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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 $U = D_{\mathbb P^2}(g)$.

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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.

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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 $Pic(\Bbb{P}^n)\to \Bbb{Z}$.

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