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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.

Asaf Karagila
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Solarflare0
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2 Answers2

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

lhf
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    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 $3 = 3\cdot 1$. – PrincessEev Apr 17 '19 at 05:28
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    @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 $\phi(1)=1$. . – ancient mathematician Apr 17 '19 at 07:11
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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.

Caleb Stanford
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