Over \(k = \bar{k}\) with \(\operatorname{char} k \neq 2\), every elliptic curve has a Legendre model \(y^2 = x(x-1)(x-\lambda)\) with \(\lambda \neq 0,1\), and \[ j = 2^8 \frac{(\lambda^2 - \lambda + 1)^3}{\lambda^2(\lambda-1)^2} \] depends only on \(E\). Two elliptic curves are isomorphic exactly when their \(j\)-invariants agree, and every element of \(k\) occurs. So \(\AA^1\) is a coarse moduli space for elliptic curves.
The construction combines Riemann–Roch, Riemann–Hurwitz, and normalization of branch points. Take \(\abs{2p_0}\): it is nonspecial, so Riemann–Roch gives \(\dim = 1\), and it is base-point free since \(E\) is not rational, giving a degree-two map \(E \to \PP^1\) with \(p_0 \mapsto \infty\). Riemann–Hurwitz forces exactly four branch points, which a Möbius transformation fixing \(\infty\) normalizes to \(0, 1, \lambda, \infty\). The residual ambiguity is the \(S_3\)-action permuting \(\{0,1,\lambda\}\), whose orbit is \(\lambda, \lambda^{-1}, 1-\lambda, (1-\lambda)^{-1}, \lambda(\lambda-1)^{-1}, \lambda^{-1}(\lambda-1)\); the formula for \(j\) is the symmetric function that kills it.
The Weierstrass model comes from \(\abs{3p_0}\): \(\ell(np_0) = n\) gives a linear relation among \(1, x, y, x^2, xy, y^2, x^3\), and completing the square puts it in Legendre form.
The space \(\AA^1\) is only a coarse moduli space: nontrivial isotrivial families show that no fine moduli space exists.
(Automorphisms)
Over \(k = \bar k\), \[ \abs{\Aut(E, p_0)} = \begin{cases} 2 & j \neq 0, 1728 \\ 4 & j = 1728, \ \operatorname{ch} k \neq 2, 3 \\ 6 & j = 0, \ \operatorname{ch} k \neq 2, 3 \\ 12 & \operatorname{ch} k = 3 \\ 24 & \operatorname{ch} k = 2 \end{cases} \] where in the last two rows \(1728 = 2^6 \cdot 3^3\) is zero in \(k\), so the conditions \(j = 0\) and \(j = 1728\) coincide and name a single curve.
Away from characteristics \(2\) and \(3\) the counts come from where the \(S_3\)-orbit of \(\lambda\) collapses: generically \(\Aut(E,p_0) = \ts{\pm 1}\); the orbit meets itself once at \(j = 1728\) (\(\lambda = -1\), the curve \(y^2 = x^3-x\)), giving order \(4\); and twice at \(j = 0\) (the Fermat cubic \(x^3+y^3=z^3\)), giving order \(6\). In all three cases the group is cyclic, generated by \((x,y) \mapsto (\zeta^2 x, \zeta^3 y)\) for an appropriate root of unity.
Characteristics \(2\) and \(3\) have different automorphism groups. In characteristic \(3\) take \(y^2 = x^3 - x\). Substituting \(x \mapsto u^2 x + r\), \(y \mapsto u^3 y\) and using that cubing is additive, the equation is preserved exactly when \(u^4 = 1\) and \(r^3 = r\), that is \(u \in \mu_4\) and \(r \in \FF_3\): twelve automorphisms. In characteristic \(2\) take \(y^2 + y = x^3\); the same computation with \(x \mapsto u^2x+r\), \(y \mapsto u^3y + u^2 s x + t\) forces \(u^3 = 1\), \(r = s^2\), \(s^4 = s\) and \(t^2 + t = s^6\), giving \(3 \cdot 4 \cdot 2 = 24\).
Two consequences follow. The automorphism group is noncommutative in these two characteristics — order \(12\) as \(\FF_3 \rtimes \mu_4\), order \(24\) as \(\SL_2(\FF_3)\) — whereas it is cyclic everywhere else. And the unique curve involved is supersingular, which is why characteristics \(2\) and \(3\) are excluded from so many statements in this chapter: the Legendre model itself requires \(\operatorname{ch} k \neq 2\).
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