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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
up vote 5 down vote accepted

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.

An analogy may be useful. As models of the theory over the empty language, any two structures of the same cardinality are isomorphic. However, such an isomorphism says nothing useful about the relationship between two groups of the same cardinality.

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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

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