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Let $\mathcal{F}, \mathcal{G}$ be quasi-coherent sheaves on a base scheme $S$. Then the $S$-schemes $\mathbb{P}(\mathcal{F}), \mathbb{P}(\mathcal{G})$ are defined by a universal property: to map from an $S$-scheme $T$ (with structure map $g: T \to S$) into $\mathbb{P}(\mathcal{F})$ is the same thing as giving a line bundle $\mathcal{L}$ on $T$ and a surjection $g^*(\mathcal{F}) \to \mathcal{L}$. The Segre embedding $\mathbb{P}(\mathcal{F}) \times_S \mathbb{P}(\mathcal{G}) \to \mathbb{P}(\mathcal{F} \otimes \mathcal{G})$ is defined in a canonical manner using this universal property. Similarly, the Plucker embedding of the Grassmannian into a projective scheme is done by giving a morphism of the functors they represent.

The Segre and Plucker embeddings are closed immersions, as is checked locally using the definitions. However, the properness of projective space $\mathbb{P}^n_{\mathbb{Z}}$ can be easily proved using only the valuative criterion, which seems easier and more elegant (though it can also be done directly). Is there a functorial way to see that the Segre and Plucker embeddings are closed immersions?

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  • $\begingroup$ The advantage of the local method is that it proves more: there is an open covering $(U_i)$ of the Grassmannian and open subsets $(V_i)$ of the projective space such that the Plücker map sends $U_i$to $V_i$ and the induced morphism has a retraction. $\endgroup$ – Laurent Moret-Bailly Jan 11 '11 at 9:23
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    $\begingroup$ One approach to the question (in the Plücker case, say) could be to apply EGA 4 (18.12.6) (or (8.11.5), under finiteness assumptions): every proper monomorphism $X\to Y$ is a closed immersion. In fact it suffices to prove properness plus the fact that for $y\in Y$ the fiber $X_y$ is isomorphic to $\mathrm{Spec}\kappa(y)$. Properness can be deduced from the valuative criterion (if $\mathcal{F}$ is finitely generated you just prove properness of the Grassmannian). The other property is over a field. $\endgroup$ – Laurent Moret-Bailly Jan 11 '11 at 9:31
  • $\begingroup$ @Akhil: I hope it's OK for you that I changed the P's into the "projective" P's. For your question, you may write down an ideal $I \subseteq \mathbb{P}(F \otimes G)$ explicitely and verify that $\mathbb{P}(F) \times_S \mathbb{P}(G)$ satisfies the same universal property as $V(I)$. I have not tried it, but this should work. $\endgroup$ – Martin Brandenburg Jan 11 '11 at 12:15
  • $\begingroup$ By the way, as with your other questions: Very interesting, 1+. $\endgroup$ – Martin Brandenburg Jan 11 '11 at 12:17
  • $\begingroup$ @Laurent Moret-Bailly: Thanks! (I would be happy to accept that as an answer.) Is there a direct way to check that $X_y$ (for $X$ the Grassmannian, $Y$ the projective space, $X \to Y$ the Plucker map) is isomorphic to $\mathrm{Spec} \kappa(y)$? $\endgroup$ – Akhil Mathew Jan 11 '11 at 14:52
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I think this is a very interesting question. It gives another proof of the Segre embedding, makes it somehow more natural and puts it into a more general context, which I explain below.

I will just talk about the Segre embedding $\mathbb{P}^n \times \mathbb{P}^m \to \mathbb{P}^{n+m+nm}$. The general case of $\mathbb{P}(F) \times \mathbb{P}(G) \to \mathbb{P}(F \otimes G)$ is similar.

Let $x_0,...,x_n$ be the homogeneous coordinates of $\mathbb{P}^n$, $y_0,...,y_m$ the ones of $\mathbb{P}^m$ and $z_{00},...,z_{nm}$ the ones of $\mathbb{P}^{n+m+nm}$. The idea of the Segre embedding is $z_{ab} = x_a y_b$. Observe that with this equation we can interchange $a,c$ and $b,d$ in $z_{ab} z_{cd}$ without changing the result. Thus, consider the homogeneous ideal generated by the $z_{ab} z_{cd} - z_{cb} z_{ad}, z_{ab} z_{cd} - z_{ad} z_{cb}$ and let $I$ denote the corresponding ideal of the structure sheaf of $\mathbb{P}^{n+m+nm}$. Now I claim that $V(I)$ and $\mathbb{P}^n \times \mathbb{P}^m$ satisfy the same universal property. So these schemes are isomorphic, and we get a closed immersion $\mathbb{P}^n \times \mathbb{P}^m \to \mathbb{P}^{n+m+nm}$.

Let $X$ be a test scheme. Then $\hom(X,\mathbb{P}^n \times \mathbb{P}^m)$ is the set of tuples $(\mathcal{L}_1,\rho,\mathcal{L}_2,\tau)$, where $\mathcal{L}_1$ (resp. $\mathcal{L}_2$) is a line bundle on $X$ and $\rho_0,...,\rho_n$ (resp. $\tau_0,...,\tau_m$) are global generators of $\mathcal{L}_1$ (resp. $\mathcal{L}_2$).

On the other hand, $\hom(X,V(I))$ is the set of all tuples $(\mathcal{L},\sigma)$, where $\mathcal{L}$ is a line bundles on $X$ and $\sigma_{00},...,\sigma_{nm}$ are global generators of $\mathcal{L}$, which satisfy the relations $\sigma_{ab} \sigma_{cd} = \sigma_{cb} \sigma_{ad}$ and $\sigma_{ab} \sigma_{cd} = \sigma_{ad} \sigma_{cb}$.

With these descriptions, there is a very natural map $\hom(X,\mathbb{P}^n \times \mathbb{P}^m) \to \hom(X,V(I))$ given by $\mathcal{L} = \mathcal{L}_1 \otimes \mathcal{L}_2$ and $\sigma_{ab} = \rho_a \otimes \tau_b$. Actually this map makes sense in every abelian tensor category, but its bijectivity seems to be a special property of the abelian tensor category of quasi-coherent sheaves on a scheme (more generally, sheaves of modules on a ringed space). The usual proof uses the local explicit description of the classifying projective schemes, but you can do it directly by writing down an inverse map:

Let $\mathcal{L},\sigma$ be given as above. Construct a line bundle $\mathcal{L}_1$ as follows: For every $i$, put $X_{\rho_i} = \cup_j X_{\sigma_{ij}}$. On this open set, let $\mathcal{L}_1$ be generated freely by a symbol $\rho_i$. The cocycles $\rho_a / \rho_i$ are defined as $\sigma_{aj} / \sigma_{ij}$ for some $j$; the choice of this $j$ does not matter because of the relations for $\sigma$. Similarily we may define $\mathcal{L}_2$ with global generators $\tau_j$. It's easy to show that $\mathcal{L} \cong \mathcal{L}_1 \otimes \mathcal{L}_2$ with $\rho_i \otimes \tau_j$ corresponding to $\sigma_{ij}$.

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  • $\begingroup$ So this is how Grothendieck defines the Segre embedding in EGA II (or the new edition of EGA I). There is of course an explicit description of it as well. (I realize this is tangential: I need to think about this a little!) $\endgroup$ – Akhil Mathew Jan 13 '11 at 4:06
  • $\begingroup$ OK, I've thought this over: this is a very nice explanation for the Segre embedding, and one which really motivates it better. Thanks! $\endgroup$ – Akhil Mathew Jan 13 '11 at 21:38

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