Do field automorphisms preserve valuation rings?

Question

I have a question about valuation fields. Does an automorphism of a discrete valuation field necessarily preserve the valuation ring? (I. e. map elements of absolute value no larger than one to elements with such property.)

is this the requirement of the definition of the automorphism of DVF? I guess this should be deduced from the algebraic definition with the help of valuation, but I don't know how to do this.

and for more general cases: nondiscrete/archimedean valuations, what general statements can we obtain? is there any reference talking about this? thx!

Answer 1

For $O_K$ a complete DVR with unique maximal ideal $(\pi_K)$ and a finite extension $L/K$, let $R$ be the integral closure of $O_K$ in $L$, let $\mathfrak{m}$ be a maximal ideal of $R$ (it contains $\pi_K$).

$S = (R-\mathfrak{m})^{-1}R$ is a DVR with uniformizer $\varpi$ that we can choose to be in the intersection of the finitely many maximal ideals of $R$, so that $\varpi^e\in \pi_K R$.

Let $a_1,\ldots,a_f\in R$ be representatives of $S/(\varpi)$. Then $$\varprojlim S/(\varpi^n)$$ is the completion of $S$. In this ring it makes sense to consider $$O_K[a_1,\ldots,a_f][[\varpi]]$$ which is complete with the same uniformizer and residue field, ie. it is the whole of $\varprojlim S/(\varpi^n)$.

But since $\varpi$ is integral over $O_K$ and $\varpi^e\in \pi_K R$ and $O_K$ is complete we have $$O_K[a_1,\ldots,a_f][[\varpi]]=O_K[a_1,\ldots,a_f,\varpi][[\pi_K]]=O_K[a_1,\ldots,a_f,\varpi]\subset R$$

Thus $$R =O_K[a_1,\ldots,a_f,\varpi]=\varprojlim S/(\varpi^n) \ is \ a\ complete \ DVR$$ Since any automorphism of $L/K$ sends $R$ to itself and since $R$ has only one maximal ideal $(\varpi)$, the automorphisms send the maximal ideal $(\varpi)$ to itself.

It stays true when $O_K$ is Henselian. But when $O_K$ is non-complete nor Henselian it may fail, try with $O_K=(\Bbb{Z}-(5))^{-1}\Bbb{Z},L=\Bbb{Q}(i),R=O_K[i]$, $S=(\Bbb{Z}[i]-(2+i))^{-1}\Bbb{Z}[i]=O_K[i,(2-i)^{-1}],\varpi=(2+i)$ ,$S_2=(\Bbb{Z}[i]-(2-i))^{-1}\Bbb{Z}[i]=O_K[i,(2+i)^{-1}],\varpi_2=2-i$,

the complex conjugaison switch $S$ and $S_2$.

Answer 2

If L/K is an algebraic extension, $A \subseteq K$ a valuation ring (i.e., for all $a \in K^*$ either $a \in A$ or $a^{-1} \in A$), and $\widetilde{A} \subseteq L$ the normalisation of $A$ in $L$, then the local rings $\widetilde{A}_\mathfrak{m}$ of $\widetilde{A}$ at maximal ideals will be precisely the extensions $A$ to $L$.

That is, there is a bijection between $$\{ \textrm{valuation rings } B \subseteq L \textrm{ such that } B \cap K = A \textrm{ and } A \to B \textrm{ is a local homomorphism} \} $$ and $$\{ \textrm{maximal ideals of } \widetilde{A} \}. $$

If moreover $L/K$ is normal, then $Aut(L/K)$ permutes the extensions / maximal ideals transitively.

In the case that there is a unique extension (e.g., if $A$ is a henselian local ring) then $Gal(L/K)$ preserves this unique extension but it doesn't in general.

Complete local rings are henselian, so this includes the situation discussed above.

In the comments you asked why field automorphisms, which are algebraic, should preserve the valuation, which appears to be an extra datum. This is because the valuation is always canonically isomorphic to the canonical projection $v: K^* \to K^* / A^* =: \Gamma$.

If $A$ is Noetherian, then $\Gamma$ is isomorphic to the integers via a choice of generator for the maximal ideal of $A$. Explicitly, if $\mathfrak{m}_A = (\pi)$ then $\mathbb{Z} \to K^* / A^*$; $n \mapsto \pi^n$ is an isomorphism.

All these facts can be found in various places. Some are on the stacks project. They are all in Bourbaki. The book Valued Fields by Engler and Prestel is also good to look at. For the number theory setting, probably there is stuff in Neukirch's book.