Exercice on a differential form Let $\omega$ be a $q$-form on $\mathbb{R}^2$ and let 


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*$Z_{\mathbb{R}^2}(dx_1)=\{p \in \mathbb{R}^2 \colon (dx_1)_{|p}=0\}$

*$Z_{S^1}(dx_1)=\{p \in S^1 \colon (dx1_{|S^1})_{|p}=0\}$


where $(dx1_{|S^1})$ is the pull-back of $q$. I have to show that $Z_{\mathbb{R}^2}(dx_1) \cap S^1 \ne Z_{S^1}(dx_1)$. 
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I try to solve it, and for me $Z_{\mathbb{R}^2}(dx_1) \cap S^1=\{(0,1),(0,-1)\}$
(Is this because $dx_1$ is zero applied on a point of type (0,c)? in this case how is the tangent vector in a single point (0,c)?) 
What about $Z_{S^1}(dx_1)$? I think it contains also the point $(1,0)$,$(-1,0)$ because in this point the tangent vector may be represent as $v=\frac {\partial}{\partial y}$ and $dx_1(v)=0$. Is it right? 

The $1$-form $dx_1$ is the differential of the coordinate function $$x_1:\>{\mathbb R}^2\to{\mathbb R}, \quad (p_1,p_2)\mapsto p_1\ .$$
For a vector $X=(X_1,X_2)\in T_p$ one has
$$x_1(p+X)-x_1(p)= X_1\ .$$
Since the right side is obviously linear in $X$ we can already conclude that $$dx_1(p).X=X_1\ .$$
This means that $dx_1(p)$ computes the first coordinate of any tangent vector $X$ attached at $p$, and this for all points $p\in{\mathbb R}^2$. It follows that $dx_1(p)\ne0\in T_p^*$ for all $p$; whence your set $Z_{{\mathbb R}^2}(dx_1)$ is empty.
When $p=(\cos\phi,\sin\phi)\in S^1$ then $(TS^1)_p$ is spanned by the vector $X=(-\sin\phi,\cos\phi)$. From $dx_1(p).X=-\sin\phi$ we conclude that $dx_1|S^1(p)=0$ iff $\sin\phi=0$, i.e. at the points $(\pm1,0)\in S^1$. These two points constitute your $Z_{S^1}(dx_1)$.
