Torsion group of elliptic curve isomorphic to $(\mathbb{Q}/\mathbb{Z})^{2}$

Question

I am reading preliminaries with the goal of learning the proof of Mazur's torsion theorem for elliptic curves.

I have read the fact: $E[m] \cong (\mathbb{Z}/m\mathbb{Z}) \oplus (\mathbb{Z}/m\mathbb{Z})$ for $E$ an elliptic curve over $K$ with $\text{char}(K) = 0$ and $m \in \mathbb{N}$. And $E[m] \cong (\mathbb{Z}/m\mathbb{Z}) \oplus (\mathbb{Z}/m'\mathbb{Z})$ when $\text{char}(K) = p$ and $m = p^{e}m'$ with $(p, m') = 1$.

There is a corollary: As an abstract group, $E_{\text{tors}} \cong (\mathbb{Q}/\mathbb{Z})\oplus(\mathbb{Q}/\mathbb{Z})$ if $\text{char}(K) = 0$, and $E_{\text{tors}} \cong (\mathbb{Z}_{(p)}/\mathbb{Z})\oplus(\mathbb{Z}_{(p)}/\mathbb{Z})$ or $E_{\text{tors}} \cong (\mathbb{Q}/\mathbb{Z}) \oplus (\mathbb{Z}_{(p)}/\mathbb{Z})$ if $\text{char}(K) = p$.

I almost understand the statement "$E_{\text{tors}} \cong (\mathbb{Q}/\mathbb{Z})\oplus(\mathbb{Q}/\mathbb{Z})$" from Prove that $E(\mathbb{C})^{\text{tor}} \cong \mathbb{Q}/\mathbb{Z} \times \mathbb{Q}/\mathbb{Z}$., although $K$ is not necessarily $\mathbb{C}$. The second part (when $\text{char}(K) = p$) I do not understand. Apparently it follows immediately from the first statement. Any help or hint is appreciated.

Answer 1

Let $K$ be algebraically closed of characteristic $p$, and let $E$ be an elliptic curve over $K$. Then it is a theorem that the $p$-torsion $E[p]$ is either isomorphic to $\Bbb Z/p\Bbb Z$ ("ordinary") or has zero order ("supersingular"). In the ordinary case the $p$-power torsion will be $E[p^\infty]\cong \Bbb Z[1/p]/\Bbb Z$ and in the supersingular case it will of course be zero. In both cases, when $p\nmid N$, the $N$-torsion is isomorphic to $(\Bbb Z/N\Bbb Z)^2$.

Let's consider the ordinary case. Then $E_{\text{tors}}$ is the direct sum of the $E[q^\infty]$ over all primes $q$. When $q\ne p$, $E[q^\infty]\cong(\Bbb Z[1/q]/\Bbb Z)\oplus(\Bbb Z[1/q]/\Bbb Z)$. So $E_{\text{tors}}$ is the direct sum of two groups $A$ and $B$ with $A\cong\bigoplus_q (\Bbb Z[1/q]/\Bbb Z)$ and $B\cong\bigoplus_{q\ne p} (\Bbb Z[1/q]/\Bbb Z)$. But $\Bbb Q/\Bbb Z\cong A$ and $\Bbb Z_{(p)}/\Bbb Z\cong B$. These are just splitting up these torsion groups as direct sums of their $q^\infty$-components.

In the supersingular case, you'll get two copies of $B$.