Let $\Omega^*$ be the algebra generated by $dx_1,\dots,dx_n$ subject to the relations $(dx_i)^2=0$ and antisymmetry.
Does this form the cotangent space $T^*$?
I thought it might, since the tangent space $T$ has basis $\left\{\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_n}\right\}$
But I wasn't sure how to see that $dx_1\frac{\partial}{\partial x_1}=1$ (assuming this behaves like a traditional dual space.)
Context: Bott and Tu says that the exterior differentiation $d$ gives us $d(0-\!\text{forms})=\text{gradient}$, and normally we write (for $3$-dim space say) $$Df=\left(\frac{\partial f}{\partial x},\dots,\frac{\partial f}{\partial z}\right)$$
Where here the exterior differentiation gives us: $$df=\frac{\partial f}{\partial x} dx+ \dots + \frac{\partial f}{\partial z} dz$$
Which in the basis $\{dx,dy,dz\}$ is the same, where I thought that this was the basis of a cotangent space perhaps.