Torsion parts of elliptic curve

Question

Given an elliptic curve $E$ over a field $k$, I know that that the kernel of the power map $[n]:E\rightarrow E$ (here $char(k)\not| n$) has the structure $ker[n]=\operatorname{Spec}(\bigsqcup_i K_i)$ where each $K_i$ is finite separable over $k$ and $\Sigma_i [K_i:k]=n^2$.

Now in the notes I am reading the following argument is used:

$ker[n](\overline{k})$ is an abelian group of order $n^2$ (here $\overline{k}$ is some algebraic closure of $k$). For all $d|n$ we have that $ker[d](\overline{k})=(ker[n])(\overline{k})[d]$. Hence, $(ker[n])(\overline{k})\cong (\mathbb{Z}/n\mathbb{Z})^2$.

The only thing that does not make sense to me is the final conclusion. I understand everything before. But, why do the arguments before have to imply that the $\overline{k}$-rational points of the kernel is not a cyclic group of order $n^2$?

Appologies if it turns out to be some silly argument that I am missing.

Answer 1

What you need is the following lemma, stated as Exercise 3.30 in Silverman's book "The Arithmetic of Elliptic Curves" or (unnumbered) Lemma p. 126 in Hindry--Silverman's book "Diophantine Geometry".

Let $A$ be a finite abelian group of order $N^r$ for some integers $N > 1, r \geq 1$. Suppose that for every divisor $D$ of $N$ we have $|A[D]| = D^r$, where $A[D]$ denotes the subgroup consisting of all elements of order dividing $D$. Then $A \cong (\Bbb Z / N \Bbb Z)^r$.

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The proof is just to decompose $A$ uniquely as a product $$\Bbb Z / d_1 \Bbb Z \times \cdots \times \Bbb Z / d_s \Bbb Z,$$ where $d_i$ divides $d_{i+1}$ and $d_i > 1$ for all $1 \leq i < s$.

You can easily see that $d_s = N$, since $A = A[d_s]$ (since $d_i$ divides $d_{i+1}$) and $A = A[N]$ (both have cardinality $N^r$).

Then you can also see that $r=s$, and finally $d_i = n$ for all $i$. I let you think about this and fill in the details.

(See also Silverman AEC: exercise 3.30 for another proof).