Assume \(\operatorname{char} k \neq 2\), and let \(f\) be a nonzero homogeneous polynomial of degree \(2\) in \(x_0, \ldots, x_n\).
Show that after a suitable linear change of variables, \(f\) can be brought into the form \(f = x_0^2 + \cdots + x_r^2\) for some \(0 \leq r \leq n\).
Show that \(f\) is irreducible if and only if \(r \geq 2\).
Assume \(r \geq 2\), and let \(Q\) be the quadric hypersurface in \(\PP^n\) defined by \(f\). Show that the singular locus \(Z = \Sing Q\) of \(Q\) is a linear variety of dimension \(n - r - 1\). In particular, \(Q\) is nonsingular if and only if \(r = n\).
In case \(r < n\), show that \(Q\) is a cone with axis \(Z\) over a nonsingular quadric hypersurface \(Q' \subseteq \PP^r\).
Here, if \(Y\) is a closed subset of \(\PP^r\) and \(Z\) is a linear subspace of dimension \(n - r - 1\) in \(\PP^n\), embed \(\PP^r\) in \(\PP^n\) so that \(\PP^r \intersect Z = \emptyset\); the cone over \(Y\) with axis \(Z\) is the union of all lines joining a point of \(Y\) to a point of \(Z\).
(The zero quadratic)
The source states part (1) for an arbitrary homogeneous quadratic while requiring \(0\le r\le n\) in the displayed normal form. The zero polynomial has rank zero and would correspond to an empty sum of squares, i.e. formally to \(r=-1\). Thus the stated range implicitly excludes \(f=0\); the card makes that hypothesis explicit.
Solution
Let \(V=k^{n+1}\). Because \(\operatorname{char}k\ne2\), the quadratic polynomial \(f\) has an associated symmetric bilinear form \[ B(u,v)=\frac12\bigl(f(u+v)-f(u)-f(v)\bigr) \] with \(f(v)=B(v,v)\).
1. There is a basis of \(V\) in which\[ \boxed{f=x_0^2+\cdots+x_r^2} \]for some \(0\le r\le n\).
Since \(f\ne0\), there is a vector \(v\) with \(f(v)\ne0\). Indeed, if \(f(v)=0\) for every \(v\), then polarization would give \[ 2B(u,v)=f(u+v)-f(u)-f(v)=0 \] for all \(u,v\), hence \(B=0\) and \(f=0\), a contradiction.
The line \(kv\) is nondegenerate for \(B\), so \[ V=kv\oplus v^\perp. \] Thus in a basis beginning with \(v\), the quadratic form is \[ a_0x_0^2+f_1(x_1,\ldots,x_n), \qquad a_0\ne0. \] Repeat on the restriction to \(v^\perp\) until the remaining restriction is zero. This gives a diagonal form \[ a_0x_0^2+\cdots+a_rx_r^2 \] with every \(a_i\ne0\). Since \(k\) is algebraically closed, choose square roots \(b_i^2=a_i\) and rescale \(x_i\) by \(b_i\). The resulting form is the displayed sum of squares. The number \(r+1\) is the rank of the symmetric matrix of \(B\), so it is independent of the choices. This proves part (1).
2. The polynomial \(f=x_0^2+\cdots+x_r^2\) is irreducible exactly when\[ \boxed{r\ge2.} \]
If \(r=0\), then \(f=x_0^2\) is reducible. If \(r=1\), algebraic closedness gives an element \(i\in k\) with \(i^2=-1\), and \[ x_0^2+x_1^2=(x_0+i x_1)(x_0-i x_1). \] The two factors are distinct because \(\operatorname{char}k\ne2\).
Conversely, suppose \(r\ge2\) and \(f\) were reducible. As a homogeneous quadratic, it would factor as a product of two nonzero linear forms \[ f=\ell m. \] Write their coefficient column vectors as \(a,b\in k^{n+1}\). The symmetric matrix of the quadratic form \(\ell m\) is \[ \frac12(ab^t+ba^t), \] whose image is contained in the span of \(a\) and \(b\) and therefore has rank at most two. But the matrix of \[ x_0^2+\cdots+x_r^2 \] has rank \(r+1\ge3\). This contradiction proves irreducibility for \(r\ge2\), establishing part (2).
3. Assume \(r\ge2\) and put\[ Q=V_+(x_0^2+\cdots+x_r^2)\subseteq\PP^n. \]Its singular locus is\[ \boxed{\Sing Q=V_+(x_0,\ldots,x_r).} \]
The partial derivatives of the defining quadratic are \[ 2x_0,\ldots,2x_r,0,\ldots,0. \] Since \(2\) is invertible, they vanish simultaneously exactly when \[ x_0=\cdots=x_r=0. \] Every such projective point automatically lies on \(Q\). By the projective Jacobian criterion P-AGH58JACOBIANRANK, this is precisely the singular locus.
If \(r<n\), it is the projectivization of the coordinate vector subspace spanned by \[ e_{r+1},\ldots,e_n, \] so it is a linear variety of dimension \[ n-r-1 \] by P-AGH211LINEAR. If \(r=n\), the displayed simultaneous equations have no projective solution, so the singular locus is empty. Consequently \[ \boxed{Q\text{ is nonsingular }\Longleftrightarrow r=n.} \] This proves part (3), with the usual convention that the empty singular locus corresponds to dimension \(-1\) in the formula.
4. If \(r<n\), let\[ Q'=V_+(x_0^2+\cdots+x_r^2)\subseteq\PP^r \]and\[ Z=V_+(x_0,\ldots,x_r)\subseteq\PP^n. \]Then \(Q\) is exactly the cone over \(Q'\) with axis \(Z\).
Embed \[ \PP^r=V_+(x_{r+1},\ldots,x_n) \] in \(\PP^n\). It is disjoint from \(Z\). The quadratic \(Q'\) has full rank \(r+1\) on this \(\PP^r\), so step 3 applied there shows that \(Q'\) is nonsingular.
Take a point \[ q=[a_0:\cdots:a_r:0:\cdots:0]\in Q' \] and a point \[ z=[0:\cdots:0:b_{r+1}:\cdots:b_n]\in Z. \] Every point on their joining line has coordinates \[ [\lambda a_0:\cdots:\lambda a_r: \mu b_{r+1}:\cdots:\mu b_n]. \] Substitution into the equation of \(Q\) gives \[ \lambda^2(a_0^2+\cdots+a_r^2)=0, \] so the whole line lies in \(Q\). Thus the cone described in the statement is contained in \(Q\).
Conversely, let \[ p=[a_0:\cdots:a_r:b_{r+1}:\cdots:b_n]\in Q. \] If all \(a_i\) vanish, then \(p\in Z\). Otherwise \[ q=[a_0:\cdots:a_r]\in Q' \] because the equation of \(Q\) only involves the \(a_i\). If all \(b_j\) vanish, then \(p=q\in Q'\). Otherwise let \[ z=[b_{r+1}:\cdots:b_n]\in Z. \] The point \(p\) lies on the line joining \(q\) and \(z\). Hence every point of \(Q\) lies in the cone. The two sets are equal, proving part (4).
5. Q.E.D.
Step 1 proves the normal form, step 2 proves irreducibility, step 3 computes the singular locus, and step 4 proves the cone description.