Is it possible (using ZFC or any other axiom system which is known to be as consistent as ZF) to prove the existence of an ultrafilter $\mathcal F$ on the set of prime numbers s.t. there is an embedding $\mathbb R\hookrightarrow \tilde{\mathbb F}=\prod_p \mathbb F_p/\mathcal F$, but without this field $\tilde{\mathbb F}$ containing a square root of $-1$? Such an ultrafilter would have to include all of the sets $\{p|p\equiv n-1\mod n\}$ for all natural $n$, because all the roots of unity exist in $\mathbb C$ but only $-1$ exists in $\mathbb R$, but I don't have any other interesting thing to say about it.
If this is indeed the case, I'm also interested in whether it is possible to say something nontrivial about the number of such embeddings of $\mathbb R$ into one specific such field (except of the fact that it is between $1$ and $2^{2^{\aleph_0}}$). Note that there is no higher trivial lower bound since, unlike $\mathbb C$, $\mathbb R$ doesn't have a nontrivial embedding into itself.
Thanks!
Roee Sinai
Edit [March 9th 2024]:
I have managed to prove the existence of such a field if you're allowed to take a countably infinite amount of copies of every $\mathbb F_p$ (or equivalently, a finite but unbounded amount), instead of just one. At first I thought this concluded the proof but turns out it does not, because, in a sense, an ultraproduct is not an associative operation: If you have an ultraproduct of an infinite collection of sets and you divide this collection to an infinite collection of infinite collections, you can't always make the original ultraproduct as an ultraproduct of some ultraproducts of the sets in the comprising collections.
This proof is also "elementary" in the sense that it does not involve relatively advanced model theoretic terms like saturation, which means I can understand it (and could come up with it in the first place). This also gives it the bonus advantage that it doesn't require the continuum hypothesis in order to work, which if I understand the comments correctly, the proposed saturation arguments do.
The proof is composed of three parts: first, we prove that the real algebraic numbers, $\bar{\mathbb Q}\cap\mathbb R$, can be embedded in an ultraproduct where all the fields appear once, then we prove that the reals can be embedded in "the nonstandard algebraic real numbers", i.e. $\mathbb R\hookrightarrow\prod(\bar{\mathbb Q}\cap\mathbb R)/\mathcal F$ for any non-principal ultrafilter $\mathcal F$ on $\mathbb N$, and finally we prove that an ultraproduct of ultraproducts is an ultraproduct, which is the statement whose general converse was remarked earlier as false, thereby completing the proof.
The first part is done similarly to this question: We look at all the Galois extensions $K/\mathbb Q$, and for each one we consider the set of primes whose Frobenius element corresponds to complex conjugation (which when the extension is real just means the identity). By Chebotarev's Theorem this set has positive density and in particular is infinite. Now, let us construct a mapping from $\mathcal O_K\cap\mathbb R$ to $\mathbb F_p$ for every prime number $p$ in this set. The way to do that is by first looking at all the ideals in $O_K$ that divide $p$. Their quotient fields are all $\mathbb F_{p^2}$ because the Frobenius element has order 2, and it's possible to take one on which the Frobenius automorphism corresponds to complex conjugation. Then, the mapping would be the quotient map into that field. Because complex conjugation keeps real numbers in place, they map to elements of $\mathbb F_{p^2}$ that are kept in place by Frobenius and thus are in $\mathbb F_p$. This mapping can be completed to $K$ itself by division, which may produce division by $0$ only on a finite number of primes for each element. Also note that when you intersect two such sets the resulting set contains another set of this kind (where the field is the Galois closure of the compositum of the two fields), and so finite intersections of these sets also have infinite cardinality and thus they can be completed to a non-principal ultrafilter. The resulting ultraproduct does not contain $i$ because, for example, $\mathbb Q[i]$'s set forces the primes to be $3$ modulo $4$. There is a fine point in this argument which is that we are free to choose a different prime ideal for each field and they may not give the same mapping. It is solved by noting that for any Galois extension $L/\mathbb Q$ for which one wants to construct a mapping from $L\cap \mathbb R$ to a large set of primes, one can first extend it by taking a generator $\alpha$ of $L\cap\mathbb R$ which WLOG is greater than all of its real conjugates and adding $\sqrt{\alpha-r}$ to it where $r$ is a rational number that is less than $\alpha$ but greater than all of its conjugates, as well as $\sqrt{r-\alpha'}$ for every real conjugate $\alpha'$ of $\alpha$, and then applying the step on the new field's Galois closure $K$. This guarantees that for any prime number $p\equiv 3\mod 4$ for which this step gives a mapping from $K\cap\mathbb R$ into $\mathbb F_p$ (except maybe for the ones that are in the denominator of $r$ and the ones in the leading coefficient of $\alpha$'s minimal polynomial) $\alpha$ will be mapped to the only root $x$ of its minimal polynomial for which $x-r$ is a square residue, whereas for all other roots $x'$, we'll have that $r-x'$ is a square residue and therefore $x'-r$ is not. This guarantees that all the mappings have the same representation for $\alpha$, and therefore, because every other element of $L\cap\mathbb R$ can be written as a polynomial in $\alpha$, a consistent representation for them as well.
