Let \(Y^r \subseteq \PP^n\) be a variety of degree \(2\). Show that \(Y\) is contained in a linear subspace \(L\) of dimension \(r + 1\) in \(\PP^n\). Thus \(Y\) is isomorphic to a quadric hypersurface in \(\PP^{r+1}\).
Solution
1. The variety \(Y\) has a nonsingular point.
By Theorem I.5.3, the singular locus of a variety is a proper closed subset. Hence its complement is a nonempty open subset of \(Y\). Choose \[ P\in\Reg Y. \]
2. Let \(X\) be the join of \(P\) with \(Y\) from Exercise I.7.7. Then\[ \dim X=r+1, \qquad \deg X=1. \]
Apply Exercise I.7.7 to the degree-two variety \(Y\) and the nonsingular point \(P\). It gives \[ \dim X=r+1 \] and \[ 0<\deg X<\deg Y=2. \] The degree of a nonempty projective variety is a positive integer by Proposition I.7.6(a), so necessarily \[ \deg X=1. \]
3. The variety \(X\) is a linear subspace\[ \boxed{L\cong\PP^{r+1}\subseteq\PP^n} \]containing \(Y\).
By definition of the join, \(Y\subseteq X\). Step 2 makes \(X\) a projective variety of degree one. By Exercise I.7.6, every pure-dimensional degree-one projective algebraic set is linear. Thus \(X\) is a linear subspace \(L\subseteq\PP^n\). Its dimension is \(r+1\) by step 2, and it contains \(Y\). This proves the first assertion.
4. Inside \(L\cong\PP^{r+1}\), the variety \(Y\) is a quadric hypersurface.
The inclusion \[ Y^r\subseteq L^{r+1} \] has codimension one. The homogeneous coordinate ring of \(L\cong\PP^{r+1}\) is a polynomial ring, hence a unique factorization domain. Therefore the homogeneous prime ideal of the irreducible codimension-one subvariety \(Y\subseteq L\) is principal, generated by an irreducible homogeneous polynomial \(F\).
For a hypersurface, Proposition I.7.6(d) identifies the degree of the hypersurface with the degree of its defining homogeneous polynomial. Since \[ \deg Y=2, \] we obtain \[ \deg F=2. \] Thus \(Y\) is a quadric hypersurface in \(L\cong\PP^{r+1}\).
5. Q.E.D.
Step 3 gives the required linear span of dimension \(r+1\), and step 4 identifies \(Y\) inside that span as a quadric hypersurface.