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 $\big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/\simeq_{\cal U}$ is an uncountable field.

**Questions.** 
1. Is there a surjective group homomorphism in either direction between the additive group of $\big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/\simeq_{\cal U}$ and the additive group of $\mathbb{R}$?

2. Same question for the multiplicative groups of $\big(\big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/\simeq_{\cal U}\big) \; \setminus\{0\}$ and $\mathbb{R}\setminus\{0\}$ respectively.