I'm studying general relativity and and learning up on tensors through a lecture series. It says that $\omega_\mu$ represents a 1-form in the $x^\mu$ coordinate system. The coordinate independent 1-form is denoted by $\omega$ as $\omega = \omega_\mu \mathrm{d}x^\mu$. However, if I assume the definition of the 1-form when using two different coordinate systems $x^\mu$ and $x^{\mu'}$, the following is true
$$\omega_{\mu'} = \frac{\partial x^\mu}{\partial x^{\mu'}} \omega_\mu$$
then I can prove the following:
$$\omega_{\mu'} \mathrm{d}x^{\mu'} = \omega_\mu \mathrm{d}x^\mu$$
However, this is how a scalar transforms. So, is $\omega = \omega_\mu \mathrm{d}x^\mu$ a scalar?
The equation $$\omega_{\mu'} \mathrm{d}x^{\mu'} = \omega_\mu \mathrm{d}x^\mu$$ does not say it transforms like a scalar; instead it says that both expressions $\omega_{\mu'} \mathrm{d}x^{\mu'}, \omega_\mu \mathrm{d}x^\mu$ are representing the same object $\omega$. The object $\omega=\omega_\mu dx^\mu$ is a differential $1$-form, not a scalar field.