Let $\mathbb{A}$ denote the field of real algebraic numbers. Let $\mathcal U$ denote a free ultrafilter. Construct $F=\prod_{\mathcal U} \mathbb{A}$. This is a field containing $\mathbb A$, and we clearly have a map, $st:F \to \mathbb R \cup \{\pm \infty\}$, given by $$ st((x_n)_n) = \lim_{\mathcal U} x_n, $$ which agrees with the identity map on $\mathbb A$.

So, $F$ looks (to me) a lot like a field of extended reals, except that there is no obvious embedding $\iota:\mathbb R \to F$. Does there exist an embedding (as a field) at all? (With the obvious properties that we would want: it agrees with the identity map on $\mathbb A$, and $st \circ \iota$ is equal to the identity map on $\mathbb R$.)

Some remarks. An axiom-of-choice argument obviously gives an embedding which might not be a field homomorphism. Also, the answer would be no if we replaced $\mathbb A$ by $\mathbb Q$, since in this case, there would be no candidate in $F$ for $\sqrt{2}$ for example; using $\mathbb A$ avoids such obstructions.

Here is a proof that it doesn't work with $\mathbb Q$. Suppose that we had field homomorphism $\phi:\mathbb R \to {}^\star\mathbb Q$. Then let $(y_n)_n$ be a representative sequence for $\phi(\sqrt{2})$. This means that $(y_n^2)_n$ is equivalent to $(2,2,\dots)$ modulo $\mathcal U$, i.e. that $$ \{n \mid y_n^2 = 2\} \in \mathcal U. $$ (See http://en.wikipedia.org/wiki/Ultraproduct). But since no rational $y$ satisfies $y^2=2$, this set is in fact empty and not in $\mathcal U$.

  • $\begingroup$ You are correct there is no embedding $\mathbb{R} \to {}^\star \mathbb{Q}$. I had gotten the impression that's not literally what you cared about: I thought you were looking for a map $\overline{\mathbb{R}} \to {}^\star \mathbb{Q} / \sim$ $\endgroup$ – Hurkyl Nov 1 '13 at 3:49

I believe such an embedding exists, and we can construct it recursively (using axiom of choice, of course).

  1. Enumerate real numbers as $\{r_\alpha\mid \alpha\in \mathfrak c\}$.
  2. Put $f_0=\operatorname{id}_A$, the identity on real algebraic numbers.
  3. Suppose we have $f_\alpha$, an elementary embedding of $\{r_\beta\mid \beta<\alpha\}^{\operatorname{acl}}$. If $r_\alpha\in \operatorname{dom} f_\alpha$, simply put $f_{\alpha+1}=f_\alpha$. Otherwise, choose any $r_\alpha'\in F$ whose standard part is $r_\alpha$ and put $f_{\alpha+1}'=f_{\alpha}\cup\{(r_\alpha,r_\alpha')\}$. Then for $f_{\alpha+1}$ take the elementary extension of $f_{\alpha+1}'$ to the algebraic closure of its domain (it exists and is unique because the algebraic closure is the same as definable closure in this case, and $f_{\alpha+1}'$ is elementary).
  4. For limit $\gamma$, put $f_\gamma=\bigcup_{\alpha<\gamma} f_\alpha$.
  5. Then $f=\bigcup_{\alpha<\mathfrak c}f_\alpha$ will be the embedding you were looking for. Note that it is not only algebraic, but also elementary (kind of trivially so: real closed fields admit q.e. in a suitable language, so any real closed subfield of a real closed field is elementary).

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