Let \(X, P_0\) be an elliptic curve defined over \(\QQ\), represented as a curve in \(\PP^2\) defined by an equation with integer coefficients. Then \(X\) can be considered as the fibre over the generic point of a scheme \(\bar{X}\) over \(\Spec \ZZ\). Let \(T \subseteq \Spec \ZZ\) be the open subset consisting of all primes \(p \neq 2\) such that the fibre \(X_{(p)}\) of \(\bar{X}\) over \(p\) is nonsingular.
For any \(n\), show that \(n_X: X \to X\) is defined over \(T\), and is a flat morphism.
Show that the kernel of \(n_X\) is also flat over \(T\).
Conclude that for any \(p \in T\), the natural map \(X(\QQ) \to X_{(p)}(\FF_p)\) induced on the groups of rational points, maps the \(n\)-torsion points of \(X(\QQ)\) injectively into the torsion subgroup of \(X_{(p)}(\FF_p)\), for any \((n, p)=1\).
By this method one can show easily that the groups \(X(\QQ)\) in (Ex. 4.17) and (Ex. 4.18) are torsion-free.
Solution
Write \[ \mathcal E=\bar X\times_{\Spec\ZZ}T \] and let \[ e:T\longrightarrow\mathcal E \] be the section extending the chosen origin \(P_0\). For an integer \(n\ge1\), write \([n]\) for multiplication by \(n\) and \[ \mathcal E[n]=\ker[n]. \]
1. The smooth model \(\mathcal E/T\) is an elliptic scheme, and\[ \boxed{[n]:\mathcal E\longrightarrow\mathcal E} \]extends \(n_X:X\to X\).
Over \(T\) every fibre of the projective plane cubic \(\bar X\) is nonsingular. Thus \(\mathcal E\to T\) is a smooth proper family of genus-one curves. The rational point \(P_0\) extends uniquely to a section over \(T\): locally at a prime of \(T\) this is the valuative criterion of properness, and uniqueness follows from separatedness. A smooth proper genus-one curve with a section is an elliptic scheme, so its group law, inverse, and zero section are all defined over \(T\).
Multiplication by \(n\) is obtained from the group law by repeated addition. Hence it is a morphism over \(T\), and its generic fibre is the original multiplication map \(n_X\).
2. The morphism\[ \boxed{[n]:\mathcal E\longrightarrow\mathcal E} \]is finite and flat over \(T\).
For every point \(s\in T\), the fibre \(\mathcal E_s\) is an elliptic curve. By Exercise IV.4.7(e), \[ \deg[n]_{\mathcal E_s}=n^2. \] Thus \([n]_{\mathcal E_s}\) is a nonconstant finite morphism of nonsingular curves. Such a morphism is flat: on local rings it gives a finite torsion-free module over a discrete valuation ring, hence a free module.
The source \(\mathcal E\) is flat over \(T\). The fibrewise flatness criterion therefore shows that \([n]\) itself is flat. It is also proper because \(\mathcal E\) is proper over \(T\) and the target is separated over \(T\). Every fibre of \([n]\) is finite, so \([n]\) is quasi-finite. A proper quasi-finite morphism is finite. Hence \([n]\) is finite and flat.
3. The kernel\[ \boxed{\mathcal E[n]\longrightarrow T} \]is finite and flat.
By definition, \[ \mathcal E[n] = T\times_{e,\mathcal E,[n]}\mathcal E. \] Thus \(\mathcal E[n]\to T\) is the base change of the finite flat morphism \([n]\) from step 2 along the zero section. Finiteness and flatness are preserved by base change.
4. Fix \(p\in T\) with \((n,p)=1\). After base change to\[ S=\Spec\ZZ_{(p)}, \]the group scheme \(\mathcal E[n]_S\) is finite etale over \(S\).
