Homomorphism of Legendre curve

Question

Let E be an elliptic curve over a finite field k (char(k) is not 2) be given by y^2 = (x-a)(x-b)(x-c) where a,b and c are distinct and are in k. Then why is (c,0) is in [2]E(k) iff c-a and c-b is a square in k-{0}?

Answer 1

Here is a conceptual explanation that applies to any $E/k$ with $\operatorname{char} k \ne 2$ and $E[2] \subseteq E(k)$, and that explains why $x-a$ and $x-b$ are the relevant rational functions (the field $k$ need not be finite).

The map $[2]\colon E \to E$ makes the function field $k(E)$ a finite extension of itself, say $L/K$, and $L$ is the unique unramified $(\mathbb{Z}/2\mathbb{Z})^2$-extension of $K$ in which the point at infinity splits completely (here we use that $E[2]$ is rational). The field $K(\sqrt{x-a},\sqrt{x-b})$ satisfies these conditions, so it is $L$. To say that a point $(x_0,y_0) \in E(k)$ lies in $2 E(k)$ means that it splits in $L/K$. When $x_0 \notin \{a,b,\infty\}$, this means simply that $x_0-a$ and $x_0-b$ should be squares in $k$.

Answer 2

For an elliptic curve over any field $K$ of characteristic different from $2$, the Kummer sequence reads

$0 \rightarrow E(K)/2E(K) \stackrel{\iota}{\rightarrow} H^1(K,E[2]) \rightarrow H^1(K,E)[2] \rightarrow 0$.

In particular, $\iota$ is an injection. Therefore $P \in 2E(K) \iff$ the image $[P]$ of $P$ in $E(K)/2E(K)$ is equal to zero $\iff \iota([P]) = 0$.

Moreover, since you have full $2$-torsion, $H^1(K,E[2]) \cong (K^{\times}/K^{\times 2})^2$ and in this case there is a well-known explicit description of the Kummer map: for any point $P = (x,y)$ different from $(a,0)$ and $(b,0)$,

$\iota(P) = (x-a,x - b) \pmod{K^{\times 2} \times K^{\times 2}}$:

see e.g. Proposition X.1.4 of Silverman's book. The result you want follows immediately from this, taking $P = (c,0)$.

Note that, as Bjorn points out in his nice answer to the question, the finiteness of $K$ is not needed or used here. In my original version of this answer, I mentioned the fact that $K$ finite implies $H^1(K,E) = 0$ -- it seemed like it could be helpful! -- but the argument does not in fact use the surjectivity of $\iota$, so is valid over any field of characteristic different from $2$ over which $E$ has full $2$-torsion.

Answer 3

Pete's is certainly the right way to look at this problem, but in this example one can argue naively using explicit calculations. One loses no generality by assuming $c=0$ (by replacing $x$ by $x+c$). Then using the duplication formula, one finds that the solutions of $[2]P = (0,0)$ are $P=(uv,uv(u+v))$ where $u$ and $v$ run through the square roots of $-a$ and $-b$ respectively. If $-a$ and $-b$ are squares in $k$ then each $P$ has coordinates in $k$. If one of the $P$ has coordinates in $k$ then they all do: so both $(uv,uv(u+v))$ and $(-uv,-uv(u-v))$ lie in $E(k)$. Thus $uv$, $u+v$ and $u-v$ lie in $k$. Hence $u\in k$ and $v\in k$ so that $-a$ and $-b$ are squares in $k$.

(Like Pete's and Bjorn's solutions, this does not require the finiteness of $k$.)