Torsion point of an elliptic curve ($E[n]$)

The $n$-torsion subgroup $E[n]$ is the kernel of multiplication by $n$. When the characteristic does not divide $n$, it is isomorphic over an algebraic closure to $(\mathbb{Z}/n\mathbb{Z}{)}^2$.

Suppose $15\nmid n$ and both $n,-n$ are nonsquares. Each elliptic curve has just one nonidentity rational 2-torsion point, $(0,0)$. The duplication formula shows that halving it requires the corresponding coefficient $\pm n$ to be a square, so neither has rational order-four points. If $3\nmid n$, both good reductions at three have order four, forcing torsion point of an elliptic curve to be exactly order two. Otherwise $5\nmid n$; the reduction orders for coefficients $1,2,3,4$ modulo five are $4,2,10,8$. Among opposite coefficients at least one has order two or four, giving the same conclusion for at least one curve by reduction of torsion points on an elliptic curve.

If $\ell$ is a prime different from the characteristic, multiplication by $\ell$ on an elliptic curve has degree ${\ell }^2$ and nonzero differential. It is a separable isogeny, so its geometric kernel has ${\ell }^2$ points, all killed by $\ell$. This gives the displayed group over an algebraic closure; the rational subgroup over the base field may be smaller.

For integral $m\ge 2$, suppose $E:y^2=x(x+1)(x+m^2)$ has good reduction at five or seven. At five, $m^2\equiv 4$; at seven, $m^2\equiv 2$ or $4$. Each resulting reduced elliptic curve has eight points. These restrictions follow from $\Delta =16m^4(m^2-1{)}^2$ and $c_4=16(m^4-m^2+1)$: if an odd prime divides $m(m^2-1)$, the j-invariant of an elliptic curve has negative valuation. The torsion-freeness of the formal group over Qp for odd p bounds the whole rational torsion by eight. The point $(m,m(m+1))$ doubles to $(0,0)$, and $(-1,0)$ is an independent 2-torsion point. These generate the stated group of order eight.

For positive integral $k$, the rational torsion subgroup of $y^2=x^3+kx$ has order dividing four. Indeed, torsion-freeness of the formal group over Qp for odd p, reduction of torsion points on an elliptic curve, and the point-count criterion for y squared equals x cubed plus k x bound its order by $p+1$ at all good primes $p\equiv 3(\mathrm{mod}4)$. The Dirichlet theorem on primes in arithmetic progressions excludes every odd torsion prime and bounds the two-primary part by four, by choosing $p\equiv 3(\mathrm{mod}8)$. There is exactly one nonzero rational 2-torsion point, namely $(0,0)$. A rational half of it must have $x^2=k$ by the elliptic-curve addition formula. Writing $k=s^2$ with $s>0$, its equation is $y^2=2s^3$, so $s=2t^2$ and $k=4t^4$. Conversely $(2t^2,4t^3)$ has order four. Thus the cyclic group of order four occurs exactly for $k=4t^4$, $t\ge 1$. The order bound itself holds for every nonzero integral $k$ of either sign: positivity was used only to classify the rational points of orders two and four.

On a Weierstrass equation $y^2=f(x)$ in characteristic other than two, the nonidentity 2-torsion points are exactly $(r,0)$ for the roots $r$ of $f$.

The division polynomial ${\psi }_n$ has roots at the nonzero $n$-torsion points of an elliptic curve, with the two points sharing an $x$-coordinate represented by the same root when $n$ is odd.

The Nagell–Lutz theorem states that a nonzero rational torsion point on an integral short Weierstrass curve has integral coordinates; if its $y$-coordinate is nonzero, then its square divides the discriminant.

For an integral torsion point $(x,y)$ with $y\ne 0$, the point $2P$ is integral. The tangent slope $(3x^2+a)/(2y)$ has integral square and is rational, hence is integral. Thus $y^2$ divides $(3x^2+a{)}^2$. Combining this with the curve equation inshows that $y^2\mid 4a^3+27b^2$.

The $n$-division field is obtained by adjoining to $K$ the coordinates of every point of $E[n]$. Over a finite field its degree is the order of Frobenius acting on $E[n]$.

Let $E/K$ have good reduction over a local field of residue characteristic $p$, with $p\nmid n$. All solutions of $nQ=P$, for all $P\in E(K)$ together, lie in a single finite unramified extension of degree at most $nf$, where $f=[K(E[n]):K]$. First lift a solution over the algebraic closure of the residue field to a finite unramified extension; the multiplication isomorphism of a formal group law uniquely corrects the error in $E_1$. This also puts $E[n]$ in the maximal unramified extension. If $\sigma$ is its Frobenius generator, ${\sigma }^f$ fixes $E[n]$. For $nQ=P$, set $T={\sigma }^{f}Q-Q\in E[n]$. Then ${\sigma }^{nf}Q=Q+nT=Q$, uniformly in $P$.

The absolute Galois group acts on $E[3]\simeq {\mathbb{F}}_3^2$, giving the mod-three Galois representation. Its invariant vectors are precisely the rational points of order dividing three.

The Weil pairing is a bilinear, alternating, nondegenerate pairingFor an isogeny $\varphi$ and its dual, it satisfies $e_n(\varphi P,Q)=e_n(P,\hat{\varphi }Q)$ whenever both sides are defined.