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At $P_1$ the tangent slope is zero. The elliptic-curve addition formula givesThe chord from $P_1$ to $P_2$ has slope $(-3-15)/(-6)=3$, so its sum has $x$-coordinate $3^2-0-(-6)=15$ and $y$-coordinate $3(0-15)-15=-60$. Thus $P_1+P_2=(15,-60)$.

Reduction at the good prime seven gives $\tilde{E}:y^2=x^3+1$. For $x=0,1,2,3,4,5,6$, the right sides are respectively $1,2,2,0,2,0,0$. There are respectively $2,2,2,1,2,1,1$ choices of $y$. Including the point at infinity gives $\#\tilde{E}({\mathbb{F}}_7)=12$. Its nonzero torsion points of an elliptic curve of order two are $(3,0),(5,0),(6,0)$, so it cannot be cyclic. Its two-primary component is $(\mathbb{Z}/2\mathbb{Z}{)}^2$ and its three-primary component is cyclic of order three; henceIn particular its exponent is six.

The reduction of an elliptic curve is a group homomorphism defined on every ${\mathbb{Q}}_7$-point by projectivity. Thus $6P$ reduces to the identity for every rational $P$. A finite point with integral coordinates reduces to an affine point, whose projective last coordinate is one, so it cannot reduce to the identity at infinity. Therefore every nonzero finite point $6P$ has nonintegral coordinates. If $6P=O$, it has no affine coordinates at all. This is the reduction exponent obstruction to integral multiples.

For good reduction at $p$, the reduction of an elliptic curve gives a mapwhose kernel is its formal group of an elliptic curve, and the torsion in that kernel is $p$-primary. Hence the prime-to-$p$ part of any torsion subgroup injects into $\tilde{E}({\mathbb{F}}_p)$.

At $p=2$, the odd part of $E(\mathbb{Q}{)}_{\mathrm{tors}}$ must divide $4$, so it is trivial. The torsion group is therefore a $2$-group. At $p=3$, all of this group has order prime to $3$ and injects into a group of order $7$, so it too is trivial. Thus

Writing $X=u$, $Y=v$, and $Z=-w$ turns the cubic into $X^3+Y^3=Z^3$. The birational coordinatesgive the Weierstrass equationThe original projective cubic has no common zero of its three partial derivatives modulo any $p\ne 3$, so it is already a smooth proper model and $C$ has good reduction at every such prime.

If $p\equiv 1(\mathrm{mod}3)$, then ${\mu }_3\subset {\mathbb{F}}_p$, all nine inflection points are rational, and $C({\mathbb{F}}_p)$ contains $(\mathbb{Z}/3\mathbb{Z}{)}^2$, so it is not cyclic. If $p\equiv 2(\mathrm{mod}3)$, cubing is a bijection on ${\mathbb{F}}_p$. Counting on the Fermat model gives $\#C({\mathbb{F}}_p)=p+1$. A finite elliptic-curve group has the form $\mathbb{Z}/d\mathbb{Z}\times \mathbb{Z}/e\mathbb{Z}$ with $d\mid e$ and $d\mid p-1$, so $d\mid \gcd (p-1,p+1)=2$. For odd $p$, the equation $x^3=432$ has exactly one root because cubing is bijective, so there is only one nonzero rational 2-torsion point and $d\ne 2$. Hence the group is cyclic. For $p=2$ it has order three and is cyclic as well.

Take $S$ to contain every finite prime dividing $n$ and every prime of bad reduction of $E$. The local theory of reduction of an elliptic curve shows that Kummer classes of rational points are unramified outside $S$. Choose a basis of the constant group $E[n]\cong (\mathbb{Z}/n\mathbb{Z}{)}^2$. Kummer theory and the Weil pairing identify the resulting two scalar coordinates ofwith power classes in $K^{\ast }/(K^{\ast }{)}^n$. The ramification statement places both coordinates in $K(S,n)$. Restriction to $\Gamma$ is injective by the nondegeneracy proved in part b, and therefore

A Minimal Weierstrass equation for $E/{\mathbb{Q}}_p$ is a Weierstrass equation of an elliptic curve with coefficients in ${\mathbb{Z}}_p$ whose discriminant has minimum $p$-adic valuation among all integral equations for $E$ related by admissible changes of variables.

Let $P\in E({\mathbb{Q}}_p)$ and choose projective coordinates $[X:Y:Z]$ with $X,Y,Z\in {\mathbb{Z}}_p$ and at least one coordinate a unit. Reducing the coordinates modulo $p$ givesMultiplying the primitive coordinates by a unit does not alter this point, so this defines the reduction map $E({\mathbb{Q}}_p)\to \bar{E}({\mathbb{F}}_p)$.

The Kummer map of an elliptic curve gives an injectionBecause $E[n]\subseteq E(K)$, the Galois action on $E[n]$ is trivial, so a cocycle in the image is a continuous homomorphism $G_K\to E[n]$. For $P\in E(K)$ and $Q$ with $nQ=P$, its kernel fixes the Kummer extension $K(Q)/K$, which is Galois of degree at most $n^2$ and exponent dividing $n$.

The local theory of reduction of an elliptic curve shows that these extensions are unramified outside the finite set consisting of primes dividing $n$, primes of bad reduction, and archimedean places. Local fields have only finitely many extensions of any bounded degree. Together with the Hermite-Minkowski finiteness theorem, this implies that only finitely many global extensions of degree at most $n^2$ with these ramification restrictions occur. Each has only finitely many homomorphisms to the finite group $E[n]$. Hence the image of $\delta$, and therefore $E(K)/nE(K)$, is finite.