# Nonstandard complex numbers and categoricity

Let ${}^*\mathbb{C}$ be a nonstandard complex number field (given, for instance, as a countable ultrapower.) By the transfer principle ${}^*\mathbb{C}$ is algebraically closed of characteristic zero, and by the construction as a quotient of $\mathbb{C}^\mathbb{N}$ we see it's of cardinality $\mathfrak{c}$. The theory of algebraically closed fields of a fixed characteristic is categorical, so this shows ${}^*\mathbb{C}$ is isomorphic to $\mathbb{C}$ in the category of fields.

I'm trying to understand how to interpret this fact in terms of the nonstandardness of ${}^*\mathbb{C}$, namely that $\exists x\in {}^*\mathbb{C} \forall r\in\mathbb{R} x \bar x<r$.

Question: Am I reading the above correctly to imply that there exists a hyperreal-valued "absolute value" on $\mathbb{C}$ which takes on infinitesimal, standard, and infinite values?

This seems impossible, because the absolute value would have to be infinite, finite, or infinitesimal on real lines in $\mathbb{C}$, and then the triangle inequality would close, for instance, the infinitesimal part under sums. Would we just get a strange decomposition of the plane into three unions of lines, according to which piece of ${}^*\mathbb{R}$ our absolute value fell into? It remains unclear to me that such a decomposition is possible.

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How odd. What exactly is the image of real infinitesimals under this isomorphism? – Conifold Aug 17 '14 at 20:25
It is $\kappa$-categorical for uncountable $\kappa$ as a field. Not as a field with distinguished subfield and absolute value function to the non-negatives in that subfield. The ultrapower construction does produce such a subfield, but the isomorphism of the categoricity does not extend. – André Nicolas Aug 17 '14 at 20:34

All ultrapowers of $\mathbb{C}$ of cardinality the cardinality of the continuum are isomorphic to $\mathbb{C}$ as fields. They are not all isomorphic to $\mathbb{C}$ as fields with additional unary function $\text{Conj}$, the conjugate function.
We can consider the full structure on $\mathbb{C}$, by adding function symbols, relation symbols for every function and relation on $\mathbb{C}$, including a unary function symbol $\text{Conj}$. The ultrapower $M$ of $\mathbb{C}$ is an elementary extension of $\mathbb{C}$ with respect to this extended language, and the elements of $M$ that satisfy $\text{Conj}(x)=x$ are a non-standard model of analysis. But of $\varphi$ is a field isomorphism of $\mathbb{C}$ onto $M$, there is little connection between what $\varphi$ maps $\mathbb{R}$ to and the non-standard model.
The point is that there are many ways to embed reals into complex numbers. ${\bf C}$ has very many automorphisms (in fact, $2^{\mathfrak c}$-many), and each of those gives a distinct embedding of ${\bf R}$ into ${\bf C}$ (because ${\bf R}$ has no nontrivial automorphisms) and each of those, in turn, gives rise to its own brand of absolute value. The intersection of those embeddings is the field of algebraic reals. – tomasz Aug 18 '14 at 14:15
Thanks to both of you. I had been counting real algebra automorphisms, rather than field automorphisms, of $\mathbb{C}$, so that helps me believe in stranger isomorphisms such as this one. If I can try to explain my current understanding: the answer to my question is "yes," but my discussion of lines is wrong because the absolute value on ${}^*\mathbb{C}$ will be realized in $\mathbb{C}$ in terms of some wild embedding of $\mathbb{R}$. If that's still off base, I'd appreciate more elaboration from either of you. – Kevin Carlson Aug 18 '14 at 20:32