# Two Projective closures of $\mathbb{A}^1\setminus\{0\}$ which are not isomorphic as varieties.

I am doing some sample problems for my upcoming Algebraic Geometry exam, and one of the questions is:

Is it true that all projective closures of an affine variety $X$ are isomorphic as varieties?

The problem gives the hint to consider various projective closures of $X=\mathbb{A}^1\setminus\{0\}$.

Here a projective closure of an affine variety $X$ is a closed subvariety $Y\subset \mathbb{P}^n$ for some $n$, together with an isomorphism $X\to U$ where $U\subset Y$ is open and dense.

## Discussion

I started by considering the natural projective closure of $X=\mathbb{A}^1\setminus\{0\}$: We consider the identification of $\mathbb{A}^1$ with $D(x_0)\subset \mathbb{P}^1$, i.e. the map $$\phi:\mathbb{A}^1\to D(x_0), \enspace p\mapsto [1:p]\in \mathbb{P}^1$$

Since $X$ is an open dense subset of $\mathbb{A}^1$, $\phi(X)$ is open and dense in $\mathbb{P}^1.$ Therefore $\mathbb{P}^1$ is a projective closure of $X$.

The aim now is to find another projective closure of $X$ which is not isomorphic to $\mathbb{P}^1$ is a variety. For this I decided to view $X$ as $Y:=\{(x,y)\in \mathbb{A}^2:xy=1\}=Z(xy-1)$. We get a projective closure for $Y$ by considering the image of $Y$ under the image of the $n=2$ analog of $\phi$:

$$\varphi:\mathbb{A}^2\to D(z)\subset \mathbb{P}^2=\{[z:x:y]\}, \enspace (x,y)\mapsto [1:x:y].$$ The image of $Y$ is $\{[1:y:x]:yx=1\}$. For example then $$\varphi(Y)=D(z)\cap V(xy-z^2)$$ and so $V(xy-z^2)$ is a projective closure for of $Y$ and thus $X$- here I have used $V(f)$ to denote the zero set of a homogeneous poly $f$ in $\mathbb{P}^2$. I can't see why this variety would not be isomorphic to $\mathbb{P}^1$- it looks like it might be.However, we also have $$\varphi(Y)=D(z)\cap V(zxy-z^3)$$

and $V(xy-z^2)\subsetneq V(zxy-z^3)$ since $xy=z^2\implies z^3=zxy$ however $[0:1:1]\in V(zxy-z^3)\setminus V(xy-z^2)$. It then should follow that $V(xy-z^2)$ and $V(zxy-z^3)$ are non-isomorphic projective closures of $X$. Is this correct? I am not completely comfortable with projective varieties yet, so any comments are appreciated!

• @RghtHndSd Actually, I do not think that I have seen this result before unfortunately – CWsl2 Dec 9 '14 at 23:05
• I removed my comment because I am no longer certain it applies, but it was essentially Georges answer. – RghtHndSd Dec 9 '14 at 23:07
• Let $Z_1 = V(xy-z^2)$ and $Z_2 = V(zxy-z^3)$ as you mention. Then $X \subset Z_1 \subset Z_2$. It follows that the projective closure is contained in $Z_1$, which must be (by Georges answer below) isomorphic to $\mathbb{P}^1$. In otherwords, $Z_2$ is completely irrelevant. – RghtHndSd Dec 9 '14 at 23:20

All smooth projective closures of $\mathbb A^1$ minus a finite number of points are isomorphic to $\mathbb P^1$ because $\mathbb P^1$ is the only smooth complete rational curve.

Edit

As @rghtHndSd very interestingly comments Curtis did not explicitly ask for smooth closures.
And indeed there are many non isomorphic closures of the proposed affine curve: you can pinch $\mathbb P^1$ at one of the added points $\infty$ and obtain a singular projective closure, for example the rational singular complete curve $y^2z-x^3=0$.
By pinching harder and harder (poor, poor $\mathbb P^1$ !) you can obtain worse and worse singular closures like, say, $y^nz-x^{n+1}=0$.

• Are all projective closures of $\mathbb{A}^1 \setminus \{0\}$ smooth? – RghtHndSd Dec 9 '14 at 23:04
• thank you for the response, but I don't think I understand the language of algebraic geometry well enough to understand your answer- I have not learned what a "smooth complete" rational curve is- but if this is in fact true then the hint in my problem was not correct – CWsl2 Dec 9 '14 at 23:11
• Dear @rghtHndSd: no they are not! I had assumed this but I have edited my text in order to take your excellent comment into account. – Georges Elencwajg Dec 9 '14 at 23:19
• Georges: Great answer! – RghtHndSd Dec 9 '14 at 23:22
• Dear @Patrick: I wonder if people realize how sadistic we algebraic geometers really are ... – Georges Elencwajg Dec 9 '14 at 23:30

Hint : Consider the blow-up of the cubic nodal curve in $\mathbb P^2$ given by $y^2 = x^3 + x^2$. The cubic nodal curve has a double point at $(0,0)$ ; if we consider its blow-up, which is a projective curve without the singularity (hence you can compute that it is isomorphic to $\mathbb P^1$), the blow-up minus the two points which get mapped to $(0,0)$ under the blow-up map is isomorphic (using this same map) to the curve $y^2 = x^3 + x^2$ in $\mathbb P^2$ minus the point $(0,0)$. Since $\mathbb A^1$ minus a point is isomorphic to the open subset of $\mathbb P^1$ minus two points, which maps isomorphically to the cubic nodal curve with the origin removed, we see that under this isomorphism, the projective closure of $\mathbb A^1 \backslash \{0\}$ is the cubic nodal curve, which is not isomorphic to $\mathbb P^1$ (because the stalk at $(0,0)$ is not regular and/or normal, and regularity/normality is preserved under isomorphism ; in other words, the cubic nodal curve is singular).

I can give more details if some part of my explanation is not all clear to you.

Hope that helps,

• How does one compute that the blowup is isomorphic to $\mathbb{P}^1$? Perhaps a genus formula? – RghtHndSd Dec 10 '14 at 0:03
• @RghtHndSd : A good "by-hand" computation with coordinates does the trick. – Patrick Da Silva Dec 10 '14 at 4:28