It's well known that the surreal numbers $\mathbf{No}$ are the largest ordered "field" (more accurately, they form a proper class with field structure, which is sometimes called a Field with capital F), in the sense that every other ordered field can be embedded in them. Since the surreal numbers are real closed, their algebraic closure is given by $\mathbf{No}[i]$, the surcomplex numbers.
My question is, is there a similar characterization of the surcomplex numbers as the largest algebraically closed Field of characteristic zero? If they aren't the largest, is it possible to find a proper class with that property?
Best Answer
Let $F$ be an algebraically closed field of characteristic zero.
Let $\kappa$ be the cardinal of a transcendance basis of $F$ over its prime subfield $\mathbb{Q}$.
Since the ${\omega_0}^{{\omega_0}^{\alpha}}$, $\alpha < \kappa$ satisfy the relation $\forall \alpha < \beta < \kappa, \forall n \in \mathbb{N}, n.({\omega_0}^{{\omega_0}^{\alpha}})^n < {\omega_0}^{{\omega_0}^{\beta}}$, they form an algebraically independant familly over $\mathbb{Q} \subset No$.
So $No[i]$, being algebraically closed, contains an isomorphic copy of $F$ as the relative algebraic closure of $\mathbb{Q}(({\omega_0}^{{\omega_0}^{\alpha}})_{\alpha < \kappa})$.
This is also true in NBG with global choice if $F$ is an algebraically closed Field of characteristic zero, and to see this one only needs to repeat the same argument with $Ord$ instead of $\kappa$.
Note that being a universal algebraically closed Field of characteristic zero caracterises $No[i]$ as a field whereas being a universal real closed Field does not characterise $No$. The theory of algebraically closed fields of a given characteristic is indeed more stable than that of real closed fields.