equations defining a subvariety

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.

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A few questions, 1) how do you define the scheme structure on $Y$? 2) are you assuming that $\overline\phi$ is everywhere defined? and a comment: denoting the zero set by $V(\ )$ when you take you sections from $V$ is not necessarily the best idea. Perhaps you could use $Z(\ )$. Also, checkout the command \dashrightarrow. Cheers. – Sándor Kovács Mar 21 '11 at 6:30
Hi Sandor, Thanks for the clarifying questions. 1) I am assuming that $X$ is embedded in a projective space such that the given line bundle $L$ is a multiple of the '$\mathcal(O)(1)$'. 2) I thought that blowing up the base locus meant that automatically $\tilde \phi$ was every where defined. I will edit later when I have more time. Thanks. Adam – aginensky Mar 21 '11 at 11:06
Perhaps this is a second question or a prequestion, but are you saying that in general if the base locus is reduced, there could be a proper subscheme, supported exactly on the base locus, such that blowing up the subscheme resolves the indeterminancies of the map? In the case I am proposing, the base locus is smooth, so that could not happen. – aginensky Mar 22 '11 at 23:31
OK, I think we're talking about the same thing and I managed to confuse things with my badly formed questions. I missed the condition that the base scheme is smooth (and hence reduced). – Sándor Kovács Mar 23 '11 at 6:17
yes I did- i've changed the question. – aginensky Mar 24 '11 at 21:11

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.

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Well, I apologize for not getting back to this sooner, but at least one of you is aware of the fact that mathematics is my avocation and not my vocation. I'm afraid I was impossibly busy with other things and could not get time to think about this. You are both right in your examples. I now think that it is unlikely to ever be true in the way I have phrased the question. Thanks for your responses. Adam – aginensky May 17 '11 at 1:53

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 $\mathcal{P} \cap \mathbb{Z}^2$.


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x-o-o-o-o-
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x-o-o-o-o-
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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


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

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This is interesting, but the main reason this works is because $f$ is finite, i.e., because $X$ is not normal. Of course, it was not assumed to be normal, but I think it would be an interesting question with that additional assumption. – Sándor Kovács Mar 26 '11 at 3:38
@ auniket & Sandor- I was specifically interested in the case I mentioned where $X$ is the secant variety to a curve $Y$. In that case what you call $T$ does indeed surject onto $S$. I am going to edit the question to reflect the fact that the base locus is of codimension 2 and we have the surjection. As for normality, I am trying to prove that under appropriate conditions of ampleness, the secant variety is normal. I want the "right" number of sections to locally define $Y$ so that we may conclude that $X$ has depth at least $s$. More in next comment. – aginensky Mar 26 '11 at 15:28
I asked the question in the most general form I could think of, because I always find that arguments based on special cases are more suspect. Nonetheless I would be happy if in general this turns out to be false, (but as Sandor said, intuitively it is plausible) but using some special properties of the secant variety set up it was true then. Nonetheless, I am going to think about auniket's example to see if I can see why it makes my statement false for secant varieties. Adam (still) – aginensky Mar 26 '11 at 15:31
@Sandor: I think the argument remains valid regardless of whether $X$ is normal or not. I have edited the answer and included an example when $X$ is a normal surface. – auniket Mar 26 '11 at 19:33
@auniket: I think you are right that $T$ cannot surject onto $S$ and also both of your examples are nice! (It took me a while to figure out what was going on, but I think I understand now). Definitely a +1! – Sándor Kovács Mar 27 '11 at 2:47