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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.

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  • $\begingroup$ @Epargyreus not a duplicate. The last paragraph of the answer states that the sheaf is free, but does not prove it or give a reference. I'm asking for a proof or a reference to a proof. $\endgroup$
    – oxeimon
    Dec 22 '15 at 7:32
  • $\begingroup$ You have an $R$-module which contains an explicit element $\omega$, and you want to know whether it is free. The formation of $\omega$ commutes with taking localizations and quotients of $R$, so (by Nakayama's Lemma) this immediately reduces to the case of a field, which you say you understand. $\endgroup$
    – Epargyreus
    Dec 22 '15 at 15:58
  • $\begingroup$ When you say $f:E\to\mathrm{Spec}\, R$ is an elliptic curve, do you mean that $f$ is smooth and has a section? $\endgroup$
    – Mohan
    Dec 22 '15 at 15:58
  • $\begingroup$ @Mohan yes that's what I mean $\endgroup$
    – oxeimon
    Dec 22 '15 at 17:38
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Ah okay, so as Epargyreus noted, we may reduce to the case of $R$ a field as follows.

There are a few steps.

  1. 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.

  2. 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.

  3. 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 $k(x)$)

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  • $\begingroup$ Given an elliptic curve $E$ over a scheme $S$, you can understand why $\omega = f_* \Omega^1_{E/S}$ is locally free of rank one in a similar way. The key point is that, by proper base change, you may reduce to the case that $S = \mathrm{Spec}(A)$ is local. Then by looking at the special fibre and generic fibres you are reduced to showing that $H^0(E,\Omega^1_{E/K})$ is one-dimensional for any field $K$, which is true almost by definition. $\endgroup$
    – Epargyreus
    Dec 23 '15 at 17:53
  • $\begingroup$ @Epargyreus It seems that proper base change (according to Wikipedia) can only give you that $f_*\Omega^1_{E/S}$ is locally free when $S$ is reduced (as well as connected). Is reducedness really necessary? Could $f_*\Omega^1_{E/S}$ fail to be locally free when $S$ is nonreduced? What reference are you using for proper base change? $\endgroup$
    – oxeimon
    Dec 24 '15 at 2:46
  • $\begingroup$ I confess I was thinking specifically about the case when $S$ is integral as in your example. The general case, as far as I remember, is a little annoying. $\endgroup$
    – Epargyreus
    Dec 24 '15 at 3:37

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