal (i.e. $I$ is a vector subspace, and for any $\alpha \in I$ and $\omega \in \Omega^*(M)$ we have $\omega \wedge \alpha \in I$). We say $I$ is a differential ideal if $d\omega \in I$ whenever $\omega \in I$.
Suppose $I$ is generated as an ideal by $\omega_1, \omega_2, \ldots, \omega_r$. Is $I$ is a differential ideal if and only if$$d\omega_i = \sum_j \omega_{ij} \wedge \omega_j$$for suitable $1$-forms $\omega_{ij}$?

Yes, this is true. First, by hypothesis $I$ is generated by the forms $\omega_1, \dots, \omega_n$, which I guess are homogeneous, thus it'll always be a graded ideal. So $I$ is a differential ideal iff $d\omega \in I$ for all $\omega \in I$.
If $I$ is a differential ideal, then necessarily $d\omega_i \in I$ for all $i$; but by hypothesis, $I$ is generated by the $\omega_j$, so $d\omega_i = \sum_j \omega_{ij} \wedge \omega_j$ for some forms $\omega_{ij}$.
Conversely, suppose $d\omega_i = \sum_j \omega_{ij} \wedge \omega_j$ for all $i$. Let $\omega \in I$; since $I$ is generated by the $\omega_i$, then $\omega = \sum_i \alpha_i \wedge \omega_i$ for some forms $\alpha_i$. Then
$$d\omega = \sum_i d(\alpha_i \wedge \omega_i) = \sum_i (d\alpha_i) \wedge \omega_i + \sum_i \sum_j \pm \alpha_i \wedge \omega_{ij} \wedge \omega_j.$$
Each term in the sum is in $I$ (because $I$ is an ideal), so $d\omega \in I$, and $I$ is a differential ideal.
