2
$\begingroup$

I have included a bolded comment in a step in the part of Gouvea's proof of Ostrowski's theorem where he shows that an archimedean absolute value on $\mathbb Q$ is equivalent to $|\ |_\infty$ (the regular absolute value). I will write my question about the comment below. Here is the relevant excerpt of the proof:


Suppose, first, that $|\ |$ is archimedean. We want to show, in this case, that it is equivalent to the $\infty-$adic absolute value. Since $|\ |$ is archimedean, $\exists n\in\mathbb Z: |n|>1$. Then $|-n| = |n| > 1$, so suppose WLOG that $n \geq 0$. In fact, we know that $n \geq 2$. Let $n_0$ be the least such positive integer $n$. Then since $n_0 > 1$ and $|n_0| > 1$, we can pick $\alpha\in\mathbb R$ such that $$ |n_0| = n_0^\alpha. $$ We claim that $\forall x\in \mathbb Q$, $|x| = |x|_\infty^\alpha$. Note that this already holds for $x = 0$ (and $x = 1$), and that if it holds for positive integers, then it holds for positive rational numbers, and then it holds for negative rational numbers, so it holds for all of $\mathbb Q$. Thus, it is sufficient to demonstrate this equality for positive integers. We will achieve this by showing that for every positive integer $n$, $|n| = n^\alpha$.

We know that the equality holds for $n = n_0$. Now take an arbitrary positive integer $n$, and write it "in base $n_0$," i.e., in the form $$ n = a_0 + a_1n_0 + a_2n_0^2 + \cdots + a_kn_0^k, $$ with $k\geq 0$, $0\leq a_i < n_0$, and $a_k\neq 0$. Now taking absolute values, we get \begin{align*} |n| & = |a_0 + a_1n_0 + a_2 n_0^2 + \cdots + a_kn_0^k| \\ & \leq |a_0| + |a_1| |n_0| + |a_2| |n_0|^2 + \cdots + |a_k| |n_0|^k \\ & = |a_0| + |a_1| n_0^\alpha + |a_2| n_0^{2\alpha} + \cdots + |a_k| n_0^{k\alpha}. \end{align*} Since we chose $n_0$ to be the smallest positive integer whose absolute value was greater than $1$, we know that $|a_i| \leq 1$, so that we get \begin{align*} |n| & \leq |a_0| + |a_1| n_0^\alpha + |a_2| n_0^{2\alpha} + \cdots + |a_k| n_0^{k\alpha} \\ & \leq 1 + n_0^\alpha + n_0^{2\alpha} + \cdots + n_0^{k\alpha} \\ & = n_0^{k\alpha}(1 + n_0^{-\alpha} + n_0^{-2\alpha} + \cdots + n_0^{-k\alpha} ) \\ & = n_0^{k\alpha} \sum_{i=0}^k n_0^{-i\alpha} \\ & \leq n_0^{k\alpha} \sum_{i=0}^\infty n_0^{-i\alpha} \textbf{ (why is this not strict?)} \\ & = n_0^{k\alpha} \frac{1}{1 - n_0^{-\alpha}} = n_0^{k\alpha} \frac{n_0^\alpha}{n_0^\alpha - 1}. \end{align*} Since $n_0^\alpha > 1$, we have that $C:= n_0^\alpha/(n_0^\alpha - 1) > 0$, and $$ |n| \leq C n_0^{k\alpha} \leq Cn^\alpha. $$ To see why we were able to tack on $Cn^\alpha$, just look at the expansion of $n$ "in base $n_0$" to see that $n_0^k \leq n$. The above formula applies for every positive integer $n$, and thus letting $n$ and $m$ be any positive integers, we can apply the formula to $n^m$, \begin{align*} |n^m| & \leq C n^{m\alpha} \\ |n| & \leq \sqrt[m]{C} n^\alpha . \end{align*} If we leave $n$ fixed and let $m$ become arbitrarily large, then since $C>0$ is fixed, we see that $\sqrt[m]{C}$ becomes arbitrarily close to $1$, and thus the above inequality implies that $|n| \leq n^\alpha$.


In the second part of this proof, Gouvea goes on to show that $|n| \geq n^\alpha$, and then he is able to conclude that $|n| = n^\alpha$. Thus, it is essential that in the excerpt above, we find that $|n| \leq n^\alpha$. However, since the terms of the sum $\sum_{i=0}^\infty n_0^{-i\alpha}$ are each nonnegative, I think it must be that $\sum_{i=0}^k n_0^{-i\alpha} < \sum_{i=0}^\infty n_0^{-i\alpha}$, and thus $$ n_0^{k\alpha} \sum_{i=0}^k n_0^{-i\alpha} < n_0^{k\alpha} \sum_{i=0}^\infty n_0^{-i\alpha}, $$ contradicting the step in Gouvea's proof. Is my logic or my reading of his proof wrong?

$\endgroup$
0
$\begingroup$

I just realized that since I can say $C > 1$, the proof still works even with a strict inequality: we end up with \begin{align*} |n^m| & < C n^{m\alpha} \\ |n| & < \sqrt[m]{C} n^\alpha . \end{align*} If we leave $n$ fixed and let $m$ become arbitrarily large, then since $C>1$ is fixed, we see that $\sqrt[m]{C}$ becomes arbitrarily close to $1$ from above, and thus the above inequality implies that $|n| \leq n^\alpha$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.