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How is it connected that scheme $X$ is separated over $Y$ and over $Spec \; \mathbb{Z}$, where $X \rightarrow Y$ is some scheme morphism?

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up vote 8 down vote accepted

I guess you're asking if there is a connection between separatedness of a morphism $f:X\rightarrow Y$ and separatedness of $X$ (and\or $Y$) over $\mathrm{Spec}(\mathbf{Z})$.

When one says that a scheme $X$ is separated, one means separated over $\mathrm{Spec}(\mathbf{Z})$, i.e., the unique morphism $X\rightarrow\mathrm{Spec}(\mathbf{Z})$ is separated. If $X$ is a separated scheme, then for any scheme $Y$, any morphism $f:X\rightarrow Y$ is separated. This is because $f$ can be factored as $\Gamma_f:X\rightarrow X\times_\mathbf{Z}Y$, the graph morphism, followed by the projection $X\times_\mathbf{Z}Y\rightarrow Y$. The first morphism is an immersion, hence separated, and the second morphism is a base change of the separated morphism $X\rightarrow\mathrm{Spec}(\mathbf{Z})$. So $f$ is a composite of separated morphisms, and therefore is itself separated.

Thus a scheme $X$ is separated if and only if for all morphisms $f:X\rightarrow Y$, $f$ is separated.

I'm not sure what else you're after. Taking $X=Y$ to be a scheme which is not separated (over $\mathrm{Spec}(\mathbf{Z})$, or equivalently, over some affine scheme, e.g. the spectrum of a field), then $\mathrm{id}_X:X\rightarrow X$ is trivially separated, but $X\rightarrow\mathrm{Spec}(\mathbf{Z})$ isn't.

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One last thing that can be said is that a scheme over an affine scheme $\text{Spec}(A)$ is separated over $A$ if and only if it is separated over $\text{Spec}(\mathbb Z)$. Or, more generally, if $f \colon Y \to S$ is a separated $S$-scheme, then $g \colon X \to Y$ is separated if and only if $fg \colon X \to S$ is separated. – Mauro Porta Dec 4 '12 at 21:02

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