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I encountered this problem while studying algebraic geometry:

Let $f \in \mathbb{C}[x, y, z]$ be a homogeneous polynomial of degree 3. The coefficients of $f$ represent a point $P_f$ in $\Bbb{P}^9$ . Show that $$\Bbb{P}^9\setminus\{P_f \,|\, \text{the variety defined by}\, f\, \text{is smooth, irreducible of degree}\, 3\}$$ is a closed subvariety of $\Bbb{P}^9$

I tried to do some computing with the Jacobian matrix but I think it is the wrong path. Someone can explain how to proceed?

reuns
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abcX
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  • Your approach sounds about right to me. The moduli space of singular hypersurfaces is given by additional equations that amount to the vanishing of the determinant of the Jacobian; this determinant can be expressed as a polynomial in the coefficients of your initial equation. – Jeroen van der Meer Dec 08 '21 at 14:06

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Your approach seems just fine to me. Consider the subvariety $X$ of $\mathbb{P}^2$ defined by a homogeneous equation $f(x,y,z)$ of degree $3$. Then $X$ is smooth if and only if, at every point $p$ of $X$, not all of the partial derivatives $\partial_{x_k} f(p)$ vanish. Notice that these partial derivatives can be expressed as a polynomial in the coefficients of the homogeneous equation that you started with, hence we're dealing with an open locus in the moduli space. The (reduced) complement, then, must be closed.

Jeroen van der Meer
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  • what do you mean by evaluating f at $p_j$? doesn't have take as input a point $p=(p_1,p_2,p_3)$? – abcX Dec 08 '21 at 16:57
  • Oops, sorry, muscle memory took over I guess. What I was alluding to is the general smoothness condition for a variety defined by multiple equations, which is that the rank of the Jacobian must be maximal. In case of a single equation, that condition just amounts to the nonvanishing of the partial derivatives. I'm on my phone right now but I'll correct my answer in a moment. – Jeroen van der Meer Dec 08 '21 at 17:10
  • Clear now! thank you so much – abcX Dec 09 '21 at 10:01