Field automorphisms

It’s worth, I suppose, talking about automorphisms in general. An automorphism is a “renaming” of stuff so that it keeps the same structure. The “auto” means self. The “morph” means shape. The “same” is suppressed, because it really comes from the word “isomorphism,” which means same shape. So this is like a self-similar shape… but not in a fractally sense.

What is required for an automorphism? Well, any special constants need to stay the same. And the operations need to be “preserved,” whatever that means. For a field, it means that 0 and 1 are fixed, and that you can move $+$ and $\cdot$ outsied the function. More specifically, $\phi$ is an automorphism of a field $F$ if it is a function $\phi:F\to F$ with

• $\phi(0)=0$
• $\phi(1)=1$
• $\phi(x+y)=\phi(x)+\phi(y)$
• $\phi(xy)=\phi(x)\phi(y)$
• $\phi$ is a bijection. That is, $\phi$ doesn’t send two different points to the same point, and it sends something to every point. (Bijection is a fancy word meaning “1-1 and onto.”)
Can you think of any field automorphisms? In any field, you always have the trivial automorphism which sends a point $x$ to $x$. It’s lame, but it counts.
What about something a bit more complicated? How about complex conjugation? What if $\phi:\mathbb C\to \mathbb C$ by $\phi(a+bi)=a-bi$ (for $a,b\in \mathbb C$)? Yes, this works. It’s not hard to check any of these rules, but I’m too lazy to do so.
How about something really crazy? Take the field $\mathbb Q(\sqrt2)$, and the automorphism $\psi$ defined by $\psi(a+b\sqrt 2)=a-b\sqrt 2$ (where $a,b\in \mathbb Q$). What? But that sends negative things to positive things? It doesn’t preserve the shape at all!
I assure you, it is a field automorphism. If you don’t believe me, check the rules yourself. The issue you’re potentially having, is that you’re thinking of $\mathbb Q(\sqrt 2)$ as sitting inside $\mathbb R$, and thinking of the ordering on $\mathbb R$. But alas, I said nothing about order. It’s damn near impossible, but try to forget everything you know about the structure of $\mathbb R$, because it will only misguide you.
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3 Responses to Field automorphisms

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