# Does this construction yield the surreal numbers?

There are two simple constructions for creating arbitrarily large non-Archimedean ordered field extensions of the reals. First given such a field one may consider rational functions over that field with $f(x)$ declared positive if $f(x)>0$ for all sufficiently large $x$. Secondly one can Dedekind complete the field by filling in all good/narrow cuts in the field as detailed in this discussion. I would like to know if iterating these two constructions over the ordinals produces the surreal numbers.

Specifically we define $F_0$ to be the real numbers. For an ordinal $\lambda$ with predecessor $\lambda-1$, define $F_\lambda$ to be the ordered field obtained by Dedekind completing the rational functions over $F_{\lambda-1}$. Otherwise if $\lambda$ is a limit ordinal, define $F_\lambda$ to be the union of all $F_\kappa$ with $\kappa<\lambda$.

No. Think about the fields as valued fields under the natural valuation induced by the order (i.e., the valuation ring consist of elements bounded by an integer). Order completion then coincides (for nonarchimedean fields) with their completion as valued fields, in particular, it preserves the value group. If $F$ has value group $\Gamma$, then $F(X)$ ordered with $X>F$ as you are doing has value group $\mathbb Z\times\Gamma$ ordered lexicographically, in particular, it contains $\Gamma$ as a convex subgroup. Thus, value groups of all $F_\alpha$ will be discretely ordered. This means none of the fields nor their union can be a real-closed field, as real-closed fields have divisible value groups. (Specifically, the $X$ from $F_1=\widehat{\mathbb R(X)}$ will have no square root in any of the fields.)

• What if Zbigniew replaced $F_{\lambda-1}(t)$ with the Puiseux field $\bigcup_{n=1}^{\infty} F_{\lambda-1}((t^{-1/n}))$? (where $t$ is larger than any element of $F_{\lambda-1}(t)$). That is real complete, assuming $F_{\lambda-1}$ is. – David E Speyer Oct 7 '14 at 13:37
• Proof: Let $R$ be real closed and let $K = R(i)/i^2+1$. By an observation of Artin, $K$ is algebraically closed. By Puiseux's theorem, $\mathcal{K}:=\bigcup_{n=1}^{\infty} K((t^{1/n}))$ is algebraically closed. Then $\mathcal{R}:=\bigcup_{n=1}^{\infty} R((t^{1/n}))$ is the fixed points of the automorphism $i \mapsto -i$, $t^{1/n} \mapsto t^{1/n}$ of $\mathcal{K}$. By Artin-Schrier, that shows $\mathcal{R}$ is real closed. – David E Speyer Oct 7 '14 at 13:37
• It is perhaps worth mentioning that if you modify the construction to also take real closure at each step, it still will not work: in terms of value groups, this makes $\Gamma_{\alpha+1}=\mathbb Q\times\Gamma_\alpha$, hence all the value groups will contain the $\mathbb Q$ from $F_1$ as a convex subgroup, untouched. What is needed is to generalize the purely transcendenttal step: one can order $F(X)$ by wedging $X$ into an arbitrary Dedekind cut in $F$ (without the “good cut” condition). – Emil Jeřábek supports Monica Oct 7 '14 at 13:37
• Emil's comment above also shows the answer to my question is "no". – David E Speyer Oct 7 '14 at 13:42
• Right. I got momentarily confused by the fact that the Puiseux series field has an infinitesimal variable, but this does not really make a difference: the field is in any case contained in the completion of the real closure of $F_{\lambda-1}(t)=F_{\lambda-1}(t^{-1})$. – Emil Jeřábek supports Monica Oct 7 '14 at 13:46

I've written up an expository paper incorporating ideas from this discussion. Here is a link: https://drive.google.com/file/d/0B1G4mOmOYMtha1JLc2tTY1d2SnM/view I welcome any comments, suggestions, etc.

Edit Based on the comments above, I would like to rephrase my question. Suppose I change the recursive procedure as follows. To proceed from $F_\alpha$ to $F_{\alpha+1}$ I perform the constructions below in the order shown.

