TITLE: equations defining a subvariety
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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.