Computing $\phi(\frac32)$ where $\phi$ is an automorphism of $\mathbb Q[\sqrt2]$ such that $\phi(1)=1$ and $\phi(\sqrt2)=\sqrt2$ This question is a followup to this question about Field Automorphisms of $\mathbb{Q}[\sqrt{2}]$.
Since $\mathbb{Q}[\sqrt{2}]$ is a vector space over $\mathbb{Q}$ with basis $\{1, \sqrt{2}\}$, I naively understand why it is the case that automorphisms $\phi$ of $\mathbb{Q}[\sqrt{2}]$ are determined wholly by the image of $1$ and $\sqrt{2}$. My problem is using this fact explicitly. For example, suppose I consider the automorphism $\phi$ such that $\phi(1) = 1$ and $\phi(\sqrt{2}) = \sqrt{2}$, and I want to compute the value of $\phi\left(\frac{3}{2}\right)$. I can do the following: 
$$ \phi\left(\frac{3}{2}\right) = \phi(3) \phi\left(\frac{1}{2}\right) = [\phi(1) + \phi(1) + \phi(1)] \phi\left(\frac{1}{2}\right) = 3\phi\left(\frac{1}{2}\right).$$
I am unsure how to proceed from here. I would assume that it is true that $$\phi\left(\frac{1}{1 + 1}\right) = \frac{\phi(1)}{\phi(1) + \phi(1)} = \frac{1}{2},$$ but I don't know what property of ring isomorphisms would allow me to do this. 
 A: $$
2\phi(\frac{3}{2}) = \phi(3) = 3\phi(1) = 3
\implies
\phi(\frac{3}{2}) =\frac{3}{2}
$$
Generalizing this argument gives $\phi(q) = q$ for all $q \in \mathbb Q$.
A: 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 $\phi\left(\frac{m}{n}\right) = \frac{m}{n}$.

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.
