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

$$\forall\,v\in V\;,\;\;x\cdot v:=Tv$$

and then we extend this action in the obvious way, meaning:

$$\sum_{j=0}^m k_jx^j\cdot v:=\sum_{j=0}^mk_jT^jv\;,\;\;k_j\in\Bbb K$$

I'll leave to you to check details.

  • $\begingroup$ "$T:A\to A$ is any linear operator$" , it think it has to be $V$ not $A$. Please correct me if I'm wrong $\endgroup$ Aug 7 '20 at 2:30
  • 1
    $\begingroup$ @LordShadow Yep, thanks. $\endgroup$
    – DonAntonio
    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 $\mathrm{End}(V)$.

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