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From: victor_meldrew_666 on 18 Jan 2010 09:53
On 16 Jan, 17:03, Timothy Murphy <gayle...(a)eircom.net> wrote:
> victor_meldrew_...(a)yahoo.co.uk wrote:
> > On 16 Jan, 13:17, Timothy Murphy <gayle...(a)eircom.net> wrote:
>
> >> This is not true.
> >> The automorphism of Q(sqrt2) under which sqrt2 -> -sqrt2
> >> can be extended to the whole of R.
>
> > Really?
>
> > In your proposed homomorphism, where would 2^(1/4) go?
>
> Sorry, you are quite right.
> In fact, an automorphism of R must preserve the order,
> since x^2 -> f(x)^2,
> and so must be continuous, and therefore the identity.

As a follow-up question yopu might consider:
Does Q_p (the field of p-adic numbers) have any automorphisms
save for the identity?
From: Timothy Murphy on 18 Jan 2010 11:26
victor_meldrew_666(a)yahoo.co.uk wrote:

>> In fact, an automorphism of R must preserve the order,
>> since x^2 -> f(x)^2,
>> and so must be continuous, and therefore the identity.
>
> As a follow-up question yopu might consider:
> Does Q_p (the field of p-adic numbers) have any automorphisms
> save for the identity?

I guess that is easier, since f(p^r x) = p^r f(x)
(assuming addition is preserved)
and so the map is again continuous.

--
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
From: victor_meldrew_666 on 18 Jan 2010 11:34
On 18 Jan, 16:26, Timothy Murphy <gayle...(a)eircom.net> wrote:
> victor_meldrew_...(a)yahoo.co.uk wrote:
> >> In fact, an automorphism of R must preserve the order,
> >> since x^2 -> f(x)^2,
> >> and so must be continuous, and therefore the identity.
>
> > As a follow-up question yopu might consider:
> > Does Q_p (the field of p-adic numbers) have any automorphisms
> > save for the identity?
>
> I guess that is easier, since f(p^r x) = p^r f(x)
> (assuming addition is preserved)
> and so the map is again continuous.

Not every additive map from Q_p to Q_p is continuous.
From: Timothy Murphy on 18 Jan 2010 16:10
victor_meldrew_666(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?
>>
>> I guess that is easier, since f(p^r x) = p^r f(x)
>> (assuming addition is preserved)
>> and so the map is again continuous.
>
> Not every additive map from Q_p to Q_p is continuous.

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

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

But I think there must be a simpler argument ...




--
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
From: Kaba on 19 Jan 2010 03:52
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.

--
http://kaba.hilvi.org
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