Conservative and divergence free algebraic vector fields
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- by E. Connell and J. Drost
- Proc. Amer. Math. Soc. 87 (1983), 607-612
- DOI: https://doi.org/10.1090/S0002-9939-1983-0687626-5
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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 $\left | {J(u)} \right | = 1$, 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.References
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Bibliographic Information
- © Copyright 1983 American Mathematical Society
- 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