How to make an ideal generated by differential forms into a differential ideal?

Question

Let $M=\mathbb{R}^4$ with standard coordinates $x_1,x_2,x_3,x_4$.

Let $\alpha=x_2dx_1+x_3dx_3+dx_4$ and $\beta=2dx_2+x_1^2dx_3+x_1dx_4$

How to find a 1-form $\gamma$ such that

the ideal generated by $\alpha,\beta,\gamma$ be a differential ideal?

Here, I think that I need to use the following form of Frobenius theorem:

Frobenius Theorem The following are equivalent.

1. An ideal generated by $\omega_1,\ldots,\omega_r$ is a differential ideal.
2. The distribution $D=\operatorname{Ker} (\omega_1)\cap\cdots \cap \operatorname{Ker} (\omega_r)$ is involutive.

However, I could not calculate $\operatorname{Ker}\alpha$ and $\operatorname{Ker}\beta$.

Answer 1

You need a $1$-form $\gamma$ so that $d\alpha$, $d\beta$, and $d\gamma$ are all in the ideal generated by $\alpha$, $\beta$, and $\gamma$. So the major HINT is: Compute $d\alpha$ and $d\beta$. Can you see an obvious $\gamma$ so that $d\alpha,d\beta\in (\gamma)$?