Chapter 1, Exercise 1.1: Classification of Affine Conics

(a) Let $Y$ be the plane curve $y=x^2$ (i.e. $Y$ is the zero set of the polynomial $f=y-x^2$). Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $k$.

(b) Let $Z$ be the plane curve $xy=1$. Show that $A(Z)$ is not isomorphic to a polynomial ring in one variable over $k$.

*(c) Let $f$ be any irreducible quadratic polynomial in $k[x,y]$, and let $W$ be the conic defined by $f$. Show that $A(W)$ is isomorphic to $A(Y)$ or $A(Z)$. Which one is it when?

(a) First note $I(Y)=(y-x^2)$ by the Nullstellensatz, then $A(Y)=k[x,y]/(y-x^2)$. Consider the surjective map $k[x,y]\to k[x]$ taking $f(x,y)\mapsto f(x,x^2)$. This has kernel $(y-x^2)$, so it induces an isomorphism $A(Y)=k[x,y]/(y-x^2)\cong k[x]$.

(b) We will show that $A(Z)\cong k[x,\frac 1 x]$. Note that this is not isomorphic to $k[x]$, for example because the polynomial $f=x$ is a nonconstant unit in $k[x,\frac 1 x]$. Once again, $I(Z)=(xy-1)$ and $A(Z)=k[x,y]/(xy-1)$. Consider the surjective map $k[x,y]\to k[x,\frac 1 x]$ taking $f(x,y)\mapsto f(x,\frac 1 x)$. This has kernel $(xy-1)$, so we induce the required isomorphism once again.

(c) This is Hartshorne's first starred exercise and as such the most elegant solution I am told requires some projective geometry. However, I have come up with an elementary solution following the classification of affine conics. Such a classification for algebraically closed fields of characteristic zero is a rather sluggish piece of algebraic manipulation, so the reader is referred to the beginning of chapter 5 of Gibson's Elementary Geometry of Algebraic Curves for the calculations. The result is simple enough, however: conics over such a field can be classified as affinely isomorphic to one of

• $y^2-x=0$ (parabolas);
• $y^2-x^2-1=0$ (conics);
• $y^2-x^2=0$ (line pairs);
• $y^2-1=0$ (repeated lines);
• $y^2=0$ (repeated line).

Irreducibility is preserved under these manipulations, and thus we can discard the three later cases in which the polynomial factors. The two remaining cases are parabolas and conics. The first is exactly $Y$ and thus of course has the same coordinate ring. The second equation can be written $$(x+y)(x-y)=1,$$ and under the algebra-preserving automorphism of $k[x,y]$ taking $x\mapsto x+y$ and $y\mapsto x-y$ this becomes $xy=1$ and thus corresponds to case b).