Let $R$ be a ring and we denote $R_M$ to be the ring $S^{-1} R$ where $S = R - M$ which is a multiplicatively closed subset of $R$, and $M$ is a maximal ideal of $R$. Suppose for every $M$, $R_M$ is a field. May we assert that $R$ is also a field?
We suppose some non unit $r$ of $R$ exists. Then it has been proven in Commutative Algebra by Atiyah and Macdonald that $r$ must be contained in a maximal ideal following an argument from Zorn's Lemma. Then $R_M$ is a field and hence there exists some $a\in R, s\in S, s'\in S$ and $u\in S$ such that $uar = uss'$ for $\displaystyle\frac{a}{s}\frac{r}{s'} = 1$ to be satisfied. Then $uar\in M$ but $uss'\in S, \notin M$. The assumption here is that $R$ contains 1 such that $M$ being a maximal ideal is prime.
The reason for this question is that in the book mentioned above (Exercise 10 Chapter 3) one has to prove that $R$ is absolutely flat if and only if $R_M$ is a field for all maximal ideals $M$ of $R$ and if the argument above is correct why did the authors impose this requirement in this form? Apologies in advance if there are any mistakes in understanding or oversight or if the question is not clear enough, as I am learning this by myself, and may pose silly questions at times.
The answer is no. Take $\Bbb{Z}/(6)$. This has two maximal ideals, $(2)$ and $(3)$. Localizing at $(2)$ means inverting $3$, so we end up with $\Bbb{Z}/(2)$. Similarly Localizing at $(3)$ gives $\Bbb{Z}/(3)$.
Your error is assuming that $r\in M$ doesn't map to $0$. Your argument shows that if $R_M$ is a field, then $R_M\cong R/M$.