Is the sheaf of differentials on an elliptic curve over $R$ with a Weierstrass equation free? Let $R$ be an integral domain and $E\stackrel{f}{\rightarrow}\text{Spec }R$ be an elliptic curve given by
$$E := \text{Proj }R[x,y,z]/(y^2z + a_1xyz + a_3yz^2 = x^3 + a_2x^2z + a_4xz^2 + a_6z^3)$$
where $a_i\in R$.
Is there a nowhere vanishing differential on $E/R$? Ie., is $f_*\Omega_{E/R}$ free? (isomorphic to $\tilde{R}$?)
If $R$ is a field, then the language of Silverman seems pretty straightforward and allows one to calculate that $\omega := \frac{dx}{2y+a_1x + a_3}$ is holomorphic and nonvanishing, hence a basis for $f_*\Omega_{E/R}$ in this case. However, I don't feel like I have the right language for discussing differentials when $R$ is not a field.
 A: Ah okay, so as Epargyreus noted, we may reduce to the case of $R$ a field as follows.
There are a few steps.


*

*First, $f_*\Omega_{E/R}$ is an invertible sheaf over $R$. This seems to be a rather nontrivial fact, which seems to come from Serre-Grothendieck duality.

*If $M$ is a locally free module of rank 1 over a ring $R$, and there exists some $m\in M$ such that for every prime $p\in\text{Spec }R$, the image of $m$ in the stalk $M_p$ is a generator for $M_p$ as a free $R$-module of rank 1, then $M$ is free. To see this, simply consider the exact sequence
$$0\rightarrow R\rightarrow M\rightarrow M/R\rightarrow 0$$
where the first map sends $1\mapsto m$. By the assumption, stalks of the quotient $M/R$ are all 0, which implies that $M/R = 0$, so $R\cong M$, ie $M$ is free.

*Let $m\in M$. For a prime $p\in\text{Spec }R$, let $m_p$ be the image of $m$ in the localization $M_p$. Then by Nakayama, if $m_p$ is nonzero in $M_p/pM_p$, then $m_p$ generates $M_p$.
Apply the above to $M := f_*\Omega_{E/R}$, and $m := $ "$\frac{dx}{2y+a_1x + a_3}$'', using the fact that $\frac{dx}{2y+a_1x+a_3}$ is a generator for the fiber of $f_*\Omega_{E/R}\otimes k(x)$ at every $x\in\text{Spec }R$ (ie, a nowhere vanishing holomorphic differential on $E$ over $k(x)$)
