Attempt to sketch a proof showing $M$ is not projective given $M$ is a $\mathbb{Q}[x]-module$ such that $dim_{\mathbb{Q}}(M) < \inf$

Question

Below is a sketch of a proof of the question in the title after a couple hours of research on the definitions. Am I headed in the right direction?

1.) $\mathbb{Q}[x]$ has an infinite basis as a $\mathbb{Q}[x]$-module.

2.) $dim_{\mathbb{Q}}(M) < \infty$ implies that the action defining the module is a map from an infinite set to a finite set.

3.) If $M$ is projective then there exist morphisms (homomorphis, epimorphism, isomorphism?) $f: M \rightarrow \mathbb{Q}[x]\oplus \cdots \oplus\mathbb{Q}[x] $ and $g:\mathbb{Q}[x] \rightarrow M\oplus\cdots\oplus M$ and for $x \in M$ we have $\star:= g(f(x)) = x$

4.) But $M$ is finitely generated as a $\mathbb{Q}[x]$-module which makes $\star$ impossible so $M$ is not projective.

Answer 1

A $\mathbb Q[x]$-module is a $\mathbb Q$-vector space together with an algebra homomorphism $\mathbb Q[x]\to\mathrm{End}_{\mathbb Q}(M)$. Such a map is completely determined by the image of $x$, so we can regard $\mathbb Q[x]$-modules as pairs $(M,f)$ consisting of a $\mathbb Q$-vector space $M$ equipped with a $\mathbb Q$-linear endomorphism $f$.

Next, if $M$ is finite dimensional, then Cayley-Hamilton says that $f$ satisfies its own characteristic polynomial, say $\chi(f)=0$.

Finally, as $M$ is finite dimensional, there is an epimorphism $\mathbb Q[x]^n\twoheadrightarrow M$. If $M$ is projective, then this is split, so $\mathbb Q[x]^n\cong M\oplus N$ for some $N$. Now multiplication by $\chi(x)$ is injective on the left, but on the right has $M$ in the kernel, a contradiction (unless $M=0$).

As for your points.

1 is false: $\mathbb Q[x]$ is free of rank one over $\mathbb Q[x]$. You probably mean that $\mathbb Q[x]$ has infinite basis over $\mathbb Q$.

2 is a bit confused: $M$ is not necessarily a quotient of $\mathbb Q[x]$, but it will be a quotient of $\mathbb Q[x]^n$ for some $n$, and as a map of $\mathbb Q$-vector spaces $\mathbb Q[x]^n$ has infinite dimension whereas $M$ has finite dimension.

3 is saying that the epimorphism $\mathbb Q[x]^n\twoheadrightarrow M$ is split, so there exists a section $M\rightarrowtail \mathbb Q[x]^n$ such that the composition yields the identity on $M$. Alternatively, $\mathbb Q[x]^n\cong M\oplus N$.

4 does not yield a contradiction when $M$ is finitely generated. The analogue of 3 holds for all finitely generated projective modules over any ring (even non-commutative).