# $K[X]$-modules are $K$-vector spaces with a linear transformation

In the Atiyah's book there is this example about $A$-modules.

Let $A=\mathbb{K}[x]$, where $\mathbb{K}$ is a field. An $A$-module is a $\mathbb{K}$-vector space with a linear transformation.

Can anyone explain me this claim?

Let $$\;V\;$$ be a $$\;\Bbb K\,-$$ vector space, and let $$\;T:V\to V\;$$ be any linear operator. Then, $$\;V\;$$ gets a structure of $$\;\Bbb K[x]\,-$$ module if we define
$$\forall\,v\in V\;,\;\;x\cdot v:=Tv$$
$$\sum_{j=0}^m k_jx^j\cdot v:=\sum_{j=0}^mk_jT^jv\;,\;\;k_j\in\Bbb K$$
• "$T:A\to A$ is any linear operator$" , it think it has to be$V$not$A$. Please correct me if I'm wrong Aug 7 '20 at 2:30 • @LordShadow Yep, thanks. Aug 7 '20 at 9:31 Say$M$is the$k[x]$module in question which is also a$k$vector space by means of the inclusion$k \hookrightarrow k[x]$. The action of$x \in k[x]$is that of a linear transformation i.e.$x(am) = a x(m)$and$x(m+m') = x(m)+x(m')$. Conversely, given any linear$L: M \to M$we can let$x(m) = L(m)$to define a$k[x]$-module structure on the vector space$M$. Note this while the action of$k$is faithful because it is a field, the action of$k[x]$may not be. For instance the linear transformation could be nilpotent or something. Working with$K$-vector spaces and (unital)$K$-algebras.... • Every$K[x]$-module is a$K$-vector space, because$K \subseteq K[x]$. • The$K[x]$-module structures on a vector space$V$are in bijection with$K$-algebra homomorphisms$K[x] \to \mathrm{End}(V)$. •$K$-algebra homomorphism$K[x] \to \mathrm{End}(V)$are in bijection with elements of$\mathrm{End}(V)\$.