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I have the following problem:

Let $R$ be a commutative ring with identity and $\phi: R \rightarrow R$ a ring automorphism. If $F=\lbrace r\in R | \phi(r)=r \rbrace$, show that $\phi^2$ being the identity map implies each element of $R$ is the root of a monic polynomial of degree two in $F[x]$.

I've tried constructing the polynomial explicitly, but I haven't had any luck. I've considered trying to use an argument using indices, but I'm not sure how to go about it.

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This is a generalization of complex conjugation. –  lhf Nov 9 '12 at 0:27
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1 Answer

up vote 2 down vote accepted

Let $r\in R$ and consider $a=r+\phi(r)$ and $b=r\phi(r)$. Then $a,b\in F$ and you can recover $r$ from $a$ and $b$ by solving a quadratic equation. More precisely, $r$ is a root of $x^2-ax+b=0$.

Note how $a,b\in F$ follows from $\phi$ being an involutive ring automorphism.

More generally, if $\phi^n$ is the identity, then every element $r\in R$ is the root of a monic polynomial of degree $n$ over $F$: $(x-r)(x-\phi(r))\cdots(x-\phi^{n-1}(r))$.

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