# The inclusion $E(K)/2E(K)\hookrightarrow K^*/(K^*)^2$ for Weak Mordell-Weil theorem

I'm reading a master's dissertation where the author proves the weak Mordell-Weil theorem for $\mathbb{Q}$ (i.e.: for any elliptic curve $E:y^2=f(x)=x^3+Ax+B$ over $\mathbb {Q}$, the group $E(\mathbb{Q})/2E(\mathbb{Q})$ is finite). He begins the proof by saying this:

Let $K:=\mathbb{Q}(\theta)=\mathbb{Q}[x]/f(x)$ and $H$ the subgroup of $K^*$ consisting of elements $a\in K^*$ such that $N_{K/\mathbb{Q}}(a)$ is a square in $\mathbb{Q}$. Define the map: \begin{align*} \mu:E(\mathbb{Q}) &\to \overline{H}<K^*/(K^*)^2\\ (x:y:1) &\mapsto \overline{x-\theta}\\ O &\mapsto \overline{1} \end{align*}

He immediatelly goes on to prove that $\mu$ is a group homomorphism among other things.t

But why is $N_{K/\mathbb{Q}}(x-\theta)$ necessarily is a square in $\mathbb{Q}$ in the first place, and why is $im(\mu)=\overline{H}$?

• For $x \in \mathbb{Q}, N(x-\theta) = \prod_{i=1}^3 \sigma_i(x-\theta))=\prod_{i=1}^3 (x-\sigma_i(\theta)) = f(x)$. – reuns Nov 19 '17 at 19:35
• I know $\theta$ is a linear combination of the roots of $f$, but why are $\sigma_i(\theta)$ exactly the roots of $f$? – rmdmc89 Nov 19 '17 at 23:26
• What is your definition of the field norm ? For $x \in \mathbb{Q}$, $N_{K/\mathbb{Q}}(x-a) = \det(xI-A)$ where $A$ is the matrix representing the multiplication by $a$ in the vector space $K$. And $\det(xI-A)$ doesn't depend on the basis we choose for $K$, in particular it stays the same when using the $n=[K:\mathbb{Q}]$ embeddings $\sigma_i : K \to \mathbb{C}$ to embed $K$ in $\mathbb{C}^n$, where $A$ becomes diagonal so that $\det(xI-A)= \prod_{i=1}^n (x-\sigma_i(a))$ – reuns Nov 19 '17 at 23:31
• If $f \in \mathbb{Q}[x]$ is irreducible with roots $\theta_1,\ldots,\theta_m$ then $\mathbb{Q}(\theta_1)\cong \mathbb{Q}(\theta_j) \cong \mathbb{Q}[x]/(f(x))$. I don't think you understand what I said about $N_{K/\mathbb{Q}}$. – reuns Nov 27 '17 at 14:45
• It is the splitting field when $disc(f) =-4A^3-27 B^2$ is a square – reuns Nov 27 '17 at 16:18

$$N_{K\Bbb Q}(x-\theta)=x^3+Ax+B=y^2$$ so is a square in $\Bbb Q$. I'm not sure that the image of $\mu$ is all of $\overline H$.
• $K^*/(K^*)^2$ is the set of quadratic extensions of $\mathbb{Q}(\theta)$, is $\overline{H}$ much smaller ? – reuns Nov 19 '17 at 19:43
• Maybe the image isn't exactly $\overline{H}$. But I've just realized that's ok, because for this proof we only need the fact that the image is a subgroup of $K^*/(K^*)^2$ – rmdmc89 Nov 27 '17 at 14:03