1. I first fill in the good/narrow gaps in $F_\alpha$ with cut points.
2. For each bad/wide gap $(A,B)$, if there is a polynomial $p(x)$ which changes sign across the gap, I algebraically extend $F_\alpha$ by a root $X_{(A,B)}$ of $p(x)$ contained within the gap (with degree of $p(x)$ minimal).
3. For each bad/wide gap $(A,B)$, such that all non-zero polynomials in $F_\alpha$ have the same sign across the gap, I add a transcendental element $X_{(A,B)}$. I declare a rational function $f(X_{(A,B)})$ over $F_\alpha$ to be positive if the rational function $f(x)$ is positive across the gap
4. Finally I add a transcendental element $X_\infty$ with a rational function $f(X_\infty)$ being declared positive if $f(x)>0$ for all sufficiently large $x\in F_\alpha$

Do I get the surreals if I recurse over all ordinals?

It seems to me that steps (1) and (2) have the effect of eventually real closing each $F_\alpha$ at some later stage.

Edit. I overlooked Emil's last comment. If I understand him correctly, then my construction produces an $Ord$-saturated field. Given sets $X<Y$, let $F_\alpha$ be the minimal field containing $X\cup Y$. Let $$A=\{a\in F_\alpha\ |\ a\le x\in X\}$$ $$B=\{b\in F_\alpha\ |\ b\ge y\in Y\}$$ Then if $(A,B)$ is not a Dedekind cut of $F_\alpha$, then there is an element $u\in F_\alpha$ with $X\subset A<u<B\supset Y$. If $(A,B)$ has a cut point $c$ in $F_\alpha$ then one of $c\pm\frac{1}{X_\infty}$ produces an appropriate $u\in F_{\alpha+1}$. Otherwise my construction explicitly produces such a $u\in F_{\alpha+1}$.

Edit 2 I realize that I need to be a little more careful in formulating my construction. I need another transfinite recursion in the passage from $F_\alpha$ to $F_{\alpha+1}$. First a couple of preliminary definitions.

Let $(A,B)$ be a Dedekind cut of an ordered field $F$. We say that $(A,B)$ is a gap if there is no cut point in $F$. We say that the gap is narrow if for any $\epsilon>0$ in $F$, we can find $a\in A$, $b\in B$ with $b-a<\epsilon$. Otherwise we say the gap is wide. We further classify the wide gaps as follows. If there is a polynomial $p(x)$ over $F$ which changes sign across the gap, we say the gap is algebraic (relative to $F$). Otherwise we say the gap is transcendental (relative to $F$).

Now given an ordered field $F_\alpha$ I want to construct an ordered field $F_{\alpha+1}$ with the property that for any Dedekind cut $(A,B)$ in $F_\alpha$ (gap or not)there is an element $u\in F_{\alpha+1}$, with $A<u<B$. I start by filling in all the narrow gaps in $F_\alpha$ by taking the standard Dedekind completion of $F_\alpha$ as detailed in this discussion. I call the resulting field $F_\alpha^*$.

Now I start another transfinite recursion. I begin by well-ordering all the wide gaps $(A_\beta,B_\beta)$ in $F_{\alpha}$. I define $F_{\alpha,0}=F_\alpha^*$. Having defined $F_{\alpha,\beta}$, I define $F_{\alpha,\beta+1}$ as follows.

Case 1. If there is already an element $u\in F_{\alpha,\beta}$ with $A_\beta<u<B_\beta$, then I take $F_{\alpha,\beta+1}= F_{\alpha,\beta}$.

If we are not in Case 1, then we can extend the wide gap $(A_\beta,B_\beta)$ in $F_\alpha$ to a wide gap $(A'_\beta,B'_\beta)$ in $F_{\alpha,\beta}$.

