A smooth Weierstrass elliptic cubic is a nonaffine algebraic group

By [F1] and [F3], $E$ is a smooth projective one-dimensional scheme over $\mathbf{C}$, hence proper: a closed subscheme of proper projective space is finite type and separated, and its projection remains closed after arbitrary base change. Its complex manifold is connected by the biholomorphism with the connected torus. The irreducible components of a smooth algebraic scheme cannot meet, since its regular local rings are domains; the finitely many components are therefore both algebraically and analytically open and closed. Connectedness leaves exactly one, so $E$ is integral and, over the algebraically closed field $\mathbf{C}$, geometrically integral.

Transported addition on $E(\mathbf{C})$ is jointly holomorphic: around any two points lift to torus coordinates $z,w$, use the holomorphic map $(z,w)\mapsto z+w$, and compose with the biholomorphism $\mathrm{\Phi }$ and its inverse from [F1]. On the dense algebraic open where $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ are affine and $x_1\mathrm{\ne }{x}_2$, set $m=(y_2-y_1)\mathrm{/}(x_2-x_1)$. The chord meets the cubic at a third point with $x$-coordinate $m^2\mathrm{/}4-x_1-x_2$, by comparison of the cubic's quadratic coefficient. Negating its $y$-coordinate gives the rational formulas $x(P+Q)=m^2\mathrm{/}4-x_1-x_2$ and $y(P+Q)=-y_1+m(x_1-x(P+Q))$. They agree with the holomorphic group law by [F1]. Apply [F2] to get a regular algebraic multiplication $E\times E\to E$. Inverse is the regular projective map $[X:Y:Z]\mapsto [X:-Y:Z]$, and the identity is the rational point $O$.

Associativity, inverse, and identity hold on every complex point by transport from the torus. They hold as scheme morphism identities: the relevant product schemes are reduced, their complex closed points are Zariski dense, and equality is a closed condition because the target is separated. Consequently $E$ is a group variety, and properness and geometric connectedness from step 1.1 make it an abelian variety by [F4]. Since $\dim E=1$, the nonaffineness criterion in [F4] proves that $E$ is not affine. This verification explicitly establishes regularity of the algebraic law; analytic uniformization alone was not treated as that supplier. AC is carried through [F2]–[F4].