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Conservative and divergence free algebraic vector fields


Authors: E. Connell and J. Drost
Journal: Proc. Amer. Math. Soc. 87 (1983), 607-612
MSC: Primary 13F20; Secondary 13B10, 13N05
DOI: https://doi.org/10.1090/S0002-9939-1983-0687626-5
MathSciNet review: 687626
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Abstract: Suppose $ k$ is a field of characteristic 0 and $ {k^{[n]}} = k[{x_1}, \ldots ,{x_n}]$. If $ {u_i}$, $ {f_j} \in {k^{[n]}}$ for $ 1 \leqslant i \leqslant n$, $ 1 \leqslant j \leqslant m$, $ u = ({u_1}, \ldots ,{u_n})$, the $ {f_j}$ are relatively prime, and each $ {f_j}u$ is conservative, then $ u$ is conservative and $ ({f_1}, \ldots ,{f_m})$ is unimodular. Given any $ u$ with $ \left\vert {J(u)} \right\vert = 1$, then each derivation $ \partial /\partial {u_i}$, has divergence 0. If $ D:{k^{[n]}} \to {k^{[n]}}$ is a $ k$-derivation with kernel of dimension $ n$ - $ - 1$, then there exists a $ g$ so that $ gD$ has divergence 0.


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DOI: https://doi.org/10.1090/S0002-9939-1983-0687626-5
Article copyright: © Copyright 1983 American Mathematical Society

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