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Let $R$ be a DVR. Let $\phi$ be an automorphism of its fraction field. Can $R$ be a proper subset of $\phi(R)$?

There definitely do exist DVRs that contain a unital subring isomorphic to themselves (e.g. $k[[x^2]]\subset k[[x]]$) but I am not sure about this question.

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Suppose $\phi$ is an automorphism such that $R\subset \phi(R)$ holds. Then it's inverse $\psi$ preserves elements of valuation $\leq 1$ (I am using the multiplicative notation for valuations here), hence $\psi$ is order-preserving. Then it's inverse $\phi$ must be order-preserving too, which implies $\phi(R)\subset R$. Therefore we must have equality, which means that no isomorphism exists such that the initial inclusion is proper.

asdq
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  • what does order-preserving mean? –  Aug 01 '19 at 18:10
  • It means that $|x|\leq |y|$ implies $|\phi(x)|\leq |\phi(y)|$ for all $x,y$ in the fraction field, where $|\cdot |$ denotes the valuation. – asdq Aug 01 '19 at 19:40