What is the insight of Quillen's proof that all projective modules over a polynomial ring are free?

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.

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 $A=k[x_1,\cdots,x_{n-1}]$, $T=x_n$, $M$ 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 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?

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.