Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2009 / B / 4

Work with the smooth projective cubic, including its points at infinity. For two points $P,Q$, draw their chord, using the tangent when they coincide, and let $R$ be its third intersection counted with intersection multiplicity. Draw the line through $R$ and $o$, again using the tangent if necessary, and let $S$ be its third intersection. Define $P+Q=S$. When $o$ is a flex, this is the familiar reflection of the first third-intersection point across the Weierstrass involution. The two-line construction also works when the chosen identity is not a flex.

To prove associativity without assuming it in the geometry, use divisor class. A nonsingular plane cubic has genus one. By the Riemann-Roch theorem, every divisor class of degree one contains exactly one effective divisor on an algebraic curve of degree one: for such a divisor on an algebraic curve $D$,and $\mathrm{deg}(K_C-D)=-1$, so $\ell (K_C-D)=0$. Thus $\ell (D)=1$, and the nonzero section is unique up to scalar. Over $K$, a degree-one effective divisor on an algebraic curve defined over $K$ is a $K$-rational point. This also proves uniqueness of that point. In particular the mapis injective and identifies these points with the corresponding degree-zero Picard group of a curve: to recover the point from a class, add $[o]$ and take its unique effective representative. This is the elliptic curve group law from Riemann-Roch.

Let $H$ denote the divisor class cut out by a line. The two lines in the chord-and-tangent group law giveTheir difference gives $P+Q\sim o+S$, henceAddition of divisor class is associative. Therefore both $(P+Q)+R$ and $P+(Q+R)$ map to $[P-o]+[Q-o]+[R-o]$. Injectivity of $\iota$ proves the chord-and-tangent group law is associative. The same argument gives identity $o$, commutativity and inverses, and covers tangencies through the use of intersection multiplicities. All constructions are defined over $K$.

The three suggested values areThere is an obstruction at $p=3$: every integer residue satisfies $x^3-x\equiv 0(\mathrm{mod}3)$, whereas $-1\equiv 2$ is not a square modulo three. Reduction of a solution in the p-adic integers would give an impossible solution modulo three.

At $p=2$, take $x=-2$. The odd 2-adic unit in a ring $-7$ is congruent to one modulo eight, so it is a square in ${\mathbb{Z}}_2$. We use the standard 2-adic square criterion that an odd unit in a ring is a square exactly when it is one modulo eight. Its sufficiency follows by successively adjusting an odd square root modulo $2^m$ by $2^{m-1}$ to obtain the next required square congruence. Thus $y^2=-7$ has a 2-adic integer solution.

For $p\ge 5$ other than $23$, the discriminant of $y^2=x^3-x-1$ isso this model has good reduction of an elliptic curve. Hasse's theorem on elliptic curves givesSince there is just one point at infinity, there is an affine point. Smoothness means that at least one partial derivative of $y^2-x^3+x+1$ is a unit in a ring modulo $p$, so Hensel's lemma lifts the affine point to ${\mathbb{Z}}_p^2$. The last exceptional odd prime is also soluble: at $p=23$, $h(-2)=-7\equiv 16$ has square roots $\pm 4$, which are simple roots of $Y^2+7$ modulo $23$, and lift by Hensel's lemma.

For small-prime checks the suggested values also give $x=0$ at $p=5$ and $x=3$ at $p=7$, where respectively $-1\equiv 2^2$ and $23\equiv 3^2$. ThereforeThe same prime list holds if one only asks for a solution modulo $p$, but the lifting above establishes the asserted p-adic integer solutions. This is local solubility of the cubic x cubed minus x minus one.

In the elliptic curve method, addition is attempted modulo the composite number. A denominator that is not a unit in a ring supplies a factor through its greatest common divisor with that number.

First double $P$ using the elliptic-curve addition formula. Its tangent slope isThese calculations are valid modulo $221$ since the only initial denominator is two, which is invertible. Thus $2P=(14,-53)$, equivalently $(14,168)$ modulo $221$.

To form $3P=P+2P$, the chord slope would requireThe denominator hasIt cannot be inverted, and thirteen is the requested proper factor. The failure has a geometric explanation: modulo thirteen, $2P=(1,-1)=-P$, so $3P$ is the point at infinity. Modulo seventeen that chord denominator remains nonzero. The differing behavior at the two primes is exactly what the elliptic curve method detects.

Choose the blocksIf $p^k\le n<p^{k+1}$, then $|x_n{|}_p=p^{-k}$, so $x_n\to 0$ in ${\mathbb{Q}}_p$. By the ultrametric inequality, for $N\ge M$,The partial sums are Cauchy sequence, and completeness of ${\mathbb{Q}}_p$ proves convergence of $\sum x_n$.

In the real series of p-adic absolute values, the block $p^k\le n<p^{k+1}$ contains $(p-1)p^k$ equal terms of size $p^{-k}$. Its sum is $p-1$, independently of $k$. Infinitely many such positive block sums force divergence. This exhibits p-adic convergence without summable absolute values: the criterion is that the terms tend to zero, rather than real absolute summability.