Automorphism of R

[Zareon]

Hi everyone!

 

This seemingly easy question is proving difficult.

I have to show that a field automorphism of R to R (real numbers)

is the identity.

 

What I've done:

Let f:R -> R

be an field automorphism.

Since f(1)=1 is a generator of Z, f is the identity on Z. Because multiplicative inverses are sent to their inverses: f(1/x)=1/f(x), it follows that f is also the identity on Q.

I`m not even sure if I`m on the right track. I obviously have to use some property of the reals now. A previous exercise which I very likely need to use asks to show that if x>0 then f(x)>0.

 

Any help is appreciated.

[matt grime]

Theorem: let F be any field, then any map F->is either zero or invertible. Proof: the kernel of a homomorphism is an *INSERT WORD HERE*, and fields only have TWO OF THESE THINGS.

 

Can't really give more hints than that without outright telling you the answer.

[Zareon]

Thanks Matt,

 

Well, the kernel is an ideal and the only ideals in a field F are the trivial ones (0) and F. Unless the map F-> maps into the trivial ring {0} (that isn't a field right?) the kernel can't be F, since 1 has to go to 1. So the kernel is {0} which means the map is injective.

 

So if the map is between fields it's always injective. I know from memory that {0} isn't a field, but I can't remember why, since the definition for a field is a commutative ring with R*=R\{0} which holds for {0}.

 

But anyway, that doesn't make F invertible right? For example, the inclusion f:Q->R is not surjective.

 

And I don't see how it helps, since it is given that f is invertible. I must show it is the identity.

[Zareon]

Ooh! Ooh! I think I got it. Could anyone please check if the reasoning is flawless?

 

I had to use some properties of the reals, so I came up with these.

Well ordering property: suppose x>0, then x=y^2 for some y, so f(x)=f(y)^2>0. So if x>0 then f(x)>0.

If b>a then b-a>= so f(b-a)=f(b)-f(a)>0, so f preserves order.

Completeness. Given a partition of Q into two disjoint subsets A,B with every element in A smaller than any element of B, there exists a unique real number 'inbetween', and vice versa. So if x is irrational, take A to be the set of rational smaller than x and B the set of rationals larger than x. Then f is the identity on A and on B and because of order preservation sup(A)=inf(B)=x is also mapped to itself. Hence f is the identity.

[matt grime]

Well ordered means any non-empty subset has a least element doesn't it? The reals are totally ordered not well ordered. Or am I misremembering this one.

 

But, you're correct, any field automorphism of R preserves the ordering, for the reason you give. And yes, this tells you there is only one automorhpism. Sorry I was no help earlier.

[Zareon]

Thanks!