Is $\mathbb{Q}$ artinian as a $\mathbb{Z}_{(2)}$-module? Let $\mathbb{Z}_{(2)}= \{ \frac ab \in \mathbb{Q} : 2\not \mid  b \}$ be the localization of $\mathbb{Z}$ at the prime ideal $(2)$.
We can consider $\mathbb{Q}$ as a $\mathbb{Z}_{(2)}$-module by the usual multiplication. Equivalently, this is the extension of scalars from $\mathbb{Z}$ to $\mathbb{Z}_{(2)}$, since $S^{-1}\mathbb{Z}⊗_\mathbb{Z} \mathbb{Q} \cong S^{-1}\mathbb{Q} =\mathbb{Q}$ holds for all submonoids $S$ of $\mathbb{Z}$.

Is $\mathbb{Q}$ artinian as a $\mathbb{Z}_{(2)}$-module?

We know that $\mathbb{Q}$ is not artinian as a $\mathbb{Z}$-module, by the same descending chain as here. However, this chain will not work here because the $\mathbb{Z}$-module $M_i=\{ \frac ab \in \mathbb{Q} : \forall j =1...,i \;\; p_j \not \mid b \}$ is no longer a module over the ring $\mathbb{Z}_{(2)}$. 
The key obstacle is that I don't know how to classify the submodules of $\mathbb{Q}$ as a $\mathbb{Z}_{(2)}$-module. Is there a result on submodules of a scalar extension?
 A: You got something wrong there. $\mathbb{Q}$ is not artinian as a $\mathbb{Z}$-module because $\mathbb{Z}\supseteq 2\mathbb{Z}\supseteq 4\mathbb{Z}\supseteq 8\mathbb{Z} \supseteq \cdots$ is a strictly descending sequence of submodules.
And similarly $\mathbb{Z}_{(2)}\supseteq 2\mathbb{Z}_{(2)}\supseteq 4\mathbb{Z}_{(2)}\supseteq 8\mathbb{Z}_{(2)} \supseteq \cdots$ is a strictly descending subsequence of $\mathbb{Z}_{(2)}$-submodules.
As for a classification: $\mathbb{Z}_{(p)}$ is what is called a discrete valuation ring and if $R$ is a DVR with maximal ideal $\mathfrak{p}=(\pi)$, then the $R$-submodules of the field of fractions $Q$ are exactly the submodules of the form $\pi^k R$ for some $k\in\mathbb{Z}\cup\{\pm\infty\}$ (where $\pi^{\infty}:=0$ and $\pi^{-\infty}R:=Q$) and those are pairwise distinct.
A: For $n\in\mathbb{N}$, let $M_n\subseteq\mathbb{Q}$ be the submodule generated by $2^n$.  Then the $M_n$ form an infinite descending chain of submodules.  (This is essentially the same example as in the linked question, just instead of using different primes, you use higher and higher powers of the only prime you have.)
