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This is a variant on the question posed here, in which the OP asks for a characterisation of the diagonalisable involutions in $\operatorname{GL}_n(A)$, where $A$ is a $k$-algebra for some field $k$ of characteristic $\neq 2$. The accepted answer shows that, if every finitely-generated projective module is free, then every involution is in fact diagonalisable.

Let us now restrict ourselves to the case where $A = k[x_1,...,x_l]$ is a polynomial ring over $k$. The Quillen-Suslin theorem says that all algebraic vector bundles over affine space are trivial, which translates to the condition that all finitely-generated projective modules over $A$ are free, so the solution above does apply. However, this feels to me like "killing a bee with a hand cannon" since the Quillen-Suslin theorem is quite a nontrivial result. Is there a more direct proof in this restricted setting?

As an aside, I'm hoping for a more explicit answer than just "use Quillen patching" as that will of course solve the problem, but I understand if that is the approach experts would take.

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    $\begingroup$ No: the answer you mention shows that the fact that involutions are diagonalisable is actually equivalent to say that every f.g. projective module is free. $\endgroup$
    – abx
    Commented Dec 7, 2021 at 5:15
  • $\begingroup$ @abx You're right, of course. In that case, may I ask: do you know if anyone has ever considered this as a possible approach to proving Quillen-Suslin? $\endgroup$ Commented Dec 7, 2021 at 6:29
  • $\begingroup$ @MartinSkilleter since the equivalence is quite straightforward, I'm not sure it could be really be viewed as an alternative approach. (And by the way, this approach forces assume $2$ is invertible, unlike in the Quillen-Suslin result.) $\endgroup$
    – YCor
    Commented Dec 7, 2021 at 12:44

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As abx pointed out in the comments, I misread the fact that this is an equivalence, so it is unlikely that there is a more elementary proof.

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