In the exercise 2.9 of the book Algebraic Geometry by Hartshone, the author questions us about the projective closure of an affine variety.

Let $Y$ be an affine variety in $\mathbb{A}^n$, identifying $\mathbb{A}^{n}$ with the open subset $U_0$ of $\mathbb{P}^n$ by the map $\varphi_{0}: (x_0,x_1,..,x_n)\mapsto (\dfrac{x_1}{x_0},...,\dfrac{x_n}{x_0})$. Then we can speak about $\bar{Y}$, the projective closure of $Y$ in $\mathbb{P}^{n}$.

a, Show that $I(\bar{Y})$ is an ideal generated by $\beta(I(Y))$

b, Prove that if $f_1,...,f_r$ generate $I(Y)$, then $\beta(f_1),...,\beta(f_r)$ do not necessarily generate $I(\bar{Y})$

My question are :

  1. From the correspondence given by the map : $\beta : f(x_1,...,x_n)\longmapsto x_{0}^{\text{deg}f}f$ where $f$ is a homogeneous polynomial, we can see that a homogeneous polynomial vanishing on $Y$ gives a homogeneous polynomial vanishing on $\bar{Y}$. But how can we give a represent an element of $I(\bar{Y})$ in term of all the element in $I(Y)$ to conclude part $a$ ? Because from part b, we get that the generators of $I(Y)$ might not have affect on the generator of $I(\bar{Y})$.
  2. Is there anyway to think about the projective closure of $Y$ geometrically ? I consider the following example, and I get confusion :

Let $f=x^2-xy$, then the zero set of $f$ in $\mathbb{A}^n$ is $Z(f)=Y=\{(t,t),(t,0)|t\in k\}$

Then $\beta(f)=z^2(x^2-xy)=F(x,y,z)$, and then the projective closure of $Y$ is $\{/(a:t:t), (a:0:t)|a,t\in k\}$

So, I think that, we can only add one more coordinate to $Y$ to get $\bar{Y}$, then my third question is : Does it make sense to think about the projective closure ? What is its importance in algebraic geometry ?


A few hints/suggestions:

i). You are working with the wrong definition of $\beta$! Please check again Hartshorne, p.11 at the very top, to get the right one. In particular, in your example you should have


and not $\beta(x^2-xy)=z^2(x^2-xy)$.

ii). For Ex.2.9.(a), it might be easier to prove that $Z(\beta(I(Y)))=\overline{Y}$. This is equivalent to $\beta(I(Y))=I(\overline{Y})$ by Ex.2.4 (note that $\beta(I(Y))$ is a prime ideal, since $I(Y)$ is).

iii). It is not true in general, that $\overline{Y} \setminus Y$ consists of only a single point. Consider $Y=(x=0)$ in $\Bbb{A}^3$. Then $\overline{Y} \setminus Y = \{[0:a:b:0] \vert a,b \in k\}$, which is isomorphic to a projective line.

  • $\begingroup$ Sorry, where was I wrong, here is what I think: Let $y_i=\dfrac{x_i}{x_0}$, let $f\in k[y_1,...,y_n]$, define $\beta : k[y_1,...,y_n] \mapsto x_{0}^{deg\text{f}}f(y_1,...,y_n)$ ? $\endgroup$ – Arsenaler Jan 10 '13 at 14:20
  • $\begingroup$ For my third question, I do not mean that $\bar{Y}\setminus Y$ consists of only a single point; I mean that we can add one more variable to the coordinate of point in $Y$, and this variable can takes any values in $k$. $\endgroup$ – Arsenaler Jan 10 '13 at 14:22
  • $\begingroup$ @msnaber: regarding your first comment: yes, that is the correct definition of $\beta$. And it gives you $\beta(x_1^2-x_1x_2)=x_1^2-x_1x_2$, not $x_0^2(x_1^2-x_1x_2)$ as you stated in the OP. regarding your second comment: it is a little bit risky to jump from a specific example to a conclusion. As an additional example, check out the twisted cubic $Y=Z(y-x^2,z-x^3)$ and its projective closure. The moral is: projective closure does not behave as nice as one would wish. $\endgroup$ – Nils Matthes Jan 10 '13 at 14:44

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