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?
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2$\begingroup$ For the smooth plane cubic, the net of conics is given by a surface of bidegree (1,2). The problem is equivalent - I think - to checking wether all nodal plane cubics are symmetric determinantal. This seems quite true.... $\endgroup$bugger– bugger2012-11-29 21:34:52 +00:00Commented Nov 29, 2012 at 21:34
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1$\begingroup$ I don't know how to edit the comment, actually it is hypersurface, not surface of course. $\endgroup$bugger– bugger2012-11-30 09:25:01 +00:00Commented Nov 30, 2012 at 9:25
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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 $$ M = \left[ \begin{array}{ccc} x & x & y \cr x & z & 0 \cr y & 0 & x \end{array} \right] $$ 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 $$ \left[ \begin{array}{ccc} 0 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 0 \end{array} \right], $$ where the conic degenerates to a double line as desired.
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1$\begingroup$ Thank you! so everything I stated stays true for nodal curves: cool. $\endgroup$bugger– bugger2012-11-30 09:26:19 +00:00Commented Nov 30, 2012 at 9:26