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Let $R$ be a commutative ring (noetherian if needed), and $P$ a finitely generated projective $R$-module (of constant rank $n$). Let $P^\vee:={\rm Hom}_R(P, R)$ be the dual. For $\varphi\in{\rm Aut}_R(P\oplus R)$, define $\varphi^*, \varphi^\vee\in{\rm Aut}_R(P^\vee\oplus R)$ by letting (where $\langle -, -\rangle$ is the natural pairing between a module and its dual) $$\langle \varphi^*(\xi), x\rangle:=\langle \xi, \varphi(x)\rangle, x\in P\oplus R, \xi\in P^\vee\oplus R$$ and $\varphi^\vee:=(\varphi^*)^{-1}$. One checkes easily that $$\langle \varphi^\vee(\xi), \varphi(x)\rangle=\langle \xi, x\rangle, x\in P\oplus R, \xi\in P^\vee\oplus R.$$ My question is:

For $x\in P\oplus R, \xi\in P^\vee\oplus R$ with $\langle \xi, x\rangle=1$, can we find $\varphi\in{\rm Aut}_R(P\oplus R)$ such that $\varphi(0, 1)=x, \varphi^\vee(0, 1)=\xi$? And if $\det P\cong R$, can we further get such a $\varphi$ preserving $\det P$ (i.e. $\det \varphi={\rm id}$)?

The following fact may be helpful:

Let $V:=\ker(P\oplus R\xrightarrow{\langle \xi, -\rangle}R)\subset P\oplus R$ and $V':=\ker(P^\vee\oplus R\xrightarrow{\langle -, x\rangle}R)\subset P^\vee\oplus R$, then it's easy to see that we have decompositions $P\oplus R=V\oplus Rx, P^\vee\oplus R=V'\oplus R\xi$. I want to show that $V\cong P$ but I can't (this is of course true if $P$ is a cancellative projective module).

I don't even know how to prove or give a counterexample when $P=R^n$. Any idea or proof (maybe in some special cases) or counterexample is welcome!

Lao-tzu
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  • What do you call $\langle, \rangle$ ? – Maxime Ramzi Feb 14 '19 at 08:02
  • The pairing between a module and its dual (or the evaluation map). – Lao-tzu Feb 14 '19 at 08:04
  • Even if you throw away the $\varphi^\vee \left(0,1\right) = \xi$ requirement, I think your $\varphi$ may not exist. Indeed, if it existed, then it would induce an $R$-module isomorphism $\left(P \oplus R\right) / Rx \to \left(P \oplus R\right) / R\left(0,1\right) = \left(P \oplus R\right) / \left(0 \oplus R\right) \cong P$. Now I claim that this is not always true. To prove this, take $R = \mathbb{R}\left[a, b, c\right] / \left(a^2 + b^2 + c^2 - 1\right)$ and $P = R^2$; this yields isomorphisms $P \oplus R \cong P^\vee \oplus R \cong R^3$ that I will use to identify all three ... – darij grinberg Feb 16 '19 at 01:13
  • ... of these $R$-modules. Take $x = \left(a,b,c\right) \in R^3 = P \oplus R$ and $\xi = \left(a,b,c\right) \in R^3 = P^\vee \oplus R$; then, $\left<\xi, x\right> = 1$. Now, if I remember correctly, the $R$-module $\left(P \oplus R\right) / Rx$ is not free (it would contradict the hairy ball theorem in some way), but the $R$-module $P$ is free (by definition), so they cannot be isomorphic. – darij grinberg Feb 16 '19 at 01:15
  • Ah yes, the non-freeness of $\left(P \oplus R\right) / Rx$ follows with some work from https://math.stackexchange.com/a/1000474/ , since $\left(P \oplus R\right) / Rx$ is isomorphic to the (stably free, but not free) $R$-module $P$ in that answer. – darij grinberg Feb 16 '19 at 01:25
  • You are right, thanks a lot! I realize my requirement is equivalent to $P$ being a cancellative projective module – Lao-tzu Feb 16 '19 at 09:01

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