What is the affine ring of $\mathbb{P}^1_K - P$ where $P$ is a closed but possibly non-$K$-rational point. Let $K$ be a field, and let $P\in\mathbb{P}^1_K$ be a closed point. If $P$ is a $K$-rational point, then $\mathbb{P}^1_K - P = \text{Spec }K[x]$.
In general I believe $\mathbb{P}^1_K - P$ is still affine. How can we describe its affine ring?
 A: I claim that $P=(F)$ for an irreducible homogeneous polynomial $F\in K[x,y]$, and that therefore $\Bbb{P}^1_K-P=D_+(F)$ is in fact a distinguished open affine.
To see this, consider the affine cones involved. The affine cone of $\Bbb{P}^1_K$ is just $\Bbb{A}^2_K$, and the affine cone of $P$ will have codimension 1 in $\Bbb{A}^2$, and is therefore an irreducible hypersurface. Thus to show that the ideal of $P$ is generated by a single homogeneous element, it suffices to show that irreducible hypersurfaces in $\Bbb{A}^2$ are cut out by a single element.
Consider the ideal $(f,g)\subseteq K[x,y]$, with $f\nmid g$ and $g\nmid f$. If $\gcd(f,g)\ne 1$, then there is an irreducible $\pi$ with $\pi \mid f$ and $\pi \mid g$, and then we have that $(f,g)\subseteq (\pi)$, however $\pi$ cuts out an irreducible hypersurface, as does $(f,g)$, so we have in fact that $(f,g)=(\pi)$. On the other hand, if $\gcd(f,g)=1$, then working in $K(x)[y]$, we have that $Af+Bg=1$ for $A,B\in K(x)[y]$, or $af+bg=p(x)$, with $a,b\in K[x,y]$ and $p\in K[x]$ by clearing denominators. Similarly, there is $a',b'\in K[x,y]$ and $q\in K[y]$ with $a'f+b'g=q(y)$. Then we have that $(K[x]/(p))[y]/(q)$ is finite dimensional over $K$, and $(p(x),q(y))\subseteq (f,g)$, so $K[x,y]/(f,g)$ is finite dimensional over $K$, hence Artinian, and hence $\operatorname{Spec} K[x,y]/(f,g)$ is $0$ dimensional, contradicting our assumption that $(f,g)$ cut out an irreducible hypersurface. Thus all irreducible hypersurfaces in $\Bbb{A}^2_K$ are cut out by a single irreducible polynomial.
As a result, this implies that the affine cone of $P$ in $\Bbb{A}^2(K)$ is cut out by a single homogeneous polynomial, $F$, and $\Bbb{P}^1_K-P=D_+(F) = \operatorname{Spec} ((K[x,y]_F)_0)$
