Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 125 / 3 / ii

In fact, $2,3,5,7$ must all be primes of bad reduction.

At a prime of good reduction of an elliptic curve, the kernel $E_1({\mathbb{Q}}_p)$ of reduction is the formal group of an elliptic curve evaluated on $p{\mathbb{Z}}_p$, using its parameter at the identity. This kernel has no prime-to-$p$ torsion: for an integer $\ell$ prime to $p$, the multiplication series is $[\ell {]}_F(T)=\ell T+$ higher-degree integral terms, so for nonzero $T\in p{\mathbb{Z}}_p$ its first term has strictly smaller valuation than the others and $[\ell {]}_F(T)\ne 0$.

For each odd good prime, the entire given torsion subgroup, of order $16$, would therefore inject into $E({\mathbb{F}}_p)$. But the Hasse theorem for elliptic curves givesNone can contain a subgroup of order $16$. Hence $3,5,7$ are bad.

For the prime-two torsion bound for good reduction, use the deeper formal subgroup corresponding to $4{\mathbb{Z}}_2$. Part (i), with $m=2>1/(2-1)$, shows that $E_2({\mathbb{Q}}_2)$ is torsion-free and has index $2$ in $E_1({\mathbb{Q}}_2)$. Therefore any finite torsion subgroup of $E_1({\mathbb{Q}}_2)$ injects into $E_1/E_2$ and has order at most $2$.

If $E$ had good reduction at $2$, its rational torsion subgroup of order $16$ would consequently have reduction image of order at least $8$. The Hasse theorem for elliptic curves instead gives $\#E({\mathbb{F}}_2)\le 5$. This contradiction proves bad reduction at $2$ as well, and thereforeThe special formal-group argument at $2$ is necessary; prime-to-$p$ injectivity alone says nothing about the given $2$-primary torsion there.