Equivalent conditions for a finitely presented module

Question

I have a few questions regarding the proof of Corollary 7.21 from here.

$(i)\implies (ii)$:

1. Why $M_\mathfrak p=M_{f_i}\otimes_{R_{f_i}}R_\mathfrak p$?
2. How does the freeness of $M_\mathfrak p$ over $R_\mathfrak p$ follow from that of $M_{f_i}$ over $R_{f_i}$ (and the above)?

$(ii)\implies (i)$:

When we localize $u:R^n\rightarrow M$ at $\mathfrak p$, we get $u_\mathfrak p: R_\mathfrak p^n\rightarrow M_\mathfrak p$, a homomorphism of $R_\mathfrak p$-modules.

3. If we apply Proposition 7.20c), shouldn't we get $f_{\mathfrak p}\in R_\mathfrak p\setminus \mathfrak p R_\mathfrak p$ instead of $f_\mathfrak p\in R\setminus \mathfrak p$ (since we have a homomorphism of $R_\mathfrak p$-modules instead of that of $R$-modules)?

4. To apply Proposition 7.20c), we also need to know that $M_\mathfrak p$ is finitely presented over $R_\mathfrak p$. Why is that so?

Answer 1

$1.$ The is because the multiplicative subset $\{f^k\mid k\in\mathbf N\}$ is contained in the multiplicative subset $S=R\smallsetminus\mathfrak p$, hence $$M_f\otimes_{R_f}R_\mathfrak{p}=(M\otimes_RR_f)\otimes_{R_f}R_\mathfrak{p}\simeq M\otimes_RR_\mathfrak{p}=M_\mathfrak{p}.$$

$2.$ Extension of scalars preserves freeness (for any $R$ algebra $A$, $R^n\otimes_RA\simeq A^n$).

$3.$ It should, but $f_\mathfrak p$ can be multiplied by a unit in $R_\mathfrak p$, and you just have to multiply by the denominator to get an element in (the canonical image of) $R$.

$4.$ Finite presentation is preserved by extension of scalars since tensor product is a right exact functor.