# $K[X]$-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 someone explain this claim to me?

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.

• "$T:A\to A$ is any linear operator$" , it think it has to be$V$not$A$. Please correct me if I'm wrong Aug 7, 2020 at 2:30 • @LordShadow Yep, thanks. Aug 7, 2020 at 9:31 Say$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 $$v\in V$$.

Proof:

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*}