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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)?

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1 Answer 1

up vote 8 down vote accepted

The isomorphism coming from duality is correct, but it is not true that the invertible sheaf $R^1f_*O_E$ on $S$ is trivial in general.

Here is a down-to-earth construction of an example where $f_*\Omega_{E/S}$ is not free. Let $S=\mathrm{Spec}(R)$ be a domain (regular of dimension $1$ if you want) such that $2\in R^*$ and $\mathrm{Pic}(S)=H^1(S, O_S^*)$ has an element of order $4$. Represent this element by a $1$-cocycle $(u_{ij})_{ij}$ with $u_{ij}\in O_S(U_{i}\cap U_{j})^*$ where $\{ U_{i} \}_i$ is some open affine covering of $S$. By hypothesis of $4$-torsion, there exist $v_i\in O_S(U_i)^*$ such that $u_{ij}^4=v_iv_j^{-1}$ for all $i,j$.

Let $E_i/U_i$ be the elliptic curve defined by the affine equation $$ y_i^2=x_i^3+ v_i x_i.$$ Over $U_{i}\cap U_{j}$, the elliptic curves $E_i, E_j$ are isomorphic by the change of variables $$y_j=u_{ij}^3y_i, \quad x_j=u_{ij}^2x_i.$$ The cocycle condition on the $\{u_{ij}\}$'s insures that the $E_i$'s glue together and gives rise to an elliptic curve $E/S$.

Now let's compute the differential sheaf on $E$. On $E_i$, a canonical basis of $\Omega_{E_i/U_i}$ is $\omega_i:=dx_i/y_i$ and we have $\omega_j=u_{ij}^{-1}\omega_i$ above $U_i\cap U_j$. If $\omega_{E/S}$ is free and generated by $\omega$, then writing $\omega_i=u_i\omega|_{E_i}$ with $u_i\in O(U_i)^*$, we get $u_{ij}=u_iu_j^{-1}$. Hence the class of $(u_{ij})$ in $H^1(S, O_S^*)$ is trivial, contradiction.

One can show that over a Dedekind domain (or any ring I think), $f_*\Omega_{E/S}$ is free if and only $E/S$ has a global smooth Weierstrass equation. It is well known (?) that this doesn't hold in general over number fields.

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Dear QiL, the phrase " ...has an element of order $4$ and $2\in R^*$ " is difficult to scan : it seems to imply that there are two orders involved and that they are in $R^*$ ! Might I humbly suggest that you write $2\in R^*$ somewhere earlier in the sentence? (I have upvoted your answer , needless to say). –  Georges Elencwajg Sep 2 '12 at 12:33
    
Dear @GeorgesElencwajg, you are right. The second condition was added while writing the example. Thanks ! –  user18119 Sep 2 '12 at 13:49
    
Thanks for the very clear, detailed answer, @QiL! –  Keenan Kidwell Sep 2 '12 at 14:05

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