Generic freeness: $M$ f.g. projective, then there is $a\notin \mathfrak p$ for which $M[a^{-1}]$ is a free $R[a^{-1}]$-module. In Jacobson's BAII, he aims to show that any finitely generated projective module over a connected ring has a rank, where he defines this as follows:
First, he shows that any finitely generated projective module over a local ring is free. Now take a ring $R$ and a module $M$, and some prime ideal $\mathfrak p$. If $M$ is generated by $n$ elments, then $M_{\mathfrak p}$ is a free $R_{\mathfrak p}$-module of rank $n_{\mathfrak p}\leqslant n$. One says $M$ has rank $m$ if $m=n_{\mathfrak p}$ for any $\mathfrak p\in{\rm Spec}\,R$, that is, the mapping $\mathfrak p\in {\rm Spec}\,M\to\Bbb Z$ is constant. He then shows this mapping is continuous. 
To do this, he shows that for any finitely generated projective modulo $M$ and $\mathfrak p$ prime there is $a\notin \mathfrak p$ for which $M[a^{-1}]$ is a free $R[a^{-1}]$-module. The proof seems natural enough to me, but can one give some motivation as to why one should think this might be true? We're saying that $R[a^{-1}]\otimes_R M$ is a free $R[a^{-1}]$ module, so somehow $M$ was away from being free by only one element, and this happens for any prime ideal of our liking. Of course, say if $R$ is a domain one can just forcefully invert all nonzero elements to get a field and hence a free module, so it seems this is a very delicate way to achieve this. 
A more precise statement is the following. Let $M$ be a finitely generated projective module, and suppose $M_{\mathfrak p}$ is free with base of $n$ elements. Then there is $a\notin \mathfrak p$ such that $f:R^n\to M$, $(a_1,\ldots,a_n)\to \sum a_ix_i$ induces an isomorphism $1\otimes f:R[a^{-1}]\otimes R^n\to R[a^{-1}]\otimes M$. The proof is very nice, and goes as follows. Let $x_1/1,\ldots,x_n/1$ be a basis of $M_{\mathfrak p}$ as $R_{\mathfrak p}$ module. We have an exact sequence $$0\longrightarrow \ker f\longrightarrow R^n\longrightarrow M\longrightarrow{\rm coker}\; f\longrightarrow 0$$
and since $f$ induces an isomorphism $R_{\mathfrak p}^n \simeq M_{\mathfrak p}$ (it sends a basis to a basis) we know that $\ker f$ and ${\rm coker}\; f$ die under localization at $\mathfrak p$. Since ${\rm coker}\; f$ is f.g., we know there is some $b\notin \mathfrak p$ for which $b({\rm coker}\; f)=0$, that is, inverting $b$ already kills ${\rm coker}\; f$. Hence we have a short exact sequence $$0\longrightarrow \ker f_b\longrightarrow R^n_b\longrightarrow M_b\longrightarrow  0$$
Since $M_b$ is projective, the sequence splits and we get that $\ker f_b$ is f.g., and since we obtain (an isomorphic copy of) $\ker f_{\mathfrak p}=0$ by further inverting in $\ker f_b$, we know there is $c\notin \mathfrak p$ such that $c (\ker f_b)=0$, that is, localizing at $a=bc\notin \mathfrak p$ kills both the kernel and cokernel, and hence gives the isomorphism. Of course we wouldn't have needed the hypothesis that $M$ was projective if we knew $\ker f$ already was f.g. 
How can one motivate this? Does this lemma have any history? Is there any example which might illuminate this construction? Is there a geometric interpretation?
Jacobson provides references at the end of the chapter: Noether (Idealtheorie in Ringbereichen), Zariski and Samuel (Comm. Alg), Atiyah and MacDonald, Samuel (Algebre Commutative), and Kaplansky (Commutative Rings). I wouldn't mind being directed to a discussion of this in any of those books (except possibly Noether's as I don't know if it has been translated). 
 A: The reasoning is that a finitely-generated projective module is "locally free at each point $\mathfrak{p} \in \mathrm{Spec}(R)$" in the sense of being free over $R_\mathfrak{p}$. Geometrically, this corresponds to looking at an 'arbitrarily small' neighborhood of $\mathfrak{p}$, i.e. the direct limit over all such neighborhoods. So in the limit of taking smaller and smaller neighborhoods, $M$ becomes free.
So we might hope that the module is also free on  an honest neighborhood (an actual open set of $\mathrm{Spec}(R)$) containing $\mathfrak{p}$.
This is what the lemma accomplishes: it shows that there exists some neighborhood $U$ -- namely, $U$ is the complement of the vanishing locus of $a \in R$, hence $U = \mathrm{Spec}(R[a^{-1}])$ -- such that the restriction of $M$ to $U$ is a free module.
As an example, let $R = \mathbb{C}[x,y] / (y^2 - x^3 - 1)$, an elliptic curve in $\mathbb{A}^2$, and let $I = (x,y-1)$. This is not a principal ideal, but $R$ is a Dedekind domain, so every ideal is locally principal. (In particular, $I$ is locally a rank-1 free module, hence $I$ is projective.)
Now, $x$ is almost a generator for $I$, since modding out by $x$ gives the desired point $(0,1)$ along with only one extra undesired point $(0,-1)$. So let's restrict to the open set where we delete the locus $y = -1$, i.e. let's invert $y+1$. Then, you can check, $I$ becomes principal, generated by $x$. (The equation you want is $(y-1)(y+1)=x^3$.)
