Equality between rank and dimension

Question

Let $M$ a finitely generated module over a noetherian commutative ring $A$. Assume that $M_{\mathfrak p}$ is a free $A_{\mathfrak p}$-module. For any prime $\mathfrak p$ and let's put $k(\mathfrak p):=A_{\mathfrak p}/\mathfrak pA_\mathfrak p$.

How can I prove the following equality:

$$\operatorname{dim}_{k(\mathfrak p)}M\otimes_A k(\mathfrak p) =\operatorname{rank}_{A_{\mathfrak p}}(M_{\mathfrak p})$$

Answer 1

Say $\Lambda$ is the rank. Then $$ \begin{aligned} M\otimes_{A}k(\mathfrak{p})=& (M\otimes_{A}A_\mathfrak{p})\otimes_{A_\mathfrak{p}}k(\mathfrak{p})= M_\mathfrak{p}\otimes_{A_\mathfrak{p}}k(\mathfrak{p})\\ =&A_\mathfrak{p}^{\oplus \Lambda}\otimes_{A_\mathfrak{p}}k(\mathfrak{p})= (A_\mathfrak{p}\otimes_{A_\mathfrak{p}}k(\mathfrak{p}))^{\oplus \Lambda}=k(\mathfrak{p})^{\oplus \Lambda} \end{aligned} $$