The next part is done by realizing the real numbers as sequences of algebraic real numbers, where the field operations all hold eventually (i.e. if $a\star b=c$ then for all sufficiently large $n$, $f(a)_n\star f(b)_n = f(c)_n$). This is done by decomposing the reals to a transfinite sequence of real closed fields $\bar{\mathbb Q}\cap\mathbb R=K_0\subset K_1\subset ...\subset K_\kappa=\mathbb R$, where each successive field is constructed by joining a new element of $\mathbb R$ to the current field and taking the real closure of the resulting field (and each limit field is constructed by union). This is possible because of the well-ordering theorem. Then, we choose in every stage a sequence that converges to the chosen joined element to represent it, and it is left to be convinced that it is possible to do that to all others in a way that respects the field operations. Elements generated by addition, subtraction and multiplication are not a problem, because one can just perform the operation elementwise, and inverse is fine as well because if a sequence converges to a nonzero number only a finite number of terms in it can be $0$. What's left to show is that one can construct real roots of polynomials, which is fine as well because there is a neighborhood of the joined element in which the polynomial behaves in a consistent way, and from some point onwards the entire sequence is inside that neighborhood. Also note that it's impossible we find two representations of the same number that disagree with each other on arbitrarily large indices, because this would mean that the number we joined is in fact contained in an algebraic extension of the previous field, which, being real, means the number is contained in the field's real closure, by contradiction to it being real closed. The step is then completed by choosing these constructed sequences as representatives of the real numbers in the ultraproduct, and noting that because the field operations work for all sufficiently large $n$, they work in the ultraproduct as well.
Finally, we prove (the probably known claim) that the ultraproduct of ultraproducts is an ultraproduct by, starting from a sequence of ultrafilters $\mathcal F_i$ and an ultrafilter $\mathcal F$, all on $\mathbb N$, constructing an ultrafilter $\tilde{\mathcal F}$ on $\mathbb N^2$ to be $\{S\in\mathcal P(\mathbb N^2)|\{n\in\mathbb N|\{m\in\mathbb N|(n,m)\in S\}\in\mathcal F_n\}\in\mathcal F\}$. It's easy to check that $\mathbb N^2\in\tilde{\mathcal F}$ and $\emptyset\notin\tilde{\mathcal F}$. If we have $S\in\tilde{\mathcal F}$ and $S\subseteq T\subseteq \mathbb N^2$ then for each $n$ there can only be more $m$s s.t. $(n,m)\in T$ than $(n,m)\in S$, and so only more $n$s s.t. $\{m\in\mathbb N|(n,m)\in T\}\in\mathcal F_n$ than for $S$, and so this set is also in $\mathcal F$ as desired, and for $S_1,S_2\in\tilde{\mathcal F}$ for all $n\in \{n\in\mathbb N|\{m\in\mathbb N|(n,m)\in S_1\}\in\mathcal F_n\}\cap \{n\in\mathbb N|\{m\in\mathbb N|(n,m)\in S_2\}\in\mathcal F_n\}$, which is a set in $\mathcal F$, $\{m\in\mathbb N|(n,m)\in S_1\cap S_2\}$ is an intersection of two sets that are inside $\mathcal F_n$, and therefore $S_1\cap S_2\in\tilde{\mathcal F}$, which shows it's indeed a filter. It's also an ultrafilter because if for some $S\in\mathcal P(\mathbb N^2)$, $S\notin\tilde{\mathcal F}$, this means that $\{n\in\mathbb N|\{m\in\mathbb N|(n,m)\in S\}\in\mathcal F_n\}\notin\mathcal F$, and therefore its complement is, i.e. $\{n\in\mathbb N|\{m\in\mathbb N|(n,m)\in S\}\notin\mathcal F_n\}\in\mathcal F$, and for each $n$ in this set $\{m\in\mathbb N|(n,m)\notin S\}\in\mathcal F_n$, which means that $\{n\in\mathbb N|\{m\in\mathbb N|(n,m)\in \mathbb N^2\setminus S\}\in\mathcal F_n\}\in\mathcal F$, and thus $\mathbb N^2\setminus S\in\tilde{\mathcal F}$, meaning it's also an ultrafilter. Finally, it's easy to see from $\tilde{\mathcal F}$'s definition that $\prod_n (\prod_m A_{n,m}/\mathcal F_n)/\mathcal F\cong \prod_{(n,m)}A_{n,m}/\tilde{\mathcal F}$ for any collection $A_{n,m}$ of sets, completing the proof.
This result is enough for what I want but I'm still interested in the original question, both out of curiosity and because I think such a solution would be more elegant, so I'll try to understand model theory and may write an answer to the original question in the future. In the meantime if you have other similarly "elementary" approaches feel free to post them in the comments below, or write a complete answer if you have one.
Finally, in this case, the number of embeddings of $\mathbb R$ into the resulting field is $2^{2^{\aleph_0}}$, the trivial upper bound, which is because in the second stage we could choose for any new element more than one sequence converging to it (in particular, we could add the sequence $\frac 1n$ to the sequence we chose, which means we get a sequence that is different from it in all places, and thus can't be the same as it in the ultraproduct).