I am trying to work out the (sketch of) proof of Ostrowski's Theorem (on classifying archimedean completely valued fields) in J.S. Milne's Algebraic Number Theory course notes (http://www.jmilne.org/math/CourseNotes/ANT.pdf, Remark 7.49, proof in footnote).

I will just recap the one spot I'm having trouble with: At this stage we have got a field $K$ complete with respect to an archimedean absolute value $\lvert \, \rvert$, with $\mathbb C\subseteq K$ and the restriction of $\lvert \, \rvert$ to $\mathbb C$ is the usual absolute value. We have taken an $x \in K\setminus \mathbb C$ and established the following inequality for all $z\in \mathbb C $ and $n \in \mathbb N$: $$\lvert x^n-z^n \rvert\geq \lvert x -z \rvert \lvert x \rvert^{n-1} .$$ The author then says ''on choosing $\lvert z \rvert <1$ and letting $n \to \infty$, we find that $\lvert x\rvert \geq \lvert x-z\rvert$''.

Now, I've divided both sides of the inequality above with $\lvert x \rvert^{n-1} $ and found $$ \lvert x \rvert + \lvert z \rvert \lvert z/x \rvert^{n-1} \geq \lvert x-z \rvert . $$

And indeed, we get what we want in the limit $n \to \infty$, if $ \lvert z \rvert< \lvert x \rvert$. But it is essential for the rest of the proof that we get this for all $\lvert z\rvert <1$.

Is there something I am missing here? I hope that I'm not overlooking something terribly obvious...

  • $\begingroup$ But how do you conclude that $\lvert x\rvert ^n \geq \lvert x-z\lvert \lvert x \rvert ^{n-1}$ (and for which $n$)? Try a proof by contradiction and assume $\lvert x \rvert ^n < \lvert x-z\lvert \lvert x \rvert ^{n-1}$ for all $n$. On the other hand I have $\lvert x\rvert ^n + \lvert z \rvert ^n \geq \lvert x-z\lvert \lvert x \rvert ^{n-1}$ and $\lvert z \rvert ^n$ does get arbitrarily small, yes, but I don't see how to derive a contradiction because the difference $\lvert x-z\lvert \lvert x \rvert ^{n-1}-\lvert x \rvert ^n$ might also get arbitrarily small. $\endgroup$ – Layer Cake Mar 6 '18 at 1:00
  • $\begingroup$ Can't you just scale your $x$ such that $\lvert x\rvert \ge 1$? $\endgroup$ – Claudius Mar 7 '18 at 9:48
  • $\begingroup$ The rest of the proof wouldn't work then. In the last line of the proof, Milne says that we replace $x$ with $x-z$ and do the same work for $x-z$ in order to get $\lvert x-2z\rvert=\lvert x-z\rvert =\lvert x \rvert$. Proceeding like this, we get a contradiction to the archimedean property. If I have to rescale in every step, then I can't get the equation $\lvert x-nz \rvert = \lvert x \rvert $ for all $n$. $\endgroup$ – Layer Cake Mar 7 '18 at 11:56
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    $\begingroup$ Why do you have to rescale at every step? Milne proves $\lvert x-z\rvert = \lvert x\rvert$ so in particular $\lvert x-z\rvert \ge 1$. $\endgroup$ – Claudius Mar 7 '18 at 12:11
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    $\begingroup$ It seems like you don't even need to scale your $x$ if you just take $z$ with $\lvert z\rvert < \lvert x\rvert$. $\endgroup$ – Claudius Mar 8 '18 at 10:02

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