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 someone explain this claim to me?


4 Answers 4


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, 2020 at 2:30
  • 1
    $\begingroup$ @LordShadow Yep, thanks. $\endgroup$
    – DonAntonio
    Aug 7, 2020 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 $\mathrm{End}(V)$.

I want to supplement DonAntonio's Answer, because his answer seems to give a converse of what OP really asked for. To make the answer whole, I prove the both direction for OP.

Fact: Let $k$ be a field, and set $A:=k[x]$. Then, an $A-$module is a $k-$vector space with a $k-$linear map $T:V\longrightarrow V$ given by $T(v):=\mu(x,v)$, where $x$ is considered as a polynomial in $k[x]$ and $\mu$ is action in the definition of modules.

Conversely, given any $k-$vector space $V$ and any $k-$linear map $T:V\longrightarrow V$, we can turn $V$ into an $A-$module using the map $\mu:A\times V\longrightarrow V$ defined by $\mu(f(x),v)=\mu(\sum_{i=0}^{n}a_{i}x^{i},v):=\sum_{i=0}^{n}a_{i}T^{i}(v)$, for any $f(x)\in k[x]$ and $v\in V$.


Let $V$ be an $A-$module with action $\mu:A\times V\longrightarrow V$. Then, $V$ is clearly a $k-$vector space, because all the elements in $k$ can be considered as constant polynomials in $k[x]$, and thus the action of $A$ on $V$ provides the scalar multiplication of $V$ by $k$.

Moreover, the axioms of $\mu$ ensure that $T$ is linear. Indeed, for any $v,w\in V$ we have $$T(v+w):=\mu(x,v+w)=\mu(x,v)+\mu(x,w)=T(v)+T(w).$$ Further, for any $a\in k$ and $v\in V$, note that the scalar multiplication is provided by $\mu$, which means that to prove $T(av)=aT(v)$, what we actually need to show is that $T(\mu(a,v))=\mu(a,T(v))$. This is true, because $$T(\mu(a,v)):=\mu(x,\mu(a,v))=\mu(xa,v)=\mu(ax,v)=\mu(a,\mu(x,v))=\mu(a,T(v)).$$

The converse direction is a trivial check of the axioms of $\mu$. The proof is concluded.

Just a remark here that, if you are new learner of module, do not write the action map as multiplication or as a dot $\cdot$. This will confuse everything. Just be heavy in notation and write $\mu$, which will make things clear.

So the axioms of the "multiplication'' are: $\mu:A\times M\longrightarrow M$ is a mapping such that for all $a,b\in A$ and $x,y\in M$, we have \begin{align*} \mu(a,x+y)&=\mu(a,x)+\mu(a,y)\\ \mu(a+b,x)&=\mu(a,x)+\mu(b,x)\\ \mu(ab,x)&=\mu(a,\mu(b,x))\\ \mu(1_{A},x)&=x. \end{align*}


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