Singular locus of analytic subvarieties In Griffiths and Harris page 21, it is proven that the singular locus, denoted $V_{s}$ is contained in an analytic subvariety of the complex manifold $M$ not equal to $V$ which is the analytic variety. We define $V_{s} = V - V^{*}$ where $V^{*}$ is the locus of smooth points of $V$.

For $p \in V$ let k be the largest integer such that there exist k functions $f_{1},...,f_{k}$ in a neighborhood $U$ of $p$ vanishing on V and such that $J(f)$ has a $k\times k$ minor not everywhere singular on V; we may assume that $|(\frac{\partial f_{i}}{\partial z_{j}})_{1 \leq i,j \leq k}| \neq 0$ on V. Let $U' \subset U$ be the locus of $|(\frac{\partial f_{i}}{\partial z_{j}})_{1 \leq i,j \leq k}| \neq 0$ and $V'$ the locus $f_{1} = ... = f_{k} = 0$. Then $V' = V \cap U'$ is a complex submanifold of $U'$, and for any holomorphic function $f$ vanishing on V the differential $df = 0$ on $V'$, i.e., $f$ is constant on $V'$. It follows that for $q \in V'$ near $p$, $V=V'$ is a manifold in a neighborhood of q and so $V_{s} \subset |(\frac{\partial f_{i}}{\partial z_{j}})_{1 \leq i,j \leq k}| = 0$

I understand the proof up until the last statement. I believe that the holomorphic function $f$ vanishes because we stated that the integer $k$ was the largest integer and so any other such functions must have a singular $k \times k$ minor. However, I don't understand how it follows that $V=V'$ is a manifold near p and how it follows from this that $V_{s}$ is contained in the locus defined by the vanishing of the $k\times k$ minor. 
Thanks
 A: I have also some trouble with this statement and the proof given. First the statement seems trivial for stupid reasons: suppose $V$ is not equal to $M$, then $M$ is an analytic subvariety of $M$ that contains $V_s$ and is not equal to $V$. The intended meaning must that the subvariety containing $V_s$ must be a proper subset of $V$ (and I think there should be a definite statement that its dimension is less than that of $V$). Also the statement before the last one in the proof does not make much sense to me : as defined $V'$ is a subset of $V$, so if $f$ vanishes on $V$, it is for sure constant on $V'$! I suspect some sort of misprint is the cause of the confusion here.
A clearer statement is suggested in the sentence following the proof, and would be :
The set $V_s$ is contained in a proper analytic subset of $V$.
To prove this, following more or less, the authors' approach, let $f_1,\dots,f_k$ be a defining set of analytic functions for $V$ in a neighborhood of $p$, and assume $k$ is minimal for this property. Let then $k'$ be the rank of $f_1,\dots,f_k$ at $p$. Note that if $k'=k$ then by the implicit function theorem $V$ is smooth at $p$ and of dimension $n-k$ (where $n$ is the dimension of $M$). Conversely if $V$ is smooth at $p$ then (by definition) there is a system $f_1,\dots,f_k$ of rank $k$ that defines $V$ in the neighborhood of $p$, and clearly $k$ is the only integer satisfying this propery, i.e. $k=n-\dim V$.
Therefore, a necessary and sufficient condition for $p$ to be in $V_s$ is that  $k' < k$ at that point. (It may help to consider the simple example $xy=0$ at $(0,0)$). Now it follows that $V_s$ must be contained in a neighborhood of $p$ in the locus of the set of functions formed the determinants of all the $k\times k$-minors of $f_1,\dots,f_k$. It remains to show that this locus intersected with $V$ is a proper analytic subset of $V$.
It seems pretty obvious that an analytic variety cannot be everywhere singular, and there ought to be a simple argument but I don't have it off-the-head - this follows from the structure theorem for these varieties, but going through this theorem was what the authors tried to avoid here.
A: I was confused about this as well. Note the following: the set $V'$ is defined as the locus of $f_1=\cdots=f_k=0$. From this it follows $V\subset V'$. Now, pick $q\in V \cap U'$. Let $V_q \subset V$ and $V'_q\subset V'$ be the connected components of $q$, in the sets $V$ and $V'$, respectively. Clearly, we have $V_q\subset V'_q$. Now, as you said, any function $f\in I(V)$ must be constant in $V'$. So, since $f(q)=0$, it follows that $f\in I(V'_q)$. In other words, $V'_q \subset V$. Since $V\subset V'$, this implies $V_q=V'_q$. Finally, remember that $V'_q\cap U'$ is a $k$-dimensional submanifold by definition, but this set is the same as $V_q\cap U'$. This shows that $q$ is a regular point of $V$.