Since \(p\nmid n\), the integer \(n\) is a unit on \(S\). The differential of \([n]\) on the relative tangent line at the identity is multiplication by \(n\), hence is an isomorphism. Translation by a section identifies the differential at any point with the differential at the identity, so \([n]\) is unramified over \(S\). Step 2 says that it is finite flat. Therefore \([n]\) is finite etale over \(S\).
The kernel \(\mathcal E[n]_S\) is its base change along the zero section, so it is finite etale over \(S\) as well.
5. For every \(p\in T\) and every \(n\) prime to \(p\), reduction induces aninjection\[ \boxed{ X(\QQ)[n]\hookrightarrow X_{(p)}(\FF_p)[n]. } \]
Again put \(S=\Spec\ZZ_{(p)}\). A point \[ P\in X(\QQ)[n] \] extends uniquely, by properness, to a section \[ s_P:S\longrightarrow\mathcal E_S. \] The two sections \([n]\circ s_P\) and \(e\) agree on the generic point of \(S\). Since \(\mathcal E_S\) is separated, they agree on all of \(S\). Hence \(s_P\) factors through the finite etale group scheme \(\mathcal E[n]_S\) of step 4.
Suppose \(P,Q\in X(\QQ)[n]\) have the same reduction modulo \(p\). Then the corresponding sections \[ s_P,s_Q:S\longrightarrow\mathcal E[n]_S \] agree at the closed point. Because \(\mathcal E[n]_S\to S\) is etale, its diagonal is open; because it is separated, its diagonal is closed. Thus the equalizer of \(s_P\) and \(s_Q\) is both open and closed in the connected scheme \(S\). It contains the closed point, so it is all of \(S\). Therefore \(s_P=s_Q\), and in particular \(P=Q\) on the generic fibre.
Reduction is a homomorphism because it is induced by the group scheme \(\mathcal E_S\). Hence it injects \(X(\QQ)[n]\) into the \(n\)-torsion of \(X_{(p)}(\FF_p)\), proving the required assertion.
6. The groups of rational points in Exercises IV.4.17 and IV.4.18 aretorsion-free.
First consider \[ E_{17}:y^2+y=x^3-x. \] By Exercise IV.4.17, its discriminant is \(37\), so it has good reduction at \(3\) and \(5\). Direct enumeration gives \[ \#E_{17}(\FF_3)=7, \qquad \#E_{17}(\FF_5)=8. \] Indeed, for \(x=0,1,2\) over \(\FF_3\) the numbers of affine \(y\)-solutions are \(2,2,2\), while over \(\FF_5\) for \(x=0,1,2,3,4\) they are \(2,2,1,0,2\); in each case one then adds the point at infinity.
Now consider \[ E_{18}:y^2=x^3-7x+10. \] Its discriminant is \[ -16\bigl(4(-7)^3+27(10)^2\bigr) =-2^8\cdot83, \] so again \(3\) and \(5\) are primes of good reduction. Direct enumeration gives \[ \#E_{18}(\FF_3)=7, \qquad \#E_{18}(\FF_5)=10. \] For \(x=0,1,2\) over \(\FF_3\) the numbers of affine \(y\)-solutions are \(2,2,2\), while over \(\FF_5\) for \(x=0,1,2,3,4\) they are \(1,2,2,2,2\).
Suppose either rational group had nonzero torsion. Then it would contain a point of some prime order \(\ell\). If \(\ell\ne3,5\), step 5 at both primes would force \(\ell\) to divide both displayed group orders, impossible because \[ \gcd(7,8)=\gcd(7,10)=1. \] If \(\ell=3\), reduction at \(5\) would force \(3\) to divide respectively \(8\) or \(10\), again impossible. If \(\ell=5\), reduction at \(3\) would force \(5\mid7\), impossible. Thus neither rational group contains a nonzero torsion point.
7. Q.E.D.
Steps <1>1–<1>2 prove that multiplication by \(n\) is defined and flat over \(T\), step 3 proves flatness of its kernel, step 5 proves the prime-to-\(p\) injectivity under reduction, and step 6 proves the stated torsion-free applications.