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. 



equations defining a subvariety
*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. 


