Hartshorne Exercise II.6.5.
Quadric Hypersurfaces. Let $\op{char} k \ne 2$, and let $X$ be the affine quadric hypersurface $\spec{ k[x_0,…,x_n]/(x_0^2+…+x_r^2)}$ cf. (I, Ex. 5.12).
(a) Show that $X$ is normal if $r\ge 2$ (use (Ex. 6.4) ).
(b) Show by a suitable linear change of coordinates that the equation of $X$ could be written as $x_0x_1 = x_2^2+…+x_r^2$. Now imitate the method of (6.5.2) to show that:
(1) If $r = 2$, then $\op{Cl} X\cong\Z/2\Z$;
(2) If $r = 3$, then $\op{Cl} X\cong\Z$ (use (6.6.1) and (Ex. 6.3) above);
(3) If $r\ge 4$ then $\op{Cl} X=0$.
(c) Now let $Q$ be the projective quadric hypersurface in $\PP{k}{n}$ defined by the same equation. Show that:
(1) If $r = 2$, $\op{Cl} Q\cong\Z$, and the class of a hyperplane section $Q.H$ is twice the generator;
(2) If $r = 3$, $\op{Cl} Q\cong\Z\bigoplus\Z$;
(3) If $r\ge 4$, $\op{Cl} X=\Z$, generated by $Q.H$.
(d) Prove Klein’s theorem, which says that if $r\ge 4$, and if $Y$ is an irreducible subvariety of codimension 1 on $Q$, then there is an irreducible hypersurface $V\subseteq\PP{k}{n}$ such that $V\cap Q=Y$, with multiplicity one. In other words, $Y$ is a complete intersection. (First show that for $r\ge 4$, the homogeneous coordinate ring $S(Q) = k[x_0,…,x_n]/(x_0^2+…+x_n^2)$ is a UFD.)
(a) Note that for any $r\ge 3$ the polynomial $f:=-x_1^2-…-x_r^2\in k[x_1,…,x_n]$ is squarefree and $x_0^2+…+x_r^2$ is irreducible. By problem 26 (Ex. 6.4), $k[x_0,…,x_n]/(x_0^2+…+x_n^2)$ is an integrally closed domain, and so a normal domain. Therefore $X$ is normal as being normal passes to localization. When $r=2$, $x_0^2+x_1^2+x_2^2$ can be written as $x_0x_1=x_2^2$ (as we will see in (b)) which is irreducible by Eisenstein’s, for example, and whose coordinate ring is again normal by problem 26 ($x_0x_1$ is squarefree).
(b) If $k$ has a fourth root of unity $i$ then the coordinate change $x_0\mapsto i\frac{x_0-x_1}{2}$ and $x_1\mapsto \frac{x_0+x_1}{2}$ does the trick.
If $r=2$, then $X=\spec{k[X,Y,Z,x_3,…,x_n]/(Z^2-XY)}$. The prime divisor $D={Y=0,Z=0}$ on X is in the support of $\op{div}Y$; the function $Y$ vanishes on $D$ with order 2 since in the local ring $(k[X,Y,Z,x_3,…,x_n]/(Z^2-XY))_{(Y,Z)}$, the maximal ideal is generated by $Z$ (in that ring, $Y=\frac{Z^2}{X}$) on which $Y=\frac{Z^2}{X}$ vanishes to order 2.
If $r=3$ then after another coordinate transform, our equation is $XY=ZW$. We have $X\cong \spec{k[X,Y,Z,W]/(XY-ZW)}\times \A{k}{n-r}$ and therefore $\op{Cl}(X)\cong \op{Cl}(\spec{k[X,Y,Z,W]/(XY-ZW)})$ by Prop. 6.6. But $\spec{k[X,Y,Z,W]/(XY-ZW)}$ is just the affine cone over the projective quadric surface in $\PP{k}{3}$, and so by Problem 29 (b) (Ex. II.6.3 (b)),
$$\op{Cl}(X)\cong\op{Cl}C(\PP{k}{1}\times\PP{k}{1})/\Z\cong\Z.$$
Now suppose $r\ge 4$. The coordinate ring of $X$ is $R=k[x_0,…,x_n]/(-x_0x_1+x_2^2+…+x_r^2)$. Consider the intersection $X\cap{x_0=0}=\spec{k[x_0,…,x_n]/(x_0,x_2^2+…+x_r^2})$. Since $x_0$ remains irreducible in $k[x_0,…,x_n]/(x_2^2+…+x_r^2)$, we see that $Y:=X\cap\{x_0=0\}$ is a prime divisor on $X$ with generic point $(x_0)$. Consider the rational function $x_0\in\Bbb K(X)$. In the local ring $R_{(x_0)}$, $x_0=\frac{1}{x_1}(x_2^2+…+x_r^2)$ generates the maximal ideal, on which $x_0$ only vanishes to order 1. that is, $\op{div}x_0=Y$. I claim that $Y$ generates $\op{Cl}X$, and then it follows that $\op{Cl}X=0$. Indeed, we have an SES,
$$\Z\xrightarrow{\cdot Y} \op{Cl}X\ra\op{Cl}(X-Y)\ra0.$$
Observe that $X-Y=\spec{R[\frac{1}{x_0}]}$. Then $R[\frac{1}{x_0}]\cong k[x_0,x_0^{-1},x_1,…,x_r]/(-x_1-\frac{1}{x_0}(x_2^2+…+x_r^2))\cong k[x_0,x_0^{-1},x_2,…,x_n]$ is a UFD. So, $\op{Cl}X$ is generated by $Y$.
(c) By Problem 29 (b), the only part that remains is for us to show that if $r=2$ then $\op{Cl}Q\cong \Z$ and $Q.H=2D$. I claim that the image of $Q.H$ in $\op{Cl}Q$
We have an SES,
$$0\ra\Z\xrightarrow{\cdot Q.H}\op{Cl}Q\ra\Z/2\Z=\{0,D\}\ra 0$$
The image of $Q.H$ in $\Z/2\Z$ is $0$. The right map is the pullback to $X$ followed by restriction to $X-P$ and inclusion in $X$… (TBC)
(d) By our previous work (ie $\op{Cl}X=0$ for $r\ge 4$), it is immediate that $S(Q)$ in this case is a UFD. Equivalently, every prime of height one in $S(Q)$ in principle. An irreducible subvariety of codimension 1 on $Q$ corresponds to a homogeneous prime of height one in $S(Q)$, thus generated by some irreducible homogeneous polynomial $\bar f\in S(Q)$. $\bar f$ is the image of some irreducible and homogeneous $f\in k[x_0,…,x_n]$, and the corresponding hypersurface $V={f=0}$ has exactly the property that $V\cap Q=Y$ with multiplicity 1.
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