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Let $R$ be an arbitrary ring and $X$ an irreducible $R$-scheme (let work for example with $X :=\mathbb{P}^m_R$) and $\mathcal{L}$ an invertible sheaf on $X$. We intend to construct a morphism $f: X \to \mathbb{P}^n_R$ to projective space.

According Hartshorne's (page 150) or Bosch's (p 450) "Algebraic geometry" to construct such morphism is equivalent to give following data:

$s_0,s_1,..., s_n \in \Gamma(X, \mathcal{L})$ global sections such that they generate $X$ in the sense $X = \bigcup_{i=0} ^n X_{s_i}$ where $X_{s_i}$ are non vanishing loci wrt $s_i$.

Then we can locally define the maps $f_i: X_{s_i} \to D_+(t_i)$ via ring maps $R[t_0/t_i, ..., t_m/t_i] \to \Gamma(\mathcal{O}_X,X_{s_i}), t_j/t_i \mapsto s_j/s_i$ and glueing them together give the desired morphism $f:X \to \mathbb{P}^n_R$. (*)

Remark: By construction we have $s_i:= f^{\#}(t_i)$.

Following question:

Let assume that $\mathcal{L}= \mathcal{O}_X$ and $\Gamma(\mathcal{O}_X,X)=R$.

Consider constant sections $s_0,s_1,..., s_n \in \Gamma(\mathcal{O}_X,X)=R$.

My question is why then the induced morphism $f: X \to \mathbb{P}^n_R$ by this data is constant?

REMARK: This question is based on following previous thread Morphism to Projective Space $\mathbb{P}^m$ of mine as also linked below and is meant to find an argument for constance of $f$ using arguments for ***general scheme morphisms**** instead of argueing like in case for morphisms between varieties.

Here I explicitely mean the following problem that occures in the answer in the related thread which hornestly doesn't satisfy me since the argument given there only covers the case when we consider $f$ as a morphism of varieties.

Indeed in this case using contravariant category equivalence between affine varieties and that of finitely generated reduced algebras over a field $k$ we can easily describe the morphism $f$ for given sections $s_i$ in following way:

We set $f: x \mapsto (s_0(x): s_1(x): ...: s_n(x))= (s_0: s_1: ...: s_n) \in \mathbb{P}^n_R$ is constant since the $s_i \in R$ are by assumption constant.

But if we leave the world of varieties then the morphism cannot just naively described as above nevertheless only using glueing technique as described in (*).

My question is how to show rigorously (without handweavings in the direction of the variety world) that in this case if the $s_i \in R$ are choosen as constants then $f$ is also a constant morphism?

btw: Does the expression $(s_0(x): s_1(x): ...: s_n(x))$ have a meaning for morphisms to $\mathbb{P}^n_R$ in category of schemes? we have $s_i(x) \in \kappa(x) = \mathcal{O}_{X,x}/m_x$ so $(s_0(x): s_1(x): ...: s_n(x))$ can't be expected to be a "point" of projective space.

user267839
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  • I think you've got to be careful about what you mean by a constant map over a general (non-point) base scheme. I'm pretty sure the correct notion isn't that the image is a single point, instead I would expect the definition to be something like if $X$, $Y$ are schemes over $S$, then $X\to Y$ is constant if it factors as $X\to S\to Y$. – jgon Mar 01 '19 at 01:31
  • Then given $s_0,\ldots,s_n\in R$, we get a morphism $\operatorname{Spec} R \to \Bbb{P}^n_R$ induced by the graded map $R[x_0,\ldots,x_n]\mapsto R[t]$ by $x_i\mapsto s_i t$. (This works, since the $s_i$ generate the unit ideal, so their images in $R[t]$ generate the irrelevant ideal properly) – jgon Mar 01 '19 at 01:36
  • This morphism $\operatorname{Spec} R \to \Bbb{P}^n_R$ is an $R$-point we can call $(s_0:\cdots :s_n)$. – jgon Mar 01 '19 at 01:36
  • @jgon:yes this was the cruical point in my considerations because since for constant $s_i$ we get $X_{s_i}=X$ so every $f_i$ and therefore by glueing also $f$ factorize throught $Spec(R)$. So with this definition we are done. Apparently I thought too topologically – user267839 Mar 01 '19 at 01:40

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I ended up answering in the comments, so I've copied the comments into a proper answer. Let me know if this doesn't address your question, but I think it should.

I think you've got to be careful about what you mean by a constant map over a general (non-point) base scheme. I'm pretty sure the correct notion isn't that the image is a single point, instead I would expect the definition to be something like if $X$, $Y$ are schemes over $S$, then a morphism $X\to Y$ is constant if it factors as $X\to S\to Y$

Then given $s_0,\ldots,s_n\in R$, we get a morphism $\operatorname{Spec} R \to \Bbb{P}^n_R$ induced by the graded map $R[x_0,\ldots,x_n]\to R[t]$ by $x_i\mapsto s_i t$. (This works, since the $s_i$ generate the unit ideal, so their images in $R[t]$ generate the irrelevant ideal properly)

This morphism $\operatorname{Spec} R \to \Bbb{P}^n_R$ is an $R$-point we can call $(s_0:\cdots :s_n)$.

jgon
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  • this exactly solves the problem. One remark: Do you mean $R[x_0,\ldots,x_n]\mapsto R[t]$ or $R[x_0,\ldots,x_n]\mapsto R$ ? – user267839 Mar 01 '19 at 01:48
  • @KarlPeter I do mean $R[x_0,\ldots,x_n]\to R[t]$, since when we take $\operatorname{Proj}$ we then get $\operatorname{Proj}R[t]\cong \operatorname{Spec} R$. That's why it's important that the $s_i$ generate the unit ideal. (Because $\operatorname{Proj}$ isn't a functor, so we need to check that it actually does induce the well defined map we claim it does) – jgon Mar 01 '19 at 02:20