Two-torsion and uniqueness of the group law on a Weierstrass cubic

In the chord-tangent law, $2P=O$ if and only if $P=-P$, i.e. if and only if $P$ is fixed by the inversion $[X:Y:Z]\mapsto [X:-Y:Z]$. For $P\mathrm{\ne }O$ write $P=[x:y:1]$; the fixed-point condition is $y=-y$, hence $y=0$ because $\mathrm{char}k\mathrm{\ne }2$, and such points satisfy $x^3+ax+b=0$. These equalities also compute the scheme-theoretic kernel: $[2]=0$ is equivalent on all tests to equality of identity and inversion. On $Z=1$ its ideal is $(2y)=(y)$. Near $O$ use the chart $Y=1$ with coordinates $u=X\mathrm{/}Y$, $v=Z\mathrm{/}Y$; inversion sends $(u,v)$ to $(-u,-v)$, so its equalizer has ideal $(u,v)$, defining the single reduced point $O$.

The polynomial $x^3+ax+b$ has three distinct roots in an algebraic closure: a common root of $x^3+ax+b$ and its derivative $3x^2+a$ would force $y=0$ in the smoothness computation, i.e. would give $3x^2=-a$, $x^3=b\mathrm{/}2$ and hence $4a^3+27b^2=0$, contrary to the hypothesis; so the discriminant $-(4a^3+27b^2)$ is nonzero. Hence $C[2](\bar{k})$ consists exactly of $O$ and the three points $[x:0:1]$ over the roots, and $C[2](k)=\{O\}$ exactly when the cubic has no $k$-root. These are two-torsion points, not inflection points; inflection points satisfy $3P=O$ instead.

For (b), let $\ast$ be the given abelian-variety law with unit $O$. The identity morphism $\mathrm{id}:(C,\cdot )\to (C,\ast )$ carries the unit of the chord-tangent law to the unit of $\ast$, and the source is a smooth geometrically integral group variety and the target an abelian variety, so by [F2] it is a group homomorphism; being an isomorphism of schemes, it is an isomorphism of group varieties, so the two laws coincide. The reverse implication is the same statement read backwards, and both laws are commutative by [F3].