I found several different ways to state Hilbert's Syzygy Theorem, one of them being:
If $k$ is a field, then $R := k[x_1,...,x_n]$ has global dimension $n$, i.e. any $R$-module $M$ has a projective resolution of length at most $n$.
(A proof of this can be found in Joseph Rotman's book on homological algebra, Theorem 8.37, p.472)
From this it is easy to deduce:
If $K$ is a field and $R=k[x_1,\cdots,x_n]$, then every finitely-generated $R$-module $M$ has a free resolution of length at most $n$.
$\Big($Proof: Pick a free resolution $\cdots F_2 \rightarrow F_1 \rightarrow F_0 \rightarrow M \rightarrow 0$ in which each $F_i$ is finitely generated (f.g.). (This resolution exists since every f.g. module is a quotient of a f.g. free module.) As the projective dimension of $M$ is $\leq n$, the $(n-1)^\text{st}$ syzygy of this resolution is projective (Rotman, Proposition 8.6, p.456). Hence there is a projective resolution $0 \rightarrow K_{n-1} \rightarrow F_{n-1} \rightarrow \cdots \rightarrow F_1 \rightarrow F_0 \rightarrow M \rightarrow 0$. As $R$ is Noetherian and $M$ is finitely generated, $K_{n-1}$ if f.g. as well. Hence the Quillen-Suslin Theorem (Rotman, Theorem 4.100, p.209) implies that $K_{n-1}$ is free.$\Big)$
$\textbf{My question:}$
Is it necessary to assume that $M$ is $\textit{finitely generated}$ to conclude that it has a free resolution of length at most $n$? All the statements of the theorem that I have seen contain the f.g. assumption (for example on nLab or Eisenbud (p.25 of the PDF-file), but I can neither come up with a counterexample for the case when $M$ is not f.g, nor with a free resolution for arbitrarily generated $M$.
Thank you very much!