plane cubics and conic bundles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:35:48Z http://mathoverflow.net/feeds/question/114927 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114927/plane-cubics-and-conic-bundles plane cubics and conic bundles bugger 2012-11-29T20:35:26Z 2012-11-30T06:05:39Z <p>It is well known that any plane cubic curve can be obtained as the discriminant locus of a conic bundle (actually even just of a net of conics). Does this hold true also for all nodal cubics (with double lines over the nodes)? How does one see this?</p> http://mathoverflow.net/questions/114927/plane-cubics-and-conic-bundles/114958#114958 Answer by Noam D. Elkies for plane cubics and conic bundles Noam D. Elkies 2012-11-30T06:05:39Z 2012-11-30T06:05:39Z <p>Yes, at least if we're not in characteristic 2. Since all nodal cubics are projectively equivalent, it is enough to find one example. Trying a few symmetric $3 \times 3$ determinants soon turns up the matrix $$M = \left[ \begin{array}{ccc} x &amp; x &amp; y \cr x &amp; z &amp; 0 \cr y &amp; 0 &amp; x \end{array} \right]$$ with determinant $x^2(z-x)-zy^2$. So the discriminant locus of the associated net of conics is $zy^2 = x^2(z-x)$, which has a node at $(x:y:z) = (0:0:1)$ [set $z=1$ to get the more familiar affine model $y^2 = x^2 - x^3$ with a node at the origin]. At that point $M$ becomes $$\left[ \begin{array}{ccc} 0 &amp; 0 &amp; 0 \cr 0 &amp; 1 &amp; 0 \cr 0 &amp; 0 &amp; 0 \end{array} \right],$$ where the conic degenerates to a double line as desired.</p>