Understanding the weak finite basis theorem I'm working my way though the proof of the weak finite basis theorem in Chapter 15 of Cassel's book "Lectures on Elliptic Curves".
I'm stuck on the proof Lemma 2, that "The kernel of $\mu$ is $2(C(\mathbb{Q}))$".
The parts I don't understand are:
(1) I don't understand the why "since $M$ has exponent 2, the kernel certainly contains $2(C(\mathbb{Q}))$". (Note $M$ is defined as a subset of $\mathbb{Q}^*(\Theta)/(\mathbb{Q}^*(\Theta)^2$). I'm guessing $M$ has "exponent 2" is referring to the $(\mathbb{Q}^*(\Theta)^2$ in the quotient of $M$? But I'm not sure why that shows the kernel contains $2(C(\mathbb{Q}))$.
(2) Further down the proof, why are we allowed to replace $\Theta$ by an 'indeterminate X'? And why does the right hand side of the equation that follows $= F(X)$?
Any help with any of these questions would be much appreciated! Thank you.
 A: Let $f: A \longrightarrow B$ be a group homomorphism. Then $B$ has exponent $2$ if and only if $2b = 0$ (if the group is written additively). In this case, $2A$ is in the kernel:
$$ f(2a) = f(a+a) = f(a) + f(a) = 2f(a) = 0 $$
since the image of $f$ has exponent $2$.
For your second question: the quadratic extension $K(\sqrt{m})$ is isomorphic to $K[x]/(x^2-m)$. Thus an equation in $K(\sqrt{m})$ is equivalent to an equation in $K[x]$ up to multiples of $x^2-m$. The same thing works for cubic extensions.
Here are the details. Let $\Theta$ be the root of an irreducible cubic monic polynomial $F$ and assume that $f(\Theta) = 0$ for some monic polynomial of degree $3$. The irreducibility of $F$ implies that $f$ is divisible by $F$. Since both polynomials have degree $3$, we must have $f(X) = c F(x)$. Since both $f$ and $F$ have leading coefficient $1$, we must have $c = 1$.
Proofs of Mordell's Theorem are a lot simpler if the roots are rational. See Silverman-Tate or any of the many introductions to elliptic curves for undergraduates.