Case 2. If $(A'_\beta,B'_\beta)$ is algebraic relative to $F_{\alpha,\beta}$, then I take $F_{\alpha,\beta+1}$ to be an algebraic extension of $F_{\alpha,\beta}$ obtained by adjoining a root $X_{(A'_\beta,B'_\beta)}$ of an irreducible polynomial $p(x)$ which changes sign across the gap. We order $F_{\alpha,\beta+1}$ so that $X_{(A'_\beta,B'_\beta)}$ lies in the gap.

Case 3. If $(A'_\beta,B'_\beta)$ is transcendental relative to $F_{\alpha,\beta}$, then I take $F_{\alpha,\beta+1}$ to be a transcendental extension of $F_{\alpha,\beta}$ obtained by adding a transcendental element $X_{(A'_\beta,B'_\beta)}$ with a rational function $f(X_{(A'_\beta,B'_\beta)})$ declared positive if the corresponding function $f:F_{\alpha,\beta}\to F_{\alpha,\beta}$ is positive across the gap.

After completing the induction $F_{\alpha,\beta}$ over $\beta$, I obtain a field $\widehat{F}_\alpha$ which contains elements spanning all gaps in $F_\alpha$. Finally I add a transcendental element $X_\infty$ to $\widehat{F}_\alpha$ with a rational function $f_(X_\infty)$ declared positive if the corresponding function $f:\widehat{F}_\alpha\to \widehat{F}_\alpha$ is positive for sufficiently large values of the argument. I call the resulting field $F_{\alpha+1}$. This last step has the effect of adding elements between Dedekind cuts in $F_\alpha$ which are not gaps, but rather have cut points.

• (2) seems potentially problematic as phrased. Let $F=\mathbb{Q}(t)$, with $t$ larger than every element of $\mathbb{Q}$. Let $A = \{ x : x < N \ \mbox{for some} \ N \in \mathbb{Q} \}$ and $B = F \setminus A$, this is the prototypical wide gap. Let $p(x) = x^2-t$ and $q(x)=x^2- 2t$. Both of these change signs across $(A,B)$. You need to adjoin two roots (call them $x_p$ and $x_q$), not just one. Moreover, your construction needs to know that $1.41421356 < x_q/x_p < 1.41421357$. – David E Speyer Oct 8 '14 at 22:02
• Or, if you mean to just choose one $p(x)$ for each $(A,B)$, how does your construction know that the root of $x^2-t$ between $1$ and $t$ cubes to the root of $x^2-t^3$ between $t$ and $t^2$? – David E Speyer Oct 8 '14 at 22:03
• In case 2, there may be more than one such polynomial (iow, more than one element of the real closure may fall into the gap). You need to take a polynomial $p$ of minimal degree that changes sign across the gap (which implies its irreducibility). Then there is a well-defined order on the extension of $F_{\alpha,\beta}$ by a root $X$ of $p$: if $f$ is a polynomial of degree less than the degree of $p$, we can put $f(X)>0$ iff $f$ is positive across the gap. There still may be more than one such minimal-degree polynomial, as shown in David Speyer’s example, hence you need to keep repeating ... – Emil Jeřábek supports Monica Oct 11 '14 at 18:41
• .. this for the same $(A_\beta,B_\beta)$ as long as there are any more algebraic elements in the original gap, it’s not enough to do it once. By the same token, you need to add these algebraic elements even if a transcendental element has been inserted in the gap, so Case 1 as formulated is incorrect. On the other hand, you are making the construction unnecessarily complicated. There is no point in distinguishing different kinds of Dedekind cuts. If you formulate (the fixed versions of) cases 2 and 3 for all cuts, the initial step of taking $F_\alpha^*$, and the $X_\infty$ step, are redundant. – Emil Jeřábek supports Monica Oct 11 '14 at 18:47
• @david,@emil Thanks for your clarifications. The reason I want a description of the surreals along these lines is because I am giving a talk on completion and non-Archimedean fields to our undergraduate math club. My target audience has a good background in analysis, eg. Dedekind completion, and some exposure to field theory. However they may not have a good background in formal set theory. Hence I plan to be vague on the precise construction of the surreals. Here is a link to my talk: link – Zbigniew Fiedorowicz Oct 15 '14 at 15:42