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Let $X\longrightarrow S$ be a morphism of (in my case: Noetherian) schemes. Then $X$ is called projective over $S$ if $X$ is isomorphic to a closed subscheme of $\mathbb{P}^{n}_{S}$.

By a slightly more general definition of projectivity, $X\longrightarrow S$ is called projective if $X$ is a closed subscheme of $\mathbb{P}(E)$ for some vector bundle $E$ over $S$.

The claim is: An example of a morphism that satisfies the more general definition can be constructed as follows: Let $S$ be a Noetherian scheme with non-trivial Picard group. Take a non-trivial line bundle $E$ over $S$ and let $X=\mathbb{P}(E)$.

Question: Although $\mathbb{P}(E)\ncong\mathbb{P}^{1}_{S}$, this does not imply (in my opinion) that $\mathbb{P}(E)$ is not isomorphic to a closed subscheme of $\mathbb{P}^{1}_{S}$.

How do I prove this? Do I need extra conditions, or does Noethericity suffice?

hm2020
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The Thin Whistler
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    If $E$ is a line bundle on $S$, $\mathbb{P}(E)$ is just $S$. – Mohan Jun 19 '21 at 15:48
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    As Mohan's pointed out, you're barking up the wrong tree a little bit here. It is true that these notions are not equivalent in general, but if $S$ admits an ample invertible sheaf they are the same. See for instance here or Stacks 087S. – KReiser Jun 20 '21 at 07:15

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Question: "Although P(E)≆P1S, this does not imply (in my opinion) that P(E) is not isomorphic to a subscheme of P1S. How do I prove this? Do I need extra conditions, or does Noethericity suffice?"

Answer: If $S$ is any scheme and $F:=\mathcal{O}_S\{e_0,..,e_n\}$ is a free rank $n+1$ module on $S$ it follows

$$\mathbb{P}(F^*) \cong \mathbb{P}^n_S$$

is projective space over $S$. If $E$ is a locally trivial (non-trivial) $\mathcal{O}_S$-module of rank $n+1$ you may construct $\pi:\mathbb{P}(E^*) \rightarrow S$. A local trivialization $U_i$ of $E$ gives $E_{U_i} \cong \mathcal{O}_{U_i}^{n+1}$ hence $\pi^{-1}(U_i) \cong \mathbb{P}^n_{U_i}$. Since $E$ is non-trivial and $F$ is trivial, certainly there is no isomorphism

$$\mathbb{P}(E^*) \cong \mathbb{P}(F^*)$$

in general. There could however be a closed immersion

$$i: \mathbb{P}(E^*) \rightarrow \mathbb{P}^N_S$$

for some large $N$. This is Theorem II.7.1 and 7.2 in Hartshorne: If $S:=Spec(A)$, there is a closed immersion $i:X:=\mathbb{P}(E^*) \rightarrow \mathbb{P}^n_S$ iff there is an invertible sheaf $L\in Pic(X)$ satisfying the two conditions $(1),(2)$ in Prop 7.2.

Example: In particular if $Pic(X)=(0)$ there can be no such embedding. For this reason there can be no closed immersion

$$i: \mathbb{A}^n_k \rightarrow \mathbb{P}^n_k$$

since there is no linebundle $L \in Pic(\mathbb{A}^n_k)$ satisfying Proposition II.7.2, (1) and (2).

Example: If $k$ is a field and if $A$ is a finitely generated $k$-algebra it follows $X:=Spec(A[x_0,..,x_n])$ is not projective, ie there is no closed immersion

$$i: X\rightarrow \mathbb{P}^n_A$$

for any $n\geq 1$. If there was such a closed embedding it would follow from HH.Thm.II.5.19 that $H^0(X, \mathcal{O}_X):=A[x_0,..,x_n]$ was a finitely generated $A$-module - a contradiction.

hm2020
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  • How do I make sure such an $X$ with trivial Picard group exists? – The Thin Whistler Jun 19 '21 at 13:52
  • And how do I know $\mathbb{A}^{n}_{k}$ is a closed subscheme of some $\mathbb{P}(E)$? – The Thin Whistler Jun 19 '21 at 14:40
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    The section starting with "example" is misleading. The real reason that there's not a closed embedding of $\Bbb A^n_k$ in to some projective scheme over $k$ for $n>0$ is that any affine scheme projective over $k$ is finite and $\Bbb A^n_k$ isn't for $n>0$. Presumably you're trying to say something about the non-existence of a very ample invertible sheaf with the remark on Picard groups, but it's easy to verify that the structure sheaf on $\Bbb A^n$ is in fact very ample: it's isomorphic to the pullback of $\mathcal{O}(1)$ under the obvious immersion $\Bbb A^n_k\to\Bbb P^n_k$. – KReiser Jun 20 '21 at 07:13
  • @KReiser - In fact $\mathbb{A}^n_A$ is "never" projective over $A$. – hm2020 Jun 20 '21 at 12:03