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.

**Edit:**
Thanks a lot for the prompt answer! Now if I add the assumption that $I$ has generators consisting of irreducible polynomials, is there any hope that the formulae hold?

**2nd edit:**
Thanks again. For your first example, as far as I know, $\mathrm{rank}J_p(I)=2$ for *general* point $p\in C$. Please note that my question is that whether the formula holds generically.

I think we can exclude your second example by adding further assumption that $I$ contains no products of linear polynomials (e.g., $I$ is contained in a prime ideal which is generated by polynomials of degrees at least 2). Actually, I am interested in homogeneous ideal $I$ whose generators are of the same degree. With all of these additional assumptions, do you think that the formulae can be true?