Let \(H_i\) and \(H_j\) be the hyperplanes in \(\PP^n\) defined by \(x_i = 0\) and \(x_j = 0\), with \(i \neq j\). Show that any regular function on \(\PP^n \sm (H_i \intersect H_j)\) is constant.
Solution
Put \[ U_i=\PP^n\setminus H_i=D_+(x_i), \qquad U_j=\PP^n\setminus H_j=D_+(x_j). \] Then \[ \PP^n\setminus(H_i\cap H_j)=U_i\cup U_j. \]
1. Every regular function on \(U_i\) has the form\[ \frac{F}{x_i^a} \]for some \(a\ge0\) and some homogeneous polynomial \(F\in k[x_0,\ldots,x_n]\) of degree \(a\), and similarly on \(U_j\).
The standard chart \(U_i\) is isomorphic to \(\AA^n\) with affine coordinates \(x_\ell/x_i\) for \(\ell\ne i\). A regular function on \(U_i\) is therefore a polynomial in these ratios. Choose \(a\) at least the total degree of that polynomial and put all terms over the common denominator \(x_i^a\). Multiplying each numerator monomial by the required power of \(x_i\) produces a homogeneous polynomial \(F\) of degree \(a\). The same argument applies to \(U_j\).
2. A regular function on \(U_i\cup U_j\) is constant.
Let \(f\) be regular on \(U_i\cup U_j\). By step 1, choose homogeneous polynomials \(F,G\) with \[ f|_{U_i}=\frac{F}{x_i^a}, \qquad f|_{U_j}=\frac{G}{x_j^b}, \] where \(\deg F=a\) and \(\deg G=b\). On the nonempty overlap \(U_i\cap U_j\), these rational functions agree, hence in the polynomial ring \[ F x_j^b=G x_i^a. \] The variables \(x_i\) and \(x_j\) are relatively prime prime elements of the UFD \(k[x_0,\ldots,x_n]\). Therefore \(x_i^a\) divides \(F\). Since \(F\) is homogeneous of degree exactly \(a\), we have \[ F=cx_i^a \] for some \(c\in k\). Thus \(f=c\) on \(U_i\). Substituting in the equality on the overlap gives \(G=cx_j^b\), so \(f=c\) on \(U_j\) as well. Hence \(f\) is constant on their union.
3. Q.E.D.
Step 2 applies to every regular function on \(\PP^n\setminus(H_i\cap H_j)\).