# Dimension of space spanned partial derivatives

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 $$dim(J(f))=n$$?

• If $n>3$, take $f=x_1^2+x_2^2+x_3^2$, which is irreducible if characteristic is not 2, but $\dim J(f)=3$. Commented Apr 4, 2019 at 13:36
• 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 $n$. Commented Apr 4, 2019 at 14:38
• @Mohan Sorry I didn't make myself clear. By $Z(f)$ I mean the zero locus in $\mathbb P^{n-1}$. Commented Apr 4, 2019 at 18:19
• My example above with $n=4$ has only isolated singularities in projective 3-space, so am not sure what you are asking. Commented Apr 4, 2019 at 18:22
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.