Question

The following question feels to me like a standard sort of 'fact' in birational geometry, but I can't seem to write down a correct set of details. Hopefully someone can point me to a reference and not a counter example!

Suppose $X$ is a variety (reduced and irreducible over an algebraically closed field, perhaps of characteristic zero) and suppose that there exist a very ample line bundle $L$ and a linear system $V \subset H^0(X,L)$ such that $Y = Bs(V)$ is the singular set of $X$ scheme theoretically, that $Y$ is smooth of codimension at at least 2, and that $\tilde X$, the blow up of $X$ along $Y$ is smooth. Further assume that $\phi_{|V|} X--> S$ birationally maps $X$ onto a smooth variety $S$. Let $\tilde \phi$ be the map from $\tilde X \to S$ induced by $V$. Further assume that, denoting by $f$ the map $\tilde X \to X$, that $f^{-1}(Y) = T$ surjects onto $S$. Let $v_1, \dots v_s$ be $s = \dim(S)$ general sections of $V$ so that the intersection $Z(v_1) \cap \dots Z(v_s) \cap S$ consists of finitely many smooth points say $p_1, \dots p_m $.

Also assume the $P = f( \tilde \phi^{-1}(\cup_{i=1:m} p_i))$ is a proper subset of $Y$. Then can one say that away from $P$, the sections $ v_1 \dots v_s$ generate the ideal of $Y$ in $X$ ?

The case I have in mind is where $Y$ is a smooth curve embedded in a sufficiently ample manner so that 1) $Y$ is defined by quadrics and 2) $X = Sec(X)$ is singular only along $Y$. Then $V$ would be the quadrics through $Y$. The point would be to use this sort of an argument to establish a minimum depth of $Sec(Y)$ along $Y$.

This is my first question, so please feel free to correct etiquette with this question as well as the mathematics.

Answer 1

Credit: This answer came out of trying to understand why auniket's answer (a.k.a. counterexample) works.

1) auniket is correct that for dimension reasons $T$ cannot surject onto $S$, so in particular my comment about $X$ being normal possibly helping is irrelevant. So is $T$.

2) It seems to me that there is a much more general problem with your desired statement. Namely I believe the following is true:

Claim: Under the conditions of the question, if in addition $\dim Y=0$ and $Y$ is reduced, then the desired statement cannot be true.

Proof: We may assume that $Y$ is a single point. Since by assumption $X$ is singular at $Y$, the local ring of $X$ at $Y$ is not a regular local ring. Therefore the ideal of $Y$ cannot be generated by $\dim X$ number of elements. On the other hand, by assumption $X$ is birational to $S$, so $\dim X=\dim S=s$. Therefore $v_1,\dots,v_s$ cannot generate the ideal of $Y$. $\square$

Note: I think this actually covers both of auniket's examples and would definitely give an arbitrary number of normal examples.

3) It seems that this still leaves a sliver of hope for you as your $Y$ is a curve (and even in the zero-dimensional case if $Y$ is non-reduced, it could work out). However, if it is reduced then you are at the absolute minimal number of generators that the singularity condition allows.

Answer 2

Note that $T := \tilde \phi(f^{-1}(Y))$ is a proper Zariski closed subset of $S$. Therefore, for generic $v_1, \ldots, v_s \in V$, $Z(v_1) \cap \cdots \cap Z(v_s) \cap T = \emptyset$. Consequently, in this case $P \cap Y = \emptyset$ and your question boils down to whether $v_1, \ldots, v_s$ generate the ideal of $Y$ (in a neighborhood of $Y$ in $X$), which should in general be false. Am I missing something?

