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I am doing Exercise II.7.7c in Hartshorne (not homework). In this problem, he asks to show that the linear system of conics in $\mathbb{P}^2$ passing through a fixed point $P$ gives an immersion of $\mathbb{P}^2-P$ into $\mathbb{P}^4$ (over $k$ algebraically closed). Considering the point $P=(1:0:0)$, this linear system has a basis ${y^2,z^2,xy,xz,yz}$ and these generate $O(2)$ except at the point $P$. Thus, they define a map $\phi:\mathbb{P}^2-P\rightarrow \mathbb{P}^4$. What I am wondering is how to show that this map is an immersion in the sense of Hartshorne (factors into an open embedding followed by a closed embedding).

In particular, I am wondering if some modified version of Hartshorne Proposition 7.3 can be used. This proposition says that if $X$ is projective variety over $k$ algebraically closed, then giving a closed immersion into projective space is the same as giving global sections of an invertible sheaf such that the space they span separates points and tangent vectors. It seems to me that separating points and tangent vectors still makes sense in this context, despite the fact that $\mathbb{P}^2-P$ is not projective. If the sections defining $\phi$ span a subspace of global sections of $O(2)_{|\mathbb{P}^2-P}$ which separates points and tangent vectors, can I then conclude that $\phi$ is an immersion in the sense above? If so, why?

I apologize in advance if this question is silly.

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Hint: Call $W=\Gamma(\mathbb P^2,\mathcal O_{\mathbb P^2}(2))$ and $V=W-P\subset W$ the $5$-dimensional subspace of homogeneous polynomials of degree $2$ vanishing at $P$. Then your $\phi=|V|$ is the composition $$\mathbb P^2\setminus P\hookrightarrow \mathbb P^2\hookrightarrow\mathbb P^4,$$ where the first arrow is an open immersion and the second arrow is the closed immersion defined by $|U|$, for some $U\subset W$ of dimension $5$. Strategy: to find the $5$ sections generating $U$ (i.e. to find $U$) one has to write $\phi$ explicitely for every point $Q\in \mathbb P^2\setminus (1:0:0)$. After that, to show that $|U|$ is a closed immersion you can use the criterior you quoted.

If I understand your last paragraph, you say: if we start with a non-projective variety $X$, can we say that a morphism $|V|:X\to \mathbb P^n$ is an immersion iff the sections generating $V$ separate points and tangent vectors? Well, it appears to me that canceling out the projectivity hypothesis with the closed immersion requirement is a bit risky. For instance, we lose the correspondence existing between morphisms to projective space and spaces of sections of line bundles.

Example: the open immersion $\mathbb A^n\to\mathbb P^n$. What line bundle on $\mathbb A^n$ does it come from?

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  • $\begingroup$ I don't understand how V defines a morphism on $\mathbb{P}^2$. The system V has a base point on $\mathbb{P}^2$, namely P. $\endgroup$
    – Cass
    Commented Sep 6, 2013 at 3:38
  • $\begingroup$ Yes, sorry. Fixed. $\endgroup$
    – Brenin
    Commented Sep 6, 2013 at 14:11

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