# Vector spaces of singular planar cubics

What is the largest dimensional linear space of singular planar cubics? Is this known?

Think of the space of planar cubics as a PP^9 (parametrized by the coefficients). The discriminant \Delta is then a degree 12 polynomial on PP^9 whose vanishing parametrizes singular cubic curves. What is the dimension of the largest linear space inside the vanishing of \Delta? I attempted doing this on a computer, but the computations were too large to terminate.

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According to Bertini's Theorem a linear system is smooth away from its base points. Thus there is a point in PP^2 contained in the singular set of every cubic in your linear system. You need three conditions to impose a singularity at a point p (f(p) = f_x(p) = f_y(p)=0). Thus maximal projective spaces inside the discriminant have dimension 6.

Edit As David pointed out in another answer, this argument uses char 0.

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The answer of jvp (above) is completely right in characteristic zero.

I'll be annoying and point out the following example, due to Serre: let K be an algebraically closed field of characteristic 2 and let F be the field with two elements inside K. Let L be the vector space of cubics in K[x,y,z] that vanish at the 7 points of P^2(F). Every cubic in this family is singular, yet there is no point at which they are all singular.

Now, L has dimension 3 (and its projectivization has dimension 2), so this is still smaller than the 7 (projectively 6) dimensional spaces of cubics which are singular at a given point. But it is a warning about the kind of issues that can come up when you ask this question in finite characteristic.

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