Let $R^3$ be euclidean 3-space and $S^2$ the 2-sphere, embedded in $R^3$ as usual. Let $A$ be the ring of all real-valued continuous functions on $S^2$. Let $T$ be the $A$-module of all $R^3$-valued continuous functions on $S^2$ (so that $T\approx A^3$ is a free $A$-module). Let $M\subset T$ consist of all those functions $f$ such that $f(x).x=0$ for all $x$ (where "dot" denotes the usual inner product in $R^3$). Let $M'\subset T$ be the submodule generated by the identity function.
Observation 1: $M\oplus M'=T$. Thus, for any prime $P\subset A$, we have $M_P\oplus M'_P\approx T_P$. But over a local ring, any direct summand of a free module is free. Therefore $M_P$ is free.
Observation 2: $M$ can't be free. If it were, it would have a basis consisting of two triples $(f_1,f_2,f_3)$ and $(g_1,g_2,g_3)$ (the entries $f_i$ and $g_i$ being real-valued functions). This basis, together with the basis consisting of the single element $(x,y,z)$ for $M'$, would form a basis for $T$. It would follow that the matrix $$\pmatrix{f_1&f_2&f_3\cr g_1&g_2&g_3\cr x&y&z\cr}$$ has unit determinant; in particular the determinant is a function on $S_3$ S^2$with no zeros. But it is a fact from topology that if$f(x).x=0$for all$x$, then there is some$x$such that$f(x)=(f_1(x),f_2(x),f_3(x))=(0,0,0)$. Thus the determinant of the displayed matrix has a zero at$x$. This contradiction shows that$M$is not free. Now let$N$be a free$A$-module of rank 2. Observation 1 shows that$M_P\approx N_P$for all primes$P$; Observation 2 shows that$M$is not isomorphic to$N$. 1 Here is an explicit counterexample: Let$R^3$be euclidean 3-space and$S^2$the 2-sphere, embedded in$R^3$as usual. Let$A$be the ring of all real-valued continuous functions on$S^2$. Let$T$be the$A$-module of all$R^3$-valued continuous functions on$S^2$(so that$T\approx A^3$is a free$A$-module). Let$M\subset T$consist of all those functions$f$such that$f(x).x=0$for all$x$(where "dot" denotes the usual inner product in$R^3$). Let$M'\subset T$be the submodule generated by the identity function. Observation 1:$M\oplus M'=T$. Thus, for any prime$P\subset A$, we have$M_P\oplus M'_P\approx T_P$. But over a local ring, any direct summand of a free module is free. Therefore$M_P$is free. Observation 2:$M$can't be free. If it were, it would have a basis consisting of two triples$(f_1,f_2,f_3)$and$(g_1,g_2,g_3)$(the entries$f_i$and$g_i$being real-valued functions). This basis, together with the basis consisting of the single element$(x,y,z)$for$M'$, would form a basis for$T$. It would follow that the matrix $$\pmatrix{f_1&f_2&f_3\cr g_1&g_2&g_3\cr x&y&z\cr}$$ has unit determinant; in particular the determinant is a function on$S_3$with no zeros. But it is a fact from topology that if$f(x).x=0$for all$x$, then there is some$x$such that$f(x)=(f_1(x),f_2(x),f_3(x))=(0,0,0)$. Thus the determinant of the displayed matrix has a zero at$x$. This contradiction shows that$M$is not free. Now let$N$be a free$A$-module of rank 2. Observation 1 shows that$M_P\approx N_P$for all primes$P$; Observation 2 shows that$M$is not isomorphic to$N\$.