Is the number of orderings the same of the number of automorphisms in a ring?

Question

Q: Given an ordered ring $A$ is the number of automorphisms of $A$ equal to the number of orderings in $A$?

An ordering on a ring is totally defined by a subset of $A$ we call $A^+$ that satisfies these conditions:

1. Its closed under addition and multiplication
2. $\forall a \in A$ one and only one of the cases holds: $a=0,a \in A^+,-a \in A^+$.

We define the relation $">"$ as: $a>b \iff a-b \in A^+$.

Let $\mathcal{A}^+$ be the set of all orderings (the set of subsets of $A$ that satisfy 1,2) and Aut$(A)$ the set of all automorphisms of $A$.

We know that $A$ is ordered so $\mathcal{A}^+$ is non-empty, let $I^+ \in \mathcal{A}^+ $ and define $f:$ Aut$(A) \rightarrow \mathcal{A}^+ $ s.t $f(\phi)=\phi(I^+)$. Note that this fucntion is well defined because if $A^+$ is an ordering on $A$ and $\phi \in $ Aut$(A)$ then $\phi(A^+)$ is also an ordering on $A$.

Its needed to prove or disprove if this function is a bijection.

Regarding the "one-to-oneness" of the function I tried proving it by contradiction assuming there are $\phi,\psi \in \text{Aut}(A) \text{ such that } f(\phi)=f(\psi) \iff \phi(I^+)= \psi(I^+)$ that aren't equal that is, there exists $a \in A$ s.t. $\phi(a) \neq \psi(a)$ then assuming without loss of generality that $\phi(a) > \psi(a)$. therefore, $\phi(a)-\psi(a) \in \phi(I^+)=\psi(I^+)$. But im having trouble continuing.

Note $f$ may not be one-to-one or onto so in that case it's necessary to come up with a counter-example.

Answer 1

No. For instance, $\operatorname{Aut}\Bbb Q(x)$ is certainly a countable set. However, there are uncountably many orders on the ring $\Bbb Q(x)$: at least one for each map $\Bbb Q(x)\hookrightarrow \Bbb Q(\alpha)\subseteq \Bbb R$ that sends $x\mapsto \alpha$, where $\alpha$ is a transcendental number. The corresponding $ A^+_\alpha$ might be devised as such: consider a rational sequence $a_n\to \alpha$. We say that a rational function $u(x)\in\Bbb Q(x)$ is in $A^+_\alpha$ if and only if there are a rational $h>0$ and some $n\in\Bbb N$ such that $u(a_m)>h$ for all $m>n$. It is clear that $ A^+_\alpha\ne A^+_\beta$ for all $\alpha\ne \beta$, because, if say $\beta>\alpha$, then $x-q\in A^+_\beta\setminus A^+_\alpha$ for some rational $q\in(\alpha,\beta)$.

Added: Also notice that these orderings encompass an uncountable family of isomorphism classes of ordered rings: an ordered isomorphism $\Phi:(\Bbb Q(x),A^+_\alpha)\to (\Bbb Q(x),A^+_\beta)$ is a fortiori an element of $\operatorname{Aut}\Bbb Q(x)$, and therefore there are at most countably many such isomorphisms. Moreover, $A^+_\beta=\Phi[A^+_\alpha]$ for any such isomorphism, therefore there can only be countably many $A^+_\beta$ in the order-isomorphism class of each $A^+_\alpha$.

Answer 2

Gae. S. has given an example of a ring with more orderings than automorphisms. In the other direction, there are also (orderable) rings with more automorphisms than orders.

Any real closed field has a unique ordering (take $A^+$ to be the set of non-zero squares), so it suffices to find a real closed field with a non-trivial automorphism.

Just as in Gae. S.'s answer, consider the field $F = \mathbb{Q}(x)$, but this time ordered so that $x$ is "infinite" (greater than every rational number). We can do this by taking the lexicographic order on $\mathbb{Q}[x]$ (coefficients of higher powers of $x$ matter more) and then taking the unique extension of this order to the fraction field. Let $R$ be a real closure of $F$ whose unique order extends the order given on $F$.

Now for any $q\in \mathbb{Q}$, there is an order automorphism $F\to F$ induced by $x\mapsto x+q$. And each of these order automorphisms extends to an automorphism of $R$.

It follows on general model theory grounds that we can do better than this: It's a theorem of Ehrenfeucht and Mostowski that if a first-order theory in a countable language has any infinite model, then it has a countably infinite model with $2^{\aleph_0}$-many automorphisms. In the specific case of real closed fields, we would do a similar construction to one above, but this time starting with the rational function field in an infinite family of variables $(x_q)_{q\in \mathbb{Q}}$ ordered as $(\mathbb{Q},\leq)$. Then any automorphism of $(\mathbb{Q},\leq)$ (of which there are continuum-many) induces an order automorphism of the field, which extends to an automorphism of the real closure.