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From: Kaba on 16 Jan 2010 07:17
Hi,

(Question at the end where the proof stumps on)

Let
* R be the set of real numbers,
* Q be the set of rational numbers,
* N be the set of natural numbers,
* Z be the set of integers, and
* f be a field automorphism in R.

Claim:
f is the identity function: f(x) = x

Proof:
1) for all x:
f(x) = f(1 * x) = f(1) * f(x)
=> f(1) = 1

2) for all x:
f(x) = f(x + 0) = f(x) + f(0)
=> f(0) = 0

3) f(2) = f(1 + 1) = f(1) + f(1) = 2
By induction, for all n in N: f(n) = n

4) for all x:
0 = f(0) = f(x + (-x)) = f(x) + f(-x)
=> f(-x) = -f(x)

5) for all z in Z, z < 0:
f(-z) = -f(z) = -z

6) for all x != 0:
1 = f(1) = f(x / x) = f(x) * f(1 / x)
=> f(1 / x) = 1 / f(x)

7) for all q = (n / m) in Q:
(n, m in Z, m != 0)
f(n / m) = f(n) * f(1 / m) = f(n) / f(m) = n / m

How do I prove f(x) = x for real numbers?

--
http://kaba.hilvi.org
From: Timothy Murphy on 16 Jan 2010 08:17
Kaba wrote:

> Let
> * R be the set of real numbers,
> * Q be the set of rational numbers,
> * N be the set of natural numbers,
> * Z be the set of integers, and
> * f be a field automorphism in R.
>
> Claim:
> f is the identity function: f(x) = x

This is not true.
The automorphism of Q(sqrt2) under which sqrt2 -> -sqrt2
can be extended to the whole of R.

--
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: Arturo Magidin on 16 Jan 2010 11:11
On Jan 16, 7:17 am, Timothy Murphy <gayle...(a)eircom.net> wrote:
> Kaba wrote:
> > Let
> > * R be the set of real numbers,
> > * Q be the set of rational numbers,
> > * N be the set of natural numbers,
> > * Z be the set of integers, and
> > * f be a field automorphism in R.
>
> > Claim:
> > f is the identity function: f(x) = x
>
> This is not true.
> The automorphism of Q(sqrt2) under which sqrt2 -> -sqrt2
> can be extended to the whole of R.

How? It can be extended to all of the algebraic closure of R, and from
there possibly to all of C, but I do not see how it can be extended to
all of R.

Suppose f:R-->R is a field automorphism, and that r in R is positive.
Then r = s^2 for some real number s; so f(r) = f(s^2) = f(s)^2. Since f
(s) is a real number, f(s)^2 is positive (it cannot be zero since r=/
=0). So an automorphism of f must send positive numbers to positive
numbers. In fact, this holds for the real algebraic closure of R (the
field of all real algebraic numbers).

Now, for the OP:

Since f must send positive numbers to positive numbers, it must
respect order: let r and s be real numbers. Then r<s if and only if s-
r>0 if and only if f(s-r) > 0 (the converse follows from the
argument above by considering f^{-1}) if and only if f(s)-f(r)>0, if
and only if f(r)<f(s).

To show your f is the identity, let x be an arbitrary real number; try
finding monotonic sequences over which you have control that converge
to x, and see what happens to their images.

--
Arturo Magidin
From: victor_meldrew_666 on 16 Jan 2010 11:44
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?
From: Timothy Murphy on 16 Jan 2010 12:03
victor_meldrew_666(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.


--
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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