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I have two questions on scheme morphisms.

  1. Is the property of a scheme morphism to be a closed immersion a local property (as it is for open immersions)?

  2. Let $X=Spec (R)$ be a noetherian scheme and $p$ a prime ideal of $R$.

    In Georges Elencwajg's very helpful answer to some Questions on scheme morphisms, he explains that one may think of the embedded scheme $g:Spec (R_p)\to X$ as of the intersection of all the neighbourhoods of $p$ in $X$.

    On the other hand, there is the closed immersion $Spec (R/p)\to X$ and as topological spaces $Spec (R/p)=V(p)$ is the closure of the ''generic point'' $p$ of the subscheme.

    As $g$ is not always a closed immersion, one can consider the topological space $V(Im(g))$, the closure of the image of $g$. Is my intuition correct that this always has to be the whole space $Spec (R)$?

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up vote 2 down vote accepted

Yes, the property of being a closed immersion for a morphism of schemes $f:Y\to X$ is local . More precisely (EGA I, Corollaire 4.2.4) :
a) If $f$ is a closed immersion, so is $f|f^{-1}U:f^{-1}U\to U$ for every open subscheme $U\subset X$.
b) If every point $x\in X$ has an open neighbourhood $x\in U \subset X$ such that $f|f^{-1}U:f^{-1}U\to U$ is a closed immersion, then $f:Y\to X$ is a closed immersion .

a) If $R$ is a domain (noetherian or not) then, yes, your intuition is correct :
The image $g(Spec(R_\mathfrak p))\subset X=Spec(R)$ is dense in $X$ for the simple reason that it contains the generic point $\eta_X=\langle 0_R\rangle =g(\langle 0_{R_\mathfrak p}\rangle )$ of $X$.
b) However density may fail if $R$ has zero divisors.
For example if $R=\mathbb C[X,Y]/\langle XY\rangle=\mathbb C[x,y]$ and $\mathfrak p=\langle y\rangle$ , consider the natural morphism $g:Spec(R_\mathfrak p)\to S=Spec(R)$
You will easily check that the image $g(Spec(R_\mathfrak p))\subset S$ has closure $\overline {g(Spec(R_\mathfrak p))}=V(\mathfrak p)\subsetneq S $,
the $x$ axis of the cross $S$.

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