Property of Quasi-Projective Scheme over Noetherian affine Scheme I have some questions about the steps used in in Liu's "Algebraic Geometry" proof of Prop. 7.1.32 (see page 261):

1.: Why $\tilde{N}$ is an invertible sheaf on $SpecB$, where for the module $N$ the sheaf  $\tilde{N}$ on $X$ is defined as $\tilde{N}(D(f)) = N_f$ for basic open sets $D(f)$ (compare with page 159)?
2.: Why $N/IN= Bv$? We have $N/IN= \oplus_i ^n B/m_i$. So shouldn't we have $n$ generators instead of the one $v$?
Why futhermore the equation $N = Bv$ and the condition from my first question imply $N \cong B$? Or in other words why $B \to N =Bv, b \mapsto bv$ is an iso?
3.: Why does  $\mathcal{L}|_V \cong \mathcal{O}_V$ hold? Indeed, since $\mathcal{L}$ invertible there always exist a neighbourhood in $W \subset U$
such that $\mathcal{L}|_W \cong \mathcal{O}_W$. But why does $V= D(t)$ this property?
 A: Disclaimer: I've never really learned any of this sheaf stuff before and have just been googling, so take what I say with a grain of salt. But I think I understand it enough now to answer the questions.


*

*The restriction $\mathcal L|_U$ is an invertible sheaf on $U$. And since $U$ is affine, we have in fact that $\mathcal L|_U \cong \widetilde{\mathcal L(U)}$ (see here). Localizing coherent is coherent (see here), so $\tilde N$ is coherent. Invertibility of $\tilde N$ boils down to the following fact: if $M,N$ are $A$-modules and $S$ is a multiplicative set in $A$, then 
$$S^{-1}(M\otimes_A N) \cong (S^{-1}M)\otimes_{S^{-1}A}(S^{-1}N)$$
as $S^{-1}A$-modules. Since $\widetilde{L(U)}$ is invertible, and $\tilde N$ is its localization, if you localize an inverse of $\widetilde{L(U)}$ (under tensor product), you get an inverse for $\tilde N$. 

*Do you see why $N/\mathfrak m_i N \cong B/\mathfrak m_i$? Tensoring an equation like $N \otimes_B M \cong B$ (remember $\tilde N$ is invertible) with $B/\mathfrak m_i$ lets you deduce so with linear algebra. So the equation $N/IN \cong \oplus_i B/\mathfrak m_i$ as $B$-modules holds, but remember $\oplus_i B/\mathfrak m_i \cong B/I$ as $B$-modules, too, so $B/I \cong N/IN$ as $B$-modules. $B/I$ is generated by the image of $1_B$ as a $B$-module, so wherever that is sent under the isomorphism $B/I \cong N/IN$, here it could be $v+IN$ with $v\in N$, generates $N/IN$ as a $B$-module, which tells you that $N=Bv+IN$, and then he uses Corollary 1.2.9 to conclude $N = Bv$. Since $\tilde N$ is invertible, $B$ must act nicely on $v$, i.e. $bv \ne 0$ for $b\ne 0$ (if $N\otimes_B M \cong B$, there's some tensor $\sum_i (b_iv)\otimes m_i \in N\otimes_B M$ which maps to $1_B$, and if $b\in B$ satisfies $bv=0$ then $b(\sum_i (b_iv)\otimes m_i) = 0$, so $b\cdot 1_B = 0$ too, so $b=0$). So the map $B\to N$ given by $b \mapsto bv$ is injective, and it's also surjective since $N = Bv$. 

*Since $\text{Hom}$ commutes with localization for finitely-presented modules, if $A$ is some Noetherian ring and if $C,D$ are finitely-presented (or at least one is finitely presented and the other finitely generated) $A$-modules and $R$ a multiplicative set in $A$, if $R^{-1}C \cong R^{-1}D$, then there is some $t\in R$ so that $C_t \cong D_t$. Our sheaf is coherent, so you can use this in our situation.
Added in response to comments:
To your first question: yes, it is because $\tilde N$ is invertible. By arguments in (1), there is a coherent $B$-module $M$ with $N \otimes_B M \cong B$ (see here, global sections of tensor product of quasi-coherent sheaves over an affine scheme is the tensor product of the global sections). Now 
$$(N/\mathfrak m_i N \otimes_{B/\mathfrak m_i} M/\mathfrak m_i M) \cong (B/\mathfrak m_i \otimes_B B/\mathfrak m_i) \otimes_B  (N \otimes_B M)$$
$$\cong (B/\mathfrak m_i \otimes_B B/\mathfrak m_i) \otimes_B B \cong B/\mathfrak m_i$$
as $B/\mathfrak m_i$-modules. Since dimension of vector spaces is multiplicative under tensor product, we must have $\dim_{B/\mathfrak m_i}(N/\mathfrak m_i N) = 1$, so that $N/\mathfrak m_i N \cong B/\mathfrak m_i$ as $B/\mathfrak m_i$-modules (and thus also as $B$-modules).
To your second question: this is super cool! Look up "hom commutes with localization" to find a proof that if $C$ is a finitely-presented $A$-module, then the natural map
$$\theta: R^{-1} \text{Hom}_A(C,D) \to \text{Hom}_{R^{-1}A}(R^{-1}C, R^{-1}D)$$
given by $\theta(f/t) = (1/t) (R^{-1}f)$, where 
$$((1/t)(R^{-1}f))[c/r] = f(c)/(tr),$$ 
is an isomorphism. We are supposing that the RHS contains an isomorphism $\phi: R^{-1}C \to R^{-1}D$. Let $t_1,\dots,t_n\in R$ and $f_1,\dots,f_n\in \text{Hom}_A(C,D)$ be so that $\theta(f_1/t_1 + \dots + f_n/t_n) = \phi$. Let $t' = t_1\cdots t_n$. Notice that $t_1,\dots,t_n$ are invertible in $A_{t'}$. Define $f':C_{t'} \to D_{t'}$ by 
$$f'(c/(t')^k) = \left( f_1(c)/t_1 + \dots + f_n(c)/t_n \right)/(t')^k.$$
You also need to know that "localization commutes with taking kernel and cokernel", which is an exactness thing, to understand this next part. (Here's an oops on my part: we need $A$ to be Noetherian for this next bit to work, which it is in your question; I fixed that above.) Now since $C$ is finitely generated, so is $C_{t'}$ over $A_{t'}$ and $A_{t'}$ is Noetherian. So $\ker f'$ and $\text{coker} f'$ are finitely generated $A_{t'}$-modules (assuming $D$ is finitely generated, for the cokernel part). And $R^{-1}\ker f'$ and $R^{-1} \text{coker} f'$ are both zero. Now I'll leave a small exercise: [if $M$ is a finitely generated $B$-module, $R \subset B$ a multiplicative set, and $R^{-1} M = 0$, then there exists $r\in R$ so that $M_r = 0$]. Take such an $r$ for $\ker f'$, take its numerator $s\in R$, set $t'' = t's$, now when you invert $t''$, $f'$ becomes $f''$ and is now injective. Do the same for the cokernel. Then you're done, and you have one element you can invert to make $f'$ an isomorphism $f: C_t \to D_t$.
