How to prove that $\mathbb{R}$ is not artinian, considered as a $\mathbb{Q}$-module How can I prove that the set $\mathbb{R}$ of real numbers as a $\mathbb{Q}$-module is not artinian?
I have tried to prove it, by the irrationals, but I could not prove it.
 A: If you are thinking about things like the irrationals then you may be missing the forest for the trees.  I suggest you take a step back and try to show the following result.

Let $K$ be a field and let $V$ be a $K$-module -- i.e., a $K$-vector space.  The following are equivalent:
  (i) $V$ is finitely generated (in other words, finite-dimensional) over $K$.
  (ii) $V$ is Noetherian.
  (iii) $V$ is Artinian.
  (iv) $V$ admits a composition series, i.e., a maximal proper chain of submodules
  $0 = V_0 \subsetneq V_1 \subsetneq \ldots \subsetneq V_n = V$.  

Note also that the fact that these equivalent conditions are not satisfied for $\mathbb{R}$ as a $\mathbb{Q}$-vector space follows from the fact that $\mathbb{R}$ is an uncountably infinite set.
Final comment: more generally, for a ring $R$, the above conditions on a left $R$-module are equivalent iff $R$ is left Artinian.  
A: Hint: The dimension of $\Bbb R$ over $\Bbb Q$ is infinite, so you can find some sequence $\langle r_i\mid i\in\Bbb N\rangle$ be linearly independent over $\Bbb Q$, use this sequence to generate a sequence of decreasing modules without a minimum.
