Jacobian criterion for projective varieties

Question

Let $P\in Y=Z(f_{1},\cdots ,f_{s})$ be a projective variety. Then $Y$ is non singular at $P$ if and only if rank of the matrix $\Vert\partial f_{i}/x_{j}(P)\Vert=n-\dim{Y}$.

I know of the statement for affine varieties, and I am trying to prove it for projective varieties. This is what I have so far.

If $P\in \mathbb{P}^{n}$ then $P$ is in some open set $U_{i}=\{(a_{0},\cdots ,1,\cdots a_{n})\in\mathbb{P}^{n}\}$ where the $1$ appears in the $i^{th}$ spot. WLOG we may assume $i=0$ is this case. So $X\cap U_{0}$ is an affine variety. Defined by $$X\cap U_{0}=Z(f_{1}(1,x_{1},\cdots x_{n}),\cdots ,f_{s}(1,x_{1},\cdots x_{n}))$$ Since this is affine we have that $X\cap U_{0}$ is nonsingular iff and only if the jacobian criterion is satisfied. I don't really see where to proceed from here since I don't know how the partials of the dehomoginzed polynomials compares to that of the homogeneous polynomials. Any suggestions would be very appreciated.

Answer 1

Use Eulers Lemma for a homogeneous polynomial $f$, of degree $d$ $$\sum_{i=0}^{n}x_{i}\dfrac{\partial f}{\partial x_{i}}=df $$

Answer 2

Consider the affine cone $C(Y)$ in $\mathbb{A}^{n+1}$; it is defined by the ideal$(f_1,\dots, f_r)$ of $k[x_0,\dots,k_{n+1}]$. Choose a point $Q$ of $C(Y)$ which is equivalent to the point of $Y$ in question. The affine Jacobi criterion applied to $C(Y)$ shows that $C(Y)$ is smooth at $Q$ if and only if the rank of the $r\times (n+1)$-Jacobi matrix $\Big(\frac{\partial f_i}{\partial x_i}(Q)\Big)$, which may be replaced with the rank of $\Big(\frac{\partial f_i}{\partial x_i}(P)\Big)$, is $(n+1)-\dim C(Y)$. One can prove (i) $\dim C(Y)=\dim Y +1$, and (ii)$\dim T_Q(C(Y))=\dim T_P(Y)+1$; in particular, $Y$ is smooth at $P$ if and only if $C(Y)$ is smooth at $Q$. This implies the desired, projective Jacobi criterion. To prove (ii), one may assume $Q=(1,1,\dots,1)$ though an automorphism, and should show, using Linear Algebra, that the rank of the Jacobi matrix above does not decrese by removing some (or eventually, any) column from the matrix.