# Do automorphisms of quotient fields preserve the underlying ring?

Suppose $$R$$ is an integral domain. Let $$\gamma: \mathrm{Frac}(R) \rightarrow \mathrm{Frac}(R)$$ be an automorphism of the quotient field. Is it true that $$\gamma(R) = R?$$ I don't think that this is true in general, but I cannot think of any examples. Are there examples for which this is not true?

• In many cases together with some counter-examples what you need is for $\sigma\in Aut(Frac(R))$ sending $S \to S$ what does it imply for the integral closure and normal closure of $S$ Commented Feb 27, 2019 at 22:06

No, a field automorphism $$\gamma: \text{Frac}(R)\to\text{Frac}(R)$$ need not map $$R$$ to $$R$$. Take $$R = \mathbb{Q}[x]$$ and $$\gamma: \mathbb{Q}(x) \to \mathbb{Q}(x)$$ defined by preserving $$\mathbb{Q}$$ and by $$\gamma(x) = 1/x$$.
• This is a counterexample, or an example where it is not true. $R$ is an integral domain, $\gamma$ is a field automorphism $\text{Frac}(R)\to\text{Frac}(R)$ with $\gamma(R) \ne R$. This is because $x$ is not invertible in $\mathbb{Q}[x]$. Commented Feb 27, 2019 at 20:39
• No worries. I stopped reading at "Is it true that $\gamma(R) = R$?" Commented Feb 27, 2019 at 20:41