Global dimension of $\mathbb Q [x]$

Question

I'm trying to show that the global dimension of $\mathbb Q [x]$ is $1$.

I have shown that $D(\mathbb Q [x]) \leq 1$ as follows. One can reduce to the case of showing that

$$\text{sup}_{\{B\}}\; \text{pd}\; B = \text{sup}_{\{I\}} \text{pd}\; R/I \leq 1$$

where $B$ runs over all cyclic $\mathbb Q [x]$-modules and $I$ runs over all ideals of $\mathbb Q [x]$. The statement above follows by $\mathbb Q [x]$ being a PID and thus $\text{pd}\; I = 0$ for all ideals $I$, and considering the exact sequence $0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0$, which gives $\text{pd}\; R/I \leq \text{pd}\; I +1 = 1$. (Since $R$ is certainly projective).

But once I've shown that $D(\mathbb Q [x]) \leq 1$, how do I show that there is a $\mathbb Q [x]$-module that is not projective (and hence that $D(\mathbb Q [x]) > 0$)?

Any help is much appreciated, my knowledge of projective modules is rather shaky...

Answer 1

Let $R$ be a PID which is not a field. Then the global dimension of $R$ is $1$.

Let $I$ be an ideal of $R$, $I\ne 0,R$. Then $I\simeq R$ (as $R$-modules), and from the exact sequence $$0\to I\to R\to R/I\to 0$$ we get $\operatorname{pd}_RR/I\le 1$. If $\operatorname{pd}_RR/I=0$, then $R/I$ is projective, in particular torsion-free, a contradiction.