Let $X$ and $Y$ be two separated schemes. I want to prove that their fiber product $X\times_{\operatorname{Spec}(\mathbb{Z})}Y$ is also a separated scheme.
Note that we have the maps $$f:X\rightarrow\operatorname{Spec}(\mathbb{Z})$$ $$g:Y\rightarrow\operatorname{Spec}(\mathbb{Z})$$ and the projection maps $p_{1}:X\times_{\operatorname{Spec}(\mathbb{Z})}Y\rightarrow X$ and $p_{2}:X\times_{\operatorname{Spec}(\mathbb{Z})}Y\rightarrow Y$such that $f\circ p_{1} = g\circ p_{2}$. I want to show that the induced morphism $\varphi:=f\circ p_{1}=g\circ p_{2}: X\times_{\operatorname{Spec}(\mathbb{Z})}Y\rightarrow \operatorname{Spec}(\mathbb{Z})$ is separated. This comes down to showing that the image of the diagonal morphism $$\Delta_{X\times_{\mathbb{Z}}Y/\operatorname{Spec}(\mathbb{Z})}:X\times_{\mathbb{Z}}Y\rightarrow (X\times_{\mathbb{Z}}\times Y)\times_{\mathbb{Z}}(X\times_{\mathbb{Z}}\times Y)$$ is closed.
By assumption we have that the $\Delta_{X/\operatorname{Spec}(\mathbb{Z})}(X)$ and $\Delta_{Y/\operatorname{Spec}(\mathbb{Z})}(Y)$ are closed.
Any help would be appreciated!
Note that the base change of a separated morphism is again separated.
Since the maps $p_{1}$ and $p_{2}$ are base changes of the given separated morphisms $f$ and $g$ respectively, we find that $p_{1}$ and $p_{2}$ are also separated morphisms.
Since the composition of two separated morphisms is again separated we find that the map $\varphi:=f\circ p_{1}=g\circ p_{2}$ is also separated.
This proves that the fiber product of two separated schemes is indeed also a separated scheme.