# Completing ordered Fields

How do these two forms of completion behave (in NBG) when fields are authorized to be proper classes?

$(i)$: Every ordered field has a real closed, algebraic extension.

$(ii$): Every ordered field has maximal dense extension.

Let's say that a real closed Field is a Field whose polynomials satisfy the intermediate value theorem. (when this can be made sense of)

There are two main problems with Fields:

-the satisfaction relation isn't necessarly defined for them

-one can't construct Field extensions easily because for instance, if $C$ is a proper class, the class of maps $C \rightarrow \{0;1\}$ doens't exist, quotients of polynomial rings needn't exist either, etc...

However, with some Fields, things can go well.

For instance, one can define the Field of fractions with one indeterminate and with coefficients in the Field of surreals. Indeed $No[X]$ can be defined since polynomials are just maps $\mathbb{N} \rightarrow No$ with finite support, so sets, and fractions $\frac{f}{g}$ can be couples of polynomials $(f,g)$ where $g$ is monic and $f,g$ have no common factor. This has a meaning in $No$ since we can restrict ourselves to finding common factors of $f,g$ in $No(\kappa)$ for $\kappa$ regular big enough.

One can define several types of order on $No(X)$: One where $X > No$.

One for instance where $F < X < G$ where $F$ the class of surreals whose sign sequence is $(+-)$ $\alpha$-times for each ordinal $\alpha$ and $G$ is that of surreals whose sign sequence is a $+$ followed by $\alpha$ $(-+)$ for each ordinal $\alpha$. It requires a few details to justify that the order is definable.

P.Ehrlich proved that every real closed Field embeds nicely in $No$.

[I claimed that the first one does not have a real closure and doens't even embed in a real closed Field, but my sketch of proof was flawed in the end so I edited.]

The second order on $No(X)$ is a dense proper extension of $No$. So $No$ is not its own completion. I feel that the completion of $No$ if it existed would have to be real closed and would embed onto $No$, but I can't prove it because I have a hard time figuring the difficulty of working with proper classes.

My questions are:

-Are these properties true for proper classes?

-Does the completion of a real closed Field have to be real closed?

-To what extent can the completeness of the theory of real closed fields can be used to derive sfor real closed Fields?