= Polynomial module from a linear operator
{title2=$p(t)\cdot v=p(T)v$}
A <linear operator> $T:V\to V$ on a <vector space> over a <field> $k$ makes $V$ a <module> over the <polynomial ring> $k[t]$: let $t$ act as $T$, let constants act as the original scalars, and define $p(t)\cdot v=p(T)v$. The identity $(pq)(T)=p(T)q(T)$ for <polynomials> $p,q$ verifies the multiplication axiom. The <submodules> are precisely the <invariant subspaces> of $T$. This construction is described in https://kconrad.math.uconn.edu/blurbs/linmultialg/moduleintro.pdf[Keith Conrad's Introductory Notes on Modules, section 6].
For a computed example, take $k=\mathbb Q$, $V=k^2$, and
$$
T=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad e_1=(1,0),\qquad e_2=(0,1).
$$
Then $Te_2=e_1$, $Te_1=0$, and $T^2=0$. Consequently
$$
(2+3t+t^2)\cdot e_2=2e_2+3e_1=(3,2).
$$
Every $(a,b)$ equals $(b+at)\cdot e_2$. The <module homomorphism> $k[t]\to V$, $p\mapsto p(T)e_2$, is therefore <surjective>, and its <kernel> is the <ideal> $(t^2)$. It induces a <module isomorphism> $V\cong k[t]/(t^2)$. This is a <cyclic module> and a <torsion module> over $k[t]$, since $t^2V=0$. Over $k$ it is a <free module> with <basis> $(e_1,e_2)$ and has no nonzero <torsion elements of a module>. Over $\mathbb Q$ its nonzero elements also have infinite additive <order of a group element>. Changing scalars changes the questions one asks about the same addition law.
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