$\mathbb{Q}$ has no proper subfields. As a result, all ordered fields elementarily equivalent to $\mathbb{Q}$ have no proper subfields which are first-order definable without parameters. And by the Tarski-Seidenberg theorem, real closed fields also have no proper subfields first-order definable without parameters.


My question is, are all the ordered fields of characteristic $0$ which have no proper subfields first-order definable without parameters known? Or failing that, are all the countable ordered fields of characteristic $0$ with this property at least known?



Is there a complete characterization of ordered fields without definable proper subfields?
This is an interesting question. We know some things about this, but we do not have a characterization of fields with this property. As Wojowu says above the restriction to countable fields doesn't help, and I don't think that restricting to ordered fields helps either. This property implies that the field cannot define the subfield $\mathbb{Q}$, and of course we are particularly interested in definability of that subfield.


First note that an imperfect field defines the image of the Frobenius, so we can just consider perfect fields. I believe that basically all known model-theoretically tame perfect fields do not admit definable subfields, but I'm not sure how much of this has been written down. We don't know which fields are tame and which aren't.


As you point out, real closed fields have this property. This is because any infinite definable subset of $R$, for $R$ real closed, has nonempty interior. So any definable subfield must have interior, but it is easy to that there are no proper subfields with interior. The last claim holds for any topological field. So we can apply this argument to other topological fields, we just need every infinite unary definable to have interior. For example for $\mathbb{Q}\_p$ this follows from a quantifier elimination result. Similar arguments should work over any local field of characteristic zero. I think it should also generalize to Henselian fields of characteristic zero, this should follow from quantifier eliminations for Henselian valued fields. So you cover things like $F((t))$ for any field $F$ of characteristic zero. To get something ordered you can take $F$ to be any ordered field.


Among the fields that people are usually interested in, there is a big divide between the large and non-large. Largeness is a field-theoretic notion introduced by Pop, it's supposed to be a field-arithmetic notion of tameness. Finite fields, number fields, and function fields are not large, most other interesting fields are. It's a theorem of Julia Robinson that any number field defines $\mathbb{Q}$. There are a number of results saying that various function fields define the field of constants. For example $\mathbb{R}(t)$ defines $\mathbb{R}$. I don't know what the state of the art is here, but for example we do know that if $F$ is an extension of $K$ with transcendence degree one, $K$ is algebraically closed in $F$, and the group of $n$th powers in $K$ has finite index for some $n \ne \mathrm{Char}(K)$, then $F$ defines $K$. See for example Fehm and Geyer's paper "A Note on Defining Transcendentals in Function Fields".


So it's worth restricting to perfect large fields. Arno Fehm has shown that a perfect large field cannot *existentially* define a proper subfield. Perfect large fields can define subfields, the fraction field of $\mathbb{R}[[x,y]]$ is large and defines $\mathbb{Q}$, this should also be orderable. We do not have a characterization of perfect large fields that do not define proper subfields.


If a field $K$ is perfect, large, and model complete (possibly after constants have been added to the language) then it cannot define any proper subfields - just because any definable set is existentially definable. For example, the theory of $n$-ordered fields has a model companion, the theory of $n$-ordered pseudo real closed fields, and these fields are model complete in the language of fields (without the orders!), so $n$-ordered pseudo real closed fields do not admit proper definable subfields.


I think it should be possible to show that the other known examples of model-theoretically tame perfect fields do not admit proper definable subfields, but I'm not sure how much is known/written down.


Edit: Fehm pointed out that Junker and Koeniggsmann show in the paper "Slim fields" that perfect PAC fields and characteristic zero Henselian fields do not define proper subfields. They study a class of fields they call "very slim", a very slim field does not define proper subfields. We don't have a classification of very slim fields, either.


