Computing $\phi(\frac32)$ where $\phi$ is an automorphism of $\mathbb Q[\sqrt2]$ such that $\phi(1)=1$ and $\phi(\sqrt2)=\sqrt2$

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

Every automorphism fixes $\mathbb{Q}$. That is, if $K$ is any field of characteristic zero, then any automorphism of $K$ fixes the unique subfield of $K$ isomorphic to $\mathbb{Q}$.

For the proof, we assume WLOG that $\mathbb{Q} \subseteq K$. 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 $n \cdot \phi\left(\frac{m}{n}\right) = \phi(m) = m$, so $\phi\left(\frac{m}{n}\right) = \frac{m}{n}$.

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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 $F = \mathbb{Q}$, since all automorphisms fix $\mathbb{Q}$, such a restriction is unnecessary.