Clarification in Silverman's proof of the descent theorem

54 Views Asked by At

In the book by Silverman called The Arithmetic of Elliptic Curves, there is the Descent Theorem (Theorem 3.1). He proves the theorem and allong the way he writes $$h(P_{n}) \leq \left( \frac{2}{m}\right)^n h(P) + \left( \frac{1}{m^2} + \frac{2}{m^2} + \frac{4}{m^2} + \cdots + \frac{2^{n-1}}{m^2} \right) (C'_{1} + C_{2}) \\ < \left( \frac{2}{m}\right)^n h(P) + \frac{C'_{1}+C_{2}}{m^2-2}$$

$C'_1 $ is a contant (but maybe it does depend on $n$?) and $C_2$ is also a contant. $m$ is a integer greater or equal to 2.

I don't get how he's gotten to this inequality based on what we know. The big term on the top right evaluates to $\frac{2^{n}-1}{m^2} (C'_{1}+ C_2 )$. I don't see why this is less than $\frac{C'_1 + C_2}{m^2 -2}$, can anyone clarify this a bit for me.

1

There are 1 best solutions below

0
On BEST ANSWER

I read the proof a bit better and it turns out Silverman made a typo in his proof. We know that for all $j$, $$h(P_j) \leq \frac{1}{m^2} \left( 2h(P_{j-1})+ C'_1 + C_2 \right).$$

He concludes the inequalty, but the correct one should be $$h(P_{n}) \leq \left( \frac{2}{m^2} \right)^n h(P) + \left( \frac{1}{m^2} + \frac{2}{(m^2)^2} + \cdots + \frac{2^{n-1}}{(m^2)^{n}} \right) (C'_1 + C_2 ).$$

This makes the second inequality true.