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Let $f\in k[x_1,\ldots,x_n]_d$ be a homogeneous polynomial of degree $d$. Then its $n$ partial derivatives span a linear space $J(f)$ in $k[x_1,\ldots,x_n]_{d-1}$. I want know that, is there something we know about the dimension of $J(f)$?

It is easy to see that $dim(J(f))=n$ if $f$ defines a smooth hypersurface. But this is not necessary. In particular I want to know the following: if $n>2$ and the zero locus $Z(f)$ contains only isolated singularity, is it true that $dim(J(f))=n$?

Thanks in advance.

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  • $\begingroup$ If $n>3$, take $f=x_1^2+x_2^2+x_3^2$, which is irreducible if characteristic is not 2, but $\dim J(f)=3$. $\endgroup$
    – Mohan
    Commented Apr 4, 2019 at 13:36
  • $\begingroup$ @Mohan Thanks for the counter example. I edited the last question. $\endgroup$
    – Akatsuki
    Commented Apr 4, 2019 at 13:46
  • $\begingroup$ Since $f$ is homogeneous, it has isolated singularities mean the origin is the only singularity. Then, the partial derivatives form a regular sequence, which is even stronger than linear independence and thus $J(f)$ has dimension $n$. $\endgroup$
    – Mohan
    Commented Apr 4, 2019 at 14:38
  • $\begingroup$ @Mohan Sorry I didn't make myself clear. By $Z(f)$ I mean the zero locus in $\mathbb P^{n-1}$. $\endgroup$
    – Akatsuki
    Commented Apr 4, 2019 at 18:19
  • $\begingroup$ My example above with $n=4$ has only isolated singularities in projective 3-space, so am not sure what you are asking. $\endgroup$
    – Mohan
    Commented Apr 4, 2019 at 18:22

1 Answer 1

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If $\dim J(f) < n$ then, after a change of basis, $\partial f/\partial x_1 = 0$, hence $Z(f) \subset \mathbb{P}^{n-1}$ is a cone. The converse is also true.

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