An integral Weierstrass equation has <good reduction of an elliptic curve>[good reduction] outside the finitely many primes dividing its nonzero discriminant. This proves finiteness of the set of bad primes. To prove finiteness of rational torsion, choose two distinct good primes. The <reduction of torsion points on an elliptic curve> injects each primary component at a good prime of different residue characteristic, so the two finite reduced point groups bound every primary component of $E(\mathbb Q)_{\rm tors}$.

For
$$
E:y^2=x^3+30x+30,
$$
the displayed equation is minimal and
$$
\Delta=-16(4\cdot30^3+27\cdot30^2)
=-2116800=-2^6\,3^3\,5^2\,7^2.
$$
Its bad primes are therefore exactly
$$
\boxed{\{2,3,5,7\}.}
$$
The good reductions at $11$ and $13$ have
$$
\#E(\mathbb F_{11})=8,
\qquad
\#E(\mathbb F_{13})=15.
$$
Their coprime orders exclude every rational torsion primary component, including the residue-characteristic components by using the other prime. Hence
$$
\boxed{E(\mathbb Q)_{\rm tors}=0.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 125 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-125/3/c/solution