Projective module of rank n and faithfully flat R-algebra

Question

Suppose P is a projective R-module of rank n.

Show that there exists a finitely generated faithfully flat R-algebra A such that tensoring with P gives a free module of rank n.

Any hint is appreciated.

Answer 1

$P$ is the image of some idempotent $e \in M_m(R)$ for some $m \ge n$. We want to find an fppf (this is very slightly stronger than what you asked for: finitely presented faithfully flat) extension $R \to A$ such that $\text{im}(e) \cong A^n$ is free when extended by scalars to $A$. This is equivalent to the condition that there exist linear maps $f : A^n \to A^m, g : A^m \to A^n$ such that $fg = e$ and $gf = \text{id}_{A^n}$.

Writing $f$ and $g$ as matrices, this is a system of $n^2 + m^2$ polynomial equations in $2mn$ variables. Now we can construct $A$ by formally adjoining to $R$ a solution to this system; that is, we take $A = R[f_{ij}, g_{ji}]/I$ where $1 \le i \le m, 1 \le j \le n$ and $I$ is the ideal generated by the conditions $fg = \text{id}_{A^m}, gf = e$. By construction $A$ is finitely presented and $P \otimes_R A$ is free.

It remains to check that $A$ is faithfully flat. To check that it's flat we can check that the localization $A \otimes_R R_P$ is flat for every prime ideal $P$ in $R$. Over a local ring $R_P$ every finitely generated projective module is free, so the system above always has solutions, or equivalently $\text{im}(e)$ is always already free. This lets us determine all possible values of $f, g$ over $R_P$ as follows: $GL_n(R_P)$ acts freely and transitively on the possible values of $f : R_P^n \to R_P^m$ (since $f$ must be an isomorphism onto $\text{im}(e)$), and given $f$ (and $e$), $g$ is uniquely determined: it satisfies

$$gfg = g = ge$$

so it must factor through $e$ and hence, as a map $\text{im}(e) \to A^n$, must be the inverse of $f : A^n \to \text{im}(e)$. Moreover this argument continues to apply over any $R_P$-algebra, since an extension of scalars of a free module remains free. It follows that $A \otimes_R R_P$ is isomorphic to the algebra of functions on the affine group scheme $GL_n$ over $R_P$, which is a localization of a polynomial algebra on $n^2$ variables over $R_P$ and hence flat.

Finally, to check that $A$ is faithfully flat given that it's flat, we can check that $A \otimes_R R/m$ is nonzero for every maximal ideal $m$ of $R$. But over a field $R/m$ it's again true as above that every finitely generated projective module is free, so again the system always has solutions.

There's probably a slightly cleaner way to say all this. Morally $A$ is the universal extension of $R$ which trivializes $P$, except that a priori the construction above might depend on the choice of $e \in M_n(R)$ presenting $P$ and I haven't checked that it doesn't.