I'm trying to prove that, if $C \subseteq \mathbb{P}^2$ is a smooth plane cubic defined over $\mathbb{Q}$, and $K/\mathbb{Q}$ is a quadratic number field, then $C(K) \ne \emptyset \implies C(\mathbb{Q}) \ne \emptyset$.
My idea was as follows:
Suppose $P \in C(K)$, and so in particular $C$ is an elliptic curve over $K$. Since is $C$ defined over $\mathbb{Q}$, if $P \in C(K)$ then the Galois-conjugate point $\bar{P} \in C(K)$ also.
Let $L$ be the line through $P$ and $\bar{P}$, which is defined over $\mathbb{Q}$ since the midpoint and gradient are rational. The elliptic curve group law gives $P + \bar{P} + Q = 0$, and since $C(K)$ is a subgroup, we have that $Q \in C(K)$ also.
Since $L$ and $C$ both defined over $\mathbb{Q}$, then $Q \in L \cap C \implies \bar{Q} \in L \cap C$, and so $Q \in \{P, \bar{P}, Q\}$. To conclude we consider what happens in each case:
- If $\bar{Q} = Q$ then $Q$ is rational, done.
- Otherwise, say $Q = P$ wlog, then $P$ intersects $L$ with multiplicity 2. But then considering the Galois action, so does $\bar{P}$. This contradicts Bezout.
My questions are:
Is the proof above correct? I'm fairly new to algebraic/projective geometry so not particularly confident in my reasoning.
Is there a standard proof? I would like to generalise this for degree $n$ extensions for larger $n$. But this approach, even if valid, seems heavily specific to quadratic extensions.
Thank you!