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I am somewhat confused by the definition of the invariant differentials in J. Silverman's book The Arithmetic of Elliptic Curves.

Let $E$ be an elliptic curve with Weierstrass equation $F(x,y)=0$. Then the invariant differential corresponding to the Weierstrass equation is defined as $$\omega = \frac{dx}{2y+a_1x+a_3},$$ where the $a_i$ are as usual the coefficients of $F(x,y)$.

However, in formulation of various theorems, for example Theorem 5.2. on page 77, the notion of an invariant differential is used for a general elliptic curve, without explicit reference to any particular Weierstrass equation.

Another example is Proposition 1.1. in the book Advanced Topics in the Arithmetic of Elliptic curves. Here is claimed that any two invariant differentials on the elliptic curve $E_\Lambda$ are scalar multiples of one another. But strictly speaking we have defined only one invariant differential on $E_\Lambda$

Thus given an elliptic curve $E$, what is meant by an invariant differential on $E$?

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  • $\begingroup$ The invariant differential is defined by Silverman for a Weierstrass equation, not for an elliptic curve. However it follows from the formulas in Table 3.1 (last line) that for any two Weierstrass equations describing a given elliptic curve $E$, the associated invariant differentials will be scalar multiples of one another. (Alternatively, it also follows immediately from Proposition III.1.5.) $\endgroup$ – RP_ Jun 20 '19 at 17:29
  • $\begingroup$ Are you aware that the space of differential forms on $E$ (relative to the base field) is 1-dimensional? "invariant" is a red herring here. $\endgroup$ – abx Jun 20 '19 at 18:32
  • $\begingroup$ I really like @abx 's comment. Since the space of differentials is 1-D, any differential is invariant under any action. $\endgroup$ – meh Jun 20 '19 at 19:34
  • $\begingroup$ The previous comments are absolutely correct but also, an invariant differential is a differential form on the elliptic curve that is invariant under translation. It turns out that any global holomorphic differential form on an abelian variety is invariant. $\endgroup$ – Asvin Jun 21 '19 at 1:59
  • $\begingroup$ Perhaps this might help: mathoverflow.net/questions/82597/… $\endgroup$ – Tom Copeland Sep 3 '19 at 3:19

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