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It apparently follows from work of Velu (MathSciNet) that every isogeny between elliptic curves in (long) Weierstrass form over $k$ can be written in the form $$ \left(\frac{u(x)}{v(x)}, \frac{s_1(x)+s_2(x)y}{t(x)}\right),\quad \text{with }u,v,s_1,s_2,t \in k[x] $$ (I was told this by Andrew Sutherland, but I cannot find a reference that discusses this over $\mathbb{Z}$ rather than a field, nor a reference that actually treats the case of arbitrary characteristic; the assumption that 6 is invertible is always made, to "simplify things". I'd also be interested to know if this was also true for nodal curves.)

Assuming the above result for schemes over $\mathbb{Z}$, that all isogenies are presented by the given data, what sort of scheme or ind-scheme does the collection of the appropriate polynomials $\{u,v,s_1,s_2,t\}$ form? Clearly one would want to restrict to coprime $u$ and $v$, and similar conditions for the others, so this cuts things down a little.

Please note that I am not an algebraic geometer, so it's not obvious to me what properties the spaces of polynomials have, much less when cutting down to non-redundant data. I do know that we don't have anything approaching quasi-compactness. Can I think of $(u,v)$ as just being the data of a rational function in one variable? Then wouldn't they form some sort of Hilbert scheme? I really don't know.

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I know how to construct the space of isogenies as a scheme. We start by noting that isogenies are exactly the elements of the $\mathrm{Hom}(E_1, E_2)$ that map the identity to the identity. Since $\mathrm{Hom}$ is representable for projective $E_1$ and $E_2$ by carving the graphs out of the Hilbert scheme of $E_1 \times E_2$, we have some kind of representation of the isogenies as a scheme, and it may be very explicit if you write down parts of the Hilbert scheme explicitly. What I don't know is what geometric properties it has that can be proven from this description.

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  • $\begingroup$ Thanks. The 'if' in the 'if you write down' is the sticking point: this is not my area of expertise... $\endgroup$
    – David Roberts
    Commented Feb 26, 2016 at 11:13
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    $\begingroup$ Nitpick: you have to remove the constant map from your scheme (since the constant map at the identity is not an isogeny) $\endgroup$ Commented Mar 27, 2016 at 12:56

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