Is a 1-form locally expressible as $dx$? Given a 1-form $\alpha$ which is non-zero at every point of a manifold $M^n$, is it true that locally I can express the form as $dx$? (that is around each point there is a coordinate neighborhood such that $\alpha$ has that form). 
I think this is false because then the form will be closed in every neighborhood and hence closed.
The problem that I have is the following reasoning:
at a point $p$ in the manifold, extend $\alpha_p$ to a basis of the cotangent space at $p$, as being linearly independent is an open condition, that implies that I can extend it to $n$ (I guess smooth) sections of the cotangent bundle linearly independent at each point of a neighborhood of $p$. Now on each point of this neighborhood take the dual basis of tangent vectors, this will produce (I guess smooth) $n$ vector fields on this neighborhood, where the dual of $\alpha$ can be written in another coordinate system as $\partial/\partial x$ and therefore $\alpha$ as $dx$.
I think the first guess is correct but not sure about the second one, can someone explain it to me? thanks 
 A: The flaw in your argument is as follows: these $n$ vector fields do not define a coordinate system (they are not $\partial/\partial x_j$ for any coordinate system) in general.
Well, let us try an example. Assume $n=2$, $M^n=R^2$, $\alpha=f(x_2)dx_1$, where $f(x_2)$ does not vanish and $f'(x_2)$ is not zero identically. Then $d\alpha=-f'(x_2)dx_1\wedge dx_2\ne0$. Let as extend $\alpha$ to a basis of the cotangent space at each point:
$$
 \alpha_1=\alpha=f(x_2)dx_1,\quad \alpha_2=dx_2.
$$
It is easy to find the dual basis of tangent vectors at every point; we obtain the vector fields
$$
 V_1=\frac 1{f(x_2)}\frac{\partial}{\partial x_1},\quad V_2=\frac{\partial}{\partial x_2}.
$$
The new coordinate $t$ along tha trajectories of $V_1$ is $t=f(x_2)(x_1-g(x_2))$, where $g$ is an arbitrary function. But $dt\ne\alpha$, sorry.
A: It's clearly necessary for $\alpha$ to be closed. Since the property in question is local, it's sufficient to prove the result for $M = \mathbb{R}^n$. Choose an arbitrary point $p\in M$. Since $M$ is contractible, we have $\alpha = df$ for some $f\in C^\infty(M)$. Since $\alpha(p)\not = 0$, there exists by the constant rank theorem some coordinates $(x_1, \dots, x_n)$ on a neighborhood $U$ centered at $p$ with $f(x_1, \dots, x_n) = x_1$. Thus $\alpha = dx_1$ on $U$.
