A theorem of Horrocks states that if $P$ is a projective $R[x]$-module ($R$ a local ring) such that $P_S$ is a free $R[x]_S$-module (where $S\subset R[x]$ is the set of all monic polynomials in $R[x]$), then $P$ is also $R[x]$-free.
My question is if this is also true if $P_f$ is a free $R[x]_f$-module for some $f\in S$.
I apreciate every answer, even though I am afraid the question is trivial.
If I understand well from his comments, the OP wants to know if one can replace in Horrocks' theorem "$P_S$ is a free $R[x]_S$-module (where $S\subset R[x]$ is the set of all monic polynomials in $R[x]$)" with "$P_f$ is a free $R[x]_f$-module for some $f\in S$". Of course we can as long as the first assertion follows from the second.
In general, if $M$ is an $R$-module, $S\subset R$ a multiplicative system, $f\in S$, and $M_f$ is $R_f$-free, then $M_S$ is $R_S$-free. This is trivial, since $M_S=R_S\otimes_{R_f}M_f$.