# Open affine subsets of projective space are principal

I have recently learned that open affine subsets of affine space $$\mathbb A_k^n$$ are principal, i.e. given by the nonvanishing locus of a single polynomial. What about open affine subsets of projective space? Are they also given by the nonvanishing locus of a homogeneous polynomial?

Take for example an open affine subset $$U \subset \mathbb P_k^2$$. Let $$U_0 = \{ [1,y_1,y_2]\} \cong \mathbb A_k^2$$, and similarly $$U_1$$ and $$U_2$$. The intersection $$U \cap U_0$$ is affine, since projective space is separated. So by KReiser's answer in the hyperlink, there exists a polynomial $$f_0(x_1,x_2)$$ such that

$$U \cap U_0 = D_{U_0}(f_0) = \{ [1,y_1,y_2] : f_0(y_1,y_2) \neq 0 \}$$ The homogenization of $$f_0$$ is $$g(y_0,y_1,y_2) = y_0^{d} f_0(\frac{y_1}{y_0}, \frac{y_2}{y_0})$$, where $$d$$ is the degree of $$f_0$$.

Now we can dehomogenize $$g$$ in two different ways, obtaining polynomials $$f_1(y_0,y_2)$$ and $$f_2(y_1,y_2)$$. The intersections $$U \cap U_1$$ and $$U \cap U_2$$ are affine varieties and therefore each is given as the nonvanishing locus of a single polynomial. I would like to say that such polynomials are respectively the dehomogenizations $$f_1$$ and $$f_2$$, since $$D_{\mathbb P^2}(g) \cap U_i = D_{U_i}(f_i)$$. This would imply

$$U \cap U_i = D_{\mathbb P^2}(g) \cap U_i$$ for $$i = 0,1,2$$, and hence $$U = D_{\mathbb P^2}(g)$$.

This looks mostly correct, but you may need to tweak your homogenization of $$f$$. Consider the case when your open set is $$D(x_0)$$: the $$f_0$$ you get on $$D(x_0)$$ is $$1$$, which doesn't homogenize correctly using your proof. Past things like that, I think your proof should work.

Alternatively, if you already know that a projective hypersurface is the zero locus of exactly one equation in $$k[x_0,\cdots,x_n]$$, then the lemma from the quoted post is enough:

Lemma: Let $$X$$ be a normal Noetherian connected separated scheme, and $$U\subset X$$ a proper nonempty affine open subset. Then $$X\setminus U$$ has pure codimension one.

The complement of your affine open will then be pure codimension 1 (a hypersurface), and one can then read off the appropriate polynomial.

The field of rational functions on $$\Bbb{P}^n$$ is $$\{ \frac{f}{g}\in Frac(k(x)), \exists d, f,g\in k[x]_d\}$$ Where $$x=(x_0,\ldots,x_n)$$ and $$k[x]_d$$ means homogeneous of degree $$d$$.

Because $$k[x]$$ is a UFD, factoring $$f,g$$ in irreducibles, we can find a representative such that $$f,g$$ have no common factor which means that $$\frac{f}{g}$$ is regular at $$a$$ iff $$g(a)\ne 0$$.

$$U$$ is an affine open subset of $$\Bbb{P}^n$$ means that $$U=Spec(O_X(U))$$ and $$O_X(U)$$ (the ring of rationals functions regular on $$U$$) is finitely generated as a $$k$$-algebra.

ie. $$O_X(U) = k[f_1/g_1,\ldots,f_m/g_m]$$ $$f_j,g_j$$ have no common factor, $$U=Spec(O_X(U))=\{ a\in \Bbb{P}^n, \text{ all the } f_j/g_j \text{ are regular at } a\}=\Bbb{P}^n- Z(\prod_{j=1}^m g_j)$$ The same argument gives that any (union of) hypersurfaces is the vanishing set of an homogeneous polynomial, and that $$\sum_j n_j Z(u_j)\to \sum_j n_j \deg(u_j)$$ is an isomorphism $$Pic(\Bbb{P}^n)\to \Bbb{Z}$$.