Let $f$ be any irreducible quadratic polynomial in $k[x,y]$, where $k$ is an algebraically closed field, and let $W$ be the conic defined by $f$. Show that the coordinate ring $A(W)$ is isomorphic to $k[x]$ or $k[x,1/x]$. Which one is it when?
The above is exercise I.1.1 from Hartshorne's Algebraic Geometry. The solution reads:
Lemma: Any nonsingular conic in $\mathbb{P}^2$ can be reduced to the form $xy+yz+zx=0$ and this curve is isomorphic to $\mathbb{P}^1$.
Proof: Choose any $3$ points on the conic, and choose coordinates so that these points are $(1:0:0),(0:1:0),(0:0:1)$, this means the conic must have the equation $cxy+ayaz+bzx=0$, with $a,b,c$ all nonzero (otherwise the conic is singular). Then multiplying $x,y,z$ by $a,b,c$ shows that the conic has equation $xy+yz+zx=0$. Hence all nonsingular conics are isomorphic to this one, and as it is easy to find one isomorphic to $\mathbb{P}^1$ they are all. Therefore, regular function on a conic = regular functions on the conic $xy+xz+yz=0$ - some hypersurface = regular functions on $\mathbb{P}^1$ -1 or 2 points = $k[x]$ or $k[x,1/x]$. The ring is $k[x]$ iff the conic $ax^2+bxy+cy^2+$ terms of degree <2 intersects the line at infinity in exactly one point, which happens iff $b^2=4ac.$
I have a few questions:
Why we can choose coordinates such that these points are the three points of infinity? For example, is it always possible to choose coordinates such that the points are the points at infinity?
Why regular function on a conic = regular functions on the conic $xy+xz+yz=0$ - some hypersurface = regular functions on $\mathbb{P}^1$ -1 or 2 points = $k[x]$ or $k[x,1/x]$.
From the proof, we know that nonsingular conics are isomorphic for $\dim =2$ (they are isomorphic to $\mathbb{P}^1 \setminus$ a point) and I know any nonsingular conics in $\mathbb{P}^3$ are isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ via the sergre embedding. This makes me wonder is it true that all nonsingular conics over an algebraically closed field are isomorphic in all dimensions?