can I use Affine cones for calculating singularities on projective quadrics?

Question

So I wanted to calculate the singularities of a quadric in the projective space. For example if you have the quadric

$$ Q=V(X_0^2+X_1^2+...+X_r^2), r<=n $$

Which lives in the projective space $\mathbb{P} ^n $.

My idea was to use the affine maps, intersect each of them with Q. Then Q is the union of these intersections and then I can calculate the singularities of the affine varieties with the jacobian criterion.

The other idea I had, was to somehow use the affine cone of the Quadric Q, but I don’t know the argument why I can use the cone.

Can anybody help me? :)

Answer 1

Original question: "For example if you have the quadric $Q=Z(X^2_0+X^2_1+X^2_2)$ Which lives in the three dimensional projective space. Can anybody help me? :)"

Answer: If you view $C:=V(x^2+y^2+z^2)\subseteq \mathbb{P}^2_k$ and consider $D(z)\cap C :=V((x/z)^2+(y/z)^2+1):=V(u^2+v^2+1)$ with $f:=u^2+v^2+1$, you find that the jacobian criterion proves ($char(k) \neq 2$) that $C\cap D(z)$ is regular. The following holds:

$$\partial f_u=2u, \partial f_v=2v$$

and there is no point $p\in C$ with $\partial f_v(p)=\partial f_u(p)=0$. Similar for $D(x),D(y)$. Hence the curve $C$ is regular in characteristic $\neq 2$. In characteristic $2$ you may write

$$x^2+y^2+z^2=(x+y+z)^2$$

hence the polynomial $f:=(x+y+z)^2$ is non-reduced and hence the curve $C:=V(f)$ is non-reduced and not regular.

In general if $X:=Proj(k[x_0,..,x_n]/I) \subseteq \mathbb{P}^n_k$ is a projective scheme defined by the homogeneous ideal $I:=(f_1,..,f_m)$ you get an affine open cover $U_i:=D(x_i)\cap X \subseteq X$ where $U_i=Spec(A_i)$ is an affine variety on the form $A_i=k[y_1,..,y_n]/J_i$. You may use the jacobian criterion to the ring $A_i$ and a set of generators of the ideal $J_i$.

Example: In your example of a quadric $Q:=x_0^2+\cdots +x_r^2$ you get on each open affine scheme $U_i$ a hypersurface $f_i \in k[y_1,..,y_n]$ and you must use the jacobi criterion to the polynomial $f_i$ to check non-singularity, similar to my example given above. It is a good exercise to generalize my calculation to more variables.

Answer 2

Yes, you can use the affine cone to check singularities. The key step is to prove the projective Jacobian criterion:

Projective Jacobian criterion: Let $Y\subset\Bbb P^n$ be a projective variety of dimension $r$ with homogeneous ideal generated by $f_1,\dots,f_t\in S=k[x_0,\cdots,x_n]$. If $P\in Y$ is a point with homogeneous coordinates $(a_0,\dots,a_n)$, $Y$ is nonsingular at $P$ if and only if the rank of the matrix $|\frac{\partial f_i}{\partial x_j}(a_0,\dots,a_n)|$ is $n-r$.

With this result, it's immediate that the affine cone on $Y$ (the variety in $\Bbb A^{n+1}$ with coordinates $x_0,\dots,x_n$ cut out by $f_1,\dots,f_t$) is nonsingular away from the vertex if and only if the projective variety is nonsingular, since the projective Jacobian criterion is exactly the usual affine Jacobian criterion for the cone away from the vertex.

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I think it's probably better if you prove the projective Jacobian criterion for yourself than if I do it for you. Here's an outline to follow:

• The projective Jacobian doesn't depend on the representation of $P$ as $(a_0,\dots,a_n)$.
• Relate the projective Jacobian to the affine Jacobian by passing to one of the standard affine opens $U_i=D_+(x_i)$ and dehomogenizing.
• Use Euler's lemma (if $f$ is a homogeneous polynomial of degree $a$ in $k[x_0,\cdots,x_n]$ then $\sum x_i\frac{\partial f}{\partial x_i} = af$) to deal with the extra row in the projective Jacobian matrix.

Feel free to leave a comment if you run in to trouble.