For example, let $X := Z(x_0^2x_2 - x_1^3) \subseteq \mathbb{P}^2$. The singular set of $X$ is $Y := \lbrace(0:0:1)\rbrace$ (with respect to homogeneous coordinates $(x_0: x_1: x_2)$ of $\mathbb{P}^2$). Let $L$ (resp. $V$) be the linear system with basis $x_0, x_1, x_2$ (resp. $x_0, x_1$). Then $S = \mathbb{P}^1$ and $T = \lbrace(0:1)\rbrace$. Therefore, if we take $v_1 := a_0x_0 + a_1x_1$ with $a_1 \neq 0$, then $Z(v_1) \cap T = \emptyset$ and consequently $P \cap Y = \emptyset$. Let $U$ be the affine neighborhood of $Y$ in $\mathbb{P}^2$ with coordinates $u_0 := x_0/x_2$ and $u_1 := x_1/x_2$. Then ideal of $Y$ on $U \cap X$ is $\mathcal{I} := \langle u_0, u_1 \rangle$ and the ideal generated by $v_1$ is $\mathcal{J} := \langle a_0u_0 + a_1u_1 \rangle$. Since the ideal in $\mathbb{C}[x,y]$ generated by $a_0u_0 + a_1u_1$ and $u_0^2 - u_1^3$ does not equal the ideal generated by $u_0$ and $u_1$, it follows that $\mathcal{I} \neq \mathcal{J}$.

Edit: The heuristics in the first paragraph remains valid in the case that $X$ is normal. Below I give an explicit example where $X$ is a normal surface. I don't know anything about secant varieties to comment about the validity of the statement in that case.

Let $X$ be the weighted projective space $\mathbb{P}^2(1, 1, 2)$. We view $X$ as the toric surface corresponding to the polygon $\mathcal{P}$ which is the triangle in $\mathbb{R}^2$ with vertices $(0,0)$, $(2,0)$ and $(0,4)$. Let me draw $\mathcal{P} \cap \mathbb{Z}^2$.

 
| | | | |
x-o-o-o-o-
| | | | |
x-o-o-o-o-
| | | | |
x-x-o-o-o-
| | | | |
x-x-o-o-o-
| | | | |
x-x-x-o-o-

Here I marked the integral points which belong to $\mathcal{P}$ by 'x' and the others by 'o' (the coordinates of the point at the bottom-left corner being $(0,0)$). Since $|\mathcal{P} \cap \mathbb{Z}^2| = 9$, it follows that $X$ is isomorphic to a subvariety of $\mathbb{P}^8$. Denote the homogeneous coordinates of $\mathbb{P}^8$ by $z_\alpha$ for all $\alpha \in \mathcal{P} \cap \mathbb{Z}^2$. Then the equations of $X$ in $\mathbb{P}^8$ determined by relations between $x_1^{\alpha_1}x_2^{\alpha_2}$ for all $\alpha := (\alpha_1, \alpha_2) \in \mathcal{P} \cap \mathbb{Z}^2$.

Let $L$ be the linear system with basis $\lbrace z_\alpha \rbrace$ and $V$ be the subspace of $L$ with basis $\lbrace z_\alpha : \alpha \neq (2,0) \rbrace$. Then you can check that $Y := BS(V)$ (as a set) consists of the only singular point of $X$ and the blow-up $\tilde X$ of $X$ along $Y$ is non-singular. Moreover, $f^{-1}(Y)$ is a curve. Finally, $S$ is the toric surface corresponding to the polygon

 
     | | | | |
     x-o-o-o-o-
     | | | | |
     x-o-o-o-o-
Q := | | | | |
     x-x-o-o-o-
     | | | | |
     x-x-o-o-o-
     | | | | |
     x-x-o-o-o-

It follows that $S$ is non-singular, and $\dim (\tilde \phi(f^{-1}(Y))) \leq 1$. Therefore, for generic $v_1, v_2 \in V$, $Z(v_1) \cap Z(v_2) \cap \tilde \phi(f^{-1}(Y)) = \emptyset$, and consequently, $P \cap Y = \emptyset$. We claim that there is a neighborhood $U$ of $Y$ such that the ideal of $Y$ on $U$ can not be generated by $2$ elements.

Indeed, let $U := X \setminus Z(z_{(2,0)})$. Then $U \cong \text{Spec}~ \mathbb{C}[x^{-1}, x^{-1}y, x^{-1}y^2] \cong \text{Spec}~ (\mathbb{C}[u,v,w]/\langle uw - v^2 \rangle)$ and $Y = Z(u,v,w) \subseteq U$. Since $uw - v^2$ is a homogeneous polynomial of degree $2$, the ideal generated by $u$, $v$ and $w$ in $\mathbb{C}[u,v,w]$ does not equal the ideal generated by $uw-v^2$, $g_1$ and $g_2$ for all $g_1, g_2 \in \mathbb{C}[u,v,w]$. This proves the claim and completes the counter example.