If $f:E\rightarrow S$ is an elliptic curve over a scheme $S$ (so $f$ is proper and smooth of relative dimension one with geometrically connected fibers of genus one, equipped with a section $0:S\rightarrow E$), then is the sheaf $\underline{\omega}_{E/S}:=f_*\Omega_{E/S}^1$ actually free of rank one? According to the statement of Grothendieck-Serre duality in Hida's book Geometric Modular Forms and Elliptic Curves, there should be a canonical isomorphism $\underline{\omega}_{E/S}\cong\mathcal{Hom}_{\mathscr{O}_S}(R^1f_*\mathscr{O}_E,\mathscr{O}_S)$, and the first result on elliptic curves in this book is that $R^1f_*\mathscr{O}_E\cong\mathscr{O}_S$. It's also not clear to me whether $S$ is being assumed (locally) Noetherian, but if that's necessary for Grothendieck-Serre duality to hold, then I'm fine with assuming it. I can't find another reference with a statement of Grothendieck-Serre duality in this generality which does not use the language of derived categories (which I unfortunately don't understand).
The reason I'm kind of skeptical about this is that in Hida's book, as well as in Katz-Mazur, it is said that $\underline{\omega}_{E/S}$ is invertible, so that an $\mathscr{O}_S$-basis $\omega$ for $\underline{\omega}_{E/S}$ can be found locally on $S$. If the invertible sheaf in question were really trivial then one would be able to choose a global $\mathscr{O}_S$-basis for $\underline{\omega}_{E/S}$, and there would be no reason to talk about doing so locally. Hida goes on to say that, choosing an $\mathscr{O}_S$-basis $\omega$ locally on $S$ allows one to regard $(\Omega_{E/S}^1,\omega)$ as a relative effective Cartier divisor in $E/S$, which also doesn't make complete sense to me because if we can only find $\omega$ locally, how are we getting a global section $\omega\in H^0(E,\Omega_{E/S}^1)=H^0(S,\underline{\omega}_{E/S})$ (unless $\underline{\omega}_{E/S}$ really is trivial)?