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From: Arturo Magidin on 19 Jan 2010 11:50
On Jan 19, 2:52 am, Kaba <n...(a)here.com> wrote:
> achille wrote:
> > How about using the fact f is order preserving [*],
> > If f(x) <> x at any point x, pick a rational number
> > r between x and f(x), then
>
> > case1) x    < f(x) => x < r AND f(x) > f(r).
> > case2) f(x) < x    => r < x AND f(r) > f(x).
>
> > both cases contradict with [*], so f(x) = x for all x.
>
> Even better.

If you try to turn this proof by contradiction into a direct proof,
you end up with essentially the argument using increasing and
decreasing sequences of rationals that converge to x.

--
Arturo Magidin

From: victor_meldrew_666 on 20 Jan 2010 03:30
On 18 Jan, 21:10, Timothy Murphy <gayle...(a)eircom.net> wrote:
> victor_meldrew_...(a)yahoo.co.uk wrote:
> >> > As a follow-up question yopu might consider:
> >> > Does Q_p (the field of p-adic numbers) have any automorphisms
> >> > save for the identity?

> Yes, I'm not doing very well at this exam ...
>
> I'm thinking, suppose x is a unit of the form 1 + py (y in Z_p).
> Then x has a q-th root in Z_p for any prime q != p
> by Hensel's Lemma, or by taking p-adic logs.

OK.

> It follows that f(x) must be in Z_p.

Why exactly is that?

> And every z in Z_p is a sum of such elements.

And so?
From: Timothy Murphy on 20 Jan 2010 08:36
victor_meldrew_666(a)yahoo.co.uk wrote:

> On 18 Jan, 21:10, Timothy Murphy <gayle...(a)eircom.net> wrote:
>> victor_meldrew_...(a)yahoo.co.uk wrote:
>> >> > As a follow-up question yopu might consider:
>> >> > Does Q_p (the field of p-adic numbers) have any automorphisms
>> >> > save for the identity?
>
>> Yes, I'm not doing very well at this exam ...
>>
>> I'm thinking, suppose x is a unit of the form 1 + py (y in Z_p).
>> Then x has a q-th root in Z_p for any prime q != p
>> by Hensel's Lemma, or by taking p-adic logs.
>
> OK.
>
>> It follows that f(x) must be in Z_p.
>
> Why exactly is that?

Suppose |f(x)|_p = p^e, ie f(x) = p^{-e}z, where z is a p-adic unit.
Then f(x) can only have a q-th root if q divides e.

>> And every z in Z_p is a sum of such elements.
>
> And so?

Well, my erroneous "proof" showed (correctly) that if f sends Z_p to Z_p
then f: Z_p -> Z_p is continuous.
Since f leaves each element of Z fixed, and Z is dense in Z_p
it follows that f leaves each element of Z_p fixed.

But as I said, I'm sure there is a simpler proof.




--
Timothy Murphy
e-mail: gayleard /at/ eircom.net
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
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