If $k$ is an ordered field, the least ordinal $s(k)$ which doesn't embed in $(k,<)$ is regular. This is because every interval of an ordered field embeds in every infinite interval so given a strictly increasing map $f: \alpha < s(k) \rightarrow s(k)$ you can define an embedding $(\sup(f(\alpha)),\in) \rightarrow (k,<)$. Moreover $s(k)$ is always uncountable since $s(\mathbb{Q}) = \omega\_1$.


Note that these results make use of the field structure of $k$. They fail even for ordered rings such as the ring obtained from $({\omega}^{\omega},\underline{+},\underline{\times})$ (natural operations), by adding additive inverses and extending the operations like one usually does.


Let $\lambda$ be an $\varepsilon$-number. Let $No(\lambda)$ denote the field of surreal numbers of length $<\lambda$. I am trying to figure out what $s(No(\lambda))$ is.


Clearly, $s(No(\lambda))$ is at least equal to ${\lambda}^+$ and at most equal to ${|No(\lambda)|}^+$. Assuming GCH, this yields $s(No(\lambda)) = {\lambda}^+$ when $\lambda$ is itself a cardinal, and ${\lambda}^+ \leq s(No(\lambda)) \leq ({\lambda}^+)^+$ otherwise.
This is just a computation of $|No(\lambda)|$ using elemenraty cardinal arithmetic and the ordered-set-focused definition of $No(\lambda)$. 


Can these results (or their "equivalent" modulo GCH where $2^{\alpha}$ replaces ${\alpha}^+$) be proven in ZFC by taking advantage of the structure of $No(\lambda)$?



Ordinals which embed in surreal subfields
I claim that $s(\text{No}(\lambda))=\lambda^+$.


To see this, first let me mention that it is a standard exercise
in elementary set theory to prove that there is no increasing or
decreasing $\lambda^+$-sequence in the set ${}^\lambda 2$ of all
binary $\lambda$-sequences, ordered in the lexical order, so that
$s<t$ just in case $s$ is below $t$ at the first point of
difference. I'll let you prove that on your own, or I can post
later if you want.


From this, I claim, it follows that $\lambda^+$ does not embed
into the set of surreal numbers born before $\lambda$. The reason
is that every surreal number born before $\lambda$ is
characterized uniquely by a $<\lambda$-sequence from $\{-1,1\}$,
depending on which side of the cut in the previously born surreals
the new number has landed. The surreal number order agrees exactly
with the lexical order on the these sequences. (A shorter sequence
is below a longer one, if the next digit is $1$, and above, if the
next digit is $-1$; alternatively, imagine adding a tail of $0$s
to compare sequences of difference lengths.) Since $\lambda^+$
does not embed into ${}^\lambda 2$, it cannot embed into
${}^{<\lambda}\{-1,1\}$.


The argument shows that you could even allow the surreals born on
day $\lambda$ itself, continuing up to any particular day
$\beta<\lambda^+$, and still $\lambda^+$ would not embed in.


