I'd like to prove (i) implies (ii) where:
(i) Whenever $f: A \to B$ is injective and $A,B$ are finitely generated then $f \otimes \operatorname{id}: A \otimes P \to B \otimes P$ is injective.
(ii) If $f: M \to N$ is injective then $f \otimes \operatorname{id}: M \otimes P \to N \otimes P$ for arbitrary $R$-modules $M,N$.
Can you tell me if this is correct (I'm spelling out all the details because I want to be sure that I understand what I'm doing):
Let $M,N$ be arbitrary modules and let $f: M \to N$ be injective. Assume that $f \otimes \operatorname{id}$ is not injective so that there is an element $u = \sum_{i=1}^n m_i \otimes p_i$ in $M \otimes P$ such that $u \neq 0$ and $(f \otimes \operatorname{id}) (u) = \sum_{i=1}^n f(m_i) \otimes p_i = 0$ in $N \otimes P$.
Let $M_0$ be the module generated by $m_1, \dots , m_n$.
$\sum_{i=1}^n f(m_i) \otimes p_i = 0$ in $N \otimes P$ hence by proposition (2.13) in Atiyah-Macdonald we get finitely generated submodules $X \subset N$,$Y \subset P$ such that $\sum_{i=1}^n f(m_i) \otimes p_i = 0$ in $X \otimes Y$. Let $S$ be the (finite) set generating $X$.
Let $Z$ be the module generated by $S$ and $f(m_1), \dots , f(m_n)$. Then $Z$ is finitely generated and contains $f(M_0)$. Since $f : M \to N$ is injective we hence have that $f\mid_{M_0} : M_0 \to Z$ is injective and well-defined. By (i) we get that $f\mid_{M_0} \otimes \operatorname{id}$ is injective.
We have $(f\mid_{M_0} \otimes \operatorname{id})(u) = \sum_{i=1}^n f(m_i) \otimes p_i = 0$ in $X \otimes Y \subset Z \otimes Y$. Hence by injectivity of $f\mid_{M_0} \otimes \operatorname{id} : M_0 \otimes Y \to Z \otimes Y$ we get that $u = 0$ in $M_0 \otimes Y \subset M \otimes P$ and hence in $M \otimes P$.