Nehtdx

This question is a follow - up to this question about Field Automorphisms of $mathbbQ[sqrt2]$.

Since $mathbbQ[sqrt2]$ is a vector space over $mathbbQ$ with basis $1, sqrt2$, I naively understand why it is the case that automorphisms $phi$ of $mathbbQ[sqrt2]$ are determined wholly by the image of $1$ and $sqrt2$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $phi$ such that $phi(1) = 1$ and $phi(sqrt2) = sqrt2$, and I want to compute the value of $phi(frac32)$. I can do the following:

$$ phi(frac32) = phi(3) phi(frac12) = [phi(1) + phi(1) + phi(1)] phi(frac12) = 3phi(frac12).$$

I am unsure how to proceed from here. I would assume that it is true that $$phi(frac11 + 1) = fracphi(1)phi(1) + phi(1) = frac12,$$ but I don't know what property of ring isomorphisms would allow me to do this.

$$
2phi(frac32) = phi(3) = 3phi(1) = 3
implies
phi(frac32) =frac32
$$
Generalizing this argument gives $phi(q) = q$ for all $q in mathbb Q$.

Every automorphism fixes $mathbbQ$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $mathbbQ$.

For the proof, we assume WLOG that $mathbbQ subseteq K$. Then:

• $phi$ fixes $0$ and $1$, by definition.




• $phi$ fixes all positive integers, since $phi(n) = phi(1 + 1 + cdots + 1) = n phi(1) = n$.




• $phi$ fixes all negative integers, since $phi(n) + phi(-n) = phi(n-n) = 0$, so $phi(-n) = -phi(n) = -n$.




• $phi$ fixes all rational numbers, since $n cdot phileft(fracmnright) = phi(m) = m$, so $phileft(fracmnright) = fracmn$.

More generally, when we consider automorphisms of a field extension $K / F$, we often restrict our attention only to automorphisms which fix the base field $F$. But when $F = mathbbQ$, since all automorphisms fix $mathbbQ$, such a restriction is unnecessary.