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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?

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For the Jacobian criterion it is fundamental to consider $I(V)$ and not just an ideal $I$ such that $V = V(I)$. For instance, take $I = ((x-1)^2)$. Then $V(I) = \{p\}\subset\mathbb{A}^1$. We have $rank (J_p(I)) = 0$. However, $dim(V) = 0\neq dim(\mathbb{A}^1)-rank (J_p(I))$.

The answer is negative also if the polynomials generating $I$ are irreducible. For instance, consider the quadric surface $Q$ given by $$ \det \left(\begin{matrix} x & y \\ y & z \end{matrix}\right) =0 $$ and the cubic surface $S$ given by $$ \det \left(\begin{matrix} x & y & z \\ y & z & w \\ z & w & x \end{matrix}\right) =0 $$ On a general point $p = [u^3:u^2v:uv^2:v^3]\in Y$ we have $Jac(Q)(p) = (uv^2,-2u^2v,u^3,0)$ and $Jac(S)(p) = (v^2(u^4-v^4),-2uv(u^4-v^4),u^2(u^4-v^4),0)$. Therefore, $\mathbb{T}_pQ = \mathbb{T}_pS$ for a general point $p\in Y$. This means that, if $Y\subset\mathbb{P}^3$ is the twisted cubic then $Q\cap S = Y$ set-theoretically. However, scheme-theoretically $Q$ and $S$ cut $Y$ twice.

Now, take the affine chart $w\neq 0$. Then, if $I =(xz-y^2,x^2z-x-y^2x+2yz-z^3)$ we have that $V(I)$ is the affine twisted cubic $C$. However, $rank(Jac_{(0,0,0)}(I)) = 1$. Then $$dim(\mathbb{A}^3)- rank(Jac_{(0,0,0)}(I)) = 2 \neq dim(C) = 1.$$

A simpler example: take the two curves $C = \{y-x^2 = 0\}$ and $L = \{y = 0\}$ in $\mathbb{A}^2$. Note that $L$ is the tangent line of $C$ in $(0,0)$. Both $C$ and $L$ are irreducible. Now, if $I = (y-x^2,y)$ then $V(I) = \{(0,0)\}$. On the other hand the scheme defined by $I$ is non-reduced. Indeed it is the origin with multiplicity two. In this case $rank(Jac_{(0,0)}(I)) = 1$, and $$dim(\mathbb{A}^2)- rank(Jac_{(0,0,0)}(I)) = 1 \neq dim(V(I)) = 0.$$

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