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:
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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?
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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 $k[x,1/x]$.
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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?
Best Answer
(1) Let $e_1 = (1:0:0)$, $e_2 = (0:1:0)$, $e_3 = (0:0:1)$. Another way to phrase this claim is: given any $3$ points $\newcommand{\P}{\mathbb{P}} \newcommand{\PP}{\mathbb{P}} P_1, P_2, P_3 \in \mathbb{P}^2$ that are not collinear, there is a linear automorphism $\varphi: \P^2 \to \P^2$ that takes $P_i \mapsto e_i$ for $i = 1, 2, 3$. This can be proved using just a little linear algebra.
Write $P_1 = (a_0:a_1:a_2), P_2 = (b_0:b_1:b_2), P_3 = (c_0:c_1:c_2)$, and let $B$ be the matrix whose columns are $P_1, P_2, P_3$: \begin{align*} B = \begin{pmatrix} | & | & |\\ P_1 & P_2 & P_3\\ | & | & | \end{pmatrix} = \begin{pmatrix} a_0 & b_0 & c_0\\ a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \end{pmatrix} \, . \end{align*} Then $B$ takes $e_i$ to $P_i$. Assuming $P_1, P_2, P_3$ are not collinear, then $B$ is invertible. Letting $A = B^{-1}$, then $A$ takes $P_i$ to $e_i$ for all $i$. Thus the map \begin{align*} \varphi: \P^2 &\to \P^2\\ \begin{pmatrix} x\\ y\\ z \end{pmatrix} &\mapsto A \begin{pmatrix} x\\ y\\ z \end{pmatrix} \, , \end{align*} or written in coordinates, $$ \varphi(x:y:z) = (A_{11} x + A_{12} y + A_{13} z : A_{21} x + A_{22} y + A_{23} z : A_{31} x + A_{32} y + A_{33} z) \, , $$ is an automorphism taking $P_i \mapsto e_i$ for $i = 1,2,3$. Or put differently, letting \begin{align*} X &= A_{11} x + A_{12} y + A_{13} z\\ Y &= A_{21} x + A_{22} y + A_{23} z\\ Z &= A_{31} x + A_{32} y + A_{33} z \end{align*} then the coordinates of $P_1, P_2, P_3$ with respect to the projective coordinates $X,Y,Z$ are $(1:0:0), (0:1:0), (0:0:1)$, respectively.
(2) We're trying to determine the coordinate ring of the affine conic $W \subseteq \mathbb{A}^2$. Considering $\mathbb{A}^2 \subseteq \P^2$ and taking the closure $\overline{W}$ of $W$ in $\P^2$, by the previous part, we have an isomorphism $\varphi: \overline{W} \overset{\sim}{\to} C$, where $C \subseteq \P^2$ is the conic $C: xy+yz+zx=0$. We can recover $W$ from $\overline{W}$ by removing the points at infinity, i.e., the points of intersection of $\overline{W}$ with the line $z=0$ at infinity. By Bézout's Theorem, $\overline{W}$ intersects $z=0$ in $2$ points, counted with multiplicity, so either $1$ or $2$ points. Thus $W = \overline{W} \setminus \{Q_1, Q_2\}$ or $W = \overline{W} \setminus \{Q_1\}$. Letting $R_i = \varphi(Q_i)$ for $i=1,2$, then restricting $\varphi$ yields an isomorphism $W \overset{\sim}{\to} C \setminus \{R_1, R_2\}$ or $W \overset{\sim}{\to} C \setminus \{R_1\}$.
Now over an algebraically closed field, every conic is isomorphic to $\P^1$. (Fix a point on the conic and consider the pencil of lines through this point, or see this post.) Denote this isomorphism by $\psi: C \overset{\sim}{\to} \P^1$. By post-composing with an automorphism of $\P^1$, we can assume that $R_1, R_2$ map to $\infty, 0$, or $R_1$ maps to $\infty$ under $\psi$. Composing the restrictions of $\varphi$ and $\psi$, we obtain either an isomorphism $\psi \circ \varphi: W \overset{\sim}{\to} \P^1 \setminus \{0, \infty\} \cong \mathbb{A}^1 \setminus \{0\}$ or $\psi \circ \varphi: W \overset{\sim}{\to} \P^1 \setminus \{\infty\} \cong \mathbb{A}^1$. These varieties have coordinate rings $k[x,1/x]$ and $k[x]$, respectively.
(3) Yes, this is true, although a degree $2$ hypersurface in $\P^n$ is usually called a quadric hypersurface, not a conic, when $n \geq 3$. See this post for a proof.