Multiplicative and additive groups of the field $(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z})/\simeq_{\cal U}$ Let ${\cal U}$ be a non-principal ultrafilter on $\omega$, and for each $n\in\omega$, let $p_n$ denote the $n$th prime, that is $p_0 = 2, p_1=3, \ldots$
Next we introduce the following standard equivalence relation on $\big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)$: we say $a \simeq_{\cal U} b$ for $a,b \in \big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)$ if and only if $$\{n\in\omega:a(n) = b(n)\}\in {\cal U}.$$
It is a standard exercise to prove that $K = \big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/\simeq_{\cal U}$ is an uncountable field.
Questions.

*

*Is there a surjective group homomorphism in either direction between the additive group of $K$ and the additive group of $\mathbb{R}$?

*Same question for the multiplicative groups of $K \setminus\{0\}$ and $\mathbb{R}\setminus\{0\}$ respectively.

 A: 1. $(K,+)$ and $(\mathbb R,+)$ are isomorphic.
The additive group of any field $K$ is a vector space over its prime field ($\mathbb F_p$ or $\mathbb Q$), hence it is determined up to isomorphism by the characteristic of $K$ and its degree over the prime field (which is just $|K|$ for uncountable $K$). Here, $K$ and $\mathbb R$ are both fields of characteristic $0$ and cardinality $2^\omega$.
2a. There exists a surjective homomorphism $K^\times\to\mathbb R^\times$.
Observe that $\mathbb R^\times\simeq\{1,-1\}\times(\mathbb R_{>0},{\times})\simeq C_2\times\mathbb Q^{(2^\omega)}$.
In any finite field of odd characteristic, squares are an index-$2$ subgroup of the multiplicative group. This is a first-order property, hence it also holds in $K$, i.e., $[K^\times:(K^\times)^2]=2$.
We start by constructing a surjective homomorphism $(K^\times)^2\to\mathbb R_{>0}$. Let $G$ be the quotient of $(K^\times)^2$ by its torsion part. Since there are only countably many roots of unity in $K$, $G$ is a torsion-free group of cardinality $2^\omega$, hence it has rank $2^\omega$, i.e., we may fix a $\mathbb Q$-linearly independent subset $\{a_r:r\in\mathbb R_{>0}\}\subseteq G$. Then $a_r\mapsto r$ extends to a surjective homomorphism $\langle a_r:r\in\mathbb R_{>0}\rangle\to\mathbb R_{>0}$. Since $\mathbb R_{>0}$ is divisible, we can extend it to a homorphism $G\to\mathbb R_{>0}$, which we compose with the quotient map to obtain $\phi\colon (K^\times)^2\to\mathbb R_{>0}$.
Finally, let us fix $a\in K^\times\smallsetminus(K^\times)^2$. Then $\phi$ extends to a surjective homomorphism $K^\times\to\mathbb R^\times$ by putting $\phi(ax)=-\sqrt{\phi(a^2)}\phi(x)$ for $x\in (K^\times)^2$.
2b. Whether there exists a surjective homomorphism (or isomorphism) $\mathbb R^\times\to K^\times$ depends on the ultrafilter.
Let
$$I_2=\{n:p_n\not\equiv1\pmod4\},$$
and for odd prime $q$,
$$I_q=\{n:p_n\not\equiv1\pmod q\}.$$
Notice that for $p,q$ odd and $p\ne q$, the fact that $\mathbb F_{p_n}^\times\simeq C_{p_n-1}$ implies
$$n\in I_q\iff\mathbb F_{p_n}\models\forall x\,\exists y\,(y^q=x),\tag{$*$}$$
and
$$n\in I_2\iff\mathbb F_{p_n}\models\forall x\,\exists y\,(x^2=y^4).\tag{$**$}$$
Also notice that by Dirichlet’s theorem on primes in arithmetic progressions, the family $\{I_q:q\text{ prime}\}$ has the strong finite intersection property, hence it is included in a nonprincipal ultrafilter.
Case I: $I_q\notin\mathcal U$ for some $q$. Then there is no surjective homomorphism $\mathbb R^\times\to K^\times$.
Indeed, then the positive first-order formulas in $(*)$ and $(**)$ hold in $\mathbb R^\times$ and in all its quotients, whereas if $I_q\notin\mathcal U$, the corresponding formula fails in $K$.
Case II: $\{I_q:q\text{ prime}\}\subseteq\mathcal U$. Then $\mathbb R^\times\simeq K^\times$.
The condition ensures that the formulas $(*)$ and $(**)$ hold in $K$, thus $(K^\times)^2$ is divisible. Moreover, the $q$-th roots for odd $q$ are unique (as this is again a first-order property), and similarly, there is no square root of $-1$. This implies that the torsion part of $K^\times$ is just $\{1,-1\}\simeq C_2$, and $K^\times\simeq C_2\times(K^\times)^2$, where $(K^\times)^2$ is a torsion-free divisible group of cardinality $2^\omega$, i.e., it is isomorphic to $\mathbb Q^{(2^\omega)}$.

The arguments above actually used very little about pseudofinite fields. They can be easily extended to get the following characterization.
Let $K$ be a field:

*

*There exists a surjective homomorphism $K^\times\to\mathbb R^\times$ iff $|K|\ge2^\omega$ and there is a nonsquare in $K$.


*There exists a surjective homomorphism $\mathbb R^\times\to K^\times$ iff $|K|\le2^{\omega}$, all elements of $K$ have $n$th roots for all odd $n$, and for each $x\in K$, $x$ or $-x$ has a square root.


*$K^\times\simeq\mathbb R^\times$ iff $|K|=2^\omega$, there are exactly two roots of unity in $K$, and $K$ satisfies the conditions in 2.
A: The answer to the second question depends on the ultrafilter. Specifically, let $q_n$ be the smallest prime divisor of $(p_n - 1)/2$. Then it depends on whether $q_n$ is equivalent to a constant.
If $q_n$ is equivalent to a constant $q$, then $K^\times$ contains an element of order $2q$, but $\mathbb{R}^\times$ does not, so there can be no surjective homomorphism $\mathbb{R}^\times \to K^\times$.
If $q_n$ is not equivalent to a constant then in particular $p_n \equiv 3 \pmod 4$ for almost all $n$, $(\mathbb{Z} / p_n\mathbb{Z})^\times \cong C_2 \times \mathbb{Z} / ((p_n-1)/2)\mathbb{Z}$, and the ultraproduct of the groups $\mathbb{Z}/((p_n-1)/2)\mathbb{Z}$ is divisible and torsion-free, so a rational vector space, so isomorphic to $\mathbb{R}$, so $K^\times$ is isomorphic to $\mathbb{R}^\times \cong C_2 \times \mathbb{R}$.
As to whether there is an epimorphism $K^\times \to \mathbb{R}^\times$, this seems less clear.
Suppose $p_n \equiv 3 \pmod 4$ for almost all $n$.
If $(p_n-1)/2$ is divisible an unbounded prime $v_n$ then one can surject onto $C_2 \times \mathbb{Z}/v_n\mathbb{Z}$, so there is a surjection $K^\times \to \mathbb{R}^\times$. Otherwise...?
