# 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.

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

• In the interest of clarity's sake it might be worth noting this is the multiplicative property of ring/field homomorphisms, i.e. $\phi(xy)=\phi(x)\phi(y)$, under the consideration $3 = 3\cdot 1$. Apr 17, 2019 at 5:28
• @EeveeTrainer, I don't think it is. In a ring $2t=t+t$ and this calculation can be justified purely by the additiveness of the homomorphism once we know $\phi(1)=1$. . Apr 17, 2019 at 7:11

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.