I have 2 questions about the dimension of a complex variety:

1) Suppose I have a variety $V$ in $\mathbb{C}^n$ and an ideal $I$ such that $V=V(I)$ ($I$ may not be radical, i.e., it may differ from $I(V)$). Can we detect the dimension of $V$ from the rank of the Jacobian matrix $J_p(I)$, where $p$ runs over $V$?

Of course, we have

$$\dim V=n-\min_p\mathrm{rank} J_p(\sqrt{I}),$$

where $p$ runs over the set of smooth points of $V$. So I wonder if we can replace $\sqrt{I}$ by $I$ in the above formula when $p$ runs over some subset of $V$.

2) With the above assumption, is it true that

$$\dim (V\cap(\mathbb{C}^*)^n)=n-\mathrm{rank} J_p(I)$$

for some $p\in V\cap(\mathbb{C}^*)^n$? (Here, of course, $\mathbb{C}^*=\mathbb{C}-\{0\}$).

I am grateful for any answer/comment.

  • $\begingroup$ For question 1, consider the hyperplane given by $V(z_1^2)$. The dimension of this variety should be $n-1$, but $rank J_p(I)=0$ at all points of $V(z_1^2)$. A similar trick might work for arbitrary $I$, but I need some time to think it through... $\endgroup$ – KReiser Sep 3 '14 at 22:38
  • $\begingroup$ Thanks a lot for your prompt reply! Now if I add the assumption that $I$ has generators consisting of irreducible polynomials, is there any hope that the formulae hold? $\endgroup$ – user173558 Sep 4 '14 at 0:55

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