# Surreal numbers and the Axiom of Choice

In ZFC and its conservative extension NBG, it can be shown that every ordered field embeds into the surreal numbers.

1. How much choice is needed to prove this?

2. Without choice, what is a simple example of an ordered field that doesn't fit into the surreals?

3. Without choice, does there exist a strictly larger ordered field that the surreals embed into?

4. Without choice, does there exist (can there exist?) a largest ordered field at all?

• How do you define the surreal without choice? – Asaf Karagila Feb 26 '19 at 18:59
• I didn't think the usual method needed choice, where each number is defined inductively as a "left set" and a "right set" of previously created numbers. – Mike Battaglia Feb 26 '19 at 20:09
• mathoverflow.net/questions/286208/… should answer (4). – Asaf Karagila Feb 26 '19 at 20:38

Notice that if an ordered field embeds into $$\mathbf{No}$$, then so does, as an ordered set, its ladder$$^{\mathbf{[a]}}$$. Since every linear order is the ladder of some ordered field, this means that a negative answer to 2. would answer the set-sized version of JDH's questions.
I assume by strictly larger ordered field you mean an ordered field which in addition does not embed in $$\mathbf{No}$$. Or do you mean that it should not inject into $$\mathbf{No}$$? In the first case, notice that if $$L$$ is a linear order which does not embed into $$\mathbf{No}$$ then considering the ordered Hahn series group $$G=\mathbb{R}[[\mathbf{No} \sqcup L]]$$, the ordered Hahn series field $$F=\mathbb{R}[[G]]$$ is an example.
[a]: The ladder $$L$$ of an ordered field $$F$$ is obtained from $$F^{\succ}:=\{x \in F: x > \mathbb{N}\}$$ with the preorder $$x\lll y$$ iff $$x^{\mathbb{N}}.