I'm reading the proof(s) of Mordell-Weil theorem using various texts. This post is to make sure what I'm reading and understanding is correct..


Rational points on Elliptic Curves by Silverman Tate gives a proof of Mordell 's theorem when elliptic curves are over rationals, $\mathbb{Q}$and have at least one rational $2$-torsion point. 


Then I noticed that Lawrence Washington's book on Elliptic curves also discuss this result this, first when all roots of cubic $f(x)$ are in $\mathbb{Q}$ and then just one of them is in $\mathbb{Q}$. And the same goes for Cassels' and Knapp's book as well. 


Then Lawrence Washington gives a remark that same can be said (i.e., Mordell-Weil holds) when we consider elliptic curves ($y^2 = f(x)$) over any number field $K$ and $f(x)$ has all roots in $K$ as we can prove Weak Mordell Weil in this case with some modification of the usual $\mathbb{Q}$ argument. But then it says nothing about the case when we just assume that $f(x)$ has at least one zero in $K$.


Now my questions, 


(1) I'm assuming that we can give a proof of Mordell-Weil when elliptic curves are over K and $f(x)$ has (just) at least one point in $K$. And I think Silverman Tate's proof in case of $\mathbb{Q}$ can be generalized, mainly Weak Mordell-Weil's (Lemma $3.4$ in the book) 


(2) What happens when elliptic curves ($y^2=f(x)) E(\mathbb{Q})$ and $E(K)$ have no $2$-torsion point or $f(x)$ has no root in $\mathbb{Q}$(respectively $K$) ? Is it the situation that is discussed in the J.S.Milne's Elliptic Curves book or J. Silverman's The Arithmetic of Elliptic Curves? 


I would appreciate if someone could help answer some or all of the questions above or maybe just refer to an appropriate source to find these proofs/ discussions. 


Thank you!



Can someone help clear up some confusion regarding the proof of Mordell-Weil theorem?
If $E$ is an elliptic curve defined over $K$ (a number field) and if $E(K)$ has a 2-torsion point $T$, let $E'=E/\langle T\rangle$ be the isogenous curve, let $f:E\to E'$ be the isogeny, and let $f':E'\to E$ be the dual isogeny. Then you always get maps
$$ E(K)/f(E'(K)) \to K^\*/{K^\*}^2\quad\text{and}\quad
E'(K)/f'(E(K)) \to K^\*/{K^\*}^2
$$
whose images are effectively calculable subgroups of $K^\*/{K^\*}^2$. However, computing these subgroups (and even proving that they are finite) requires knowing that the ring of integers of $K$ has finite class number and finitely generated unit group. So that's why people often take $K=\mathbb{Q}$, since then there's no issue with non-principal ideals or non-trivial units. (To be precisely, for this argument one only needs to know that the 2-part of the class group is finite.)


If $E$ does not have a $K$-rational 2-torsion point, you can go to an extension field where it does, which will be a cubic extension of $K$; but then, of course, the unit group and class group may increase.


The computations are simpler, in some ways, if all of the $2$-torsion of $E$ is $K$-rational, since then one can ignore 2-isogenies and simply use a map
$$ E(K)/2E(K) \to K^\*/{K^\*}^2 \times K^\*/{K^\*}^2 $$
whose image lies in a finite, effectively computable, subgroup. 


