Isomorphism between projective varieties $\mathbf{P}^{1}$ and a conic in $\mathbf{P}^{2}$

Question

I'm trying to establish an isomorphism between the projective line $\mathbf{P}^{1}$ and the conic in $\mathbf{P}^{2}$ defined by $Y=Z(g)$, where $g=x^2+y^2-z^2$. This is part of exercise I.3.1 in Hartshorne's Algebraic Geometry.

I defined $\varphi : \mathbf{P}^{1} \rightarrow Y$ by $\varphi (a,b) = (a^2 -b^2,2ab,a^2+b^2)$ and showed that it's a morphism of varieties. I'm having trouble defining the inverse map. First, I tried using $\psi : Y \rightarrow \mathbf{P}^{1}$ defined by $\psi (a,b,c) = (b,c-a)$ to get $$\psi\varphi(a,b)=\psi(a^2 -b^2,2ab,a^2+b^2)=(2ab,2b^2)=(a,b) \text{ if }b\neq0$$ $$\varphi\psi(a,b,c)=\varphi(b,c-a)=(2a(c-a),2b(c-a),2c(c-a))=(a,b,c)\text{ if }a\neq c$$ This is problematic because of the restrictions on $a,b,c$, and also because $(1,0,1)\in Y$ but $\psi(1,0,1)=(0,0)\notin \mathbf{P}^{1}$. Then I tried using $\psi : Y \rightarrow \mathbf{P}^{1}$ defined by $$\psi (a,b,c) = \begin{cases}(b,c-a) \text{ if } a\neq c \\ (1,0) \text{ if } a=c \end{cases}$$ This is troublesome because now I can't show continuity: if $Z(T)$ is a closed set in $\mathbf{P}^{1}$, then $\psi^{-1}(Z(T))$ is a closed set in $Y$. (I need $\psi^{-1}(Z(T))$ to be the zero set of some homogeneous polynomials in $k[x,y,z]$ of positive degree, where $k$ is a field of characteristic zero.)

Does anyone know how to establish the definition and continuity of the inverse map to $\varphi$? I will only be able to understand answers within the scope and context of Hartshorne up to chapter I section 3. Thanks.

Answer 1

1) The isomorphism $\phi$ is between $\mathbb P^1$ and a conic $Y\subset\mathbb P^2$, not a cone.
(This conic is related to the cone in $\mathbb A^3$ with the same equation, but you should not confuse these two algebraic varieties).
2) As in all categories there are morphisms between varieties but you must keep in mind a most important point: those morphisms are defined locally and there is no hope that a morphism can be given by a single "formula": see the "Morality" below.
3) Your question beautifully illustrates the difficulty in 2) and allows for a pedagogical illustration of how to correctly define the inverse morphism $\psi=\phi^{-1}:Y\to \mathbb P^1$, namely:

a) The conic $Y$ has an open (in the Zariski topology!) covering $\{U,V\}$, where $U=Y\setminus \{[-1:0:1]\}$ consists of the points $p=[x:y:z]\in Y$ satisfying $y\neq 0$ or $x+z\neq0$ and $V=Y\setminus \{[1:0:1]\}$ consists of the points $q=[x:y:z]\in Y$ satisfying $z-x\neq 0$ or $y\neq 0$.
b) We have a morphism $\psi_ U:U\to \mathbb P^1:[x:y:z]\mapsto [y:x+z]$
and a morphism $\psi_ V:V\to \mathbb P^1:[x:y:z]\mapsto [z-x:y]$
c) The morphisms $\psi_ U$ and $\psi_ V$ coincide in the (huge!) open intersection $U\cap V\subset Y$and thus together define the required isomorphism $\psi=\phi^{-1}:Y\to \mathbb P^1$.
d) Notice that our maps $\psi_ U$ and $\psi_ V$ are morphisms (and thus continuous) but to state merely that they are continuous in the Zariski topology is worthless: for example any set-theoretic bijection between curves is a homeomorphism but is absurdly far from being an isomorphism in general.

