What is the insight of Quillen's proof that all projective modules over a polynomial ring are free? One of the more misleadingly difficult theorems in mathematics is that all finitely generated projective modules over a polynomial ring are free.  It involves some of the most basic notions in commutative algebra, and really sounds as though it should be easy (the graded case, for example, is easy), but it's not.  The question at least goes as far back as Serre's FAC, but it wasn't proved until 1976, by Quillen in Projective modules over polynomial rings EDIT: and also independently by Suslin.
I decided that this is the sort of fact that I should know a rough outline of how to prove, but the paper was not very helpful.  Usually when someone kills off a famous conjecture in 5 pages, it's because they've developed some fantastic new piece of machinery people didn't have before.  And, indeed, Quillen is famous for inventing some fancy and wonderful machinery, and the paper is only 5 pages long, but as far as I can tell, none of that fancy machinery actually appears in the proof.

So, what was it that Quillen saw, that Serre missed?

 A: Here is a summary of what I learned from a nice expository account by Eisenbud (written in French), can be found as number 27 here.1
First, one studies a more general problem: Let $A$ be a Noetherian ring, $M$ a finite presented projective $A[T]$-module. When is $M$ extended from $A$, meaning there is $A$-module $N$ such that $M = A[T]\otimes_AN$?
The proof can be broken down to 2 punches:
Theorem 1 (Horrocks) If $A$ is local and there is a monic $f \in A[T]$ such that $M_f$ is free over $A_f$, then $M$ is $A$-free
(this statement is much more elementary than what was stated in Quillen's paper).
Theorem 2 (Quillen) If for each maximal ideal $m \subset A$, $M_m$ is extended from $A_m[T]$, then  $M$ is extended from $A$
(on $A$, locally extended implies globally extended).
So the proof of Serre's conjecture goes as follows: Let $A=k[x_1,\dotsc,x_{n-1}]$, $T=x_n$, $M$ projective over $A[T]$. Induction (invert all monic polynomials in $k[T]$ to reduce the dimension) + further localizing at maximal ideals of $A$ +  Theorem 1 show that $M$ is locally extended. Theorem 2 shows that $M$ is actually extended from $A$, so by induction must be free.
Eisenbud's note also provides a very elementary proof of Horrocks's Theorem, basically using linear algebra, due to Swan and Lindel (Horrocks's original proof was quite a bit more geometric).
As Lieven wrote, the key contribution by Quillen was Theorem 2: patching. Actually the proof is fairly natural, there is only one candidate for $N$, namely $N=M/TM$, so let $M'=A[T]\otimes_AN$ and build an isomorphism $M \to M'$ from the known isomorphism locally.
It is hard to answer your question: what did Serre miss (-:? I don't know what he tried. Anyone knows?
1Eisenbud, David, Solution du problème de Serre par Quillen–Suslin, Semin. d’Algebre, Paul Dubreil, Paris 1975-76 (29eme Annee), Lect. Notes Math. 586, 9-19 (1977). DOI: 10.1007/BFb008711, PDF on the MSRI website, ZBL0352.13005, MR568878.
A: Actually, Quillen's proof did contain a fantastic new idea : Quillen patching.
It states that if $P$ is a fg projective $R[t]$-module then the set of all f in $R$, such that (for localizations at the multiplicative systems $ \left\{ 1,f,f^2,... \right\}$ $P_f$ is extended from $R_f$ (that is $P_f = Q \otimes_{R_f} R_f[t]$ for a projective $R_f$-module $Q$, is an ideal of $R$.
Apply this to $R=\mathbb k[x_1,...,x_d]$ then the localization of $P$ at the set of all monic polynomials in $R[t]$ is a projective $k(t)[x_0,...,x_d]$ module whence free by induction. But then $P_g$ is free for some monic poly $g$ in $t$.
Now take a maximal ideal $\mathfrak m$ of $R$ and consider the extension $P\mathfrak m$ of $P$ to $R_\mathfrak{m}[t]$ (the localization at $\mathfrak m$). Because $(P\mathfrak m)_g$ is a free $(R_\mathfrak{m}[t])_g$ module it follows from Horrocks result that $P\mathfrak m$ is a free $R_\mathfrak{m}[t]$-module and extended from $R_m$. Whence the Quillen-ideal for $P$ equals $R$. Serre's conjecture follows by considering $1$ and induction.
