Multiplicative and additive groups of the field $(prod_{ninomega}mathbb{Z}/p_nmathbb{Z})/simeq_{cal U}$

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 $K^timestomathbb R^times$.

Observe that $mathbb R^timessimeq{1,-1}times(mathbb R_{>0},{times})simeq C_2timesmathbb Q^{(2^omega)}$.

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 $(K^times)^2tomathbb R_{>0}$. 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 ${a_r:rinmathbb R_{>0}}subseteq G$. Then $a_rmapsto r$ extends to a surjective homomorphism $langle a_r:rinmathbb R_{>0}rangletomathbb R_{>0}$. Since $mathbb R_{>0}$ is divisible, we can extend it to a homorphism $Gtomathbb R_{>0}$, which we compose with the quotient map to obtain $phicolon (K^times)^2tomathbb R_{>0}$.

Finally, let us fix $ain K^timessmallsetminus(K^times)^2$. Then $phi$ extends to a surjective homomorphism $K^timestomathbb R^times$ by putting $phi(ax)=-sqrt{phi(a^2)}phi(x)$ for $xin (K^times)^2$.

2b. Whether there exists a surjective homomorphism (or isomorphism) $mathbb R^timesto K^times$ depends on the ultrafilter.

Let $$I_2={n:p_nnotequiv1pmod4},$$ and for odd prime $q$, $$I_q={n:p_nnotequiv1pmod q}.$$ Notice that for $p,q$ odd and $pne q$, the fact that $mathbb F_{p_n}^timessimeq C_{p_n-1}$ implies $$nin I_qiffmathbb F_{p_n}modelsforall x,exists y,(y^q=x),tag{$*$}$$ and $$nin I_2iffmathbb F_{p_n}modelsforall x,exists y,(x^2=y^4).tag{$**$}$$

Also notice that by Dirichlet’s theorem on primes in arithmetic progressions, the family ${I_q:qtext{ prime}}$ has the strong finite intersection property, hence it is included in a nonprincipal ultrafilter.

Case I: $I_qnotinmathcal U$ for some $q$. Then there is no surjective homomorphism $mathbb R^timesto K^times$.

Indeed, then the positive first-order formulas in $(*)$ and $(**)$ hold in $mathbb R^times$ and in all its quotients, whereas if $I_qnotinmathcal U$, the corresponding formula fails in $K$.

Case II: ${I_q:qtext{ prime}}subseteqmathcal U$. Then $mathbb R^timessimeq K^times$.

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 $K^timessimeq C_2times(K^times)^2$, 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 $K^timestomathbb R^times$ iff $|K|ge2^omega$ and there is a nonsquare in $K$.

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

3. $K^timessimeqmathbb R^times$ iff $|K|=2^omega$, there are exactly two roots of unity in $K$, and $K$ satisfies the conditions in 2.

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 $n$, $(mathbb{Z} / p_nmathbb{Z})^times cong C_2 times mathbb{Z} / ((p_n-1)/2)mathbb{Z}$, 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 $C_2 times mathbb{Z}/v_nmathbb{Z}$, so there is a surjection $K^times to mathbb{R}^times$. Otherwise...?