Morality
The morphism $\psi$ is a perfectly defined inverse of $\phi$, but it is impossible to write it in the form $\psi([x:y:z])=[f(x,y,z):g(x,y,z):h(x,y,z)]$ with homogeneous polynomials of the same degree $f(x,y,z),g(x,y,z), h(x,y,z)\in k[x,y,z]$.

Answer 2

I don't know how you got your map $\varphi: \mathbb{P}^1 \to Y$, but here's how I would've gotten it. In the affine chart where $Z \neq 0$, the conic is the circle $C: x^2 + y^2 = 1$. Consider the pencil of lines $y = t(x+1)$ through the point $(-1,0)$. Each line intersects the circle in exactly one other point (by Bézout's Theorem), and each point (except for $(-1,0)$) is hit by exactly one line. Substituting $y = t(x+1)$ into $x^2 + y^2 = 1$ and doing some algebra gives us the parametrization $\varphi: \mathbb{A}^1 \to C$, $t \mapsto \left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right)$. Homogenizing this gives the map $\varphi$ you've defined above.

How do we go the other way? Well, we have $y = t(x+1)$, which allows us to solve $t = \frac{y}{x+1}$. Homogenizing yields the map $\psi: Y \to \mathbb{P}^1$ given by $[S:T] = [Y: X + Z]$, or $(b : a + c)$ in your notation.

To see that this map is well-defined and continuous at the point $[-1 : 0 : 1]$, we can rewrite it in a slightly different form, using the equation $X^2 + Y^2 = Z^2$: \begin{align*} [Y : X+Z] &= [Y^2 : (X+Z)Y] = [Z^2 - X^2 : (X+Z)Y]\\ &= [(Z+X)(Z-X) : (X+Z)Y] = [Z-X : Y] \, . \end{align*} From this definition we see that the point $[-1 : 0 : 1]$ maps to $[2 : 0] = [1: 0]$. This map is given by homogeneous polynomials, so it is continuous, and the two definitions agree on the intersection of the open sets on which they are defined.

Answer 3

Let $C$ be an irreducible conic in $\mathbb{P}^2$. Upto change of coordinates, we can assume that it is given by $xz = y^2$.

Now we set $f:\mathbb{P}^1 \rightarrow C\subset \mathbb{P}^2, \: [s:t]\mapsto [s^2:st:t^2]$. The inverse $g:C \rightarrow \mathbb{P}^1$ is defined as follows: \begin{equation} [x:y:z] \mapsto \begin{cases} [x:y], & (x,y) \neq (0,0)\\ [y:z], & (y,z) \neq (0,0)\\ \end{cases} \end{equation}

Answer 4

You must define $\psi(1,0,1)=(1,0)$. This allows to remove the condition $b\ne 0$ in the first composition. This allows also to remove the condition $a\ne c$ in the second composition, because, on $Y$, $a=c\implies (a,b,c)=(1,0,1)$.

I remains to prove that this extended definition of $\psi$ is continuous on $Y$, which is not difficult. In fact, if one want the continuity for the Zariski topology, the closed sets are the finite subsets, and continuity results from the fact that we have bijections. Slightly more difficult is to show that the extended definition of $\psi$ is a morphism. The proof depends on your definition of a morphism. Generally the definition relies on homomorphisms of quotients of polynomial rings. This means that $a, b, c$ have to be viewed as indeterminates, and thus one can divide by $b$ and $a-c$ (which are nonzero polynomials) without taking care of any specific value.

For the continuity for the usual topology, one can use the geometric interpretation that follows.

Geometric interpretation: Let $L(a,b)$ be the line of equation $a(x-z)+by=0$. This line passes through $(1,0,1)$ and, if $b\ne 0$, intersects $Y$ in exactly another point, which is $\varphi(a, b)$. Conversely, any point of $Y$, different from $(1,0,1)$, defines a unique line passing through $(1,0,1)$; this defines $(a, b)$, and this is the definition of $\psi$. If $b=0$, the line $L(a,b)$ is tangent to $Y$ at $(1,0,1)$. That is, it is the limit of the secant $L(a,b)$ when $b\to 0$. The implies that the limit of the point $\varphi(a, b)$ is $(1,0,1)$. This explains why the extended definition of $\psi$ is continuous. Again, a complete algebraic proof depends on the algebraic background that you use.