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.**


1. Is there a surjective group homomorphism in either direction between the additive group of $K$ and the additive group of $\mathbb{R}$?
2. Same question for the multiplicative groups of $K \setminus\{0\}$ and $\mathbb{R}\setminus\{0\}$ respectively.



Multiplicative and additive groups of the field $(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z})/\simeq_{\cal U}$
**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 $\mathbb Q^{(2^\omega)}$.




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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:


1. There exists a surjective homomorphism $K^\times\to\mathbb R^\times$ iff $|K|\ge2^\omega$ and there is a nonsquare in $K$.
2. 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.
3. $K^\times\simeq\mathbb R^\times$ iff $|K|=2^\omega$, there are exactly two roots of unity in $K$, and $K$ satisfies the conditions in 2.


