For a surreal number $X = \{X_L|X_R\}$, its birthday is defined recursively to be the smallest ordinal bigger than the birthdays of all $x \in X_L \cup X_R$. If $X$ can be written as $\{X_L|X_R\}$ in several way, then the birthday is the minimal of the possible birthdays.
Let $\alpha$ be some uncountable regular cardinal, and consider the set of surreal numbers with birthday less than $\alpha$. Does this set form a field?
For $\alpha = \omega$ this is not the case, since 3 has birthday 3, while $\frac13$ has birthday $\omega$. It seems to me that division is the main obstruction, but my thinking is that on day $\omega$ each number has a reciprocal so perhaps this won't be a problem for uncountable cardinals.
Best Answer
The specific result you are asking about is true and was proved for any uncountable ordinal (irrespective of it being regular or not) by Harry Gonshor in his book An introduction to the theory of surreal numbers.
The exact characterization of ordinals $\lambda$ such that the set $\mathbf{No}(\lambda)$ of numbers with birth day $<\lambda$ is a subfield was given by Lou van den Dries and Philip Ehrlich in Fields of surreal numbers and exponentiation. Their result is that $\mathbf{No}(\lambda)$ is a subfield if and only if $\lambda$ satisfies $\omega^{\lambda}=\lambda$. Those ordinal numbers are sometimes called $\varepsilon$-numbers, and can also be defined as numbers $\lambda>2$ which are closed under ordinal exponentiation.
In particular, there are many small countable ordinals $\lambda$ for which $\mathbf{No}(\lambda)$ is a field.