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.

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.

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.

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.

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.

Thanks!