Let \(k\) be an algebraically closed field of characteristic zero, and let \[ X=V(xy-z^2)\subseteq\AA^3_k. \] Compute the divisor class group \(\Cl(X)\).
Solution
Put \[ A=k[x,y,z]/(xy-z^2), \qquad X=\Spec A, \] and let \[ P=V(x,z)\subseteq X. \]
1. The ring \(A\) is a normal domain of dimension \(2\), and \(P\) is a primedivisor on \(X\).
The polynomial \[ xy-z^2 \] is irreducible, so \(A\) is a domain. Since \(A\) is a hypersurface in a three-dimensional polynomial ring, \[ \dim A=2, \] and \(A\) is Cohen–Macaulay, hence satisfies Serre’s condition \(S_2\).
The partial derivatives of \(xy-z^2\) are \[ y,\qquad x,\qquad -2z. \] Because \(\operatorname{char}k=0\), they vanish simultaneously only at the vertex \[ \mfm=(x,y,z). \] Thus the singular locus has codimension \(2\) in \(X\), so \(A\) is regular in codimension \(1\), i.e. it satisfies \(R_1\). Serre’s criterion \(R_1+S_2\) now shows that \(A\) is normal.
Finally, \[ A/(x,z)\cong k[y], \] so \((x,z)\) is a height-one prime. Hence \(P\) is a prime divisor.
2. The class group \(\Cl(X)\) is generated by \([P]\).
On the distinguished open set \[ D(x)\subseteq X, \] the relation \(xy=z^2\) gives \[ A_x \cong k[x,x^{-1},z], \qquad y=\frac{z^2}{x}. \] This is a UFD, so every Weil divisor on \(D(x)\) is principal.
Let \(D\) be any Weil divisor on \(X\). Its restriction to \(D(x)\) is therefore the divisor of some rational function \(g\in k(X)^\times\). Hence \[ D-\div_X(g) \] is supported on \[ V(x). \] But \[ \sqrt{(x)}=(x,z), \] so \(P\) is the only prime divisor contained in \(V(x)\). Consequently \[ D\sim nP \] for some \(n\in\ZZ\). Thus \([P]\) generates \(\Cl(X)\).
3. The principal divisor of \(x\) is\[ \boxed{\div_X(x)=2P.} \]
At the generic point of \(P\), the element \(y\) is a unit. Localizing first at \(y\) eliminates \(x\): \[ A_y \cong k[y,y^{-1},z], \qquad x=\frac{z^2}{y}. \] The prime \(P\) becomes the principal prime \((z)\) in this ring. Hence \(z\) is a uniformizer at \(P\), so \[ \operatorname{ord}_P(z)=1 \] and \[ \operatorname{ord}_P(x) = \operatorname{ord}_P(z^2/y) =2. \] Step 2 showed that \(P\) is the only prime divisor in \(V(x)\), so \(x\) has no other divisorial zeros. Therefore \(\div_X(x)=2P\).
4. The divisor \(P\) is not Cartier at the vertex\[ \mfm=(x,y,z). \]
The ideal of \(P\) is \[ I=(x,z)\subseteq A. \] If \(P\) were Cartier at the vertex, the localized ideal \(I_\mfm\) would be principal.
The ring \(A\) is standard graded with \[ \deg x=\deg y=\deg z=1. \] Every element of \(\mfm I\) has degree at least \(2\). Hence the degree-one classes of \(x\) and \(z\) are linearly independent in \[ I/\mfm I, \] and they span it. Thus \[ \dim_k I/\mfm I=2. \] By Nakayama’s lemma, \(I_\mfm\) needs at least two generators and is not principal. Therefore \(P\) is not Cartier at the vertex.
5. The class \([P]\) is nonzero and has order exactly \(2\).
Step 3 gives \[ 2[P]=0 \] because \(2P\) is principal.
4. Hence\[ [P]\neq0. \]Therefore \([P]\) has order exactly \(2\).
6. Consequently,\[ \boxed{ \Cl(X)\cong\ZZ/2\ZZ, } \]generated by the class of the ruling \(P=V(x,z)\).
By step 2, \([P]\) generates the class group. By step 5, it has order exactly \(2\). This gives the displayed isomorphism.
7. Q.E.D.
Step 6 is the requested class-group computation.