Can you give me some example of finitely generated Projective A module P with rank 2 s.t. $P\oplus R[x] \cong P'\oplus R[x]$ and P is not isomorphic to P' where R is a local ring of dimension 3 and A=R[x]?
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Can you give some more information about this problem? I am wondering the solution. – Bob Dobbs Oct 20 '22 at 19:47
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I am studying cancellation of projective modules and I would like to know the cases where it does not hold.So i am searching for examples. – Dgarg12 Oct 21 '22 at 20:09
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I know only the example Matthe gived. If you add a 1-dim trivial bundle to the tangent bundle of the sphere it becomes 3-dim trivial bundle. – Bob Dobbs Oct 26 '22 at 00:17
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The following form of Bass' Cancellation Theorem shows that this is impossible.
Let $P,P'$ and $Q$ be finitely generated projective modules over a commutative ring $A$, with $\mathrm{rk}(P)$ $>$ $\mathrm{dim}(A/\mathrm{rad}(A))$. Then, if $P\oplus Q$ $\cong$ $P'\oplus Q$, one has $P$ $\cong$ $P'$.
The proof of (an even slightly stronger form of) the theorem is in Lombardi & Quitté, Ch. XIV, (3.11). Unlike in Bass' original theorem, Noetherianity of $A$ is not required.
In your setup, $Q=A$, $\mathrm{rk}(P)=2$, and $A/\mathrm{rad}(A)$ $=$ $R[X]/\mathfrak{m}[X]$ $=$ $k[X]$ is one-dimensional. The Krull dimension of the local ring $R$ is immaterial here.
Edit A more concise exposition of the above theorem can be found here (in French). I don't believe there are many examples around. The standard one, due to Hochster, is a module $P$ with $P\oplus A$ $\cong$ $A^2\oplus A$ but $P$ $\not\cong$ $A^2$, where $A$ is the coordinate ring of the real unit sphere. It is described in these notes by Keith Conrad.
