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I'm attempting to solve Exercise VIII.8.8b in Silverman's Arithmetic of Elliptic Curves:

Prove that $H(x_1+\cdots+x_N)\leq NH(x_1)\cdots H(x_N)$.

This should be elementary, but I'm having trouble figuring out what to do with the $N$ on the right. Plugging in definitions from each end, I imagine the proof should go like

$$H(x_1+\cdots+x_N)=H_K([x_1+\cdots+x_N,1])^{1/[K:\mathbb{Q}]}=\prod_{v\in M_K}\max\{|x_1+\cdots+x_N|_v,1\}^{n_v/[K:\mathbb{Q}]}$$ $$\leq\prod_{v\in M_K}\max\{|x_1|_v+\cdots+|x_N|_v,1\}^{n_v/[K:\mathbb{Q}]}\text{ by the triangle inequality}$$ $$\leq\cdots\text{ (I don't know what goes here) }\cdots\leq$$ $$N\prod_{v\in M_K}\left(\prod_{i=1}^N\max\{|x_i|_v,1|\}^{n_v/[K:\mathbb{Q}]}\right)=N\,\,\prod_{i=1}^N\prod_{v\in M_K}\max\{|x_i|_v,1\}^{n_v/[K:\mathbb{Q}]}=NH(x_1)\cdots H(x_n)$$

($K$ can be any field containing the $x_i$, e.g. $K=\mathbb{Q}(x_0,\ldots,x_N)$, but this doesn't matter)

I don't see what to do with this $N$ - I can't think of any way to pass it into the product over $v\in M_K$, because that product is infinite, and the only way I can think of to compare these two quantities is to compare term-by-term for each $v\in M_K$. These two ideas for attacking the problem seem to conflict with each other. Can anyone give me a tiny nudge in the right direction? I'm sure I'm missing something obvious.

P.S. I considered induction on $N$, but that requires that I actually prove the result for $N=2$, which I'm unable to do for the above reason.

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For the nonarchimedean places, you can actually replace $|x_1+\cdots+x_N|_v$ with $\max|x_i|_v$, without the addition (by 'replace' I mean with the inequality between the two expressions). For the archimedean places, you can replace $|x_1+\cdots+x_N|_v$ with $N\max|x_1|_v$. You'll only get the $N$ from the archimedean places, not the nonarchimedean ones. Also, although the product is infinite, all but finitely many of the factors are $1$, so it's "really" finite. –  Arturo Magidin May 4 '11 at 4:45
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@Arturo: Ach, I'm smacking my forehead quite vigorously for not thinking of using the nonarchimedean triangle inequality. However, I'm still a bit confused; if there is more than one archimedean place, then mightn't there be multiple factors of $N$ introduced? I can see that this must be fixed by the exponents $n_v/[K:\mathbb{Q}]$, but I'm not seeing how, exactly. –  Zev Chonoles May 4 '11 at 5:02
    
I know that $n_v$ for archimedean places is always either 1 or 2 (since it's either $[\mathbb{R}:\mathbb{R}]$ or $[\mathbb{C}:\mathbb{R}]$), but I'm not sure how to connect this with the matter of whether $N\text{max}|x_i|_v\leq 1$ or not, to ensure that the total exponent of $N$ that is introduced in the product is exactly 1. –  Zev Chonoles May 4 '11 at 5:07
    
@Zev: I confess I'm not sure right now (pretty late here, and I'm about to go to bed). I'll see if something occurs tomorrow morning. –  Arturo Magidin May 4 '11 at 5:08
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@Zev, what Arturo says solves the problem already, by noting that $\sum n_v = [K:\mathbb{Q}]$ for archimedean places. (degree of extension = number of real embeddings + 2 * number of complex embeddings up to conjugates). –  Soarer May 4 '11 at 5:41

1 Answer 1

up vote 4 down vote accepted

Have you though about the case of rational numbers? Doing so may give you a hint as to how to do the general case.

If $x_i = a_i/b_i$ (a ratio of coprime integers), then $H(x_i) = \max\{a_i,b_i\}$. Now $$\sum x_i = \dfrac{a_1 b_2\cdots b_N + \cdots + b_1\cdots b_{N-1} a_N} {b_1\cdots b_n}.$$ Now this expression may not be in terms of coprime integers, but cancelling common factors will only reduce the height, so it suffices to bound each of the numerator and denominator of this expression by the $N$ times the products of the $H(x_i)$. In fact the denominator is bounded just by the product. The factor of $N$ is needed, on the other hand, to get an upper bound on the numerator.

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