I'm sure this is really simple, and I might be right; it's just that I'm not sure. I'm asked to prove that the Klein-Gordon equation it's invariant under global gauge transformations.
In Greiner's book "Relativistic Quantum Mechanics" there's a section where it explains how the K-G equations it's invariant under gauge transformations, "1.10 Gauge invariance of the Coupling", where he uses the most general form of a gauge transformation $$A'_{\mu}(x) = A_{\mu}(x) + \frac{\partial \chi(x)}{\partial x^{\mu}}$$ and the wave function transformed it's $$\psi' = \exp \left(\frac{ie}{\hbar c}\chi\right)\psi$$ now it's of my understanding that a global gauge transformation when the $\chi \neq \chi(x)$, in other words, the $\chi$ it's just a constant, doesn't depend on the coordinates.
In a local gauge transformation $\chi = \chi (x)$, it does depend on where you are standing in the space, so here Greiner uses a local gauge transformation because $\chi$ depends on the coordinates, and when substituting on the K-G equation the form of the equation its preserved, and because of the derivative of the wave function, now as I said earlier, I'm asked to prove that the Klein-Gordon equation it's invariant under global gauge transformations. My question is wouldn't it be trivial? Because if $\chi$ doesn't depend on $x$ then: $$A'_{\mu}(x) = A_{\mu}(x) + \frac{\partial \chi}{\partial x^{\mu}} = A_{\mu}(x) + 0 = A_{\mu}(x)$$ and $$\psi' = \exp \left(\frac{ie}{\hbar c}\chi\right)\psi$$ would only be a constant phase which doesn't really affect the equation so it's global invariant? I suppose it's not that easy and I'm doing something wrong, please help.