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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.

Bo Tielman
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1 Answers1

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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.

Bo Tielman
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