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Suppose $V$ is a vector space over a field $k$. Fixing a linear transformation $T$, it is common to make $V$ a $k[x]$-module by defining $f(x)\cdot v=f(T)(v)$.

Is every possible $k[x]$-module structure over $V$ necessarily induced by some $T\in L(V)$?

If $V$ is some $k[x]$-module, we can define a map $T$ on $V$ by $T(v)=x\cdot v$. Then $T$ is additive. But for scalars, $$T(cv)=x\cdot(cv)=(xc)\cdot v=(cx)\cdot v=c\cdot(x\cdot v)=c\cdot T(v)$$ but I don't think we can assume that multiplication by scalars in $k$ over $V$ as a $k$-vector space needs to be the same as multiplication by scalars in $k$ when viewing $V$ as a $k[x]$-module.

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When you say "$k[x]$-module structure on $V$" are you requiring a compatibility relation between the $k[x]$-module structure and the vector space structure? – Qiaochu Yuan Mar 3 '13 at 23:20
Yes, it is assumed to be the same, as far as I know, and then this method indeed answers your question, affirmatively. – Berci Mar 3 '13 at 23:20
@Berci: that isn't clear to me (read the last line). But then I wonder what's the point of introducing the vector space structure at all. – Qiaochu Yuan Mar 3 '13 at 23:21
You guys are right. I think introducing a vector space structure first is silly if I don't require compatibility. So I guess it is true if we only know of a $k[x]$ module structure first, define $T$ as a above, and then consider $T$ as a $k$ vector space linear transformation on $V$ where the structure is that induced by restricting the action of $k[x]$ on $V$ to the subfield $k$ only. – Selina Kyle Mar 3 '13 at 23:32

If $R$ is a ring (commutative or not), then the category of $R[x]$-modules is equivalent to the category of pairs $(M,f)$, where $M$ is an $R$-module and $f : M \to M$ is an endomorphism. This is a consequence of the universal property of the polynomial algebra, no calculations are needed. In particular, $R[x]$-modules extending a given $R$-module structure correspond to endomorphisms of this $R$-module.

When $M$ is a right $R$-module, then a left $R[x]$-module structure on the underlying abelian group (thus forgetting the module structure) corresponds to a left $R$-module structure on $M$ and an endomorphism of this module structure. A large class of examples (but not all) is given by $(R,R)$-bimodules together with an endomorphism. Of course the two actions of $R$ don't have to coincide here. Hagen's answer refers to the $(\mathbb{C},\mathbb{C})$-bimodule $\mathbb{C}$, where on the right the action is given by complex conjugation.

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