[Math] Finite morphism that is not projective

algebraic-geometry

There are two definitions of a projective morphism.

Hartshorne: A morphism $f: X\to Y$ is projective if it factors as $f=gi$, where $i: X\to P_Y^n$ is closed imbedding and $g: P_Y^n\to Y$ is canonical.

Ravi's notes: Morphisms of the form $Proj(L)\to Y$ where $L$ is a quasicoherent sheaf of graded $O_Y$-algebras, finitely generated in degree 1.

In Ravi's notes, it is shown that finite morphisms are projective. The corresponding algebraic fact is that $Proj(A\oplus B\oplus B\oplus B\cdots)=Spec(B)$ for a finite morphism $A\to B$ of rings. In Ravi's notes, it is also mentioned that a finite morphism need not be projective in Hartshorne's sense. Is there an easy counterexample showing this?

I apologize if this turns out to be a simple question.

best,
minimax

Best Answer

The following seems to hold: Let $\mathcal{L}:=\oplus_{i\geq 0}\mathcal{L}_i$ with $\mathcal{L}_0:=\mathcal{O}_Y$ and $\mathcal{L}$ generated by $\mathcal{L}_1$ as sheaf of graded $\mathcal{O}_Y$-algebras. Let $U:=Spec(A)\subseteq Y$ be an open subscheme with $\mathcal{L}_1(U):=L$ and let $z_1,..,z_n\in L$ be elements generating $\mathcal{L}(U)$ as graded $A$-algebra. It follows $\mathcal{L}(U)\cong B:=A[x_1,..,x_n]/I$ where $I$ is a homogeneous ideal in $A[x_1,..,x_n]$. Let $\pi: \mathbb{P}(\mathcal{L})\rightarrow Y$ be the projection morphism. It follows there is an isomorphism $\pi^{-1}(U)\cong \mathbb{P}(\mathcal{L}_U) \cong \operatorname{Proj}(B)$. Here $\mathcal{L}_U$ is the restriciton of $\mathcal{L}$ to the open set $U$. There is over $U$ a closed embedding

$i: \pi^{-1}(U)\cong \operatorname{Proj}(B) \rightarrow \mathbb{P}^n_U.$

Hence locally you can embed the scheme $\mathbb{P}(\mathcal{L})$ into a projective space over $U$, but not globally.

Hartshorne starts with a trivial sheaf $\mathcal{T}:=\mathcal{O}_Y^{n+1}$ of rank $n+1$, and constructs $\mathbb{P}^n_Y:=\operatorname{Proj}(Sym_{\mathcal{O}_Y}^*(\mathcal{T}^*))$. Then defines a morphism $f:X\rightarrow Y$ to be projective if $X$ globally has a closed embedding $i: X\rightarrow \mathbb{P}^n_Y$ with $p\circ i=f$. Hence it seems Vakils definition generalize Hartshornes definition to the case when $\mathcal{L}$ is no longer on the form $Sym_{\mathcal{O}_Y}^*(\mathcal{E}^*)$ for a trivial rank $n+1$ $\mathcal{O}_Y$-module $\mathcal{E}$.

Your question: "In Ravi's notes, it is also mentioned that a finite morphism need not be projective in Hartshorne's sense. Is there an easy counterexample showing this?"

Answer to your question: It seems Vakils definition is more general: the sheaf of graded algebras $\mathcal{L}$ need not be on the form $Sym_{\mathcal{O}_Y}^*(\mathcal{E}^*)$ for a trivial $\mathcal{O}_Y$-module $\mathcal{E}$ of rank $n+1$.

Here I define $Sym_{\mathcal{O}_Y}^*(\mathcal{E}^*)$ as follows:

$$Sym_{\mathcal{O}_Y}^*(\mathcal{E}^*):=\oplus_{i \geq 0} Sym_{\mathcal{O}_Y}^i(\mathcal{E}^*).$$