Talk:Twisted cubic

Algebraically closed field?
Its not clear to me if all of the properties (as listed in the "properties" section) hold in the case of a field not being algebraically closed. If they do, then a note stating this should be explicitly added, viz. "these properties do not require the field over which the projective space is defined to be algebraically closed". I suppose this might be obvious if one "did one's homework", but I'm lazy and dumb... linas 22:26, 23 December 2005 (UTC)


 * I've linked to Degree of an algebraic variety, where algebraic closure is certainly used (and is mentioned there). Otherwise, if you start at the parametric end, there may not be any such issues. Charles Matthews 22:31, 23 December 2005 (UTC)