Let $k$ be a field. Let $R$ be the set of infinite upper triangular matrices (columns and row are indexed by $\mathbb{Z}$) with only finitely many nonzero entries in any row or column. Then let $V$ be the vector space of infinite column vectors (indexed by $\mathbb{Z}$) with finitely many nonzero entries. Then we can view $V$ as an $R$-module with action given by matrix multiplication.
Then I would like to show that $V$ has no simple submodules.
However, to me, this is not intuitively correct. For example, let $W$ consists the vectors with some entries at the first row and zeros elsewhere. Then $W$ has only finitely many nonzero entries (1 or 0 entry) and it is closed under addition. Also for any $A\in R$, $AW\subseteq W$. So it appears to me that $W$ is a submodule, and it is one-dimensional. So it must be simple.
What the role of infinity here?
Because if we consider the finite case (finite upper triangular matrix $R$, finite column vectors space $V$, the submodule $W$ I mentioned before is the only simple submodule of $V$. right?) Why this is not true in infinite case? What I missed?
Thanks!
I'm doing everything assuming that the sides of $R$ are indexed by $\mathbb N$, and so are the positions of $V$, and that is how matrix multiplication is defined.
The subspace $\{(\lambda, 0, 0\ldots)^T\mid \lambda \in k\}$ appears to be a simple submodule of $_RV$.
(N.B. At the time of this answer, the indexing by $\mathbb Z$ was not mentioned.)