# Localization of an infinite product of fields at its maximal ideals

Let $k$ be a field. Let $R = \prod_{n \in \mathbb{N}} k$. Due to the answer in this question Infinite product of fields, we know that $R$ is zero dimensional, and the localization $R_m$ at every maximal ideal $m \subset R$ is a field. In fact, it is $R/m$. Is $R_m = k$ or is this not true in general for an arbitrary field?

If the statement is not true in general, what is an example where it fails to be true?

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I thinks this is false in general if you assume the Axiom of Choice. For example, the hyperreal numbers can be constructed as an ultrapower of $\mathbb{R}$; that is, taking $R$ to be the set of all sequences on $R$, and moding out by the maximal ideal determined by a particular ultrafilter on $\mathbb{N}$. But the hyperreals are not isomorphic to the reals, at least as an ordered field. –  Arturo Magidin Jun 20 '12 at 21:53
$R$ is a commutative von Neumann regular ring, and this shows that $\dim R=0$ and $R_m$ is a field for any maximal ideal $m$. I guess that for $k=\mathbb{Q}$ and $m$ the maximal ideal containing $\oplus\mathbb{Q}$, the quotient ring $R/m$ has cardinality $c=|\mathbb{R}|$, and therefore it is not isomorphic to $k$. –  user26857 Jun 21 '12 at 22:14