# “Archimedeanising” an ordered field

If $K$ is an ordered field, let $B$ be the subring comprising the $x \in K$ such that $|x| \le n$ for some $n \in N$, and let $I$ be the ideal of $B$ comprising the infinitesimal elements (i.e. the $x \in K$ such that $|x| \le 1/n$ for every non-zero $n \in N$). Then $I$ is a maximal ideal and the order on $K$ induces an order on $A = B/I$ making it into an archimedean ordered field.

Has this construction been studied? More specifically, when does the natural projection of $B$ onto $A$ split (in the category of rings or, better still, ordered rings)? I believe it always has an order-preserving splitting if $K$ is real closed.

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Can you just choose a transcendence basis for $A$ over $\mathbb Q$, for each one pick an arbitrary point in their inverse image, and then for every other number pick the unique point in its inverse image which satisfies its minimal polynomial over $\mathbb Q$ adjoin the transcendence basis? – Will Sawin Feb 3 '12 at 18:51
That doesn't work in general. Let $K$ be the splitting field for the polynomial $X^2 - 2 - \epsilon$ over the field $\mathbb{Q}(\epsilon)$ of rational functionals with rational coefficients of an an infinitesimal indeterminate $\epsilon$. Then $A$ is isomorphic to $\mathbb{Q}[\sqrt{2}]$, but 2 is not a square in $K$. – Rob Arthan Feb 4 '12 at 9:33

Yes, the map is sometimes called the standard part map, and it turns out that one map define it in general for a group in an o-minimal structure. See here, for example. Regarding the splitting, if it should only respect the additive group structure, I think it exists since everything is a vector space over $\mathbb{Q}$ (as explained in the comment)
@Moshe: Thanks for a useful reference. I think you will find the existence of a standard part map is a hypothesis rather than a theorem in the work you refer to e.g., see the paper by Marikova. I should have said that the splittings I am interested in are to be ring homomorphism, or, even better, ordered ring homomorphisms. As you say the projection always splits when you just view $K$ and $A$ as vector spaces over $\mathbb{Q}$ (given AC to construct a basis, in general, as Gerald points out). – Rob Arthan Feb 4 '12 at 9:46