Let $V/k$ be an algebraic variety defined over a field $k$ and let $S$ be the set of subfields $k_0\subseteq k$ for which there exists an algebraic variety $V_0/k_0$ whose base change to $k$ is isomorphic to $V$.
Any field $k_0\in S$ that contains no other elements of $S$ is a minimal field of definition for $V$.
In general, an algebraic variety may have more than one minimal field of definition; this does not occur for elliptic curves but it does occur for curves of genus 2.
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- Last edited by Andrew Sutherland on 2020-10-10 14:54:54
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- 2020-10-10 14:54:54 by Andrew Sutherland (Reviewed)
- 2020-10-10 14:42:28 by Andrew Sutherland
- 2020-10-10 14:40:31 by Andrew Sutherland
- 2020-10-10 14:39:57 by Andrew Sutherland
- 2020-10-10 14:25:09 by Andrew Sutherland