torsion free. Therefore $\omega$ determines the same foliation of codimension $2$ as $v$. If we now look at the $1$-forms annihilated by $v$ we get the subsheaf generated by the forms $$\omega_1= f_3 dx_2+f_2dx_3,\quad \omega_2=f_3 dx_1- f_1 dx_3\ \text{ and } 
\omega_3=f_2 dx_1 + f_1 dx_2.$$ These generators satisfy the relation $f_1\omega_1 + f_2 \omega_2= f_3\omega_3$. The subsheaf $\mathcal{E}=(\omega_1,\omega_2,\omega_3)$ is generically of rank $2$ outside the zeros of the ideal $(f_1,f_2,f_3)$ but $\mathcal{E}\otimes k({\mathfrak{p}})$ is of rank $3$ when ${\mathfrak{p}}$ is in the zeros of this ideal. Therefore $\mathcal{E}$ is *not* locally free. Moreover when we compute the determinant of $\mathcal{E}$ we get $\wedge^2\mathcal{E}= (f_1,f_2,f_3)\cdot(\omega)\subseteq \Omega^2_X$, $$\omega_1\wedge\omega_2=f_3\omega,\quad
\omega_3\wedge\omega_1=f_2\omega,\quad
\omega_2\wedge\omega_3=f_1\omega.$$ In particular $\omega$ is not in $\wedge^2\mathcal{E}$. But by [@GH Lemma, p. 210], if $\omega$ is locally decomposable, then $\omega\in \wedge^2\mathcal{E}$. Then $\omega$ is *not* locally decomposable around the zeros of the ideal $(f_1,f_2,f_3)$.

Composing a morphism $\omega:{\mathcal{L}}\to \Omega^q$ with the contraction of forms with vector fields give us a morphism $$\bigwedge^q TX\otimes {\mathcal{L}}\to{\mathcal{O}}_X.$$

The ideal sheaf ${\mathscr{J}}(\omega)$ is defined to be the sheaf-theoretic image of the morphism $\bigwedge^q TX\otimes {\mathcal{L}}\to{\mathcal{O}}_X$. The subscheme it defines is called *the singular set* of $\omega$ and denoted ${\mathrm{Sing}(\omega)}\subseteq X$. We will denote it just 