TITLE: show that $\mathbb{C}[x]$ is prime but not primitive
QUESTION [0 upvotes]: A ring $R$ is (left) primitive if there is a left $R$-module $M$ with the following two properties:


$M$ is simple (i.e. the only submodules of $M$ are $(0)$ and $M$).
$M$ is faithful (i.e. if $r\cdot m = 0\,\forall m\in M, r = 0$).

Given a ring $R$, a proper two-sided ideal $P$ is prime if whenever $a,b \in R$ are such that $aRb \subseteq P, a\in P$ or $b\in P.$ A prime ring is a ring such that $(0)$ is a prime ideal of $R$.

Let $R=\mathbb{C}[x].$ Show that $R$ is prime but not primitive.

To show that $R$ is prime, it suffices to show that $(0)$ is a prime ideal of $R,$ which holds because if $xRy \subseteq (0), x=0$ or $y=0$ as $x\neq 0, y\neq 0\Rightarrow xy \neq 0$ by a degree argument and the fact that $\mathbb{C}$ is an integral domain.
To show that $R$ is not primitive, I need to show that there is no simple faithful left $R$-module $M$.  Suppose there is such a module. Then there is a map $\cdot : R\times M\to M$ satisfying $(rs)\cdot m = r\cdot (s\cdot m), r\cdot (m+n) = r\cdot m + r\cdot n, (r+s)\cdot m = r\cdot m +s\cdot m$ and $1\cdot m = m$ for $r,s \in R$ and $m,n \in M$ and if $rM= (0)$ for some $r\in R,$ then $r=0.$ I think there's a relationship between simple modules and maximal ideals.

REPLY [2 votes]: A commutative ring is primitive if and only if it is a field.
Indeed, if $I$ is a nonzero ideal of the commutative ring $A$, we can see that $I$ is contained in the annihilator of $A/I$.
In particular, if $I$ is maximal and nonzero, the simple module $A/I$ is not faithful.
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