Weak Mordell-Weil and Minkowski

225 Views Asked by At

I'm reading J.H. Silverman's The Arithmetic of Elliptic Curves, chapter VIII section $1$, where he proves the weak Mordell-Weil theorem.

Right after proving proposition $1.6$, he makes a remark:

enter image description here

enter image description here

My questions are:

1) The Minkowski theorem I know (from here) says that there are finitely many number fields whose discriminants are $\leq N$ (constant). Does the fact that $[K(Q):K]\leq m^2$ imply that the discriminant $\Delta_{K(Q)/K}$ is bounded? Why? Besides, what is the role of $S$ when applying Minkowski?

(maybe he's refering to a different Minkowski theorem, I don't know)

2) I understand that the Galois conjugates of $Q$ is of the form $Q+T$ for some $T\in E[m]$, but I can't see why this justifies the claim that $[K(Q):K]\leq m^2$.

1

There are 1 best solutions below

5
On BEST ANSWER

Remember that the primes dividing the relative discriminant of $L/K$ are exactly the primes of $K$ that are ramified in $L$. If we furthermore bound the degree of $L/K$, then we can bound the power with respect to which any particular prime divides the discriminant. Thus if we fix a finite set of ramified primes and bound the degree, we bound the discriminant, and so bound the number of fields extensions $L/K$.

As for your second question, there are $m^2$ elements $T$ of $E[m]$, and so $Q$ has at most $m^2$ Galois conjugates. Thus its minimal polynomial has degree at most $m^2$, and hence so does the extension $K(Q)$.