A class of counterexamples to the Cancellation Problem for arbitrary rings (Q2724121)
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scientific article; zbMATH DE number 1615678
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | A class of counterexamples to the Cancellation Problem for arbitrary rings |
scientific article; zbMATH DE number 1615678 |
Statements
A class of counterexamples to the Cancellation Problem for arbitrary rings (English)
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9 July 2001
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general cancellation problem
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locally nilpotent derivations
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unimodular rows
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The ``cancellation problem'' asks the following question: Let \(B\) be an affine domain over a field \(k\) and \(n\in {\mathbb N}\). Assume that \(B[T]\cong_k k[X_1,\dots,X_n]\). Does it then follow that \(B\cong_k k[Y_1,\dots,Y_{n-1}]\)? For background on this and other cancellation problems in algebraic geometry see: \textit{H.~Kraft} in: Topological Methods in Algebraic Transformation Groups, Prog. Math. {80}, 81--105 (1989; Zbl 0719.14030).NEWLINENEWLINENow let \(A\) be a commutative ring, \(B\) an \(A\)-domain, and \(n\in {\mathbb N}\). Assume that \(B[ T] \cong_A A[ X_1,\dots,X_n]\). The ``general cancellation problem'' asks: Does it then follow that \(B\cong_A A[Y_1,\dots,Y_{n-1}]\)?NEWLINENEWLINEIn the present paper a class of counterexamples to the last question is constructed. Namely, let \(k\) be a field of characteristic zero and let \(A\) be a finitely generated commutative \(k\)-algebra without zero divisors. Suppose that there exists a unimodular row \((a_1,\dots,a_n)\), i.e. \(a_1b_1+\dots+a_nb_n=1\) for some \(b_1,\dots,b_n\in A\), which cannot be completed to an invertible square matrix. [ For example, it was shown by Raynaud and Suslin that this is the case for \(A={\mathbb C}\,[a,b,c,x,y,z]/(ax+by+cz-1)\) and the row \((\overline x, \overline y, \overline z)\).]{} Let \(B\) be the kernel of the \(k\)-derivation NEWLINE\[NEWLINE D=b_1\frac{\partial}{\partial X_1}+\dots+b_n\frac{\partial}{\partial X_n} NEWLINE\]NEWLINE of the ring \(A[X_1,\dots,X_n]\). The authors prove that \(B[a_1X_1+\dots+a_nX_n]= A[X_1,\dots,X_n]\), but \(B\) is not isomorphic to \(A[Y_1,\dots, Y_{n-1}]\).
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