Faithful module is infinitely dimensional as a vector space.

Question

Let $M$ be a $\mathbb{C}[x]$-module. Since $\mathbb{C} \subset \mathbb{C}[x]$, we may consider $M$ as a $\mathbb{C}$-vector space. If $M$ is faithful, must $M$ be infinite dimensional as a $\mathbb{C}$-vector space?

I have tried messing around a bit with Cayley-Hamilton to show this, but to no avail. Any help would be appreciated.

Answer 1

Suppose $M$ were a finite-dimensional vector space over $\mathbb{C}$. Let $f : M \to M$ be the action of $x$, ie $f(m) = x \cdot m$. Then by the Cayley–Hamilton theorem, there is a polynomial $p \in \mathbb{C}[x]$, $p \neq 0$ (eg. the characteristic polynomial of $f$) such that $p(f) = 0$; in other words, for all $m \in M$, $(p(f))(m) = 0 = p(x) \cdot m$. So $M$ cannot be faithful as a $\mathbb{C}[x]$-module.

If you want even more detail: let $p(x) = a_0 + a_1 x + \dots + a_n x^n$. Then $p(f) = a_0 \operatorname{id} + a_1 f + \dots + a_n f^{\circ n}$, and $p(f) = 0$ means that for all $m$, $$a_0 m + a_1 f(m) + \dots + a_n f(f(\dots f(m)\dots)) = 0 = (a_0 + a_1 x + \dots + a_n x^n) \cdot m$$

Answer 2

Suppose $\dim_{\Bbb{C}} M = n$ is finite. Then we can identify $M$ with $\Bbb{C}^n$. Since $M$ is a $\Bbb{C}[x]$-module, $x$ represents an $n \times n$ complex matrix $A$ (an endomorphism of $\Bbb{C}^n$).

Let $p_A(x)$ be the characteristic polynomial of $A$. By Cayley-Hamilton theorem, $p_A(A)=0$. So $M$ cannot be a faithful $\Bbb{C}[x]$-module, because $p_A \in \operatorname{Ann}_{\Bbb{C}[x]}(M)$.