Let \(k\) be an algebraically closed field of characteristic zero and let \(n\geq1\). Compute \(\Pic(\PP^n\slice{k})\).
Solution
Put \[ X=\PP^n_k \] and let \[ H=V(x_0)\subseteq X \] be a hyperplane.
1. Since \(X\) is smooth,\[ \Pic(X)\cong\Cl(X). \]
Projective space is smooth over \(k\). Hence every local ring of \(X\) is regular and every regular local ring is a UFD. Thus \(X\) is locally factorial, every Weil divisor is Cartier, and the natural map from Cartier divisor classes to Weil divisor classes is an isomorphism: \[ \Pic(X)\cong\Cl(X). \]
2. Every prime divisor \(D\subseteq X\) is a hypersurface\[ D=V(F) \]for an irreducible homogeneous polynomial\[ F\in k[x_0,\ldots,x_n]. \]
A prime divisor has codimension one. Its homogeneous ideal \[ I(D)\subseteq k[x_0,\ldots,x_n] \] is therefore a homogeneous prime ideal of height one. The polynomial ring is a UFD, so every height-one prime ideal is principal: \[ I(D)=(F) \] for an irreducible polynomial \(F\). Since the ideal is homogeneous, \(F\) may be chosen homogeneous.
3. If\[ D=V(F) \]with\[ d=\deg F, \]then\[ \boxed{D\sim dH.} \]
The rational function \[ \frac{F}{x_0^d} \] on projective space is well defined because numerator and denominator are homogeneous of the same degree. Its divisor is \[ \div\left(\frac{F}{x_0^d}\right) = D-dH. \] Thus \(D-dH\) is principal, so \[ D\sim dH. \]
4. The class \([H]\) generates \(\Cl(X)\).
Every Weil divisor is a finite integral linear combination of prime divisors. By step 3, each prime divisor is linearly equivalent to an integral multiple of \(H\). Hence every divisor class is an integral multiple of \([H]\).
5. Degree defines a homomorphism\[ \deg:\Cl(X)\longrightarrow\ZZ \]satisfying\[ \deg[H]=1. \]
For a prime divisor \[ D=V(F), \] define \[ \deg D=\deg F, \] and extend additively to all Weil divisors.
It remains to check that principal divisors have degree zero. Let \[ r\in k(X)^\times. \] Write \[ r=\frac{F}{G} \] with \(F\) and \(G\) homogeneous of the same degree \(d\). Factoring them into irreducible homogeneous polynomials gives \[ \deg\div(F)=d \qquad\text{and}\qquad \deg\div(G)=d. \] Therefore \[ \deg\div(r) = \deg\div(F)-\deg\div(G) =0. \] So degree descends to \(\Cl(X)\).
Since \(H=V(x_0)\) is a degree-one hypersurface, \[ \deg[H]=1. \]
6. The divisor class group is\[ \boxed{ \Cl(\PP^n_k)\cong\ZZ, } \]generated by \([H]\).
By step 4, the homomorphism \[ \ZZ\longrightarrow\Cl(X), \qquad m\longmapsto m[H] \] is surjective. By step 5, composing it with degree gives the identity on \(\ZZ\). Hence the homomorphism is also injective and therefore an isomorphism.
7. Consequently,\[ \boxed{ \Pic(\PP^n_k) \cong \ZZ, \qquad 1\longleftrightarrow\OO_{\PP^n}(1). } \]
Step 1 identifies \(\Pic(X)\) with \(\Cl(X)\), and step 6 identifies the latter with \(\ZZ\) generated by the hyperplane class. The line bundle associated with the hyperplane divisor is \[ \OO_{\PP^n}(H)=\OO_{\PP^n}(1). \] Thus every line bundle on \(\PP^n_k\) is uniquely of the form \[ \OO_{\PP^n}(m), \qquad m\in\ZZ. \]
8. Q.E.D.
Step 7 is the requested Picard-group computation.