Does there exist a module structure over $\mathbb{C}^2$ as a $\mathbb{C}[x]$-module?

Question

Let $R = \mathbb{C}[x]$, $M = \mathbb{C}^2$. Does there exist a module structure over $M$ as an $R$-module or not?

Answer 1

There exist a lot of ways to show this. I think that the simplest is to say that we can define a $\Bbb{C}[x]$-module structure on $M$ by composing $$\Bbb{C}[x] \longrightarrow \Bbb{C} \longrightarrow \operatorname{End}(M)$$ where the map $\Bbb{C}[x] \longrightarrow \Bbb{C}$ is any ring morphism (for example $f \mapsto f(0)$).

This can be generalized, saying that if there exists a ring morphism $R \longrightarrow S$ (so that $S$ is an $R$-algebra), then every $S$-module can be given a structure of $R$-module.

Answer 2

To make an abelian group $M$ into a module over $R$ we need a ring homomorphism from $R$ into the (non-commutative) ring of group endomorphisms of $R$. To make $\mathbf{C^2}$, a module we can construct a homomorphism that goes into the subring of vector space endomorphisms.

Here we can easily specify a ring homomorphism $\mathbf{C}[X]\to \mathrm{End\,}(\mathbf{C}^2)$ by sending constants to (the same) constants and $x$ to a fixed $2\times2$ matrix.

Answer 3

If $R$ is a commutative ring and $M$ is some $R$-module, then $R[x]$-module structures on $M$ extending the given $R$-module structure correspond 1:1 to $R$-linear endomorphisms of $M$.

So, any $\mathbb{C}$-linear map $\mathbb{C}^2 \to \mathbb{C}^2$ (and surely, there are plenty of them!) corresponds to a $\mathbb{C}[x]$-module structure on $\mathbb{C}^2$.