Maximal ideal in local ring The maximal ideal in $\mathbb{Z}_{(2)}$ should be $(2)$, but I don't understand this well. Suppose I take $\frac35\in \mathbb{Z}_{(2)}$. It is not in $(2)$ but in $(3).$ But what is the ideal between $(3)$ and $\mathbb{Z}_{(2)}$? Does the maximal ideal of a local ring have to "dominate" one chain of ideals or it can "dominate" many chains?
 A: In $\mathbb{Z}_{(2)}$, $\frac{3}{5}$ is a unit, and hence the ideal generated by it is the whole ring. It is a unit because it is invertible (with inverse $\frac{5}{3}$).
In general, suppose a proper ideal $I$ is generated by $\frac{a}{b}$. Notice that all the odd primes are invertible in $\mathbb{Z}_{(2)}$ (when we localize at a prime $P$ we invert any element which is not in $P$). Really $a$ generates $I$ since I can multiply a generator by a unit and still get a generator. Similarly, if $p\neq 2$ appears in the factorization of $a$, then I can multiply it by $\frac{1}{p}$ to get another generator of $I$. Doing this for all odd primes leaves me with $a=0$ or $a=2^n$ for $n>0$.  I think this answers your question about chains.  Any chain of ideals will look like $... \subset (2^{k_2}) \subset (2^{k_1})$ where $k_i \leq k_{i+1}$. In other words the set of ideals forms a totally ordered set under inclusion.
This is not true in general. For example, look at the ideal $J=(x,y)$ in $R = \mathbb{R}[x,y]$. In $R_J$ we have two different chains $(0)\subset (x)\subset (x,y)$ and $(0)\subset (y)\subset (x,y)$, but $(x)$ and $(y)$ are not comparible. That is, the set of ideal is a poset but not a linear order.
