Let $A$ be a commutative ring with identity and let $M$ be an $A$-module. The fiber of $M$ at $P \in \text{Spec}A$ is the module $M(P):=M_P / PM_P$, which is a vector space over the residue field $A(P)=A_P / PA_P$. My question is, how does this scalar multiplication look like? I believe it is induced by the multiplication as we consider $M_P$ as an $A_P$-module?
Later I am supposed to prove that if $M$ is finite, then $rk_M(P)$ is finite where $rk_M$ is the function $$rk_M : \text{Spec} A \to \mathbb{N} \cup \{ \infty \}$$ given by $$P \mapsto \dim_{A(P)} M(P)$$ Before trying to prove this, I am trying to understand the structures $M(P)$ and $A(P)$ and how the vector space thing works, can anyone bring clarity to this?
Let $R$ be a commutative ring, let $M$ be an $R$-module, and let $I \subseteq R$ be an ideal. Then $M/IM$ is naturally an $R/I$-module, with scalar multiplication given by $(r + I)(m + IM) = rm + IM$ for all $r \in R$ and $m \in M$. (Easy exercise: check that this is well-defined and makes $M/IM$ into an $R/I$-module.)
Similarly, let $U$ be a multiplicatively closed subset of $R$, and let $R[U^{-1}]$ and $M[U^{-1}]$ be the localizations at $U$. Then $M[U^{-1}]$ is naturally an $R[U^{-1}]$-module, with scalar multiplication given by $\frac{r}{u} \cdot \frac{m}{v} = \frac{rm}{uv}$ for all $r \in R$, $m \in M$, and $u, v \in V$. (Since $U$ is multiplicatively closed, $uv \in V$, so $\frac{rm}{uv}$ is an element of $M[U^{-1}]$.)
In fact, these are both special cases of a general construction: For any ring homomorphism $f: R \to S$ and any $R$-module $M$, the tensor product $M \otimes_R S$ is an $S$-module with scalar multiplication given on simple tensors by $s (m \otimes t) = m \otimes st$. This is called extension of scalars. (There are natural $R$-module isomorphisms $M/IM \cong M \otimes_R R/I$ and $M[U^{-1}] \cong M \otimes_R R[U^{-1}]$, so these are the extensions of scalars along the quotient map $R \to R/I$ and the localization map $R \to R[U^{-1}]$, respectively.)