At the wikipedia page for the Frobenius theorem link, it is said that
If $D$ is a smooth tangent distribution on $M$, then the annihilator of $D$, $I(D)$ consists of all forms $\alpha \in \Omega^k(M)$ (for any $k \in {1, \cdots, \dim M}$) such that $$ \alpha(v_1, \cdots, v_k) = 0 $$ for all $v_1, \cdots, v_k \in D$. The set $I(D)$ forms a subring and, in fact, an ideal in $\Omega(M)$.
However, I cannot understand why it is an ideal. I consider the simple case at $\mathbb{R}^3$, then the smooth distribution can be chosen as $$ D = \left\{\frac{\partial}{\partial z} \right\}. $$ The corresponding annihilator for this distribution can be considered as $$ I(D) = \text{Span} \{ \mathrm{d}x, \mathrm{d}y, \mathrm{d}x \wedge \mathrm{d}y \}. $$ Is it right? Then, I find that $ \mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z \notin I(D)$, because $$ \left\langle \mathrm{d}x \wedge \mathrm{d}y \wedge \mathrm{d}z; \frac{\partial}{\partial z} \right\rangle = \mathrm{d}x \wedge \mathrm{d}y \neq 0. $$ Therefore, I think $I(D)$ isn't the ideal of the exterior algebra formed by differential forms.
So, what mistakes I have made? Thank you.
The error is subtle and hidden in the notation. The tangent vector $\frac{\partial}{\partial z}$ can't be paired with a 3-form, which is something that wants to eat three tangent vectors.
Looking at your definition, we see that (for instance) for a 3-form to be in the annihilator, it must be the case that if you feed the 3-form three things in $D$, you get zero. The counterexample disappears, since your 3-form does indeed annihilate the tuple $(\frac{\partial}{\partial z}, \frac{\partial}{\partial z}, \frac{\partial}{\partial z})$.