I am reading the Rotman's book "An Introduction to Homological Algebra" and I do not understand the part of the proof of theorem 8.47 in page 481.
Theorem 8.47 Let R be a commutative noetherian ring. If every finitely generated R-module has Finite Free Resolution (FFR), then every finitely generated R[x]-module has Finite Free Resolution (FFR).
One step of the proof is based on proving the following:
if $M$ is a finitely generated $R[x]$-module such that $ann(M) \cap R \neq \{0\}$, then $M$ has FFR.
Here is the proof in Rotman's book: Let $m\in M$ be nonzero and let $I= ann(m) \cap R\neq 0$ be the R-ideal. We have that $R/I\simeq \langle m \rangle_R$ the R-submodule of $M$ generated by $m$. Because R[x] is a flat R-module, we have the following exact sequence: $$0 \to R[x] \otimes_R I \to R[x] \to R[x] \otimes_R \langle m\rangle_R \to 0 $$
Rotman concludes that the R[x]-module $R[x] \otimes_R \langle m\rangle_R \simeq R[x]/(R[x]\otimes_R I) \simeq R[x]/(R[x]I)$ is a cyclic submodule of $M$.
It is clear that it is cyclic, but why is it a submodule of $M$?
If we define a $R[x]$-homomorphism $\phi:R[x] \otimes_R \langle m\rangle_R \to M$ by $\phi(p(x)\otimes_R m) = p(x)m$, I think that this may not be injective since $ann(m)\neq 0$
It is immediate that $R[x] \otimes_R \langle m\rangle_R$ has FFR, but would not it conclude the same if $I=0$. I do not see in which part of the proof the condition $I\neq 0$ is used.